Math for Biology Workbook

c J¨urgenGerlach Department of Mathematics Radford University

September 7, 2015

Contents 3.2 U.S. and Metric Units . . . . . 18 3.3 Conversion Methods ...... 18 1 Scientific Notation 5 3.3.1 Substitution ...... 18 1.1 Scientific Notation ...... 5 3.3.2 Conversion Ratios . . . 19 1.2 Significant Figures ...... 5 3.3.3 Conversion Factors . . . 19 1.3 Rounding ...... 6 3.4 Worked Problems ...... 19 1.4 Unit Prefixes ...... 6 3.5 Exercises ...... 20 1.5 Calculator ...... 6 1.6 Worked Problems ...... 7 4 A Little Geometry 22 1.7 Exercises ...... 8 4.1 Two Dimensions: Area and Perime- ter ...... 22 2 Metric System 9 4.2 Three Dimensions: Volume and 2.1 Basic Physical Units ...... 9 Surface Area ...... 23 2.2 Metric Measurements . . . . . 9 4.3 Doubling Lengths and Surface 2.2.1 Length ...... 9 Area to Volume Ratios . . . . . 24 2.2.2 Area ...... 10 4.4 Angular Speed and Centrifuges 25 2.2.3 Volume ...... 10 4.5 Worked Problems ...... 26 2.2.4 Velocity and Acceleration 10 4.6 Exercises ...... 30 2.2.5 Flux ...... 11 2.2.6 Mass ...... 11 5 A Little Algebra 32 2.2.7 Density ...... 11 5.1 General Laws ...... 32 2.2.8 Force ...... 11 5.1.1 Fractions ...... 34 2.2.9 Work ...... 11 5.1.2 Hierarchy of Operations 34 2.2.10 Power ...... 12 5.2 FOIL, Binomial Formulas and 2.2.11 Pressure ...... 12 the Pascal Triangle ...... 35 2.2.12 Temperature ...... 12 5.2.1 Pascal’s Triangle . . . . 36 2.3 Worked Problems ...... 12 5.3 Hardy-Weinberg Equilibrium . 37 2.4 Exercises ...... 14 5.4 Exponents ...... 39 5.4.1 Fractional Exponents . 40 3 Unit Conversion 17 5.5 Worked Problems ...... 41 3.1 Customary U.S. Units . . . . . 17 5.6 Exercises ...... 43

1 6 Graphs 46 10 The Binomial Distribution 87 6.1 Scatter Plots ...... 46 10.1 Bernoulli Trials ...... 87 6.2 Function-Style Graphs . . . . . 47 10.2 Binomial Probability Distribution 88 6.3 Watch the Scales ...... 48 10.2.1 Calculators and Comput- 6.3.1 Log Plots ...... 48 ers ...... 90 6.3.2 Loglog Plots ...... 49 10.2.2 Cumulative Density Dis- 6.4 Worked Problems ...... 50 tribution ...... 90 6.5 Exercises ...... 51 10.3 Worked Problems ...... 91 10.4 Exercises ...... 95 7 Solving Equations 54 7.1 Equations and Solutions . . . . 54 11 Functions 96 7.2 Solution Strategies ...... 54 11.1 Graphs of Functions ...... 97 7.2.1 Algebraic Methods . . . 54 11.2 Composition and Inverse Func- 7.2.2 Discussion of Pitfalls . . 55 tions ...... 98 7.2.3 Graphical Solutions . . 56 11.3 Worked Problems ...... 101 7.2.4 Quadratic Equations . . 57 11.4 Exercises ...... 103 7.3 Worked Problems ...... 59 7.4 Exercises ...... 63 12 Linear Functions 104 12.1 Lines ...... 104 8 Ratios and Proportion 65 12.2 Point-Slope Form ...... 106 8.1 Ratios ...... 65 12.3 Almost Linear Data ...... 107 8.2 Percent ...... 65 12.4 Worked Problems ...... 108 8.3 Percentlike Quantities . . . . . 66 12.5 Exercises ...... 113 8.3.1 Parts per Thousand . . 66 8.3.2 Parts per Million, Parts 13 The Common Logarithm 114 per Billion ...... 66 13.1 Introduction ...... 114 8.4 Percentage Change ...... 67 13.2 Deﬁnition ...... 114 8.5 Proportion ...... 68 13.2.1 Estimation of Logarithms 115 8.6 Mark and Recapture ...... 68 13.3 Properties ...... 115 8.7 Dilution and Mixtures . . . . . 69 13.3.1 Scientiﬁc Notation . . . 117 8.7.1 Serial Dilutions . . . . . 71 13.3.2 The Function f(x) = log x117 8.8 Worked Problems ...... 71 13.4 Solving Exponential Equations 118 8.9 Exercises ...... 74 13.5 Applications ...... 119 13.5.1 Log Plots ...... 119 9 Elements of Probability 77 13.5.2 pH Values ...... 119 9.1 Probability ...... 77 13.5.3 Moment Magnitude Scale 120 9.2 Laws of Probability ...... 78 13.6 Worked Problems ...... 121 9.3 Conditional Probability and In- 13.7 Exercises ...... 124 dependent Events ...... 79 9.4 Hamilton’s Rule ...... 81 14 Exponential Functions 125 9.5 The Birthday Problem . . . . . 82 14.1 Introduction ...... 125 9.6 Worked Problems ...... 82 14.2 Exponential Growth and Decay 126 9.7 Exercises ...... 85 14.3 Doubling Time ...... 128

2 14.4 Half-Life ...... 131 14.5 Almost Exponential Data . . . 133 14.6 Worked Problems ...... 134 14.7 Exercises ...... 139

15 Power Functions 141 15.1 Deﬁnition ...... 141 15.2 Approximation with Power Func- tions ...... 142 15.3 Worked Problems ...... 143 15.4 Exercises ...... 146

16 Difference Equations 149 16.1 Introduction ...... 149 16.2 First Order Difference Equations 151 16.2.1 Computers and Calcula- tors ...... 154 16.3 Steady States and Stability . . 155 16.3.1 Affine Equations . . . . 158 16.4 Logistic Growth ...... 161 16.4.1 Logistic Approximations 163 16.5 Worked Problems ...... 164 16.6 Exercises ...... 169

3 This workbook is designed for MATH 119, This is not a recipe book (as in ”take the the Mathematics for Biology course at Radford square root, double it, and add three to the an- University. swer”). We emphasize the thoughts and ideas It is expected that the students have a work- which go into a particular concept, along with ing knowledge of basic mathematics at a high pitfalls and common errors. In many cases we school level, but calculus, pre-calculus or trigonom- will present different methods to solve a prob- etry are not required. The topics have been lem, and in the end the student has to select selected based on their relevance for biology. the approach with which he or she is most com- Logarithms and exponential are extremely use- fortable. ful in many areas of biology, and probability On occasion the workbook goes past the and the binomial distribution are included for bare minimum of the course. These parts could their use in genetics. be considered further reading material for the Difference equations are included because interested student, or they may serve as refer- they can be easily implemented in EXCEL or ence material. other computing environments. This opens to In its first run the MATH 119 course was door to first simulations of population growth taught using the text ”Quantitative Reason- without the Calculus machinery. Of course ing and the Environment” by Langkamp and Calculus or Linear Algebra are inevitable once Hull. Some of their ideas, examples and ex- stability issues arise. Chapter 16 looks at first ercises have been incorporated into this work- order difference equations; another chapter with book. examples of higher order difference equations The graphs were produced with EXCEL, (Fibonacci numbers, Leslie population mod- Maple and Paint. els) and systems of difference equations (SIR, Many thanks to the biology faculty who predator-prey) is planned. Examples, mind have let me audit their classes, and to those you, not a detailed analysis, which would be who have contributed teaching material, and well beyond the scope of this workbook. to those who served as my sounding board for this endeavor. This workbook is not designed like an ordi- nary math book where the concepts build log- ically on top of each other, and one chapter is a prerequisite for the next. For instance, most students know how to cancel common factors, expand an expression or solve simple simple equations; and we use them from the onset, while a detailed discussion of algebra operations or strategies for solving an equation factoring will be presented much later. Calculations are usually carried out with highest possible precision, but intermediate results (to help the student follow the steps) and final answers are rounded to a few decimal places, because most of the time two or three significant digits matter.

4 1 Scientiﬁc Notation Examples:

In biology, we often have to deal with very 256 = 2.56 102 3 significant digits large or very small quantities. For instance, −0.45 = −4.5 10−1 2 significant digits the adult human body consists of approximately 37 trillion cells; rod shaped bacteria, bacilli, Usually you can determine the significant are between 1 to 4 micrometers (microns) long; digits by counting the digits left to right, ig- drinking water may contain up to 10 parts per noring leading zeros. billion of arsenic by EPA standards, and so on. Trailing zeros are a little tricky. Undoubt- Scientific notation is a convenient way to ex- edly, press such numbers. 256.0 = 256

1.1 Scientific Notation The first number contains four significant digits, the second only three. In scientific notation Definition (Scientific Notation): Every pos- we would write 2.560 102, and 2.56 102, respec- itive real number x > 0 can be written in the tively. The first mantissa contains four digits, form the second only three. n x = a 10 Without scientific notation there remains a bit of ambiguity when dealing with large num- where the exponent n is an integer (whole numbers. For example, it is not possible to tell how ber). The number 1 ≤ a < 10 is called the many significant digits the number mantissa, and the number n indicates the order of magnitude. 42, 600, 000, 000 Examples: contains. We have 256 = 2.56 102 0.0035 = 3.5 10−3 42, 600, 000, 000 = 4.26 109 = 4.260 109 6.9 = 6.9 100 The number has at least three significant fig- The requirement that 1 ≤ a < 10 ensures ures, but it could be more. that scientific notation is unique. Otherwise As we work applied problems, we should we could have 22 = 22 100 = 2.2 101 = 0.22 102. keep the number of significant figures about 423 For negative numbers we can just include a the same. For instance, in 22.7 , both, nu- negative sign, as in −414 = −4.14 102. If you merator and denominator have three signifi- 423 want to write x = 0 in scientific notation, you cant digits. It makes no sense to write 22.7 = have to allow for a = 0. 18.63436123. This is pretending that we have very high accuracy, and that we stand behind the correctness of all of these ten digits. In- 1.2 Significant Figures stead we should round to three places and write

Signiﬁcant Digits (Figures): This is the 423 = 18.6 number of digits of the mantissa. 22.7

5 1.3 Rounding 1.4 Unit Preﬁxes

Rounding: Always round to the nearest. The major prefixes are listed in the Figure 1. tera T one trillion 1012 giga G one billion 109 mega M one million 106 kilo k one thousand 103 hecto h one hundred 102 deca da ten 101 Examples: one 100 Round to two significant figures: deci d one tenth 10−1 −2 2.336 ≈ 2.3 centi c one hundredth 10 milli m one thousandth 10−3 2.383 ≈ 2.4 micro µ one millionth 10−6 Round to three significant figures: nano n one billionth 10−9 pico p one trillionth 10−12 −22.469 ≈ −22.5 8 ≈ 2.67 Figure 1: Unit Prefixes 3 When rounding to the nearest, the cutoff As we go through the table, the powers of is five. Digits in the range 0 − 4 are rounded ten change in increments of three. This is sim- down, digits in the range 5 − 9 are rounded ilar to the way we group digits in blocks of up. If the digit in question is exactly equal to three. Example: five, we need a tie breaker. For instance, the number 2.35 is exactly in the middle between 23, 400, 000, 000, 000 2.3 and 2.4. A recommended strategy is to = 23.4 1012 = 23.4 trillion round to even. Example: There is also a fine tuning around unity. This 2.35 ≈ 2.4 accommodates things like centimeter, or hec- 2.45 ≈ 2.4 toliter, et cetera. 2.55 ≈ 2.6 Caution: A billion may not equal a bil- 2.65 ≈ 2.6 lion! The short scale, where a billion equals 109, is used in the U.S., Great Britain and We rounded to 2.4 or 2.6, rather than to 2.5. in the Arabic world. Continental Europe and In a more complicated calculation which in- Latin America use the long scale, where a bil- volves several steps, it is a good a idea to keep lion equals 1012, and 109 is called a milliard in the highest precision possible throughout, and that number scale. wait with rounding until the final answer is calculated. 1.5 Calculator For simplicity’s sake we shall use ”=” from here on, even when we are rounding, as in π = The descriptions are given for TI calculators. 10 3.142, or 7 = 1.4286. Use common sense when The labels of the keys may be different for deciding the number of significant figures! other brands, but the ideas are the same.

6 Use the EE key to enter numbers given in (c) From the table we obtain that a tril- scientiﬁc notation. For example lion equals 109, and thus

2.34 10−5 becomes 2.34EE − 5 63.9 trillion = 63.9 109 = 6.39 1010 Calculators will switch to scientific notation when numbers become too large or too The last step was necessary because small. On the TI-84+ the number 0.000234 is 63.9 is not between 1 and 10. shown as (d) 2.34E − 4 0.25 1011 = 2.5 1010 which is to be interpreted as 2.34 10−4. 2. Round the numbers to four significant You can also change the MODE to SCI. figures. In this case all numbers will be displayed in scientific notation. For instance (a) 345, 678 25, 000 becomes 2.5E4 (b) 0.0691 (c) 22.715 The engineering mode ENG is very useful. (d) π Here the powers of ten are incremented in steps 22 of three, and we can use the unit prefix table (e) directly. For example, 7 Solutions: 25, 000, 000 becomes 25E6 (a) 345, 678 ≈ 345, 700 and the number is 25 million. 678 is rounded to 700. (b) 0.0691 ≈ 0.06910 1.6 Worked Problems This problem is questionable, as the original number only has three sig- 1. Convert the given numbers to scientific nificant figures. notation (c) 22.715 ≈ 22.72 (a) 8, 630, 000 Round to even (2), rather than down (b) 0.004, 56 to 22.71. (c) 63.9 trillion (d) π = 3.141592 ... ≈ 3.142 11 22 (d) 0.25 10 (e) = 3.14285 ... ≈ 3.143 7 Solutions: 3. Compute the expression and express the (a) 8, 630, 000 = 8.63 106 result in scientific notation. Move the decimal point to the left (a) 3.521 107 − 2.63 105 by six spaces. (b) 6.9 10−4 × 2.34 106 (b) 0.004, 56 = 4.56 10−3 Move the decimal point to the right Solutions: Calculators will solve these prob- by three spaces. lems just fine. Here we give some detail.

7 (a) 3.521 107 − 2.63 105 (b) 6.4 · 10−7 × 3.75 · 103 We cannot subtract 2.63 from 3.521 5.7 104 (c) directly, because the exponents are 7.7 10−5 different. Instead we change the last 6 · 105 + 7.2 · 10−2 term so that the exponent becomes (d) 2.5 · 102 + 3 · 10−5 7 as well. (e) (5 · 104)2 7 5 3.521 10 − 2.63 10 (f) (5 · 105)2 7 7 = 3.521 10 − 0.0263 10 (g) (5 · 10−4)2 7 √ = (3.521 − 0.0263) 10 (h) 4.9 · 10−5 = 3.4947 107 √ (i) 4.9 · 10−4 ≈ 3.495 107 3. How many seconds are in a year? Use The larger number had just four sig- scientific notation and round the result nificant figures, and we shouldn’t put to four significant digits. any faith in the fifth digit. (b) 6.9 10−4 × 2.34 106 Answers This is straight multiplication. Mul- 1. (a) 4.3 · 1010 tiply mantissas and powers of ten −5 separately: (b) 4 · 10 (c) 1.2 · 101 6.9 10−4 × 2.34 106 (d) 6 · 100 = 6.9 × 10−4 × 2.34 × 106 (e) 2.5 · 10−1 = 16.146 × 102 ≈ 1.6 103 (f) 7.56 1013 1.7 Exercises (g) 5.3 10−5 1. Convert to scientific notation: 2. (a) 107 −3 (a) 43, 000, 000, 000 (b) 2.4 · 10 8 (b) 0.000, 04 (c) 7.043 · 10 (c) 12 (d) 2.4 · 103 (d) 6 (e) 2.5 · 109 1 (f) 2.5 · 1011 (e) 4 (f) 75.6 trillion (g) 2.5 · 10−7 (g) 53 millionth (h) 7 · 10−3 −2 2. Compute the expressions, and express the (i) 2.214 · 10 results in scientific notation. Round to 3. 3.154 · 107 seconds four significant digits, if necessary.

(a) 1.8 · 107 − 8 · 106

8 2 Metric System The definitions have since been updated to obtain higher precision. The metric system, or more specifically SI units Compound Units are made up from the ba- (SystèmeInternational d’Unités), dominates the sic units by multiplication or division. The scientific literature. An adult human is about most popular terms are summarized in the ta- 1.7 meters tall; the green tree frog (Hyla cinerea) ble. is about 4 to 5 centimeters long; the common vampire bat (Desmodus rotundus) weighs about Area length × length L2 55 to 60 gram; blood sugar levels are measured 3 in milligram per deciliter, to name a few exam- Volume area × length L ples. displacement L Velocity time T 2.1 Basic Physical Units change of velocity L Acceleration time T 2 Most physical units are composed of the three volume L3 basic units: length (L), time (T) and mass (M). Flux In the metric system these are measured in time T mass M meters (m), seconds (sec) and gram (g), re- Density 3 spectively. Other SI units are Ampere (A) for volume L ML electric current, Kelvin (K) for temperature, Force mass × acceleration mole (mol) for amount of substance and can- T 2 dela (cd) for luminous intensity. ML2 Work force × displacement T 2 Historical Background: work ML2 Power 1. Meter: The distance from the pole to the time T 3 force M equator is set as Pressure area LT 2 10, 000 km = 107 m Figure 2: Compound Units which makes one meter a ten millionth of that distance. 2.2 Metric Measurements 2. A second is based on the rotation of the 2.2.1 Length Earth. One revolution about its own axis takes 24 hours, which is equivalent to Length is measured in meters (m). For large distances use kilometers, where 24 × 60 × 60 sec = 86, 400 sec 1 m = 103 m = 1, 000 m

3. A kilogram is the mass of one liter (L) of For small lengths centimeters (cm), millimeters water (a liter is set as a one thousandth (mm) or microns (µm) are commonly used of a cubic meter). A gram then is a one thousandth of this quantity. 1 cm = 10−2 m = 0.01 m

9 1 mm = 10−3 m = 0.001 m 2.2.3 Volume 1 µm = 10−6 m = 0.000, 001 m The natural unit for volume is the cubic me- 3 This boils down to knowing your prefixes. ter (m ), which is the volume of a cube whose sides are all one meter long. This unit is rather 2.2.2 Area large, and most useful applications use liters (L), or even smaller quantities such as milliliters The basic unit is the square meter (sq. m or (mL). 2 m ), which is the area of a square where both A liter is defined to the volume of the cube sides are one meter long. whose sides are 10 cm long. In terms of cubic Large areas are expressed in square kilome- meters we find (10 cm = 0.1 m) ters, the area of 1 km by 1 km square. Specif- ically, 1 L = 0.1 m × 0.1 m × 0.1 m 1 km2 = 1 km × 1 km = 0.001 m3 = 1, 000 m × 1, 000 m 3 = 1, 000, 000 m2 = 106 m2 and therefore 1 m = 1, 000L. In reference cubic centimeters we see that Beware that a kilo of square meters (1,000 m2) is NOT equivalent to a square kilometer! 1 L = 10 cm × 10 cm × 10 cm A hectare (ha) is the area of a square of = 1, 000 cm3 size 100 m by 100 m. Therefore

1 ha = 100 m × 100 m Therefore 2 4 2 = 10, 000 m = 10 m 1 mL = 1 cm3 1 km2 = 100 ha that is, a milliliter is the equivalent of a cubic Small areas can be expressed in square cen- centimeter! timeters or square millimeters:

2 1 cm = 1 cm × 1 cm 2.2.4 Velocity and Acceleration = 0.01 m × 0.01 m Velocity is measured in meters per second (m/sec) = 0.000, 1 m2 = 10−4 m2 2 or in kilometers per hour (kmh or km/h), ac- 1 mm = 1 mm × 1 mm celeration is measured in meters per second = 0.001 m × 0.001 m squared (m/sec2). This is meters per second = 0.000, 001 m2 = 10−6 m2 per second.

It follows that Important velocities: The speed of sound is 344 m/sec. 1 m2 = 10, 000 cm2 The speed of light is 299,792,458 m/sec. 2 2 1 m = 1, 000, 000 mm The acceleration due to gravity on Earth is 1 cm2 = 100 mm2 9.807 m/sec2.

10 2.2.5 Flux 2.2.7 Density This quantity1 is the product of area and veloc- This is mass per volume, and usually density ity, or equivalently, the ratio volume per time. is denoted by ρ. A common unit is gram per cubic centimeter (g/cm3). L flux = area × velocity = L2 × Notice that T L3 volume g kg tonne = = 3 = = 3 T time cm L m 3 Flux can be measured in liters per second. The density of water is 1 g/cm = 1 kg/L, the density of air is about 1.2 mg/L and the Example: A small river with cross sectional density of the Earth is 5.52 g/cm3. area of 20 m2, and average velocity of 2 m/sec transports 2.2.8 Force m3 20 m2 × 2 m/sec = 40 Force is the product of mass and acceleration. sec Its SI unit is Newton (N), which is defined as L = 40, 000 sec 1 kg · 1 m 1 N = 1 sec2 2.2.6 Mass A much smaller unit for force is dyne, which is The basic unit for mass is a kilogram (kg), defined as which, under the appropriate conditions, equals 1 g · 1 cm the mass of one liter of water. 1 dyne = 1 sec2 A kilogram can be further partitioned into grams (g) or milligrams (mg), where It follows that

1 g = 0.001 kg 1 N = 100, 000 dyne 1 mg = 0.001 g The force of 1 kg on Earth is 9.81 N. or equivalently 2.2.9 Work 1 kg = 1, 000 g Work is the product of force and displacement. 1 g = 1, 000 mg It is equivalent to energy, and to momentum (mass times the square of the velocity). Com- For large masses we use the metric ton (tonne) mon units for work are Joule (J), erg, Watt hours (Wh) or calories. Definitions: 1 tonne = 1, 000 kg 1 J = 1 N × 1 m (Newton meter) 1In the sequel we will use the term ”flux” in the sense of a volumetric flow rate. In general terms, flux 1 erg = 1 dyne × 1 cm is defined as the product of density and velocity. For 1 Wh = 3, 600 J incompressible fluids (water, blood) these definitions are equivalent. 1 cal = 4.1840 J

11 Joule is considered the oﬃcial SI unit, the 2.2.12 Temperature others are listed for convenience. Biologists usually measure temperature2 in de- A calorie (cal) is the energy required to grees Celsius. It was originally set so that wa- raise the temperature of one gram of water by ter freezes at 0oC and it boils at 100oC. The one degree Celsius. relation to Kelvin (K) is that Beware that the calories listed on food items are kilo calories (kcal), and they are sometimes K = oC + 273.15 written as Cal, with an upper case C. 2.3 Worked Problems 2.2.10 Power 1. (a) How many millimeters make up a Power is the ratio of work per time. The SI kilometer? unit for power is 1 Watt (W) = 1 Joule/sec. (b) Convert 0.000,000,044 km to meters One horsepower is the equivalent of 745.7 and microns. Watt. While Watt is a measurement of power, Solutions: Watt hours (Wh) measure energy (power mul- (a) 1 km = 1, 000 m = 1, 000, 000 mm tiplied by time results in work). (b) 0.000, 000, 044 km = 0.000, 044 m Example: If you use a 25 Watt light bulb for = 44 10−6 m = 44µm one hour, you use up 25 Watt hours, and the energy consumption is 2. (a) Convert 25,000 square meters to hectare. (b) Is it reasonable that a leaf has area 25 W × 1 h 30 million square millimeters? What 25 J = × 3, 600 sec = 90, 000 J about 30 thousand square centime- sec ters? (c) How many cubic millimeters make 2.2.11 Pressure up a milliliter?

Pressure is the ratio force per area, and its SI Solutions: unit is Pascal (Pa), defined as (a) 25, 000 m2 = 2.5 × 10, 000 m2 1 N = 2.5 ha 1 Pa = 2 m (b) 30 million square millimeters is the area of a rectangle with sides 6,000 In the medical field blood pressure is mea- mm (= 6 m) and 5,000 mm (5 m). sured in mmHg (millimeter of mercury) and 1 This is unreasonably large for a leaf. mmHg = 133 Pa. 30, 000 mm2 = 150 mm × 200 mm, The tire pressure in your car should be about which is a 15 cm by 20 cm rectangle, 32 PSI = 221,000 Pa = 221 kPa, and the at- a reasonable size for a big leaf. mospheric pressure at sea level is about 101 kPa. 2The official SI unit for temperature is Kelvin.

12 Yes, leaves are not rectangular, but (b) 5,709,000,000 mg = 5,709,000 g we are just comparing sizes of leaves = 5,709 kg = 5.709 tonnes to sizes of rectangles. (c) 1 mm = 0.1 cm. Therefore 6. What work is required to lift 75 kg by 5 cm? 1 mm3 = (0.1 cm)3 Solution: = 0.001 cm3 = 0.001 mL The force exerted by 75 kg on Earth is

3. What is the ﬂux trough a blood vessel m 2 F = 75kg × 9.81 = 736 N whose cross sectional area is 4 mm when sec2 the blood passes through the vessel at 7 mm/sec? Hence, the resulting work is

Solution: W = 736 N × 5 cm 4 mm2 × 7 mm/sec = 736 N × 0.05 m = 36.8 J mm3 mL = 28 = 0.028 sec sec 7. A bird eats a worm containing 150 J of and it will take about 36 seconds for one energy. How many kcals (food) is that? milliliter of blood to pass through the Solution: 1 cal = 4.1840 J. Therefore vessel. 150 4. (a) What is the mass of 12 mL of water? 150 J = cal 4.1840 (b) What is the mass of one cubic meter = 35.85 cal ≈ 0.036 kcal of water?

Solutions: 8. (a) What physical quantity is obtained (a) The mass of one liter of water is 1 when you divide velocity by accel- kg = 1,000 g. Multiply by 0.012 to eration? obtain 12 mL, and the mass is 12 g. (b) What physical quantity is obtained (b) The mass of one liter of water is 1 when you multiply mass and the square kg = 1,000 g. Multiply by 1,000 to of a velocity? obtain a cubic meter; the resulting Solutions: mass is 1,000 kg = 1 tonne. velocity L/T 5. How many tonnes are in (a) = = T acceleration L/T 2 (a) 23 kg, The answer is time. (b) 5,709,000,000 mg? L 2 (b) mass × velocity2 = M × Solutions: T ML2 = = work (a) 23 kg = 0.025 tonnes T 2

13 2.4 Exercises 7. A base pair of DNA is approximately 0.340 nm long (1 nm = 1 nanometer = 0.000,000,001 1. (a) Convert 2,000 mm to cm, to m and m). The human genome is about 3.2 bil- to km. lion base pairs long. What is the length (b) Convert 2.12 cm to meters. of human DNA measured in meters? (c) Convert 0.000,021 m to µm.

2. A ﬁsh measures 37.5 cm in length. Con- 8. How many square kilometers are the equiv- vert the length to mm and to m. alent of 4,000 hectares?

3. Convert to meters and express the results in scientific notation. Some values are 9. How many liters does a rectangular tank averages or crude approximations. with dimension 40 cm x 50 cm x 80 cm hold? (a) 53 pm, the atomic radius of hydrogen, 10. How many milliliter of water fit into a (b) 600 nm, the wavelength of visible container of size 25mm x 25mm x 10mm? light, (c) 2 µm, size of bacteria, 11. A leaf has area 25 cm2. Convert the area (d) 100 µm, the thickness of human hair, to square millimeters. (e) 0.76 mm, the thickness of ID cards, (f) 3 cm, the size of an acorn, 12. How many cubic millimeters fit into a (g) 1.75 m, the height of a human, solid with volume 0.03 L? (h) 42.2 km, the length of a marathon, (i) 6,371 km, the radius of the Earth, 13. How many cubic millimeters are contained (j) 384,000 km, the distance to the moon, in one milliliter? and (k) 150 million km, the distance to the 14. A roll of bandage is 10cm wide and 2.5m sun. long. What is its area? 4. How many kilometers are covered by a 4 x 400m relay? 15. How many 2.5 mL doses of a vaccine can be obtained from 1.5 L? 5. A particle moves 14 cm, and then 9 km, and then 455 µm, and finally 87mm. Ex- press the total distance in meters. Do not 16. How many liters of water fell on 5 hectares round the result. of land when the rain was measured at 4cm? 6. Hesperoyucca whipplei is among the fastest growing plants on Earth, and it was recorded to have grown 3.7m in 14 days. What is 17. Estimate the area of the leaf pictured in the growth rate in mm per hour? the figure.

14 (c) cm per minute squared (cm/sec2)? kg (d) kg per meter per hour squared ( m hr2 )? 27. (a) What work is required to raise 55 kg by 6 m? (b) How much power is required to raise 55 kg by 6 m in 12 seconds?

28. Calculate how much energy in Joules and in megajoules your body has eﬀectively released into the environment if you go for a run and burn 600 Calories. (Re- 18. The cross sectional area of a small stream member Calorie 6= calorie) is 0.25 m2, and it ﬂows at 0.2 m/sec. How much water pases though the creek each 29. A cubic meter of gold has a mass of 19.3 second? tonnes. What is the density of gold in g/cm3 (gram per cubic centimeter)? 19. True/False: The mass of one milliliter of water is one gram. 30. An engineering website records the density of diamonds as 3,500 kg/m3. 20. How many grams are the equivalent of 0.045 tonnes? (a) Convert this value to g/cm3.

21. How much do 4 cubic centimeters of wa- (b) What is the volume of a 7 carat di- ter weigh? amond? Use: 1 carat = 200 mg.

22. An African elephant weighed in at 6.8 31. What is the pressure exerted by 100 dyne 2 tonnes. Convert the weight to kilogram on 5 cm ? and to gram. 32. Gasoline has a density of about 0.74 kg/L. 23. A liver cell has mass 0.000,000,003,5 g. What is the volume of 10 grams of gasoline? (a) Convert this mass to mg and µg. 33. Use scientiﬁc notation to express (b) How many cells make up one gram of liver tissue? (a) 12 milligram in gram 24. What (compound) unit is obtained when (b) 42.2 kilometers in meters (distance force is divided by pressure? of a marathon) (c) 29 tonnes in gram 25. What (compound) unit is obtained when force is multiplied by velocity? (d) 0.062 mL in liters (e) 20 µm in meters 26. What physical unit could be measured in 34. Find convenient units for the quantities (a) kilogram per liter? below. Example: 4.8·10−7 m = 0.48µm = (b) millimeter per day? 480 nm.

15 (a) 3.2 · 108g 12. 30,000 mm3 (b) 0.5 · 10−4L 13. 1000 (c) 8.8 · 108µm 14. 2,500 cm2 = 0.25 m2 (d) 2.5 · 107m2 (e) 4 · 10−10m3 15. 600

Answers 16. 2,000,000 L 2 1. (a) 2,000 mm = 200 cm = 2 m = 0.002 17. About 21 cm km 18. 50 L (b) 0.0212 m (c) 21 µm 19. True

2. 375 mm = 0.37 5m 20. 45,000 g

3. (a) 5.3 · 10−11 m 21. 4 g (b) 6 · 10−7 m 22. 6,800 kg = 6,800,000 g (c) 2 · 10−6 m 23. (a) 0.000,003,5 mg = 0.003,5 µg (d) 10−4 m (b) 286 million cells (e) 7.6 · 10−4 m (f) 3 · 10−2 m 24. area (g) 1.75 · 100 m 25. power (h) 4.22 · 104 m 26. (a) density (i) 6.371 · 106 m (b) velocity (j) 3.84 · 108 m (c) acceleration (k) 1.5 · 1011 m (d) pressure 4. 1.6 km 27. (a) 3,237 J 5. 9,000.227,455 m (b) 270 W

6. 11 mm/hr 28. 2,510,400 J = 2.5104 MJ

7. ≈ 1.1 m 29. 19.3 g/cm3

8. 40 sq km 30. (a) 3.5 g/cm3 9. 160 L (b) 0.4 mL

10. 6.25 mL 31. 2 Pa

11. 2,500 mm2 32. 13.5 mL

16 33. (a) 1.2 · 10−3g 3 Unit Conversion (b) 4.22 · 104m Unit conversion is routine in biological work. 6 (c) 2.9 · 10 g For instance, it is reported that sailfish (Istio- (d) 6.2 · 10−2L phorus) can swim 100 m in 4.8 seconds. How (e) 2 · 10−5 fast is this measured in miles per hour? Or, in a field study you determine that the aver- 34. (a) 320 tonnes age weight of tree squirrels in a state park is (b) 0.05 mL is 1 lb and 5 oz. What is the equivalent value in gram? Or, you may have to figure out how (c) 880m = 0.88 km many 4 mg doses of a vaccine can be obtain (d) 25 km2 from a 1.5 gal supply. (e) 0.4 mm3 This section will cover unit conversion methods. Before we get to the techniques, we will summarize the important relationships in tables. Conversion within the metric system was touched upon in the previous section and mas- tery of unit prefixes goes a long way for such problems.

3.1 Customary U.S. Units Historically, the common units used in the U.S. today were derived from British (imperial) units. They are closely related, but not quite identical. 1 ft = 12 in 1 yd = 3 ft 1 mi = 5,280 ft 1 mi = 1,760 yd 1 acre = 43,560 ft2 1 mi2 = 640 acres 1 gal = 4 qts 1 qt = 32 oz 1 ft3 = 7.480,519 gal 1 barrel = 42 gal 1 lb = 16 oz 1 ton = 2,000 lb

Figure 3: U.S. Units

Notice that the term ounce is used as a volume unit and as a mass unit. We speak of

17 ﬂuid ounces when talking about volumes, and 1 ft = 0.3048 m of dry ounces for masses. 1 yd = 0.9144 m Other customary units in the U.S. include 1 mi = 1.609,344 km horsepower (hp) and British thermal units (BTU). 1 in = 2.54 cm One horsepower (hp) is the power required to 1 ft2 = 0.092,903 m2 raise 550 lb at the rate of 1 ft/sec against grav- 1 mi2 = 2.589,988 km2 ity, and one BTU is the energy needed to raise 1 acre = 0.404,686 ha the temperature of 1 lb of water by one degree 1 qt = 0.946,353 L Fahrenheit. 1 gal = 3.785,412 L 1 lb = 0.453,592 kg 1 oz = 28.349,523 g 3.2 U.S. and Metric Units 1 ton = 0.907,185 tonnes The data are summarized in two tables. Figure 5: Metric to U.S. Things that are commonly expressed in feet or yards should be converted to meters, miles should be converted to kilometers, and inches Finally, we note that to centimeters. Areas of the size of acres should be expressed in hectares, a liter is about a 1 BTU = 1, 055.056 J quart, a pound is about one half of a kilogram, 1 cal = 4.184 J and imperial and metric tonnes are about the 1 hp = 745.7 W same.

1 m = 3.280,840 ft BTU stands for British thermal unit and hp is 1 m = 1.093,613 yd horsepower. 1 km = 0.621,371 mi 1 cm = 0.393,701 in 3.3 Conversion Methods 1 m2 = 10.764 ft2 1 km2 = 0.386,102 mi2 There are basically three methods that can be 1 ha = 2.471,054 acres applied to convert between units. We illustrate 1 L = 1.056,688 qt all three for the problem of converting 72 km/h 1 L = 0.264,172 gal to m/sec. Further examples are given in the 1 kg = 2.204,623 lb next section. 1 g = 0.035,274 oz 1 tonne = 1.102,311 tons 3.3.1 Substitution

Figure 4: U.S. to Metric Example:

For temperature conversions we have 72 km 72 × 1, 000 m = = 20 m/sec hr 3, 600 sec 5 oC = ( oF − 32o) 9 As the name indicates, 1 km was substituted o o o F = 1.8 C + 32 by 1,000 m, and one hour was replaced by 3,600 K = oC + 273.15 seconds.

18 3.3.2 Conversion Ratios 3.4 Worked Problems This is probably the most popular conversion 1. Sports: How many acres are in an NFL method. football ﬁeld (playing area plus end zones)?

Example: Solution (Conversion Ratios): A football 72 km ﬁeld is 100 yards long and 160 feet wide, hr and the end zones are 10 yards each. 72 km 1, 000 m 1 hr 1 min = × × × (100 + 2 × 10) yards × 160 feet hr km 60 min 60 sec 3 ft acres 72 × 1, 000 m = 19, 200 · yd · ft · · = = 20 m/sec yd 43, 560 ft2 60 × 60 sec = 1.32 acres The same calculation is frequently expressed in box form: 2. Highway: How many miles are the equivalent of 500 feet? 72 km 1, 000 m 1 hr 1 min hr km 60 min 60 sec Solution (Conversion Ratios): mi The idea behind this method is multiplica- 500 ft = 500 ft × tion by 1. For instance, 5, 280 ft 500 = mi = 0.0947 mi 1 km = 1, 000 m 5, 280 ≈ 0.1 mi and therefore 3. Land Management: Convert square 1, 000 m = 1 miles to hectares. What is the conver- 1 km sion factor? The same goes for Solution: 1 hour 1 min = 1 or = 1 2 2 2 2 60 min 60 sec (1 mi) = (1, 609 m) = 1.609 m ha = 2, 588, 881 m2 · 3.3.3 Conversion Factors 10, 000 m2 ≈ 259 ha All values found in the conversion tables are conversion factors. Thus, to convert from square miles to In our example we have to calculate factor hectares, you should multiply by 259. for conversion from km/h to m/sec ourselves 4. Volume: Convert 0.3 cubic inches to (substitution). milliliters. km 1, 000 m 5 m = = = 0.278 m/sec Solution: hr 3, 600 sec 18 sec 0.3 in2 = 0.3 × (2.54 cm)3 Once the factor is established, we can calculate = 0.3 × 2.543 cm3 72 km/h = 72 × 0.278 m/sec = 20 m/sec = 4.91 mL

19 Recall that cubic centimeters and milliliters The answer is 16,480 Watts, which is the are identical. equivalent of 22.1 horsepower.

5. Pediatrics: A baby is born at 7 lb. and 8. Medical: A person’s blood sugar is mea- 9 oz. Express the birth weight in gram. sured at 6 mmol/L (millimol per liter). Convert this value to mg/dL (milligram Solution: per deciliter). 7 lb + 9 oz = (7 × 16 + 9) oz 28.35 g Solution: mg/dL is the customary unit = 121 oz × = 3, 430 g for blood sugar in the medical field. A oz deciliter is one tenth of a liter. 6. Temperature Conversion We need a little chemistry background (a) Pediatrics: A child is running a to solve this problem. Glucose C6H12O6 temperature of 102.5 degrees Fahren- has atomic mass 180 Da3. Therefore one heit. What is the equivalent Celsius mol of glucose has mass 180 g (=180,000 value? mg), and substitution results in (b) Culinary: A recipe for a loaf of 6 mmol 0.006 mol 180, 000 mg bread calls for 220 degrees Celsius = · L 10 dL mol baking temperature. What is the 1080 mg American stove stetting? = = 108 mg/dL 10 dL Solutions: 3.5 Exercises (a) C = 5 (102.5 − 32) = 39.2 9 Use scientific notation, if necessary. (b) F = 1.8 · 220 + 32 = 428 1. How many fluid ounces are in a cubic 7. Engineering: How much power is re- foot? quired to raise two tons by five feet in two seconds? 2. How many miles are in 1500 feet?

Solution: The force is 3. Convert 4 million inches to miles. F = 2 tons · 9.81 m/sec2 4. Your lab animals need 3 ounces of a liq- kg m uid food supplement each day. How long = 2 · 1, 102.3 · 9.81 2 sec will a ﬁve gallon supply last? = 21, 627 N The distance is 5 ft = 1.524 m. Hence, 5. How many dry ounces make up 0.7 tons? the power is given by 6. Convert 0.3 acre feet to gallons. Acre 21, 627 N · 1.524 m foot is a volume unit commonly used for P = 2 sec large amounts of water. = 16, 480 W 3The respective atomic mass values are 12 Da for 1 hp = 16, 480 W · carbon, 1 Da for hydrogen and 16 Da for oxygen. Thus, 745.7 W glucose has atomic mass 6 × 12 Da +12 × 1 Da +6 × 16 = 22.1 hp Da = 180 Da.

20 7. Federal law has reserved 100 acres of land 17. A particular chemical has been shown to to be left unharvested. If the trees grow have beneﬁciary eﬀects on human cog- at a density of 3 trees/1000 sq feet, how nitive ability at dosages of 60µg per kg many trees have been protected from log- of body weight. However, this very same ging? chemical is harmful at dosages over 100µg per kg of body weight. What is a healthy 8. Doctors suggest to drink 8 cups of water dosage for a patient weighing 120 lbs? per day. How many milliliters is this? What dosage is harmful?

9. A 12 inch ruler is how many centimeters 18. (a) Convert 32oF, 70oF, 0oF and 1000oF long? to Celsius. o o o o 10. Buﬀalo Mountain in Floyd County is 3,971 (b) Convert 25 C, 37 C, −12 C and 300 C feet above sea level. What is its elevation to Fahrenheit. expressed in meters? (c) At what temperature do you get the same value on the Fahrenheit and 11. How many milliliters are in a gallon of on the Celsius scale? milk? 19. (a) Find the conversion factor from mmol/L 12. How many acres are in a 1 kilometer by to mg/dL for glucose. I. e. convert 3 kilometer enclosure? 1 mmol/L to mg/dL. 13. The average density of the Earth is 5.52 (b) Use the result from part (a) to ex- g/cm3. What is this density in (a) kg/m3, press (b) lb/ft3? i. 7.5 mmol/L ii. 4 mmol/L 14. Bob has a maximum speed of 14 mph, Eric has a maximum speed of 8 m/sec, in mg/dL. John’s maximum speed is 100 ft/min, and (c) What A patient’s glucose level is 153 Ted has a maximum speed of 800 km/day. mg/dL. What is the equivalent in Who is the fastest? mmol/L?

15. The current land speed record was set Answers by Andy Green with a Thrust SSS, and he reached 1224 kmh, and thus he be- 1. 957.5 oz. came the ﬁrst driver of a land vehicle to 2. 0.284 miles break the sound barrier. Suppose you could sustain this speed, and the roads 3. 63.1 miles were cleared oﬀ, how long would it take to drive the 2,790 miles from L.A. to New 4. 213 days York City? 5. 22,400 oz.

16. Find the largest value of the following 6. 97,755 gal. masses: 0.1 pounds, 2 ounces, 30 grams, and 0.02 kilograms. 7. 13,068 trees

21 8. 1,893 mL 4 A Little Geometry

9. 30.48 cm The surface area to volume ratio is an impor- 10. 1,210 m tant concept in biology. Leafs have a large surface area in relation 11. 3,785.4 mL to their volume, the surface area of apples is small compared to their volumes. A large sur- 12. 741.3 acres face area means a lot of interaction to the sur- 13. (a) 5,520 kg/m3 (b) 345 lbs/ft3 rounding environment, which is beneﬁcial for photosynthesis (sun light), but it also means a 14. Ted lot of exposure to the elements such as heat or extreme cold. 15. 3 hr 40 min 6 sec On a cellular level all nutrients must en- 16. 2 oz ter though the plasma membrane, and waste leaves that way. The size of the surface area 17. 3.266 mg and 5.443 mg limits the size (volume) of the cell: A cell cannot survive if it doesn’t receive enough nour- 18. (a) 0oC, 21.6oC, -17.8oC and 537.80oC. ishment. o o o o (b) 77 F, 98.6 F, 10.4 F and 572 F In this section we review some basics of ge- (c) -40oC = -40oF ometry.

19. (a) 1 mmol/L = 18 mg/dL 4.1 Two Dimensions: Area and Perime- (b) (i) 135 mg/dL (ii) 27 mg/dL ter (c) 8.5 mmol/L

The simplest two-dimensional object is a rectangle. If the sides are denoted by a and b, respectively, the area A becomes

A = ab

and the perimeter p is

p = 2a + 2b

22 For the special case of a square (a = b = x) we ﬁnd that

A = x2 and p = 4x

The area A of a circle with radius r is

A = πr2 The area A of a triangle is one half of the product of base and height: and its circumference c is given by 1 c = 2πr A = ch 2 If the diameter d is used, we ﬁnd that and the perimeter p is d = 2r p = a + b + c c = πd π d2 For right triangles, and for those triangles A = 4 only, we have the famous Pythagorean formula The units for areas must be the square of a c2 = a2 + b2 length, such as square foot, square centimeters, et cetera. where c is the hypothenuse (the leg opposite the right angle). For such triangles one can 4.2 Three Dimensions: Volume and use the remaining sides as base and height re- Surface Area spectively, and the area becomes ab A = 2 The number π is ubiquitous when dealing with circles and spheres. Is is deﬁned as the ratio between circumference and diameter of a circle circumference π = = 3.141, 592, 653,... The three-dimensional counterpart of a rect- diameter angle is a box. Its volume V is given by the which remains the same, regardless of the size product of the circle. V = abc

23 and the surface area consists of six rectangular and its surface area S is faces with total area S = 2πr2 + 2πrh = 2πr(r + h) S = 2(ab + bc + ca) The term 2πr2 accounts for the areas of the For the special case of a cube (a = b = c = bottom piece and the lid, and the term 2πrh x) we have measures the lateral surface area. V = x3 and S = 6x2

A sphere with radius r has volume Finally we look at a cone. Its has a cir- 4 cular base with radius r, and a height h. The V = πr3 3 lateral length l from the tip to the base perime- and surface area ter is connected to the other quantities (right triangle, Pythagoras) S = 4πr2 l2 = r2 + h2

The volume of such a cone is given by 1 V = πr2h 3 and its surface area is

S = πr2 + πrl

Volumes can be measured in cubic feet, cubic meters, and more importantly liters. Look for the third powers of length units. A circular cylinder is shaped like a can. It is determined by the radius r of the circles 4.3 Doubling Lengths and Surface Area on top and bottom, and by its height h. The to Volume Ratios volume v of the cylinder is given by It is easy to see that the area of a square grows V = πr2h by a factor of 4 if the sides are doubled.

24 This can also be formally veriﬁed. We use x In an abstract calculation we ﬁnd that for the original length of the sides and x0 = 2x V 0 = (x0)3 = (2x)3 for the new length. Then the respective areas 3 are = 8x = 8V

A = x2 and a similar computation can be used to see that the volume of a sphere also raises by an A0 = (x0)2 = (2x)2 = 4x2 = 4A eightfold when the radius is doubled. This is a feature of three-dimensional objects (8 = 23). The surface area to volume ratio of a three-dimensional object is deﬁned as surface area S SVR = = volume V For a cube whose sides have length x, we have V = x3 and S = 6x2. Hence For circles it is not that obvious to tell by S 6x2 6 SVR = = = what factor the area is increased when the ra- V x3 x dius (or the diameter) is doubled. A formal and we can calculate the surface area to volume 0 calculation with r and r = 2r as radii shows ratio directly. that When the sides of the cube are doubled, 0 A0 = π(r0)2 = π(2r)2 the new surface area is S = 4S and the new volume becomes V 0 = 8V , as we saw above. = π4r2 = 4A Therefore, the new surface area to volume ratio As a rule of thumb, we see that the area is quadruples when the sides or the radius of an S0 4S SVR0 = = object are doubled. This is a characteristic of V 0 8V the two dimensional space (note that 4 = 22). 1 S 1 = = SVR Things change in three dimensions. When 2 V 2 we double the sides of a cube, the volume will and which is one half of the original ratio. become an eightfold of its original size. For instance, if you have a set of dice, all the same 4.4 Angular Speed and Centrifuges size, and you want to build a die one whose sides are twice as long, you need eight dice for The angular speed measures how fast some- your construction. thing is revolving. This could be a search light

25 at an airport, the crankshaft in a car, an old This is more than double the gravitational force, vinyl record or a centrifuge, to name a few ex- which is enough to keep the water from leaving amples. the bucket in an up-side down position. One full rotation is 360o, which in radians equals 2π = 6.283.... We shall denote angular Centrifuges are used to separate cell mate- speed by ω, and we will express it in radians rial, and in order to isolate a certain substance, per time. you have to adjust the rpm setting to obtain the correct RCF. Suppose that the rotor has Example: A search light makes one full rev- radius r cm, and that rpm is the rpm setting. olution every 6 seconds. The angular speed Then becomes 2π rpm · 2π rpm · π ω = = 1.257/sec ω = = 5 sec 60 sec 30 sec Example: The engine in a car revolves at and 3,500 rpm (revolutions per minute). The an- 2 2 rpm ·π · r cm gular speed becomes 900 sec2 RCF = 9.8 m 3, 500 · 2π sec2 ω = = 366.5/sec π2 rpm2 r 60 sec = 900 · 9.8 · 100 In centrifuges we generate an acceleration = 1.119 10−5 rpm2r (which becomes a force when applied to a mass). The parameters of importance are the radius Therefore of the rotor r, the angular speed ω and the rel- RCF RCF ative centrifugal force RCF , which relates the rpm2 = 89365.3 ≈ 90, 000 r r acceleration of the centrifuge to that of gravity. It is computed as and s ω2r RCF RCF = rpm ≈ 300 g r 2 where g = 9.8 m/sec is the constant of gravity. Example: The radius of the JA-20 rotor in a Example: A bucket of water is attached to a Beckman J2-21 centrifuge is 10.8 cm. Isolation rope so that the combined length of arm and of chloroplasts requires a RCF of 1,300. The rope is r = 1.5 m. You spin the bucket such proper rpm selection is that one revolution takes 1.5 seconds. Here the r1, 300 angular speed is rpm = 300 = 3, 291 ≈ 3, 300 10.8 2π 4π ω = = / sec 1.5 sec 3 4.5 Worked Problems and the relative centrifugal force becomes 2 1. The diameter of a quarter, as in a 25 ω2r 16π · 1.5 m RCF = = 9 sec2 cent coin, is 24.26 mm, its thickness is g 9.8 m sec2 1.75 mm, and its mass is 5.670 g. 16 · π2 · 1.5 = = 2.69 (a) What is the area of each of the faces? 9 · 9.8

26 (b) What is the volume of a quarter? (a) We have r = 0.5µm and h = 3µm. (c) What is its density? Thus 2 3 Solutions: V = π · 0.5 · 3µm = 2.356µm3 (a) We are looking for the area of a circle. The diameter is 24.26 mm; thus and the radius is 12.13 mm, and the area becomes S = 2π · 0.5(0.5 + 3)µm2 = 11.00µm2 A = π(12.13 mm)2 = 462.2 mm2 = 4.622cm2 (b)

11.00µm2 (b) This is the volume of a cylinder. We SVR = = 4.67/µm have just computed the area of its 2.356µm3 base, and multiplication by the height yields the desired result (1.75 mm = (c) Mass = density × volume. Thus 0.175 cm) m = 1.1 g/mL × 2.356 µm3 V = 4.622cm2 × 0.175 cm µm3 = 2.6 g = 0.809 cm3 = 0.809 mL mL The tricky part is to simplify the (c) Density is mass per volume, and we ratio of cubic microns to milliliters obtain (cubic centimeters). Since 5.670 g ρ = = 7.01 g/mL 1µm = 0.001 mm 0.809 mL = 0.000, 1 cm = 10−4 cm 2. The bacterium Escherichia coli (E coli) is shaped like a rod (bacillus) and it is it follows that about 3 µm long and 1 µm in diame- 3 3 −4 3 ter. Use a cylindrical model to answer 1µm = 10 cm the questions below. = 10−12 cm3

(a) What are the volume and the sur- and that face area? µm3 (b) What is the surface area to volume = 10−12 ratio? mL (c) What is the mass of a single cell if Therefore the mass of the cell is we assume that the density is ρ = −12 1.1 g/mL? m = 2.6 10 g = 2.6 pg

Solutions: Here pg stands for picogram.

27 3. Repeat the exercise for a eukaryotic cell, Here we are dealing in nanograms. assuming that the cell is shaped like a sphere with radius 24µm. 4. Find a formula for the surface area to volume ratio for a sphere. (a) What are the volume and the surface area? Solution: We apply the formulas for a (b) What is the surface area to volume sphere and obtain ratio? (c) What is the mass of a single cell if 4πr2 3 SVR = = 4 3 we assume that the density is ρ = 3 πr r 1.1 g/mL? Here are some useful applications of this Solution: result: (a) Here r = 12µm and from the formulas for spheres we ﬁnd (a) For the eukaryotic cell with r = 12µm we get SVR = 3 = 0.25/µm, 4 12µm V = π(12µm)3 which conﬁrms the result of the pre- 3 vious exercise. = 7, 238 µm3 (b) The formula shows that doubling the and radius will reduce the surface area S = 4π(12µm)2 to volume ratio by 50%. 2 = 1, 810µm (c) Suppose you have an estimate of the (b) minimum surface to volume ratio, which a cell can handle. Then you 2 1, 810µm can calculate the maximum radius SVR = 3 = 0.25/µm 7, 238 µm of a spherical cell from In comparison to the bacterium we 3 see that the surface area to volume r = ratio is much smaller for our eukary- SVR ote. This requires a higher degree sophistication on part of eukaryotic 5. A dump truck can carry about 10 cubic cell in order move matter around in- yards of dirt. side the cell, and to pass it through the plasma membrane. (a) Given that dirt has a density of 1.6 g/cm3, what is the weight (mass) of (c) The mass of the cell becomes (mul- one truck load? tiply mass and density) (b) Excavation for a building site re- m = 1.1 g/mL × 7, 238 µm3 quires the removal of µm3 = 7, 962 g (42ft)x(29ft)x(8ft) of dirt. How many mL truck loads are required? = 7, 962 × 10−12 g = 7.962 10−9 g ≈ 8 ng Solutions:

28 (a) First we convert 10 cubic yards to Therefore cubic centimeters: 24 mi2 r2 = = 7.64 mi2 10 yd2 = 10 × (91.44 cm)3 π = 7.646 106 cm3 and p Hence, the mass becomes (volume r = 7.64 mi2 = 2.76 mi times density) The diameter is twice the radius: g m = 7.646 106 cm3 × 1.6 d = 5.5 miles. cm3 = 12.2 106 g = 12.2 103 kg (b) Asking for gallons means that we are looking for a volume. We are = 12.2 tonnes = 13.5 tons dealing with a huge circle covered with a layer of oil. Thus the for- (b) The volume of dirt on our site is mula for the volume of a cylinder applies (this cylinder is very flat). V = 42 × 29 × 6 ft3 yd3 1 = 7, 308 ft3 × V = 24 mi2 × in 27 ft2 8 = 270.67 yd3 We still have to convert the product Thus, approximately 27 loads are of square miles and inches to gal- required. It’s up to the driver to lons. We use feet as a common com- decide what to do about the extra parison unit for miles and inches, 0.67 cubic yards. and then convert to gallons and barrels. 6. Oil Spill. A circular oil spill in the Gulf 24 of Mexico covers 24 square miles. V = mi2 × in 8 1 (a) What is the diameter of the pol- = 3 × (5, 280 ft)2 × ft luted area? 12 7.48 gal = 6.97 106 ft3 × (b) Suppose that on average the oil layer ft3 is an eighth of an inch thick, how = 52.1 106 gal many gallons have been spilled? = 1.24 106 barrels Solutions: = 1.24 million barrels With wind and ocean currents, a real oil spill will never look like a circle; we just One barrel equals 42 gallons. use this model as an approximate shape. 7. Water Resources. A river is 1,800 feet (a) We know the area, and we want to wide and the average depth is 20 feet. determine radius and diameter. If the water flows at 2.5 mph on average, then how many gallons flow down 24 mi2 = A = πr2 the river each second?

29 Solution: drum slows down and gravity takes over, The ﬂux is the area of the cross section and the riders slide down the wall slowly. (rectangle) multiplied by the velocity. Suppose that you stand 3 meters from mi the center and that the barrel makes 30 F = 1, 800 ft × 20 ft × 2.5 hour revolutions per minute. mi = 90, 000 ft2 (a) How fast are you moving, and hr (b) what relative centrifugal force (RCF) Substituting 1 mi = 5,280 ft and 1 hr = do you experience? 3,600 sec we obtain

90, 000 × 5, 280 ft3 Solution: F = 3, 600 sec (a) The circumference of the rotor is ft3 7.48 gal 2π ·3 m and you travel this distance = 132, 000 × sec ft3 30 times in a minute, which makes gal for a speed of = 987, 000 × sec 2π · 3 · 30 m s = = 3π m/sec Roughly, this river transports a million 60 sec gallons of water each second! The speed is 9.425 m/sec, which con- 8. An artiﬁcial pond in a zoo has a circu- verted to miles per hour becomes lar boundary with a 40 meter diameter. 21.1 mph. The depth decreases linearly from zero to 30·2π (b) The angular speed is ω = 60 sec = 5 meters in the center. How much water π sec, and therefore does it contain? Assume that it is shaped like a cone (up-side down). π2 sec2 · 3 m RCF = 9.8 m = 3.018 Solution: sec2 The data for the cone are r = 20 m and Three times the acceleration of grav- h = 5 m. Thus, the volume becomes ity will keep the riders glued to the 1 wall! V = π(20 m)2 5 m 3 = 2, 094 m3 4.6 Exercises

This is the equivalent of about 2 million 1. A circular plot at a garden show has cir- liters. cumference 75 ft. What is its area?

9. The Rotor is an amusement park ride. 2. (a) By how much does the area of a It consists of a large upright barrel spin- square change, if all sides are in- ning at fast rates. Once the barrel has creased by a factor of 5? attained full speed, the ﬂoor is retracted, (b) By how much does the volume of leaving the riders stuck to the wall of the a cube change, if all sides are in- drum. At the end of the ride cycle, the creased by a factor of 5?

30 3. A farmer puts up a fence with perimeter 11. Suppose that a eukaryotic cell has the 2,500 ft to enclose a pasture. Assuming shape of a cube whose sides are 24 µm that the pasture is shaped like a square, each. how many acres is it? (a) What are the volume and the sur- 4. Use 1 L = 1,000 cm3. face area?

(a) A cube is to hold one liter, how long (b) What is the surface to volume ratio? are the sides? (c) What is the mass of the cell, assuming that the cell density is ρ = (b) A sphere is to hold one liter, what 1.1 g/mL? is its diameter? 12. The cross sectional area of a small creek 5. Find a formula for the surface area to is 2,500 cm2, and its average flow rate is volume ratio of a circular cylinder. 0.4 m/sec. How many liters of water pass 6. An ant travels along the perimeter of a through the creek each hour? circle at a speed of 1.6 cm per second. 13. How long will it take for 0.1L of blood to What is the area of the circle if it takes pass through a vessel with diameter 3mm the ant 3 minutes to complete one loop? if the average flow rate is 12mm/sec? Express the result in square meters. 14. How fast do second hand, the minute 7. A cylindrical test tube has diameter 1 hand and the hour hand move on a clock? cm. How tall is the column of liquid, Express your answers in revolutions per measured from the bottom, if you fill the minute and in radians per hour. test tube with 6 mL of a liquid? 15. An object attached to a rope is spun with 8. The distance from the equator to the pole radius 10 m, and such that it takes four is 10,000 km. What is the surface area of seconds for one revolution. the Earth, assuming that it is a perfect sphere? (a) What is the relative centrifugal force? 9. 40 million gallons of oil were spilled into (b) What is the RCF if the object spins the ocean during the Odyssey Oil Spill twice as fast? off the coast of Nova Scotia in 1988. As- (c) What is the RFC if the radius is in- suming that the layer of oil was 1/2 inch creased to 20 m (and it still takes 4 thick on average, what area did it cover? seconds for one revolution)?

10. A pond is shaped like a cone. Its diam- 16. The radius of the rotor in your centrifuge eter across is 56m and the depth in the is 12.3 cm. It takes 10,000 RCF to isolate center is 5m. mitochondria. What is the required rpm setting? (a) What is the surface area of the the water? (b) How many liters does the pond contain?

31 Answers 5 A Little Algebra

1. 447.6 sq. ft. The computations in the ﬁrst sections involved some routine algebra. In this section we will 2. (a) 25 times as large review some basic algebra. Most of it should (b) 125 times more be familiar material. The binomial formula 3. 9 acres 2 2 2 4. (a) 10 cm (b) 6.2 cm (p + q) = p + 2pq + q 2(r + h) 1 1 is related to the Hardy-Weinberg Law from ge- 5. SVR = = 2 + rh r h netics. 6. 0.66 m2 5.1 General Laws 7. 7.6 cm The commutative and associative laws are the 8. 509.3 million km2 basic rules of arithmetic. They work for addition 9. 4.603 square miles a + b = b + a 10. (a) 2,463 m2 (b) 4.105 million liters (a + b) + c = a + (b + c) 11. (a) 13,824 µm3 and 3,456 µm2 and for multiplication (b) 0.25/µm (c) 15.2 ng a · b = b · a (a · b) · c = a · (b · c) 12. 360,000 L

13. 19 min 39 sec There is nothing diﬃcult here (3 + 8 = 8 + 3).

14. Second hand: 1 rpm and 377 radians per hour. Minute hand: 1/60 rpm and 6.28 radians per hour. The associative laws allow us to group terms Hour hand: 1/720 rpm and 0.524 radians anyway we like it. Since (a+b)+c = a+(b+c), per hour. we might just omit the parentheses and write a + b + c. The same goes for multiplication: 15. (a) 2.515 (a · b) · c = a · (b · c) = a · b · c (b) 10.1 The distributive law connects addition and (c) 5.03 multiplication:

16. 8,553 rpm a · (b + c) = a · b + a · c

In all formulas the letters a, b, c are variables. You may replace them by any numbers

32 Examples: 1. You add 18% tip to a $ 45 restaurant bill. What is the total amount due? 45 + 0.18 · 45 = (1 + 0.18) · 45 = 1.18 · 45 The point is not so much that the total bill is $53.10, but that we can find this Figure 6: a · (b + c) = a · b + a · c amount directly by multiplying the tab by 1.18 (rather than taking 18% first and than adding it to the total). of your liking, and the equations will always work out. 2. A $300 recliner is reduced by 40%. How The multiplication symbol · is usually omit- much do you have to pay? ted. Everybody knows that 3x means 3 · x. Rather taking 40% of 300 and then sub- The equality symbol (=) means that two tract it from 300, you can find the sales expressions have the same value. It should price by taking 60% of 300: never be used as a mathematical punctuation 300−0.4·300 = (1−0.4)·300 = 0.6·300 symbol, separating different trains of thought! A hanging equality sign usually makes refer- The sales price is $180, of course. ence to a previous expression, as in 3. A population of 800 bugs increases by x2 − 49 = x2 − 7x + 7x − 49 25%. The new population is = x(x − 7) + 7(x − 7) 800 + 0.25 · 800 = 1.25 · 800 = 1, 000 = (x + 7)(x − 7) The growth factor is 1 + r = 1.25 The chain of equalities is to long to be fitted into a single line, and we have to break it up. Subtraction is the inverse of addition. Specif- We return to the distributive law. If you ically, if a and b are given numbers, then the read it from left to right, it tells you how to number x with the property that expand an expression; when read from right to a = b + x left, it is the factoring rule. The equality sign is a two-way street! For example, going left to is defined to be x = a − b. Similarly, division right in is the inverse of multiplication. If a and b are given, then x = a ÷ b is the number such that 4y(x + 2) = 4xy + 8y a = b · x we distribute the 4y with x and 2, while going right to left we extract the common term y This shows that division by zero is not permit- from both expressions. ted. Because if a 6= 0 and b = 0, we get the As a very common special case of the dis- contradiction a = 0 · x = 0, no matter what tributive law we note the formula value we take for x, and if both are zero, i.e. if a = 0 and b = 0, then any number x will do a + ra = (1 + r) a the trick.

33 5.1.1 Fractions Examples: 8 6 48 4 · 12 A fraction is the result of a division: 1. · = = = 4 a 3 4 12 12 = a ÷ b b 4x x2 4x · 6 8 2. · = = 3 6 3x2 x We have two important special cases 3 2 9 8 1 a a 3. − = − = = a and = 1 4 3 12 12 12 1 a x − 1 x2 − 1 (x − 1)(x + 1) Addition (or subtraction) of fractions de- 4. x = = x + 1 x(x + 1) x(x + 1) mands that the denominators are identical, in x − 1 1 = = 1 − which case x x a b a + b + = c c c 5.1.2 Hierarchy of Operations Multiplication and division of fractions abide Hypothetically the quantity 3 + 5 · 6 could be by the following rules either a c ac (3 + 5) × 6 = 8 × 6 = 48 · = b d bd a c a b ad or ÷ = = b d c d bc 3 + (5 × 6) = 3 + 30 = 33 Which is correct, 48 or 33? You multiply fractions by multiplication of the There is a convention about the order in respective numerators and denominators; divi- which algebraic operations are carried out. In sion of fractions follows a cross-multiplication. Again, there are important special cases: terms of priority we have the hierarchy Number times fraction 1. exponents b ab a · = 2. multiplication, division c c 3. addition, subtraction Cancelation of common factors Use parentheses if you want to break the order. ac a Operations with the same priority are usually = bc b carried out left to right. You may know this principle by the acronym You may also interpret this law as stating that PEMDAS (parentheses, exponents, multiplica- multiplication of a fraction by the same num- tion, division, addition, subtraction). ber in numerator and denominator does not In our example the answer 33 is correct. If change its value. we mean the calculation, which leads to 48, we Three locations of a minus sign have to write (3 + 5) · 6. −a a a We have used PEMDAS before without mak- = = − b −b b ing a big fuss about it. For instance, in the distributive law the expression a · b + a · c means

34 (a · b) + (a · c), but knowing the order of oper- 5.2 FOIL, Binomial Formulas and the ations, the parentheses are superﬂuous. Pascal Triangle Your calculator knows the order of op- Repeated application of the distributive law re- erations. If you enter sults in the famous foil formula: 3 + 5 ∗ 6 = (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd it will respond with 33. = F + O + I + L But there are pitfalls! The calculator will compute what you type in, and not what you If a = c and b = d, the FOIL-result be- mean. Here are some typical errors: comes

20 (a + b)2 = a2 + 2ab + b2 1. 2 + 3 If you type and if b is replaced by −b it follows that (a − b)2 = a2 + 2a(−b) + (−b)2 20 ÷ 2 + 3 = = a2 − 2ab + b2 the calculator interprets this as

(20 ÷ 2) + 3 = 10 + 3 = 13

as division has priority over addition. The correct way is to use parentheses

20 ÷ (2 + 3) =

which will produce the right answer.

42 Figure 7: (a + b)2 = a2 + 2ab + b2 2. 3 · 7 42 42 Obviously, 3·7 = 21 = 2. If you enter If in the FOIL formula a = c and d = −b, the middle terms will cancel and we ﬁnd that 42 ÷ 3 · 7 = (a + b)(a − b) = a2 − b2 the calculator will respond with 98, because (equal priority, left to right) The last three results are known as binomial formulas. If you are comfortable with 42 ÷ 3 · 7 = 14 · 7 = 98 them, you can avoid lengthy FOIL-exercises:

The correct way to enter the expression (x + 5)2 = x2 + 10x + 25 is (8 − x)2 = 64 − 16x + x2 42 ÷ (3 · 7) = 9 − x2 = 32 − x2 = (3 − x)(3 + x)

35 Figure 9: Pascal’s Triangle Figure 8: a2 − b2 = (a − b)(a + b)

easily write a formula for (a+b)n for any power Higher order binomial formulas can be con- n, and Pascal’s triangle will do just that. ﬁrmed by repeated multiplication and FOIL- The Pascal triangle looks like a pyramid, expansions. The third order binomial formulas with ones on the outside. The numbers on are the inside are obtained by adding the numbers which are directly above the space in the pre- (a + b)3 = a3 + 3a2b + 3ab2 + b3 vious line. 3 3 2 2 3 (a − b) = a − 3a b + 3ab − b For instance, if you look at the ﬁgure and a3 − b3 = (a − b)(a2 + ab + b2) you take the line beginning with

1 6 15 20 ... 5.2.1 Pascal’s Triangle Pascal’s triangle is a useful tool when working you see that 6 is obtained by adding 1 and 5, with binomial formulas. which are located above the 6, and then take If you look at the formula 15=5+10, 20=10+10, and so on. Just add the two numbers, which are directly above. (a + b)2 = a2 + 2ab + b2 you should notice that the powers of a decrease as you move from left to right, while those of b increase. When you add the exponents, the sum always equals 2 (use b0 = 1 and a0 = 1), Inspection of the Pascal triangle reveals that 2 and the pattern of coefficients 1-2-1 describes the coefficients of (a + b) show up in the third 3 the formula completely. line, those of (a+b) in the fourth. As it turns out, the coefficients of (a+b)n are the values in The same can be observed in the n-th line, provided that you start counting (a + b)3 = a3 + 3a2b + 3ab2 + b3 with zero. For instance (a + b)6 Here the exponents add to 3, and the coeffi- 6 5 4 2 3 3 cients exhibit a 1-3-3-1 pattern. Hence, if we = a + 6a b + 15a b + 20a b knew the pattern of the coefficients, we could +15a2b4 + 6ab5 + b6

36 or that a gamete has genotype AA is 36% (60% from the egg and 60% from the sperm). The 4 (x − 3) genotype Aa can occur in two ways: Allele A = x4 + 4x3(−3) + 6x2(−3)2 comes from the egg (60% probability), and a +4x(−3)3 + (−3)4 comes from the sperm (40% probability), or vice versa. In each case the probability is 0.24 = = x4 − 12x3 + 54x2 − 108x + 81 0.6 · 0.4 and overall, the genotype Aa has fre- FOIL it, if you don’t trust these results! quency 42% . Finally, genotype aa has frequency 16% = 0.42. This reasoning works for any values of p 5.3 Hardy-Weinberg Equilibrium and q, and we can summarize the result as We divert from the algebra review for a mo- Genotype Frequency ment, and look at a biological application. AA p2 The Hardy-Weinberg Law has its origin in Aa 2pq population genetics. Instead of studying ge- aa q2 netics for just one individual organism, we keep track of all alleles for a particular gene in the Since p + q = 1, it follows from the binomial population. formula that Suppose that a gene has alleles A and a. p2 + 2pq + q2 = (p + q)2 = 12 = 1 We denote by p the proportion of the allele A in the population, and by q the proportion of This is the Hardy-Weinberg principle. If the the allele a. Then, by design, we must have relative frequencies of genotypes follow this equation, we say that the population is in Hardy- p + q = 1 Weinberg equilibrium. Recall, that the distribution of genotypes Let’s say that for example the allele A has was based on probabilities alone. This pre- sumes that there are no mutations, there is no natural selection and no gene flow, that mating is random and that the population is large enough to justify probabilistic reasoning. When the population is in Hardy-Weinberg equilibrium it is not evolving with respect to that gene. Example: Wildflowers. A study of wildflowers on a meadow finds 320 red flowers (genotype RR), 160 pink flowers (Rr), and 20 white flowers (rr). Figure 10: Punnett Square Genotype Phenotype Freq. Perc. RR red 320 64% frequency p = 0.6 = 60% and that a has fre- Rr pink 160 32% quency q = 0.4 = 40%. If mating was solely rr white 20 4% based on the laws of probability, than the chances 500 100%

37 The data are summarized in the table. Here We compare observation and predictions in the alleles R and W determine the color. First a table. we calculate the frequency of the alleles in the observed H W equilib. population. Red flowers carry two R alleles, BB brown 373 49.9% 53.0% 397.5 pink flowers only have one R allele. Therefore Bb tan 344 45.9% 39.6% 297.0 bb white 32 4.2% 7.4% 55.5 R : 320 · 2 + 160 = 800 750 100% 100% 750 r : 160 + 2 · 20 = 200 The side-by-side comparison shows that the The values for p and q become observed tan phenotype is over-represented, com- 800 pared to the Hardy Weinberg predictions, at p = = 0.8 and 1000 the expense of the other phenotypes, and we 200 have a heterozygous advantage. q = = 0.2 1000 Further analysis is required to determine It now follows that whether the findings are statistically significant, or whether they can be explained by chance p2 = 0.82 = 0.64 = 64% alone. 2pq = 2 · 0.8 · 0.2 = 0.32 = 32% Example: PKU. Phenylketonuria (PKU) is q2 = 0.22 = 0.04 = 4% a recessive disease. When left untreated it can lead to brain damage, or even death. In the Since the predicted percentages match the ob- United States one in 15,000 babies is born with servations, we conclude that the population is PKU (for Caucasian babies the ratio is worse, in Hardy Weinberg equilibrium. it is 1:10,000). Example: Rock Mice. In a study of rock We assume that the population is in Hardy mice researchers counted 374 brown mice (BB), Weinberg equilibrium in regard to PKU. If A 344 tan mice (Bb) and 32 white mice (bb). is the dominant allele, and a is the recessive First we determine p and q. We have allele, we know that aa has frequency 1 B : 374 · 2 + 344 = 1092 q2 = 15, 000 b : 3440 + 2 · 32 = 408 Thus, With 750 mice we have a total of 1,500 alleles s 1 and the frequencies are q = = 0.008, 165 15, 000 1092 p = = 0.728 and 5000 and consequently, 408 q = = 0.272 1500 p = 1 − q = 0.991, 835 The equilibrium percentages are The relative frequency of Aa then becomes

p2 = 0.7282 = 53.0% 2pq = 0.0162 2pq = 39.6% and we conclude that about 1.62% of the pop- q2 = 7.4% ulation are heterozygotes, they carry the PKU

38 allele without showing symptoms of the dis- Routine applications of these rules are cal- ease. culations such as

6 9 15 5.4 Exponents 10 · 10 = 10 or (x2)2 = x4 For any number a and any positive integer we have the familiar definition Our definition of an did require that n must be a positive integer. If we want to use n = 0, an = a · a · a ··· a | {z } or allow for negative exponents, we need to n factors expand our definition, and we should do it in For example, such a way, that the basic laws remain valid. We begin with a0 and set x = a0. Then 54 = 5 · 5 · 5 · 5 = 625 ax = a · a0 = a1+0 = a or Division by a shows that x = 1, and we get 53 5 5 5 − = − × − × − a0 = x = 1. This only works if a 6= 0 and we 4 4 4 4 say that 00 is undefined4. 3 5 125 Now we look at negative exponents. Let = − 3 = − = −1.953, 125 4 64 a 6= 0 and set x = a−n. Then The special case a1 = a is obvious, but worth xan = a−nan = a−n+n = a0 = 1 mentioning. There are two fundamental rules for expo- Therefore, nents, and we begin by looking at examples. 1 a−n = x = First we have an The important special case n = −1 results in 53 · 54 = (5 · 5 · 5 · 5) · (5 · 5 · 5 · 5) = 58 1 a−1 b and we see that in a product and we add the −1 a = n and = exponents. Secondly a b a

4 There is a host of other rules governing op- 3 5 erations with exponents. But they can be eas- = (·5 · 5 · 5) · (5 · 5 · 5) · (5 · 5 · 5) · (5 · 5 · 5) ily derived from the basic concepts. For in- = 512 stance, We have am 1 tell us that when the exponents are stacked, m m −n m−n n = a · n = a · a = a we multiply them. a a The associative law lets us disregard the 4If a = 0, the equation becomes parentheses, and there is nothing special about 0 · x = 0 the numbers 3, 4 and 5. The underlying basic laws for exponents are and x could be any real number. For this reason we say that 00 is undeﬁned. Many formulas and results in 0 aman = am+n Calculus become easier to state if the convention 0 = 1 is used. See whether your calculator has an opinion on (am)n = amn this.

39 as well as If we were interested in 124/3, we could compute (ab)n = (ab) · (ab) ··· (ab) √ 124/3 = (121/3)4 = ( 3 12)4 = (a · a ··· a) · (b · b ··· b) = anbn There is nothing special about 12, 3 and 4, and and as a general rule we ﬁnd that n √ a a a a 1/n n = · ··· a = a b b b b a · a ··· a an and that = = b · b ··· b bn √ √ am/n = n am = n am

Examples: If n = 2, we are taking a square root, and we know that in this case a cannot be negative. 1. (4 · 105)2 = 42 · (105)2 = 16 · 1010 On the other hand, it is possible to take cubed = 1.6 · 10 · 1010 = 1.6 · 1011 roots of negative numbers as in √ −4 3 −64 = −4 3 −4−8 −12 2. 8 = 3 = 3 3 because (−4)3 = −64. We don’t want to get 3 0 too technical here, and we adopt the philoso- 3. = 1 5x2 phy that when in doubt we take a ≥ 0. It is worthwhile to point out that there is 3 4 34 81 no ± when you compute square roots. For in- 4. = = √ √ x2 (x2)4 x8 stance, 25 = 5 and not 25 = ±5. Undoubt- edly, the equation 5.4.1 Fractional Exponents x2 = 25 It is not uncommon to encounter expressions is solved by either x = 5 or by x = −5, but this like 124/3, 101.2 or 0.25−1/2. Your calculator is a different story. When we talk about the will give you numerical values for any of these. square root, we mean the principal root, and Why? How does the calculator evaluate these this is the non-negative answer. Your calcula- expressions? tor sees it this way too, there is no ± popping Again, we need to come up with a definition √ up when you compute things like 25. which is consistent with the basic rules. For the other introductory examples we find First we look at an example, and we set that x = 121/3. Then √ 6 1.2 6/5 5 6 3 10 = 10 = 10 = 1.585 3 1/3 1 ·3 1 x = 12 = 12 3 = 12 = 12 = 15.85

Thus, x3 = 12, and therefore x must be the and cubed root of 12: √ 0.25−1/2 = (0.25−1)1/2 = 41/2 = 4 √ 121/3 = x = 3 12 = 2

40 Most calculators use caret key ∧ to take 4. Expand and simplify the expression exponents. If your calculator does not have a 2 (convenient) key to take the n-th order root (x − 2)(x + 2)(x + 4) of a number, you can always use the 1/n-th power as an alternative, as in Solution: √ 5 10 = 101/5 = 100.2 = 1.585 (x − 2)(x + 2)(x2 + 4) = [(x − 2)(x + 2)] (x2 + 4) h i 5.5 Worked Problems = x2 + 2x − 2x − 4 (x2 + 4) 1. A population of 16,000 bees declines by = (x2 − 4)(x2 + 4) 12.5%. What is the new population? = x4 + 4x2 − 4x2 − 16 Solution: = x4 − 16

16, 000 − 0.125 · 16, 000 The solution becomes much shorter if you 2 2 = (1 − 0.125) · 16, 000 apply the formula (a−b)(a+b) = a −b (twice). = 0.875 · 16, 000 = 14, 000 5. Simplify the expression 3 1 The new population is 14,000 bees and (a) − 8 6 the multiplier is 0.875. 25x3 + 15x (b) 2. Expand the expression −3(4 − x). 10x2 x + 1 1 (c) x − . Solution: x x2

−3(4 − x) = −12 + 3x Solutions Don’t forget to distribute the negative (a) Here the least common denomina- sign. tor is 24, and we have 3. Factor the expression x2y2 − 3xy3 3 1 3 · 3 1 · 4 − = − 8 6 8 · 3 6 · 4 Solution: Both terms have one factor of 9 4 9 − 4 5 = − = = x and two factors of y in common: 24 24 24 24 x2y2 − 3xy3 (b) The numerator has the common fac- = x · x · y · y − 3 · x · y · y · y · y tor 5x, and we take advantage of it. = x · y · y(x − 3 · y · y) 25x3 + 15x = xy(x − 3y) 10x2 5x(5x2 + 3) 5x2 + 3 = = You don’t need to show this much detail (5x)(2x) 2x in your work! We only did this once to 5x2 3 5x 3 illustrate the details. = + = + 2x 2x 2 2x

41 5x2+3 5x 3 4 Both answers, 2x and 2 + 2x are (c) (10 + x) acceptable solutions of the problem. = 104 + 4 103x + 6 102x2 (c) As a ﬁrst step, we multiply numer- +4 10x3 + x4 ator and denominator in the ﬁrst = 10, 000 + 4, 000x + 600x2 term by x. This will get rid of the +10x3 + x4 double fraction, and it also generates a common denominator for the 7. Simplify two fractions. The rest is straightforward. (a) (5 · 10−1)−2 x5y−2z x + 1 1 (b) x − x3y5z3 x x2 3 2 1 18x 9z x + x x 1 (c) ÷ = − yz x2y x2 x2 x2 + 1 1 x2 + 1 − 1 = − = Solutions: x2 x2 x2 = 1 (a) (5 · 10−1)−2 = 5−2 · (10−1)−2 1 2 100 = 52 · 10 = 25 = 4 The problem is even simpler once −1 5 1 After all, 5 · 10 = 10 = 2 and you realize that 1 −2 2 ( 2 ) = 2 = 4. x + 1 1 (b) Collect the powers for each variable x = 1 + x x2 separately:

6. Expand the expressions. x5y−2z = x5−3y−2−5z1−3 x3y5z3 2 (a) (2.4 + x) x2 = x2y−7z−2 = (b) (2x − 1)3 y7z2 (c) (10 + x)4 (c) Cross-multiply and simplify:

Solutions: We apply the binomial formu- 18x3 9z2 18x3 · x2y 2x5 ÷ = = las. Feel free to FOIL and check the re- yz x2y yz · 9z2 z3 sults. 8. Compute the values (a) (2.4 + x)2 4/3 = 2.42 + 2 · 2.4 · x + x2 (a) 8 −3/2 = 5.76 + 4.8x + x2 (b) 9 −2.5 (b) (2x − 1)3 (c) 0.01 = (2x)3 − 3(2x)2 + 3(2x) − 1 (d) 2560.25 √ = 8x3 − 12x2 + 6x − 1 (e) ( 125)4/3

42 You are encouraged to conﬁrm all results Since on your calculator. 2 · p · q · 500 = 247.2 and Solutions: q2 · 500 = 152.8 √ (a) 84/3 = ( 3 8)4 = 24 = 16 we expect to have about 247 brown and 1 1 1 (b) 9−3/2 = = √ = about 153 yellow dogs. 93/2 ( 9)3 33 1 = 27 5.6 Exercises −2.5 (c) 0.01−2.5 = 10−2 1. Expand the expressions (−2)·(−2.5) 5 = 10 = 10 = 100, 000 (a) 3(4 + x) 3 (d) If we use that 125 = 5 , we obtain (b) −15(6 + x) √ 4/3 1/24/3 (c) −4(10 − x) 125 = 53 (d) (2 − x)(2 + x) 3· 1 · 4 2 = 5 2 3 = 5 = 25 (e) 3x(3 − x + y) 9. Allele A is dominant over a. Plants of (f) (5 + 2x)2 genotype A+ (this means AA or Aa) have (g) (x − 3)2 round seeds, those with genotype aa have (h) x(x − 2)2 wrinkled seeds. If the relative frequency of A in the population is 70%, then what 2. Factor the expressions percentage of plants will have wrinkled seeds? (a) 21 − 7x (b) 2xy − x2 Solution: We know that p = 70% = 0.7. 2 Therefore q = 1−p = 0.3, and q2 = 0.09. (c) 144 − x Thus, the relative frequency of aa is q2 = (d) 4x2 − 12x + 9 9%. 3. Express as a single fraction and simplify 10. A population of 500 dogs is in Hardy- 4 3 Weinberg equilibrium in regard to coat (a) − 3 4 color. If 100 dogs are black (BB), then 1 how many dogs should be brown (Bb), (b) x + x and how many are yellow (bb). 3 (c) × x2 Solution: We know that x x + 1 x − 1 100 1 (d) − p2 = = = 0.2 x − 1 x + 1 500 5 x2 6 Therefore (e) × √ 8 x p = 0.2 = 0.447 and x2 6 (f) ÷ q = 1 − 0.447 = 0.553 8 x

43 5000 L/sec 4. SuperSale! All items are 45% off! What (b) should you expect to pay for a $139.95 25 m2 500 cal camera? (c) 20 Watt 5. The rat population in a city increases by 13. Compute and simplify. Work using the 13% per year. Beginning with 2,300 rats, pertinent formulas and confirm your re- how many rats do you expect the follow- sults on a calculator. ing year? (a) 163/4 6. The population of African lions is currently estimated at 32,000. What is their (b) 0.253/2 population next year, if it declines by 8% (c) 125−2/3 annually? −3/2 4 (d) 9 7. A circle of radius r = 25 meters is changed −0.5 by x meters (the new radius is r = 25 + (e) 225 x meters). What is the resulting area? (f) 0.000, 320−1.2 Find a general expression involving x, and check your result when x = 2m, x = 14. A population has alleles A and B. Tests −1m and x = 10cm. show that 80% of all gametes have genotype AA. What percentage of the popu- 8. A circular pool in a zoo has a diameter lation has genotype BB if the population of 40 meters, and the spectators are sep- is in Hardy-Weinberg equilibrium? arated from the pool by a fence 2 meters away from the edge of the pool. How 15. A species has a gene with alleles A and much area is between the pool and the B. In a specific population of 4,670 you fence? count 3,781 individuals with genotype AA. How many have genotype AB and how 9. Simplify many have genotype BB, if the population is in Hardy Weinberg equilibrium? (a) 107 · 103 (b) (107)3 16. There are 22 black mice (BB), 34 grey mice (Bb) and 15 white mice (bb) in a 10−7 x8 (c) habitat. After the next breeding sea- 10−8 x9 son the population will reach 120 mice. (d) (2x3)8 Given that the population is in Hardy- 10. Expand and simplify the expression Weinberg equilibrium, estimate the num- (x + 4 · 103)2. Express all numerical val- ber of mice for each phenotype. ues in scientific notation.

11. Simplify (x−1 + y−1)−1

12. Compute and simplify 45 km (a) 60 m/sec

44 Answers (b) 1021 10 1. (a) 12 + 3x (c) x (b) −90 − 15x (d) 256x24 (c) −40 + 4x 10. x2 + 8 · 103x + 1.6 · 107 (d) 4 − x2 xy 2 11. (e) 9x − 3x + 3xy x + y (f) 25 + 20x + 4x2 12. (a) 750 seconds 2 (g) x − 6x + 9 (b) 0.2 m/sec 3 2 (h) x − 4x + 4x (c) 104.6 seconds

2. (a) 7(3 − x) 13. (a) 8 (b) x(2y − x) (b) 0.125 (c) (12 − x)(12 + x) (c) 0.04 (d) (2x − 3)2 27 (d) 7 8 3. (a) 1 12 (e) x2 + 1 15 (b) (f) 15, 625 x (c) 3x 14. 1.1% 4x 4x (d) = (x − 1)(x + 1) x2 − 1 15. 842 have genotype AB, 47 have genotype 3x BB. (e) 4 x3 16. 36.2 black, 59.4 grey and 24.4 white (round (f) 48 to whole numbers). 4. $ 76.97

5. 2599

6. 29,440

7. π(25 + x)2 m2 = π(625 + 50x + x2) m2 729π m2 = 2290 m2, 576π m2 = 1, 810 m2 and 630.01π m2 = 1, 979 m2

8. 264 m2

9. (a) 1010

45 6 Graphs order, and we do not connect dots.

Graphs are a splendid way to summarize data, Example: Eight second graders had a physi- and to present them visually. There are nu- cal. Here are the results. merous ways to represent biological data. In this section we will focus on the mathematical Name Height (cm) Weight (kg) side of graphing for scatter plots and functions; Allie 115 20.1 pie charts, frequency histograms or bar charts Brian 125 23.3 will not be considered. We look primarily at Caleb 122 24.8 continuous variables (as opposed to discrete or Dante 130 25.2 qualitative variables). We will use mathemati- Emma 119 24.7 cal examples, as well as illustrations from pop- Frank 122 23.0 ulation growth, allometry, medical data and Garth 116 22.1 performance in college. Haley 121 22.3

6.1 Scatter Plots Although not necessary, the names have When you graph data from a two-column ta- been included in the graph in order to em- ble, you interpret the numbers as respective phasize that in a scatter plot each data pair x and y coordinates, and then you plot the becomes a single point in the graph. points. In this fashion you can translate numerical data into points in a graph. Example: x y 1 5 3 8 3 3 6 4 −1 2 The ﬁgure shows the resulting graph obtained in EXCEL. Example: Scatter points look more impres- sive for larger data sets. The ﬁgure below relates student’s grades to their attendance records.

Each of the dots represents one pair on the list. There is no need to list the data in a particular

46 The horizontal axis shows the percentage score, can change that using the WINDOW button. and the vertical axis denotes the number of If you switch to the TABLE setting, the display missed classes 5 for students in a MATH 119 will give you data for selected x values. class. Students in the upper left corner quit As it was the case for scatter plots, pairs of coming to class, gave up on homework and did x and y are being displayed, but there are two not take exams. None of the students with a major differences: perfect attendance record failed the class (one struggled). Attaching names would break con- (a) We have infinitely many x values, at least 6 fidentiality, and it would clutter up the graph in theory . without giving new meaningful information. (b) For each x value there is only one y, which 7 Scatter plots can also be done on graph- results from the formula . In fact, the ing calculators using the ”LIST” environment. formula may fail to produce an answer, But data entry is a daunting task, and the if for example you are trying to divide by plots usually do not look very appealing. A zero, or take the square root of a negative computer is a much better tool for this task. number. This graphing method is tied to the notion 6.2 Function-Style Graphs of functions. A function is an input-output This is the type of graph, which you obtain relationship: To each (admissible) input x, we on a calculator. It requires that you type in assign exactly one output y. The letter f is a formula for y, and let the calculator do the frequently used to denote a function, and we rest. write Take, for example, the problem y = f(x) 2x pronounced ”f of x”. x is the input, and y = y = 1 + 2x−3 f(x) is the output. A more detailed discussion of functions is presented in Chapter 11. A graphing calculator will show a picture like Many examples for function-style graphs in the one given in the figure. biology use time t as the input. The outputs could be things like populations, temperatures, oxygen concentrations, and so on, measured at time t. But time is not the only conceivable variable. One could display photosynthetic energy production y as a function of the size of a leaf x, or the density of algae y as a function the depth x below the water surface, and so on. 6A calculator, or any other graphing device for that matter, will only display a finite number of points, and Typically, the x values and the y values will connect them by line segments. But the points are so range from −10 to 10 on a calculator, but you close together that they appear like a smooth curve. 7In the 2nd grader example Caleb and Frank are 5This is an example of a finite discrete variable; it both 122 cm tall, but Caleb weighs more. This results can take on the values 0, 1, 2 ... 42 only. in different y values for the same x-coordinate.

47 Graphs can also be used to conﬁrm the cor- Year U. S. Population rectness of an algebraic calculation. For exam- 1790 3,929,214 ple, in the last section we found that (simplify 1800 5,236,631 the expression ...) 1810 7,239,881 1820 9,638,453 x − 1 1 1830 12,866,020 x = 1 − x + 1 x 1840 17,069,453 1850 23,191,876 If we graph both sides of this equation on a 1860 31,443,321 calculator, we only see one graph, the reason 1870 38,558,371 being, that for any x-value, the values on the 8 1880 49,371,340 right and on the left are equal . After all, sim- 1890 62,979,766 plifying means that we don’t change the value 1900 76,212,168 of an expression, we just make it easier to han- 1910 92,228,531 dle. 1920 106,021,568 1930 123,202,660 1940 132,165,129 1950 151,325,798 1960 179,323,175 1970 203,211,926 1980 226,545,805 1990 248,709,873 2000 281,421,906 2010 308,745,538

Figure 12: U.S. Historical Data

1 x− x 1 Figure 11: x+1 = 1 − x advantage of this display is that the large (re- cent) population ﬁgures dominate the graph, and we can barely make out diﬀerences for the 6.3 Watch the Scales data from the early 1800s. 6.3.1 Log Plots We begin with an example. The U.S. census data are given in Figure 12.

The scatter plot is shown below. It has the typical shape of an exponential growth. A dis-

8Maybe you have noticed a minor technical diﬀer- ence. If we take x = −1, the value on the left is unde- ﬁned, but the value on the right is 2. A further inves- In this situation you will often see a display tigation would take us right into Calculus. with a logarithmic scale on the y-axis. Here

48 the powers of 10 are equally spaced, and ad- table below. justments are made in between. Lot size(m2) No. of Species 0.25 2 1 4 12 7 50 11 120 13 600 21 2, 500 31

In a regular scatter plot we see that a fairly nice curve emerges, but the graph is dominated by the largest lot, and the data for the smaller As we go down on the y-axis, the values get lots are cluttered near the vertical axis. closer and closer to zero (0.01, 0.001, 0.0001 ...), but they are never equal to zero. It is impossible to represent zero, or negative numbers on such a scale. A glimpse of the logarithmic scale is shown in Figure 13 below. The distance between 1 and 10 is the same as the distance between 2 and 20, or between 0.8 and 8 on this scale.

Figure 14: Area Species Curve

Figure 13: Logarithmic Scale By changing to a loglog scale we can show more detail on both ends of the scale, small and large. Moreover, a fairly linear graph emerges, which caught the attention of biologists. 6.3.2 Loglog Plots

Graphs with logarithmic scales on both axes are called loglog plots, and they are extremely useful in applications. Example: You investigate how the diversity of grasses on a meadow changes with area. On the smallest plot of 50 cm × 50 cm you ﬁnd just two species. As you increase the lot size you detect more and more species. On the largest plot (50 m × 50 m) you count 31 diﬀer- Figure 15: Area Species Curve in Loglog Scale ent species. The data are summarized in the

49 6.4 Worked Problems (a) How many animals where present at time zero? EXCEL or other graphing software are very useful for these exercises. (b) After how many generations will the population reach 2,500 individuals? 1. Temperature Graph. Answer the ques- (c) At what number will the population tions based on the graph. level oﬀ (carrying capacity)?

Solution: (a) The original population is about 300. (b) The population breaks the 2,500 mark after about 15 generations. (c) The population levels oﬀ at about (a) What were the high and the low 7,000. temperatures in Radford on Febru- 3. Consider the curve deﬁned by the equa- ary 25? tion (b) When were the temperatures above 4 2 freezing? y = x − 6x + 5x + 6 (c) When did the temperatures change (a) Complete the table below most rapidly? x y 0 Solution: 1 (a) The low at 21o F between 6 am and 2 7 am; the high was 34o F at 4 pm. 5 (b) The temperature remained above freez- 10 ing between 1 pm and about 5:30 (b) Graph the curve on a regular scale pm. for x between 0 and 10. (c) The fastest increase occurred between (c) Graph the function for x between 0 8 am and 9 am. and 10 on a log scale. 2. Animal Population. The graph of a (d) Graph the function for x between hypothetical animal population is shown 0.1 and 10 in a loglog plot. below. Solutions: x y 0 6 1 6 (a) 2 8 5 506 10 9, 456

50 1−1 4x The graphs of x−1 + and 4 4 + x are shown below.

(b) 6.5 Exercises 1. Construct a scatter plot for the data below. x y 4 7 2 6 8 5 3 3 0 4 7 2 (c) 2. The Life Table for Belding’s Ground Squir- rels at Tioga Pass in Nevada (see Camp- bell, p. 1173) contains the following data. These are the number of surviving animals of a cohort in the beginning of the year. Age Females Males 0 − 1 337 349 1 − 2 252 248 2 − 3 127 108 (d) 3 − 4 67 34 1−1 4 − 5 35 11 4. Simplify the expression x−1 + and 4 5 − 6 19 2 conﬁrm the result graphically. 6 − 7 9 7 − 8 5 Solution: 8 − 9 4 9 − 10 1 1−1 1 4x x−1 + = · −1 1 (a) Graph the data for females and males 4 x + 4 4x 4x in a common plot with regular scal- = 4 + x ing.

51 (b) Repeat part (a) with a logarithmic 8. Factor the expression x4−16 and conﬁrm scale on the y-axis. the result graphically. 3. The table below contains the heart rate at rest for selected mammals (Kong, NIU). Answers

Mass (g) Pulse (beats/minute) Mouse 25 670 Rat 200 420 Guinea pig 300 300 Rabbit 2,000 205 1. Small dog 5,000 120 Large dog 30,000 85 Man 70,000 72 Horse 450,000 38

Graph the data with a scatter plot on a loglog scale. It is not necessary to label the data points. 2. 4. Let y = x2 −3x+5. Complete the table and sketch the graph. x y −1 0 1 2 3 4

5. Sketch the graph of 2x y = 2x + 1 for −4 ≤ x ≤ 4. 3. 2 6. Sketch the graph of y = x − 3x + 5 for x y 0.1 ≤ x ≤ 100 on a loglog scale. −1 9 7. Sketch the graph of 0 5 4. 1 3 4x + 3x 2 3 y = x x 5 + 2 3 5 for −10 ≤ x ≤ 10 on a logarithmic scale. 4 9

52 5.

8. (x − 2)(x + 2)(x2 + 4)

53 7 Solving Equations solutions, it can have infinitely many solutions, or it may not have a solution at all. Equations are everywhere. In this section we Examples: look at the basics of solving equations, and the techniques discussed here will be applied in the 4x + 7 = x + 22 x = 5 remainder of this workbook in a variety of bi- x2 + 4 = 5x x = 1 and x = 4 ological applications. √ x2 = x all x ≥ 0 4x + 7 = 4x − 12 no solution 7.1 Equations and Solutions In the last section we dealt with expressions. 7.2 Solution Strategies They can be evaluated, simplified, graphed, ex- Before you get overwhelmed by technicalities, panded, factored, but they cannot be solved, keep the overall goal in mind: You want to find because they lack an equality symbol. all x so that the equality holds true. Some- An equation is a statement which claims times you can tell the answer by inspection (or that two expressions are equal to each other. by a lucky guess), sometimes there is a formula For example, in to find the solutions, sometimes it may take

2 many hours pursuing a systematic approach. x + 4 = 5x In this section we will ﬁrst look at the pencil and paper method (algebra), then we look we are looking for values of the variable x such for graphical solutions, and we conclude with that the expressions x2 + 4 and 5x have the the special case of quadratic equations. same value. x2 +4 and 5x taken by themselves are just expressions; we created an equation by connecting them with an ”=”. 7.2.1 Algebraic Methods A solution of an equation is a value of the When you use algebra, you try to isolate the variable, such that equality is achieved. For variable. There are two operations which do instance, the equation not change the solution set:

x2 + 4 = 5x 1. add (subtract) the same quantity to both sides of the equation, has solutions x = 1 and x = 4 because 2. multiply (divide) both sides of the equation by a non-zero quantity. x = 1 : 12 + 4 = 5 and 5 · 1 = 5 x = 4 : 42 + 4 = 20 and 5 · 4 = 20 It is left to you to decide which steps should be taken to solve the problem eﬃciently. 2 x = 3 is not a solution, because 3 + 4 = 13, Example: while 5 · 3 = 15. 1 x2 = x + 3 To solve an equation means to ﬁnd all pos- 4 sible solutions. First, we get rid of the fraction and multiply How many solutions does an equation have? both sides by 4: The answer depends on the equation. It can have exactly one solution, it can have several x2 = 4x + 12

54 Now we subtract 4x + 12 from both sides 7.2.2 Discussion of Pitfalls In this section we look at typical examples where x2 − 4x − 12 = 0 after careful calculations we end up with an- At this point factoring comes in handy swers which do not satisfy the original equations, or where we miss some of the solutions. (x − 6)(x + 2) = 0 Squaring both sides, or multiplication or division by zero are often at the heart of the From here we conclude that either x = 6 or problem. that x = −2, since these are the only two Squaring both Sides. If two quantities choices for x which make the expression on the are equal, then their squares must be the same. left equal to zero. True, but two numbers which are not equal can 2 2 It is usually a good idea to double-check have the same square, as in 2 = 4 = (−2) , and substitute the values for x into the original and this may be at the source of the problem. equation. When x = 6 we have Example: √ 3x − 5 = 5 62 = 9 = 6 + 3 4 When we take the square on both sides, we find that and both sides equal 9. When we substitute 3x − 5 = 25 x = −2 we find that and therefore 3x = 30 and x = 10. Substi- (−2)2 tution confirms that = 1 = −2 + 3 √ √ 4 3 · 10 − 5 = 25 = 5 and both sides equal 1. No problem here. The Zero-Product-Principle is a power- Example: Here we show how squaring both ful tool to solve equations. It states that if the sides might introduce extraneous solutions. We product of numbers is zero, then one of the consider the equation factors must be zero. x − 1 = 3 Example: If Obviously, x = 4 is the only solution, and 2 x(x − 1)(x + 2)(x − 3) = 0 squaring both sides is not necessary. But, for the sake of making a point, let’s say we square then x = 0, or x = 1, or x = −2 or x = 3, and both sides, and then take it from there. We we see that this equation has four solutions. obtain It is essential that one side of the equation is zero. We cannot tell much about the solution x − 1 = 3 set from the equation (x − 1)2 = 9 x2 − 2x + 1 = 9 x(x − 1)(x + 2)(x − 3)2 = 8 x2 − 2x − 9 = 0 because the value on the right is not zero. (x − 4)(x + 2) = 0

55 with solutions x = 4 and x = −2. The first an- Example: In this example we multiply both swer is the original solution. x = −2 became sides by x−2 in order to get rid of the fractions, a legitimate solutions after we took the square and then take the usual simplification steps. on both sides. This illustrates the danger of 6 3x squaring both sides, and it is good practice to x + = x − 2 x − 2 confirm your final results for the original equa- x2 − 2x + 6 3x tion. = x − 2 x − 2 Multiplication or Division by Zero. x2 − 2x + 6 = 3x Nobody in their right mind would do this. Di- x2 − 5x + 6 = 0 vision be zero is generally prohibited, and mul- (x − 2)(x − 3) = 0 tiplication by zero makes both sides of the equation zero, and we lose all information. The The solutions are solutions x = 3 and x = 2. crux is that multiplication or division by zero x = 3 solves the original equation, because are usually hidden by the use of variables. 6 3 · 3 3 + = 9 = Example: Consider this string of calculations 1 1 but x = 2 is not a solution, because the orig- x2 − 4 = x − 2 inal equation is not defined (not meaningful) (x − 2)(x + 2) = x − 2 when x = 2. It became a viable answer after x + 2 = 1 we multiplied both sides by (x − 2). Here multiplication by zero was disguised as the factor x = −1 (x − 2). x = −1 is a solution to the original equation, 7.2.3 Graphical Solutions because (−1)2 − 4 = 1 − 4 = −3 = (−1) − 2, and x = −1 makes both sides equal −3,. no Solving equations graphically is always an op- doubt about that. tion, no matter how hard or easy the problem But x = 2 works as well in the original might be! Efficient use of your graphing calcu- problem (4 − 4 = 0 = 2 − 2 and both sides are lator will yield useful numerical results! 0). So, where did x = 2 disappear? In the step In the example from the second to the third line we divided x2 + 4 = 5x by the quantity (x − 2). But when x = 2 we have (x − 2) = 0, and the division by (x − 2) we are looking for values x so that both sides knocked out the viable solution x = 2. of the equation have the same value. The table shows values on both sides for different choices Example: This example is even more obvious: of x: x x2 + 4 5x 2 x = 4x 0 4 0 x = 4 1 5 5 2 8 10 The original equation has solutions x = 4 and 3 13 15 x = 0. The latter disappeared as an option 4 20 20 after division by x. 5 29 25

56 and it is easy to see that both sides match when x = 1 and when x = 4. Instead of setting up tables, we can have a calculator graph the two expressions. We are now looking for choices of x with identical y- values. This means that we are interested in the points of intersection of the graphs. Again, there is a quick way to do this on your calculator. Look for things like ”zero” or ”root” on a graphing menu.

7.2.4 Quadratic Equations Mathematicians classify equations into various types. Equations of the form ax = b are called linear, and the solution is x = b/a, provided From the graph we see that the intersec- that a 6= 0. tions occur at the points (1, 5) and (4, 20). The Increasing the power of x results in a quadratic x-coordinate is the desired solution. The y co- equation, which takes the general form ordinate is a byproduct, as it shows the matching value of the expressions on either side. ax2 + bx + c = 0 In textbook pencil and paper problems the solutions often work out nicely as whole num- It is understood that a 6= 0, otherwise the bers or simple fractions, but in most realistic equation would be linear. Quadratic equations problems this is not the case. The calculator is can have two, one or no solutions. not afraid of messy numbers, and most graph- There are three common ways to solve a ing calculators have a way to identify points of quadratic equation. directly intersections. Look for a CALC menu 1. Factoring. This is a fabulous short-cut! under graphing, select ”intersect”, and be pre- But it does not always work. For in- pared to answer a few questions (which curves stance, are involved, what is the interval of interest). Detailed instructions depend on the brand of x2 − 5x + 4 = 0 the calculator. (x − 1)(x − 4) = 0 As an alternative to looking for intersection points you can also rewrite the equation can be worked by factoring and the solu- so that one side is equal to zero. Now only one tions are x = 1 and x = 4. Done! expression is involved, and you are looking for The equation the x-intercepts of the curve. 2 For the problem x2 + 4 = 5x you can x − 4x + 1 = 0 switch to does not factor easily, but it still has so- x2 + 4 − 5x = 0 lutions (see below). Just because we can- and the graph of y = x2 + 4 − 5x shows x not factor an expression doesn’t mean intercepts at x = 1 and x = 4. that there are no solutions.

57 2. Complete the Square. This technique order for perfect squares to emerge (com- is not popular with students. But in the plete the square), for instance simple problem x2 − 4x + 1 = 0 2 x = 16 x2 − 4x + 4 = 3 2 factoring is not required. We have a per- (x − 2) = 3 fect square already, and the solutions are √ and the solutions are x = 2 − 3 and x = ±4. √ x = 2 + 3. In the problem 3. Quadratic Formula. Fortunately, it 2 (x − 3) = 16 is possible to solve the quadratic equation for any combination of the parame- we have a perfect square again, and now ters a, b and c by completing the square, the solutions are x − 3 = ±4. This and the final result is summarized in the implies that quadratic formula x = 3 ± 4 , √ −b ± b2 − 4ac x = and the solutions are x = 7 or x = −1. 2a If the number on the right is not a square The term b2 − 4ac is called the discrim- number, square roots must be used. For inant. If b2 − 4ac > 0, the equation has example, two distinct solutions, if b2 − 4ac = 0 it has just one solution (namely x = − b ), 2 2a (x − 3) = 5 and if b2 − 4ac < 0 solutions cannot be √ found because we cannot take the square has solutions√ x = 3 + 5 = 5.236 or x = 3 − 5 = 0.764. root of a negative number. The beauty of this technique lies in the Example: x2 − 4x + 1 = 0 fact that we can tell the number of solu- Here a = 1, b = −4 and c = 1, the dis- tions by the sign of the number on the criminant is (−4)2 −4 = 16−4 = 12 > 0, right. For example and the solutions are 2 √ (x − 3) = 16 4 ± 12 √ x = = 2 ± 3 (x − 3)2 = 0 2 (x − 3)2 = −5 √ √ In√ the last step we used that 12 = 4 · 3 = The first equation has two solutions, as 2 3. Now the fraction has a common we saw above. The second equation has factor of 2, and after cancelation we ar- the solution x = 3 only, and the last rive at the final result. equation does not have any solutions, be- Most of the time, equations are not served cause squares cannot be negative. to us on a silver platter (standard form), and What makes the method so unpopular, is it is necessary to rewrite and to simplify them that we have to rewrite the equation in so that the formula can be applied. Examples

58 are given in the next section. Keep in mind Using the quadratic formula we see that that solving equations graphically is always an √ option, especially when the algebra gets out of 1 ± 5 ϕ = control. 2 We close this part with an application: The The golden ratio is usually associated with the Golden Ratio is observed in nature in plenty larger number of forms. Just search the internet for ”golden √ ratio in nature”, and you will ﬁnd a variety 1 + 5 of applications, among them the ratio of the ϕ = = 1.618, 034 2 bones in our ﬁngers or arms, or proportions of facial features, or the ratio of sections in bees Notice, that the smaller solution is (head, thorax, abdomen), and many more. The √ golden ratio also shows up in art and architec- 1 − 5 = −0.618, 034 = 1 − ϕ ture. 2

7.3 Worked Problems 1. Solve the given equation x (a) − 2 = 4 3 x 3 (b) = x − 3 2 √ (c) 3x − 2 = 5 Interestingly, the Golden Ratio is the solu- (d) x(x − 1) = (x − 2)(x + 3) tion of a quadratic equation. If the sides of a rectangle are denoted by a and b, then we have Solutions: a golden rectangle, if a is to b what a+b is to a. We denote the golden ratio by ϕ, and put the (a) This is straightforward information into equation form and to obtain x − 2 = 4 a a + b 3 ϕ = = x b a = 6 But then 3 x = 18 a + b b ϕ = = 1 + a a (b) First we multiply both sides by 1 = 1 + 2(x − 3), which gets rid of the frac- ϕ tions. The rest is routine. and multiplication by ϕ on both sides results x 3 in = x − 3 2 ϕ2 = ϕ + 1 2x = 3(x − 3) = 3x − 9 ϕ2 − ϕ − 1 = 0 9 = 3x − 2x = x

59 and we are done. The solution is Solutions: x = 9. Substitution into the original equation conﬁrms the result: (a) Factoring looks challenging, but the quadratic formula will do the job. 9 9 3 = = We have a = 2, b = −5 and c = 3. 9 − 3 6 2 The discriminant is (c) Squaring both sides will get us started: b2 − 4ac = (−5)2 − 4 · 2 · 3 = 1 √ 3x − 2 = 5 and thus 3x − 2 = 25 √ 3x = 27 5 ± 1 x = x = 9 4

Substitution into the original equa- The solutions are tion conﬁrms the result: 5 + 1 6 √ √ √ x = = = 1.5 3 · 9 − 2 = 27 − 2 = 25 = 5 4 4 5 − 1 x = = 1 (d) We expand both sides ﬁrst, and af- 4 ter subtracting x2 from both sides, By the way, once you have the solu- we end up with a linear equation. tions, factoring is easy x(x − 1) = (x − 2)(x + 3) 2x2 − 5x + 3 = (x − 1)(2x − 3) x2 − x = x2 + x − 6 6 = 2x (b) We take advantage of the perfect x = 3 square (do not expand!): Check: (x − 1)2 − 9 = 0 3 · 2 = 6 = 1 · 6 (x − 1)2 = 9 x − 1 = ±3 2. Solve the equations x = 1 ± 3 (a) 2x2 − 5x + 3 = 0 and the solutions are x = 4 and x = (b) (x − 1)2 − 9 = 0 −2. (c) x2 − 4x + 6 = 0 (c) This quadratic equation does not have 1 (d) x + = 5 a solution, because the discriminant x is

b2 − 4ac = (−4)2 − 4 · 1 · 6 = −8 < 0

60 1 (d) Here we have to rearrange the equa- (d) x + = 5 tion before we can apply the quadratic x formula. Solutions 1 x + = 5 (a) The graph of y = x has a vertical x x−3 asymptote at x = 3, but most of x2 + 1 = 5x 2 all, it intersects with the horizontal x − 5x + 1 = 0 3 line y = 2 = 1.5 at the point where x = 3. Therefore (a = 1, b = −5 , c = 1) √ 5 ± 21 x = 2

The conﬁrmation of this√ result is 5+ 21 challenging. For x = 2 we ﬁnd that (multiply top and bottom by the ”conjugate” - a routine step in (b) This quadratic equation has two so- algebra) lutions. One at x = 1, the other at √ x = 1.5 1 2 5 − 21 = √ · √ x 5 + 21 5 − 21 √ √ 2(5 − 21) 5 − 21 = = 25 − 21 2 This shows that the two solutions are reciprocals of each other. More- over, (c) This quadratic equation has no so- 1 2 x + lution. The parabola y = x −4x+6 √x √ never crosses the x-axis. 5 + 21 5 − 21 = + 2√ √2 5 + 21 + 5 − 21 = = 5 2 3. Solve the equations graphically. The equations are taken from Problems 1 and 2,

and we can confirm our algebraic results 1 by inspection of the resulting graphs. (d) Here we see that the curve y = x+ x intersects with the line y = 5 at x 3 (a) = two different locations. The first√ in- x − 3 2 tersection occurs at x = 5− 21 = 2 √ (b) 2x2 − 5x + 3 = 0 5+ 21 0.209, and the second at x = 2 = (c) x2 − 4x + 6 = 0 4.701.

61 2 4. Solve x(x − 1)(x + 2)(x − 3) = 8 5. The length of a rectangular garden plot is four meters longer than its width. The Solution: This equation cannot be solved area is 165 square meters. What are the by factoring, because the value on the dimensions? right is not zero. Expansion of the expression on the left leads to

x5 − 5x4 + x3 + 21x2 − 18x = 8

and there is no convenient method to solve such an equation explicitly. Graphing (or other numerical methods) are about the only recourse.

Solution: Denote the width by x, then the length is x+4, and the area becomes x(x + 4) = x2 + 4x. Hence, the problem at hand is

x2 + 4x = 165

We subtract 165 from both sides and then factor the result:

x2 + 4x − 165 = 0 We ﬁnd the solutions (x + 15)(x − 11) = 0

x1 = −1.943 with solutions x = 11 and x = −15. We x2 = −0.326 disregard the solution x = −15, as the sides of a rectangle cannot be negative. x3 = 1.847 Thus the width becomes 11 m and the x = 2.000 4 length is 15 m. x5 = 3.422 6. The height h of a projectile above the from the graph. The solutions near x = 2 ground at time t is given by the equation are clustered, but zooming shows that we have two solutions in this vicinity. h = 80 + 64t − 16t2

62 √ where h is measured in feet, and time is 5. 4 − x = 3 measured in seconds. After how many seconds will the object hit the ground? 6. (x − 5)(x − 1) = (x − 4)(x − 2)

Solution: These type of questions are com- 7. (x − 5)(x − 4) = (x − 2)(x − 1) monplace in math books. In the feet- x2 − 4 second system the term −16t2 accounts 8. = x + 2 x + 2 for the loss of height due to gravity. x2 − 4 In our case the projectile has height h = 9. = x − 2 80 ft above the ground when t = 0, and x + 2 after one second we are at h = 80 + x + 2 x + 3 10. = 64 − 16 = 128 ft, and so on. Hitting the x − 1 x − 3 ground means that h = 0. This results √ √ in the quadratic equation 11. x + 56 = 4 + x 1 3 80 + 64t − 16t2 = 0 12. + 1 = x x and the quadratic formula yields 13. x2 − 2x − 99 = 0 p −64 ± 642 − 4 · (−16) · 80 14. x2 + 25 = 10x t = −32 √ 2 9216 96 15. 2x − x − 3 = 0 = 2 ± = 2 ± −32 −32 16. x2 − 2x − 2 = 0 = 2 ± 3 17. x2 + 2x + 4 = 0 The solutions are t = 2 + 3 = 5 seconds and t = 2 − 3 = −1 sec. The negative 18. (x − 2)2 − 4 = 0 solution is irrelevant (we don’t go back 3 in time), and the ﬁnal result is that the 19. x + = 4 projectile will hit the ground after ﬁve x √ seconds. 20. x + x = 12

7.4 Exercises 21. The sides of a rectangle are in golden ratio. Find the length of the remaining side Solve for x in problems 1 - 20. when

1. 2x − 3 = 7 (a) the longer side is 55 cm, 2x − 1 (b) the short side is 55 cm. 2. = 5 5 22. The length of a rectangular garden is three x + 2 3. = 4 meters longer than the width. x − 1 x + 2 (a) What are the dimensions if the area 4. = 1 x − 1 is 108 square meters?

63 (b) What are the dimensions if the area 19. x = 1 and x = 3 is 100 square meters? 20. x = 9 23. The height h above ground measured of a baseball t seconds after it is hit is given 21. (a) 33.992 cm ≈ 34 cm by the formula (b) 88, 992 cm ≈ 89 cm

h = 4 + 40t − 16t2 22. (a) 12 × 9 (b) 11.612 × 8.612

where h is measured in feet. After how 23. 2.6 seconds (2.5963) many seconds will it hit the ground?

Answers 1. x = 5

2. x = 13

3. x = 2

4. no solution

5. x = −5

6. no solution

7. x = 3

8. no solution

9. all x 6= −2

10. x = −1

11. x = 25

12. x = 2

13. x = 11 and x = −9

14. x = 5

3 15. x = 2 and x = −1 √ 16. x = 1 ± 3, that is, x = 2.732 and x = −0.732

17. no solution

18. x = 0 and x = 4

64 8 Ratios and Proportion This is the deﬁnition of π. It is the ratio of the perimeter of a circle to its diameter. In this part covers ratios and proportion, and we will review percentages and related topics. Example (Energy Payback Ratio): This Important biological applications include di- ratio relates the energy output of a power plant lutions of chemicals, the mark and recapture to the energy input. When 2,800,000 GJ (giga- method, as well problems from medicine, ecol- loules) are required to produce 11,350,000 GJ ogy and epidemiology. by a natural gas powered plant, the payback ratio becomes energy output 11, 350, 000 GJ 8.1 Ratios = = 4.053 energy input 2, 800, 000 GJ Ratios are typically the division of terms with the same units. The result is a number which Each unit of energy input results in a fourfold does not have any physical units. Such a quan- of energy output. tity is called dimensionless. Incidentally, the data are based on a study Example: The RU Factbook 2014 reports 411 by Meier and Kulcinski of the University of biology majors and 75 mathematics major. The Wisconsin. The same authors report payback ratio of majors is ratios of 11:1 for coal, 27:1 for fusion and 23:1 for wind energy. 411 students = 5.48 75 students 8.2 Percent The units in numerator and denominator are Percents usually appear in conjunction with ”students”. They cancel, and the result is a the ratio number. This ratio is about 5.5, and we con- parts clude that the ratio biology:math majors is whole 11 11:2 ( 2 = 5.5). The result will be a number between 0 and 1, Example (Circles): For all circles, large or and to make the number more user-friendly it small, the ratio of the perimeter (circumfer- is multiplied by 100. ence) to the diameter is the same. The term percent goes back the the Latin per centum, and means by the hundred. We Diameter Perimeter Ratio use the symbol % to express percents. Mathe- 4.2 in 13.2 in 3.143 matically we have 16 miles 50 miles 3.125 1 78 mm 245 mm 3.141 % = = 0.01 and 100% = 1 2.3 m 7.2 m 3.130 100 55 µm 173 µm 3.145 For example,

The slight variations of the ratios are due to 1 45 45% = 45 · = = 0.45 rounding. Using the formulas from geometry 100 100 we ﬁnd that perimeter 2πr Example: A class has 20 female and 16 male = = π students for a total of 36 students. In this case diameter 2r

65 the percentage of female students is computed In comparison, in 2012 the CDC found that from the infant mortality rate in the United States was 5.98 . 20 female students 20 5 = = 36 total students 36 9 Example:h The legal limit of a drivers blood alcohol concentration in Virginia is 0.08%. This = 0.55556 = 55.6% is equivalent to 0.8 . h while the ratio becomes Both examples show that changing from 20 female students 20 5 percent to per mille, the decimal place is shifted = = by one place. 16 male students 16 4 that is, the ratio female:male students is 5:4. 8.3.2 Parts per Million, Parts per Bil- Example (HIV in Nigeria): 3.1% of the lion adult population of Nigeria was infected with HIV in 2012 (CIA World Factbook). When quantities get really small, we resort to When 3.1% of the population being infected, parts per million (ppm) or parts per billion then 96.9% are not infected, and for the ratio (ppb). we ﬁnd that Example: The abundance of hydrogen iso- not infected 96.9% topes in the atmosphere are = = 31.258 infected 3.1% 1H 99.985% Roughly speaking, we have one HIV infected 2H 0.015% person per 31 HIV-free people. Thus, the abundance of 2H is 8.3 Percentlike Quantities 0.015% = 0.15 = 150 ppm 8.3.1 Parts per Thousand h 0.015 0.15 150 When the reference is changed from one hun- = = 100 1, 000 1, 000, 000 dred to one thousand, we speak of parts per thousand, sometimes also called per mille. Com- mon notations are that is, 150 of a million hydrogen atoms are the 2H isotope. or ppt Example: Radford drinking water contains h 1.52 ppm of chlorine (Radford City Drinking Example: The United Nations reported in Water Report, 2013). 2010 that the infant mortality9 in Japan was Examination of the ﬁne print reveals that 2.62 deaths per 1,000 life births. The infant ppm is used as milligram per liter (mg/L), and mortality rate then becomes it follows that one liter contains 1,52 mg of 2.62 deaths chlorine. Conversion to gallons yields = 0.002, 62 = 2.62 1, 000 births h 1.52 mg 3.785 L 5.75 mg 9 · = Death of a child one year or younger. L gal gal

66 Using ppm as mg/L is a violation of the There are diﬀerent ways to look at a rela- condition that things like percentages should tive change, or to calculate it. The quantity be free of physical units. However, it is common practice to use the density of water ∆ value = ﬁnal value − initial value

ρ = 1 kg/L = 1, 000 g/L is the absolute change. This quantity usually becomes more meaningful when we relate it to when dealing with aqueous solutions. Then the beginning value. one ppm becomes one milligram per liter: Example: The population of the United States ρ 1000 g mg = = grew by 20.6 million during the years 2005- 1, 000, 000 1, 000, 000 L L 2013, while that of Norway increased only by 461 thousand during the same time span. Com- 8.4 Percentage Change parison to the actual populations in 2005 re- Now we look at relative changes, expressed as sults in a percentage. We hear things like ”cost of liv- 20.6 million = 7.0% ing increased by 2%”, or ”India’s tiger pop- 295.5 million ulation has increased by 30%” (BBC News, 1/25/2015) on a daily basis. What does it for the United States and in mean, and how is it calculated? 0.461 million = 10.0% In order to ﬁnd a relative change we com- 4.623 million pute the quantity for Norway. This shows that in relative terms ﬁnal value − initial value Norway’s population outpaced that of the United initial value States during 2005 - 2013 time span.

Multiplication of the result by 100% yields the The relative growth formula can also be percentage change. broken down into Example: Bacteria colonies in a laboratory final value − initial value increased from 720 colonies per liter to 1260 r = initial value colonies per liter. Here the relative change be- final value comes = − 1 = M − 1 initial value 1260 − 720 540 = = 0.75 = 75% The quantity 720 720 final value Example: India’s tiger population has risen M = ( = 1 + r) initial value from 1,706 in 2011 to 2,226 in 2014. The percentage change is is the multiplier (growth factor) and r is the relative growth rate. This will become a major 2226 − 1706 520 = = 0.305 = 30.5% theme in the study of exponential growth. 1706 1706 Example: We return to India’s tigers. We This is an increase of 30% over three years, have which accounts for an annual growth of about 2, 226 M = = 1.305 10%. 1, 706

67 that is, the population grew by a factor of or equivalently, 1.305, and the relative growth is actual distance r = M − 1 = 0.305 = 30.5% = 200, 000 × map distance as before. On such a map 8 cm represent

8.5 Proportion 200, 000 · 8 cm = 1, 600, 000 cm = 16 km Two quantities x and y are proportional, if in reality, or conversely, 600 m in reality are their ratio remains constant, and we write shown as x ∝ y 600 m = 0.003 m = 3mm In this case we say that the quantities vary 200, 000 directly with each other. on this map. A constant ratio also implies that Example (Unit Conversion): Blood sugar y = c values can be measured in mmol/L (millimol x per liter), but it customary to use mg/dL (mil- and thus, y = cx, where c is the constant of ligram per deciliter10) readings. proportionality. The atomic mass of glucose O6C12H6 is 180 Example: Time and distance are proportional Da (12·6 Da for the six carbon atoms, 12 Da for when you travel at a constant speed on a high- hydrogen, and 16 · 6 Da from oxygen). Hence, way. If you spend twice as much time, you will one mol of glucose has mass 180 g. The values go twice as far; if plan to travel only one third on the two scales will remain proportional and of the distance, it will take one third of the we find that time. We have 1 mmol 0.001 mol 180 g = · distance ∝ time L 10 dL mol 0.018 g mg = = 18 The constant of proportionality becomes dL dL distance Therefore, multiplication by 18 takes you from c = time mmol/L to mg/dL. For instance, 7.5 mmol/L is equivalent to 7.5 · 18 = 135 mg/dL. which is the speed. We see that when x ∝ y, where x and y have different physical units, then the units of c balance the equation. 8.6 Mark and Recapture Example: Maps use the principle of propor- This is a technique which is used to estimate tionality: The ratio of map distance to dis- populations in the wild. It is also known as tance in reality always remains the same. As capture-recapture, as mark-release-recapture or an example we look at the scale 1:200,000, which as sight-resight. The basic idea is to capture is widely used in aviation. This means that a few animals and to mark them in a way so that they can be identified later. The animals map distance 1 = 10 actual distance 200, 000 One deciliter equals 0.1 liter.

68 are then released. A second capture follows at 2. The estimates are very sensitive to the a later date. It is assumed that the percentage data, because we extrapolate. of tagged animals in the second catch is repre- If for instance, one more bird without a sentative of the percentage of marked animals tag had been captured, n would change in the entire population. to n = 84, and the population estimate We use the following notation becomes N total population 71 · 84 N = = 259.3 T tagged in the first catch 23 n number of second catch If , on the other hand, if the additional t tagged in second catch bird had a tag, we now would have n = 84 and t = 24 and the estimate changes If we argue that percentages (ratios) remain to identical we find that 71 · 84 N = = 248.5 t T 24 = n N This shows we should take population One could also argue that t/T = n/N, but in estimates based on this method with a any case it follows that grain of salt. T n The mark-recapture method assumes that N = t the animals move around and that we have ran- and we have a population estimate. dom samples each time. If the animals all hang out at their respective preferred locations, we Example: In a study of birds on an island 71 do not get a good mix, and the population esti- birds are captured, tagged and released. Three mates are tainted. The time between captures days later 83 birds are caught, 23 of which are needs to be long enough so that we get a good tagged from the first capture. Thus we have mix, but it shouldn’t be too long. Otherwise the values some tagged animals might die, or other indi- T = 71 n = 83 t = 23 viduals might move into or out of the region of study. The formula yields 71 · 83 N = = 256.2 8.7 Dilution and Mixtures 23 and we have a population estimate of about We begin with an example: The instructions 256 birds. on a can of frozen pink lemonade concentrate Further discussion: ask to add four parts of water for each part of frozen concentrate. This tells us that no mat- 1. In the second sample, 23/83=27.7% of ter how much or how little we want to fix, one the birds had been tagged. fifth of the mixture will be concentrate, and On the other hand, with our estimate of the remainder will be water. In this example, N = 256 birds and a total of 71 tagged the solute (the frozen concentrate) is available birds around, the ratio percentage of tagged at 100% concentration, while in the mixture it birds is 71/256=27.7% . These percent- is reduced to 1/5 = 20% concentration, and we age match, which is encouraging. call 5 the dilution factor.

69 In general mixture problems we follow the We now take another look at the notion of amount of the solute (the substance to be dis- a dilution factor and we define it as solved) in each solvent (medium in which the C process takes place). A concentration (C) is δ = 1 C measured as solute (s) per volume (V ), and 2 we have It is the stronger concentration divided by the s weaker concentration. C = or, equivalently s = CV V But we can obtain the same factor by looking at volumes only. With the use of formula If we denote the initial values (stock solution) (1) we find that with subscripts 1, and the final product (sam- C C V C V V ple) with subscript 2, we obtain the formula δ = 1 = 1 1 = 2 2 = 2 C2 C2V1 C2V1 V1 C V = C V (= s) (1) 1 1 2 2 and the dilution factor is the ratio of the larger volume to the smaller volume. because the solute s must be the same in both mixtures. Example: We return to the last case. V1 = 10 mL were dissolved in V = 500 mL. Thus, the Example: We return to the pink lemonade 2 dilution factor is problem, and suppose that we want to prepare two gallons. V 500 δ = 2 = = 50 We begin with pure concentrate, and thus V1 10 C1 = 1, and V1 is unknown. As for the final and the concentrations behave accordingly: product we know that C2 = 1/5 = 0.2 = 20% (four parts of water per one part of concen- C 25 mg/mL trate, 1 ), and we need V = 2 gal. Therefore C = 1 = = 0.5 mg/mL 1+4 2 2 δ 50 C V V = 2 2 = 0.2 · 2 gal = 0.4 gal Example: A stock solution contains 18 mg/mL 1 C 1 of fictosin and we want to prepare a 30 mL Since frozen concentrate usually comes is 12 mixture with concentration 3 mg/mL. oz. containers, we should get at least five cans We solve this problem by looking at dilu- (four is not quite enough, why?). tion factors. By inspection of the concentrations we see that Example: A stock solution contains 25 mg/L of imaginol, we add 10 mL of the stock to 490 18 mg/mL δ = = 6 mL of water, and we are interested in the con- 3 mg/mL centration of imaginol in the mixture. Therefore, using that V2 = 30 mL, we find that Here C1 = 25 mg/mL, V1 = 10 mL and V2 = 500 mL and we obtain V 30 mL V = 2 = = 5mL 1 δ 6 C1V1 25 mg/mL · 10 mL C2 = = V2 500 mL Thus, you need to mix 5 mL of the stock solu- = 0.5 mg/mL tion with 25 mL of water.

70 8.7.1 Serial Dilutions 8.8 Worked Problems Sometimes it is necessary to dilute a mixture 1. In a experiment the Moravian geneticist by a tremendous amount, like putting a droplet Gregor Johann Mendel (1822-1884) in- into a full bathtub. This is not practical and terbred true yellow round seed peas with lab supplies can be very expensive. In order true green wrinkled seed peas. The F2 to save cost, such a dilutions are performed in progeny were distributed as shown in the sequence. Dilute the stock solution, mix well, table then take a portion of the mixture and dilute it again. 315 yellow round seeds 108 green round seeds 101 yellow wrinkled seeds 32 green wrinkled seeds

Approximate the ratio of phenotypes with whole numbers as A : B : C : D so that A + B + C + D = 16.

Solution: The total number of plants is Figure 16: Four 1:10 Serial Dilutions 315 + 108 + 101 + 32 = 556, and we can easily compute the percentages. The Each dilution step will have its own dilu- sum of the percentages will be one, and if tion factor, and the cumulative dilution factor we multiply by 16, we obtain the desired is the product of the individual factors. For result. instance, suppose you perform four 1:10 dilutions. The dilution factor is 10 each time, Seeds Freq. Perc. ×16 and four such dilutions result in the factor δ = Yellow round 315 56.7% 9.06 104 = 10, 000 overall. Green rounds 108 19.4% 3.11 Example: After three 1:200 serial dilutions, a Yellow wrinkled 101 18.1% 2.91 10 µL droplet contained 53 bacteria. What is Green wrinkled 32 5.8% 0.92 the original bacteria concentration? Sum 556 100% 16 Here the dilution factor is 200 in each step The phenotypes follow about a 9:3:3:1 ra- (one part bacteria mixture, 199 parts broth), tio, which is prevalent in genetics. and overall the factor becomes δ = 2003 = 8, 000, 000 2. The National Forest Headquarters in Roanoke manage the George Washington National The current concentration is Forest (up I-81 North either towards West 53 bacteria Virginia or towards the Blue Ridge Park- C2 = = 5.3 bacteria/µL 10 µL way) with 1,065,398 acres, and 723,350 and therefore acres of the Jeﬀerson National Forest (mostly west of Roanoke, including the Mount C = δ · C = 4.24 107 bacteria/µL 1 2 Rogers area and the forest land north of which is the equivalent of 42.4 trillion bacteria Blacksburg, including the Cascades and per liter. Mountain Lake). What percentage of the

71 entire area is the Jeﬀerson National For- Solution: Recall that the volume of a 4πr3 est? What is the ratio of the acreages? sphere is given by V = and the 3 2 Solution: The combined area is 1,788,748 surface area is S = 4πr . We do not have acres (just add) and the percentages be- any speciﬁc numbers to work with, which come makes the problem more challenging. 1, 065, 398 We denote the original radius by r, the = 59.6% 1, 788, 748 original volume by V , and for the new, bigger sphere we use r0 and V 0, respec- for the George Washington National For- tively. Then, because V 0 = 1.5V , we est, and obtain

723, 350 V 0 4π(r0)3/3 = 40.4% 1.5 = = = (r0)3 1, 788, 748 V 4π/3 for the Jefferson National Forest. The √ Thus, r0 = 3 1.5 = 1.145, and we see ratio is roughly 3:2 because that the radius has increased by 14.5%. 1, 065, 398 3 = 1.473 ≈ 1.5 = At this point we branch off, and offer two 723, 350 2 solution strategies. You could also round the percentages to Solution One: Assume that the length 60% : 40%. units are miraculously chosen so that the 3. A food inspector detected 3.5 mg of ar- original sphere has radius r = 1. Then 4π senic in one kilogram of fish. What is the the original volume is V = 3 , the sur- concentration? face area is S = 4π, and the original surface area to volume ratio is SVR = 4π Solution: The ratio arsenic per total mass 4π/3 = 3. becomes The new sphere has radius r0, and the 3.5 mg 3.5 mg surface area to volume ratio becomes = = 3.5 ppm 1 kg 1, 000, 000 mg 4π(r0)2 3 SVR0 = = The arsenic content in this sample is 4π(r0)2/3 r0 3.5 ppm. (Example adapted from Lang- kamp/Hull). Comparing both ratios we get

4. The volume (and not the radius!) of a SVR0 3/r0 1 = = = 0.874 sphere increases by 50%. How do radius SVR 3 r0 and surface to volume ratio change? and ee conclude that the surface area to This scenario could come up in the study volume ratio has dropped11 by 12.6%. of cell growth, where ultimately the changes of the surface area to volume ratio are of 11Recall that M = ﬁnal value = 1 + r, and there- initial value interest. fore r = M − 1 = −0.126.

72 Solution Two: In the geometry chapter Add 4 × 72 oz of water for a total of 360 we have seen that the surface area to vol- oz, which is the equivalent of 2.8125 gal- ume ratio for a sphere of an arbitrary ra- lons of pink lemonade. dius r is 3 8. You dilute 5 mL of a fantasium stock so- SVR = r lution with 75 mL of water. The concen- and if we compare spheres with respec- tration of the mixture is 6 mg/L, what is tive radii r and r0, we ﬁnd that the concentration of the stock solution? SVR0 3/r0 r = = Solution: We oﬀer two options. SVR 3/r r0 Solution One: We have V = 5 mL, V = But we have already calculated the quan- 1 2 r0 1/3 75 mL and C2 = 6 mg/L. Therefore tity r = 1.5 , and therefore C V 6 · 75 SVR0 r C = 2 2 = mg/L = = 1.5−1/3 = 0.874 1 V 5 SVR r0 1 = 150 mg/L 5. What actual distance appears as 5/8 inches on a 1:24,000 map? Solution Two: The dilution factor is Solution: The distance in reality is δ = V2/V1 = 75/5 = 15 5 24, 000 · in = 15, 000 in = 1, 250 ft 8 Therefore the stock concentration is

6. Connelly’s run is a small stream in Rad- C1 = 15 · 6 mg/L = 150 mg/L ford. On a first capture students caught, marked and released 123 crayfish, on a 9. You want to prepare a mixture contain- second capture 9 of 120 fish were marked. ing 1 g sodium hydroxide (NaOH) from Estimate the crayfish population from the a 0.2 M stock solution. How much stock data. solution is required. Use that the molar mass of NaOH is 40 g/mol. Solution: We have T = 123, n = 120 and t − 9, therefore Solution: Our mixture must contain s = 123 · 120 1 g of NaOH. Since s = CV , we get N = = 1640 9 s 1 g V = = C 0.2 M 7. How much lemonade can be made with 1 g mol 1 six 12-oz cans of concentrate, when we = · = L 0.2 mol/L 40 g 8 still require to use four parts of water of = 125 mL each part of concentrate?

Solution: This one is easy. The total con- 10. A broth has bacterial density of 200 mil- centrate is lion bacteria per liter. You need a 20 µL 12 oz droplet with about 5 bacteria. How do 6 cans · = 72 oz you proceed? can

73 Solution: The given bacterial density is 6. The natural abundance of 16O in the at- 8 18 C1 = 2 · 10 bacteria/liter, the desired mosphere is 99.762%, that of O is 0.2% concentration is and the natural abundance of 17O is 0.0377%. 16 17 18 5 bact. 1000 µL 1000 mL What is the ratio O : O : O. Ex- C = · · press the abundance of 17O in ppm (parts 2 20 µL mL L per million)? = 250, 000 bacteria/L 7. The EPA sets the maximum contaminant Therefore the dilution factor becomes level of arsenic in drinking water at 10 8 C1 2 · 10 ppb (parts per billion), and that of fluo- δ = = 5 = 800 rides at 4 ppm (parts per million). C2 2.5 · 10 Serial dilutions appear to be in order. (a) How many grams of One could, for instance, use two 1:10 di- i. arsenic lutions followed by a 1:8 dilution. ii. fluorides are allowed in one cubic meter (1000 8.9 Exercises L) of water? 1. Radford University reports that women (b) How much uncontaminated water is make up 56% of its student body in the required to dilute one gram of fall of 2014. What is the ratio female:male i. arsenic students? ii. fluorides 2. It was observed that the sex ratio fe- safely to EPA standards? male:male for the Alligator mississippi- ensis is 5:1. What percentage of alliga- 8. The U.S. population was 282.16 million tors will be male? in 2000 and 309.35 million in 2010 (both for July 1). The national debt was $5,674 3. You got 20 answers correct on a test of billion in 2000 and $13,562 billion in 2010. 24 questions. What percentage score did you receive? How many would you have (a) Find the percentage increase of the to get correct to get higher than a 90%? population for the decade. (b) Find the percentage increase of the 4. In a genetic experiment with summer squash national debt for the decade. 171 plants had a white fruit color, 42 were yellow and 15 were green. Find the (c) Compute the debt per capita for 2000 white:yellow:green ratio so that the total and for 2010. adds to 16. (white+yellow+green = 16). (d) Compute the percentage change of the per capita debt for the decade. 5. The United States had a population of 304 million in 2008, making up 4.55% of 9. You determine that the ambient temper- the world population. At the same time ature outside is 274 K. You walk inside Brazil accounted for 2.87% of the world your apartment, and the thermometer jumps population. What was the population of up to 290 K. Find the percent change of Brazil? the temperatures.

74 10. In a test of water quality an automated (a) Three days later you capture 83 birds, sample measured 1,020 ppb of phospho- 23 of which are tagged. Estimate rus, a cheaper siphon test registered only the bird population on the island. 880 ppb. Assuming that the automated (b) Just to be sure, you repeat the ex- test is correct, what is the percentage er- periment on the fourth day. This ror of the cheaper test? time you find that 18 of 59 birds 11. (a) The volume of a spherical cell in- are tagged. What is your popula- creases by 12%; how much does the tion estimate? diameter increase? 19. A BIOL 131 class trapped and marked 45 (b) The radius of a spherical cell increases crayfish in Wildwood Park. Three days by 2%; how much does the surface later, they captured 30. Of these 10 were to volume ratio change? marked. How many fish would you esti- 12. The blood sugar level of a person increases mate are in this population? from 92 mg/dL before lunch to 116 mg/dL 20. You have a stock solution with concen- after lunch. What is the percentage in- tration 40mg/mL of fantasium. You need crease? 50 mL with 6mg/mL of fantasium. How 13. On a map with scale 1:50,000 two points much stock should be mixed with how are 14 cm apart. What is their actual much water? distance? 21. It takes three parts of water for each part 14. On the Virginia Highway Map one inch of frozen orange juice. How much frozen on the map is approximately 13 miles. orange juice is required for one gallon of What is the scale of this map. juice?

15. Find the conversion factor when lb/gal is 22. Your stock solution contains 45 mg/mL changed to g/mL. of fictosin, you need 12mL with concen- 16. The CDC reports that 480,000 Ameri- tration 3mg/mL for a patient. How much cans die from cigarette smoking each year. of the stock should you mix with water? Given that the U.S. population is 320 23. You add 25 mL of a stock solution with million and that of Virginia is 8.4 million, concentration 18 mg/mL of fictosin to estimate the annual deaths from smoking 225 mL of water. What is the fictosin in Virginia. concentration in the mixture? 17. On the first day biologists capture 67 fish, mark them and put them back in the 24. You have a 5 M stock solution of sucrose. pond. The next day they catch 42 fish, You want to make 100 mL of a 0.4 M so- 12 of which are marked. Estimate the lution. How much stock solution should fish in the pond. you mix with water? 18. You observe birds on an island. On the 25. You have 4 mL of a bacterial culture. Af- first day you capture, tag and release 71 ter performing three 1:80 dilutions you birds. count 192 bacteria in a 10 µL sample.

75 What is the original bacteria concentra- 17. 234.5 tion measured in bacteria per milliliter? How many bacteria does your 4 mL cul- 18. (a) 256.2 (b) 232.7 ture contain? 19. 135

26. You begin with 5 mL of a bacterial cul- 20. 7.5 mL stock, 42.5 mL water ture. After three 1:50 serial dilutions you ﬁnd that 1 mL contains 84 bacteria. How 21. 1 quart many bacteria were in the original sample? 22. 0.8 mL 23. 1.8 mg/mL Answers: 24. 8 mL 1. 14 : 11 ≈ 5 : 4 25. 9.83 billion per milliliter 2. 16.7% and 39.3 billion

3. 88.3%; 22 (21.6 if partial credit is given) 26. 52.5 million 4. 12 : 3 : 1

5. 192 million

6. 2, 625 : 1 : 5.3 or 498.8 : 0.19 : 1 and 377 ppm

7. (a) (i) 10 mg (ii) 4 g (b) (i) 100 m3 (ii) 250 L

8. (a) 9.6% (b) 139 % (c) $20, 109.16 and $43,840.31 (d) 118%

9. 5.8 % (5.8394%)

10. 13.7 % (13.7255%)

11. (a) 3.85% ≈ 4% (b) decreases by 1.96% ≈ 2%

12. 26.1%

13. 7 km

14. 1 : 823, 680 ≈ 1 : 825, 000

15. 8.345

16. 12, 600

76 9 Elements of Probability and the event of rolling an even number is the set Processes in nature have a certain degree of E = {2, 4, 6} randomness and non-predictability. While on for humans the ratio of female to male progeny Definition: The probability of an event E is is about equal, you can find couples with four girls and no boys at all, or families where all n(E) P (E) = children are male. An understanding of such n(S) randomness is vital in biology, particularly in the study of genetics. As another application where n(S) denotes the size of the sample space, we will look at Hamilton’s Rule from evolution. that is, the number of possible outcomes, and n(E) denotes the number of outcomes that make up the event E. In this definition it is im- 9.1 Probability perative that there are only finitely many and For simplicity we assume that our observation equally likely outcomes. (experiment) has only finitely many outcomes, Examples: which are all equally likely. Typical examples are rolling a (fair) die, flipping a coin or draw- 1. The probability of rolling a ”3” on a fair ing a card. The sample space S consists of all die is possible outcomes, and an event E is a subset 1 of the sample space. P (E) = = 0.167 , 6 because n(E) = 1 and n(S) = 6.

2. A standard deck has 52 cards, two of which are red threes (♥ 3 and ♦ 3). We make drawing a red three our event E. Then the probability of E is

2 P (E) = = 0.0385 = 3.85% 52 Figure 17: Venn Diagram because n(S) = 52 and n(E) = 2.

Events are often depicted by Venn diagrams. 3. In this example we are rolling two dice, The big blue box depicts the sample space con- a red and a blue one, and we want to taining g all possible outcomes, while the yel- ﬁnd the probability of getting an eight low region singles out a speciﬁc event. or higher. Example: When rolling a die, the outcomes Here an outcome is a pair of numbers are the numbers one through six, the sample (r, b), with r being the value on the red space is the set die, and b the value on the blue die. We have 36 possible outcomes as illustrated S = {1, 2, 3, 4, 5, 6} in Figure 18.

77 1. P (S) = 1

2. P (E) ≥ 0

3. If A and B are exclusive events, that is, if A and B cannot occur simultaneously, then12

P (A or B) = P (A) + P (B) Figure 18: Outcomes for a Pair of Dice As a consequence of these axioms we find There are 15 possible cases where the that outcome is eight or higher; hence the prob- 0 ≤ P (E) ≤ 1 ability of the event is holds for any event E. We say that E is a 15 certain event if P (E) = 1, and that E is an P (E) = = 0.417 36 impossible event if P (E) = 0. 9.2 Laws of Probability Moreover, the events The concept of probability can be extended to E occurs spaces with infinitely many outcomes, and to E does not occur situations where not all outcomes are equally likely. are mutually exclusive and they cover the en- Empirical probabilities are based on col- tire sample space (all possibilities). Therefore lected data. For instance, in the United States the blood types are distributed as follows 1 = P (S) O A B AB = P (E or (E does not occur)) + 37.4% 35.7% 8.5% 3.4% = P (E) + P (E does not occur) - 6.6% 6.3% 1.5% 0.6% and it follows that (negation rule) Therefore, the probability that a randomly selected person has blood type B+ is 0.085 = P (E does not occur) = 1 − P (E) 8.5%. Rolling dice or drawing cards are examples Examples: We roll a single fair die. theoretical probabilities. For example, the chances of rolling a ”3” are 1/6 in theory, but 1. The probability of rolling a ”7” is zero. there is no guaranty that when you roll a die It is an impossible event. 30 times you will get exactly five threes. The Law of Large Numbers states that in the long 2. The probability of rolling a number be- run the empirical probabilities will approach tween one and six is one. It is a certain the theoretical probabilities. event, all possible outcomes are covered. For an arbitrary sample space S and a set 12In probability theory one would require that this of events E, A, B we require that the proba- statement also holds for infinitely many exclusive bilities observe the following three rules: events.

78 3. The probability of not getting a ”3” is Again, Figure 18 reveals that 5 P (do not roll a 3) P (A) = 36 = 1 − P (roll a 3) 18 1 1 5 P (B) = = = 1 − = = 0.833 36 2 6 6 20 P (A or B) = by the negation rule. 36 23 P (A) + P (B) = 36 The third property of probabilities 5+18 20 The numbers do not add up, 36 6= 36 , P (A or B) = P (A) + P (B) but this is not in violation of the OR-Rule, because the events are not exclusive! A and B is known a the additive property, or the ”OR- occur can at the same time, for instance, when Rule”. It requires that the events are mutually the red die has a ”2” and the blue die shows exclusive. a ”6”. Three outcomes are doubly counted in 5+18 23 the sum 36 = 36 .

9.3 Conditional Probability and In- dependent Events Very often we want to know the probability of an event A, given that an event B has occurred. We express this value as P (A|B). In this case B plays the role of the sample space, and the conditional probability is computed as

In the example of rolling two dice, the prob- P (A and B) P (A|B) = (2) ability of rolling a ”7” or a ”10” is P (B)

P (roll a 7 or a 10) = P (roll a 7) + P (roll a 10) 6 3 9 = + = = 0.25 36 36 36 The values are taken from the table in Figure 18. The addition principle applies, because you cannot roll a ”7” and a ”10” at the same time. The story is diﬀerent when you look at the events Example: We are rolling a red and a blue die again, and let A be the events of rolling A roll an ”8” an eight or higher. We have seen before that 15 B the red die shows an even number. P (A) = 36 .

79 But we peek, and we see that the red die This is called the multiplicative rule (or AND- shows a ”3”. This changes our odds, because rule) for independent events. we now need a ”5” or a ”6” on the blue die to Example: When we roll two dice, the re- make at least eight points. The probability of sult on the blue die is not impacted by the 2 1 this is happening is 6 = 3 . We make B the value on the red die, and the outcomes are in- event of getting a ”3” on the red die, and in dependent. Therefore the probability of get- light of our new notation, we have ting an even number on the red die and a five or a six on the blue die equals P (roll 8 or higher | red die shows ”3”) 1 P ((even on red) and (5 or 6 on blue)) = P (A|B) = 3 = P (even on red) · P (5 or 6 on blue) 1 1 1 We now show how the formula (2) leads to = · = the same result. Inspection of Figure 18 reveals 2 3 6 that This result can be confirmed by looking at Fig- ure 18. There are six outcomes for which the P (roll 8 or higher and red die is ”3”) blue die shows a five or a six, and the red 2 = P (A and B) = and die has an even number and the probability 36 is 6 = 1 , as expected. P (red die shows ”3”) 36 6 1 = P (B) = 6 and formula (2) results in

P (A and B) 2/36 1 P (A|B) = = = P (B) 1/6 3

If the events A and B are exclusive, the two events cannot occur at the same time, which makes ”A and B” an impossible event with probability zero, and Figure 19: Tree Diagram P (A and B) 0 P (A|B) = = = 0 Complex situations are often sketched with P (B) P (B) tree diagrams. In Figure 19 the events are de- If the probability of the event A is not af- noted by upper case letters, and the probabili- fected by the event B, we call the events A and ties are shown on the branches. In the ﬁgure it B independent. Then is imperative that A and B are exclusive, and that p+q = 1 from the OR-rule. Moreover, C, P (A and B) P (A) = P (A|B) = D and E must be exclusive and u + v + w = 1, P (B) and so on. and therefore As we go down a certain path, we multiply the probabilities. For instance, the probability P (A and B) = P (A) P (B) that A and C occur is the product p · u (AND

80 rule), and the probability that A, E and H 9.4 Hamilton’s Rule occur is p · w · x. Altruism is a behavior where the an individual Example: A disease is linked to a recessive (the altruist) makes sacrifices to benefit its kin. allele a. Your uncle (mother’s brother) has the Parenting is one such example. The guiding illness, but neither do your mother or your ma- principle is to promote the success of its own ternal grandparents. No one on your father’s genes. side of the family carries the disease, and we as- In Hamilton’s rule we measure the benefit sume that your father has genotype AA. What B in terms of average additional offspring for is the probability that the allele, and poten- the kin, and the cost C of an altruistic act tially pass it on to your children? in terms of the fewer offspring by the altruist. The relatedness factor r is crucial. Siblings, on average, share 50% of the genes and r = 0.5. In an aunt/nice relationship, the aunt shares 50% of the genes with the mother, who in turn as 50% of the genes in common with the daughter, which puts the aunt/nice relatedness at r = 0.5 · 0.5 = 0.25 (multiplication principle). For First of all, since your father has genotype cousins the relatedness factor is r = 0.125. AA, it is impossible that you have the illness. r Relationship Secondly, maternal grandparents must both be heterozygotes, and since your mother does not 0.5 siblings, parent and child have the disease, the possibility that your mother 0.25 aunt or uncle with nice or nephew is aa can be ruled out. 0.125 cousins For the remaining cases, the probabilities Hamilton’s Rule states that altruism is fa- 1 for your mother’s genotype are P (AA) = 3 vored if 2 and P (Aa) = 3 . The probability that your rB > C carry the a allele is 1/2 if your mother is Aa. Therefore the probability that you have geno- Example: A monkey is under attack by a type Aa is predator, but it will have a 75% chance of survival if the predator is distracted, else it will 2 1 1 get killed. The monkey’s brother has a 25% P = · = 3 2 3 chance of being killed if he distracts the predator. Both would have an average number of The tree diagram will clarify the situation. four offspring later in life. Should the brother help? The benefit in this scenario is B = 4 · 0.75 = 3 that is, on average three future offspring would be saved (this is an expected value). The cost for the brother is C = 4 · 0.25 = 1

81 that is, one future offspring is put at risk. The 3. N = 3. Now a third person enters the monkeys are siblings and therefore r = 0.5. We rink. The probability of all different birth- find that rB = 0.5 · 3 = 1.5, which is greater days becomes (AND-rule) than C = 1, and therefore the brother should 364 363 distract the predator. P (E) = · = 99.18% 365 365 More examples are given in the ”Worked Problems” section. 4. Adding a forth person leads to 364 363 362 P (E) = · · = 98.36% 9.5 The Birthday Problem 365 365 365 This is a somewhat famous problem, and we 5. If we continue in this fashion, the proba- include it at this point, although there is no bility of different birthdays in a group of immediate biological connection. N people becomes Problem: What is the probability that in a group of N people at least two people have 364 · 363 · 362 ··· (366 − N) P (E) = a common birthday? For simplicity, we ignore 365N−1 leap years and assume that the year has 365 days. The table and the graph in Figure 20 summa- In a group of two or three people the odds rize the results. The break even point is at for a shared birthday are fairly slim, in larger N = 23 people. In a group of N = 50 people, groups the chances become better, and as soon the chances of at least one common birthday as we have 366 or more people we are guaran- are at 97%, for a group of 100 people it is al- teed at least one common birthday13. most find to find at least one matching birth- The problem is easier to solve when we ask day. Test it your favorite group or club, or at for the probability that all birthdays are dif- a large party! ferent, and we make this our event E. We will now go over the probability of E for increasing 9.6 Worked Problems values of N, the size of the group. 1. You flip a coin, what is the probability 1. N = 1. With just one person the cannot that you get heads twice. be a matching birthday, and P (E) = 1 Solution: We denote heads by H and get- 2. N = 2. If we think of the first person’s ting tails by T . Then the sample space birthday as a given, then there are 364 becomes different ways for the second person to ( ) HHHHHTHTHHTT avoid having the same birthday. There- S = 364 THHTHTTTHTTT fore P (E) = 365 = 0.9973 = 99.73% and it is very likely that the birthdays are different. The sample space has eight possible outcomes, three of which have two heads and 13 This is and application of the Pigeonhole Princi- one tail. Therefore ple: We have 365 possible birthdays (pigeonholes) and more than 366 people (pigeons). Therefore, at least two 3 P (E) = = 0.375 = 37.5% people (pigeons) must share a birthday (pigeonhole). 8

82 erozygotes. What is the probability that a progeny has genotype aa?

Solution: The possible outcomes can be expressed in a Punnett square, and we see that in one of four cases we obtain genotype aa. Thus,

1 P (aa) = = 0.25 = 25% 4

3. What is the probability of rolling a ”7” or a ”10” with a pair of dice?

Solution: In reference to Figure 18 we see 6 that P (”7”) = 36 and that P (”10”) = 3 36 . The events are exclusive. Therefore

Figure 20: The Birthday Problem P (”7” or ”10”) = P (”7”) + P (”10”) 6 + 3 This problem can also be worked with a = = 0.25 tree diagram, where all probabilities are 36 1 p = 2 . 4. A family has three children. What is the probability that they have at least one girl.

Solution: The ”at least” statement is the stumbling block here.

Question: How can they not have at least one girl? Answer: If all kids are boys. We compute the probability that all children are boys ﬁrst. This can be done by a tree diagram similar to the coin toss 2. Suppose that for one gene we have alle- problem, or we can argue that the sex les A and a, and that we cross two het- of the ﬁrst, second or third child are all

83 independent events, and thus the multiplication rule applies. In either case we ﬁnd

1 1 1 1 P (all boys) = · · = 2 2 2 8 The diagram confirms the result that short By the negation rule, the probability of and red progeny has probability 3/16, it having at least one girl is also explains why 9:3:3:1 is such an important ratio in genetics. P (at least girl) 6. Population Genetics. Cystic fibrosis = P (not all boys) (CF) is linked to a recessive allele. Sup- 1 = 1 − P (all boys) = 1 − pose that one of every 2,500 newborns is 8 7 affected. What is the probability that a = = 87.5% randomly selected person carries the al- 8 lele without having the disease?

5. Dihybrid Cross. We are looking at two Solution: Our population gene pool has traits. alleles A and a. We set p = P (A) the probability of selecting gene A, and q = Allele R is dominant over r; the pheno- P (a). type for R− (this means RR or Rr) is red, and that for rr is white. What is the probability of picking allele a twice? By the multiplication rule (AND Allele T is dominant over t, the pheno- rule) we get type for T − is tall, and that for tt is 2 short. P (aa) = q , What is the probability of getting short but we also know that red progeny when we cross RrT t×RrT t? 1 P (aa) = 2, 500 Solution: The results for each trait are 3 Thus q2 = 1 and q = 1 . The nega- independent and P (R−) = 4 while P (tt) = 2,500 50 1 tion rule implies that p = P (A) = 49 . 4 . Therefore 50 The problem calls for P (Aa). A person P (red and short) can pick up the a-allele either as the ﬁrst = P (R−) · P (tt) or as the second allele. This results in 3 1 3 = · = P (Aa) = 2 P (A) P (a) = 2pq 4 4 16 2 · 49 = 0.1875 = 18.75% = = 0.0392 502 As an alternative we can set up a tree This is essentially a Hardy-Weinberg prob- diagram. lem phrased in terms of probabilities, or

84 you could argue that Hardy-Weinberg is (d) of rolling a pair of sixes? really an application of the rules of prob- (e) that both dice show even numbers? ability. (f) that the sum is an even number? 7. Hamilton’s Rule. A young boy is close (g) of not getting a ten? to drowning in a heavy surf. Should his sister swim out and try to rescue him if 3. A family has four children. What is the she has a 25% chance of drowning her- probability that self? Assume that both would have two (a) all of them are girls? children each later in life. (Problem adapted from Campbell.) (b) at least one is a girl? (c) of having two boys and two girls? Solution: The benefit of the rescue is B = 2, the two children the boy might 4. A city plants 50 oak trees, 75 ash trees have. The cost for the sister is C = and 100 maple trees randomly scattered 2·25% = 0.5. They are siblings, therefore in a newly designed park. A trail cuts r = 0.5 and we get trough the park so that 90 trees are to the north of the trail, the remaining trees rB = 1 > 0.5 = C are to the south. A pair of wrens settles in one of the trees (randomly chooses one and the sister should attempt to rescue of the trees). her brother. This is based on Hamilton’s Rule alone. For real people in an emer- (a) What is the probability that the birds gency situation there are many other fac- nest to the north of the trail? tors to consider. (b) What is the probability that the birds nest on a maple tree? 9.7 Exercises (c) What is the probability that the birds 1. You flip a coin four times. What is the nest on a maple tree north of the probability of getting trail? (d) What is the probability that the birds (a) all heads? do not nest in an oak tree? (b) heads on the first toss, and tails on the other three tosses? 5. A disease is linked to a recessive allele a. (c) heads and tails exactly twice (any (a) If both parents have genotype Aa, order)? what is the probability that the child will have the disease? 2. You roll a pair of dice. What is the probability (b) If one parent has the disease, what is the probability that a child will (a) of rolling a six or higher? have the disease if the other parent (b) that at least one of the dice shows has a three? i. genotype AA? (c) of rolling a pair? ii. genotype Aa?

85 6. Dihybrid Cross. We are looking at two Would the answer change if the woman traits. Allele R is dominant over r; the only had a 80% chance of survival. Use phenotype for R- (RR or Rr) is red, the Hamilton’s Rule. phenotype for rr is white. Allele T is dominant over t, the phenotype for T- 11. A bird foraging in a flock with two aunts is tall, and for tt it is short. We cross sees a predator. It has a 20% chance RrTt x RrTt. What is the probability of of dying if it gives an alarm call. How- getting tall white progeny? ever, the two aunts have a 75% chance of living if they fly off immediately. Birds 7. Crossing mice. When yellow mice in the population have an average of six are bread together, a phenotypic ratio offspring, but the aunts are already two of about two yellow mice for each non- years old and each has had half of those yellow mouse is observed. But when yel- offspring. Should the young bird give an low mice are bred with non-yellow mice, alarm call? the progeny show a 1:1 ratio of yellow to non-yellow mice. Can you explain what’s Answers going on here? Use Y and y for the alleles, and use that 1. (a) 1/16 Y is dominant over y, that is, yellow mice (b) 1/16 have genotype Y −, while non-yellow mice have genotype yy. (c) 3/8

8. What is the relatedness factor r for grand- 2. (a) 13/18 parents and grandchildren in Hamilton’s (b) 11/36 rule? (c) 1/6 9. A group of three monkeys - all siblings- (d) 1/36 is under attack by predators. A fourth (e) 1/4 monkey, a brother of the three, notices (f) 1/2 the predators. He has a 80% chance of getting killed if he distracts the preda- (g) 11/12 tors, but each of the siblings will have 3. (a) 1/16 a 90% chance of survival if he gives an alarm call. All would have an average (b) 15/16 number of ﬁve oﬀspring in the future. (c) 3/8 Should the brother help? 4. (a) 40% 10. A young woman needs a kidney trans- (b) 44.4% plant due to a rare genetic defect. She has a 90% chance of survival, if she does. (c) 17.8% Her cousin is a match, but she has a 10% (d) 77.8% chance of dying if she donates her kidney. Both would have an average of two kids 5. (a) 25% if they live. Should she help her cousin? (b) (i) 0 (ii) 50%

86 6. 3/16 10 The Binomial Distribution

7. The genotype YY is lethal. In many cases the observations are ”black” or ”white”, ”true” or ’false”, ”either - or” with 8. r = 0.25 no in-between. This leads to Bernoulli trials 9. Yes, 6, 75 > 4 and the binomial distribution. Applications include genetics, medical research and many 10. Yes, 0.225 > 0.2 others. 11. No, 1.125 < 1.2 10.1 Bernoulli Trials

We begin with an example. A disease is linked to a recessive allele a, and both parents have genotype Aa. In this case the probability that a child will have the disease is 25%. Now suppose that they have three children. What is the probability that none of the children has the disease? What is the probability that one or two have the illness? How likely is it that all have the disease? Here is another scenario: A basketball player has a 80% free-throw percentage. During a game he steps to the line 12 times. Aside from fatigue, crowd noise, or other factors, what is the probability that he sinks 10 shots? A Bernoulli14 trial is an experiment whose outcome is either success or failure. There is no ethical or moral value attributed to the term ”success”, it just means that the event in question occurred. The probability of success denoted by p, that of failure is q. We then perform a sequence of n identical such trials, and it is assumed that the outcomes are independent, and that p and q remain the same throughout. The question becomes to determine the probability of k successes In the recessive disease example, the number of experiments is the number of children, namely n = 3, and p = 0.25, if having the disease is the outcome of interest (success). For

14Named for the Swiss mathematician Jacob Bernoulli, 1655-1705.

87 each new child the cards are shuﬄed again, and This 1 − 3 − 3 − 1 pattern is linked to the the outcomes are independent. computation For the basketball player we have n = 12 3 3 2 2 3 and p = 0.8. It is assumed that the exper- (p + q) = p + 3p q + 3pq + q iments are independent, and that making or The term p3 is the probability of three suc- missing the basket has no bearing on the re- cesses, the term 3p2q represents the probability sult of the next shot. of two successes, the term 3pq2 is linked to one success, and the last term q3 is the probability 10.2 Binomial Probability Distribu- of no success at all. tion Similar patterns emerge for other values of We now give some background15 for the for- n. If only n = 2 trials are considered, the mula to compute the probability of k successes probabilities are tied to in n trials when the probability of success is p. (p + q)2 = p2 + 2pq + q2

For n = 5 the tree diagram would get much bigger, but the probabilities would again be related to

(p+q)5 = p5+5p4q+10p3q2+10p2q3+5pq4+q5

For instance, having three successes and two failures has probability 10p3q2, and so on. No- tice that p + q = 1, and considering all cases, the probabilities will always add up to one, because (p + q)n = 1n = 1. We have argued before that the coeﬃcients in the expansion of (p + q)n can be found from Figure 21: Three Bernoulli Trials Pascal’s triangle, and if we want a formula for the probabilities, we need a formula for these Figure 21 shows the possible results for three numbers. Bernoulli trials. Successes are indicated by S, First we deﬁne the factorial for positive failures by F . The column of numbers counts integers: the successes on each branch, and the last column denotes the probability for each case. We n! = 1 · 2 · 3 ··· (n − 1) · n see that there is only one way to get three successes, and this happens with probability p3. For instance, There are three ways to get two successes, each 5! = 1 · 2 · 3 · 4 · 5 = 120 has probability p2q. The three ways of having 25 only one success have probability pq2, and the 25! = 1 · 2 · 3 ··· 24 · 25 = 1.55 × 10 one outcome will all failures has probability q3. The convention 15It is a proof for n = 3 and we do a lot of hand waving for the general case. 0! = 1

88 is useful, although it doesn’t quite fit the defi- the row, and k indicates how far you go down nition of factorials. a particular row - left to right, or right to left, The binomial coefficient is defined as it doesn’t matter because the triangle is sym- ! metric. For both, n and k, it is essential to n n! = C(n, k) = start counting with zero. k (n − k)! k! for integers 0 ≤ k ≤ n. Notice that there is no bar between n and k in the parentheses (it is not a fraction). The binomial coefficient is pronounced as ”n choose k”, because it stands for the number of ways to select k items from a set of n elements16, which motivates the ”C” in the notation C(n, k). Example: Let n = 5 and k = 3, then ! 5 5! 120 = = = 10 3 2! 3! 2 · 6 Figure 22: Pascal’s Triangle There is a more practical way to compute this expression, without the evaluation of fac- Example: We confirm the highlighted num- torials: bers in Figure 22. ! The ”6” is the third number in the 5th row, 5 5! 5 · 4 · 3 · 2 · 1 5 · 4 = = = = 10 but since we start counting at zero, we get n = 3 2! 3! 2 · 1 · 3 · 2 · 1 2 · 1 4 and k = 2. It now follows that ! 4 4 · 3 If we apply the strategy of the example = = 6 to any C(n, k), we should proceed as follows: 2 2 · 1 First take the smaller of k! or (n−k)! and enter For the ”35” we find that n = 7 and k = 3 and the terms in decreasing order in the denomina- ! tor. Then begin to write n! in decreasing order 7 7 · 6 · 5 = = 35 in the numerator, and align each number on 3 3 · 2 · 1 top with one below, but stop when you run out of matching entries. The value of the resulting And finally for ”28” we obtain n = 8 and k = 6 fraction is the binomial coefficient. The exam- and ple below will illustrate this technique some ! 8 8 · 7 · 6 · 5 · 4 · 3 8 · 7 more. = = = 28 It can be shown that all entries in Pascal’s 6 6 · 5 · 4 · 3 · 2 · 1 2 · 1 ! n triangle have the form . Here n counts k We are now ready for the big formula. The 16Consider the set {a, b, c, d} of n = 4 elements. Now probability of k successes in n Bernoulli trials pick any k = 2 of these. This gives you the options is {a, b}, {a, c}, {a, d}, {b, c}, {b, d} or {c, d}. We have n! b(n, p, k) = pkqn−k (3) 6 = C(4, 2) ways to do this. (n − k)! k!

89 where p is the probability of success, and q = nomial coefficient. For instance C(12, 10) can 1−p is the probability of failure. This is called be computed by ”12 nCr 10”. the binomial probability distribution. The binomial distribution is found under The first fraction in (3) is the binomial co- DISTR, and it is called ”binompdf”. In order efficient from Pascal’s triangle; it counts in how to compute b(12, 0.8, 10), just enter many ways k successes can be achieved. The last term involves the probabilities. Each suc- binompdf(12,0.8,10) cess comes with probability p, and thus k suc- k cesses lead to the factor p . If we have k suc- and your calculator will produce 0.2835. The cesses in n trials, then we also have (n − k) command has the structure ”binompdf(n,p,k)”, n−k failures, and these introduce the factor q . and if k is omitted, ”binompdf(n,p)” will dis- The tree diagram in Figure 21 gives an illus- plays all values for k from 0 to n. tration for n = 3. EXCEL has a lot of built-in functions. Example: Back to the basketball case. What ”=FACT(n)” will compute a factorial, the com- is the probability of 10 successes in 12 trials mand ”=COMBIN(n,k)” will calculate C(n, k) when the probability of success is 80%? and ”=BINOM.DIST(k,n,p,false)” can be used Here n = 12, k = 10 and p = 0.8. The for binomial distributions. probability of failure17 is q = 1 − p = 0.2, and 10 successes in 12 trials means that we have 10.2.2 Cumulative Density Distribution n − k = 2 failures. For the binomial coefficient we find Very often we are interested in more than just ! ! the probability for a particular value of k. For 12 12 12 · 11 = = = 66 instance, in the basketball problem we might 10 2 2 · 1 ask for 10 or more baskets in 12 attempts. The cumulative probability density function takes The full binomial distribution formula (3) re- care of this. It adds all the probabilities from sults in k = 0 up to a certain threshold. b(12, 0.8, 10) The formulas become very complicated, and 12! we illustrate the concept with an example. Let’s = 0.8p10 0.22 = 66 · 0.1074 · 0.04 10! 2! say we take n = 5 trials and the probability of = 0.2835 success is p = 0.4. In the table below the column labeled ”pdf” contains all probabilities for 10.2.1 Calculators and Computers k from 0 to 5, computed with formula (3) as usual. Graphing calculators can help a great deal with the computations. The following instructions k pdf cdf work for TI calculators, and similar steps can 0 0.078 0.078 be applied on other brands. 1 0.259 0.337 The MATH button has a PRB option. Here 2 0.346 0.683 you find ”!” for factorials and ”nCr” for the bi- 3 0.230 0.913 17If you make a basket 80% of the time, you miss 4 0.077 0.990 20% of the time. 5 0.010 1.000

90 The column ”cdf” keeps the running total. For = P (not 9 or less baskets instance, at k = 2 it contains the sum = 1 − 0.442 = 0.558 0.078 + 0.259 + 0.346 = 0.683 by the negation rule. When we reach k = 5, all cases are covered and the probabilities add to one. 10.3 Worked Problems 1. Compute (a) 16! 16! (b) 11! 16! (c) 5! 11! Solution: (a) 16! = 1 · 2 · 3 ··· 15 · 16 = 20, 922, 789, 888, 000 The graph shows both functions. The den- 16! 1 · 2 ··· 10 · 11 · 12 ··· 15 · 16 (b) = sity function peaks for k = 2. This is not sur- 11! 1 · 2 · 3 ··· 10 · 11 prising, because with n = 5 trials and a 40% = 12 · 13 · 14 · 14 · 15 · 16 success rate one would expect 0.4 · 5 = 2 suc- = 524, 160 cesses. The cumulative function adds the pdf 16! 16 · 15 · 14 · 13 · 12 (c) = values, it increases and reaches 1 when k = 5. 5! 11! 5 · 4 · 3 · 2 · 1 The command ”binomcdf(n,p,k)” will com- = 4, 368 pute cumulative densities on a calculator, and in EXCEL the command is 2. Sickle-cell anaemia (SCA) is linked to a ”=BINOM.DIST(k,n,p,true)”. recessive allele. Suppose that the par- Example: A basketball player with a 80% free ents are both heterozygotes with respect throw percentage steps to the line 12 times. to this trait. What is the probability that What is the probability of making 10 or more two of their three children have the dis- points? ease? We need to be careful here. We are inter- Solution: Here n = 3 (each child is an ex- ested in 10, 11 or 12 successes, but the cumu- periment!), and we are looking for k = 2 lative density function will start counting with occurrences of the SCA. We also know k = 0. Hence, and this is a calculator result, that the probability of acquiring the dis- binomcdf(12, 0.8, 9) = 0.442 ease is p = 0.25. The binomial distribution formula implies that is the probability of making anywhere from 3! none to nine baskets, and therefore the proba- b(3, 0.25, 2) = 0.252 0.75 1! 2! bility of sinking 10 to 12 shots is = 3 · 0.0625 · 0.75 = 0.140625 P (at least 10 baskets) ≈ 14%

91 3. Given that 30% of the students are smok- (d) If the group has at least one smoker, ers, what is the probability that a ran- they cannot be all non-smokers. Us- domly selected group of six students ing the negation rule and the result from part (a) we obtain (a) consists entirely of non-smokers? (b) consists entirely of smokers? P (at least one smoker) (c) is evenly split between smokers and = 1 − P (all non-smokers) non-smokers? = 1 − 0.118 = 0.882 (d) has at least one smoker? (e) has two or more smokers? (e) This problem calls for adding the probabilities for k = 2, 3, 4, 5 and Solution: For this problem n = 6 in all 6. This becomes a little easier, if we cases, and p = 0.3, if we look for smokers negate the statement ﬁrst and look (success). Selecting a non-smoker then for the probability of having at most has probability q = 0.7. one smoker in the group. We know from part (a) that the probability (a) Selecting a non-smoker six times in for k = 0 smokers in the group is a row has probability 0.118 and for k = 1 we obtain 6! P (all non-smokers) b(6, 0.3, 1) = 0.31 0.75 = 0.76 = 0.118 5! 1! = 6 · 0.3 · 0.168 = 0.303 by the multiplication rule for independent events. One can also apply and having none or just one smoker formula (3) with k = 0 (recall that in the group has probability 0! = 1 and that p0 = 1 for any number p). We ﬁnd that 0.118 + 0.303 = 0.421 = 42.1%

6! By the way, the same result can be b(6, 0.3, 0) = 0.30 0.76 6! 0! obtained using the cumulative dis- = 0.76 = 0.118 tribution, and the calculator shows18 that (b) Here k = 6 and binomcdf(6, 0.3, 1) = 0.420, 175 6! b(6, 0.3, 6) = 0.36 0.70 0! 6! Thus, having two or more smokers = 0.36 = 0.000, 729 = 0.07% in in group occurs with probability (negation principle) (c) In this case k = 3 and it follows that 1 − 0.420 = 0.580 = 58.0% 6! b(6, 0.3, 3) = 0.33 0.73 3! 3! 18There is a little rounding discrepancy; 42.0175% is = 20 · 0.027 · 0.0.343 = 0.185 correct.

92 4. In a 1995 study the CDC reported that (a) ”At least one” victim rules out that 20% of college women have been a vic- none has been raped. The probabil- tim of rape during their lifetime. Some ity of not having been victimized is critics question the validity of this ﬁgure, q = 1 − p = 0.8, and 0.815 = 0.035. while newer studies seem to support sim- Therefore the probability that none ilar values19. For the sake of argument, has been abused is 3.5%, and the let’s stick to the 20% ﬁgure for this prob- probability that at least one has been lem. raped is (negation rule)

1 − 0.035 = 0.965 = 96.5%

(b) Here n = 15, p = 0.2 and k = 1 and we are looking for

b(15, 0.2, 1) 15! = 0.21 0.814 14! 1! = 15 · 0.2 · 0.044 The graph shows the probabilities for hav- = 0.132 = 13.2% ing k victims in the group. In a group of 15 one would expect having about three (c) This excludes none, one or two rape victims in the group20, and this has the victims. The cumulative density dis- highest probability. But by chance it could tribution yields (calculator and/or very well happen that we have fewer or the graph above) more victims in the group. However, having eight or more in a randomly selected binomcdf(15, 0.2, 2) = 0.398 group is highly unlikely, as the graph shows. that is, having up to two victims Here are the questions: What is the prob- in the group has probability 39.8%, ability that in a group of 15 randomly and therefore having three or more selected college women has probability 0.602 = 60.2%

(a) at least one has been a victim of 5. Given that 4% of the population carry rape? the cystic ﬁbrosis allele. What is the probability that in a group of 50 people (b) exactly one of them has been a victim? (a) nobody, (c) three of more have been raped? (b) one person, (c) two people, Solutions: (d) three or more 19A CDC study in 2014 reports that 19.3% of women have been victim of rape during their lifetime, about carry the CF allele? 78% of the rape victims were 25 years old or younger. 2020% of 15 is 3, and statisticians call this the ex- Solution: Here n = 50, p = 0.04, and pected value. we are looking for the probabilities with

93 k = 0, k = 1 and k = 2. Routine compu- 7. A student takes a multiple choice test by tations show that random guesses. There are ﬁve questions on the test, each with four choices. What binompdf(50, 0.04, 0) = 0.130 is the probability of getting binompdf(50, 0.04, 1) = 0.271 (a) four correct answers? binompdf(50, 0.04, 2) = 0.276 (b) at least four correct answers?

Follow up: How do the results change which answers parts (a) - (c). If we add when all questions ”true or false”? the values, we obtain 0.677 = 67.7%, which is the probability of having up to Solution: Here n = 5, and for the first two carriers. Hence, the probability of questions we have a p = 0.25 chance of having three or more carriers is success. (a) k = 4 and 1 − 0.677 = 0.323 = 32.3% b(5, 0.25, 4) 6. An experimental medical procedure is 86% 5! = · 0.254 · 0.75 successful. What is the probability that 4! 1! 5 · 3 15 it fails for tree of twelve patients in a hos- = = = 0.0147 pital? What is the probability that it 45 1024 fails for three or more patients? and the chance of getting four correct answers is about 1.5%. Solution: With an 86% success rate and 12 patients, we expect 0.86 · 12 = 10.3 (b) This case asks for four or five cor- successes, and thus 1.7 failures. But this rect answers. The probability of get- is not the question. We need the prob- ting all answers correct is ability of k = 9 successes in n = 12 1 0.255 = = 0.000, 98 trials when the probability of success is 1024 p = 0.86, and we find that and addition shows that the proba- b(12, 0.86, 9) = 0.155 bility of getting four or more correct answers is 15 + 1 16 When we ask for three or more failures, = we are looking at 9 or less successes. The 1024 1024 1 cumulative density function yields = = 0.0156 = 1.56% 64 binomcdf(12, 0.86, 9) = 0.230 In the follow-up the probability of success changes to p = 0.5, and similar com- Our hospital is not doing too well. Its putations show that the probability of success rate of 75% lags behind the av- getting four problems correct is erage, and about the bottom quarter of hospitals have similar dismal success rates. b(5, 0.5, 4) = 0.15625

94 and the probability of getting four or more with random guesses? correct is 7. An experimental medical procedure is 90% b(5, 0.5, 4) + b(5, 0.5, 5) successful. What is the probability that it fails on three of twelve patients in a = 0.15625 + 0.03125 = 0.1875 hospital?

10.4 Exercises 8. Given that 4% of the population carry the recessive cystic ﬁbrosis allele. What 1. Compute is the probability that in a group of 25 (a) 12! people (b) 9! (a) nobody 12! (c) (b) one person 9! (c) two people 12! (d) 9! 3! carry the allele? 2. You ﬂip a coin four times. What is the probability of getting Answers

(a) all tails? 1. (a) 479,001,600 (b) tails three times? (b) 362,880 (c) heads and tails exactly twice (any (c) 1,320 order)? (d) 220

3. You roll a die ﬁve times. What is the 2. (a) 1/16 probability of rolling sixes twice? (b) 1/4 4. A family has ﬁve children. What is the (c) 3/8 probability that they have four girls and one boy? 3. 16.1%

5. Suppose that Democrats and Republicans 4. 15.6% are evenly split on campus. What is the 5. 22.6% probability that 12 randomly selected students are split evenly into six supporters 6. (a) 1.465% , (b) 1.5625% of each party? 7. 8.5% 6. A multiple choice test has 5 questions with four choices each. What is the prob- 8. (a) 36.0%, (b) 37.5%, ability that you get (c) 18.8%

(a) four correct answers, (b) at least four correct answers

95 11 Functions in the output y = 2, the input x = 25 generates the output y = 5, and so on. Functions are a very important concept in math- For any input x, the output y from the √ ematics, and they are usually introduced in function will be x. In function notation we Pre-Calculus. In the next chapters we will look write at linear, exponential, logarithmic and power √ f(x) = x functions which play an important part in biological modeling. The purpose of this section is and the initial examples become to introduce the function concept. We touched loosely on this topic in the context of graphing f(4) = 2 and f(25) = 5 already. Deﬁnition: A function f is a rule which as- We cannot take square roots of negative signs to each admissible input x exactly one numbers. Thus, the domain of this function output y. The domain of a function is the set is the interval 0 ≤ x < ∞, and since taking of all admissible inputs, and the range consists square roots will never produce a negative re- of all possible output values. We write21 sult, the range is the interval 0 ≤ y < ∞ as well. y = f(x) Example (An Exponential Function): As- sign the output y = 2x to every input x. For pronounced as ”f of x”, where y is the output instance, the input x = 3 results in the out- assigned to the input x. put y = 23 = 8, for x = −2 we get y = −2 1 2 = 4 = 0.25 and x = 0 produces the output y = 20 = 1. In function notation we have

f(x) = 2x

and our initial examples can be expressed as The black box concept is often used to vi- 1 sualize functions. An input x is submitted to f(3) = 8 f(−2) = f(0) = 1 the black box, and the rule will assign an out- 4 put y. The notation f(x) emphasizes that the The domain of this function consists of all real result is the output of the function f from the numbers22, and the range are the positive num- input x. bers (y > 0). It is customary say that the input is the independent variable, and the output is the de- Example: A Function Defined by a Ta- pendent variable, as it depends on the input. ble. In this example we define a function by Example (Square Roots): Here the rule (f) is taking square roots. The input x = 4 results 22In the Algebra chapter we have looked at ax for positive integers x, and then we extended the meaning 21f is the function, f(x) is the formula for the output. of ax to negative integers x, and later on to fractional There is a fine difference here, and many a math major exponents. It requires some serious Calculus to define struggles with that too. ax when x cannot be written as a fraction.

96 listing all inputs and their associated outputs. useful to interpret those kinds of examples in terms of functions, sometimes using functions Input Output makes simple things unnecessarily abstract and x f(x) complicated. In the sequel we will restrict our- 1 4 selves to functions tied to numbers. 2 6 4 9 11.1 Graphs of Functions 5 4 8 5 From your experience with graphing calculators you know that the graph of a function This means that x = 1 is mapped to f(1) = 4, should be some kind of a curve. This is true x = 2 is mapped to f(2) = 6 and so on. in many cases, because most of the time the The requirement that each admissible input output can be described by a formula. The has exactly one output is met. The domain functions of this function is the set of numbers D = √ f(x) = x {1, 2, 4, 5, 8}. The function is not defined for 4 − x any other values. The range of the function f(x) = are the numbers R = {4, 5, 6, 9}, and the fact 2 f(x) = 2x that x = 1 and x = 5 have the same output is not in violation the definition of a function. serve as examples. In this case you can enter the formula, and the graphing calculator23 will Several Variables. There is really no rea- produce a nice curve. son to restrict the input of a function to single In these graphs the inputs lie on the x- numbers. For example, the binomial distribu- axis and the outputs are located on the y- tion formula axis. In order to illustrate the input-output n! relation, select an input x on the x-axis, move b(n, p, k) = pkqn−k (n − k)! k! up or down to the curve, and then over to the y axis. This number is the output y. Fig- can be viewed as a function of three variables. ure 23 illustrates this notion for the function The input consists of the triple (n, p, k), where √ f(x)√ = x with input x = 6.25 and output n is a positive integer, 0 ≤ p ≤ 1 is a real y = 6.25 = 2.5. number, and 0 ≤ k ≤ n is an integer (this is You can visualize the domain of a func- a description of the domain). The output is tion by projecting the graph of the function the value computed from the formula on the down onto the x-axis, and its range by projec- right, and it can be shown that the range of tion onto the y-axis. For example, the function this function is the interval 0 ≤ y ≤ 1. given in Figure 24 has domain 1 ≤ x ≤ 4 and General Functions. The function con- range 1 ≤ y ≤ 5. cept is not restricted to numbers only. For You may have noticed than in any of those example, you can construct a function by as- graphs you never have a situation where one signing the date of birth (output) to each per- 23 son (input). When you classify organisms by Before the advent of graphing calculators people would compute f(x) for selected values of x (T-table), species, you assign to each organism (input) plot the points, and then connect the dots by a smooth its scientific name (output). Sometimes it is curve.

97 The composition of functions is needed to explain inverse functions. Pre-Calculus usually covers many aspects of functions in great detail; here we limit ourselves to the composition and to inverse functions. Composition. In the composition of functions, the output of one function is used as an input for another function, as illustrated in ﬁg- √ ure below. Figure 23: f(x) = x

The input x is fed into the function g and it produces the output y = g(x). This value is then used as input for the function f. The output becomes

Figure 24: Domain/Range z = f(y) = f(g(x))

Overall, the input x results in the output z = and the same x value comes with two differ- f(g(x)). By combining the two functions we ent points on the graph. This does not occur, have created an new input-output relationship, because one input cannot have two different that is, we have formed a new function h. This outputs. This idea is called the Vertical Line function is called the composition of f and g, 24 Test . Moreover, any curve which passes the and its is commonly denoted as h = f ◦ g. In Vertical Line Test can be used to define a func- particular, tion graphically. Just apply the principle of Figure 23. h(x) = (f ◦ g)(x) = f(g(x)) If a function is given in form of a table, √ a scatter plot is the most appropriate way to Example: Take f(x) = x and g(x) = 2x−3. display its graph. The function f takes square roots of the input, therefore if the input is g(x), the output becomes 11.2 Composition and Inverse Func- q tions f(g(x)) = g(x) This section is included because exponential But we have a formula for g(x), and once we functions and logarithmic functions, which will substitute we obtain be covered later on, are inverses of each other. q √ h(x) = f(g(x)) = g(x) = 2x − 3 24Any vertical line intersects with the graph at most √ once. and the composition is h(x) = 2x − 3.

98 Example: If f(x) = 1/x and g(x) = 1 − x2, Definition: A function g is called the inverse then function of f if 1 1 f(g(x)) = = g(f(x)) = x and f(g(y)) = y (4) g(x) 1 − x2 is valid for all x in the domain of f and for all and y in the range of f. 1 g(f(x)) = 1 − f(x)2 = 1 − x2 Clearly, the results are different, and f ◦ g 6= g ◦ f is the rule, rather than the exception. Example: The temperature T in degree Cel- sius x meters above the ground is given by A few remarks are in order: T (x) = 20 − 0.01x 1. The notation f −1 is frequently used to denote the inverse of a function. But this i.e. at the ground level the temperature is can be misleading, because 25oC, and it decreases by one degree Celsius for every elevation gain of 100 m. The height in 1 f −1(x) 6= meters of a projectile t seconds after its launch f(x)) is given by that is, the inverse of a function is not its h(t) = 80t − 4.9t2 reciprocal. For this reason we denote an inverse by g. In order the find the air temperature at the 2. Suppose that f takes the number a to b, location of the projectile as a function of time, then the inverse will take b back to a. In we have to first determine the height and then function notation we have compute the temperature. This results in f(a) = b and g(b) = a T (h(t)) = T (80t − 4.9t2) = 20 − 0.01 80t − 4.9t2 If we combine these statements, we get = 20 − 0.8t + 0.049t2 a = g(b) = g(f(a))

This is a composition of functions. The out- and we see that the composition of input of the height function is the input to the verse functions results in the identity func- temperature function. tion.

Inverse Functions. When we construct 3. Most of the time, if g(f(x)) = x works the inverse of a function, we aim to ﬁnd a func- out, the other requirements for inverse tion which ”undoes” the eﬀects of the original functions will fall into place. But in some function. For instance, if f takes the input 5 cases it is important to pay attention to to 8, then the inverse function should take 8 domain and range of f and g in order to back to 5. avoid pitfalls.

99 4. Not every function has an inverse. In the that x ≥ 0. Result: For positive numbers the example of tabular data we had f(1) = 4 square root of a square is the original number! and f(5) = 4. It is impossible to con- For the converse we have struct an inverse function, because we √ √ cannot map the value 4 back to 1 and f(g(y)) = f( y) = ( y)2 = y to 5 at the same time (a function g can only have one output). No problem here. y ≥ 0 is required for the In mathematics we call functions for which domain of g. different inputs always result in differ- Notice that the parabola f(x) = x2 is not ent outputs one-to-one, and this is the one-to-one. For instance, f(4) = 16 and f(−4) = class of functions which have an inverse. 16, and we can only go back from y = 16 to We do not want to get into further de- x = 4, if we exclude the negative x = −4. tail; usually we will encounter some ”red 1 Example: f(x) = is its own inverse: flags” as we attempt to compute an in- x verse. 1 1 f(f(x)) = f = = x Example: The functions x 1/x x + 5 f(x) = 2x − 5 and g(x) = The reciprocal of the reciprocal is the original 2 number. are inverses of each other. The relationship Usually, inverse function are nicely paired on a calculator and connected with the ”2ND” g(f(x)) = g(2x − 5) √ key. x2 comes with , LOG with 10x, LN (2x − 5) + 5 = with ex, and the same goes with the trigono- 2 metric functions. If you experiment with your 2x = = x calculator, you will observe that in most cases 2 no matter what messy number the original func- holds for all x, and f(g(y)) = y works in a tion produces, the inverse will take you back to similar fashion, and there are no issues with where you started: domains and ranges. log 412.7 = 2.615634469 Example: The functions Ans √ 10 = 412.7 f(x) = x2 and g(x) = x The construction of an inverse requires to are inverses, provided that, and this is impor- recover x from y = f(x). In other words, we tant, we exclude negative numbers. Verifica- have to solve the equation y = f(x) for x. tion: 2x q √ Example: Let f(x) = . to find the g(f(x)) = f(x) = x2 = |x| = x x − 1 inverse we have to solve √ 2 The equation x = |x| holds for all numbers, 2x y = positive or negative, but the last step requires x − 1

100 It’s a long process, but it can be done (the (e) Does the function have an inverse? steps should be evident): Solutions: 2x y = x − 1 (a) The function takes the sum of the (x − 1)y = 2x input and its reciprocal. xy − y = 2x (b) xy − 2x = y (y − 2)x = y f(1) = 1 + 1 = 2 y 1 x = f(2) = 2 + = 2.5 y − 2 2 1 y f(−2) = −2 + = −2.5 and the inverse is x = g(y) = , or by −2 y − 2 1 renaming the variable, it becomes f(4) = 4 + = 4.25 4 x 1 1 1 g(x) = f( 4 ) = + 1 = 4.25 x − 2 4 4 Technically, still we have to verify that g (c) The graph is meets all the requirements of an inverse function. This includes to conﬁrm that

2x x−1 g(f(x)) = 2x = x x−1 − 2 Secondly, g is not defined for x = 2, the reason being that there is not a single value for x which makes f(x) = 2. Conversely, f is undefined for x = 1, and g(x) 6= 1 for any selection (d) x = 0 has to be excluded from the of x. domain, as its reciprocal is undefined. All other real numbers are 11.3 Worked Problems permitted. 1. A function is defined by the formula (e) The function cannot have an inverse. 1 1 For instance, x = 4 and x = 4 have f(x) = x + the same output y = 4.25, and it is x impossible to go back from 4.25. In (a) Describe the the function verbally. math terms: The inverse does not (b) What are the respective outputs for exist, because the function is not the inputs x = 1, x = 2, x = −2, one-to-one. x = 4 and x = 1 ? 4 2. A function is defined as (c) Graph the function for −4 ≤ x ≤ 4. √ (d) What is the domain of the function? f(x) = x − 1 − 4

101 (a) Sketch the graph of this function. 3. A function is deﬁned by the data below.

(b) What are the domain and the range Input Output of the function? x f(x) −3 16 (c) Construct the inverse of the func- 0 12 tion. 2 8 5 10 Solutions: 10 0 (a) Compute f(0), f(2) and f(f(5)). (b) Sketch the graph of the function.

Solutions:

(a) Directly from the table we obtain that f(0) = 12 and f(2) = 8. Since (a) f(5) = 10, it follows that f(f(5)) = f(10) = 0. (b) x ≥ 1 is required, so that the square (b) A scatter plot is most appropriate root exits. Therefore, the domain is for tabular data. {x ≥ 1}. y = −4 is the smallest possible value (it occurs for x = 1). Therefore, the range is {y ≥ −4}. √ (c) We begin with y = x − 1 − 4 and solve the equation for x:

√ y = x − 1 − 4 4. Let f(x) = x(1−2x), g(x) = 1 (1−x) √ 2 x − 1 = y + 4 and h(x) = 1 − 2x. Determine 2 x − 1 = (y + 4) (a) f(g(x)) 2 x = 1 + (y + 4) (b) g(h(x)) (c) h(h(x)) Thus, after we switch x and y, the inverse function becomes Solutions:

g(x) = 1 + (x + 4)2 (a) First we formally substitute g(x) for x in the definition of f, then we apply the definition of g. Simplifica- The condition y ≥ −4 is implied in tion will do the rest. the first step. Therefore the domain of g is the set {x ≥ −4}. f(g(x)) = g(x)(1 − 2g(x))

102 1 1 = (1 − x)(1 − 2 (1 − x)) (d) Find x such that f(x) = 0 2 2 (e) Construct the inverse for f. 1 = (1 − x)(1 − (1 − x)) x 2 3. Let f(x) = 1 x + 1 = x (1 − x) 2 (a) Compute f(0), f(1), f(99) and f(−3) (b) (b) Sketch the graph of f. (c) Construct the inverse for f. 1 g(h(x)) = (1 − h(x)) 2 4. A function is defined by the data below. 1 = (1 − (1 − 2x)) Input Output 2 x f(x) 1 = 2x = x −2 3 2 0 1 which shows that g and h are in- 3 6 verses. 6 6 (c) Sometimes it is beneficial to apply 10 5 a function twice. Here we obtain (a) Compute f(0), f(10) and f(f(f(6))). (b) Sketch the graph of the function. h(h(x)) = 1 − 2h(x) √ = 1 − 2(1 − 2x) 5. Let f(x) = x, and let g(x) = x2 + 1. = 1 − 2 + 4x Form = 4x − 1 (a) f(g(x)) and (b) g(f(x)) 11.4 Exercises 1 6. Confirm that the functions f(x) = 1. Let f(x) = x2 − 4. x − 1 1 and g(x) = 1 + x are inverses of each (a) Describe the the function verbally. other. (b) What are the respective outputs for the inputs x = 1, x = 2, x = −2, Answers x = 4? 1. (a) The function subtracts 4 from the (c) Graph the function for −5 ≤ x ≤ 5. square of the input. (d) What is the domain of the function? (b) −3, 0, 0, 12, respectively. (e) Does the function have an inverse? x 2. Let f(x) = + 4 10 (a) Describe the function verbally. (b) Compute f(0), f(20), f(−10). (c) Sketch the graph of f. (c)

103 (d) All real numbers. 12 Linear Functions (e) No. Linear models are the most natural approach 2. (a) The function takes one tenth of the in many problem solving situations. input and then adds 4 to it. For instance, if the cost for a wedding increases by $200 for eight additional guests, then (b) 4, 6 and 3, respectively. twelve more guests will raise the cost by $300. It’s natural, it’s logical and we haven’t even said anything about linear relationships. Linear models are often used as a ﬁrst step in complex situations, and biology is no exception. We will touch upon examples from cli- mate change, population growth, temperature (c) regulation and anatomy. (d) x = −40 12.1 Lines (e) g(x) = 10x − 40 Most everybody is familiar with 1 99 3 3. (a) 0, 2 = 0.5, 100 = 0.99 and 2 = 1.5, respectively. y = mx + b

as an equation for a line. Here ∆y change in y m = = ∆x change in x rise = run (b) is the slope of the line. It is also called the x x gradient, although in mathematics this term is (c) g(x) = = − 1 − x x − 1 usually reserved for multi-variate functions. In 4. (a) 1, 5 and 6, respectively. most applications the slope represents a rate. What distinguishes a (straight) line from all other curves, is that the slope remains the same throughout. It does not matter which points P and Q are selected on the line, the ra- (b) tio of the diﬀerences of the y-coordinates (∆y) √ to the diﬀerences of the x-coordinates (∆x) al- 5. (a) f(g(x)) = x2 + 1 ways remains the same. (b) g(f(x)) = x + 1 The term b is the y-intercept. It can be found by setting x = 0. In a graph it is value 6. Form compositions to obtain f(g(x)) = x where the line crosses the y-axis. and g(f(x)) = x. Example (Temperature Conversion): The relation between the Celsius (C) and Fahren-

104 Example (Cost of a Wedding): If x is the number of guests and y is the cost of a wedding, the statement that eight more guests increase the cost by $200 translates into ∆y = 200 when ∆x = 8. The slope becomes

∆y 200 m = = = 25 ∆x 8 This is a rate, it measures the cost per person. ∆y The relationship m = ∆x can be rephrased as Figure 25: Line ∆y = m ∆x and with m = 25, we find that ∆y = 25∆x, heit (F) temperature is given by the equation which tells us that if the number of attendees changes by ∆x, the cost will change by ∆y = F = 1.8C + 32 25 ∆x. We have no information about the fixed When we set y = F and x = C we have an cost, i.e. expenses which are independent of equation of a line the number of guests. Thus, b is unknown and y = 1.8x + 32 we cannot set up an equation of the cost of a wedding in terms of the number of guests. The slope is m = 1.8; it tells us that for each 1o Linear Functions. When we express a increment on the Celsius scale, the Fahrenheit line in function form, temperature will go up by 1.8o. In order to illustrate the ∆ notation, we f(x) = mx + b have to pick two points. So, for instance x = o o 10 C is the equivalent of y = 50 F and x = we call it a linear function. Examples include 35oC corresponds to y = 95oF . For these two data we get f(x) = 4x − 9 f(x) = 6 − 0.5x ∆C = ∆x = 35 − 10 = 25 4 − x f(x) = ∆F = ∆y = 95 − 50 = 45 2 that is, a 25 degree increase in the Celsius scale There is really no significant difference becomes with a 45 degree raise of the Fahrenheit tween f(x) = 2x−3 and y = 2x−3. The first temperature, and the ratio becomes notation emphasizes the function concept, the second is more appropriate for graphs in the ∆y 45 m = = = 1.8 , xy-plane. If you want to find the value of a ∆x 25 linear function for a specific point x, the func- as expected. tion notation comes in handy. For example, The y-intercept is b = 32. Water freezes at let x = 0oC, which is the equivalent of 32oF . f(x) = 2x − 3

105 then the value for x = 2 is 12.2 Point-Slope Form

f(2) = 2 · 2 − 3 = 1 A line is completely determined if we know two points which belong to the line. It is also de- The alternative would be writing termined if we know just one point, as well as the slope. y = 2x − 3 Let’s begin with the latter. Suppose that the slope of a line is m, and that the point and then P (x0, y0) belongs to the line. Then the equation of this line is y|x=2 = 2 · 2 − 3 = 1

This is all very technical, and hybrids, such as y − y0 = m(x − x0) (5) y(2) = 1 are acceptable. Tables. Consider the data in the table This equation is known as the point-slope form of a line, as it requires the knowledge of a point x y and the slope. −2 14 0 20 Example: A line has slope m = 3, and it 2 26 contains the point P (2, 1). Its equation is 4 32 6 38 y − 1 = 3(x − 2)

We notice that the data on the left increase by which can be rewritten as 2, and those on the right increase in steps of 6. This is a clear indication that we are dealing y = 3x − 5 with a line! On the left we have ∆x = 2 and on the right we get ∆y = 6. Therefore, Let’s confirm the result: Obviously, the slope ∆y 6 is m = 3, and when we set x = 2, we find that m = = = 3 ∆x 2 y = 3 · 2 − 5 = 1, as desired. There is an alternative solution. We know When x = 0 we see that y = 20 = b, which is that m = 3, therefore the equation becomes the y-intercept, and the equation for the data in the table is y = 3x + b y = 3x + 20 When we set x = 2 and y = 1, the substitution A graph confirms the linear structure of the yields data. 1 = 3 · 2 + b

and solving for b results in b = −5. Hence, the equation is y = 3x − 5, as before.

Two Points. Two points also determine a line, but the corresponding formula becomes

106 more involved25. Usually it is easiest to use the When we compute the slope between P and Q, two points to determine the slope, and then we find that apply the point-slope formula (5) using any of ∆y y − y0 the two points. m = = ∆x x − x0 Example: Let P (1, 4) and Q(5, 2). Then the Multiplication yields formula (5) slope of the line connecting P and Q is given by y − y0 = m(x − x0) ∆y 2 − 4 1 m = = = − ∆x 5 − 1 2 12.3 Almost Linear Data Once m is found, we can apply formula (5) and Data in real life are not perfect. There is vari- obtain that ation in nature and there are measurement errors. Sometimes you have a lot of data, and 1 y − 4 = − (x − 1) there appears to be a linear relationship, but 2 the numbers don’t fit perfectly. The goal then In slope-intercept form the line becomes becomes to find an approximating line. This topic is carefully studied in regression analysis. 1 9 This workbook is not the place for a detailed y = − x + = −0.5x + 4.5 2 2 analysis. Instead, we shall use an intuitive approach, the straight-edge method. If we use the point Q in the point slope formula The steps are simple: instead, we obtain 1. Graph the data points carefully. 1 y − 2 = − (x − 5) 2. Trust your intuition, and draw a line which 2 best fits the data. which is equivalent y = −0.5x+4.5. This illus- 3. Identify two points on the line - near op- trates that we can use either point the point- posite ends for the sake of stability. slope formula. 4. Use the points from Step 3 to write an We test our result: If x = 1, then y = equation for the approximating line. −0.5 + 4.5 = 4 and for x = 5 we find that y = −0.5·5+4.5 = −2.5+4.5 = 2, as expected. Example: The data and with a graph are given below. Derivation of the Point-Slope Formula. Suppose that P (x0, y0) is a point on the line, x 1 3 5 6 8 8 and that Q(x, y) is any other point on this line. y 2 2 4 5 5 7

25 The line containing the points P (x1, y1) and Q(x2, y2) can be deﬁned by y − y y − y 1 = 2 x − x1 x − x2 or by y2 − y1 y − y2 = (x − x2) x2 − x1

107 The estimated interpolating line contains go to STAT and CALC, and select the Lin- the points P (2, 2) and Q(8, 6). The slope of Reg(ax+b) option. the interpolating line is In the above example, the calculator will produce a = 0.6395 and b = 0.8627. The data 6 − 2 4 2 m = = = can be displayed with the STAT PLOT op- 8 − 2 6 3 tions (turn the plots on), and the line can be and the line becomes included using the usual ”Y=” key for graphing. 2 y − 2 = (x − 2) In EXCEL you should list the data in columns 3 and produce a scatter plot. If you right-click on a data point, the ”Add Trendline” option which can be rewritten as will show up. Select ”Linear”, and if you opt 2 2 for ”Display Equation on chart”, an equation y = x + (6) 3 3 of the line will show up on the graph. Calculator and EXCEL will produce iden- We see that the line contains one data point tical results, because both are the least squares exactly, otherwise it overshoots or undershoots solution, and if you compare the computer an- the data. The table below lists data, predic- swer tions and errors. y = 0.6395x + 0.8627 x y predicted error to our estimation (6) 1 2 1.33 −0.67 y = 0.6667x + 0.6667 3 2 2.67 0.67 5 4 4 0.00 you will see that both lines are fairly close, and 6 5 4.67 −0.33 that our estimation was pretty good, particu- 8 5 6 1.00 larly at the upper end. 8 7 6 −1.00 x y least squares our estimate 1 2 1.50 1.33 The predictions are obtained by substituting 3 2 2.78 2.67 the x-values into the formula (6). These are 5 4 4.06 4.00 the corresponding points on the line. The er- 6 5 4.70 4.67 rors (residuals) subtract the actual data from 8 5 5.98 6 the predictions. For instance, the observation 8 7 5.98 6 for x = 1 is y = 2, the prediction by formula is y = 1.33, the residual is 1.33 − 2 = −0.67. 12.4 Worked Problems Our approach is ambiguous. Different peo- 1. The global mean temperature has risen ple may come up different lines, and many pretty from 14.00oC in the 1970s to 14.51oC in good answers are possible. Regression analysis the 2000s. Estimate the mean tempera- identifies the line which minimizes the squares ture for the 2020s. of the errors. This technique is well understood and built into many of computational devices. Solution: The temperature has increased On a TI calculator go to STAT and Edit to by 0.51oC in three decades, which trans- o enter the data into the lists L1 and L2. Then lates into a 0.17 C increase per decade.

108 Two decades down the road we expect a the elevation is raised by 600m, which rise of another 0.34oC, which brings the is equivalent to a loss of 0.25m for ev- total average up to 14.85oC. ery 100m increase in elevation. Chang- The problem is solved, and it shows that ing from 2,400m to 2,800m raises the el- using linear models comes naturally. With- evation by 400m, it thus decreases the out making it explicit, we have used the height of the plant by 1m, and we expect techniques of lines and linear functions: the willow to be 2.8m tall. Given are two temperature data, three Solution Two: A graph says it all. The decades apart. This results in the slope given information can be translated into ∆ temp the points (2400, 3.8) and (3000, 2.3). Con- m = ∆ time nect the points by a line, and identify the 14.51oC − 14.00oC height for 2,800m. = 3 decades = 0.17oC/decade

If we denote temperature x decades past the 2000s by T (x), our points become (−3, 14.00) and (0, 14.51), and since we have determined m already, we can apply the point-slope form of a line.

T (x) − 14.51 = 0.17x Solution Three: We denote the height of the plant by y and the elevation above Therefore sea level by x. Then our data are P (2400, 3.8) T (x) = 0.17x + 14.51 and Q(3000, 2.3). The slope of the line is ∆y 2.3 − 3.8 −1.5 and two decades past the 2000s the tem- m = = = ∆x 3000 − 2400 600 perature will be T (2) = 14.51 + 0.34 = = −0.0025 14.85. The point-slope formula using m and the 2. At an elevation of 2,400m above sea level point P yields a salix drammondiana plant (Drummond’s willow) grew 3.8m tall, while at 3000m y − 3.8 = −0.0025(x − 2, 400) the height of the same species was only 2.3m. Estimate the height at 2,800m above which is equivalent to (no reason to dis- sea level. tribute the −0.0025)

Solution: We oﬀer three solution strate- y = 3.8 − 0.0025 (x − 2, 400) gies, which all lead to the same (correct) answer. When x = 2, 800 we obtain

Solution One: The height of the plant y = 3.8 − 0.0025 (2, 800 − 2, 400) drops by 3.8 − 2.3 = 1.5 meters when = 3.8 − 1 = 2.8

109 3. A population of 3,000 fish increases by (b) 160 fish annually. What is the popula- x y tion three years from now? When will it −1 3.1 reach 4,000 fish? 0 2.8 1 2.5 Solution: We denote the population by 2 2.2 p, and the time measured in years from 3 1.9 by t. The current (t = 0) population is 3,000, which makes this number the p- (c) intercept. The annual increase sets the x y slope at m = 160 fish per year. There- 25 17 fore, 30 21 35 25 p = 160t + 3, 000 40 29 In three years (t = 3) we have 45 33

p = 160 · 3 + 3, 000 = 3, 480 Solution: (a) The data for x increase by one (∆x = and our population will be 3,480 ﬁsh. ∆y 1), and in this case m = 1 = ∆y, To answer the second question we set p = and the y-increments are automati- 4000 and solve for t: cally the slope. Thus we have m = 4. The y intercept is found from 4, 000 = 160t + 3, 000 x = 0, and we see that b = 10. Hence the equation becomes 160t = 1000 1, 000 t = = 6.25 y = 4x + 10 160

and it will take about years and three (b) Again, ∆x = 1, and since the y- month until the population reaches 4,000 values decrease by 0.3 each time, ﬁsh. we have m = ∆y = −0.3. The y- intercept is located at 2.8, which re- 4. Find an equation of the line form the sults in the equation data in the table. y = −0.3x + 2.8 (a) (c) Here ∆x = 5 and ∆y = 4. Thus x y ∆y 4 m = ∆x = 5 = 0.8. −1 6 We could backtrack in the table and 0 10 ﬁnd y-values for x = 20, x = 15 and 1 14 so on until we reach x = 0 and then 2 18 identify the y-intercept that way. As 3 22 an alternative we use the point-slope

110 formula with m = 0.8 and P (30, 21) Solution: It is understood that we use a and obtain linear model in each case. Secondly, the x-increments are constant in all tables, y − 21 = 0.8 (x − 30) and we just need to focus on the numbers on the right. which can be rewritten as (a) Here y increases by 0.6 for the ﬁrst y = 0.8x − 3 values. If we continue this pattern we obtain In any of these exercises we can check the results by comparison of the origi- x y nal data to the formula. For instance, in 0 0.2 part (c) substitute any of the x-values, 2 0.8 say x = 40, into the formula and calcu- 4 1.4 late y. Here we obtain 6 2.0 8 2.6 y = 0.8 · 40 − 3 = 32 − 3 = 29 Follow-up: It can be shown that which agrees with the value in the table. y = 0.3x + 0.2 (b) Here we see that in two steps y de- 5. Complete the table. creases by 64. Thus in each step y should decrease by 32, which is car- (a) ried out below. x y 0 0.2 x y 2 0.8 1980 444 4 1990 412 6 2000 380 8 2010 348 2020 316 (b) x y Follow-up: y = 380 − 3.2(x − 2000), 1980 if you are curious about a formula. 1990 412 (c) Let’s add variety and solve this prob- 2000 lem by a diﬀerent method. We are 2010 348 given the points P (600, 0.86) and 2020 Q(900, 0.74). Therefore the slope is 0.74 − 0.86 −0.12 (c) m = = x y 900 − 300 300 1 500 = − = −0.000, 4 600 0.86 2, 500 700 The point-slope formula yields 800 x − 600 900 0.74 y − 0.86 = − 2, 500

111 and therefore The slope of the approximating line is 6 5 x − 600 m = 7.2 = 6 = 0.83, and we conclude y = 0.86 − 2, 500 that for each one degree increase of the air temperature, the body temperature Now we can plug in values for x and of the lizards will increase by about 0.83oC. determine y. For instance x = 800 Note: The least squares line found by leads to calculator is y = 0.81x + 5.66. 800 − 600 y = 0.86 − 2, 500 7. A study of heavy metal concentrations in 200 ﬁsh had the results listed below (Spokane = 0.86 − River, Langkamp/Hull, some data omit- 2, 500 ted). Find an approximating line. = 0.86 − 0.08 = 0.78 Proceeding similarly for the other Lead (ppm) 0.73 1.14 0.60 1.59 values we obtain Zinc (pp) 45.3 50.3 40.2 64.0 x y 500 0.90 Solution: 600 0.86 700 0.82 800 0.78 900 0.74

6. The body temperatures of Anolis lizards have been recorded in a shady environment at different temperatures with the results given in the table below. Graph Using the points P (0.4, 35) and Q(1.8, 70) the data, determine an approximating line we find that and estimate its slope. What is the bio- ∆y 70 − 35 35 logical significance of the slope? m = = = = 25 ∆x 1.8 − 0.4 1.4 Air Temp 24 26 28 28 29 Thus, y − 35 = 25(x − 0.4), which leads Body Temp 25 27 28 29 29 to Both temperatures are measured in de- y = 25x + 25 gree Celsius. Note: The least squares regression line Solution: (calculator) is y = 22.5x + 27.1. One would expect that, depending on the size of the fish, lead and zinc concentrations would remain proportional. If that were the case, the estimating line would have to pass through the origin, which is not the case in our example. More data should clarify the situation.

112 12.5 Exercises Answers 1. Assume that y changes linearly with x. 1 1. y = − 2 x + 48 Complete the table below and ﬁnd an equation of the line. x y 16 40 x y 20 38 16 40 24 36 20 28 34 24 32 32 28 34 1 32 2. y = 4 − 3 x 2. Find an equation for the line with x-intercept 3. y = −2x + 25 at x = 12 and y-intercept at y = 4. 4. y = 0.3x + 9, b = 9 3. Find an equation of the line containing the points P (4, 17) and Q(10, 5). 5. 4 weeks

4. Write an equation for the line with slope 6. 6 a.m. the next day 0.3 and containing the point P (120, 45). 7. about 2020 What is the y-intercept of this line? 8. y = 3x + 77, 5. A plant is 72 cm tall and it grows at 3 answers may vary cm per week. How long will it take until it is 84 cm tall?

6. At 10:00 a.m. the water level is 2.5 feet below ﬂood stage, and it rises at 1.5 inches per hour. At what time will the water level reach ﬂood stage?

7. The farming of aquatic organisms, to- taled 45.7 million metric tonnes (Mmt) worldwide in 2000. The amount is increasing by 2.7 Mmt each year. Use a linear model to predict when the yield will reach 100 Mmt.

8. Use the straightedge method to approximate the data in the graph by a linear function. x 0 1 3 4 4 y 78 80 86 90 88

113 √ 1 13 The Common Logarithm and therefore, log 10 = 2 , or in decimals we have log 3.162 = 0.5. Logarithms play a very important role in math- The challenge is working with numbers that ematics, mainly in the context of solving expo- are not easily identified as powers of 10. For nential equations. We already came across log- instance, if x = 75, it takes some work to come arithmic scales in the context of graphing, and up with the result26 in the application section we will look at the pH value from chemistry and at earth quake 75 = 101.875 scales. We will focus on the common loga- and therefore log 75 = 1.875. rithm, which is the logarithm with base 10. 13.2 Definition 13.1 Introduction From the introduction it should be clear that Suppose that x > 0 is a positive number. Our logarithms are closely linked to exponents, and goal is to find a number y such that the basic laws for exponents x = 10y an am = an+m (7) This number y is the logarithm of x, and we (an)m = anm (8) will write y = log x. For some numbers x this is a no-brainer: will be frequently used in this chapter. Here is a compact form of the definition of Examples: the logarithm: 1. Let x = 100. Since 100 = 102, we see Definition: Let x > 0, then that y = 2, and we have log 100 = 2. y = log x ⇔ x = 10y (9) 2. Let x = 100, 000, 000, 000. We see that x = 1011, and therefore where ⇔ means ”if and only if”; both state- log 100, 000, 000, 000 = 11. ments are equivalent.

3. Take x = 0.001. Then x = 10−3 and The equivalence in the deﬁnition allows to log 0.001 = −3. transform statements from logarithmic form to exponential form and vice versa. For example, For integer powers of 10, the logarithm counts by how many places we have moved the dec- 12 = 10x imal point starting with 1.0 - positive if we move it to the right, negative otherwise. is a statement in exponential form, its logarithmic equivalent is x 0.001 0.01 0.1 1 10 100 1000 log x −3 −2 −1 0 1 2 3 x = log 12 0 Whenever the number x is a power of 10, If we convert the statement that 10 = 1 integer or not, it is straightforward to deter- to logarithmic form, we ﬁnd that log 1 = 0, mine the logarithm. For instance, take 26Here and in other places we will round to about √ three decimal places, and all numerical statements will x = 10 = 101/2 have small rounding errors.

114 and conversion of 101 = 10 leads to log 10 = 1. In summary, we have

log 1 = 0 log 10 = 1 Figure 26: Logarithmic Scale which are two very important logarithms.

0.3 1 1.3 13.2.1 Estimation of Logarithms 6. 20 = 2·10 ≈ 10 10 = 10 , which implies that log 20 ≈ 1.3 The observation that The logarithm is a build-in function on most 210 = 1, 024 ≈ 1, 000 = 103 calculators, and comparison to calculator results shows that our estimates are fairly good. allows us to estimate log 2 conveniently. We 1 take the power 10 on both sides and use the log 2 = 0.301, 030 ≈ 0.3 second basic law (8): log 4 = 0.602, 060 ≈ 0.6 1/10 1/10 log 5 = 0.698, 970 ≈ 0.7 2 = 210 ≈ 103 = 103/10 log 8 = 0.903, 090 ≈ 0.9 Therefore, converting to logarithmic form, we conclude that A graph of the logarithm function is shown in Figure 27. Notice, that it is deﬁned for pos- 3 log 2 ≈ = 0.3 itive x only, and that logarithms are negative 10 when 0 < x < 1. The estimate that 2 ≈ 100.3 has numerous corollaries: 2 1. 4 = 22 ≈ 100.3 = 100.6 and thus log 4 ≈ 0.6. 3 2. 8 = 23 ≈ 100.3 = 100.9 and therefore log 8 ≈ 0.9.

3. Division yields Figure 27: Graph of y = log x 10 10 5 = ≈ = 100.7 2 100.3 13.3 Properties and we ﬁnd that log 5 ≈ 0.7. In working the examples and exercises you may 1 −1 already have noticed some properties of the 4. = 2−1 ≈ 100.3 = 10−0.3 and 2 logarithm. 1 thus log 2 ≈ −0.3, or in decimals we have Identities. The deﬁnition of the logarithm log 0.5 ≈ −0.3. is an equivalence, and we can take one side and 1 −1 substitute it into the other. By doing so we 5. 0.2 = = 5−1 ≈ 100.7 = 5 obtain 10−0.7 and log 0.2 ≈ −0.7. x = 10y = 10log x

115 as well as Then, by switching over to exponential form, we get the equivalent relationships y = log x = log 10y a = 10x and b = 10y After renaming the variables we obtain the two identities Now we multiply and apply (7) to obtain log a a = 10 (10) ab = 10x · 10y = 10x+y a = log 10a (11) which in logarithmic form becomes The first identity requires that a > 0, the second works for any number a. x + y = log(a · b) Examples: Thus we have shown that 7.5 = 10log 7.5 = 100.875 log(a · b) = x + y = log a + log b −0.6 = log 10−0.6 = log 0.251 which is the first basic law for logarithms (12). Basic Rules. Logarithms, like exponen- In order to confirm (13), we form tial expressions, follow a number of rules. The two rules which parallel the basic rules of ex- an = (10x)n = 10nx ponentials are Then log(a · b) = log a + log b (12) log an = nx = n log a log an = n log a (13) Other Identities. Two more identities are useful with logarithms Examples: 1 log 15 = log(3 · 5) = log 3 + log 5 log = − log a a = 0.477 + 0.699 = 1.176 a log = log a − log b b log 800 = log(8 · 100) = log 8 + log 100 1 −1 ≈ 0.9 + 2 = 2.9 The first rule works because a = a , and (13) implies that log 64 = log 26 = 6 log 2 1 ≈ 6 · 0.3 = 1.8 log = log a−1 = − log a a √ 3 1/3 1 log 4 = log 4 = log 4 For the second rule we note that a = a · 1 , and 3 b b 1 thus ≈ 0.6 = 0.2 3 a 1 log = log a + log = log a − log b Verification of the Basic Rules for Log- b b arithms. For positive numbers a and b we set Example: We can estimate log 0.25 by either x = log a and y = log b rule.

116 The first rule, along with log 4 ≈ 0.6 im- Example: Let x = 0.000, 145. Then plies that log x = −3.839 = −4 + 0.161 1 log 0.25 = log = − log 4 4 Thus, n = −4 and log a = 0.161. We find that ≈ −0.6 a = 100.161 = 1.45, and x = 1.45 · 10−4. while the second rule, along with Here it was necessary to subtract an extra ”1”, because the tail −0.839 does not lie be- log 25 = log 52 = 2 log 5 ≈ 2 · 0.7 = 1.4 tween zero and one. Actually, shows that 10−0.839 = 0.145 25 log 0.25 = log = log 25 − log 100 100 and we have to multiply by 10 in order to ob- ≈ 1.4 − 2 = −0.6 tain the required range for a. Example: If log x = 8.45, we can tell without 13.3.1 Scientific Notation a calculator that x is on the order of 108, that This section is not intended as an efficient method is, x equals a few hundred million. On the to represent a number in scientific notation. It other hand, if log x = −5.45, then x is on the −6 rather shows a connection between the loga- order of 10 , and we are talking about a few rithm and scientific notation. millionth. Recall that in scientific notation we express a number x > 0 in the form 13.3.2 The Function f(x) = log x

x = a · 10n Here we take a brief look at the logarithm from a function perspective, and set27 where 1 ≤ a < 10 is the mantissa, and where the exponent n is an integer. Using our new f(x) = log x rules (12) and (13), we find that For every input x > 0 we find exactly one out- n log x = log a + log 10 = n + log a put y, which can be any real number. The graph of f is shown in Figure 27, and it If n is positive, matters are straightforward. supports that the domain of f are the positive The integer part of the logarithm is the expo- numbers, and the range are all real numbers. nent n, and the decimals can be used to com- When we define pute the mantissa. If n is negative, we have to subtract an extra ”1”. g(x) = 10x Example: Take x = 5, 636. Then 27As we said before, there is not much difference be- log 5, 636 = 3.751 tween y = log x and f(x) = log x. The notation has changed, and the point of view is slightly different. This tells us that n = 3 and log a = 0.751, By the way, a calculator responds with ”log(” when you use the LOG key, supporting the notion of a func- 0.751 which implies that a = 10 = 5.636, and in tion, and when we use log x we are talking about a scientific notation we find that function, whose name comprises three letters, rather than the customary ”f” or ”g”, and it is customary to x = a 10n = 5.636 · 103 omit the parentheses.

117 the identity (10) implies that We apply identity (13) to the ﬁrst expression and then solve for x: g(f(x)) = 10f(x) = 10log x = x x log 6 = log 24 while (11) leads to 0.778x = 1.380 1.380 x x = = 1.774 f(g(x)) = log g(x) = log 10 = x 0.778 1.774 Hence, the logarithm and 10x are inverse func- And indeed, 6 = 24. tions! There is nothing special about 6 and 24. If The basic rules for logarithms become we go through the same process for arbitrary (positive) numbers a and b we ﬁnd f(a · b) = f(a) + f(b) f(an) = n f(a) ax = b log ax = log b in function notation. Moreover, f has the prop- log b erties that x = log a f(1) = 0 and f(10) = 1 Result: log b and that ax = b ⇔ x = (14) 1 log a f = −f(a) a Examples: The equation 5x = 16 has solution and a log 16 f = f(a) − f(b) x = = 1.204 , b log 5 x 13.4 Solving Exponential Equations and 3 = 0.1 has solution log 0.1 −1 The power of logarithms come into play when x = = = −2.096 log 3 0.447 we want to solve equations of the type

ax = b Logarithms for any base. Logarithms can be defined with an arbitrary base b. The These are called exponential equations, as the common logarithm uses base b = 10. The unknown is in the exponent. 6x = 24 is an ex- most convenient logarithm in Calculus is the ample, and the problem is much different from natural logarithm, which uses Euler’s number solving 6x = 24. e = 2.7182 ... as a base. Computer scientists have use for the logarithm with base 2. The Example: We solve 6x = 24. Since the quan- general definition of a logarithm is tities are the same, their respective logarithms must be equal (”take the logarithm on both y y = logb x ⇔ x = b sides”) and we obtain that In this sense log x = log10 x and the natural x log 6 = log 12 logarithm becomes ln x = loge x.

118 The equation x = by can be solved for y by To remedy this problem we graph the log- formula (14), and we ﬁnd that arithms of the data. x y log y log x y = logb x = 0 15 1.176 log b 1 12 1.079 2 20 1.301 If you know one logarithm, you know them all! 3 450 2.653 For example, y = log2 20 is the number 4 3, 200 3.505 which solves 2y = 20. But the solution of this log 20 5 2, 900 3.462 equation, it is y = log 2 = 4.322, and we have just computed that

log2 20 = 4.322

Calculators usually have the LOG key for the common logarithm, and the LN key for the natural logarithm.

13.5 Applications But the average user does not care for loga- 13.5.1 Log Plots rithms. So we display 10y on the y-axis, and the original magnitudes are restored. This is We have already looked at logarithmic scales the basic idea behind graphs on a logarithmic in the graphing chapter. The big advantage of scale. logarithmic scales is that it is possible to show We will see further applications of this con- large and small quantities in the same graph cept in the next chapters. in a meaningful way. A data set with its scatter plot is shown 13.5.2 pH Values below. The term pH stands for power of hydrogen. If the molar concentration of hydrogen of an aqueous solution is denoted by [H+], then the pH is deﬁned as pH = − log[H+] (15)

Example: At 25oC distilled water has molar concentration We see that the details for the small data have 10−7 moles of hydrogen [H+] = = 10−7 M been lost, because the large value of 3,200 dom- Liter inates the graph. The 67% increase from 12 to Therefore the pH of water is 20 becomes insigniﬁcant when other values lie in the thousands. pH = − log 10−7 = −(−7) = 7

119 Water is considered to be pH neutral. Solu- Here m is the moment released by the earth- tions with pH values less than 7 are acidic, quake measured in dyne-cm (erg), and the con- those with higher pH are alkaline (basal). Be stants are chosen in such a way that the values aware that a high pH represents a low hydro- of Mw resemble those of the Richter scale. gen concentration and vice versa. Example: The energy of the Tohaku earth- We observe, that a concentration must al- quake (Fukushima, March 11, 2011) was 29 ways be less than one, and therefore its loga- 3.5 · 10 dyne-cm. Thus, the moment magni- rithm will always be negative (see Figure 27). tude was The purpose of the minus sign in the defini- 2 M = log(3.5 · 1029) − 10.7 tion (15) is to make the pH a positive quan- w 3 tity. Commonly, pH values are given with one 2 = · 29.5 − 10.7 decimal place accuracy, and there is no need 3 for higher precision. = 19.7 − 10.7 = 9.0 If the pH is known, we can use formula (15) The aftershock of the same quake had mo- to calculate the hydrogen concentration as ment magnitude Mw = 7.1, and we can use the [H+] = 10−pH M definition to calculate its energy: 2 7.1 = log m − 10.7 Example: The pH of coffee is 5.5. Therefore 3 2 [H+] = 10−pH M = 10−5.5 M 17.8 = log m −6 3 = 3.2 · 10 M 3 log m = · 17.8 = 26.7 2 More examples can be found worked prob- m = 1026.7 = 5.0 · 1026 lems below. We see that the moment of the aftershock was 5.0 · 1026 dyne-cm, which is significantly less 13.5.3 Moment Magnitude Scale than the energy of the original quake (by a The strength of earthquakes are commonly given factor of about 700). by values on the Richter scale. It is named in honor of C. F Richter, a physicist at CalTech Sometimes the energy is given in seismic in Pasadena, CA, in the 1930ies. The measure- moments M, which is the equivalent of Nm ment is based on the amplitude of the waves (Newton-meter) or Joule (J). In this case the shown by a specific seismograph located 100 formula needs to be adjusted. Since 7 km from the epicenter of the quake. 1 J = 10 dyne-cm, it follows that The deflection of a needle in a scientific in- m = M · 107 strument is not a meaningful physical quantity. The energy (moment) released by an earth- and therefore quake is much more useful. In 1979, Hanks log m = log(M · 107) = log M + 7 and Kanamori also of CalTech, developed the moment magnitude scale Mw; it is defined as Substitution yields 2 2 M = log m − 10.7 M = log m − 10.7 w 3 w 3

120 2 2/3 = (log M + 7) − 10.7 (f) x = 10 (= 4.642) 3 2 14 3. Estimate log 3. = log M + − 10.7 3 3 2 Solution: Here we need a convenient ap- = log M − 6.03 3 proximation of 3n, similar to 210 ≈ 1, 000.

4 13.6 Worked Problems Solution One: We use 3 = 81 ≈ 80 along with the already established result 1. Without using a calculator determine that 8 ≈ 100.9. Then

(a) log 10, 000, 000 34 = 81 ≈ 80 = 10 · 8 ≈ 10 · 100.9 (b) log 0.1 = 101.9 1 (c) log 100 Therefore

Solutions: 1/4 1/4 3 = 34 ≈ 101.9 = 100.475 (a) 10, 000, 000 = 107, therefore log 10, 000, 000 = 7 and thus log 3 ≈ 0.475. Comparison (b) 0.1 = 10−1, therefore log 0.1 = −1 to the calculator values shows that our result is too low, with a relative error28 1 −2 1 (c) 100 = 10 , therefore log 100 = −2 of 0.4%.

2. Convert from exponential to logarithmic Solution Two. We start from scratch with- form, or vice versa. out previous estimates. Patient trial and (a) 10x = 1, 000 error experiments lead to x (b) 10 = 25 321 = 1.046 · 1010 ≈ 1010 (c) x = 103.7 Therefore, (d) log x = −2 1/21 (e) log x = 3.7 3 ≈ 1010 = 1010/21 2 (f) log x = 3 10 and log 3 ≈ 21 = 0.4762, with relative Solutions: The question just asks for con- error 0.2%. version, and giving the values for x is an added bonus. 4. Solve for x x (a) x = log 1, 000 (= 3) (a) 100 = 1, 000 x (b) x = log 25 (= 1.398) (b) 2 = 0.125 (c) 12x = 1 (c) log x = 3.7 3 x+1 x (d) x = 10−2 (= 0.01) (d) 2 = 3 3.7 28 estimate - exact = −0.0021 = −0.0044 (e) x = 10 (= 5, 011.9) exact log 3

121 Solutions: Graphical solutions are always an option, either to ﬁnd solutions or to conﬁrm known (a) Solution One: The equation has the results. Plots for all four parts are given form ax = b with a = 100 and b = in Figure 28. 1, 000, and formula (14) implies that log 1000 3 x = = = 1.5 log 100 2

Solution Two: This problem can be done without logarithms. 1, 000 = 103, 100 = 102 and 100x = (102)x = 102x, and the problem becomes 102x = 103 Comparing the powers yields 3 2x = 3 and therefore x = 2 = 1.5. (b) With formula (14) we ﬁnd that log 0.125 x = = −3 Figure 28: Graphs for Problem 4 log 2 This result is not surprising, because 5. What is the pH of blood, given that the 1 1 0.125 = = = 2−3 hydrogen concentration is 4 · 10−8 M? 8 23 (c) Logarithms are required here, and Solution: Apply the deﬁnition (15) to ob- formula (14) implies that tain

log 1 − log 3 −8 x = 3 = pH = − log(4·10 ) = −(−7.4) = 7.4 log 12 log 12 −0.477 = = −0.442 6. For a vinegar we ﬁnd that 1.079 [H+] = 7.5 · 10−3. What is the pH? (d) Take the logarithm on both sides and solve for x: Solution: 2x+1 = 3x pH = − log(7.5 · 10−3) = 2.1 log 2x+1 = log 3x (x + 1) log 2 = x log 3 7. What is the hydrogen concentration of milk, given that its pH is 6.4? 0.301(x + 1) = 0.477x 0.301x + 0.301 = 0.477x Solution: Solve for [H+] in the deﬁnition. 0.301 = 0.176x 0.301 [H+] = 10−pH M = 10−6.4 M x = = 1.710 0.176 = 4.0 · 10−7 M

122 8. The pH of a beer is 4.0. Is its hydrogen Now we apply the definition concentration higher or lower than that 2 M = log 1027 − 10.7 of water? w 3 2 Solution: A smaller pH means a higher = · 27 − 10.7 3 hydrogen concentration. Since the pH = 18 − 10.7 = 7.3 values differ by 3, their hydrogen concentrations differ 1, 000 = 103. 11. On Feb. 17, 2015, an earthquake with If you don’t trust this line of reasoning, magnitude Mw = 2.4 was reported near you can calculate hydrogen concentration Bluefield, VW. How much energy was re- directly: leased?

[H+] = 10−4 M Solution: We set Mw = 2.4 and solve the equation for m. 10−4 is 1,000 times larger than 10−7. 2 2.4 = log m − 10.7 3 9. You come across 40 mL of an unknown 2 mixture and you detect that it contains 13.1 = log m −5 3 3 · 10 moles of hydrogen. What is the 3 log m = · 13.1 = 19.65 pH? Is is alkaline or acidic? 2 19.65 19 Solution: As a first step we have to find m = 10 = 4.5 · 10 the molar concentration: The moment was 4.5 · 1019 dyne-cm. 3 · 10−5 moles 25 12. We are comparing two earthquakes. One × 40 mL 25 releases ten times as much energy as the 7.5 · 10−4 moles other. How do the moment magnitudes = 1, 000 mL differ? 7.5 · 10−4 moles = Solution: Let us say that the smaller quake Liter has moment m and the larger has mo- = 7.5 · 10−4 M ment m0 = 10m. Then the moment mag- Now we calculate the pH nitude of the bigger earthquake is 0 2 0 −4 M = log m − 10.7 pH = − log(7.5 · 10 ) = 3.1 w 3 2 = log 10m − 10.7 and we see that this mixture is acidic. 3 2 10. What is the moment magnitude of a earth- = (1 + log m) − 10.7 20 3 quake of strength 10 J? 2 = Mw + = Mw + 0.666, 666,... Solution: First we convert to dyne-cm: 3 that is, the moment magnitude of the 1020 J = 1020 · 107 dyne-cm stronger earthquake is about 0.7 units = 1027 dyne-cm bigger.

123 13.7 Exercises 8. You discover 3L of an unknown solution and you measure a total of 8×10−7 moles 1. Without using a calculator determine of H+ in this mixture. Calculate the pH. (a) log 10, 000 9. You change a mixture so that its hydro- (b) log 0.000, 001 gen concentration is doubled. How will 1 the pH change? (c) log 1,000 2. Convert from exponential to logarithmic 10. The 1906 San Francisco earthquake had form, or vice versa. moment magnitude 7.9. How much energy was released by the quake? (a) 10x = 1.6 11. What is the moment magnitude Mw of (b) log x = 1.6 an earthquake when the moment is 3.9 × 27 3. Estimate log 125 and log 1.25 10 dyne-cm?

4. Solve for x: Answers (a) 10x = 10, 000 1. (a) 4 (b) 10x = 0.000, 1 (b) −6 (c) 10x = 5 (c) −3 (d) 10x = 0.2 2. (a) x = log 1.6 (e) log x = 3 (b) x = 101.6 (f) log x = −2 (g) log x = 0.4 3. 2.1 and 0.1, respectively. (h) log x = 1.4 4. (a) 4 (i) 5x = 20 (b) −4 (j) 1.08x = 2 (c) 0.6990 (k) 2x+2 = 4x (d) −0.6990 (l) log x + log 4 = log 40 (e) 1, 000 (m) log(x − 2) + log(x + 2) = log 32 (f) 0.01

5. The pH of Sprite is 3.3. What is its hy- (g) 2.5119 drogen concentration? (h) 25.119 (i) 1.8614 6. The hydrogen concentration of blood is [H+] = 4.0 × 10−8M. What is its pH? (j) 9.006 (k) 2 7. The pH of milk is 6.4. Is the hydrogen concentration of milk higher or lower (l) 10 than that of water? And by what factor? (m) 6

124 5. 5 × 10−4 M 14 Exponential Functions

6. 7.4 In everyday language we often attach the term exponential when things change rapidly. In 7. 4 × 10−7; the H+ concentration of milk an inﬂation prices increase exponentially, in an is four times higher. epidemic a disease grows exponentially, kudzu 8. 6.574 ≈ 6.6 is spreading exponentially, and so on. In mathematics, this term is reserved for a very special 9. The pH will decrease by 0.3, the mixture setting, which is the topic of this section. is more acidic. Population models are arguably the most prominent biological application of exponen- 10. 7.943 × 1027 dyne-cm tial growth; this applies to easily observed and 11. 7.7 counted animals, as well as bacteria and other cell growth. Radioactive decay is the prime example of exponential decay.

14.1 Introduction Here we introduce the basic concepts of growth rates and multipliers in a hypothetical example. Suppose that the population of monkeys on an island increases by 6% annually. Beginning with 455 animals in 2015, predict the population for future years, and in particular, estimate the population in 2050. We begin by calculating the population in 2016. 6% of 455 equals 27.3, and we expect 482.3 monkeys on the island in the following year. In terms of mathematics, the calculation looks like

482.3 = 455 + 0.06 · 455 = (1 + 0.06) · 455 = 1.06 · 455

In order to derive a pattern it helps tremen- dously to factor the term 1.06. When we look at the population for 2017, we have to repeat the calculation with 455 being replaced by 482.3.

482.3 + 0.06 · 482.3 = (1 + 0.06) · 482.3

125 = 1.06 · 482.3 = 1.06 · (1.06 · 455) Here the variables are 2 = 1.06 · 455 = 511.2 x independent variable, usually time and the population in 2017 should be around y dependent variable, often population 511 animals. y0 initial value We notice that by stepping from one year to r growth or decay rate, r > −1 the next, the population is multiplied by 1.06 M multiplier M = 1 + r , M > 0 each time. One year after 2016, the population In the introductory example we had y0 = will be 1.06 · 455, two years after 2015 it will 2 455, r = 0.06 and M = 1.06. be 1.06 · 455, and in 2050, which is 35 years Our model covers exponential decay as well, after 2015, we expect the population to be at 35 in which case r is negative. But a decline can- 1.06 · 455 = 3, 497 monkeys. not be faster than 100%, and if the decline The formula of the population x years after equals 100%, everything vanishes immediately, 2015 becomes and a mathematical model is no longer needed. y = 455 · 1.06x For this reason we require that r > −1. The multiplier and the rate are related by There are three important quantities involved. First, we need to know the initial pop- M = 1 + r ⇔ r = M − 1 ulation y0, which was 455 in the example. Sec- ondly, we need to know the growth rate r, and r > −1 implies that M > 0. Multipliers which was 0.06=6%. And finally, there is a must be positive! r = 0 is equivalent to M = multiplier M, which in our case was M = 1.06. 1, and in this case the quantity of interest y The multiplier and the growth rate are related remains constant. by M = 1 + r. For x = 0 the formula (16) reduces to In our computations we allow decimal val- y = y M 0 = y ues in population numbers. After all, this is 0 0 just a model. Round to the nearest integer if This is why y0 is called the initial condition. It 482.3 monkeys bothers you. An EXCEL sim- is the value of y when x = 0. ulation for the first 15 years is shown below Example: A fish population increases from 357 to 411 fish within one year. We assume that the population continues to grow the same rate, and we want to know the population five years later. The information can be summarized as x y 0 357 1 411 and since 357 is associated with x = 0 it follows 14.2 Exponential Growth and Decay that y0 = 357. If we substitute the data for An exponential model has the form x = 1 into the definition (16), we find that x x 1 y = y0M = y0(1 + r) (16) 411 = 357 · M = 357M

126 and thus by 54 fish, and that each year we would have 411 additional 54 fish, which results in the model M = = 1.151 357 y = 357 + 54x This means that in order to get from one year to the next, we have to multiply the population The two models agree for x = 0 and for x = 1, by 1.151. If we want to look five years ahead, other than that they are much different, espe- we have to multiply by M five times, and our cially when x becomes large. prediction is A side by side comparison of the two models is given in Figure 29. 357 · 1.1515 = 722.0 fish

The general model is

y = 357 · 1.151x

Knowing M means that we also know the growth rate:

r = M − 1 = 0.151 = 15.1%

In this example r = 15.1%, and we observe that our population about doubled in size during this ﬁve year period.

Linear versus Exponential Growth. There is a huge difference between linear and exponential models. In linear models the quantity of interest y changes in fixed increments, and this change is responsible for the slope. In exponential models the quantity y changes Figure 29: Linear vs. Exponential at a fixed rate, which leads to the multiplier M = 1 + r. Graphs of Exponential Functions have Example: We return to the fish population, a very characteristic shape. The plots for y = which increased from 357 to 411 fish from one 48 · 1.2x and y = 75 · 0.8x are shown in Figure year to the next. 30 below. The growth curve increases toward In the exponential model we argued that infinity, while the decay curve approaches the the change represented a 15.1% increase, that x-axis as x increases. Notice that the curves the population will continue to grow at this look somewhat alike, with the x-directions re- rate, which in turn resulted in the equation versed. When exponential functions are graphed on x y = 357 · 1.151 a logarithmic scale they become linear! Re- call29 that graphs on a logarithmic scale are For a linear model on the other hand, we would argue that the number of fish increased 29This process was outlined in Section 13.5.1.

127 and we see that stepping forward by one time unit is equivalent to multiplication by M. Step- ping forward by T time units is equivalent to multiplication by M T because x+T x T f(x + T ) = y0M = y0M M = M T f(x) x x Figure 30: y = 48 · 1.2 and y = 75 · 0.8 We will use this principle when we determine doubling time and half-life (M T = 2 and M T = just a plot the logarithms of the expressions, 1 2 , respectively). along with a manipulation of the vertical axis. In our case we ﬁnd that 14.3 Doubling Time x log y = log(y0 · M ) We begin with an example and look at the ex- x = log y0 + log M ponential function = x log M + log y 0 y = 48 · 1.2x But this is an equation of a line with slope m = log M and y-intercept b = log y0. The graphs of y = 48 · 1.2x and y = 75 · 0.8x on a logarithmic scale are shown below.

The notion of converting exponential curves The graph shows the x values for y = 12.5, into lines via logarithms is exploited in the ap- y = 25, y = 50, y = 100 and y = 200, and proximation of almost exponential data. we see these points are about equally spaced Exponentials in Function Notation. We along the x-axis. This tells us that doubling can also write the y-value always takes the same amount of time. f(x) = y M x 0 We can also observe this fact numerically. and express the exponential growth model as x 48 · 10x a function. It satisﬁes 0 48 f(0) = y0 1 57.3 2 69.1 which conﬁrms that y0 is the initial value (associated with x = 0). Moreover 3 82.9 4 99.5 x+1 x f(x + 1) = y0M = y0M M 5 119 = Mf(x) 6 143

128 When x = 0 we have y = 48, and y reaches Examples: The doubling time for the growth 96 = 2·48 a little bit before x = 4. But we can model y = 25 · 1.05x is also start with x = 2, where y ≈ 70 and when log 2 x = 6 we have already passed the y = 140 T = = 14.2 mark. Again, it took a little less than four log 1.05 time units to double the y-value. and that for y = 5x is We can determine the doubling time ana- log 2 lytically. At x = 0 we have y = 48 and we T = = 0.431 need to find x such that log 5 96 = 48 · 1.2x Example: A bacteria culture grows exponen- Division by 48 results in tially and it increased from 20 bacteria initially 1.2x = 2 to 1,000 bacteria within two hours. Let us first estimate the doubling time, and which is an exponential equation. Fortunately, then compute it analytically. we have already discussed equations of this Doubling once results in 40 bacteria, dou- form in Section 13.4, and formula (14) shows bling again leads to 80 bacteria and so on. that Doubling five times results in log 2 x = = 3.802 log 1.2 20 · 2 · 2 · 2 · 2 · 2 = 20 · 25 = 640 which confirms our observation of a doubling time a little less than four. bacteria, and doubling one more time yields We move on and we seek to find the dou- 1,280 bacteria. Thus, our bacteria population bling time T for the general exponential model doubles between 5 and 6 times within two hours, and cell divisions should occur about every 20 x 30 y = y0 · M to 24 minutes . The string of inequalities We pick any x, then at time x + T the y-value should have doubled. This translates into 1, 000 25 = 32 ≤ = 50 ≤ 64 = 26 x+T x 20 y0 M = 2y0 M is another way to approach this problem. The We divide by y0, which shows that the start- population increased by a factor of 50, which ing values are immaterial in this context, and requires between 5 and 6 duplications. obtain For the analytic solution we measure time M x+T = M x M T = 2 M x x in hours. Then the model becomes Further division by M x results in y = 20 · M x

M T = 2 (17) and substitution of x = 2 and y = 1, 000 results in which relates the doubling time T and the mul- 1, 000 = 20 · M 2 tiplier M, and we ﬁnd that 30 2 hours 1 hour log 2 We have 6 = 3 = 20 min and T = (18) 2 hours = 0.4 hours = 24 min. log M 5

129 and we deduce that In the example with y = 48 · 1.2x, we have M = 1.2 and hence r = M − 1 = 0.2 = 20%. r1000 √ M = = 50 The Rule of 72 estimates a doubling time of 20 72/20 = 3.6, which is close to the exact value The formula for the doubling time (18) then T = 3.802 with an error margin of 6%. implies that How does this rule work? With the tools of Calculus it can be shown log 2 that T = √ = 0.352 log 50 log(1 + r) ≈ 0.434r

Conversion to minutes shows a doubling time is a very good approximation for small num- of 21.3 minutes. bers r. But 1 + r = M, and the formula (18) for T implies that Example (Rule of 72): This rule is used in log 2 = T · log M business to estimate the time it takes to double an investment at a given interest rate. It is a = T · log(1 + r) ≈ 0.434rT rule of thumb: Therefore, Interest Rate × Doubling Time = 72 log 2 r · T ≈ = 0.693 0.434 At 3% interest it will take an investment 24 years to double in value, because 3 × 24 = 72. In the Rule of 72 we entered the percentages If an investment was doubled in 8 years, the directly (8% as 8 and not as 0.08 as we nor- interest rate was 9% (9 × 8 = 72), and so on. mally do), and thus This is an approximate rule, because if an investment doubles just in one year, the interest 100 r · T ≈ 69.3 rate is 100% and not 72%. This doesn’t make for a good rule. But change Compound interest follows the exponential 69.3 to 72, which is divisible by 2, 3, 4, 6, 8, 9, model (16) as well. $455 invested at 6% inter- 12, 18, 24 and 36, and you have a convenient est uses the same mathematics as monkey pop- estimation tool! ulation of 455 animals growing at a 6% annual rate, and we can use the Rule of 72 to estimate There is an alternate, equivalent way to ex- doubling times in population examples as well. press exponential growth using the doubling In the opening example we started with 455 time. In (17) we established that M T = 2. monkeys in 2007, and in the simulation it took Thus, until 2027 for the population to reach 915.5. Doubling took about 12 years at 6% growth M = 21/T rate, as predicted by the Rule of 72. M x = 2x/T In the ﬁsh population example we had a x x/T y = y0M = y0 2 growth rate of 15.1%, and the Rule of 72 estimates a doubling time of a little under ﬁve and we have years (T ≈ 4.8), which is consistent with the x/T data in Figure 29. y = y0 2 (19)

130 The multiplier no longer appears explicitly, but With this terminology we have x = nT , it can be recovered from M = 21/T . and thus x/T = n and the exponential model (19) simplifies to Example: A lung cancer cell doubles by mi- n tosis about every three months. Single cancer y = y0 · 2 cells cannot be detected by X-rays and it takes about one billion cells for a tumor to become Example: We review the bacterial growth from visible. Beginning with a single cell we want to above: 20 bacteria grew to 1,000 bacteria in determine how long it will take until the can- two hours. Since initially y0 = 20 bacteria are cer can be detected by X-rays. This example present, the model becomes is adapted from Langkamp/Hull. y = 20 · 2n We measure time in years. Then the doubling time is T = 0.25 = 1/4, and using for- Setting y = 1, 000, we find that mula (19) our model for the number of tumor 1, 000 = 20 · 2n cells becomes 1, 000 2n = = 50 x 20 y = 2x/0.25 = 24x = 24 = 16x log 50 n = = 5.644 log 2 9 since y0 = 1. We now have to set y = 10 and and we see that we went through 5.644 gener- solve the resulting exponential equation. ations in 2 hours, which accounts for a generation time of x 9 16 = 10 120 minutes log 109 9 T = = 21.262 min x = = ≈ 7.5 5.644 log 16 1.204 which confirms the result we found before. and we see that it takes about seven and a half years until the tumor is visible. 14.4 Half-Life The notion of doubling time is linked to expo- Generation Time. For cell (bacterial) nential growth. Its counterpart for exponential growth the time between cell divisions is called decay is half-life. It is the time it takes for a generation time (time per generation), and it is substance to reduce to one half of its original the equivalent of the doubling time. If we de- size. note the number of generations by n, the total elapsed time becomes x = nT . Example: If the generation time is T = 24 minutes, and we go though n = 8 generations, then the total time is x = 24·8 = 192 minutes, or 3 hours and 12 minutes. If, on the other hand, we observe cell growth for two hours (120 minutes), and we know that the time per generation is 32 minutes, then we x 120 have gone through n = T = 32 = 3.75 generations.

131 The graph shows the curve y = 75 · 0.8x, and The reader is encouraged to plot y = 75 · 0.8x we marked off the points corresponding to y = and y = 75 · 2−0.322x in the common graph 400, y = 200, . . . y = 12.5. The spacing be- to confirm that they completely overlap. The tween the points is uniform along the x-axis, connection between the curves results from about 3.1 units apart, and we call this the half- 2−0.322 = 0.8. life. We still work with the exponential function Example (Radio Carbon Dating): The x 14 y = y0M , where M < 1, resulting in the abundance of the radioactive isotope C is decay. We denote the half-life by H; then value about one part per one trillion 12C isotopes. of y at time x+H must be one half of the value All living organisms exhibit this ratio. But at time x. We translate this statement into an when an organism dies, the radioactive isotope equation, which we will then solve for H: decays with a half-life of 5,730 years, and the ratio becomes smaller. 1 y M x+H = y M x 0 2 0 1 M x M H = M x 2 1 M H = 2 log 1/2 log 2 H = = − log M log M and we have a formula for H. M < 1, and therefore its logarithm is negative, and the mi- Figure 31: Decay of 14C nus in the formula makes H a positive quantity. As we did for the doubling time, we can After 5,730 years only one half of the nat- 14 12 use the half-life in an alternative description ural C : C ratio is present, after another for exponential decay: 5,730 years (11,460 years from the beginning) the ratio is down to one quarter of the natural −x/H y = y0 · 2 (20) ratio, and so on, and the ratio after x years can be described by Example: Let y = 75 · 0.8x. Here M = 0.8 y = 2−x/5,730 and the half-life becomes Scientists can measure the 14C :12 C ratio log 2 0.301 of artifacts, and they use it for age determina- H = − = − = 3.106 log 0.8 −0.969 tion. Suppose that, for instance, a piece of wood If we want to express y in the alternate form contains only 20% of the natural 14C :12 C (20), it is easiest to look at the exponent first: ratio. Then substitution results in x x 0.2 = 2−x/5,730 = = 0.322x H 3.106 and we find that (exponential equation method) and thus x log 0.2 −0.322x − = = −2.322 y = 75 · 2 5, 730 log 2

132 x = (−2.322) × (−5, 730) We continue with the example and take log- = 13, 304 arithms. x y log y 0 4 0.602 and we conclude that the piece of wood is about 1 15 1.176 13,300 years old. This result agrees well with 2 30 1.477 the graph in Figure 31. y = 0.2 occurs a lit- 3 90 1.954 tle past the half way mark between 10,000 and 4 210 2.322 15,000 years. 5 450 2.653 A graph of the logarithmic data with an 14.5 Almost Exponential Data approximating line are shown below. In biological observations, the collected data may indicate an exponential trend, but the ﬁt is not perfect. In this case we are looking for an approximating exponential curve. We explain the method by example.

We select the points (0, 0.7) and (5, 2.7) for the approximating line. Then 2.7 − 0.7 2 m = = = 0.4 5 − 0 5 But we are modeling the logarithm of the data, and the ”line” becomes

The data and their graph indicate an exponen- log y = 0.4x + 0.7 tial trend. Finally, we convert back to exponential form: In order to ﬁnd an approximating curve we take advantage of the fact that exponen- y = 10log y = 100.7+0.4x tial functions become linear when we take log- x = 100.7 · 100.4 = 5.012 · 2.512x arithms. This leads to the following procedure: The graph of the data with the exponential 1. Take the logarithms of the y-values. approximation are shown below.

2. Carefully graph x against the logarithms.

3. Find an approximating line be the straight- edge method (see Section 12.3).

4. Convert the approximating line to exponential form.

133 Exponential regression is important and most x y graphing calculators or software have built-in 0 10 functions for this task. 1 (b) In EXCEL plot the data and ask for an for 2 an ”Exponential” trendline. There is an op- 3 tion to display the equation. In our example 4 50 EXCEL would give the exponential approxi- x y mation as 0 500 1 (c) y = 4.8446 e0.9323x 2 400 3 The number e is Euler’s constant, and e0.9323 = 4 2.540, and the EXCEL approximation is equivalent to Solutions: y = 4.845 · 2.540x (a) When x = 0 we have y = 20. There- fore y0 = 20. We also see that Our eye-balled estimation y = 5.012 · 2.512x 25 = 20M is not far from the computer result! 25 ⇒ M = = 1.25 On a graphing calculator the data have to 20 be written to lists using the STAT key, fol- and we have found the multiplier. lowed by EDIT. Once they are entered, use For r we obtain STAT followed by CALC and ExpReg. The calculator should respond with y = a ∗ b ∧ x, r = M − 1 = 0.25 = 25% where a = 4.8446 and b = 2.540. The details and we use may vary with the brand and model of the cal- x culator, but the resulting approximation will y = 20 · 1.25 be identical to the one found in EXCEL, as all to complete the table. use the least squares regression technique. x y 0 20 14.6 Worked Problems 1 25 2 31.25 1. Find an exponential model of the form 3 39.06 x y = yoM for the data below, and com- 4 48.83 plete the tables. What are M and r in (b) Again, we have the initial informa- each case? tion that y0 = 10 when x = 0. We also have a value when x = 4, and x y substitution into (16) yields 0 20 1 25 50 = 10 · M 4 (a) 50 2 M 4 = = 5 3 10 √4 4 M = 5 = 51/4 = 1.495

134 It follows that starling population in North America was estimated at 200 million (stanford.edu). r = M − 1 = 0.495 = 49.5% Assuming the accuracy of the data and an exponential model, estimate year when The model becomes the population reaches ten billion. y = 10 · 1.495x Solution: We use x to count the years and the completed table takes the after 1890. Then y0 = 60 and for x = 100 form we have x y 200, 000, 000 = 60 · M 100 0 10 1 14.95 Now we solve this equation for M 2 22.36 200, 000, 000 3 33.44 M 100 = 4 50 60 200, 000, 0000.01 M = (c) This is very similar to the previous 60 part. y0 = 500 and = 1.162 2 400 = 500 · M and the exponential model becomes 2 400 M = = 0.8 x √500 y = 60 · 1.162 M = 0.8 = 0.894 Finally, we set y = 10 billion and solve The values are decreasing and the the resulting exponential equation for x. rate is negative. 1010 = 60 · 1.162x 1010 r = M − 1 = −0.106 = −10.6% 1.162x = = 1.67 · 108 60 The formula for y is log(1.67 · 108) x = log 1.162 x y = 500 · 0.894 8.222 = = 126 and the completed table follows. 0.0652 Therefore, the population will reach ten x y billion 126 years after 1890, which makes 0 500 it the year 2016. 1 447 2 400 3. Elephantdatabase.org reports that the ele- 3 358 phant population in Africa declined from 4 320 474,134 animals in 2007 to 421,955 ele- phants in 2012. Assuming an exponen- 2. In the year 1890 Shakespeare enthusiasts tial decay model, predict when the ele- released 60 European starlings in New phant population will drop below 350,000 York’s Central Park. A century later the animals.

135 Solution: The data are five years apart, (b) We know that T = 4, and we have therefore the multiplier satisfies to find M first, and then determine r. It follows from (17) that 421, 955 = 472, 134M 5 421, 955 M = 21/T = 21/4 = 1.189 M 5 = = 0.8937 472, 134 and thus r = M − 1 = 0.189 = M = 0.89371/5 = 0.9778 18.9%. This is a daily growth rate. 72 r = M − 1 = −0.0222. This tells us that The Rule of 72 predicts r = 4 = the population declines at an annual rate 18%. of 2.2%. The population model becomes 5. A culture of bacteria grows from 200 to P = 472, 134 · 0.9778x 15,000 bacteria in three hours. How many generations have evolved and what is the where x counts the years after 2007. Set- generation time? ting P = 350, 000 yields Solution: We use the model 350, 000 = 472, 134 · 0.9778x 350, 000 y = 200 · 2n 0.9778x = = 0.7413 472, 134 Setting y = 15, 000 we find that log 0.7413 x = log 0.9778 15, 000 = 200 · 2n −0.1300 15, 000 = = 13.32 2n = = 75 −0.009760 200 log 75 The model predicts that the population n = = 6.229 drops below 350,000 in the year 2020 (13 log 2 years after 2007). Thus, we have gone through 6.229 gener- 4. (a) A population grows according to the ations in 3 hours (180 minutes), and the law y = 450 · 1.1x. What is the time per generation is doubling time? 180min T = = 28.9 min (b) What is the growth rate for a pop- 6.229 ulation which doubles in four days? 6. Iodine-123 is a radioactive isotope which Solutions: is used in thyroid therapy. It has a half- life of 13.22 hours. If 2 mg are being (a) Here M = 1.1, and formula (18) used, then how log will it take until only yields 0.1 mg is present? log 2 T = = 7.27 Solution: We use the alternate form (20) log 1.1 of the exponential model, and the amount Note that r = 0.1 = 10%, and that y of iodine-123 in the body after x hours the Rule of 72 estimates a doubling is time of 7.2. y = 2 · 2−x/13.22

136 The question asks for the time x such (b) We begin with an estimate: Origi- that y = 0.1. Substitution results in nally we have 1,000g. After 40 days there are 500g in the soil, after 80 0.1 = 2 · 2−x/13.22 days the amount will be reduced to −x/13.22 2 = 0.05 250g, after 120 days we have 125g x log 0.05 − = = −4.322 left, but after 160 days we are down 13.22 log 2 to 62.5g. The result should be near x = (−13.22) × (−4.322) 140 days. = 57.1 And now for the analytical solution: Thus, it takes 57.1 hours (two days and We set y = 100 and solve for x: about nine hours) until the amount of −x/40 radioactive iodine is below one tenth of 100 = 1, 000 · 2 100 1 a milligram. 2−x/40 = = 1, 000 10 7. Glyphosate is an ingredient in many her- 2x/40 = 10 bicides. When it lands on a plant it is x 40 quickly absorbed and broken down, but 2 = 10 log(1040) 40 when it lands on soil it breaks down slowly x = = and kills beneficial microbes in the pro- log 2 0.301 cess. The average half-life in soil is 40 = 132.9 days (adapted from Landkamp/Hull). In the third step we took recipro- (a) Determine the daily rate of decay. cals on both sides (get rid of minus (b) If 4 kg are used on one hectare, with signs), and in the following step we 1 kg landing on soil, then how long raised both sides to the 40th power will it take until only 100 g are left? (get rid of the fraction). As a result we see that it takes about 133 Solution: We measure the glyphosate con- days (4.4 months) until only 100g tent in gram and time in days. Then the are left. amount in the soil after x days is (1kg = Check the result: When we substi- 1,000g) tute x = 132.9 into the model, we −132.9/40 y = 1, 000 · 2−x/40 find that 1, 000 · 2 = 100. A graph of y = 1, 000·2−x/40 reveals (a) The multiplier is M = 2−1/40, be- that y = 100 for x ≈ 133, which cause x confirms our result as well. −1/40 − 1 x −x/40 2 = 2 40 = 2 and therefore the rate is r = M − 1 = 2−1/40 − 1 = 2−0.025 − 1 = −0.0172 and we conclude that the substance decays at a rate of 1.7% per day.

137 8. The population of the United States since We do not need the full equation of the 1950 is given in the table below line. Recall that x Year Population y = y0 M 1950 151, 325, 798 ⇔ log y = x log M + log y0 1960 179, 323, 175 and therefore 1970 203, 211, 926 1980 226, 545, 805 log M = m = 0.00517 1990 248, 709, 873 This implies that 2000 281, 421, 906 M = 100.00517 = 1.01197 2010 308, 745, 538 and we see that the annual growth rate Use an exponential model and the straight- of the U.S. for that period was edge method to approximate the annual growth rate. r = 0.01197 = 1.2% Least squares regression by computer yields Solution: We use the variable x to count a growth rate of 1.167%. the years since 1950 and y for the population data. Taking the logarithms results 9. Construct the inverse function of in the table below. x f(x) = y0 M x log y 0 8.180 Solution: Recall that in order to ﬁnd the 10 8.254 inverse function, we have to solve the 20 8.308 equation y = f(x) for x. In our case 30 8.355 we obtain 40 8.396 x y = f(x) = y0 M 50 8.449 y M x = 60 8.490 y0 log y In the next step we graph this informa- x = y0 log M tion and ﬁnd an approximating line. where in the last step we used the formula for exponential equations (14). Af- ter renaming the variables we obtain log x g(x) = y0 log M as the inverse function of f.

In the very special case where y0 = 1 and M = 10, we see that Using the points (0, 8.19) and (60, 8.5), log x we see that the slope is g(x) = = log x log 10 8.5 − 8.19 0.31 m = = = 0.00517 is the inverse function of f(x) = 10x. 60 − 0 60

138 14.7 Exercises 6. A substance grows according to the law y = 15 · 1.05x. What is the doubling 1. Assume that y changes exponentially with time? x. Complete the tables below and ﬁnd an x equations of the form y = y0M . 7. A bacteria culture doubles in size every 45 minutes. x y 0 6 (a) What is the hourly growth rate? 1 (a) (b) Initially 250 bacteria are present. 2 How long will it take until the colony 3 750 has one million bacteria? 4 8. A bacteria culture grows from 600 bacte- x y ria to 13,000 bacteria in two hours. What 0 500 is the generation time (the time it takes 1 (b) to double in size)? 2 125 3 9. Scientists discovered a new radioactive 4 substance which decayed by 20% within one hour. What is its half-life? x y 0 10. A pollutant has a half-life of 4 months. 1 4 (c) 2 (a) What is the monthly rate of decay? 3 5 (b) What is the annual decay rate. 4 (c) If initially 600 g of the substance are 2. Sketch the graph of the functions present, how long will it take until y = 5 · 2x and y = 250 · 0.8x on a the amount is reduced to 1 g? logarithmic scale. 11. The half life of the radioactive isotope 14C is 5730 years. A piece of wood mea- 3. Express y = 5 · 2−x/4 in the form sures only 20% of the common 14C:12C y = y M x. What are y , M and r? 0 0 ratio. How old is it?

4. The population of the City of Radford 12. A pain killer has a half life of eight hours, grew from 15,859 in 2000 to 16,408 in i.e. the body absorbs one half of the drug 2010. Assuming an exponential model, within eight hours. what is the annual growth rate? (a) A patient is given a 12mg dose. How 5. A population of 200 deer grows at an an- much is left in the body after one nual rate of 8%. day? (b) What dosage is required so that af- (a) What is the population after 5 years? ter six hours there are still 6 mg of (b) What is the doubling time? the drug left in the body?

139 Answers 5. (a) 294 (293.8656) (b) 9 years (9.0065) 1. (a) y = 6 · 5x x y 6. 14.2 (14.207) 0 6 1 30 7. (a) 152% 2 150 (b) 9 hours (8.974) 3 750 4 3750 8. 27 minutes (b) y = 500 · 0.5x = 500 · 2−x 9. 3.1 hours (3.1063) x y 0 500 10. (a) 15.9% (r = −0.159) 1 250 (b) 87.5% (r = −0.875) 2 125 (c) 3 years (3.076) 3 62.5 4 31.25 11. 3, 300 years (13,304.6) √ x x/2 (c) y = 3.578 · 1.118 = 12.8 · 1.25 12. (a) 1.5 mg x y 0 3.578 (b) 10 mg (10.091) 1 4 2 4.472 3 5 4 5.590

Figure 32: Problem 2

x 3. y = 5 · 2−1/4 = 5 · 0.8409x y0 = 5, M = 0.8409, r = −0.1591 = −15.91%

4. r = 0.3409%

140 15 Power Functions The reason is quite simple: Take the logarithm of y and simplify; then Power functions are very popular with biolo- a a gists, in particular in the study of allometry. log y = log(c x ) = log c + log x Kleiber’s Law, which relates body mass and = a log x + log c basal metabolic rates of mammals, is probably the most celebrated example. Species-area If we introduce the new variables curves (the number of species in a given area) u = log x and v = log y are another application of power functions. we have the equation 15.1 Deﬁnition v = ax + log c

Deﬁnition: Power functions have the form which is an equation of a line with slope a and v-intercept log c. y = c xa (21) Example: Consider the power function √ y = 10 x. Then Examples of power functions include √ v = log y = log(10 x) 2 y = 3x 1/2 1 √ = log 10 + log x = 1 + log x y = 10 x = 10x1/2 2 2 1 y = = 2x−1 = u + 1 x 2 In a loglog plot we graph the line v = 1 u + 1, In a way, it is much easier to work with power 2 and ”cheat” when we label the axes. functions than with polynomials, because they only have one term. However, there is no re- striction on the power a. Any number is permitted, not just positive integers.

√ This graph still represents y = 10 x. The points for √ x = 1 y = 10 1 = 10 √ At ﬁrst glace, there is nothing special about x = 100 y = 10 100 = 100 √ the graphs of power functions, but we obtain x = 0.01 y = 10 0.01 = 1 lines when we plot power functions on a loglog scale (logarithms on both axes). are highlighted on the graph.

141 Matching a power function to data is a lit- in order to find the exponent. tle tricky. We begin with an example. Check the worked problems for more examples and applications. x y 2 20 5 125 15.2 Approximation with Power Func- tions Because y = cxa, this table translates into the set of equations Data are never perfect, and approximations are needed. We can adapt the straight-edge 20 = c 2a method to power functions and use the follow- 125 = c 5a ing steps: with a and c as unknowns. Division yields 1. Take the logarithms of the x- and y-values 125 c 5a of the data. = a 20 c 2 2. Carefully plot the logarithmic data. 5a 5a 6.25 = = = 2.5a 2a 2 3. Find an approximating line be the straight- edge method (see Section 12.3). Unless you realize that 2.52 = 6.25, you need logarithms to solve this exponential equation, 4. Convert the approximating line back into and obtain a power function. log 6.25 a = = 2 log 2.5 Example: The data and the respective loga- Once a = 2 has been identified, we substitute rithms are given in the table. this value into the first equation and solve for x y u = log x v = log y c: 0.5 0.05 −0.301 −1.301 20 = c 22 = 4c ⇒ c = 5 2 1 0.301 0 and the underlying power function becomes 10 10 1 1 y = 5x2 40 50 1.602 1.699 150 400 2.176 2.602 Usually the exponent a is the more relevant The graph below shows the logarithms in part of a power function, and if data (x1, y1) the uv-plane. The points (u, v) = (0, −0.7) and (x2, y2) are known, the exponent a can be computed from and (u, v) = (2, 2.4) were selected for the approximating line. log y /y a = 2 1 (22) log x2/x1 The verification of this result is a routine exercise with logarithms. For our last example we would calculate log 125/20 log 6.25 a = = = 2 log 5/2 log 2.5

142 In the final step we have to convert the line to a power function with the original variables x and y. Since u = log x and v = log y, it follows that x = 10u and y = 10v There are two strategies to complete Step 4: Option One: Convert the points back to the xy-system and construct the power func- Graphing calculators or computer software tion. have built-in functions for this task, based upon u v x y the least squares approximation principle. 0 −0.7 ⇒ 1 0.2 On a graphing calculator enter the points 2 2.4 100 251 in lists with x- and y-coordinates (STAT menu and Edit option). Once this is completed, use We get c = 0.2 directly, because the value for the STAT menu, followed by CALC and se- x = 1 is known. For the parameter a we use lect the PwrReg option. For the data from our formula (22) and find that example, the calculator should respond with log 251/0.2 3.1 y = a ∗ x ∧ b where a = 0.223 and b = 1.52 . a = = = 1.55 log 100/1 2 In EXCEL one should graph the points in a scatter plot first and then select ”Power” and the interpolating power function is for the trendline option. If you ask to dis-

1.55 play the equation on the chart, the formula y = 0.2 · x y = 0.223x1.52 will appear. Option Two: Determine the equation for By the way, our straight-edge approxima- 1.55 the interpolating line in the uv-plane, and con- tion y = 0.2x is fairly close to the least vert it to the original variables. squares approximation of the computing de- The v-intercept (0, −0.7) is a given point, vices. and the slope becomes 15.3 Worked Problems 2.4 − (−0.7) 3.1 m = = = 1.55 2 − 0 2 1. Find a power function y = c xa for the Therefore the line in the uv-plane is data below. x y v = 1.55u − 0.7 (a) 1 3 It now follows that 2 48 x y y = 10v = 101.55u−0.7 (b) −0.7 u 1.55 1.55 1 3 = 10 · (10 ) = 0.2 · x 2 30 Of course either option yields the same ap- x y proximation. A graph of the points along with (c) 2 6 the power approximation is shown below. 5 20

143 x y (c) Here we obtain from (22) (d) 10 1, 000 log y /y log 20/6 20 25 a = 2 1 = log x2/x1 log 5/2 log 3.333 Solutions: = = 1.314 log 2.5 If we substitute x = 1 into the formula for a power function (21) we always ob- We use the first point and substitu- tain y = c 1a = c, regardless of the tion in order to find c value of a. We will take advantage of 1.314 this shortcut in problems (a) and (b). 6 = c 2 = 2.486 c 6 c = = 2.413 In part (b) we will show detailed steps 2.486 to find a and c, otherwise we will apply 1.314 formula (22) for brevity. and thus y = 2.413 · x . (d) In this exercise we see notice that (a) Formula (22) implies that the y-data are decreasing, which forces a negative exponent. From (22) we log y2/y1 log 48/3 find that a = = log x /x log 2/1 2 1 log y /y log 25/1, 000 log 16 a = 2 1 = = = 4 log x /x log 20/10 log 2 2 1 log 0.025 = = −5.322 c = 3 because we have information log 2 at x = 1, and the power function Substitution with the first point yields becomes y = 3x4. −5.322 (b) The given information translates into 1, 000 = c · 10 1, 000 103 c = = 3 = c 1a 10−5.322 10−5.322 3+5.322 0.322 8 30 = c 2a = 10 = 10 · 10 = 2.099 · 108 The first equation becomes c = 3, and after substitution we see that and the underlying power function is a 30 2 = = 10 3 −5.322 3 y = 2.099 · 10 · x

The solution of this exponential equa- This function can be manipulated 5.322 tion is 10 into y = 1, 000 · x . Try it. log 10 1 a = = = 3.322 2. Interpolate the data log 2 0.301 x 5 16 100 300 1, 500 and the power function is y = 3x3.322. y 75 50 20 5 2.5

144 by a power function. Use the straight- This was Option Two. The steps for Op- edge method and compare this estimate tion One require to convert the points to the least squares fit from a computer to xy-coordinates first, the result being or calculator. (3.16, 100) and (10, 000, 1.26), and then to construct the interpolating power func- Solution: The first step calls for taking tion. Answers will be identical for either the logarithm of the data. option. u = log x v = log y The least squares method yields the ap- 0.699 1.875 proximation 1.204 1.699 y = 251.67 x−0.631 2.000 1.301 2.477 0.699 3. Basal Metabolic Rates. This is a clas- 3.176 0.398 sical problem for power functions. The data go back to research done by Max A graph in the uv-plane (Step 2) and an Kleiber in the 1930ies who compared body approximating line (Step 3) are shown in mass of mammals and the respective basal the figure. metabolic rates (oxygen consumption at rest). Some of the data are listed below. The mass is measured in gram, the metabolic rate in milliliter (mL) of ogy- gen per hour.

Mass BMR Moose 325000 51419 Lion 98000 16954 Cougar 37200 8842 The line passes through the points (0.5, 2) Coyote 10000 2687 and (4, 0.1). The slope of this line is Racoon 5075 1599 Red fox 4400 2442 0.1 − 2 −1.9 m = = = −0.543 Rock hyrax 2400 660 4 − 0.5 3.5 Possum 2005 731.6 and its equation becomes (see Point-Slope Meerkat 850 310 Form (5)) Ground Squirrel 205.4 140.4 Indian Gerbil 87 75.7 v − 2 = −0.543 (u − 0.5) Chipmunk 45.8 72.7 v = −0.543u + 2.271 Naked mole rat 32 20.5 Bat 27 8.3 Now we convert to xy-variables: Shrew 17.1 54.7 Problem: Use the straight-edge method y = 10v = 10−0.543u+2.271 to ﬁnd an approximating power function 2.271 u −0.543 = 10 · (10 ) with x representing body mass and y rep- = 187 x−0.543 resenting oxygen consumption.

145 Solution: First we have to calculate the exponent: logarithms of the data. 5 log 10 a = 10 = 0.8 106 u v log 10 Moose 5.51 4.71 Then we substitute the ﬁrst point and Lion 4.99 4.23 a = 0.8 into the equation y = c · xa and Cougar 4.57 3.95 obtain Coyote 4.00 3.43 10 = c · 100.8 Racoon 3.71 3.20 Red fox 3.64 3.39 which implies that Rock hyrax 3.38 2.82 10 c = = 100.2 = 1.6 Possum 3.30 2.86 100.8 Meerkat 2.93 2.49 and our approximation is Ground Squirrel 2.31 2.15 Indian Gerbil 1.94 1.88 y = 1.6 x0.8 Chipmunk 1.66 1.86 Naked mole rat 1.51 1.31 With Option Two we would use the equa- Bat 1.43 0.92 tion v = 0.8u + 0.2 and take it from Shrew 1.23 1.74 there. The next task is to graph these points The least squares result (computer/calculator) and to eyeball an interpolating line. for these data yields y = 1.9726 x0.7918

15.4 Exercises √ 1. Sketch the function y = 4 x on a loglog scale. Try these options: (a) Sketch you own loglog graphing paper (1, 10, 100, ... equally spaced on both axes) and graph the func- For simplicity we use the line contain- tion. ing the points (1, 1) and (6, 5) in the uv- (b) Compute u = log x and v = log y plane. We choose Option One to com- for selected points and graph u ver- plete the problem, and we convert the sus v. points back to the xy-plane. Using x = 10u and y = 10v, we ﬁnd the points 2. Assume that y = cxa is a power function. Find c and a and complete the tables. x y x y 10 10 1 0.1 106 105 (a) 2 5 Now we use formula (22) to construct the 10 10

146 x y x y 1 3 1 425 (b) 3 1 4 250 6 (c) 15 140 60 80 x y 250 48 1 3 (c) 1,000 25 4 8 12 6. You investigate how the diversity of grasses on a meadow changes with area. On the 3. Sketch the functions smallest plot of 50 cm × 50 cm you ﬁnd just two species. As you increase the lot (a) y = 0.01x3 size you detect more and more species. 10 (b) y = On the largest plot (50 m × 50 m) you x2 count 31 diﬀerent species. The data are (c) y = 2.3x0.7 summarized in the table below.

2 on a regular grid and on a loglog grid. Lot size(m ) No. of Species 0.25 2 4. Let y = 4x1.5. 1 4 12 7 (a) Find x such that y = 500. 50 11 120 13 (b) Find x such that y = 100. 600 21 2, 500 31 5. Find a power function approximation for the data in the table. Use the straight- Use the straight edge method to ﬁnd an edge method for power functions and com- approximating power function with x rep- pare the approximations to the given data. resenting area, and y for the number of In addition, use a calculator and compare species. the calculator results to your straight- edge results. 7. The table below contains the heart rate at rest for selected mammals (Kong, NIU). x 1 2 5 10 (a) y 2 20 77 250 Mass (g) Pulse x y Mouse 25 670 Rat 200 420 0.2 1.1 Guinea pig 300 300 4 40 Rabbit 2,000 205 (b) 10 120 Small dog 5,000 120 25 360 Large dog 30,000 85 60 1,000 Man 70,000 72 100 1,900 Horse 450,000 38

147 Find an approximating power function (c) y = 3x2/3 by the straight edge method. Use x for x y the mass and y for the heart rate. 1 3 √ 4 7.56 = 6 3 2 Answers 8 12

1. 3.

Figure 33: Problem 1 (a) Figure 35: Problem 3

Figure 36: Problem 3 Figure 34: Problem 1 (b)

4. (a) x = 25 2. (a) y = 0.1x2 (b) x = 252/3 = 8.550 x y 5. Answers may vary. 1 0.1 2 0.4 (a) y ≈ 2.5x2, use (u, v) = (0, 0.4) and 5 2.5 (u, v) = (1, 2.4) 10 10 x 1 2 5 10 y 2 20 77 250 3 (b) y = est. 2.5 10 62.5 250 x x y Calculator: y = 2.92x2.01 1 3 (b) y = 7x1.2, 3 1 use (u, v) = (−0.7, 0) and 6 0.5 (u, v) = (2, 3.3) and some rounding.

148 x y est. 16 Difference Equations 0.2 1.1 1.015 4 40 36.95 Difference equations are a powerful modeling 10 120 110.9 tool, which can be easily investigated with com- 25 360 333.1 puter simulations. Population growth, predator- 60 1000 952.5 prey, host-parasite or disease modeling are typ- 100 1900 1758 ical applications. This section makes heavy use of the func- Calculator: y = 7.581x1.197 tion notation, and a review of Chapter 11 is −0.4 (c) y = 400x , advisable. use (u, v) = (0, 2.6) and (u, v) = (3, 1.4) 16.1 Introduction x y est. 1 425 400.0 In this introductory section we study an exam- 4 250 229.7 ple which illustrates the main ideas and con- 15 140 135.4 cepts behind difference equations. 60 80 77.77 The Problem: We monitor the deer popu- 250 48 43.94 lation in a large park, which currently consists 1000 25 25.24 of 1,200 deer. We know that the population in-

−0.407 creases at 10% annually, and hunting reduces Calculator: y = 429.8x the herd by 100 animals each year. Predict the 6. Answer may vary. population in the future. 0.3 This looks a bit like the discussion of the y = 3.4 x the monkey population in the beginning of Chap- ter 14. We start with a population of 1,200. The 10% increase results in 120 more deer in the following year, but we also have to subtract the 100 deer killed by hunters. The new population becomes

1, 200 + 120 − 100 = 1, 220

7. Answer may vary. In the next year we go through the same y = 2, 000 x−0.3 process, starting with 1,220 deer, and we obtain 1, 200 + 122 − 100 = 1, 242

Repeating the steps one more time leads to31

1, 242 + 124.2 − 100 = 1, 266.2

31Round to the nearest integer if the decimal bothers you.

149 The results are summarized in the table be- This is an example of a diﬀerence equation. low. We need a starting value, and everything else Year Population follows by a domino eﬀect. When n = 1, the 0 1, 200 equation (23) becomes 1 1, 220 2 1, 242 p(1) = 1.1 p(0) − 100 3 1, 266.2 = 1.1 · 1, 200 − 100 = 1, 220

Our next step is to derive a notation so that Substituting n = 2 into (23) leads to we can express our model mathematically. We let n be the number of years from the original p(2) = 1.1 p(1) − 100 32 count, and we use p(n) for the population in = 1.1 · 1, 220 − 100 = 1, 242 year n. With this notation out table is to be inter- and when n = 3 the diﬀerence equation (23) preted as becomes

Year Population p(3) = 1.1 p(2) − 100 n p(n) = 1.1 · 1, 242 − 100 = 1, 266.2 n = 0 1, 200 = p(0) n = 1 1, 220 = p(1) and so on. n = 2 1, 242 = p(2) It is a unique feature of difference equations n = 3 1, 266.2 = p(3) that we can determine p(n) only for integers. p(2) = 1, 242 and p(3) = 1, 266.2, but p(2.4) is p(2) means the population in the second unspecified. year, and p(10) stands for the population ten This is in stark contrast to exponential mod- years later, etc. els. If we set hunting aside, the difference equa- What can we say about our model? We tion becomes know that p(0) = 1, 200, since this is the original deer count, but we don’t have an explicit p(n) = 1.1 p(n − 1) formula for the deer population in following In an equivalent exponential model we would years. Instead, we have a recipe how to move use r = 0.1, M = 1 + r = 1.1 and y = 1, 200, from one year to the next. If we want the deer 0 with solution population in the year n, which is written as p(n), we have to add 10% to last year’s popu- y = 1, 200 · 1.1x (24) lation, denoted by p(n − 1), and subtract 100. This results in the equation It can be shown that p(n) matches the exponential growth formula33 (24), but it is only p(n) = p(n − 1) + 0.1 p(n − 1) − 100 defined for integers. On the other hand, there is no problem with setting x = 2.4 in (24) to which shortens to obtain y = 1, 200 · 1.12.4 = 1, 508.4. Graphi- p(n) = 1.1 p(n − 1) − 100 (23) cally, the exponential model leads to a smooth curve (red), while the difference equation pro- 32We use the function notation. p(n) is pronounced duces a scatter plot (blue). as ”p of n”, and in this example it means the population in year n. 33p(n) = 1, 200 · 1.1n

150 n ≥ 0 counting index, must be an integer, usually time (years, days, seconds or even generations) p(n) state of the system at time n, frequently a population p(0) initial condition f(p) update function, describes how p changes

A difference equation has the characteris- This graph also contrasts the difference be- tics of a feed-back loop: For the given state tween discrete and continuous variables. The p(n − 1), we identify the update f(p(n − 1)), p(n) are defined at the positive integers only, and then change the state of the system in the which makes n a discrete variable, while y is next time step to p(n) = p(n−1)+f(p(n−1)). defined for all x, and x is a continuous variable. The update function f(p), along with the It is usually very difficult and not very use- initial condition p(0), determine the behavior ful to find explicit solutions to difference equa- of the system in the future. When p(0) is given, tions. Instead, we let computers do numerical we can use n = 1 in (25) to obtain calculations, and focus on qualitative features, such as p(1) = p(0) + f(p(0)) will the points grow indefinitely? will the population become extinct? which makes p(1) a known quantity. Now set will the system settle for a steady state? n = 2, then and so on. p(2) = p(1) + f(p(1))

is known, and continuing in this fashion, we 16.2 First Order Difference Equations can compute p(n) for any number n (we just have to be patient if n is very big, or use a We will use the letter p to denote the state computer to speed things up). variable, and the state of the system at a par- After this abstract and fairly general intro- ticular time n is denoted by p(n). In difference duction, and we turn to specific examples. equations, the state of the system changes as Example: The opening example we had 1,200 time progresses. This change is driven by an deer to begin with. This makes for update function f, which depends on the state of the system itself. p(0) = 1, 200 In its most general form, a first order difference equation has the structure as initial condition. The change from one year to the next was the 10% population increase p(n) = p(n − 1) + f(p(n − 1)) (25) minus the deer lost to hunting. Therefore the update function becomes where f(p) = 0.1p − 100

151 and we can now assemble the difference equa- Example: Consider tion according to (25): a(n) = a(n − 1) + 12 p(n) = p(n − 1) + 0.1p(n − 1) − 100 a(0) = 100 which is equivalent to (23). Here a increases by 12 each time (f(a) = 12 is a Let’s take another look at the f(p) term. constant function), beginning with a(0) = 100. Initially we have p = 1, 200 deer, and terefore The next terms are (substitute) a(1) = a(0) + 12 = 100 + 12 = 112 f(1, 200) = 0.1 · 1, 200 − 100 = 20 a(2) = a(1) + 12 = 112 + 12 = 124 Thus, the population will increase by 20 an- a(3) = a(2) + 12 = 124 + 12 = 136 imals, and the new population becomes p = 1, 220. When we substitute this number into and so on. f, we find that Example (New Car Loan): This example is not biological, but the same ideas apply: You f(1, 220) = 0.1 · 1, 220 − 100 = 22 borrow $10,000 for a new car. The interest rate is 4.8%, and you make monthly payments and we will have 1, 220 + 22 = 1, 242 deer in of $400. the next year. If for some reason the deer pop- Let B(n) stand for the balance on your loan ulation becomes 2,000, then after n months. Clearly, B(0) = 10, 000, which is the original loan amount. From one month f(2, 000) = 0.1 · 2, 000 − 100 = 100 4.8% to the next, your debt increases by 12 = and the following year will see 2, 100 deer. 0.4%, while your payment lowers the debt by Slogan: If you want to calculate the new $400. The balance B on your loan will change value p(n), plug the current value p(n−1) into by the formula for f, and add the result to p(n − f(B) = 0.004 B − 400 1). and the difference equation becomes Example: Research suggests that a migratory bird population decreases by 8% annually. The B(n) = B(n − 1) + 0.004B(n − 1) − 400 last count in 2010 showed 7,600 birds. = 1.004B(n − 1) − 400 We construct a difference equation for this scenario: If n stands for the years after 2010, After one month your outstanding debt is we have p(0) = 7, 600 as initial condition. The 8% decrease can be modeled by B(1) = 10, 000 + 0.004 · 10, 000 − 400 = 10, 000 + 40 − 400 = 9, 640 f(p) = 0.92p In words: You owe $10,000, you accumulate and the difference equation becomes $40 in interest, and your payment reduces the debt by $400. p(n) = p(n − 1) − 0.08p(n − 1) It is very tedious (and dull) to continue = 0.92p(n − 1) these computations by hand. The resulting

152 data are known as an amortization schedule and the new population is p(2) = 40 + 17.14 = in the business world. As it turns out, the 57.14. loan will be paid off in just 27 months. In the At this point the computations become messy. real world, customers take out four year loans, We summarize our results for the first five it- and the monthly payments in this case would erations in a table. be $229.39, rather than $400. This is done by n p(n) f(p(n)) adjusting the payments so that B(48) ≈ 0. 0 25 15 As the models get more sophisticated, the 1 40 17.14 update function f will be more complex, and 2 57.14 15.58 the difference equation may become very clut- 3 72.73 11.48 tered. When you experiment with difference 4 84.21 7.22 equations and you calculate iterates without a 5 91.43 computer, it may be advisable to break up the steps and compute the update f(p(n−1)) first, Graphs. As mentioned in the introduc- and then form p(n) = p(n − 1) + f(p(n − 1). tion, scatter plots are a natural way to depict the results of difference equations. The inputs Example: We take p(0) = 25 as initial pop- are integers, and the gaps between the points ulation and use the update function are a characteristic feature. The graph below p · (100 − p) shows the solution to f(p) = p + 100 a(n) = 0.6a(n − 1) + 24 This is a special case of the Beverton-Holt model, a(0) = 0 which in turn is a variation of logistic growth, which we will study below. The resulting equation becomes

p(n) p(n − 1) (100 − p(n − 1)) = p(n − 1) + p(n − 1) + 100 is not terribly confusing, nonetheless, we will compute the update ﬁrst, and then form p(n). We have p(0) = 25. Now we form f(25), Terminology and Notation. A diﬀer- that is, we substitute 25 for p, and obtain ence equation is usually written as 25 · (100 − 25) 25 · 75 f(25) = = = 15 p(n) − p(n − 1) = f(p(n − 1)) 25 + 100 125 This is the update, and the new population is which emphasizes the change of the state p. If p(1) = 25 + 15 = 40. we set Now we repeat the process with p = 40. ∆p = p(n) − p(n − 1) Then it becomes 40 · (100 − 40) 120 f(40) = = = 17.14 40 + 100 7 ∆p = f(p(n − 1))

153 and we see that the change of p is determined or by the update function f. This approach al- pn+1 = pn + f(pn) = g(pn) lows for a nice connection to the ”differential equation” All of these describe the same process; they dp are packaged differently, but the mathematics = f(p) dx remains the same. which the continuous variable counterpart to a difference equation. Differential equations 16.2.1 Computers and Calculators are a Calculus based field of mathematics, and Explicit solutions to difference equation are the they are widely used in science and engineer- exception. There is a class of special cases (the ing, and they are becoming an increasingly im- affine equations) which admit solution formu- portant tool in the life sciences. las, but in most cases we have to settle for Equations of the form numerical results. We have seen before that p(n) = g(p(n − 1)) these computations can be very tedious and time consuming and we turn to computing de- are called recurrence relations or iterative equa- vices for a remedy. tions, with g being the iteration function. The The MODE setting for graphing calcula- two concepts are completely equivalent when tors34 often contains a SEQ option (SEQ stands we use for sequence). It is designed to construct a se- g(p) = p + f(p) quence of numbers from the iterative formula but the perspective is different. For example, the difference equation u(n) = g(u(n − 1))

p(n) − p(n − 1) = 0.2p(n − 1) Once you are in the SEQ mode, and you select ”Y=” you will be asked for a number of things: emphasizes that p increases by 0.2 = 20% each time, while the same process written as nMin starting value of index. u(n) implement the formula here; p(n) = 1.2p(n − 1) use 2ND u (above the 7) for u and the X,T,Θ,n key for n. indicates that p grows by a factor of 1.2 in u(nMin) initial condition. every step. Both are equivalent but they are expressed from diﬀerent points of view. We A plot of the solution can be obtained with will use the terms diﬀerence equation, iteration the GRAPH key, and TABLE returns the so- process or recurrence relation interchangeably. lution in tabular form. Some authors like to up the index by 1 and Example: We solve write a(n) = a(n − 1) p(n + 1) = p(n) + f(p(n)) +0.25a(n − 1)(4 − a(n − 1)) while others prefer subscripts for indexing, and a(0) = 0.1 the equations become 34This is in reference to TI calculators. Check user pn = pn−1 + f(pn−1) = g(pn−1) manuals for other brands.

154 numerically. Here the iteration function is The cell B2 contains the initial value a(0) = 0.1. The cell B3 directly below it implements g(a) = a + 0.25a(1 − a) the recurrence formula and calculates a(1). It Once SEQ has been selected, the entries in starts with an ”=”, like all formulas in EXCEL ”Y=” should be do, and then it makes repeated reference to cell B2. In fact, we are entering g(B2), that is, any nMin = 0 occurrence of a in the formula for g is replaced u(n) = u(n-1) + 0.25*u(n-1)*(4-u(n-1)) by B2. u(nMin) = { 0.1 } The next cell in the column is B4 and it The name of the variable is immaterial, and contains a(2). Its computation requires calcu- we have to live with the fact that we have to lation of g(B3), and we redo the last step with use u in place of a. B2 replaced by B3. As we go down the col- The TABLE command returns umn, we have to repeat this process over and over. Fortunately, we only have to enter the n u(n) formula only once in B3; for the rest we use 0 .1 the ﬁll feature35. The cell references will be 1 .1975 updated automatically. 2 .38525 EXCEL makes it very easy to display the 3 .73339 solutions graphically as a scatter plot. 4 1.3323 . . . . and with GRAPH we can visualize the result.

In EXCEL we can use relative cell references and the quick ﬁll handle to obtain the same table. We stick to the same example, and the result is shown below.

16.3 Steady States and Stability A steady state of a diﬀerence equation is a value of the state variable which does not change anymore. Example: Consider (we have seen this example before)

a(n) = 0.6a(n − 1) + 24

The left part contains the EXCEL output, the If a(0) = 60, then right part displays the actual cell entries. The column with the index n is included for con- a(1) = 0.6 · 60 + 24 = 36 + 24 = 60 venience, and the values of a(n) are shown in 35Go to the lower right of the cell until a ”+” appears. column B. Then hold and drag.

155 and we see that a(n) = 60 for all subsequent Here f(p) = 0.25 p (4 − p), and p∗ = 0 and terms36. p∗ = 4 result in f(p∗) = 0.

A steady state is also called a fixed point Fixed points can be stable or unstable. Fig- or an equilibrium. It is a value p∗ of the state ure 37 shows solutions to variable such that p(n) = p(n − 1) p(n) = p(n − 1) = p∗ +0.25p(n − 1)(4 − p(n − 1)) To find a steady state, replace both, p(n) and p(n−1), by p and solve the resulting equation. with different starting values. When started The solutions p∗ are the desired fixed points. exactly at p(0) = 0 or p(0) = 4, the trajec- tories remain constant. The solutions starting Example: Consider (we have used this exam- at p(0) = ±0.1 move away from zero, which ple in the computer section) makes p∗ = 0 an unstable equilibrium. On the p(n) = p(n − 1) other hand, solutions appear to be pulled toward p∗ = 4, and we call this a stable steady +0.25p(n − 1)(4 − p(n − 1)) state. Setting p(n) = p(n − 1) = p leads to

p = p + 0.25 p (4 − p) 0 = 0.25 p (4 − p) and we have two equilibria, namely p∗ = 0 and p∗ = 4. Check: If p(0) = 0, then

p(1) = 0 + 0.25 · 0 · (4 − 0) = 0 Figure 37: Stability Example and if p(0) = 4, then

p(1) = 0 + 0.25 · 4 · (4 − 4) = 4 Can we predict stability without calculating iterates and turning to graphs? To answer this question, we go back to the If the difference equation has the form (25) standard form of a difference equation we can argue that the update f(p) must vanish for steady states, and we can find fixed points p(n) = p(n − 1) + f(p(n − 1)) by setting f(p) = 0. Example: If f(p(n − 1)) > 0, then

p(n) = p(n − 1) p(n) > p(n − 1) +0.25p(n − 1)(4 − p(n − 1)) and the values of p will increase. Conversely, 36Diﬀerence equations follow a domino eﬀect. If there is no change in the step from n = 0 to n = 1, then there if f(p(n − 1)) < 0, then are no changes in the later steps and a(n) will remain constant. p(n) < p(n − 1)

156 from it, however, on the other side from where we started. As it turns out, p∗ is stable, if f(p) has a negative slope at p∗, and the slope is between −2 and 0. A careful sketch of f will reveal the slope at any point, and as a rule of thumb, be cautious Figure 38: An Update Function if the update function is steep at a fixed point. Example: We return to and p decreases. For f(p(n − 1)) = 0, we find p(n) = p(n − 1) that p(n) = p(n − 1) = p∗, and we are at a fixed point. +0.25p(n − 1)(4 − p(n − 1)) In the graph in Figure 38 depicts an exam- where f(p) = 0.25 p (4 − p). The graph of f ple of an update function. The states p make becomes up the horizontal axis (in our previous graphs, p was shown along the vertical axis). When the curve lies above the p-axis, the iterates p(n) become bigger, and the flow goes to the right; when the curve lies below the p-axis, the situation is reversed. The p-intercepts make up the steady states. The two fixed points are located at the the Still in reference to Figure 38, the steady points where the curve intersects with the hor- states on the outside must be unstable: If p > izontal axis (p∗ = 0 and p∗ = 4). f(p) is posi- p∗, the update is positive and p becomes even tive between p = 0 and p = 4. This is the re- bigger and moves away from p∗. Conversely, if gion where the points p(n) are increasing (look p < p∗, the update will be negative, and again at the labels on the vertical axis in Figure 37). p moves away from p∗. Therefore, a steady Outside that range f is negative and the iter- state will be unstable if f(p) changes from neg- ates decrease. ative to positive at p∗, or in other words, a At the point p∗ = 0, the graph of f switches steady state p∗ is unstable, if f has a positive from negative to positive, and the equilibrium slope37 at p∗. is unstable. At p∗ = 4, f changes from positive ∗ It is tempting to say that p is stable, if to negative, and the slope is −1, as a careful ∗ f(p) has a negative slope at p . But it is not sketch reveals. Thus, p∗ = 4 is stable. that simple. If p < p∗ and f(p) > 0, then p will increase. But if f is very steep, and f(p) is Example: very large, the increase may be so large that we p(n) = 1.1p(n − 1) − 100 overshoot p∗ by a lot and move further away This is the opening example (23) of the deer 37 We use this term loosely. Straight lines have the population. same slope everywhere, for other curves the slopes vary. Calculus defines the slope of a function f(x) at a par- First we have to identify f. Subtracting ticular point x as the derivative, denoted by f 0(x), and p(n − 1) on both sides results in a major portion of Calculus is devoted to the study of derivatives. p(n) − p(n − 1) = 0.1p(n − 1) − 100

157 Hence, (replace p(n − 1) by p) Suppose that we start very close to p∗ with p(0) = 3.9. Then f(p) = 0.1p − 100 f(3.9) = −2.5 · 3.9 + 10 = 0.25 The equation f(p) = 0 has solution p∗ = p(1) = 3.9 + f(3.9) = 4.15 1, 000. When 1, 000 deer are present, a 10% f(4.15) = −2.5 · 4.15 + 10 = −0.375 increase leads to 100 more animals, which is p(2) = 4.15 − 0.375 = 3.775 exactly the amount lost to hunting, and the population will remain constant. and we see that the iterates move away from p∗ = 4 in an alternating fashion.

A lot more can be said about stability with the tools of Calculus. But this is beyond the scope of this workbook.

16.3.1 Affine Equations f is a line with a positive slope, and therefore This section is all math. Its main purpose is the steady state is unstable. to give a little background for an important In the original example we had the initial class of difference equations. It can be used as count p(0) = 1, 200, and the iterates move reference material or as additional reading for away from p∗ = 1, 000 toward positive infin- the interested student. The remainder of the ity. workbook will not make use of the material in If we use p(0) = 800 as initial condition, we this section. find that It is possible to find exact solution formulas for the class of affine difference equations. p(1) = 1.1 · 800 − 100 = 780 We shall present these formulas, and show how they can be derived. p(2) = 1.1 · 780 − 100 = 758 p(3) = 1.1 · 758 − 100 = 733.8 Definition: Difference equations of the form p(n) = p(n − 1) + r p(n − 1) + k (26) and we see that the iterates move away from ∗ p = 1, 000, this time toward negative infinity, are called affine. Here r and k are constants. and the model becomes meaningless once the In this case the associated update function population falls below zero (extinction). Example: Here we explore what happens when f(p) = r p + k f has a negative slope, but it is too steep. Let is linear, with slope r and intercept k.

f(p) = −2.5p + 10 Special Case: Linear Growth Here r = 0, and the diﬀerence equation ∗ f(p) = 0 leads to p = 4, and because the becomes slope of f is less than −2, this steady state is unstable. p(n) = p(n − 1) + k (27)

158 that is, we add k at each iteration. solution and simplify: The solution is given by p(n − 1) + r p(n − 1) p(n) = kn + p(0) = (1 + r) p(n − 1) n−1 which is a linear function for the variable n = (1 + r) p(0) (1 + r) with slope k and intercept p(0). = p(0) (1 + r)n The verification of this formula requires two = p(n) steps. First we check the initial condition, and it is plain that for n = 0 we obtain p(0) = p(0). This shows that the function (29) satisfies the Next, we test the difference equation. We be- difference equation (28). gin on the right of (27), substitute the solution The update function is with n replaced by n − 1, and simplify: f(p) = r p p(n − 1) + k The case r = 0 is redundant. For r 6= 0 it = k(n − 1) + p(0) + k follows that p∗ = 0 is the only steady state. It = kn − k + p(0) + k = kn + p(0) is stable if and only if the slope satisfies −2 < = p(n) r < 0. The general affine equation (26) has the and we are done. update function f(p) = rp + k, and setting Steady states are redundant. Since f(p) = f(p) = 0 leads to k, equilibria are only possible if k = 0. But k then p∗ = − p(n) = p(n − 1) r and any value is an equilibrium, because noth- as the only fixed point. It is stable if −2 < r < ing changes. 0, because then the slope of f is in the desired range. Special Case: Exponential Growth. Our next goal is to derive a solution for- Here we consider the case where k = 0, mula, and we use a common technique (trick) which results in of applied math: Instead of using the population values p directly, we use a new variable y p(n) = p(n − 1) + r p(n − 1) (28) to measure by how much we deviate from the ∗ The update function becomes f(p) = r p, which equilibrium p . The relationship becomes is a line of slope r passing though the origin. p = y + p∗ or y = p − p∗ The solution is given by and we use whichever formula is more suiting. p(n) = p(0)(1 + r)n (29) p follows a difference equation, and we con- which is an exponential function with multi- struct the corresponding equation for y: plier M = 1 + r. y(n) For n = 0 we have p(0) on both sides, and ∗ the initial condition is confirmed. Again, be- = p(n) − p ginning on the right of (28), we substitute the = p(n − 1) + r p(n − 1) + k − p∗

159 = y(n − 1) + p∗ It is unstable, because r > 0. +r (y(n − 1) + p∗) + k − p∗ We had the initial condition p(0) = 1, 200 k and formula (30) yields = y(n − 1) + r y(n − 1) + r − + k r p(n) = 1, 200 · 1.1n + 1, 000 · (1 − 1.1n) = y(n − 1) + r y(n − 1) For example, if we are interested in p(3) we Thus, y(n) follows an exponential growth find that model, and we already found a solution formula for this case. Using (29) we have p(3) = 1, 200 · 1.13 + 1, 000 · (1 − 1.13) = 1597.2 − 331 = 1, 266.2 y(n) = y(0)(1 + r)n without having to compute p(1), p(2) and p(3) and back-substitution yields successively. If the initial condition is changed to p(0) = p(n) 800, as we did later in the stability discussion, ∗ = y(n) + p and we still are interested in p(3), we get = y(0)(1 + r)n + p∗ p(3) = 800 · 1.13 + 1, 000 · (1 − 1.13) = (p(0) − p∗)(1 + r)n + p∗ = 1597.2 − 331 = 733.8 = p(0)(1 + r)n + p∗(1 − (1 + r)n) and thus Example: The new car loan study falls into the category of affine equations. Here we had p(n) = p(0)(1 + r)n + p∗(1 − (1 + r)n) (30) B(n) = B(n − 1) + 0.004B(n − 1) − 400 When −2 < r < 0, we have −1 < (1 + r) < 1, and high powers of this number will become with B(0) = 10, 000. In this case n very small. In this case, the terms (1 + r) f(B) = 0.004B − 400 become negligible, and formula (30) shows that p(n) ≈ p∗, no matter where we start initially. and r = 0.004 and k = −400. ∗ −400 Again, this confirms that the steady state is The steady state is B = − 0.004 = 100, 000 stable. and we have We review older examples using the new n n formula (30). B(n) = 10, 000·1.004 +100, 000·(1−1.004 ) Example: In the introductory section we looked So, for instance, the remaining debt after one at deer populations following the difference equa- year is tion B(12) p(n) = p(n − 1) + 0.1p(n − 1) − 100 = 10, 000 · 1.00412 + 100, 000 · (1 − 1.00412) = 10, 490.70 − 4907.02 = 5, 583.68 This is an affine equation with r = 0.1 and k = −100. The steady state is What is the significance of the steady state? For one, it is unstable, because r > 0. Sec- ∗ k p = − = 1, 000 ondly, if the balance for some reason equals r

160 $100,000 exactly, then a $400 payment cov- 16.4 Logistic Growth ers the interest only, and the debt remains the same. The person who owes more that $100,000, The exponential model will get deeper in debt with just $400 pay- p(n) = p(n − 1) + r p(n − 1) ments; the person who owes less than $100,000 will eventually be debt-free with regular $400 is a good way to describe population growth, installments. but it suffers from a big handicap: The popula- Example: tions will always increase, at the same constant rate, and they will grow without bound. This a(n) = 0.6 a(n − 1) + 24 a(0) = 0 is not realistic. As a population gets larger, the fight for resources will become more intense, This example was used in graphing and in the and the growth should slow down. stability discussion. The logistic growth model addresses this When we subtract a(n − 1) on both sides, situation. It uses a population dependent growth we find that rate with two important parameters, the maximum growth rate rm and the carrying capacity a(n) − a(n − 1) = −0.4 a(n − 1) + 24 K. The maximum growth rate rm is the rate associated with unlimited resources. This rate Therefore, applies when the population is rather small. The other parameter is the carrying capacity f(a) = −0.4a + 24 K. This is the maximum population size which the environment can support. If the popula- and r = −0.4 and k = 24. The equilibrium tion gets past this point, it will encounter a negative growth rate and and it will decline. 24 a∗ = − = 60 −0.4 is stable, because −2 < r < 0. Formula (30) implies that

a(n) = 60 (1 − 0.6n)

In the graphing example the value a(10) was found in a successive computation. Our formula yields A linear function is the simplest way to ac- commodate this description. We use a(10) = 60 (1 − 0.610) r p r = r − m p = r 1 − = 60 (1 − 0.006, 065) = 59.64 m K m K which, of course, coincides with the value we When p = 0, we get r = rm, and at carrying have found before. capacity we have p = K and r = 0.

161 Upon substitution, the former exponential model now becomes the logistic model

p(n) = p(n − 1) + p(n − 1) r 1 − p(n − 1) (31) m K

Example: With rm = 0.1 and K = 1, 000, the logistic model (31) has the form We can clearly see that the exponential model grows toward infinity, while the logistic model p(n) approaches the carrying capacity K = 500. p(n − 1) = p(n − 1) + 0.1 1 − p(n − 1) In the early stages, until about n = 20, both 1, 000 curves are pretty much identical, at least it ap- The graph shows the solution with p(0) = 75 pears as such in the graph, and we call it the as initial condition exponential growth phase of logistic growth. Here the population size p is fairly small, compared to the carrying capacity, and the population grows as if the resources were unlimited. The fastest growth in the logistic model occurs when the population reaches p = 250 = 1 2 K. This is called the inflection point. There- after the growth decelerates and it will approach zero as the population approaches the The points are so close that the scatter plot steady state K = 500. looks like a smooth curve. We also observe the typical S-shape of logistic growth curves. It follows from formula (31), that the update function for logistic growth is Example: Here we contrast the exponential p f(p) = r 1 − p versus the logistic model. We use rm = 0.1, m K K = 500 and p(0) = 10. The exponential 2 rm p model is = rmp − K p(n) = p(n − 1) + 0.1 p(n − 1) just replace p(n − 1) by the variable p. In this regard, logistic growth is exponential growth while the logistic model becomes (f(p) = rmp) with a quadratic correction term 2 (− rm p ). If p is fairly small compared to the p(n) K carrying capacity, we do not have to worry p(n − 1) = p(n − 1) + 0.1 1 − p(n − 1) about this term, but as the populations in- 500 creases, the quadratic term becomes more and p(n − 1)2 = p(n − 1) + 0.01 p(n − 1) − more relevant. 5, 000 Stability. We see from the factored form Both solutions are displayed in the figure be- of the update function that low. f(0) = 0 and f(K) = 0

162 which shows that logistic growth has steady for some constant r. This is logistic growth for states p∗ = 0 and p∗ = K. The graph of f is K = 1 = 100%, and the iterations become a parabola opening downward. It increases at p∗ = 0, which makes this steady state unstable. p(n) f decreases at p∗ = K, and the this steady = p(n − 1) + r (1 − p(n − 1)) p(n − 1) state is stable, unless f is too steep38. In this model all individuals will eventually become infected, because p will always approach carrying capacity in logistic growth. The model assumes that interactions between individuals are random, it does not allow for re- covery from the disease, and it does not dis- tinguish between susceptible individuals and those who have built up an immunity. But, de- spite of all of its shortfalls, it is a good starting point for the study of epidemics. The fastest growth occurs when p is one half of the carrying capacity. 16.4.1 Logistic Approximations At carrying capacity the population has reached Collected data may show a logistic trend, but the maximum attainable population size. This since we do not have an explicit formula for is a state of fierce competition and fight for the logistic iterates, it makes it difficult to de- survival, as the environment cannot support termine a suited approximation. additional organisms. Calculus uses the curve Epidemics. Logistic growth is not limited K p = to population models. In our next example we 1 + ae−rx look at a model for the spread of a contagious to describe logistic growth. Here a has to be disease. By p we denote the proportion of in- selected to match the initial condition at x = 0 fected individuals. Then the uninfected pro- K portion is q = 1 − p, because the infected plus p(0) = the uninfected proportions have to add to one. 1 + a The disease is spreading whenever infected in- and e is the Euler constant e = 2.718, 281,.... dividuals come into contact with not infected The formula for p is found by solving the individuals, and the increase of infected indi- logistic differential equation39. While not an viduals is proportional to the product pq. This, exact solution of the logistic difference equa- by the way, is a from of the mass-action law. tion (31), it is a fairly good approximation. The update function then becomes Unfortunately, logistic approximation is not a trendline option in EXCEL, but the graph- f(p) = r p q = r (1 − p) p ing calculator has this capability: Enter the

39 38 This is the equation (Calculus required) The carrying capacity K is stable, if 0 < rmax < 2. Periodic, yet stable, patterns appear for 1 < r < 2 dp p = r 1 − p and cyclic behavior and chaotic patterns evolve if 2 ≤ dx K rmax < 3.

163 data into lists with STAT and EDIT, as usual. and with x = 0 we see that p(0) becomes Then use STAT and CALC, and select ”Logis- 24.669 tic”. The calculator will provide values for K, p(0) = = 1.87 a and r in the logistic function. 1 + 12.179 Example: Consider the data The solution to the associated diﬀerence n p equation 0 2 p(n) 1 3 p(n − 1) 2 5 = p(n − 1) + 0.593 1 − p(n − 1) 4 12 24.7 6 18 with p(0) = 1.87, along with the original data 10 24 is given below. The approximations are a good We enter the values into two lists on a calcu- match at the ends, but there are some gaps in lator, and select the logistic approximation as the center. described above. The calculator responds with n p(n) data y = c/(1 + ae ∧ (−bx)) 0 1.87 2 a = 12.17918098 1 2.89 3 b = .5934057215 2 4.41 5 3 6.56 c = 24.66931882 4 9.42 12 and the logistic approximation of the data be- 5 12.87 comes 6 16.53 18 24.669 7 19.77 p = 8 22.11 1 + 12.179e−0.59341x 9 23.48 A graph of the points with the approximating 10 24.17 24 curve is given below. 16.5 Worked Problems

1. Find a diﬀerence equation for the given situation, and calculate the ﬁrst three iterates. Find steady states, if possible, and address their stability.

(a) You work through a math book at 6 pages per day, beginning on page 23. Rounded to three signiﬁcant digits, the cal- (b) Beginning with $200, a child spends culator found K = c = 24.669 as carrying ca- half of its money in any given week, pacity; for the rate we have rm = b = 0.593, while also getting a $10 allowance.

164 (c) A fish population in a lake decreases The money disappears quickly, but at an annual rate of 5%. To combat the steady influx of $10 keeps the this decline, 400 fish are released child from being broke. The steady each year. In the beginning there state is found from f(p) = 0, and are 5,000 fish in the lake. we see that

(d) Bacteria grow at a maximum rate ∗ of 3% per minute. We start with 25 p = 20 bacteria, and the carrying capacity At this level the child spends $10 is 10,000. Use a logistic model. each week, and get the money back in form of the allowance. f(p) is a Solutions: line with a negative slope, which is (a) If p denotes the starting page, we between −2 and 0, and the steady know that p(0) = 23, and the up- state is stable. date function is f(p) = 6. The dif- (c) Let p(n) be the ﬁsh population in ference equation then takes the form year n. Then p(0) = 5, 000. The p(n) = p(n − 1) + 6 update function is

and the first iterates become f(p) = −0.05p + 400 p(1) = p(0) + 6 = 23 + 6 = 29 The equation f(p) = 0 yields p(2) = p(1) + 6 = 29 + 6 = 35 400 p(3) = p(2) + 6 = 35 + 6 = 41 p∗ = = 8, 000 0.05 There are no steady states, because f(p) = 6 6= 0. When 8,000 fish are present, a 5% (b) We let p(n) be the available funds decline means the loss of 400 fish, at the beginning of week n. Then which is balanced by restocking. The p(0) = 200. For the update we find equilibrium is stable, because f is a line with a negative slope between 1 f(p) = − p + 10 −2 and 0. 2 The difference equation becomes This results in p(n − 1) p(n) p(n) = p(n − 1) − + 10 2 = p(n − 1) − 0.05p(n − 1) + 400 p(n − 1) = + 10 = 0.95p(n − 1) + 400 2 The story of the first three weeks is and for the first iterations we have 200 p(1) = + 10 = 110 2 p(1) = 0.95 · 5, 000 + 400 = 5, 150 110 p(2) = + 10 = 65 p(2) = 0.95 · 5, 515 + 400 = 5, 292.5 2 p(3) = 0.95 · 5, 292.5 + 400 = 5, 427.9 65 p(3) = + 10 = 42.50 2

165 (d) This is a logistic growth problem. (a) Here we multiply by 0.8 repeatedly. The information translates into a(1) = 0.8 · 250 = 200

rm = 0.03 a(2) = 0.8 · 200 = 160 K = 10, 000 a(3) = 0.8 · 160 = 128 p(0) = 25 a(4) = 0.8 · 128 = 102.4 a(5) = 0.8 · 102.4 = 81.92 and the equation becomes The formula for the iterates is p(n) = p(n − 1) + a(n) = 250 · 0.8n p(n − 1) 0.03 1 − p(n − 1) 10, 000 This follows from formula (29) with r = −0.2. The next iterates are (calculator) (b) Here the rule is: Beginning with 80, take half of the current value and p(1) = 25.748 add 20. This results in p(2) = 26.519 80 a(1) = + 20 = 60 p(3) = 27.312 2 60 a(2) = + 20 = 50 This is still in the exponential growth 2 phase (25 is much less than 10,000), 50 a(3) = + 20 = 45 and the bacteria grow at 3% per 2 minute. Incidentally, p(24) = 50.7 45 a(4) = + 20 = 42.5 and initially the doubling time is a 2 little under 24 minutes. 42.5 a(5) = + 20 = 41.25 The growth function is 2 By looking at the data, we notice p f(p) = 0.03 1 − p that in each step we cut the distance 10, 000 to a∗ = 40 in half, and the general with steady states p∗ = 0 (unstable) formula becomes ∗ 40 and p = 10, 000 (stable). a(n) = 40 + 2n 2. Compute a(5) from We could also use formula (30) with ∗ (a) a(n) = 0.8 · a(n − 1) a = 40, r = −0.5 and a(0) = 80: a(0) = 250 a(n) a(n − 1) 1n 1n (b) a(n) = + 20 = 80 · + 40 · 1 − 2 2 2 a(0) = 80 1n = 40 + (80 − 40) · and ﬁnd a general formula for a(n). 2 40 = 40 + Solutions 2n

166 3. Find a difference equation for the data in The update is f(p) = −p/3, be- the table. cause if 2/3 are left, then 1/3 must n p(n) have been taken out. 0 65 4. Consider the update function 1 69 (a) 2 73 p (p − 4) (10 − p) f(p) = 3 77 50 4 81 Compute p(3) for the initial values n p(n) 0 80 (a) p(0) = 3 1 120 (b) (b) p(0) = 4 2 180 (c) p(0) = 5 3 270 4 405 Solutions: In all cases we compute the n p(n) update f(p) first, and then determine the 0 81 next p. 1 54 (c) 3·(−1)·7 21 2 36 (a) f(3) = 50 = − 40 = −0.525 3 24 Therefore 4 16 p(1) = 3 − 0.525 = 2.475 Solutions: In this type of problem you have to identify a pattern in the table, Following the same pattern we ob- and then translate it into an update func- tain tion and a difference equation. Nothing f(2.475) = −0.710 can be done if you don’t see the pattern. p(2) = 2.475 − 0.710 = 1.765 (a) Here the values go up by 4 in each f(1.765) = −0.812 step. The update is f(p) = 4 and the difference equation becomes p(3) = 1.765 − 0.812 = 0.953 p(n) = p(n − 1) + 4 (b) We have f(4) = 0, which makes ∗ (b) We add one half of the number each p = 4 an equilibrium point, and time. This makes f(p) = p/2 and a(0) = a(1) = a(2) = a(3) = 4 we have p(n − 1) p(n) = p(n − 1) + (c) Routine calculations result in 2 = 1.5 p(n − 1) n p(n) f(p(n)) 0 5 0.625 (c) Here we take 2/3 of the current num- 1 5.625 1.000 ber, and the equation becomes 2 6.625 1.467 2 p(n) = p(n − 1) 3 8.092 3

167 5. Find the equilibria when In logistic growth, p∗ = 0 always is an unstable steady state, because if (a) p(n) = 0.25p(n − 1) + 300 only very few organisms are present, (b) p(n) = 1.2p(n − 1) − 80 the population will start growing and ∗ (c) p(n) = p(n − 1) thus move away from p = 0. p(n−1) The maximum growth rate in this +0.4 1 − 250 p(n − 1) example is rmax = 0.4, and the car- and discuss their stability. rying capacity is stable, because 0 < rmax < 2. Solutions: 6. Find the steady states for the update func- (a) If we substitute p = p(n) = p(n−1), tion we ﬁnd that p (p − 4) (10 − p) f(p) = p = 0.25p + 300 50 and discuss stability. which results in the solution p∗ = 400 as steady state. Is it stable? Solution: Steady states are solutions of ∗ ∗ First, we determine update function. f(p) = 0, and thus we have p = 0, p = ∗ Subtraction yields 4 and p = 10 as equilibria. The graph of f is given below. p(n) − p(n − 1) = −0.75p(n − 1) + 300

Hence,

f(p) = −0.75p + 300

The update function is linear with The slopes at p∗ = 0 and p∗ = 10 are slope between −2 and 0, and there- negative, and above −2. Therefore these fore the steady state is stable. points are stable. At p∗ = 4 we have (b) We begin by ﬁnding the update: a positive slope, and this steady state is unstable. p(n) − p(n − 1) This update function was used above. We = 0.2p(n − 1) − 80 saw that p∗ = 4 was a steady state, and when the iterations were started at p(0) = Thus, f(p) = 0.2p−80. The steady 3, the iterates moved away toward 0, while state is found from f(p) = 0, with for p(0) = 5 the iterates moved toward ∗ solution p = 400. f has a positive 10. This illuminates the unstable nature slope, therefore the steady state is of the equilibrium. unstable. 7. Epidemics. An epidemic spreads ac- (c) This is a logistic growth problem. cording to the law The steady states are p∗ = 0 and p∗ = 250. p(n)

168 = p(n − 1) + 0.6(1 − p(n − 1))p(n − 1) rmax = 1.8 > 1 we will overshoot the carrying capacity, then bounce back below where n measures time in months, and it, and keep oscillating about K = 1, 000 p(n) is the proportion of infected indi- and gradually close in on this value. viduals in the n-th month. Initially 3% of the population is infected. How many will be infected after 4 month? How many after one year?

Solution: This problem is purely computational. p(0) = 0.03 is known, and we are asked to compute p(4) and p(12).

n p(n) 16.6 Exercises 0 0.030 1 0.047 1. Find p(5) from 2 0.075 (a) p(n) = p(n − 1) + 4, p(0) = 25 3 0.116 4 0.178 (b) p(n) = 1.1p(n − 1) , p(0) = 60 5 0.265 (c) p(n) = 2p(n − 1) − 10, p(0) = 12 6 0.382 1 7 0.524 2. Find p(5) from p(n) = 2 p(n − 1) + 3 8 0.673 and 9 0.805 (a) p(0) = 36 10 0.899 11 0.954 (b) p(0) = 0 12 0.980 (c) p(0) = 6

We see that after four months about 18% 3. Find a difference equation for the values are infected, and that after one year 98% in the table and fill in the missing values. of the population carry the disease. n p(n) n p(n) 8. It was stated in the discussion of logis- 0 28 0 81 tic growth that the carrying capacity is 1 25 1 108 stable if 0 < rmax < 2. In this problem 2 22 2 144 we compute a few iterations when rmax 3 19 3 192 is close to 2. 4 4 Compute p(1) though p(10) for logistic 5 5 growth with rmax = 1.8, K = 1000 and 4. Find the equilibria for the difference equa- p(0) = 50, and graph the result. tions, and discuss their stability. Again, this is a purely computational problem. The first values are p(1) = 135.5, (a) p(n) = 1.05p(n − 1) − 200 p(2) = 346.4 and p(3) = 753.9. Because (b) p(n) = 0.75p(n − 1) + 68

169 (c) p(n) = p(n − 1) n p(n) p(n−1) 0 28 +0.03 1 − 12 p(n − 1) 1 25 5. Spread of a Disease. p(n) is the per- 2 22 centage of infected people at time n, the 3 19 quantity 1 − p(n) is the percentage of 4 16 not-infected people. The disease spreads 5 13 whenever infected and uninfected people p(n) = 4 p(n − 1) come in contact; this is represented by 3 the product p(n − 1)(1 − p(n − 1)), and n p(n) we arrive at the diﬀerence equation 0 81 1 108 p(n) = p(n − 1) + p(n − 1)(1 − p(n − 1)) 2 144 3 192 Given that initially 1% of the population 4 256 are infected, what percentage is infected 5 341.3 at time n = 5, what percentage when n = 10? 4. (a) p∗ = 4, 000, stable (b) p∗ = 272, stable 6. Use a calculator to ﬁnd a logistic approx- ∗ imation for the data (c) p = 0, unstable and p∗ = 12, stable x p 5. 27.5% and 99.997% 0 3 125 3 15 6. p = 8 80 1 + 35.4e−0.518x 12 117 rmax = 0.5183, K = 125.0, p(0) = 3.434 15 123

Answers

1. (a) 45 (b) 96.6 (c) 74

2. (a) 6.9375 (b) 5.8125 (c) 6

3. p(n) = p(n − 1) − 3

170