These summarize what Blecher did in class. A few more details and plenty of pictures were given in class. See also http://www.math.uh.edu/ pamb/m1432.htm for Pam Balthazars online Calc II notes, which the tests are probably based more∼ on (I do not make the tests). Chapter 7 of II. 7.1: Inverse functions.

Functions: If X and Y are sets, then a f : X Y is a rule that assigns to each • element x X, one and only one element f(x) Y . [Picture.]→ ∈ ∈ X is the domain of f. It is the of inputs we can put into f. • The range of f is f(x) : x X , this is a of Y . It is the set of outputs of f. • { ∈ } Example. X equals the students in this class, Y = 0, 1, 2, , 9 , and f(x) is the first digit • of the social security number of student x. Range{ of f is ·0 ·, · 1, 4}, 5, 6, 7, 9 . { } This function is many-to-one (since many had social security number 4) [Picture.] • A function is one-to-one (written 1-1) if f(x )= f(x ) implies x = x . • 1 2 1 2 [Picture drawn in class.] • Example. A = temperatures in Celsius; and B =temperatures in Fahrenheit. Let f : A B • 9 → be the ‘conversion’ from Celsius to Fahrenheit: f(x)= 5 x + 32. Check that f is one-to-one. 9 9 9 9 Solution. f(x1) = f(x2) means that 5 x1 +32 = 5 x2 + 32. So 5 x1 = 5 x2, so x1 = x2. We’ve shown that f(x1)= f(x2) implies x1 = x2, so f is one-to-one. If f : X Y is one-to-one then it has an inverse function g : E X, where E is the range • of f [Picture.]→ Explicitly, g(b)= a if and only if f(a)= b. This inverse→ function g is written 1 as f − . For example in the Celsius to Fahrenheit example, the inverse function is the ‘conversion’ • from Fahrenheit to Celsius: f 1(y)= 5 (y 32), or if you like (changing names of variables) − 9 − f 1(x)= 5 (x 32). − 9 − How to get this? Method: Step 1. Write y = f(x). Step 2. Solve for x in terms of y. Step • 1 3. Write f − (y) for the expression found in Step 2. Lets apply this to the Celsius to Fahrenheit function... Step 1: y = 9 x + 32. Step 2: • 5 9 x = y 32, so x = 5 (y 32). Step 3: f 1(y)= 5 (y 32). 5 − 9 − − 9 − 1 1 1 5 Beware: dont confuse f − with f . In the Celsius to Fahrenheit example, f − (x)= 9 (x 32), • 1 1 5 1 1 − but f = 9 = 9x+160 , and f − = f . 5 x+32 6 f 1(f(x)) = x for all x in the domain of f; and f(f 1(y)) = y for all y in the range of f. • − − 1 1 Note that f − is one-to-one, so it has an inverse, namely f. The graphs of f and f − are • symmetric (that is, mirror images of each other) about the line y = x. [Picture drawn in class.] If f is strictly increasing or strictly decreasing then f is one-to-one, and so has an inverse • function. Recall that if f ′ > 0 on an interval (except maybe at a finite number of points) then f is strictly increasing there, and if f ′ < 0 on an interval (except maybe at a finite number of points) then f is strictly decreasing there.

1 9 So in the Celsius to Fahrenheit example, f ′ = 5 > 0, so f is strictly increasing, so has an • inverse function. Show that sin x is invertible on [ π/2, π/2]. • d − Solution: dx (sin x) = cos x > 0 on [ π/2, π/2] except at the two endpoints. So by Calc I we know sin x is strictly increasing on− [ π/2, π/2]. Hence it is invertible on [ π/2, π/2]. − − For what K is x3 Kx2 + 2x a one-to-one function (and so invertible). • − d 3 2 2 Solution: dx (x Kx +2x) = 3x 2Kx+2. This is a quadratic with positive first term, so its graph has three− possibilities [Picture− drawn in class]. In the first two cases (pictures) this is > 0 except maybe at one point, which implies that f(x)= x3 Kx2 +2x is strictly increasing, so one-to-one. In the third case the derivative is negative in− places so f is NOT strictly increasing. By College Algebra, the first two pictures correspond to where the discriminant ∆ = b2 4ac is 0. But the discriminant ∆ = b2 4ac = 4K2 4 3 2= 4(K2 6) 0 when K2−< 6. So≤ if √6 K √6 then x3 Kx−2 + 2x is one-to-one,− · · so invertible.− ≤ − ≤ ≤ − 1 FACT: If f is continuous and one-to-one on an interval then it has an inverse, and f − is • also continuous. KEY FACT IN THIS CHAPTER: If f is one-to-one and differentiable on an open interval • 1 containing a point a, and if f ′(a) = 0 then f − is differentiable at the number b = f(a), 1 1 6 and (f − )′(b)= . f ′(a) 1 Example 1. Let f(x)= x + √x. Find (f − )′(2). • 1 1 2 1 3 Solution: We ask: f(?) = 2, so ? = 1. Since f ′ =1+ 2 x− , we have f ′(1) = 1 + 2 = 2 , 1 3 2 so (f − )′(2) = 1/( 2 )= 3 .] 5 3 1 Example 2. If f(x)= x + x +5 find (f − )′(3). Also find an for the tangent line • 1 to the graph of y = f − (x) at the number x = 3. 4 2 Solution: We ask: f(?)=3, so ?= 1. Also f ′(x) = 5x + 3x , so f ′( 1)=5+3=8 1 1 − − and (f − )′(3) = 8 . The tangent line is y ( 1) = 1 (x 3) or y = 1 x 11 . − − 8 − 8 − 8 Example 3. If f(x)= x cos x find (f 1) . In particular, find (f 1) ( 1). • − − ′ − ′ − Solution: f = 1+sin x, so (f 1) (y) = 1 , where f(x) = y. Since f(0) = 1, we ′ − ′ 1+sin x − have (f 1) ( 1) = 1 = 1. − ′ − 1+sin 0 The KEY FACT IN THIS CHAPTER above may be rewritten: • 1 1 (f − )′(f(x)) = . f ′(x) or 1 1 (f − )′(y)= 1 . f ′(f − (y))

Writing y = f(x) so x = f 1(y), the last formula reads: • − dx 1 = . dy dy dx

Example 4. If y = 2x7 + 3x find the derivative of the inverse function, and evaluate it at • y = 5. dy 6 dx 1 1 Solution: dx = 14x + 3, so dy = dy = 14x6+3 . When y = 5 then x = 1, so here dx dx 1 1 dy = 14+3 = 17 . 1 Proof of the KEY FACT IN THIS CHAPTER: f − (f(x)) = x, so by the we have • 1 1 1 (f − )′(f(x)) f ′(x)=1. So(f − )′(f(x)) = . · f ′(x) 7.2: The function. x 1 In this course we define ln x = 1 t dt for x > 0. This is called the • function. [Picture as an area underR graph.] ln x We define loga x = ln a for x > 0 and a > 0, a = 1. This is called the logarithm base a. • 6 x Sometimes log10 is written as log. In this course we define e to be the inverse of the natural logarithm function ln x. We define ax = ex ln a (for a> 0). 1 1 Clearly ln 1 = 1 t dt = 0. • R From the picture above as an area under graph we see that ln x > 0 if x > 1. Note that • ln x< 0 if 0

d d x 1 1 Thus dx (ln x)= dx ( 1 t dt)= x by the second fundamental theorem of Calculus. • R d 1 Thus dx (ln x)= x > 0 for x> 0, so ln x is strictly increasing by Calc I, hence it is 1-1 and • invertible.

d2 d 1 1 2 (ln x)= ( )= 2 < 0 for x> 0, so ln x is concave down by Calc I. • dx dx x − x (Picture of ln x in class.) We define e to be the number so that ln e = 1 [Picture.] • Properties of ln: • ln(ab) = ln(a) + ln(b), a,b> 0 a ln( ) = ln(a) ln(b), a,b> 0 b − ln(ar)= r ln a, a> 0 (the text proves the last formula first for r rational, later for any r.)

d 1 1 These are proved in the textbook. I’ll just prove the first one: dx (ln(ax)) = ax a = x by • d 1 1 the chain rule. So dx (ln(ax) ln(x)) = x x = 0. Thus by Calc I, ln(ax) ln(x) = C, a , so ln(ax) = ln(x)+C−. Put x = 1:− ln(a) = ln(1)+C = C. So ln(ax)− = ln(x)+ln(a).

1 Example. Solve 2 ln x = ln(2x 1). • 1 − 1 Solution: ln x 2 = ln(2x 1). Since ln is 1-1 we have x 2 = 2x 1. Squaring, x = (2x 1)2 = 4x2 4x + 1. So− 4x2 5x +1=(x 1)(4x 1) = 0. Hence− x = 1 or 1/4. The latter− is impossible− since ln(2x −1) doesnt make− sense if− 2x 1 0. So x = 1. − − ≤ limx ln x =+ , and limx 0+ ln x = . • →∞ ∞ → −∞ Proofs: Note that ln 2 > 0, so ln(2n)= n ln(2) as n . So since ln x is strictly 1 → ∞ n → ∞ increasing, limx ln x =+ . Similarly, ln( n ) = ln(2− ) = n ln(2) as n . →∞ ∞ 2 − → −∞ → ∞ So since ln x is strictly increasing, limx 0+ ln x = . → −∞ Putting these facts above together gives the graph of ln above. [Picture above.] • 2 (x+3) 1 Example. Expand ln 3 . [Done in class, answer = 2 ln(x + 3) ln x ln(x 2)]. x √x 2 3 • − − − − Example. What is the domain of log √3 4x 3. • 10 − Solution: The log doesnt make sense if √3 4x 3 0. So 4x 3 > 0 or x> 3/4. Domain: (3/4, ). − ≤ − ∞ 2 1 4 Example. Combine into one expression: ln x 2 ln(x + 1) 2 ln(2 x ). [Done in class, • answer = ln x ]. − − − (x2+1)2√2 x4 − Example. If ln 10 2.3 find ln 10.3 approximately using differentials. • ≈ Solution: We use the Calc I formula f(x + h) f(x)+ hf ′(x) with x = 10, h = ∆x = 0.3,f(x) =ln x. So f (x)= 1 = 1 , and ln 10.3 ≈ln 10 + (0.3) 1 2.3 + 0.03 = 2.33. ′ x 10 ≈ 10 ≈ Example. Find d (ln(1 + x2)). • dx 1 2x Solution: By the chain rule this is 1+x2 2x = 1+x2 .

4 Example. Find dy if y = (x+3) by ‘logarithmic differentiation’. • dx (x4+12)√x2+5 4 4 √ 2 4 1 2 Solution: ln y = ln((x+3) ) ln((x +12) x + 5) = 4 ln(x+3) ln(x +12) 2 ln(x +5). − − 3 − y′ 4 4x 2x Differentiating both sides with respect to x we have = 4 2 . So y x+3 − x +12 − 2(x +5) 4 4x3 x (x + 3)4 4 4x3 x y′ = y( )= ( ). x + 3 − x4 + 12 − x2 + 5 (x4 + 12)√x2 + 5 x + 3 − x4 + 12 − x2 + 5

7.3: The logarithm function II. d 1 1 Since dx (ln x)= x , we have x dx = ln x + C if x> 0. But what if x< 0? • R Example. Find d (ln x ). • dx | | d 1 Solution: If x > 0 then this is dx (ln x) = x . If x < 0 then by the chain rule this is d 1 1 d 1 dx (ln( x)) = x ( 1) = x . So dx (ln x )= x . − − − | | We conclude from the last example that • 1 dx = ln x + C. Z x | |

By the chain rule, d (ln u(x)) = 1 u (x)= u′(x) . Similarly, d (ln u(x) )= u′(x) . • dx u(x) · ′ u(x) dx | | u(x) So • u (x) ′ dx = ln u(x) + C. Z u(x) | | d Example: Find dx (ln(ln x)). • d 1 dx ln x x 1 Solution: ln x = ln x = x ln x . Examples worked in class (if you did not write down the details in class you can also find • them in e.g. Pam B’s online notes): d x d d ( ), (ln(2x) ln(5x)), (sin(ln(2x))). dx 1+ln x dx dx

2 6x2 2 Find 1 x3 x−+1 dx. • − R 2 6x2 2 2 3x2 1 3 2 Solution. 3 − dx = 2 3 − dx = 2ln x x + 1 ] = 2 ln 7 2ln1= 2ln7= 1 x x+1 1 x x+1 | − | 1 − ln 72 = ln 49.R − R −

sin x sin x 1 tan x dx = cos x dx = −cos x dx = ln cos x + C = ln cos x + C = ln sec x + C. • R R − R − | | | | | | Similarly, cot x dx = ln sin x + C. • R | | Find sec x dx. • R sec x+tan x sec2 x+sec x tan x Solution: Trick: = sec x sec x+tan x dx = sec x+tan x dx = ln sec x + tan x + C. R d 2 R d | | We used the Cal I facts that dx (tan x) = sec x and dx (sec x) = sec x tan x.

Find π/8 sec(2x) dx. • 0 Solution:R Let u = 2x, du = 2dx, so integral becomes π/4 1 1 π/4 1 π π 1 sec u du = ln sec u + tan u ]0 = ln sec + tan ln sec 0 tan 0 . 2 Z0 2 | | 2 | 4 4 |− 2 | − | 1 √ But this equals 2 ln( 2 + 1). Also memorize: csc x dx = ln csc x cot x + C. This is derived similarly to sec x dx. • R | − | R Examples worked in class (if you did not write down the details in class you can also find • them in e.g. Pam B’s online notes): π/2 2 e 2 csc x ln x dx (1 + csc x) dx; dx; (Q46) dx; 2 . Zπ/4 Z 2 + cot x Z1 x Z x(ln x)

7.4: The . The exponential function ex is the inverse function of ln x. [Picture of mirror of • y = ln x]. So ln(ex)= x; and eln x = x if x> 0. • Thus since ln 1 = 0 we must have e0 = 1. Since ln e = 1 we must have e1 = e. • From the picture, ex > 0 for all x • x x From the picture, limx e = and limx e = 0. • →∞ ∞ →−∞ From our KEY FACT IN THIS CHAPTER (see page 2 above), if y = f(x) = ln x then • x = ey and d y 1 1 1 y (e ) = (f − )′(y)= = 1 = x = e . dy f ′(x) x So the derivative of ex is ex. This makes ex one of the most important functions in Calculus. • FACT: ex+y = exey. • [Proof: Let a = ex+y, b = ex, c = ey. Then ln a = ln ex+y = x + y = ln b + ln c = ln(bc). So a = bc. Similar proofs show: • x x y e x y xy x 1 e − = , (e ) = e , e− = . ey ex

By the chain rule • d u(x) u(x) (e )= e u′(x). dx ·

u(x) u(x) Thus u′(x)e dx = e + C. • R 2 Example. Find d (e3x ). • dx 3x2 d 2 3x2 Solution. = e dx (3x ) = 6xe .

√x Example. Find e dx. • √x R 1 1 2 1 2 u u Solution. Let u = x then du = 2 x− , and integral becomes 2e du = 2e + C = 2e√x + C. R

9 e√x Example. Find 1 √x dx. • R Solution. = [2e√x]9 = 2e√9 2e√1 = 2e3 2e. 1 − − Examples worked in class (if you did not write down the details in class you can also find • them in e.g. Pam B’s online notes): x x d x2+sin x d 3/x d 2 2x x ln x d e + e− (e ); (e− ); (x e e ); (ln( )). dx dx dx − dx 2

2 x x Sketch (a) y = xe− and (b) y = e− 2 . • x x x x Solution. (a) y′ = e− xe− = (1 x)e− by the product rule. Also y′′ = e− x x x − − x − − e− + xe− = (x 2)e− . Critical points: y′ = (1 x)e− = 0 occur when x = 1, and here 1 1 − −x y = e− = e . When x > 1 we have y′ = (1 x)e− < 0 so f is decreasing. When x < 1 x − x we have y′ = (1 x)e− > 0 so f is increasing. We have y′′ = (x 2)e− = 0 when x = 2. This is an inflection− point (Calc I). By L’Hospitals rule − x d x 1 lim = lim dx = lim = 0, x ex x d x x ex →∞ →∞ dx e →∞ and x x x lim xe− = ( lim x)( lim e− ) = ( lim x)( lim e )= . x x x x x →−∞ →−∞ →−∞ →−∞ →∞ −∞ Graph drawn in class. x2 (b) y = e− 2 is similar, but notice that this function is EVEN (f(x) = f( x)). It is x2 − always positive. When x = 0 then y = 1. We have y′ = e− 2 ( x) < 0 if x > 0. So y is 2 2 · − 2 x x 2 x 2 decreasing for x> 0. We have y = e− 2 x( x)e− 2 = (x 1)e− 2 = 0 when x = 1, ′′ − − − − x2 or x = 1. These are points of inflection point (Calc I). We have limx e− 2 = 0, since x2 ± u →∞ 2 as x , and e 0 as u . − Graph→ −∞ drawn in→∞ class (the bell→ ).→ −∞

2 Example. Find the equation of the normal line to y = ln ex at the point ( 2, 4). • − Solution. This is y = x2, which has derivative 2x = 2( 2) = 4. So the normal line has slope 1/( 4) = 1/4. The normal line is y 4= 1 (x (− 2)) =−1 x + 1 ; or y = 1 x + 9 . − − − 4 − − 4 2 4 2 7.5: Powers and general . Recall ab = eb ln a for a> 0. • Here b is called the exponent or power, and a is called the base. • 0 0 1 ln a 2 2 ln a ln(a a) 1 ln a ln(1/a) So a = e = 1; a = e = a; a = e = e · = a a. Also, a− = e− = e = • 1/a. · Similarly, an is the product of a with itself n times, if n = 1, 2, 3, , and is the product of • 1 with itself n times, if n = 1, 2, 3, . · · · a | | − − − · · · 1 a 2 is the square root of a, since • 1 2 1 ln a 2 2 1 ln a ln a (a 2 ) = (e 2 ) = e · 2 = e = a.

1 Similarly, a n is the nth root of a if n = 1, 2, 3, . • · · · m Similarly, a n = ( √n a)m if m and n are positive . • Rules for powers: • b b+c b c b c a b 1 b c bc 1 c 1 a = a a , a − = , a− = , (a ) = a , ( ) = . ac ab ab abc

These are proved in the text. We prove the first: • ab+c = e(b+c) ln a = eb ln a+c ln a = eb ln aec ln a = abac.

π2 2 √5 Example. = π − . • π√5

• b d b d b ln x b ln x b bx b 1 (x )= (e )= e = = bx − , b = 0. dx dx x x 6

b 1 1 b So x − dx = b x + C. • R d (u(x)b)= b u(x)b 1 u (x). • dx − ′ Example. d ((x2 + 3)√2)= √2(x2 + 3)√2 12x = 2√2x(x2 + 3)√2 1. • dx − − 1 π 1 π 1 1 1 Example. 0 x dx = π+1 x − ]0 = π+1 . • R Picture drawn in class of y = xb. • The general exponential function y = ax, for a> 0: • d (ax)= d (ex ln a)= ex ln a ln a = ax ln a. • dx dx x 1 x So a dx = ln a a + C. • R d (au(x))= au(x) (ln a) u (x). • dx ′ d2 x d x 2 x 2 (a )= (a ln a) = (ln a) a . • dx dx Picture drawn in class of y = ax. • d 1 d 1 x 1 x 1 1 1 Example. ( 2 )= (( 2 ) ) = ( 2 ) ln( 2 )= 2 ln( 2 ). • dx π x dx π π · π π x π 2 Find x5 x dx. • − Solution:R Let u = x2 so du = 2xdx and the integral becomes − − 1 u 1 1 u 1 x2 5 du = 5 + C = 5− + C. −2 Z −2 ln 5 −2 ln 5

Find d ((ln x)ln x). • dx Solution: Lets do this by logarithmic differentiation (I did it by two different methods in class). If y = (ln x)ln x then ln y = ln((ln x)ln x) = (ln x) ln(ln x). Differentiating both sides: 1 1 1 1 y′ = ln(ln x)+ln x = (ln(ln x) + 1). y x x ln x x ln x 1 So y′ = (ln x) x (ln(ln x) + 1). Examples worked in class (if you did not write down the details in class you can also find • them in e.g. Pam B’s online notes):

2 2 d (53x ) , (Q48) d ((cos x)x +1) the last by logarithmic differentiation. • dx dx ln x General logarithms: Recall loga x = ln a for a > 0, a = 1,x > 0. We call a the base of the • logarithm. 6 log 1 = 0 and log a = ln a = 1. • a a ln a x ln x loga x is the inverse function to a . To prove this, note that if y = loga x = ln a then • ln x y y ln a ln a ln x a = e = e ln a = e = x. Thus the graph of log x is the mirror image of the graph of ax about the line y = x. Picture • a drawn in class of y = loga x. Properties of log : • a b 1 log (bc) = log b + log c , log = log b log c , log (bc)= c log b , log = log b. a a a a c a − a a a a b − a

x Example. Simplify log 1 (2 ). • 4 ln 2 ln 2 ln 2 x Solution: = x log 1 2= x 1 = x 2 = x = . ln ln 2 2ln2 2 4 4 − − − d (log x)= d ( ln x )= 1 . • dx a dx ln a x ln a d (log u(x)) = 1 u (x)= u′(x) . • dx a u(x) ln a ′ u(x) ln a d 2 6x Example. (log (3x + 1)) = 2 . • dx 2 (3x +1) ln 2 Examples worked in class (if you did not write down the details in class you can also find • d them in e.g. Pam B’s online notes): dx (log10 cos x). And: What is the equation of the tangent line to y = 2 3sin x at x = 0. [Answer: y 1= (ln 3)x.] − − − 7.6: Exponential growth and decay

Exponential growth and decay models are used in: Population Growth, Radioactive Decay, Investments, Mixing problems, Newtons Law of Cooling, etc. In nature, certain quantities grow according to, or proportional to, its size. • For example, compound interest: money in an interest bearing account. • Or population growth: if p(t) is the size of the population at time t, then the rate at which • p(t) increases is proportional to the number p(t). In math, ‘proportional to’ means a ‘constant multiple of’. So the last bullet, in math, reads • dp = Kp(t), dt for a constant K. The latter is a ‘differential equation’, which we will solve. Indeed the solution of the • differential equation y′ = Ky is: y = C eKt for a constant C. Here is the proof: First note that y = CeKt does satisfy the differential Kt equation y′ = Ky, since y′ = CKe = Ky. Next, if g(t) is any function which satisfies the Kt differential equation y′ = Ky, then g′ = Kg. Let h(t)= e− g(t). Then Kt Kt Kt Kt h′(t)= Ke− g(t)+ e− g′(t)= Ke− g(t)+ e− Kg(t) = 0. − − Kt Kt So h(t) is a constant, C say. So e− g(t)= C and hence g(t)= Ce . Summarizing: in nature, any quantity y which grows according to its size, must be of the • form y = CeKt. So its graph is basically the exponential graph y = ax sketched in the last section [Picture]. Growing: K > 0. Shrinking/decaying: K < 0. • Example. Solve the differential equation y′ = 2y,y(0) = 3. • − 2t 0 Solution. Here K = 2, so the solution is y = Ce− . We have 3 = y(0) = Ce− = C. 2t − So y = 3e− .

In population growth we usually write C as P0, this is p(0) the initial population (at time • 0). Example. The population of a town grows unhampered. Suppose that the population in • 1970 was 197, and the population in 1990 was 412. What will the population be in 2060. Solution. Let t be the number of years after 1970 in decades. Let p(t) be the population Kt after the tth decade after 1970. So p(0) = 197,p(2) = 412. We know p(t) = P0e . So 0 2K 2K 412 1 412 197 = P0e = P0, and 412 = P0e = 197e . Thus 2K = ln 197 so K = 2 ln 197 0.369. 9K 3.321 ≈ We want to know p(9) = P0e = 197e = 5450. So the population will be 5450 in 2060. Similarly, the compound interest example in the second bullet of this section. If A(t) is • the amount of dollars we have in the account at time t, invested at an annual (compound) interest rate of r%, compounded continually, satisfies the same differential equation

dA r = A(t), dt 100

rt see p. 419 in text for a proof. So it has solution A(t) = Ce 100 . We usually write C as A0, this is A(0), the initial investment. Example. I have $6000. I need $10000 in 5 years. At what interest rate must I invest my • money to achieve my financial goal. rt Solution. Interest rate = r%, and A(t) = A0e 100 as above. We have A(0) = 6000 = 5r 5r 5r 0 100 100 100 10000 5 A0e = A0. We need A(5) = 10000 = A0e = 6000e . So e = 6000 = 3 , and 5r 5 100 5 5 100 = ln 3 . Hence r = 5 ln 3 = 20 ln 3 = 10.217%. The doubling time DT is the time it takes my population to double. If y = CeKt then C • is the initial amount (at time t = 0), and we clearly have at the doubling time y(DT ) = K(DT ) ln 2 ln 2 Ce = 2C. So K(DT ) = ln 2, or DT = K . If you like, K = DT . The half life HL is the time it takes a decaying item/population to halve in size. If y = CeKt • then we clearly have at the halving time that C/2= CeK(HL). So K(HL)=ln 1 = ln 2. 2 − Hence HL = ln 2 , or if you prefer K = ln 2 . − K − HL Example. Suppose a culture of bacteria is growing in such a way that the change in the • number of bacteria is proportional to the number present. The number of bacteria double every 200 minutes and there are currently 5000 bacteria in the culture. How many bacteria were present 2 hours ago? ln 2 ln 2 Solution. Here DT = 200 = K so K = 200 . If C is the number of bacteria present 2 t ln 2 hours ago, then at t minutes from then the population is p = CeKt = Ce 200 . So now there are 3 120 ln 2 3ln2 5 3 5000 = Ce 200 = Ce 5 = Celn 2 = C2 5 .

5000 Hence C = 3 = 3298.77. 2 5 Example. After 3 days a sample of radon-222 decayed to 58 % of its original amount. What • is the half-life of radon-222? Solution. As usual y = CeKt. If we started with a batch of C = 100, after 3 days we have 3K 58 1 58 = 100e . So3K = ln 100 = ln 0.58, and K = 3 ln 0.58. So the half-life of radon-222 is ln 2 ln 2 HL = K = 1 3.817 days. − − 3 ln 0.58 ≈ ln 3 K(T T ) K(T T ) Tripling time: TT = K . [To see this notice that 3C = Ce so e = 3, so • ln 3 K(T T ) = ln 3. Hence TT = K .] Example. Begin with 200 bacteria in a culture, when t = 4 we have 300. Find how long it • will take for the number of bacteria to triple in size. Solution. As usual y = CeKt = 200eKt, and 300 = 200e4K so e4K = 3/2 and so 1 3 ln 3 K = 4 ln 2 . Thus tripling time TT = 3 4. ln 2 · 7.7: Inverse .

. [Picture of the graph of drawn in class.] It is clear from the graph of sin that sin • π π has no inverse (it is not 1-1). However on [ 2 , 2 ] it is 1-1, so has an inverse there that we 1 − π π call arcsin, or sin− , a function from [ 1, 1] to [ 2 , 2 ] (since [ 1, 1] is the range of sin)). [Picture of the graph of arcsin drawn in− class.] − −

1 π π sin− y = x is the same as saying that sin x = y, and x . • − 2 ≤ ≤ 2 1 π π sin− (sin x)= x if 2 x 2 . • 1 − ≤ ≤ sin(sin− (y)) = y if 1 y 1. − ≤ ≤ Example. If y = arcsin x find cos y. • Solution. Method 1: cos y = 1 sin2 y = √1 x2 q − − x opp Method 2: Draw a right . Since sin y = x = 1 = hyp , we draw a right triangle with hypotenuse 1 and opp = x. By Pythagoras the adj = √1 x2, so cos x = adj = √1 x2. − hyp − Example. Find cos(2 arcsin( 3 )). • 5 Solution. We use the double angle formula cos(2θ) = cos2 θ sin2 θ = 1 2sin2 θ = 2 cos2 θ 1. So cos(2 arcsin( 3 )) = 1 2sin2(arcsin( 3 )) = 1 2( 3 )−2 = 7 . − − 5 − 5 − 5 25 7.7–7.8 Continued in handwritten form.