15.Applications of Differentiation

15: APPLICATIONS OF DIFFERENTIATION

Stationary Points zero. Therefore, we first obtain the gradient function, dy Stationary points are points on a graph where the then, equate it to zero. Next, we solve for x. gradient is zero. A stationary point can be any one of dx , a maximum, minimum or a point of inflexion. These The values of x obtained will be the x coordinates of are illustrated below. the stationary points. We can substitute these values of x into the equation of the curve to determine the corresponding y coordinates of the stationary points.

Example 1 Find the stationary points on the graph of y = 4x3 − 3x2 − 6x .

Solution yx=43632-- x x dy =12xx2 -- 6 6 dx dy Letting = 0 Let us examine more closely the maximum and dx minimum points on a curve. We notice that a tangent 12x2 − 6x − 6 = 0 to the curve, drawn at a maximum point, is a 2 horizontal line and hence its gradient is zero. This is 2x − x −1 = 0 also true for the tangent drawn at a minimum point (2x +1)(x −1) = 0 and at a point of inflexion (at a higher level of 1 xx=- ,1= mathematics, we would discover inflexion points 2 where the tangent is not horizontal). Just before and 3 2 1 ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ 3 just after a maximum or a minimum point, the At x = − , y = 4 − − 3 − − 6 − = 1 2 ⎝⎜ 2⎠⎟ ⎝⎜ 2⎠⎟ ⎝⎜ 2⎠⎟ 4 gradient of a curve changes to the opposite sign. In 3 2 the case of a point of inflexion, though, the gradient At x = 1, y = 4(1) − 3(1) − 6(1) = −5 does not change sign. Hence, there are two stationary points on the curve

with coordinates, (−½, 1¾) and (1, −5).

+ +

+ The nature of the stationary points

+ + To determine whether a point is a maximum or a

- + minimum point or inflexion point, we must examine what happens to the gradient of the curve in the www.faspassmaths.com vicinity of these points. In order to do so, we need not sketch the curve. We can use differential calculus to If we know the equation of a curve, we can determine investigate what happens. There are two methods for the stationary points. This is because at no other carrying out this step and both methods will be point on the graph is the gradient zero, except at a described. stationary point. The converse is also true. That is, if the gradient at any point is zero, then the point is a Method 1: Use of the first derivative stationary point. We can use the first derivative or the gradient To obtain the coordinates of the stationary point, we function to determine how the gradient changes in the need to find out where on the curve the gradient is vicinity of the stationary point. At any stationary

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 1 of 9 point, the gradient is zero. By drawing tangents ‘just In using the first differential to determine the nature before’ and ‘just after’ the stationary point we of a stationary point, we use the procedure outlined observe that the changes in gradient differ in sign below: depending on the type of stationary point. 1. Choose a value of x, ‘just smaller’ and one that is Figure A: Maximum Point ‘just larger’ than the x value of the stationary point. For example, if there is a stationary point at x = 2, we may choose x = 1.9 and x = 2.1.

2. Calculate dy for each of these two values of x. dx 3. By observing the sign change of dy , we can dx deduce the nature of the stationary point.

Figure B; Minimum Point 4. If the gradient changes from positive to negative before and after the stationary point, the point is a maximum.

A change from positive to zero to negative indicates 5. If the gradient changes from negative to positive that the point is a maximum, this is illustrated in before and after the stationary point, the point is a Figure A. On the other hand, a change from negative minimum. to zero to positive indicates that the point is a minimum as illustrated by the Figure B.

When there is no change in the sign, it indicates an inflexion point. As shown in Figure C, these points, the signs of the gradient can either be both positive or both negative, as shown below. This is because they change from positive, to zero at the point, and then to 6. If there is no change in the sign of the gradient, positive again OR from negative, to zero at the point, the stationary point is a point of inflexion. and then to negative again. Example 2 Figure C: Points of inflexion www.faspassmaths.comA curve whose equation is y = 4x3 − 3x2 − 6x has stationary points at (−0.5, 1.75) and (1, −5). Determine the nature of these points.

Solution Both Positive Both Negative We apply the procedure outlined above, noting that

For the stationary point, (−0.5, 1.75), the x-coordinate is −0.5. 1. We can choose x = −0.6 as the value ‘just

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 2 of 9 smaller’ than −0.5, and we can choose x = −0.4 as the value ‘just larger’ than −0.5. dy 2. Calculating for each of these two values, we dx have

æödy 2 ç÷ =12(- 0.6) --- 6( 0.6) 6= 1.92 > 0 èødx x=-0.6

æödy 2 Minimum point ç÷ =12(- 0.4) --- 6( 0.4) 6=- 1.68< 0 èødx x=-0.4 The tangent at A has a negative slope, the tangent B has a zero slope and the tangent at C has a 3. The gradient changes from positive to zero to positive slope. As we move from left to right, the negative. Hence, (−0.5, 1.75) is a maximum slope of the tangent increases, so the second point. derivative is positive.

Repeating these steps for the point (1, −5), we can choose x = 0.9 as the value ‘just smaller’ than 1, and If the second derivative is negative then the gradient choose x = 1.1 as the value ‘just larger’ than 1. decreases as we move along the curve with the gradient decreasing from positive just before the point to zero at the point and to negative just after the æödy 2 ç÷ =12(0.9)-- 6(0.9) 6=- 1.68< 0 point. This pattern indicates that the point is a dx èøx=0.9 maximum point. æödy 2 ç÷ =12(1.1)-- 6(1.1) 6=> 1.92 0 èødx x=1.1 The gradient changes from negative to zero to positive. Hence, (1,−5) is a minimum point.

(Remember, if the gradient ever changes from negative, to zero, at the point, then to negative again OR from positive, to zero at the point, then to positive again, then the point would have been an inflexion point.)

Method 2: Use of the second derivative To determine the nature of the stationary points, we can obtain the second derivative and examine its sign The tangent at A has a positive slope, the tangent at the turning point. The second differential is also a at B has a zero slope and the tangent at C has a function and obtained by differentiating the first negative slope. As we move from left to right, the derivative. It represents the derivative of the gradient slope of the tangent decreases, so the second function and itwww.faspassmaths.com specifies the rate of change of the derivative is negative. gradient function.

While this method is quite efficient, it can become If the second derivative is positive then the gradient quite a task to find the second derivative when the increases as you move along the curve with the first derivative is an extremely complex function. In gradient increasing from negative just before the such a case we use Method 1 to determine the nature point to zero at the point and to positive just after the of the stationary points. point. This pattern indicates that the point is a minimum point.

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 3 of 9 Alternative Notation For a decreasing function, the rate at which y changes So far, we have used to refer to the first with respect to x is negative. �� differential. We can also refer to the first differential < � �� as � (�). Figure4 shows the function �(�) = � − 12�. The intervals on which the function are increasing and The second differential is written as or as � (�). decreasing intervals are indicated.

Example 3 Figure 4: Increasing and decreasing functions A curve whose equation is y = 4x3 − 3x2 − 6x has stationary points at (−0.5, 1.75) and (1, −5). Determine the nature of these points.

Solution We solved this problem by using method 1 already. Now we will use the second differential to determine the nature of the stationary points. yx=43632-- x x dy =12xx2 -- 6 6 dx dy2 =24x - 6 dx2 At (−0.5, 1.75), dy2 =24(--1 ) 6 dx2 2 =-18< 0 Þ Maximum The function �(�) = � − 12� is increasing in the interval (−∞,−2)∪(2,∞) and is decreasing in the At (1, −5), interval (−2,2). dy2 =24(1)- 6 In summary, for any function, �(�) 2 dx If > 0 in an interval, then the function is =>18 0 increasing in the interval. Þ Minimum If < 0 in an interval, then the function is Increasing and decreasing functions decreasing in the interval. A sketch of a curve gives us not just the coordinates of the stationary but we can also observe intervals for Maxima and minima which the curve is increasing or decreasing. We say Differential calculus can be applied to many that a function is increasing in an interval if the situations in real life where it is necessary to function valueswww.faspassmaths.com increase as the input values increase maximise or minimize functions. within that interval. Similarly, a function is decreasing in an interval if the function values Example 5 decrease as the input values increase over that A farmer wishes to fence off a rectangular sheep interval. pen and uses all of 100 m of fencing. An adjoining For an increasing function, the rate at which y hedge is to be used as one side of the pen. changes with respect to x is positive, or Determine the maximum area of the pen. �� > � ��

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 4 of 9 Solution Example 6 We create a diagram based on the data, for a A rectangular block has base dimensions 2x cm clearer understanding of the information. by 3x cm. Given that its volume is 1 800 cm3, prove that the total surface area, A cm2, is 3000 Ax=+12 2 . x Calculate the value of x for which A has a stationary value. Find this value of A and determine whether A is a maximum or a minimum.

Solution

Let the width of the pen be x m Since there are three sides to this fence, and the perimeter is 100 m \The length of the pen will be: =100 -()xx+ =100- 2x The area of the pen, Ax=()100- 2 x= 100 x- 2 x2 A is now a function of x For a stationary value of A, we must equate Let the height of the block be h cm. Hence, dA = 0 23xxh´´= 1800 dx 2 dA 61800xh= =100- 2() 2xx= 100- 4 1800 300 dx h == 6xx22 Let 100- 4x = 0 \x =25

In our example, we know that a stationary value of The total surface area, A cm2 = Area of the front A occurs at x = 25 . and back faces + Area of left and right faces + We need to find the second derivative to determine Area of the top and base. or confirm its nature. Axhxhxx=22()()()´+ 23´+ 22´ 3 dA2 =-4 (which is negative) 2 dx2 A=++4 hx 6 hx 12 x 2 \A is a maximum value when x = 25 . A=+10 hx 12 x 300 Substitute h = , we get x2

æö100 223000 Axxx=+=+10ç÷2 12 12 www.faspassmaths.comèøx x

At a stationary value of A, = 0 Axx=+3000-12 12

dA -223000 The maximum area of the pen is: =3000()- 1xx+= 24 -+12 x dx x 25´ 50 = 1250 m2.

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 5 of 9 dA -3000 We also use the Greek symbol δ, (common delta), to =+=24x 0 dx x2 mean a small change in the value of a variable. So, if 3000 x changes from a value of x1, to a new slightly = 24x x2 different value, x2, then we say 24x3 = 3000 δx = x2 – x1 3 x = 125 Small changes and gradient x ==3 125 5 We recall that the gradient of a curve at a point is the When x = 5 , the value of æödy gradient of the tangent at the point or ç÷ . We 2 3000 èødx A =+=+=12() 5 12(25) 600 900 cm2 xa= 5 also recall that gradient is the change of y with Dy d y respect to x or or . dA =-3000xx-2 + 24 Dx d x dx We will now illustrate how we can use calculus to dA2 =+6000x-3 24 calculate the change in one variable when we know dx2 the small change in another related variable. 6000 Consider the points Pxy(, ) and Sx(,+D xy+D y ) on =+24 3 x the curve shown below. When x = 5 , Dy The gradient of a chord, PS = . dA2 6000 =+24 > 0 Dx dx23()5 Note that the point S is some distance away from P. \ A is a minimum when x = 5 . But since we are interested in very small changes, let us move S closer and closer to P, such that it almost

touches P. When this happens, the gradient of the Small Changes chord, PS is approximately equal to the gradient of the tangent PT. If two variables are related then we can use calculus to find the approximate change in one variable if there is a small change in the other. It is important to emphasise that the method used is valid only for small changes and also that the result is only approximate. T If two variables are related, then a change in x will S produce a change in y. So too, a change in y will also y produce a change in x. +Dy A change could either be an increase (indicated by a Dy positive sign) or a decrease (indicated by a negative P y sign). www.faspassmaths.comD Notation x The change in any variable is defined as the x x difference between the final value and the initial +Dx value of the variable. We use the Greek symbol, D, Dxdy (Capital Delta) to denote a change in the value of a Hence, » at the point P. This relationship is variable. If, say, x changes from an initial value of Dydx x1, to a final value, x2, then, we write used to calculate Dy when given Dx or to calculate Dx when given . Dx = x2 – x1 Dy

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 6 of 9 Example 8 For infinitely small changes The radius, r cm, of a circle changes from 5 cm to Dydy dy 5.1 cm. Find the approximate change in the area A » or D»yx D 2 Dxdx dx cm , leaving the result in terms of π. State whether this change is an increase or decrease. If the variables are different from x and from y, the principle still holds and we may adjust them Solution accordingly. We recall the formula for the area of a circle, 2 Ar= p Since A is a function of r then, If V is a function of r, then we can calculate the dA change in V (Δ�), given the change in r (Δ�) using D»Ar ´D, where Dr = a small change in r dr dV D»Vr D. and DA is the corresponding change in A, dr Dr =5.1- 5= 0.1 If A is a function of r, then we can calculate the dA = 2p r change in A (Δ�), given the change in r (Δ�) dr dA D»Ar D. \DAr »2p ´ 0.1 dr \DA »250.1p () ´ (Substitute r = 5 , the original Example 7 value) D»A p cm2 > 0, this indicates there is an The variables x and y are related by y = . If x increase in A. changes from 8 to 7.9, find (i) the approximate Example 9 change in y, stating whether it is an increase or a Given, yx=3452 - x+, Calculate decrease (ii) the new value of y (iii) the percentage dy change in y. a. when x =-2 dx Solution b. the approximate change in y when x 1 increases from – 2 to – 1.97. (i) y = x Dx =7.9- 8=- 0.1 Solution 2 dy dy -2 1 a. , =-x =- yx=345- x+ =64x - 2 dx dx x dy When x =-2 D»yx D dx dy =6(-- 2) 4=-- 12 4=- 16 -1 -3 dx D»y 2 () -0.1 » 1.6 ´ 10 [Substitute the original ()8 dy b. If y is a function of x then, D»yx ´Dwhere value of x] dx Since the value is positive, there is an increase. Dx = small change in x and Dy = corresponding www.faspassmaths.comchange in y (ii) The new value of y is the original value of y D»y {}62() - - 40.03 ´ added to the corresponding change in y. D»-y 0.48 When x = 8 , y = = 0.125 -3 New value of y ≈ 0.125 + 1.6 × 10 ≈ 0.1266 Since Dy <0 this indicates a decrease.

(iii) The percentage change is 1.6´ 10-3 »´100% ≈ 1.28 % 0.125

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 7 of 9 Differentiation as a Rate of Change rule’, to find the corresponding rate of change of one Rate of change refers to the rate at which any variable, given the change in the other variable. variable, say x, changes with respect to time, t. The dx Example 10 gradient function, , is therefore a rate of change. dt The radius, r cm, of a circle is increasing at the rate of 0.2 cms-1. Calculate the corresponding rate of The symbol, ��, refers to the change in the variable change of the area, A cm2, at the instant when the that is studied and the symboldt refers to the change radius is 5 cm. Ar= p 2 in time. () dV The rate of change of the volume, V, …is . dt Solution We wish to find the rate of change of the area, , dA The rate of change of the area, A. …is . given the rate of change of the radius, . dt 2-dr 1 Ar==p ++0.2 cms( veÞ increase) Positive and negative rates of change dt dA = 2p r A positive sign denotes an increasing rate. dr Sometimes the word increasing or the word dA dA dr 2-1 decreasing may not be used. In these cases, we are =´=2pp() 5´ 0.2= 2 cm s( + veÞ increase) dt dr dt expected to deduce whether the rate is increasing or It is sometimes possible to obtain the rate of change decreasing by the words of the question. of an area, A cm2, to be say, −6 cm2s-1. The negative For example, sign indicates a decreasing rate. However, if we were A length, l, of elastic is stretched at the rate of 0.5 asked for the rate of decrease of the area, the answer cms-1… will be stated as 6 cm2s-1. dl Stretched Þ increasing length \=+0.5 cms-1 dt Example 11 In this case, the unit of measure indicated that length, The side of a metal cube, x cm, increases from 9.9 l, is the variable under study. cm at a constant rate of 0.005 cms-1 as it is heated. Find the rate at which the volume is increasing A fire is spreading at the rate of 5 m2s-1… when t = 20 seconds.

SpreadingÞ increasing rateü dA 2-1 ý\=+5ms (Let Solution 2-1 dt Units of 5 m sÞ Area þ The volume of the cube increases as it is heated. A be area) Side increases at the rate of 0.005 cms-1. dV A negative sign denotes a decreasing rate. Required to calculate: at t = 20 . dt Water is leaking from a tank at the rate of Let volume of the cube be V cm3. 4 cm3s-1… dV 3 dV 2 Leaking Þ decreasing rate \=-4cms3-1. \Vx= and by differentiation,= 3x dt dx www.faspassmaths.comdV dV dx dV The units indicated that volume, V, was the variable. 2 =´ =30.005x ´ dt dx dt dx Related rates of change It is common to have a situation involving two When t = 20 x =+9.9 20() 0.005 = 10 changing quantities and we would likely be required because after 20 seconds the value of x is the to find the rate at which one is changing, given the original value plus the increase after the duration rate at which the other is changing. of the 20 s.

In such a situation, we need to have an equation that dV 2 Hence, =310() ´ 0.005= 1.5 cm3s-1 connects the two quantities. Then we use the ‘chain dt

Copyright © 2019. Some Rights Reserved. www.faspassmaths.com Pg 8 of 9 Example 12 A spherical balloon is filled with air at the constant rate of 200 cm3s-1. Calculate the rate at which the radius is increasing, when the radius = 10 cm.

Solution Let V = Volume and r = radius of the balloon. 4 Vr= p 3 3 dV =+200 cm3-1 s (increasing rate) dt dr dV dr =´ dt dt dV dV 4 ==pp()34rr22 dr 3 dr 11 == dV dV 4p r 2 dr When � = 10 cm, dr 11 =200´ =cms-1 dt 4p () 10 2 2p