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A guide for teachers - Years 11 and 12

1 2 3 4 5 6 7 8 9 1 0 1 1 1

2Supporting Australian Mathematics ProjectCalculus: Module 12

Applications of di?erentiation

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Editor: Dr Jane Pitkethly, La Trobe University

Illustrations and web design: Catherine Tan, Michael Shaw Applications of dierentiation - A guide for teachers (Years 11-12)

Principal author:? Dr Michael Evans, AMSI

Peter Brown, University of NSW

Associate Professor David Hunt, University of NSW

Dr Daniel Mathews, Monash University

Assumed knowledge. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4 Motivation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4 Content. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5 Graph sketching. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5 More graph sketching. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .18 Maxima and minima problems. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .26 Related rates of change. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .37 Links forward. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .41 Critical points. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .41 Finding gradients on a parametric curve. . . . . . . . . . . . . . . . . . . . . . . .42 History and applications. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .45 History. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .45 Applications. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .46 Answers to exercises. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .49

A guide for teachers - Years 11 and 12

•{5}

In this module, we consider three topics:

•graph sketching •maxima and minima problems •related rates. We will mainly focus on nicely behaved functions which are differentiable at each point of their domains. Some of the examples are very straightforward, while others are more difficult and require technical skills to arrive at a solution.

Content

Graph sketching

Increasing and decreasing functions

Letfbe some function defined on an interval.

Definition

The functionfisincreasingover this interval if, for all pointsx1andx2in the interval, x

1·x2AE)f(x1)·f(x2).

This means that the value of the function at a larger number is greater than or equal to the value of the function at a smaller number. The graph on the left shows a differentiable function. The graph on the right shows a piecewise-defined continuous function. Both these functions are increasing.y x 0y x

0Examples of increasing functions.

{6}•Applications of differentiation

Definition

The functionfisdecreasingover this interval if, for all pointsx1andx2in the interval, x

1·x2AE)f(x1)¸f(x2).

The following graph shows an example of a decreasing function.y x0Example of a decreasing function. Note that a function that is constant on the interval is both increasing and decreasing overthisinterval. Ifwewanttoexcludesuchcases,thenweomittheequalitycomponent in our definition, and we add the wordstrictly: •A function isstrictly increasingifx1Çx2impliesf(x1)Çf(x2). •A function isstrictly decreasingifx1Çx2impliesf(x1)Èf(x2). We will use the following results. These results refer to intervals where the function is differentiable. Issues such as endpoints have to be treated separately. •Iff0(x)È0 for allxin the interval, then the functionfis strictly increasing. •Iff0(x)Ç0 for allxin the interval, then the functionfis strictly decreasing. •Iff0(x)AE0 for allxin the interval, then the functionfis constant.

Stationary points

Definitions

Letfbe a differentiable function.

•Astationary pointoffis a numberxsuch thatf0(x)AE0. •The pointcis amaximum pointof the functionfif and only iff(c)¸f(x), for allx in the domain off. The valuef(c) of the function atcis called themaximum value of the function. •The pointcis aminimumpointof the functionfif and only iff(c)·f(x), for allxin the domain off. The valuef(c) of the function atcis called theminimum valueof the function.

A guide for teachers - Years 11 and 12

•{7}

Local maxima and minima

In the following diagram, the pointalooks like a maximum provided we stay close to it, and the pointblooks like a minimum provided we stay close to it.y x0 a bDefinitions •Thepointcisalocalmaximumpointofthefunctionfifthereexistsaninterval(a,b) withc2(a,b) such thatf(c)¸f(x), for allx2(a,b). •The pointcis alocalminimumpointof the functionfif there exists an interval (a,b) withc2(a,b) such thatf(c)·f(x), for allx2(a,b). These are sometimes calledrelative maximumandrelative minimumpoints. Local maxima and minima are often referred to asturning points. The following diagram shows the graph ofyAEf(x), wherefis a differentiable function. It appears from the diagram that the tangents to the graph at the points which are local maxima or minima are horizontal. That is, at a local maximum or minimum pointc, we havef0(c)AE0, and hence each local maximum or minimum point is a stationary point. y x0 a bThe result appears graphically obvious, but we will present a formal proof in the case of a local maximum. {8}•Applications of differentiation

Theorem

Letfbe a differentiable function. Ifcis a local maximum point, thenf0(c)AE0. Proof Consider the interval (c¡±,cű), with±È0 chosen so thatf(c)¸f(x) for all x2(c¡±,cű). For all positivehsuch that 0ÇhDZ, we havef(c)¸f(cÅh) and therefore f(cÅh)¡f(c)h

·0.

Hence,

f

0(c)AElimh!0Åf(cÅh)¡f(c)h

·0. (1)

For all negativehsuch that¡±ÇhÇ0, we havef(c)¸f(cÅh) and therefore f(cÅh)¡f(c)h

¸0.

Hence,

f

0(c)AElimh!0¡f(cÅh)¡f(c)h

¸0. (2)

From (1) and (2), it follows thatf0(c)AE0.The first derivative test for local maxima and minima The derivative of the function can be used to determine when a local maximum or local minimum occurs.

Theorem(First derivative test)

Letfbe a differentiable function. Suppose thatcis a stationary point, that is,f0(c)AE0. aIf there exists±È0such thatf0(x)È0, for allx2(c¡±,c), andf0(x)Ç0, for all x2(c,cű), thencis a local maximum point. bIf there exists±È0such thatf0(x)Ç0, for allx2(c¡±,c), andf0(x)È0, for all x2(c,cű), thencis a local minimum point. Proof aThe function is increasing on the interval (c¡±,c), and decreasing on the in- terval (c,cű). Hence,f(c)¸f(x) for allx2(c¡±,cű). bThe function is decreasing on the interval (c¡±,c), and increasing on the in- terval (c,cű). Hence,f(c)·f(x) for allx2(c¡±,cű).

A guide for teachers - Years 11 and 12

•{9} In simple language, the first derivative test says: •Iff0(c)AE0 withf0(x)È0 immediately to the left ofcandf0(x)Ç0 immediately to the right ofc, thencis a local maximum point.y x0 ( c , f ( c ))We can also illustrate this with a gradient diagram.

Value ofxc

Sign off0(x)Å0¡

Slope of graphyAEf(x)-

•Iff0(c)AE0 withf0(x)Ç0 immediately to the left ofcandf0(x)È0 immediately to the right ofc, thencis a local minimum point. y x 0 ( c , f ( c ))We can also illustrate this with a gradient diagram.

Value ofxc

Sign off0(x)¡0Å

Slope of graphyAEf(x)-

{10}•Applications of differentiation There is another important type of stationary point: •Iff0(c)AE0 withf0(x)È0 on both sides ofc, thencis astationary point of inflexion.y x 0 ( c , f ( c ))Here is a gradient diagram for this situation.

Value ofxc

Sign off0(x)Å0Å

Slope of graphyAEf(x)-

•Iff0(c)AE0 withf0(x)Ç0 on both sides ofc, thencis astationary point of inflexion. y x 0 ( c , f ( c ))Here is a gradient diagram for this situation.

Value ofxc

Sign off0(x)¡0¡

Slope of graphyAEf(x)-

A guide for teachers - Years 11 and 12

•{11}Example Find the stationary points off(x)AE3x4Å16x3Å24x2Å3, and determine their nature.

Solution

The derivative offis

f

0(x)AE12x3Å48x2Å48x

AE12x(x2Å4xÅ4)

AE12x(xÅ2)2.

Sof0(x)AE0 impliesxAE0 orxAE¡2.

•IfxÇ¡2, thenf0(x)Ç0. •If¡2ÇxÇ0, thenf0(x)Ç0. •IfxÈ0, thenf0(x)È0. We can represent this in a gradient diagram.Value ofx¡20

Sign off0(x)¡0¡0Å

Slope of graphyAEf(x)--

Hence, there are stationary points atxAE0 andxAE¡2: there is a local minimum atxAE0, and a stationary point of inflexion atxAE¡2. The graph ofyAEf(x) is shown in the following diagram, but not all the features of the graph have been carefully considered at this stage.y x ( ? 2,19) ( 0 ,3)y = 3x⎷ + 16x² + 24x + 3 {12}•Applications of differentiationExercise 1 Assume that the derivative of the functionfis given byf0(x)AE(x¡1)2(x¡3). Find the values ofxwhich are stationary points off, and state their nature.Exercise 2

Find the stationary points off(x)AEx3¡5x2Å3xÅ2, and determine their nature.Use of the second derivative

The second derivative is introduced in the moduleIntroduction to differential calculus. Using functional notation, the second derivative of the functionfis written asf00. Using Leibniz notation, the second derivative is written as d2ydx

2, whereyis a function ofx.

In the moduleMotion in a straight line, it is shown that the acceleration of a particle is the second derivative of its position with respect to time. That is, if the position of the particle at timetis denoted byx(t), then the acceleration of the particle is¨x(t).

Recall that, in kinematics,

xmeansdxdt and¨xmeansd2xdt 2.

Concave up and concave down

Letfbe a function defined on the interval (a,b), and assume thatf0andf00exist at all points in (a,b). We consider the shape of the curveyAEf(x). Iff00(x)È0, for allx2(a,b), then the slope of the curve is increasing in the interval (a,b).

The curve is said to beconcave up.y

x0 y x0 y x0Examples of concave-up curves.

A guide for teachers - Years 11 and 12

•{13} Iff00(x)Ç0,forallx2(a,b),thentheslopeofthecurveisdecreasingintheinterval(a,b).

The curve isconcave down.y

x0Example of a concave-down curve.

Inflexion points

A point where the curve changes from concave up to concave down, or from concave down to concave up, is called apoint of inflexion. In the following diagram, there are points of inflexion atxAEcandxAEd. y x ( c , f ( c ))(d,f(d))Examples of points of inflexion. The graphs ofyAE(x¡2)3Å1 andyAE ¡(x¡2)3Å1 are shown below. The point (2,1) is a point of inflexion for each of these graphs. In fact, the point (2,1) is a stationary point of inflexion for each of these graphs. y x 0y x 0 (2,1)y = (x ? 2) 3 + 1y = ?(x ? 2) 3 + 1 (2,1) {14}•Applications of differentiation The graph ofyAEx3¡3x2Å4xÅ4 is as follows. It has a point of inflexion at (1,6), but this is not a stationary point. In fact, this function has dydx

È0, for allx.y

x y = x ? ⎷ 3 x ² + 4 x + 4 (1,6)Note.Clearly, a necessary condition for a twice-differentiable functionfto have a point of inflexion atxAEcis thatf00(c)AE0. We will see that this isnota sufficient condition, and care must be taken when using it to find an inflexion point. For there to be a point of inflexion, there must be achange of concavity.Example Find the inflexion point of the cubic functionf(x)AEx3¡3x2¡144x.

Solution

We find the first and second derivatives:

f

0(x)AE3x2¡6x¡144 andf00(x)AE6x¡6.

Thusf00(x)AE0 impliesxAE1. ForxÇ1, we havef00(x)Ç0. ForxÈ1, we havef00(x)È0. The curve changes fromconcave downtoconcave upatxAE1. Hence, there is a point of inflexion atxAE1. y x (1, ?

146)It is not hard to show that there is a local maximum atxAE ¡6 and a local minimum at

xAE8.

A guide for teachers - Years 11 and 12

•{15}Exercise 3 Find the inflexion points of the functionf(x)AEx4Å28x3Å10x.The second derivative test We have seen how to use the first derivative test to determine whether a stationary point isalocalmaximum, alocalminimumorneitherofthese. Thesecondderivativeprovides an alternative test.

Theorem(Second derivative test)

Supposefis twice differentiable at a stationary pointc. aIff00(c)Ç0, thenfhas a local maximum atc.y x0 y = f ( x ) ( c , f ( c ))bIff00(c)È0, thenfhas a local minimum atc. y x 0 y = f ( x ) ( c , f ( c ))We will prove the first part of this theorem. The proof of the second part is similar. {16}•Applications of differentiation Proof Assume thatf00(c)Ç0. Thenf0(x) is strictly decreasing over some interval aroundc. Hence, there is a positive number±such thatf0(x) is strictly decreasing on the interval (c¡±,cű).

We now use the first derivative test.

•To the left ofc: For anyh2(c¡±,c), we havef0(h)Èf0(c)AE0. •To the right ofc: For anyk2(c,cű), we havef0(k)Çf0(c)AE0.

It follows from the first derivative test thatcis a local maximum point.Ifcis a stationary point withf00(c)AE0, then we cannot use the second derivative test to

determine if it is a local maximum, a local minimum or neither of these. For example, iff(x)AEx3, thenf0(0)AE0 andf00(0)AE0. In this case, there is a stationary point of inflexion at 0.y 0x y = x ?Iff(x)AEx4, thenf0(0)AE0 andf00(0)AE0. In this case, there is a local minimum at 0. y 0x y = x ?Thus the second derivative cannot be used as a test whenf0(c)AEf00(c)AE0.

A guide for teachers - Years 11 and 12

•{17}Example Locate and describe the stationary points off(x)AEx4¡8x2.

Solution

The first derivative isf0(x)AE4x3¡16x, and the second derivative isf00(x)AE12x2¡16.

We find the stationary points by solvingf0(x)AE0:

4x3¡16xAE0

4x(x2¡4)AE0.

Hence, the stationary points are atxAE0,xAE2 andxAE ¡2. We use the first derivative test by considering a gradient diagram.Value ofx¡202

Sign off0(x)¡0Å0¡0Å

Slope of graphyAEf(x)---

•There is a local minimum atxAE¡2. •There is a local maximum atxAE0. •There is a local minimum atxAE2. We can also check the local minimum and maximum points using the second derivative test, by considering the values of the second derivative: •f00(¡2)AE32È0, so there is a local minimum atxAE¡2 •f00(0)AE¡16Ç0, so there is a local maximum atxAE0 •f00(2)AE32È0, so there is a local minimum atxAE2. The question has not required us to sketch the graph. But we have enough information to complete a good sketch if we also observe that thex-intercepts are 0, 2p2 and¡2p2.y 0x y = x ? ⎷ 8 x ² ⎷

2222

( ⎷ 2, ⎷

16)(2,⎷16)

{18}•Applications of differentiationExercise 4

Locate and describe the stationary points of the graph ofyAEf(x) iff0(x)AEx3(x2¡5).More graph sketching

Sketching graphs of polynomials

The modulePolynomialsintroduces drawing the graphs of polynomials, but does not consider finding the coordinates of the points of inflexion. This is now possible. By going through the following steps, we can find the important features of the graph of a given polynomial:

1thex- andy-intercepts

2stationary points

3where the graph is increasing and where it is decreasing

4local maxima and local minima

5behaviour asxbecomes very large positive and very large negative

6points of inflexion.

Sometimes it is useful to find the coordinates of a few extra points on the graph. Note.These steps are only a guide. Some flexibility in approach to these questions is desirable. In this module, we adhere to these steps as an aid to readability.Example

Sketch the graph ofyAEx3Åx2¡8x¡12.

Solution

1We first find thex-intercepts andy-intercepts.

WhenxAE0, we haveyAE¡12, and so they-intercept is¡12.

WhenyAE0,

0AEx3Åx2¡8x¡12

AE(xÅ2)2(x¡3),

soxAE¡2 orxAE3. Thex-intercepts are 3 and¡2.

A guide for teachers - Years 11 and 12

•{19}2The derivative is dydx

AE3x2Å2x¡8AE(3x¡4)(xÅ2).

So dydx

AE0 impliesxAE¡2 orxAE43

. The coordinates of the stationary points are (¡2,0) and (43 ,¡50027 ).

3The graph ofdydx

againstxis a parabola with a positive coefficient ofx2. We can see: • dydx

AE0 if and only ifxAE¡2 orxAE43

• dydx

È0 if and only ifxÇ¡2 orxÈ43

• dydx

Ç0 if and only if¡2ÇxÇ43

.

So the graph is increasing on (¡1,¡2)[(43

,1) and decreasing on (¡2,43 ).Value ofx¡24 3

Sign of

dydxÅ0¡0Å

Slope of graph--

4From the investigation of the sign ofdydx

, we see that there is a local maximum at xAE¡2, and a local minimum atxAE43 .

5Asx!1,y!1, and asx!¡1,y!¡1.

6The second derivative is

d 2ydx

2AE6xÅ2.

IfxÇ ¡13

, thend2ydx

2Ç0, and ifxÈ ¡13

, thend2ydx

2È0. There is an inflexion point

wherexAE¡13 .y 0 ? 2 ? 1 23x
y = x ⎷ + x ² ? 8 x ? 12 ( ? , ) 1

3?25027

( , ) 4

3?50027

{20}•Applications of differentiationExercise 5 Sketch the graph ofyAE3x4¡44x3Å144x2.Exercise 6 Sketch the graph ofyAE4x3¡18x2Å48x¡290.Sketching graphs of other functions The six steps used to help us sketch graphs of polynomials can be used to help sketch other graphs. In theLinks forwardsection, we look briefly at graphs where there are local maxima or minima at which the function is not differentiable.Example

Sketch the graph ofyAEpx¡x2, forx¸0.

Solution

The first and second derivatives are

dydx AE12 px

¡2x,d2ydx

2AE¡14

x¡32

¡2, forxÈ0.

We follow the six steps for sketching polynomial graphs, which are still useful for this type of graph.

1We first find thex-intercepts:

px¡x2AE0 px(1¡x32 )AE0, which givesxAE0 orxAE1.

2We find the stationary points by solvingdydx

AE0: 12 px

¡2xAE0

1AE4x32

x 32
AE14 xAE³14 ´ 23
.

So there is a stationary point atxAE¡14

¢ 23

AE2¡43

.

A guide for teachers - Years 11 and 12

•{21}3If 0ÇxÇ2¡43 , thendydx

È0, and ifxÈ2¡43

, thendydx

Ç0.

We can summarise this in a gradient diagram.Value ofx2

¡43

Sign of

dydxÅ0¡

Slope of graph-

Hence, there is a local maximum atxAE2¡43

. The correspondingy-value is yAE¡2¡43 ¢ 12

¡¡2¡43

¢2AE2¡23

¡2¡83

AE2¡83

¡22¡1¢AE3£2¡83

.

4We can also determine that there is a local maximum atxAE2¡43

by carrying out the second derivative test. WhenxAE2¡43 , d 2ydx

2AE¡14

¡2¡43

¢¡32

¡2AE¡14

¡22¢¡2AE¡3.

Since dydx

AE0 andd2ydx

2Ç0 atxAE2¡43

, the pointxAE2¡43 is a local maximum.

5Asx!1,y!¡1. We do not considerx!¡1, since the domain of the function is

[0,1).

6For allxÈ0, we have

d 2ydx

2AE¡14

x¡32

¡2Ç0.

Hence, there are no points of inflexion.

We can now sketch the graph.y

0123451

? 1 ? 2 ? 3 ? 4 ? 5 2 x y = ⎷ x ? x ² ( , ) 1 2   3 2   {22}•Applications of differentiationExample

Sketch the graph off(x)AEexÅe¡2x.

Solution

The domain offis all real numbers. The first two derivatives offare f

0(x)AEex¡2e¡2x

f

00(x)AEexÅ4e¡2x.

1For allx, we haveexÈ0 ande¡2xÈ0, and thereforef(x)È0. Hence, there are no

x-intercepts. We havef(0)AE2, and so they-intercept is 2.

2We find the stationary points by solvingf0(x)AE0:

e x¡2e¡2xAE0 e 3xAE2 xAE13 loge2.

So there is a stationary point atxAE13

loge2. The value of the function at this point is f(13 loge2)AE3£2¡23 .

3IfxÇ13

loge2, the gradient is negative and the graph ofyAEf(x) is decreasing.

IfxÈ13

loge2, the gradient is positive and the graph ofyAEf(x) is increasing.

4We havef00(x)AEexÅ4e¡2xÈ0, for allx. So there is a local minimum atxAE13

loge2.

5Asx!1,f(x)!1, and asx!¡1,f(x)!1.

6Sincef00(x)AEexÅ4e¡2xÈ0, for allx, there are no inflexion points.( , )

y 0

510152025

 224x
loge (2)33 2 2/3 y = e x + e  2 x (0,2)

A guide for teachers - Years 11 and 12

•{23}Example

Sketch the graph off(x)AEe¡xÅx.

Solution

The domain offis all real numbers. The first two derivatives aref0(x)AE ¡e¡xÅ1 and f

00(x)AEe¡x.

1We havef(0)AE1, so they-intercept is 1. We will see later that there is a local mini-

mum atxAE0. Indeed, the minimumvalue ofthe function occurs whenxAE0. Hence, there are nox-intercepts.

2Solvingf0(x)AE0 gives

¡e¡xÅ1AE0()e¡xAE1()xAE0.

Hence, there is a stationary point atxAE0.

3IfxÇ0, the gradient is negative and the graph ofyAEf(x) is decreasing.

IfxÈ0, the gradient is positive and the graph ofyAEf(x) is increasing.

4We havef00(x)AEe¡xÈ0, for allx. Hence, there is a local minimum whenxAE0.

5Ifx! ¡1, thenf(x)! 1. Ifx! 1, then the lineyAExis an asymptote, since

e

¡x!0.y

(0,1)5 ? 5 ? 1010
?

5?105010xExercise 7

Sketch the functionf: (0,1)!Rgiven byf(x)AEx2logex.Maxima and minima at end points It may be that the maximum or minimum value of a graph occurs at the endpoints of that graph. Here are two examples to illustrate the need to check endpoints when trying to find the maximum and minimum values of a function. {24}•Applications of differentiationExample Sketch the graph of the functionf: [1,3]!Rgiven byf(x)AEx2(x¡4), and find the maximum and minimum values of the function.

Solution

Iff(x)AE0, thenxAE0 orxAE4, which are outside the domain. Therefore the graph has nox-intercepts. We next find the derivative and the second derivative off: f

0(x)AE3x2¡8xAEx(3x¡8)

f

00(x)AE6x¡8.

Hence, there is a stationary point atxAE83

, and a point of inflexion atxAE43 .

To determine ifxAE83

is a local maximum or minimum, we use the second derivative test: f

00³83

´

AE8È0,

and hence there is a local minimum atxAE83 . We now find the values offat the two endpoints and the local minimum: f

³83

´

AE¡25627

,f(1)AE¡3,f(3)AE¡9. Hence, the maximum value offis¡3, and it occurs whenxAE1. The minimum value offis¡25627 , and it occurs whenxAE83 .y (1, ? 3) (3, ? 9)0 x ( , ) 8

3?25627

( , ) 4

3?12827

y = x 2 ( x ? 4)

A guide for teachers - Years 11 and 12

•{25}Example Definef: [¡7,6]!Rbyf(x)AE3x4Å8x3¡174x2¡360x. Find the maximum and mini- mum values of the function.

Solution

The first and second derivatives are

f

0(x)AE12x3Å24x2¡348x¡360

AE12(x3Å2x2¡29x¡30)AE12(xÅ1)(x¡5)(xÅ6) f

00(x)AE36x2Å48x¡348AE12(3x2Å4x¡29).

Sof0(x)AE0 impliesxAE5 orxAE ¡1 orxAE ¡6. There are stationary points at these three values. We use the second derivative test: f

00(5)AE792È0,f00(¡1)AE¡360Ç0,f00(¡6)AE660È0.

Hence, there is a local minimum atxAE5, a local maximum atxAE ¡1, and a local mini- mum atxAE¡6.

We have the following points on the graph:

•the left endpoint (¡7,¡1547) •a local minimum (¡6,¡1944) •a local maximum (¡1,181) •a local minimum (5,¡3275) •the right endpoint (6,¡2808).y ( ?

1,181)

(?7,?1547) ( ? 6, ? 1944)
(6, ? 2808)
(5, ? 3275)

0xThe maximum value of the functionfis 181, which occurs atxAE¡1, and the minimum

value offis¡3275, which occurs atxAE5. {26}•Applications of differentiationExercise 8 Sketch the graph of the functionf: [¡1,2]!Rgiven byf(x)AEx2(xÅ4), and find the maximum and minimum values of this function.Maxima and minima problems The need to find local maxima and minima arises in many situations. The first example we will look at is very familiar, and can also be solved without using calculus. Exam- ples of solving such problems without the use of calculus can be found in the module

Quadratics.Example

Find the dimensions of a rectangle with perimeter 1000 metres so that the area of the rectangle is a maximum.

Solution

Let the length of the rectangle bexm, the width beym, and the area beAm2.y xThe perimeter of the rectangle is 1000 metres. So

1000AE2xÅ2y,

and hence yAE500¡x.

The area is given byAAExy. Thus

A(x)AEx(500¡x)AE500x¡x2. (1)

Becausexandyare lengths, we must have 0·x·500. The problem now reduces to finding the value ofxin [0,500] for whichAis a maximum. SinceAis differentiable, the maximum must occur at an endpoint or a stationary point.

A guide for teachers - Years 11 and 12

•{27}From (1), we have dAdx

AE500¡2x.

Setting

dAdx

AE0 givesxAE250.

Hence, the possible values forAto be a maximum arexAE0,xAE250 andxAE500. Since A(0)AEA(500)AE0, the maximum value ofAoccurs whenxAE250. The rectangle is a square with side lengths 250 metres. The maximum area is 62 500 square metres.

Notes.

1 dAdx

È0, for 0·xÇ250, anddAdx

Ç0, for 250Çx·500. Hence, there is a local maxi- mum atxAE250. 2 d2Adx

2AE¡2Ç0. This is a second way to see thatxAE250 is a local maximum.

3The graph ofA(x)AE500x¡x2is a parabola with a negative coefficient ofx2and a

turning point atxAE250. This is a third way of establishing the local maximum.

4It is worth looking at the graph ofA(x) againstx.A(x)

x0 (250, 62 500 ) A ( x ) = 500 x ? x 2 100 200 300 400 500 60060 000

50 000

40 000

30 000

20 000

10 000Exercise 9

A farmer has 8 km of fencing wire, and wishes to fence a rectangular piece of land. One boundary of the land is the bank of a straight river. What are the dimensions of the rect- angle so that the area is maximised? {28}•Applications of differentiation The following steps provide a general procedure which you can follow to solve maxima and minima problems.

Steps for solving maxima and minima problems

Step 1.Where possible draw a diagram to illustrate the problem. Label the diagram and designate your variables and constants. Note any restrictions on the values of the variables. Step 2.Writeanexpressionforthequantitythatisgoingtobemaximisedorminimised. Eliminate some of the variables. Form an equation for this quantity in terms of a single independent variable. This may require some algebraic manipulation. Step 3.IfyAEf(x) is the quantity to be maximised or minimised, find the values ofx for whichf0(x)AE0. Step 4.Test each point for whichf0(x)AE0 to determine if it is a local maximum, a local minimum or neither. Step 5.If the functionyAEf(x) is defined on an interval, such as [a,b] or [0,1), check the values of the function at the end points.Example A square sheet of cardboard with each sideacentimetres is to be used to make an open- top box by cutting a small square of cardboard from each of the corners and bending up the sides. What is the side length of the small squares if the box is to have as large a volume as possible?

Solution

Step 1.x

x xxx x x x a ? 2 x a ? 2 x a

A guide for teachers - Years 11 and 12

•{29}Let the side length of the small squares bexcm. The side length of the open box is (a¡2x) cm, and the height isxcm. Hereais a constant, andxis the variable we will work with. We must have

0·x·a2

.

Step 2.The volumeVcm3of the box is given by

V(x)AEx(a¡2x)2AE4x3¡4ax2Åa2x.

Step 3.We have

dVdx

AE12x2¡8axÅa2AE(2x¡a)(6x¡a).

Thus dVdx

AE0 impliesxAEa2

orxAEa6 .

Step 4.We note thatxAEa2

is an endpoint and thatV¡a2

¢AE0. We will use the second

derivative test forxAEa6 . We have d 2Vdx

2AE24x¡8aAE8(3x¡a).

WhenxAEa6

, we get d 2Vdx

2AE8³

3£a6

¡a´

AE¡4aÇ0.

Hence,xAEa6

is a local maximum. Step 5.The maximum value of the function is atxAEa6 , asV(0)AEV¡a2

¢AE0. The

maximum volume is V

³a6

´

AE2a327

. The following diagram shows the graph ofVagainstx.V x 0 ( , ) a

62a?27

a 2 V ( x ) = x ( a ⎷ 2 x ) 2 {30}•Applications of differentiation The following example illustrates a number of issues that can occur.Example A company wants to run a pipeline from a pointAon the shore to a pointBon an island which is 6 km from the shore. It costs $Pper kilometre to run the pipeline on shore, and $Qper kilometre to run it underwater. There is a pointB0on the shore so thatBB0is at right angles toAB0. The straight shoreline is the lineAB0. The distanceAB0is 9 km. Find how the pipeline should be laid to minimise the cost if

1PAE4000 andQAE5000

2PAE5000 andQAE13 000

3PAE24 000 andQAE25 000.B

9 km 6 km B" ASolution We will work through most of the problem without assigning values toPandQ.

Step 1.

B 6 km B" x km C (9 ? x) km A

A guide for teachers - Years 11 and 12

•{31}Suppose that the pipeline leaves the shorexkm fromB0at a pointCbetweenB0andA. The distanceACis (9¡x) km. By Pythagoras" theorem, the distanceCBisp36Åx2km.

It is important to note that

0·x·9.

Step 2.Let $Tbe the total cost. Then

T(x)AEP(9¡x)ÅQp36Åx2. (1)

Step 3.We have

dTdx

AEQxp36Åx2¡P.

Hence, solving

dTdx

AE0 gives

Qxp36Åx2¡PAE0

QxAEPp36Åx2

Q

2x2AEP2(36Åx2)

(Q2¡P2)x2AE36P2 xAEs36P2Q

2¡P2AE6PpQ

2¡P2. (2)

Note that we needQÈPfor this solutionxto exist, and we also need 0·x·9. IfQ·P, the pipeline should go directly fromAtoB, with minimum cost $3p13Q.

Step 4.Using the second derivative test:

d 2Tdx

2AE36Q(36Åx2)32

È0,

for allx. Hence, there is a local minimum atxAE6PpQ

2¡P2for the function with rule

T(x). Such a local minimum may occur outside the interval [0,9].

Step 5.

•IfxAE0, thenTAE9PÅ6Q. •IfxAE9, thenTAE3p13Q. {32}•Applications of differentiation•IfxAE6PpQ

2¡P2, then from (1) we have

TAEP³

9¡6PpQ

2¡P2´

ÅQs36Å36P2Q

2¡P2

AE9P¡6P2pQ

2¡P2ÅQs36(Q2¡P2)Å36P2Q

2¡P2

AE9P¡6P2pQ

2¡P2Å6Q2pQ

2¡P2

AE9PÅ6qQ

2¡P2. (3)

The local minimum occurs in the interval [0,9] if and only if 6PpQ

2¡P2·9.

We now solve this inequality for the ratio

PQ , assuming thatQÈP: 6PpQ

2¡P2·9()36P2Q

2¡P2·81

()36P2·81(Q2¡P2) ()117P2·81Q2 () P2Q

2·81117

. Thus the local minimum occurs in the interval [0,9] if and only if PQ

·3p13

. We now consider the particular values ofPandQspecified in the question.

1PAE4000 andQAE5000.

By equation (1), we have

TAE4000(9¡x)Å5000p36Åx2, for 0·x·9.

Note that

T(0)AE36000Å30000AE66000

T(9)AE15000p13¼54083.

A guide for teachers - Years 11 and 12

•{33}By equation (2), the local minimum point isxAE8 and in this case, by equation (3), the minimum cost is T minAE9£4000Å6£3000AE$54000.T x0 (0,66 000 ) (8,54 000 )(9,54 083) 2 4 6 8 1080 000

60 000

40 000

20 0002PAE5000 andQAE13000.

By equation (1), we have

TAE5000(9¡x)Å13000p36Åx2, for 0·x·9.

We note that

T(0)AE123000,T(9)AE39000p13¼140616.

By equation (2), the local minimum point isxAE52

and in this case, by equation (3), the minimum cost is T minAE9£5000Å6£12000AE$117000. T x0 (0,123 000 )(9,140 616) 2 4 6 8 10

150 000

100 000

50 000

( ,117 000 ) 5

23PAE24000 andQAE25000.

By equation (1), we have

TAE24000(9¡x)Å25000p36Åx2, for 0·x·9. {34}•Applications of differentiationWe note that

T(0)AE366000,T(9)AE75000p13¼270416.

By equation (2), the local minimum occurs atxAE1447 , which is outside the required domain. In fact, we have dTdx

Ç0, for allx2[0,9]. The minimum cost is

T(9)AE75000p13¼$270416.T

x0 (0,366 000 ) (9,270 416 ) 2 4 6 8 10400 000

300 000

200 000

100 000Note.In parts 1 and 2, the minimum occurs at a local minimum. But, in part 3, the

minimum occurs at an endpoint.The following example has reasonably demanding algebra and involves some geometry,

but the result is surprisingly neat.Example A right cone is circumscribed around a given sphere. Find when its volume is a mini- mum.

Solution

Step 1.The following diagram shows a vertical cross-section of the cone and sphere. h Ar O RR C A" B

A guide for teachers - Years 11 and 12

•{35}The sphere has radiusR, which we treat as a constant. The cone has radiusrand heighth. These are variables. From the geometry, we must havehÈ2RÈ0 andrÈRÈ0. The centre of the sphere is marked byO. The radiusOA0is drawn perpendicular toCB.

Step 2.We will findhin terms ofrandR.

We begin by noting thatOCAEh¡R. By using Pythagoras" theorem in4OCA0, we get CA 0AEph

2¡2hR. Since4CA0Ois similar to4CAB(AAA), we can write

CA 0CA

AEOA0BA

.

Hence,

ph

2¡2hRh

AERr .

Solving forh, we obtain

h

2¡2hRh

2AER2r

2(square both sides)

r

2(h2¡2hR)AEh2R2(cross-multiply)

r

2h2¡2hRr2AEh2R2

r

2h¡2Rr2AEhR2(ash6AE0)

h(r2¡R2)AE2r2R hAE2r2Rr

2¡R2.

The volume of the cone is given byVAE13

¼r2h. Substituting forh, we obtain

VAE2¼r4R3(r2¡R2).

We have now expressed the volume in terms of the one variabler.

Step 3.We have

dVdr

AE4¼Rr3(r2¡2R2)3(r2¡R2)2.

So dVdr AE0 implies thatr3(r2¡2R2)AE0, which implies thatrAE0 orrAEp2R.

Clearly,rAEp2Ris the solution we want.

{36}•Applications of differentiationStep 4.Using dVdr

AE4¼Rr3(r2¡2R2)3(r2¡R2)2,

we can complete the following gradient diagram.Value ofrp2RSign of dVdr¡0Å

Slope of graph-

Alternatively, we can use the second derivative test. We have d 2Vdr

2AE4¼R(r6¡3r4R2Å6r2R4)3(r2¡R2)3.

SubstitutingrAEp2Rgives

d 2Vdr

2AE32¼R3

È0.

Hence, we have a local minimum atrAEp2R.

The graph ofVagainstris as follows. There is a vertical asymptote atrAER, and the graph approaches a parabola with equationVAE2¼R3 r2asrbecomes very large.V rr = R 0 ( ? 2R, 8 ߨ R 3 ) 3 2r 4 R 3( r 2 - R 2 ) V =

A guide for teachers - Years 11 and 12

•{37}Exercise 10 aFindthemaximumareaofarectanglethatcanbeinscribedintheellipsex216

Åy29

AE1. Assume that the sides of the rectangle are parallel to the axes. bFindthemaximumareaofarectanglethatcanbeinscribedintheellipsex2a

2Åy2b

2AE1. Assume that the sides of the rectangle are parallel to the axes.Exercise 11 A hollow cone has base radiusRand heightH. What is the volume of the largest cylinder that can be placed under it?Related rates of change Related rates of change are simply an application of the chain rule. In related-rate prob- lems, you find the rate at which some quantity is changing by relating it to other quanti- ties for which the rate of change is known. In the following examples, we repeatedly use the result dxdy AE1dy dx , which is established and discussed in the moduleIntroduction to differential calculus.Example

Anupturnedconewithsemiverticalangle45

±isbeingfilledwithwaterataconstantrate

of 30 cm

3per second.r

hWhen the depth of the water is 60 cm, find the rate at which

1h, the depth of the water, is increasing

2r, the radius of the surface of the water, is increasing

3S, the area of the water surface, is increasing.

{38}•Applications of differentiationSolutionr

45 º

hThe volumeVof the water in the cone is given by VAE13

¼r2h.

The cross-section of the cone is a right-angled isosceles triangle, and thereforerAEh.

Hence,

VAE13

¼r3AE13

¼h3.

1We seekdhdt

. NowVAE13

¼h3, and thereforedVdh

AE¼h2. Hence,

dhdt

AEdhdV

£dVdt

AE

1¼h2£30

AE

30¼h2.

WhenhAE60, we havedhdt

AE1120¼cm/s.

2We have seen thatr(t)AEh(t). Thus, whenhAE60, we havedrdt

AE1120¼cm/s.

3We seekdSdt

. The area of the water"s surface is

SAE¼r2.

Therefore

dSdt

AEdSdr

£drdt

AE2¼r£30¼r2

AE 60r
.

WhenhAE60, we haverAE60, and sodSdt

AE1 cm2/s.

A guide for teachers - Years 11 and 12

•{39}Example Variablesxandyare related by the equationyAE2¡4x . Given thatxandyare functions oftand thatdxdt

AE10, finddydt

in terms ofx.

Solution

IfyAE2¡4x

, thendydx AE4x

2. Therefore

dydt

AEdydx

£dxdt

AE 4x

2£10

AE 40x

2.Related rates questions arise in many situations. The following demanding example in-

volving the flow of liquids relates the rate of change of depth to the rate of change of volume.Example A vessel of water is in the form of a frustum of a cone with semivertical angle 45

±. The

bottomcircleofthevesselisaholeofradiusrcm. Waterflowsfromthisholeatavelocity ofcp2ghcm/s, wherecis a constant andhcm is the height of the surface of the water above the hole.

1Find the rate in cm3/s at which water flows from the vessel.

2Find the rate in cm/s at whichhis increasing.H

R h h ? r {40}•Applications of differentiationSolution WefirstfindthevolumeVofwaterinthefrustum. Considerremovingasmallconefrom the 'tip" of a cone. This forms the hole. LetHbe the height of the larger cone (up to the water surface) and leth1be the height of the smaller cone (which is removed). LetRbe the radius of the larger cone and letrbe the radius of the smaller cone.

The height of the water ishAEH¡h1. Therefore

VAE13

¼(R2H¡r2h1)AE13

¼¡R2H¡r2(H¡h)¢.

Because the semivertical angle is 45

±, we haveRAEHandR¡rAEh. Hence,RAErÅh.

Substituting forRandHin the equation forV, we have VAE13

¼¡R2H¡r2(H¡h)¢

AE 13

¼¡R3¡r2(R¡h)¢

AE 13

¼¡(rÅh)3¡r3¢

AE 13

¼¡3r2hÅ3rh2Åh3¢.

1Letvcm/s be the velocity of the flow of water from the hole at timet. We are given

thatvAEcp2gh, wherehis the height of the water above the hole at timet. Since dVdt AEarea of the cross-section of the hole£velocity of the flow of water from the hole, we have dVdt

AE¼r2vAE¼r2cq2ghcm3/s.

2We found that the volume of waterVcm3is given by

VAE13

¼(3r2hÅ3rh2Åh3),

and so dVdh

AE¼(rÅh)2.

Using the chain rule, we have

dhdt

AEdhdV

£dVdt

AEcr2p2gh(rÅh)2cm/s.

A guide for teachers - Years 11 and 12

•{41}Exercise 12 A pointPis moving along the curve whose equation isyAEpx

3Å56. WhenPis at (2,8),

yis increasing at the rate of 2 units per second. How fast isxchanging?Exercise 13 A meteor enters the earth"s atmosphere and burns at a rate that at each instant is pro- portional to its surface area. Assuming that the meteor is always spherical, show that the radius is decreasing at a constant rate.Links forward

Critical points

In the sectionGraph sketching, we looked at stationary points of a functionf, that is, points wheref0(x)AE0. We gave conditions for a stationary point to be local minimum or a local maximum. Local maxima and minima also occur in other cases. The point (c,f(c)) on the left-hand graph is a local minimum. The point (d,f(d)) on the right-hand graph is a local maxi- mum. These are examples of critical points. In each case, the function is not differen- tiable at that point.y x 0 y x0 ( c , f ( c ))(d,f(d))Examples of critical points.

Definition

Acritical pointfor a functionfis any value ofxin the domain offat whichf0(x)AE0 or at whichfis not differentiable.

Every stationary point is a critical point.

{42}•Applications of differentiation Not every critical point is a local minimum or maximum point. In the following diagram, the function is not differentiable at the point (c,f(c)), since the tangent is vertical. There is a point of inflexion at (c,f(c)).y x0 ( c , f ( c ))Finding gradients on a parametric curve

We start with a simple example.

Consider the parametric curve defined by

xAEt2 yAEt3.

We will find

dydx in terms oftby using the chain rule and the result dtdx AE1dx dt .

We have

dxdt

AE2tanddydt

AE3t2. Hence,

dydx

AEdydt

£dtdx

AE 3t22t AE 3t2 . 5 5 1 2 3 4 y x0

A guide for teachers - Years 11 and 12

•{43}

Projectile motion

The motion of a particle projected at an angle®to the horizontal with an initial velocity ofum/s can be described by the parametric equations xAEucos(®)t yAEusin(®)t¡12 gt2, wheregis the acceleration due to gravity. A method for deriving these equations is given in the moduleMotion in a straight line. The velocities in the horizontal and vertical directions are dxdt

AEucos(®) anddydt

AEusin(®)¡gt.

Using the chain rule, we have

dydx

AEdydt

£dtdx

AEusin(®)¡gtucos(®).

Thisgivesthegradientofthepathoftheprojectileintermsoft, fort¸0. Thisalsoallows us to find the angle of inclination of the path of the projectile at any timet. We use the fact that tanµAEdydx , whereµistheangleofinclinationofthepathtothehorizontal. Forexample, if®AE¼4 and uAE10, then tanµAE10p2

¡gt10p2

AE1¡p2gt10

.

Cycloids

Acycloidis the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line. For a wheel of radius 1, the parametric equations of the cycloid are xAEµ¡sinµ yAE1¡cosµ. The graph ofyagainstxfor 0·µ·6¼is as follows. {44}•Applications of differentiationy x0 2 ߨ 4ߨ 6ߨ

4321We can find the gradient at a point on the cycloid by using the chain rule. We have

dxdµAE1¡cosµanddydµAEsinµ.

The chain rule gives

dydx

AEdydµ£dµdx

AE sinµ1¡cosµ AE

2sinµ2

cosµ2 2sin

2µ2

AEcotµ2

. Also, d 2ydx

2AEddx

³ dydx ´ AE ddx ³ cotµ2 ´ AE ddµ³ cotµ2 ´

£dµdx

AE¡

14sin

4(µ2

).

Notes.

•There are stationary points where cotµ2

AE0, which is equivalent to cosµ2

AE0. This oc-

curs whenµis an odd multiple of¼, that is, whenµ2{...,¡5¼,¡3¼,¡¼,¼,3¼,5¼,...}.

Note that

d2ydx

2Ç0 for these values ofµ, and so there are local maxima whenµis an

odd multiple of¼. •The function is not differentiable whenµis even multiple of¼. That is, these are critical points of the function. •There are no points of inflexion. •Thex-intercepts occur when cosµAE1, that is, whenµis an even multiple of¼.

A guide for teachers - Years 11 and 12

•{45}

Cardioidsy

?2 ?1.5 ?1 ?0.5 0 x 1 0.5 ? 0.5 ?

1Acardioidis the curve traced by a point on the

perimeter of a circle that is rolling around a fixed cir- cle of the same radius. The cardioid shown in the graph has parametric equations xAE(1¡cost)cost yAE(1¡cost)sint.

It is plotted fort2[0,2¼].

We can calculate the gradient of the cardioid for a particular value oftas follows: dxdt

AE¡sintÅ2sintcostAEsin2t¡sint

dydt

AEcost¡cos2tÅsin2tAEcost¡cos2t,

and so dydx

AEdydt

£dtdx

AEcost¡cos2tsin2t¡sint.

The gradient is defined for sin2t¡sint6AE0. Fort2[0,2¼], this means that gradient is not defined fortAE0,¼3 ,¼,5¼3 ,2¼. The correspondingx-values are 0,14 ,¡2,14 ,0.

There are points of zero gradient whentAE2¼3

,4¼3 , which corresponds toxAE¡34 .

History and applications

History

The history of calculus is discussed in the modules: •Introduction to differential calculus •Motion in a straight line. {46}•Applications of differentiation

Applications

Application in biologyExample:Chemotherapy

Malignanttumoursrespondtoradiationtherapyandchemotherapy. Consideramedical experiment in which mice with tumours are given a chemotherapeutic drug. At the time of the drug being administered, the average tumour size is about 0.5 cm

3. The tumour

volumeV(t) aftertdays is modelled by

V(t)AE0.005e0.24tÅ0.495e¡0.12t,

for 0·t·18. Let us find the minimum point. The first and second derivatives are V

0(t)AE0.0012e0.24t¡0.0594e¡0.12t

V

00(t)AE0.000288e0.24tÅ0.007128e¡0.12t.

Note thatV00(t)È0, for alltin the domain. By solvingV0(t)AE0, we find that the min- imum point occurs whent¼10.84 days. The volume of the tumour is approximately

0.20 cm

3at this time.V(t) cm?

t days 0 0.6 0.5 0.4 0.3 0.2 0.1 5 10 15 20 (10.84,0.20)Applications in physics Calculus is essential in many areas of physics. Some additional applications of calculus to physics are given in the modulesMotion in a straight lineandThe calculus of trigono- metric functions.

A guide for teachers - Years 11 and 12

•{47}Example:Damped simple harmonic motion The displacement of a body of massmundergoing damped harmonic motion is given by the formula xAEAe¡bt2mcos(!t), where !AEsk m

¡³b2m´

2. Herekandbare positive constants. (Note that damped simple harmonic motion re- duces to simple harmonic motion whenbAE0.)x tThe motion can be investigated by using calculus to find the turning points.

SincexAEAe¡bt2mcos(!t), we have

dxdt

AE¡b2mAe¡bt2mcos(!t)¡A!e¡bt2msin(!t)

AE¡Ae¡bt2m³b2mcos(!t)Å!sin(!t)´

.

The stationary points are found by putting

dxdt

AE0. This implies

tan(!t)AE¡b2m!. {48}•Applications of differentiation Here is example from optics using related rates of change.Example:Thin lens formula

The thin lens formula in physics is

1s

Å1S

AE1f , wheresis the distance of an object from the lens,Sis the distance of the image from the lens, andfis the focal length of the lens. Herefis a constant, andsandSare variables. Suppose the object is moving away from the lens at a rate of 3 cm/s. How fast and in which direction will the image be moving?

We are given

dsdt

AE3. The question is: What isdSdt

?

We know by the chain rule that

dSdt

AEdSds

£dsdt

AEdSds

£3.

MakingSthesubjectofthethinlensformula, wehaveSAEf ss¡f. ThederivativeofSwith respect tosis dSds

AE¡f2(s¡f)2.

Thus dSdt

AE¡f2(s¡f)2£3AE¡3f2(s¡f)2.

The image is moving towards the lens at

3f2(s¡f)2cm/s.

A guide for teachers - Years 11 and 12

•{49}

Answers to exercises

Exercise 1

We can form a gradient diagram for this function.Value ofx13

Sign off0(x)¡0¡0Å

Slope of graphyAEf(x)--

There is a stationary point of inflexion atxAE1, and a local minimum atxAE3.

Exercise 2

Letf(x)AEx3¡5x2Å3xÅ2. The derivative isf0(x)AE3x2¡10xÅ3AE(3x¡1)(x¡3). So the stationary points arexAE13 andxAE3. •WhenxÇ13 ,f0(x)È0. •When13

ÇxÇ3,f0(x)Ç0.

•WhenxÈ3,f0(x)È0.

Hence, there is a local maximum atxAE13

, and a local minimum atxAE3.

Exercise 3

The inflexion points arexAE0 andxAE¡14.

Exercise 4

We are givenf0(x)AEx3(x2¡5)AEx5¡5x3. So the stationary points arexAE0,xAEp5 and xAE¡p5. The second derivative isf00(x)AE5x4¡15x2AE5x2(x2¡3). •f00(0)AE0, so we cannot use the second derivative test forxAE0. However, note that f

0(¡1)AE4 andf0(1)AE ¡4. The gradient changes from positive to negative atxAE0.

Hence, there is a local maximum atxAE0.

•f00(p5)AE50È0, so there is a local minimum atxAEp5. •f00(¡p5)AE50È0, so there is a local minimum atxAE¡p5. {50}•Applications of differentiation

Exercise 5

LetyAE3x4¡44x3Å144x2. Then the first and second derivatives are dydx AE12x3¡132x2Å288xAE12x(x2¡11xÅ24)AE12x(x¡3)(x¡8) d 2ydx

2AE36x2¡264xÅ288AE12(3x2¡22xÅ24)AE12(3x¡4)(x¡6).

1We first find thex-intercepts:

3x4¡44x3Å144x2AE0

x

2(3x2¡44xÅ144)AE0,

which givesxAE0 orxAE23 (11¡p13) orxAE23 (11Åp13).

2We havedydx

AE12x(x¡3)(x¡8). Sodydx

AE0 impliesxAE0 orxAE3 orxAE8.

3Value ofx038

Sign of

dydx¡0Å0¡0Å

Slope of graph---

4We use the second derivative test:

•atxAE0, we haved2ydx

2AE288È0, so there is a local minimum

•atxAE3, we haved2ydx

2AE¡180Ç0, so there is a local maximum

•atxAE8, we haved2ydx

2AE480È0, so there is a local minimum.

5Asx!1,y!1, and asx!¡1,y!1.

6 d2ydx

2AE0 impliesxAE43

orxAE6. The second derivative changes from positive to neg- ative atxAE43 , and from negative to positive atxAE6. So there are points of inflexion atxAE43 andxAE6.y x0 (3,351) 2 3 (11 ? ⎷ 13) 2 3 (11 +⎷ 13) (8, ?

1024)(6,?432)y = 3x² ? 44x + 144x

( 4 4352

3 , 27

)

A guide for teachers - Years 11 and 12

•{51}

Exercise 6

LetyAE4x3¡18x2Å48x¡290. The first and second derivatives are dydx

AE12x2¡36xÅ48AE12(x2¡3xÅ4)

d 2ydx

2AE12(2x¡3).

1They-intercept isyAE¡290. We now find thex-intercepts:

4x3¡18x2Å48x¡290AE0

2x3¡9x2Å24x¡145AE0

(x¡5)(2x2ÅxÅ29)AE0. The quadratic has no real solutions, and so the onlyx-intercept isxAE5. 2 dydx AE12(x2¡3xÅ4). The discriminant is 12(9¡16)Ç0. Hence,dydx

È0, for allx, and

there are no stationary points.

3The function is increasing, for allx.

4There are no local maxima or minima.

5Asx!1,y!1, and asx!¡1,y!¡1.

6 d2ydx

2AE12(2x¡3). The second derivative changes from negative to positive atxAE32

.

So there is a point of inflexion atxAE32

.y x05 y = 4 x ? ⎷ 18 x ² + 48 x ⎷ 290
⎷ 290
3 ( 2 , ⎷ 245)
{52}•Applications of differentiation

Exercise 7

Definef: (0,1)!Rbyf(x)AEx2logex. Then

f

0(x)AE2xlogexÅx2£1x

AE2xlogexÅxAEx(2logexÅ1)

f

00(x)AE2logexÅ2Å1AE2logexÅ3.

1ForxÈ0, we havef(x)AE0()logexAE0()xAE1. So thex-intercept is 1.

2We havef0(x)AEx(2logexÅ1). Sof0(x)AE0 impliesxAEe¡12

. There is a stationary point atxAEe¡12 .

3We have

f

0(x)È0()x(2logexÅ1)È0()2logexÅ1È0()xÈe¡12

, and sof0(x)Ç0()0ÇxÇe¡12 .

4f00(e¡12

)AE2È0. Hence, there is a local minimum atxAEe¡12 .

5Asx!1,f(x)!1.

6We havef00(x)AE2logexÅ3. So

f

00(x)È0()xÈe¡32

,f00(x)Ç0()0ÇxÇe¡32 .

There is a point of inflexion atxAEe¡32

.y x4 3 2 1 0 0.5 1.0 1.5 2.0 2.5 3.0y = x?log ީ y x0.2 0.1 ? 0.1 ? 0.2 ?

0.3 0 0.2 0.4 0.6 0.8 1.0 1.2

?3 2 ( ݁ , ) ? 3 2 ݁ ⎷ (݁ , ) ?1 2 ? 1 2 ݁

A guide for teachers - Years 11 and 12

•{53}

Exercise 8

Definef: [¡1,2]!Rbyf(x)AEx2(xÅ4). The first and second derivatives are f

0(x)AE3x2Å8x,f00(x)AE6xÅ8.

Sof0(x)AE0 whenxAE0 orxAE ¡83

. The second value is outside the required domain. Sincef00(0)AE8È0, there is a local minimum atxAE0, withf(0)AE0. We now find the value of the function at the endpoints:f(¡1)AE3 andf(2)AE24.y x ( ? 1,3) ( 0 ,0) ( 2 ,24)The minimum value of the function is 0 and the maximum value is 24.

Exercise 9

Letxkm,xkm andykm be the lengths of fencing of the three sides of the rectangle to be enclosed. ThenyAE8¡2x. LetA(x) km2be the area of the enclosed land. ThenA(x)AEx(8¡2x)AE8x¡2x2. The first two derivatives are A

0(x)AE8¡4xandA00(x)AE¡4.

SoA0(x)AE0 impliesxAE2. SinceA00(2)AE ¡4Ç0, there is a local maximum atxAE2. The rectangle of maximum area has dimensions 2 km£4 km, and the maximum area is 8 km 2. {54}•Applications of differentiation

Exercise 10

aThe ellipse has equationx216

Åy29

AE1.y Y Z W0 ? 3 ? 443
x X ( x, 3 4 ⎷16?x²)Assume the top-right corner of the rectangle is atX(x,34 p16¡x2). The areaAof the rectangleXY ZWis given by

AAEXY£XWAE2x£32

p16¡x2AE3xp16¡x2.

We have

dAdx

AE3³

p16¡x2¡x2p16¡x2´

AE3³16¡x2¡x2p16¡x2´

AE3³16¡2x2p16¡x2´

. So dAdx

AE0 impliesxAE2p2. Furthermore, we have

d 2Adx

2AE6x(¡24Åx2)(16¡x2)32

, and so d2Adx

2Ç0 whenxAE2p2. The maximum area is 3£2p2£q16¡(2p2)

2AE24.

bThe ellipse has equation x 2a

2Åy2b

2AE1, wherea,bÈ0. Assume that the top-right corner of the rectangle is at (x,y), where

0·x·aand 0·y·b. LetAbe the area of the rectangle. Then

AAE4ba

xpa

2¡x2

dAdx

AE4b(a2¡2x2)a

pa

2¡x2.

So dAdx

AE0 impliesxAEap2

. The maximum area is 2ab.

A guide for teachers - Years 11 and 12

•{55}

Exercise 11

Let the cylinder have radiusrand heighth. HererÈ0 andhÈ0, and they are variables. Let the cone have radiusRand heightH. They are constants.H R rhThe volumeVof the cylinder is given by

VAE¼r2h.

Using similar triangles, we have

H¡hr

AEHR

HR¡hRAErH

hAEH¡rHR .

Substitute forhin the formula forV:

VAE¼r2³

H¡rHR

´

AE¼HR

¡Rr2¡r3¢.

Differentiate with respect tor:

dVdr

AE¼HR

¡2Rr¡3r2¢AE¼HrR

¡2R¡3r¢.

Check for stationary points:

dVdr

AE0 impliesrAE2R3

, sincer6AE0. The second derivative is d 2Vdr

2AE¼HR

¡2R¡6r¢.

So, whenrAE2R3

, we haved2Vdr

2AE¡2¼HÇ0. A local maximum occurs whenrAE2R3

. This gives the maximum volume of the cylinder, which isVAE4¼R2H27 . {56}•Applications of differentiation

Exercise 12

The pointPis moving along the curveyAEpx

3Å56. We have

dydx

AE3x22

px

3Å56,

and so dxdt

AEdxdy

£dydt

AE2px

3Å563x2£dydt

.

WhenxAE2, we are given thatdydt

AE2, and so

dxdt AE2p2

3Å563£22£2AE83

.

Exercise 13

Assume the meteor is a sphere of radiusr. Its surface area isSAE4¼r2, and its volume is VAE43 ¼r3. The volume is decreasing at a rate proportional to the surface area. That is, dVdt

AE¡kSAE¡4¼kr2,

for some positive constantk. Using the chain rule, we have drdt

AEdrdV

£dVdt

AE14¼r2£¡4¼kr2AE¡k.

So the radius is decreasing at a constant rate.

0 1 2 3 4 5 6 7 8 9 10 11 12

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