P1: FXS/ABE P2: FXS CHAPTER 10downloads.cambridge.edu.au/education/extra/209... · P1: FXS/ABE P2: FXS 0521842344c10.xml CUAU030-EVANS August 26, 2008 6:26 348 Essential Mathematical

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0521842344c10.xml CUAU030-EVANS August 26, 2008 6:26

C H A P T E R

10Applications ofdifferentiation

ObjectivesTo be able to find the equation of the tangent and the normal at a given point on a

curve.

To be able to calculate the angles between straight lines and curves.

To be able to find the stationary points on the curves of certain polynomial

functions and to state the nature of such points.

To use differentiation techniques to sketch graphs of rational functions.

To solve maxima and minima problems.

To use the derivative of a function in rates of change problems.

To use the chain rule in the solution of related rates problems.

10.1 Tangents and normalsThe derivative of a function is a new function that gives the measure of the gradient at each

point of the curve. Having the gradient, we can find the equation of the tangent for a given

point on the curve. Suppose (x1, y1) is a point on the curve y = f (x). Then if f is differentiable

for x = x1, the equation of the tangent at (x1, y1) is given by y − y1 = f ′(x1)(x − x1).

Example 1

Find the equation of the tangent of the curve y = x3 + 1

2x2 at the point x = 1.

Solution

When x = 1, y = 3

2so

(1,

3

2

)is a point on the tangent.

Further,dy

dx= 3x2 + x

347Cambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

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348 Essential Mathematical Methods 3 & 4 CAS

Thus the gradient of the tangent to the curve at x = 1 is 4 and the equation of the

tangent is y − 3

2= 4(x − 1), which becomes y = 4x − 5

2.

The normal to a curve at a point on the curve is the line that passes through the point and is

perpendicular to the tangent at that point.

From earlier work you should know that lines with gradients m1 and m2 are perpendicular if,

and only if, m1m2 = −1.

Thus if a tangent has a gradient m, the normal has gradient − 1

m.

Example 2

Find the equation of the normal to the curve with equation y = x3 − 2x2 at the point (1, −1).

Solution

The point (1, −1) is on the normal.

Furtherdy

dx= 3x2 − 4x

Thus the gradient of the normal at x = 1 is−1

−1= +1.

Hence the equation of the normal is y + 1 = 1(x − 1)

i.e., the equation of the normal is y = x − 2

Example 3

Use a CAS calculator to find the equation of the tangent to the curve with equation

y = x32 − 4x

12 at the point on the graph where x = 4.

Solution

Using the TI-NspireUse Tangent Line (b 4 8) as

shown.

Cambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

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Chapter 10 — Applications of differentiation 349

Using the Casio ClassPadThis can be done in the menu. Enter the

function, select the tick box and tap to produce

the graph. Ensure the graph window is selected

(bold border) and tap Analysis > Sketch >

Tangent. Press the number 4 and OK to sketch the

tangent at x = 4. The equation of the tangent is

shown in the formula line.

Note that the results in are approximate

calculations, not algebraic calculations. Different

screen resolutions will give slightly different

answers. Scrolling in the formula line should

convince you that the answer should be read as

y = 2x − 8.

Example 4

Find the equation of the tangent of:

a f (x) = x13 where x = 0 b f (x) = x

23 where x = 0

Solution

a The derivative of f is not defined at x = 0.

For x ∈ R\{0}, f ′(x) = 1

3x− 2

3 . It is

clear that f is continuous at x = 0 and

|f ′(x)| → ∞ as x → 0. It can be said

that the graph of y = f (x) has a vertical

tangent at x = 0.

x

y

x = 0

0


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b The derivative of f is not defined at x = 0.

For x ∈ R\{0}, f ′(x) = 2

3x− 1

3 . It is clear

that f is continuous at x = 0 and f ′(x) →∞as x → 0+ and f ′(x) → −∞ as x → 0−.

There is a cusp at x = 0 and the graph of

y = f (x) has a vertical tangent at x = 0.

x

y

0x = 0

Exercise 10A

1 Find the equation of the tangent and of the normal of the curve y = x2 − 1 at the point

(2, 3).

2 Find the equation of the normal to the curve y = x2 + 3x − 1 at the point where the curve

cuts the y-axis.

3 Find the equations of the normals to the curve y = x2 − 5x + 6 at the points where it cuts

the x-axis.

4 Find the equation of the tangent and of the normal of the curve y = (2x + 1)9 at the point

(0, 1).

5 Find the coordinates of the point on y = x2 − 5 at which the curve has gradient 3. Hence

find the value of c for which the line y = 3x + c is tangent to y = x2 − 5

6 Find the equation to i the tangent and ii the normal at the point corresponding to the given

x-value on each of the following curves:

a y = x2 − 2; x = 1 b y = x2 − 3x − 1; x = 0

c y = 1

x; x = −1 d y = (x − 2)(x2 + 1); x = −1

e y = √3x + 1; x = 0 f y = √

x ; x = 1

g y = x23 + 1; x = 1 h y = x3 − 8x ; x = 2

i y = x3 − 3x2 + 2; x = 2 j y = 2x3 + x2 − 4x + 1; x = 1

7 Find the equation to the tangent at the point corresponding to the given x-value on each of

the following curves:

a y = x2 − 1

x2 + 1; x = 0 b y = √

3x2 + 1; x = 1

c y = 1

2x − 1; x = 0 d y = 1

(2x − 1)2 ; x = 1

8 Find the equation to the tangent at the point where y = 0 for each of the following curves:

a y = (x − 4)13 b y = (x + 5)

23 c y = (2x + 1)

13 d y = (x + 5)

45


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10.2 Angles between curves∗

Angle between two straight linesTwo straight lines l1 and l2 make angles �1 and �2

with the positive direction of the x-axis. We write

gradient of l1 = m1 = tan �1

gradient of l2 = m2 = tan �2

The angle, �, between the lines is given by:

� = �2 − �1 (see the diagram)

We note that tan � = tan (�2 − �1)

= tan �2 − tan �1

1 + tan �1 tan �2

l1

x

y

0

θ1

α

θ2

l2

Example 5

Two straight lines l1 and l2 have equations 3x − 2y = 5 and 4x + 5y = 1. Find the angle

between l1 and l2.

Solution

tan �2 = −4

5and tan �1 = 3

2∴ �2 = 180 − 38.66 �1 = 56.31

= 141.34

∴ the angle between the two lines is

85.03◦ = 85◦2′x

y

0

l1

3x – 2y = 5

4x + 5y = 1

l2

θ1

θ2

α

Alternative solution

Gradient of l1 = m1

= 3

2

Gradient of l2 = m2

= −4

5

∗ This topic is not listed in the study design, but can be included for the understanding of other topics that are included.


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Thus if � is the angle between l1 and l2:

tan � =3

2−

(−4

5

)

1 + 3

2× −4

5

= −23

2

In this calculation we have found �, the obtuse angle between the lines l1 and l2. The

acute angle � between the lines satisfies tan � = 23

2and as before:

� = 85.03◦

= 85◦2′

The process described above to find the angle between two straight lines may be used to find

the angle between two curves as we define the angle between curves to be the angle between

the tangents to the curves at the point of intersection.

Example 6

Find the angle between the curve y = x2 and the line y = x + 12 at the point of their

intersection, (4, 16).

Solution

The derivative of x2 is 2x.

Therefore the gradient of y = x2 at x = 4 is 8.

tan �1 = 1 and tan �2 = 8

�2 = 82.87

∴ �1 = 45

∴ � = �2 − �1

= 82.87 − 45

= 37.87

i.e. the acute angle between the curve and the line is 37◦52′.

x

α

0

(4, 16)

θ2θ1

y

Example 7

Find the angle between the curve f (x) = 4 − x2 and the curve g(x) = x2 − 4 at the point (2, 0).


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Solution

f ′(x) = −2x and g′(x) = 2x and therefore

f ′(2) = −4 and g′(2) = 4

From the diagram:

� = �2 − �1

∴ � = 104.04 − 75.96 as tan �2 = −4

and tan �1 = 4

= 28.07

α

0 2

–4

–2

θ2

θ1x

y

4

∴ the angle between the two parabolas is 28.07◦.

Alternative solution

tan � = tan (�2 − �1)

= tan �2 − tan �1

1 + tan �2 tan �1

= −4 − 4

1 + −4 × 4

= −8

−15= 8

15∴ � = 28.07◦

Exercise 10B

1 Find the magnitude of the acute angle between the lines 2y + 3x = 4 and x + y = 5

2 The graphs with equations y = x2 − 2x and y = x intersect at the point (3, 3). Find the acute

angle lying between them at this point.

3 The curves with equations y = 9 − x2 and y = x2 − 9 intersect at the point (3, 0).

Find the obtuse angle lying between the curves at this point.

4 Find the acute angles lying between the following pairs of lines. (Express your answers in

the form tan−1( ).

a y = x − 6, y = 3x + 1 b 2x + y + 3 = 0, x + y = 0

c 3x + 4y − 2 = 0, 2x − 3y − 4 = 0

5 a Find the coordinates of the points of intersection of the curves with equations y = x3

and y = x2.

b Find the angle of intersection at each of these points.

6 Find the acute angle of intersection of the curves with equations y = x2 and y = 1

x

7 Find the acute angles of intersection of the given pairs of curves:

a y = x3, y = x b y = 2x, y = x + x3

c y = x2, y = 1 d x = 2, y = x2


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10.3 Linear approximationIn the graph l is the tangent to the curve at

the point (x, f (x)).

f ′(x) is the gradient of the tangent.

x

y

x + h

y = f (x)

hf '(x)P(x, f (x))

((x + h), f (x + h))

hθ

0

l

From the diagram it can be seen that when

h is small f (x + h) ≈ f (x) + h f ′(x)

This can also be seen by considering the

definition of derivative:

f ′(x) = limh→0

f (x + h) − f (x)

h

∴ f ′(x) ≈ f (x + h) − f (x)

hwhere h is small

∴ f (x + h) ≈ h f ′(x) + f (x) where h is small

The approximate increase in f (x) is given by f (x + h) − f (x) = h f ′(x)

With the alternative notation:

�y ≈ dy

dx�x

The linear approximation of a function may be considered as using the tangent at a point to

approximate the curve in an immediate neighbourhood of the point at which the tangent is

determined.

Example 8

Given that f (x) = x4 − x3, find in terms of p the approximate increase in f (x) as x increases

from 2 to 2 + p, where p is small.

Solution

f (2 + p) ≈ p f ′(2) + f (2) where f ′(x) = 4x3 − 3x2

∴ f (2 + p) ≈ p(4 × 23 − 3 × 22) + (24 − 23)

= p(20) + 8

= 20p + 8

The approximate increase = f (2 + p) − f (2)

= 20p + 8 − 8

∴ the increase in f (x) is 20p.

Using Liebniz notation:

�y

�x≈ dy

dx

∴ �y ≈ dy

dx�x

When x = 2

�y ≈ 20p


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The following table of values indicates the accuracy of the approximation:

p x f (x) f (2) + 20p

−0.01 1.99 7.8018 7.8

0 2 8 8

0.01 2.01 8.2018 8.2

0.02 2.02 8.4073 8.4

0.03 2.03 8.6164 8.6

↓

0.1 2.1 10.187 10

In practical problems you are often required to find the percentage change in a quantity

resulting from a given change in another quantity. For a function f the percentage change

between a and a + h is defined to be

100

(f (a + h) − f (a)

f (a)

)

provided f (a) = 0.

Using the result that f (a + h) ≈ h f ′(a) + f (a):

the percentage change is approximately equal to100h f ′(a)

f (a)

Example 9

The time for a pendulum of length l cm to complete one swing is given by the function with

rule f (l) = k√

l, where k is a constant. If an error is made in the measurement of the length so

that the measured length is 21

2% greater than the actual length, find the approximate

percentage error if the function f (l) is used to calculate the time of a swing.

Solution

Let l1 be the actual length of the pendulum.

By the above, the percentage error = 100h f ′(l1)

f (l1)

where f (l1) = kl121 , f ′(l1) = 1

2kl

− 12

1 and h = l1

40


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∴ the percentage error =

100 × l1

40× k

2l121

÷ kl

121

=

5l

121 k

4

× 1

kl121

= 5

4

Example 10

Given that y = 10 − 5

xand that the value of y increases from 5 by a small amount

p

10, find in

terms of p:

a the approximate change in x b the corresponding percentage change in x

Solution

a For y = 10 − 5

x,

dy

dx= 5

x2

and when y = 5, x = 1.

∴ �y

�x≈ 5

1= 5, close to x = 1

i.e.1

5�y ≈ �x

Here �y = p

10and so �x ≈ p

50

b Percentage change ≈ �x

x× 100%

=( p

50× 100

)%

= 2p%

Example 11

Differentiate1√x

with respect to x and use the result to find an approximate value for1√

100.5.

Solution

y = x− 12 and

dy

dx= −1

2x− 3

2

When x = 100, y = 1

10and

dy

dx= −1

2(100)−

32

= −1

2000

Usingdy

dx≈ �y

�x

�y ≈ dy

dx�x

= −1

2000× 0.5 when x = 100 and �x = 0.5

∴ �y = − 1

4000 (cont’d.)


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and1√

100.5≈ 1

10− 1

4000

= 0.099 75

Exercise 10C

1 The equation of a curve is y = 4x3 − 8x2 + 10

a Finddy

dx. b Find the value of

dy

dxwhen x = 2.

c Find the approximate change in y as x increases from 2 to 2.02.

2 For y = x− 12 :

a Finddy

dx. b Determine the value of

dy

dxfor x = 100.

c Find the change in y as x changes from 100 to 103.

d Find an approximate value of1√103

.

e i Find the change in y as x changes from a to a + h where h is small.

ii Find an expression for1√

a + hwhere h is small.

3 Given that y = x4 − 5x3, find in terms of q the approximate increase in y as x increases

from 2 to 2 + q where q is small.

4 Given that y = 6 − 4x + 5x2 and that the value of x increases from 5 by a small amountp

20, determine in terms of p:

a the approximate change in y b the corresponding percentage change in y

5 A curve has equation y = 5x2 + 8

x. Find an expression for

dy

dxand hence find, in terms of

p, where p is small, the approximate increase in y as x increases from 2 to 2 + p.

6 For each of the following, write down an expression for the approximate change, �y, in y

when x changes from a to a + p where p is small:

a y = 3x + 7 b y = √x c y = 2

3x + 1

d y = (2x + 1)4 e y = (6x2 − 1)3 f y = x2 + 2x + 6

x2 + 1

g y = 3√x2 + 10 h y = 6x + 1

x + 1

7 The area of a circular disc increases from 100� cm2 to 101� cm2. Find the corresponding

increase in the radius.

8 Given that y = x− 13 use calculus to determine an approximate value for

13√0.9

.


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9 Given thatdy

dx= 6x − 4, calculate the approximate change in y when x increases from

−2 to −1.97.

10 The radius r of a circle is 5 cm. Find the increase in the area, A cm2, of the circle when the

radius expands by p cm where p is small.

11 The time for a swing, T seconds, of a pendulum of length l m is given by the rule

T = 2�

√l

g, where g is a constant.

a FinddT

dl.

b Find the approximate increase in T when l is increased from 1.6 to 1.7. Give the

answer in terms of g.

12 A 2% error is made in measuring the radius of a sphere. Find the percentage error in the

surface area. (The surface area of a sphere is given by A = 4�r2.)

13 One side of a rectangle is three times the other. If the perimeter increases by 2%, what is

the percentage increase in the area?

14 a For f (x) = 1

1 − x, find:

i f ′(x) ii f ′(0)

b Show that f (h) ≈ 1 + h when h is small. [Use f (0 + h) ≈ f ′(0)h + f (0)]

15 a For f (x) = √1 + x , find:

i f ′(x) ii f ′(0)

b Show that f (h) ≈ 1 + h

2when h is small.

16 Show that f (x) = 1

2 − xcan be approximated by f (x) = 1

2+ x

4for x close to 0.

17 The radius of a sphere is measured as 4 cm with a possible error of 0.05. What is the

approximate error for:

a the surface area? (Use S = 4�r2) b the volume?

(Use V = 4

3�r3

)

10.4 Stationary pointsIn the previous chapter we have seen that the

gradient at a point (a, g(a)) of the curve with

rule y = g(x) is given by g′(a). A point

(a, g(a)) on a curve y = g(x) is said to be a

stationary point if g′(a) = 0. (Equivalently:

for y = g(x),dy

dx= 0 when x = a

implies ((a, g(a)) is a stationary point.)

x

y

C

B

A

0


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For example in the graph shown, there are stationary points at A, B and C. At such points the

tangents are parallel to the x-axis (illustrated as dotted lines).

The reason for the name stationary points becomes clear if we look at an application to

motion of a particle.

Example 12

The displacement, x metres, of a particle moving in a straight line is given by x = 9t − 1

3t3,

where 0 ≤ t ≤ 3√

3 and t seconds is the time taken. Find its maximum displacement.

Solution

dx

dt= 9 − t2, and maximum displacement

occurs whendx

dt= 0

So t = 3 or t = −3 (but t = −3 lies outside

the domain).

At t = 3, x = 18.

Thus the stationary point is (3, 18) and the

maximum displacement is 18 metres.

Note that the stationary point occurs when the

rate of change of displacement with respect to

time (velocity) is zero. The particle stopped

moving forward at t = 3.

t

x

18

0 3

Example 13

Find the stationary points of the following functions:

a y = 9 + 12x − 2x2 b p = 2t3 − 5t2 − 4t + 13 for t > 0

c y = 4 + 3x − x3 d p = |12x2 − 6|

Solution

a y = 9 + 12x − 2x2

dy

dx= 12 − 4x

A stationary point occurs whendy

dx= 0, i.e. when 12 − 4x = 0

∴ x = 3

When x = 3, y = 9 + 12 × 3 − 2 × 32

= 27

Thus the stationary point is at (3, 27).


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b P = 2t3 − 5t2 − 4t + 13dp

dt= 6t2 − 10t − 4, t > 0

dp

dt= 0 implies 2(3t2 − 5t − 2) = 0

∴ (3t + 1)(t − 2) = 0

∴ t = −1

3or t = 2

but t > 0 and therefore the only

acceptable solution is t = 2.

When t = 2, p = 16 − 20 − 8 + 13 = 1.

So the corresponding stationary point

is (2, 1).

c y = 4 + 3x − x3

dy

dx= 3 − 3x2

dy

dx= 0 implies 3(1 − x2) = 0

∴ x = ±1

∴ stationary points occur at (1, 6)

and (−1, 2).

d First consider the graph of y = 12x2 − 6.

It has a stationary point wheredy

dx= 0,

which is where x = 0. The coordinates of the

stationary point are (0, −6).

The graph of y = |12x2 − 6| is as shown. It

is clear that the stationary point has

coordinates (0, 6).

x

y

–6

0

6

Example 14

The curve with equation y = x3 + ax2 + bx + c passes through (0, 5) with stationary point

(2, 7). Find a, b, c.

Solution

When x = 0, y = 5

Thus 5 = cdy

dx= 3x2 + 2ax + b and at x = 2,

dy

dx= 0

Therefore 0 = 12 + 4a + b (1)

The point (2, 7) is on the curve and therefore:

7 = 23 + 22a + 2b + 5

2 = 8 + 4a + 2b

∴ 4a + 2b + 6 = 0 (2)

Subtract (1) from (2):

−b + 6 = 0

∴ b = 6

Substitute in (1):

0 = 12 + 4a + 6

−18 = 4a

−9

2= a

∴ a = −9

2, b = 6, c = 5


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Using the TI-NspireDefine (b 4 1)

f (x) = x3 + ax2 + bx + c

Find the derivative (b 4 1) of f (x)

and store (/ ) this derivative as d f (x)

as shown.

Then use solve ( f (0) = 5 and

f (2) = 7 and d f (2) = 0, {a, b, c}) to find

a, b and c.

Using the Casio ClassPadUsing Interactive > Define, define

f (x) = x3 + ax2 + bx + c and

g(x) = 3x2 + 2ax + b.

Then turn on the soft keyboard and tap

from the menu twice to produce a

simultaneous equation template with three

lines.

Enter the equations and variables as shown.

Exercise 10D

1 Find the stationary points for each of the following:

a f (x) = x3 − 12x b g(x) = 2x2 − 4x

c h(x) = 5x4 − 4x5 d f (t) = 8t + 5t2 − t3 for t > 0

e g(z) = 8z2 − 3z4 f f (x) = 5 − 2x + 3x2

g h(x) = x3 − 4x2 − 3x + 20, x > 0 h f (x) = 3x4 − 16x3 + 24x2 − 10

2 a The curve with equation f (x) = x2 − ax + 9 has a stationary point when x = 3.

Find a.

b The curve with equation h(x) = x3 − bx2 − 9x + 7 has a stationary point when

x = −1. Find b.


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3 The tangent to the curve of y = ax2 + bx + c at the point where x = 2 is parallel to the line

y = 4x. There is a stationary point at (1, −3). Find the value of a, b and c.

4 The graph of y = ax3 + bx2 + cx + d touches the line 2y + 6x = 15 at the point A

(0, 7

1

2

)and has a stationary point at B(3, −6). Find the values of a, b, c, and d.

5 The curve with equation y = ax + b

2x − 1has a stationary point at (2, 7). Find:

a the value of a and b b the coordinates of the other stationary point

6 Find the x-coordinates, in terms of n, of the stationary points of the curve with equation

y = (2x − 1)n(x + 2), where n is a natural number.

7 Find the x-coordinates of the stationary points of the curve with equation y = (x2 − 1)n

where n is an integer greater than 1.

8 Find the coordinates of the stationary points of the curve with equation y = x

x2 + 1

9 Find the coordinates of the stationary points of the curve with equation:

a y = |x2 − 4x| b y = |x3 − 4x| c y = |x3 − 4x2| d y = |x4 − 4x2|

10.5 Types of stationary pointsThe graph below has three stationary points A, B, C.

x

y

0

B

CA

A The point A is called a local maximum point. Notice

that immediately to the left of A, the gradient is

positive and immediately to right the gradient is

negative. A diagram to represent this is:

+ −0

gradient

B The point B is called a local minimum point. Notice

that immediately to the left of B, the gradient is

negative and immediately to the right the gradient is

positive. A diagram to represent this is:

− +0

gradient


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C The point C is called a stationary point of inflexion.

A diagram for this is:+ +0

gradient

Clearly it is also possible to have stationary points of

inflexion for which our diagram would be:

Stationary points of type A and B are referred to as turning points.

− −0

gradient

Example 15

For the function f : R → R, f (x) = 3x3 − 4x + 1:

a Find the stationary points and state their nature.

b Sketch the graph.

Solution

a y = f (x) has stationary points where f ′(x) = 0 and f ′(x) = 9x2 − 4 = 0

which implies x = ±2

3.

There are stationary points at

(−2

3, f

(−2

3

)), and

(2

3, f

(2

3

));

that is, at

(−2

3, 2

7

9

)and

(2

3, −7

9

).

f ′(x) is of constant sign on each of:{x: x < −2

3

},

{x: −2

3< x <

2

3

}and

{x: x >

2

3

}

To calculate the sign of f ′(x) on each of these sets, simply choose a

representative number in each.

Thus f ′(−1) = 9 − 4 = 5 > 0

f ′(0) = 0 − 4 = −4 < 0

f ′(1) = 9 − 4 = −5 > 0

We can thus put together the following table:

shape of f

0 0+ +−

−23

23

f '(x)

x

∴ there is a local maximum at

(−2

3, 2

7

9

)and a local minimum at

(2

3, −7

9

).


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b To sketch the graph of this function we need to find the axis intercepts and

investigate the behaviour of the graph for:

x >2

3and x < −2

3

Firstly, f (0) = 1, the y-intercept is 1.

Consider f (x) = 0, which implies 3x3 − 4x + 1 = 0

By inspection (factor theorem) (x − 1) is a factor and by division

3x3 − 4x + 1 = (x − 1)(3x2 + 3x − 1)

Now (x − 1)(3x2 + 3x − 1) = 0 implies that x = 1 or 3x2 + 3x − 1 = 0

3x2 + 3x − 1 = 3

[(x + 1

2

)2

− 1

4− 1

3

]

= 3

[(x + 1

2

)2

− 21

36

]

= 3

(x + 1

2− 1

6

√21

).

(x + 1

2+ 1

6

√21

)

Thus the x-intercepts are at:

x = −(

1

2+ 1

6

√21

), x = 1

6

√21 − 1

2, x = 1

For x >2

3, f (x) becomes larger.

For x < −2

3, f (x) becomes smaller.

x

y

–2 –1

–1

0

1

2

1 2

23

79

– , 2

23

79

, –


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Using the TI-NspireOnce you have plotted the graph of a function, the CAS calculator can be used to

determine a number of key values for that graph that include, where appropriate:

its value at any point the value of its derivative at any point

its zeros its local maxima its local minima.

Example 16

Plot the graph of y = x3 − 19x + 20 and determine:

a the value of y when x = −4 b the values of x when y = 0

c the value ofdy

dxwhen x = −1

d the coordinates of the local maximum.

Solution

Graph the equation of

y = x3 − 19x + 20 in an appropriate

Window (b 4).

Define f (x) = x3 − 19x + 20

a f (−4) = 32

b Solve ( f (x) = 0, x)


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c Find the derivative of f (x) at

x = 1 as shown.

d Solve

(d

dx( f (x)) = 0, x

)and

substitute to find the y-coordinate.

Using the Casio ClasspadDefine y = x3 − 19x + 20.

Plot the graph.

a f (−x) = 32

b Enter and highlight f (x) = 0 then tap

Interactive > Equation/inequality > solve

and set the variable to x.

c In , tap and enter x as the

variable (solid square) and f (x) as the

function (empty square). Enter the value

x = −1 as shown.

d In the screen the derivative has first been

found then copied to a new entry line and

solved. The y value can be found by

substitution of the solutions into f (x). (Copy

and paste for accuracy).

Example 17

Sketch the graph of y = |x3 − 36x|


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Solution

First consider the graph of y = |x3 − 36x|Intercepts

The x-axis intercepts occur where x3 − 36x = 0

x(x2 − 4) = 0 which implies x = 0 or x = 6 or x = −6

Turning points

dy

dx= 0 implies 3x2 − 36 = 0

Therefore there are turning points where x = ±√12 = ±2

√3

When x = 2√

3, y = (2√

3)3 − 36(2√

3)

= 24√

3 − 72√

3

= −48√

3

When x = −2√

3, y = (−2√

3)3 − 36(−2√

3)

= −24√

3 + 72√

3

= 48√

3

The turning points have

coordinates (2√

3, −48√

3)

and (−2√

3, 48√

3).

It is clear that (2√

3, −48√

3)

is a local minimum and

(−2√

3, 48√

3) is a local

maximum.

x

y

(–2√3, 48√3)

(2√3, –48√3)

0–6 6

The graph of y = |x3 − 36x|may now be drawn.

It is clear that (2√

3, 48√

3) and

(−2√

3, 48√

3) are local maxima.

x

y

(–2√3, 48√3) (2√3, 48√3)

0–6 6


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Exercise 10E

1 Write down the values of x for which the following derived functions are zero and prepare

in each case a gradient table, as in Example 15, showing whether the corresponding

points on the graph are local maxima, local minima or stationary points of inflexion:

a f ′(x) = 4x2 b f ′(x) = (x − 2)(x + 5)

c f ′(x) = (x + 1)(2x − 1) d f ′(x) = −x2 + x + 12

e f ′(x) = x2 − x − 12 f f ′(x) = 5x4 − 27x3

g f ′(x) = (x − 1)(x − 3) h f ′(x) = −(x − 1)(x − 3)

2 Find the stationary points on each of the following curves and state their nature:

a y = x(x2 − 12) b y = x2(3 − x)

c y = x3 − 5x2 + 3x + 2 d y = 3 − x3

e y = 3x4 + 16x3 + 24x2 + 3 f y = x(x2 − 1)

3 Sketch the graphs of each of the following, finding:

i axes intercepts ii stationary points

a y = 4x3 − 3x4 b y = x3 − 6x2

c y = 3x2 − x3 d y = x3 + 6x2 + 9x + 4

e y = (x2 − 1)5 f y = (x2 − 1)4

4 a Find the stationary points of the graph y = 2x3 + 3x2 − 12x + 7, stating the nature of

each.

b Show that the graph passes through (1, 0).

c Find the other axes intercepts. d Sketch the graph.

5 a Show that the polynomial P(x) = x3 + ax2 + b has a stationary point at x = 0 for all a

and for all b.

b Given that P(x) has a second stationary point at (−2, 6), find the values of a and b and

the nature of both stationary points.

6 Sketch the graph of f (x) = (2x − 1)5(2x − 4)4. Clearly state the coordinates of:

a axes intercepts b stationary points

State the nature of each of the stationary points.

7 a Sketch the graph of f (x) = (4x2 − 1)6 and g(x) = (4x2 − 1)5 on the one set of axes.

b Find:

i {x: (4x2 − 1)6 > (4x2 − 1)5} ii {x: f ′(x) > g′(x)}


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8 Sketch the graph of each of the following. State the axes intercepts and coordinates of

stationary points.

a y = x3 + x2 − 8x − 12 b y = 4x3 − 18x2 + 48x − 290

9 Find the coordinates of the stationary points of each of the following and determine their

nature:

a f (x) = 3x4 + 4x3 b f (x) = x4 + 2x3 − 1 c f (x) = 3x3 − 3x2 + 12x + 9

10 Consider the function f defined by:

f (x) = 1

8(x − 1)3(8 − 3x) + 1

a Show that f (0) = 0 and f (3) = 0.

b Show that f ′(x) = 3

8(x − 1)2(9 − 4x) and specify the value of x for which f ′(x) ≥ 0.

c Sketch the graph of y = f (x)

11 Sketch the graph of y = 3x4 − 44x3 + 144x2, finding the coordinates of all turning

points.

12 The graphs below show the graph of f ′ of a function f. Find the values of x for

which the graph of y = f (x) has a stationary point and state the nature of the stationary

point.

a

y = f '(x)

x

y

–1 1

0

5

b

y = f '(x)

x

y

20

c

y = f '(x)

x

y

0–4

d

y = f '(x)

x

y

2–3 0

13 Find the coordinates of the stationary points, and state the nature of each, for the curve

with equation:

a y = x4 − 16x2 b y = |x4 − 16x2|c y = |x2m − 16x2m−2| where m is a positive integer greater than or equal to 2


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14 Let f : R → R, f (x) = x3 − 3x2 − 9x + 11. Sketch the graph of:

a y = f (x) b y = 2 f (x) c y = f (x + 2)

d y = f (x − 2) e y = − f (x)

15 Let f : R → R, f (x) = 2 + 3x − x3. Sketch the graph of:

a y = f (x) b y = −2 f (x) c y = 2 f (x − 1)

d y = f (x) − 3 e y = 3 f (x + 1)

16 The graph shown opposite has equation

y = f (x). Suppose a dilation of factor p

from the x-axis followed by a translation

of l units in the positive direction of the

x-axis is applied to the graph.

For the graph of the image, state:x

y

A(a, 0) B(b, 0)

P(h, k)

0

a the axes intercepts

b the coordinates of the turning point.

17 It is found that the graph of quartic function passes through the points with coordinates

(1, 21), (2, 96), (5, 645), (6, 816), (7, 861).

a Find the rule of the quartic and plot the graph. Determine the turning points and axes

intercepts.

b Plot the derivative graph on the same screen.

c Find the value of the function when x = 10.

d For what value(s) of x is the value of the function 500?

10.6 Absolute maxima and minimaIn the last section local maxima and minima were discussed. These are often not the actual

maximum and minimum values of the function. The actual maximum value for a function

defined on an interval is called the absolute maximum. The corresponding point on the graph

of the function is not necessarily a stationary point. The actual minimum value for a function

defined on an interval is called the absolute minimum. The corresponding point on the graph

of the function is not necessarily a stationary point.

M is the absolute maximum value of a continuous function f for an interval [a, b] if

f (x) ≤ M for all x ∈ [a, b] .

N is the absolute minimum value of a continuous function f for an interval [a, b] if

f (x) ≥ N for all x ∈ [a, b].

Example 18

Let f : [−2, 4] → R, f (x) = x2 + 2. Find the absolute maximum and the absolute minimum

value of the function.


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Solution

The minimum value occurs when x = 0

and is 2.

The maximum value occurs when x = 4

and is 18.

The minimum value occurs at a stationary

point of the graph but the endpoint (4, 18)

is not a stationary point.

The absolute maximum value is 18 and

the absolute minimum value is 2.

y(4, 18)

(–2, 6)

0

(0, 2)x

Example 19

Let f : [−2, 1] → R, f (x) = x3 + 2. Find the maximum and minimum value of the function.

Solution

The minimum value occurs when x = −2

and is −6.

The maximum value occurs when x = 1

and is 3.

The absolute minimum value and the absolute

maximum value do not occur at stationary points.

(1, 3)

(–2, –6)

(0, 2)

y

0x

Example 20

From a square piece of metal of side length 2 m,

four squares are removed as shown in the figure opposite.

The metal is then folded about the dotted lines to

give an open box with sides of height x m.

a Show that the volume, V m3, is given by

V = 4x3 − 8x2 + 4x

b Sketch the graph of V against x for a suitable domain.

c If the height of the box must be less than 0.3 m,

i.e. x ≤ 0.3, what will be the maximum volume of the box?

Solution

a Length of box = (2 − 2x) metres, height = x metres

∴ volume = (2 − 2x)2x

= (4 − 8x + 4x2)x

= 4x3 − 8x2 + 4x cubic metres


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b Let V = 4x3 − 8x2 + 4x

Local maximum point will occur whendV

dx= 0

dV

dx= 12x2 − 16x + 4

anddV

dx= 0 implies

12x2 − 16x + 4 = 0

∴ 3x2 − 4x + 1 = 0

∴ (3x − 1)(x − 1) = 0

∴ x = 1

3or x = 1

Note: When x = 1, the length of the box = 2 − 2 × 1 which is zero.

∴ the only value to be considered is x = 1

3

∴ a local maximum occurs when x = 1

3

and volume =(

2 − 2 × 1

3

)2

× 1

3

= 16

9× 1

3

= 16

27m3

0 1

V(m3)

x (cm)

16

2713

,

c The local maximum of V (x) defined on [0, 1] was at

(1

3,

16

27

).

But for the new problem V ′ (x) > 0 for all x ∈ [0, 0.3] and1

3is not in this

interval. Therefore the maximum volume occurs when x = 0.3 and is 0.588.

Example 21

For the function f : [−1, 3] → R, f(x) = 2 − |x |, sketch the graph and state the absolute

maximum and minimum value of the function.

Solution

There are no stationary points, but the

function has an absolute maximum value of 2

when x = 0. It has a minimum value

of −1 when x = 3.( –1, 1)

(3, –1)

(0, 2)

y

0x


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Exercise 10F

1 Let f : [−3, 2] → R, f(x) = x3 + 2x + 3. Find the absolute maximum and absolute

minimum value of the function for its domain.

2 Let f : [−1.5, 2.5] → R, f(x) = 2x3 − 6x2. Find the absolute maximum and absolute

minimum value of the function.

3 Let f : [−2, 6] → R, f(x) = 2x4 − 8x2. Find the absolute maximum and absolute


4 Let f : [−3, 3] → R, f(x) = 2 − 8x2. Find the absolute maximum and absolute


5 A rectangular block is such that the sides of its base are of length x cm and 3x cm. The

sum of the lengths of all its edges is 20 cm.

a Show that the volume V cm3 is given by V = 15x2 − 12x3

b FinddV

dx.

c Find the coordinates of the local maximum of the graph of V against x for

x ∈ [0, 1.25].

d If x ∈ [0, 0.8] find the absolute maximum value of V and the value of x for which this

occurs.

e If x ∈ [0, 1] find the absolute maximum value of V and the value of x for which this

occurs.

6 For the variables x, y and z it is known that x + y = 30 and z = xy

a If x ∈ [2, 5] find the possible values of y.

b Find the absolute maximum and minimum values of z.

7 A piece of string 10 metres long is cut into two pieces to form two squares.

a If one piece of string has length x metres show that the combined area of the two

squares is given by A = 1

8(x2 − 10x + 50)

b Findd A

dx.

c Find the value of x that makes A a minimum.

d If two squares are formed but x ∈ [0, 11], find the maximum possible area of the two

squares.

8 Find the absolute maximum and minimum values of the function

g: [2.1, 8] → R, g(x) = x + 1

x − 2

9 Consider the function f : [2, 3] → R, f(x) = 1

x − 1+ 1

4 − x.

a Find f ′(x).

b Find the coordinates of the stationary point of the graph of y = f(x).

c Find the absolute maximum and absolute minimum of the function.Cambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

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10 Consider the function f : [0, 3] → R, f (x) = 1

x + 1+ 1

4 − x.

a Find f ′(x).

b Find the coordinates of the stationary point of the graph of y = f(x)

c Find the absolute maximum and absolute minimum of the function.

11 For the function f : [−1, 8] → R, f(x) = 2 − x23 , sketch the graph and state the absolute

maximum and minimum value of the function.

10.7 Maxima and minima problemsMany practical problem require that some quantity (for example, cost of manufacture or fuel

consumption) be minimised, that is, be made as small as possible. Other problems require that

some quantity (for example, profit on sales or attendance at a concert) be maximised, that is,

be made as large as possible. We can use differential calculus to solve many of these problems.

Example 22

A farmer has sufficient fencing to make a rectangular pen of perimeter 200 metres. What

dimensions will give an enclosure of maximum area?

Solution

Let the required length of fencing be x metres. Then the width is 100 − x metres and

the area is A square metres where A = x(100 − x) = 100x − x2

The maximum value of A occurs whendA

dx= 0

dA

dx= 100 − 2x

anddA

dx= 0 implies x = 50

x

A

0 100To see that x = 50 gives a maximum we may

use the gradient chart

slope

0+ −

50

dAdx

x

∴ maximum when x = 50 and A = 50(100 − 50) = 2500

∴ maximum area = 2500 square metres

∴ the corresponding dimensions are 50 m by 50 m.


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Example 23

Two variables x and y are such that x4y = 8. A third variable z is defined by z = x + y. Find the

values of x and y that give z a stationary value and show that this value of z is a minimum.

Solution

Obtain y in terms of x from the equation x4y = 8

y = 8x−4

Substitute in the equation z = x + y

z = x + 8x−4 (1)

Therefore z is in terms of one variable, x.

Differentiate with respect to x:

dz

dx= 1 − 32x−5

Stationary point occurs wheredz

dx= 0

1 − 32x−5 = 0

∴ 32x−5 = 1

∴ x5 = 32

∴ x = 2 and the corresponding value of y is 8 × 2−4 = 1

2Now substitute in equation (1) to find z:

z = 2 + 8

16

= 21

2

Determine nature of stationary point by gradient chart:

− +

2

dzdx

This gives a minimum at

(2, 2

1

2

).


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Example 24

A cylindrical tin canister closed at both ends has a surface area of 100 cm2. Find, correct to

two decimal places, the greatest volume it can have. If the radius of the canister can be at most

2 cm, find the greatest volume it can have.

Solution

Let the radius of the circular end of the tin be r cm and the height of the tin h cm. Also

let the volume of the tin be V m3.

Obtain equations for the surface area and the volume.

Surface area: 2�r2 + 2�rh = 100 (1)

Volume: V = �r2h (2)

The process we follow now is very similar to Example 23.

Obtain h in terms of r from equation (1):

h = 1

2�r(100 − 2�r2)

Substitute in equation (2):

V = �r2 × 1

2�r(100 − 2�r2)

∴ V = 50r − �r3 (3)

The stationary point of the graph of V = 50r − �r3 occurs whendV

dr= 0

dV

dr= 0 implies 50 − 3�r2 = 0

∴ r = ±√

50

3�≈ 2.3

But r = −√

50

3�does not fit the practical situation.

Substitute in equation (3) to find V:

V ≈ 76.78

The maximum volume is 76.78 cm3 correct to two decimal places.


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Determine the nature of the stationary point.

0+ −

2.3

giving a maximum at (2.3, 76.8).

It can be observed that the volume is given by

a function f with rule f (r ) = 50r − �r3 and

domain

(0,

√50

�

), giving the graph.

r

V

0

(2.3, 76.8)

π50

If the greatest radius the canister can have is 2 cm, then the function f has domain

[0, 2]. It has been seen that f ′(r ) > 0 for all r ∈ [0, 2]. The maximum value occurs

when r = 2. The maximum volume in this case is f (2) = 100 − 8� ≈ 74.87 cm3.

In some situations the variables may not be continuous. For instance one or the other of them

may only take integer values. In such cases it is not strictly valid to use techniques of

differentiation to solve the problem. However, in some problems we may model the

non-continuous case with a continuous function so that the techniques of differential calculus

may be used. Worked examples 25 and 26 illustrate this.

Example 25

A TV cable company has 1000 subscribers who are paying $5 per month. It can get 100 more

subscribers for each $0.10 decrease in the monthly fee. What rate will yield maximum revenue

and what will this revenue be?

Solution

Let x denote the rate.

Then the number of subscribers = 1000 + 100

(5 − x

0.1

)(Note that we are treating a discrete situation with a continuous function.)

And the revenue = x(1000 + 1000(5 − x))

= 1000(6x − x2)

Let R denote the revenue.

ThendR

dx= 1000(6 − 2x)

dR

dx= 0 implies 6 − 2x = 0

Hence x = 3

The gradient chart is:the shape

of f

0+ −

3

f '(x)

x

Thus for maximum revenue the rate should be $3 and this gives a total

revenue of $9000.Cambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

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Example 26

A manufacturer annually produces and sells 10 000 shirts. Sales are uniformly distributed

throughout the year. The production cost of each shirt is $23 and other costs (storage,

insurance, interest) depend on the total number of shirts in a production run. (A production run

is the number, x, of shirts which are under production at a given time.)

The other costs (annually) are $x32 . The set-up costs for a production run are $40.

Find the size of a production run which will minimise the other costs for a year.

Solution

The number of production runs per year = 10 000

x

The set-up costs for these production runs = 40

(10 000

x

)

The total set-up and carrying costs, C = x32 + 400 000

x, x > 0

∴ C = x32 + 400 000x−1

dC

dx= 3

2x

12 − 400 000

x2

dC

dx= 0 implies

3

2x

12 = 400 000

x2

∴ x52 = 400 000 × 2

3

∴ x ≈ 148.04shape

0− +

148.04

dCdx

x

Each production run should be 148 items.

x

20000

15000

10000

5000

100

23

C = x + 400 000x–1

200

Number of shirts in production run

300 400

C ($)


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Exercise 10G

1 Find two positive numbers whose sum is 4 and such that the sum of the cube of the first and

the square of the second is as small as possible.

2 Find the point on the parabola y = x2 that is closest to the point (3, 0).

3 The number of salmon swimming upstream in a river to spawn is approximated by

s(x) = −x3+ 3x2 + 360x + 5000 with x representing the temperature of the water in

degrees (◦C). (This function is valid only if 6 ≤ x ≤ 20.) Find the water temperature that

produces the maximum number of salmon swimming upstream.

4 The number of mosquitoes, M(x) in millions, in a certain area depends on the September

rainfall, x, measured in mm, and is approximately given by:

M(x) = 1

30(50 − 32x + 14x2 − x3)

Find the rainfall that will produce the maximum and the minimum number of

mosquitoes.

5 Find the maximum area of a rectangular piece of ground that can be enclosed by 100 m of

fencing.

6 For x + y = 100 prove that the product P = xy is a maximum when x = y and find the

maximum value of P.

7 A farmer has 4 km of fencing wire and wishes to fence a rectangular piece of land through

which flows a straight river, which is to be utilised as one side of the enclosure. How can

this be done to enclose as much land as possible?

8 Two positive quantities p and q vary in such a way that p3q = 9. Another quantity z is

defined by z = 16p + 3q. Find values of p and q that make z a minimum.

9 A cuboid has a total surface area of 150 cm2 with a square base of side x cm.

a Show that the height, h cm, of the cuboid is given by h = 75 − x2

2xb Express the volume of the cuboid in terms of x.

c Hence determine its maximum volume as x varies.

10.8 Rates of changeThe derivative of a function has been used to find the gradient of the corresponding curve. It is

clear that the process of differentiation may be used to tackle problems involving rates of

change of many kinds.

The derivative of y with respect to x,dy

dx, gives the rate of change of y with respect to x. Of

course ifdy

dx> 0 the change is an increase in the value of y corresponding to an increase in x,

and ifdy

dx< 0 the change is a decrease in the value of y corresponding to an increase in x.


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Example 27

A balloon which develops a microscopic leak will decrease in volume. Its volume V cm3 at

time t seconds is V = 600 − 10t − 1

100t2, t > 0.

a Find the rate of change of volume after:

i 10 seconds ii 20 seconds

b For how long could the model be valid?

Solution

adV

dt= −10 − t

50i When t = 10,

dV

dt= −10 − 1

5

= −101

5

i.e. the volume is decreasing at a

rate of 101

5cm3 per second.

ii When t = 20,

dV

dt= −10 − 2

5

= −102

5

i.e. the volume is decreasing at a

rate of 102

5cm3 per second.

b The model will not be meaningful when V < 0. Consider V = 0.

600 − 10t − 1

100t2 = 0

∴ t =10 ±

√100 + 1

100× 600 × 4

−0.02∴ t = −1056.78 or t = 56.78 (to two decimal places)

∴ the model may be suitable for 0 < t < 56.78

Displacement, velocity and acceleration were introduced for a body moving in a straight line

in earlier work.* Displacement was specified with respect to a reference point O on that line.

For velocity (v m/s):

v = ds

dt

and acceleration (a m/s2)

a = dv

dt

Example 28

A point moves along a straight line so that its distance x cm from a point O at time t seconds is

given by the formula x = t3 − 6t2 + 9t. Find:

a at what times and in what positions the point will have zero velocity

b its acceleration at those instants c its velocity when its acceleration is zero

* Essential Mathematical Methods 1 & 2 CASCambridge University Press • Uncorrected Sample Pages • 2008 © Evans, Lipson, Jones, Avery, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

SAMPLE

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Solution

Velocity = v

= dx

dt

= 3t2 − 12t + 9

a When v = 0,

3(t2 − 4t + 3) = 0

(t − 1)(t − 3) = 0

t = 1 or t = 3

i.e. the velocity is zero when

t = 1 and t = 3 and where

x = 4 and x = 0

b Acceleration = dv

dt= 6t − 12

∴ acceleration = −6 m/s2 when t = 1 and

acceleration = 6 m/s2 when t = 3

c The acceleration is zero when 6t − 12 = 0, i.e. when t = 2

When t = 2, the velocity v = 3 × 4 − 24 + 9

= −3 m/s

Exercise 10H

1 Express each of following in symbols:

a the rate of change of volume (V ) with respect to time (t)

b the rate of change of surface area (S ) of a sphere with respect to radius (r)

c the rate of change of volume (V ) of a cube with respect to edge length (x)

d the rate of change of area (A ) with respect to time (t)

e the rate of change of volume (V ) of water in a glass, with respect of depth of

water (h)

2 If your interest (I) in Essential Mathematical Methods can be expressed as I = 4

(t + 1)2 ,

where t is the time in days measured from the first day of Term 1, how fast is your interest

waning when t = 10?

3 A reservoir is being emptied, and the quantity of water, V m3, remaining in the reservoir t

days after it starts to empty is given by V(t) = 103(90 − t)3.

a At what rate is the reservoir being emptied at time t?

b How long does it take to empty the reservoir?

c What is the volume of water in the reservoir when t = 0?

d After what time is the reservoir being emptied at 3 × 105 m3/day?

e Sketch the graph of V(t) against t.

f Sketch the graph of V ′(t) against t.

4 A coffee percolator allows 1000 mL of water to flow into a filter in 20 minutes. The

volume which has flowed into the filter t minutes from the start is given by:

V(t) = 1

160

(5t4 − t5

5

), 0 ≤ t ≤ 20


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a At what rate is the water flowing into the filter at time t minutes?

b Sketch the graph ofdV

dtagainst t for 0 ≤ t ≤ 20.

c When is the rate of flow greatest?

5 The graph shown below is that of the volume, V m3, of water in a reservoir at time t days.

Time(days)

550500

12345678

100 150 200 250 300 350 400 450 500

V (× 107 m3)

a At what times is the rate of flow

from the reservoir 0 m3/day?

b Find an estimate for the flow at

t = 200.

c Find the average rate of flow for

the interval [100, 250].

d State the times for which there is

net flow into the reservoir.

6 A point moves along a straight line so that its distance, x cm, from a point O at

time t seconds is given by x = 2t3 − 9t2 + 12t. Find:

a the velocity v as a function of t

b at what times and in what positions the point will have zero velocity

c its acceleration at those instants d its velocity when its acceleration is zero

7 A particle moves in a straight line such that its position x cm from a point O at time t

seconds is given by the equation x = 8 + 2t − t2. Find:

a its initial position b its initial velocity

c when and where the velocity is zero d its acceleration at time t

8 A particle is moving in a straight line such that its position x cm from a point O at time t

seconds is given by x = √2t2 + 2. Find:

a the velocity as a function of t b the acceleration as a function of t

c the velocity and acceleration when t = 1

10.9 Related rates of changeConsider the situation of a container, which is a right

circular cone, being filled from a tap.

At time t seconds:

there are V cm3of water in the cone

the height of the water in the cone is h cm

the radius of the circular water surface is r cm.

r cm

10 cm

30 cm

h cmV, h and r change as the water flows in:

dV

dtis the rate of change of volume with respect to time.

dh

dtis the rate of change of height with respect to time.

dr

dtis the rate of change of radius with respect to time.


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It is clear that these rates are related to each other. The chain rule is used to establish these

relationships.

For example, if the height of the cone is 30 cm and the radius of the cone is 10 cm, similar

triangles yield

r

h= 10

30

and h = 3r

Then the chain rule is used.

dh

dt= dh

dr

dr

dt

= 3 · dr

dt

r cm

10 cm

30 cm

h cm

The volume of a cone is given in general by V = 1

3�r2h and in this case V = �r3 since

h = 3r. Therefore by using the chain rule again:

dV

dt= dV

dr

dr

dt

= 3�r2 · dr

dt

The relationships between the rates have been established.

Example 29

The radius of a circular disc is increasing at a constant rate of 0.005 cm/s. Find the rate at

which the area of the disc is increasing when the radius is 20 cm.

Solution

Let A cm2 be the area of the disc and r cm the radius of the disc.

A = �r2 anddA

dr= 2�r

ThendA

dt= dA

dr× dr

dt= 2�r × 0.005

= �r

100

When r = 20,dA

dt= 20�

100

= �

5cm2/s

The area of the disc is increasing at a rate of�

5cm2/s.


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Example 30

Variables x and y are related by the equation y = 2x − 4

x. Given that x and y are functions of t

and thatdx

dt= 4, find:

ady

dtin terms of x b

dy

dtwhen x = 4

Solution

a For y = 2x − 4

x, y = 2 − 4

xand

dy

dx= 4

x2

Thereforedy

dt= dy

dx× dx

dt

= 4

x2× 4

= 16

x2

b When x = 4,dy

dt= 16

42

= 1

Example 31

The diagram shows a block of ice. Dimensions are in centimetres.

a Find an expression for the surface area, S cm2, in terms of x.

b FinddS

dx.

c The ice is melting so that S is decreasing at a rate of 4 cm2/s.

Find the rate at which x is changing when x = 5. 9xx

x

Solution

a S = 9x2 + 9x2 + 9x2 + 9x2 + x2 + x2 bdS

dx= 76x2

= 38x2

cdS

dt= dS

dx× dx

dt

−4 = 76x × dx

dt−4

76x= dx

dt

When x = 5,dx

dt= −4

76 × 5

= −1

95

x is decreasing at a rate of1

95cm/s.


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Exercise 10I

1 A circular patch of oil spreads out from a point on a lake so that the area of the patch

grows at a rate of 3 cm2/s. At what rate is the radius increasing when the radius of the oil

patch is 5 cm?

2 The edges of a square are increasing in length at a rate of 0.2 cm/s. Find the rate at which

the area of the square is increasing when the length of each side of the square is 20 cm.

3 A circular slick is increasing at a rate of 5 cm2/s. Find the rate of increase of the radius of

the slick when the area is 400� cm2.

4 When the depth of liquid in a container is x cm the volume is x(x2 + 36) cm3. Liquid is

added to the container at a rate of 3 cm3/s. Find the rate of change of the depth of liquid at

the instant when x = 11.

5 The surface area of a cube is changing at the rate of 8 cm2/s. How fast is the volume

changing when the surface area is 60 cm?

6 A vessel has such a shape that when the depth of the water in it is x cm, the volume,

V cm3, is given by the rule V = 108x + x2. Water is poured in at a constant rate of

30 cm3/s. At what rate is the level of water rising when the depth is 8 cm?

7 A melting snowball which is always spherical in shape is decreasing in volume at a

constant rate of 8 cm3/min. Find the rate at which the radius is changing when the

snowball has a radius of 4 cm. (The volume of a sphere of radius r is4

3�r3.)

8 Variables x and y are related by the equation y = (x2 − 5)10. Given that x and y are

functions of t and thatdx

dt= 1, find:

ady

dtin terms of x b

dy

dtwhen x = 4

9 A species of tree has a perfectly cylindrical trunk. It grows so that the radius of the trunk

increases uniformly at a rate of 0.01 metre/year. The height of the trunk remains constant.

One particular tree has radius 0.2 metres and height 5.0 metres. Find the rate at which the

volume of its trunk is increasing.

10 The diagram shows a container in the form of an inverted

cone of height 80 cm and diameter 40 cm. Water is poured

into the container at the rate of 20 cm3/s.

a Find the volume, V cm3, of water in the container when

the depth is x cm.

b Find the rate at which the depth is increasing when x = 5.

40 cm

80 cm


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10.10 Families of functionsExample 32

Consider the function f (x) = (x − a)2(x − b) where a and b are positive constants with

b > a.

a Find the derivative of f(x) with respect to x.

b Find the coordinates of the stationary points of the graph of y = f (x)

c Show that the stationary point at (a, 0) is always a local maximum.

d Find the values of a and b if the stationary points occur where x = 3 and x = 4

Solution

a Use a CAS calculator to find that f ′(x) = (x − a)(3x − a − 2b)

b The coordinates of the stationary points are (a, 0) and

(a + 2b

3,

4(a − b)3

27

)

c If x < a then f ′(x) > 0 and if x > a and x <a + 2b

3then f ′(x) < 0, therefore

local maximum.

d a = 3 as a < b anda + 2b

3= 4 implies b = 9

2

Example 33

The graph of the function y = x3 − 3x2 is translated by a units in the positive direction of the

x-axis and b units in the positive direction of the y-axis (a and b are positive constants).

a Find the coordinates of the turning points of the graph of y = x3 − 3x2

b Find the coordinates of the turning points of its image.

Solution

a The turning points have coordinates (0, 0) and (2, −4).

b The turning points of the image are (a, b) and (2 + a, −4 + b).

Example 34

A cubic function has rule y = ax3 + bx2 + cx + d. It passes through the points (1, 6) and

(10, 8) and has turning points where x = −1 and x = 2.

a Find using matrix methods the values of a, b, c and d.

b Find the equation of the image of the curve under the transformation defined by the matrix

equation:

T(X + B) = X′ where T =[

0 −3

2 0

]and B =

[1

2

]


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Solution

a A CAS calculator has been used for matrix calculations.

The equations obtained are:

6 = a + b + c + d

8 = 1000a + 100b + 10c + d

And asdy

dx= 3ax2 + 2bx + c

0 = 3a − 2b + c

and

0 = 12a + 4b + c

These can be written as a matrix equation.

1 1 1 1

1000 100 10 1

3 −2 1 0

12 4 1 0

a

b

c

d

=

6

8

0

0

Therefore

a

b

c

d

=

1 1 1 1

1000 100 10 1

3 −2 1 0

12 4 1 0

−1

6

8

0

0

=

4

1593

−2

531

−8

531

9584

1593

Therefore a = 4

1593b = −2

531c = −8

531d = 9584

1593

b First solve the matrix equation for X.

T−1 T(X + B) = T−1X′

X + B = T−1 X′

and X = T−1 X′ − B

Therefore

[x

y

]=

0

1

2−1

30

[x ′

y′

]−

[1

2

]=

y′

2

− x ′

3

−

[1

2

]

Therefore x = y′

2− 1 and y = − x ′

3− 2


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The image curve has equation

− x ′

3= a

(y′

2− 1

)3

+ b

(y′

2− 1

)2

+ c

(y′

2− 1

)+ d + 2

where a, b, c and d have the values given above.

Example 35

A cubic function f has rule f (x) = ax3 + bx2 + cx . The graph has a stationary point at

(1, 6).

a Find a and b in terms of c.

b Find the value of c for which the graph has a stationary point at the point where x = 2.

Solution

a Two equations are obtained.

6 = a + b + c as f (1) = 6

f ′(x) = 3ax2 + 2bx + c and therefore

3a + 2b + c = 0

The solution is a = c − 12 and b = 18 − 2c

The equation is f (x) = (c − 12)x3 + (18 − 2c)x2 + cx

b f ′(x) = 3(c − 12)x2 + 2(18 − 2c)x + c

f ′(2) = 0

Therefore 12(c − 12) + 4(18 − 2c) + c = 0

5c = 72

c = 72

5

Exercise 10J

1 Consider the function f (x) = (x − 1)2(x − b) where b > 1.

a Find the derivative of f(x) with respect to x.

b Find the coordinates of the stationary points of the graph of y = f (x)

c Show that the stationary point at (1, 0) is always a local maximum.

d Find the value of b if the stationary points occur where x = 1 and x = 4.

2 Consider the function f : [0, ∞) → R, defined by

f (x) = x2 − ax3

where a is a real number, a > 0.

a Determine intervals on which f is a decreasing function and the intervals on which f is

an increasing function.

b Find the equation of the tangent to the graph of f at the point (1

a, 0).

c Find the equation of the normal to the graph of f at the point (1

a, 0).

d What is the range of f ?


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3 A line with equation y = mx + c is a tangent to the curve y = (x − 3)2 at a point P where

x = a, such that 0 < a < 3.

(3, 0)

y

0x

P

a

a i Find the gradient of the curve where x = a, for 0 < a < 3.

ii Hence express m in terms of a.

b State the coordinates of the point P, expressing your answer in terms of a.

c Find the equation of the tangent where x = a.

d Find the x-axis intercept of the tangent.

4 a The graph of f (x) = x4 is translated to the graph of y = f (x + h). Find the possible

values of h if f (1 + h) = 16.

b The graph of f (x) = x3 is transformed to the graph of y = f (ax). Find the possible

value of a if the graph of y = f (ax) passes through the point with coordinates (1, 8).

c The quartic function with equation y = ax4 − bx3 has a turning point with

coordinates (1, 16). Find the values of a and b.

5 The graph of the function y = x4 − 4x2 is translated by a units in the positive direction of

the x-axis and b units in the positive direction of the y-axis (a and b are positive

constants).

a Find the coordinates of the turning points of the graph of y = x4 − 4x2

b Find the coordinates of the turning points of its image.

6 Consider the cubic function with rule f (x) = (x − a)2(x − 1) where a > 1.

a Find the coordinates of the turning points of the graph of y = f (x)

b State the nature of each of the turning points.

c Find the equation of the tangent to the curve at the point where:

i x = 1 ii x = a iii x = a + 1

2

7 Consider the quartic function with rule f (x) = (x − 1)2(x − b)2 with b > 1.

a Find the derivative of f.

b Find the coordinates of the turning points of f.

c Find the value of b such that the graph of y = f (x) has a turning point at (2, 1).


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8 A cubic function has rule y = ax3 + bx2 + cx + d. It passes through the points (1, 6) and

(10, 8) and has turning points where x = −1 and x = 1.

a Find using matrix methods the values of a, b, c and d.

b Find the equation of the image of the curve under the transformation defined by the

matrix equation:

T(X + B) = X′ where T =[

0 −2

1 0

]and B =

[1

3

]

9 A cubic function f has rule f (x) = ax3 + bx2 + cx . The graph has a stationary point at

(1, 10).

a Find a and b in terms of c.

b Find the value of c for which the graph has a stationary point at the point where x = 3.

10 A quartic function f has rule f (x) = ax4 + bx3 + cx2 + dx . The graph has a stationary

point at (1, 1) and passes through the point (−1, 4).

a Find a, b and c in terms of d.

b Find the value of d for which the graph has a stationary point at the point where x = 4.


SAMPLE

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Review


Chapter summary

For point (x, y) on the curve with equation y = f (x), the equation of the tangent to the

curve at (x1, y1) is

y − y1 = f ′(x1)(x − x1)

and the equation of the normal to the curve at (x1, y1) is

y − y1 = − 1

f ′(x1)(x − x1)

Two curves with equations y = f (x) and y = g(x) intersect at (x1, y1) and have gradients

m1 and m2 respectively.

Let � be the angle between the curves at (x1, y1).

Then tan � = m1 − m2

1 + m1m2

The relationship f (x + h) ≈ f (x) + h f ′(x) for a small value of h is used to estimate a

value of f (x + h) close to a known value f (x).

The curve with equation y = f (x) has stationary points where f ′(x) = 0.

The point (a, f (a)) is a local maximum if

f ′(a) = 0

and immediately to the left the gradient is positive and immediately to the right the gradient

is negative.

The point (b, f (b)) is a local minimum if

f ′(b) = 0

and immediately to the left the gradient is negative and immediately to the right the

gradient is positive.

The point (c, f (c)) is a stationary point of inflexion if

f ′(c) = 0

and immediately to the left and the right the gradient is positive or immediately to the left

and the right the gradient is negative.

M is the absolute maximum value of a function f in an interval [a, b] if f (x) ≤ M for

all x ∈ [a, b].

N is the absolute minimum value of a function f in an interval [a, b] if f (x) ≥ N for all

x ∈ [a, b].

Multiple-choice questions

1 The line with equation y = 4x + c is a tangent to the curve with equation y = x2 − x – 5.

The value of c is:

A−45

4B −1 + 2

√2 C 2 D

5

2E −2

5


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Rev

iew


2 The equation of the tangent to the curve of the function with equation y = x4 at the point

where x = 1 is:

A y = −4x − 3 B y = 1

4x − 3 C y = −4x D y = 1

4x + 5

4E y = 4x − 3

3 For the function with rule f (x), f ′(a) = f ′(b) = 0, f ′(x) > 0 for x ∈ [a, b], f ′(x) < 0 for

x < a and f ′(x) > 0 for x > b. The nature of the stationary points of the graph of y = f (x)

at the points with coordinates (a, f (a)) and (b, f (b)) is:

A local maximum at (a, f (a)) and local minimum at (b, f (b))

B local minimum at (a, f (a)) and local maximum at (b, f (b))

C stationary point of inflexion at (a, f (a)) and local minimum at (b, f (b))

D stationary point of inflexion at (a, f (a)) and local maximum at (b, f (b))

E local minimum at (a, f (a)) and stationary point of inflexion at (b, f (b))

4 Water is draining from a cone-shaped funnel at a rate of 500 cm3/min. The cone has a base

radius of 20 cm and a height of 100 cm. Let h cm be the depth of water in the funnel at time

t minutes. The rate of decrease of h in cm/min is given by:

A3750

�h2B

12 500

�h2C 30�h2 D 10 E

24

�

5 The radius of a sphere is increasing at a rate of 5 cm/min. When the radius is 10 cm, the

rate of increase, in cm3/min, of the volume of the sphere is:

A 2000� B2000�

3C

4000�

3D

40�

3E

500�

3

6 The graph of the function with rule y = f (x) has a local maximum at the point with

coordinates (a, f (a)). The graph also has a local minimum at the origin but no other

stationary points. The graph of the function with rule y = −2 f( x

2

)+ k where k is a

positive real number has:

A a local maximum at the point with coordinates (2a, −2 f (a) + k)

B a local minimum at the point with coordinates(a

2, 2 f (a) + k

)C a local maximum at the point with coordinates

(a

2, −2 f (a) + k

)D a local maximum at the point with coordinates (2a, −2 f (a) − k)

E a local minimum at the point with coordinates (2a, −2 f (a) + k)

7 For f (x) = x3 − x2 − 1, the values of x for which the graph of y = f (x) has stationary

points are:

A2

3only B 0 and

2

3C 0 and −2

3D −1

3and 1 E

1

3and −1

8 Let f be differentiable for all values of x in [0, 6]. The graph with equation y = f (x) has a

local minimum point at (2, 4). The equation of the tangent at the point with coordinates

(2, 4) is:

A y = 2x B x = 2 C y = 4 D 2x − 4y = 0 E 4x − 2y = 0

9 The volume V cm3 of a solid is given by the formula V = −10x(2x2 − 6) where x cm is a

particular measurement. The value of x for which the volume is a maximum is:

A 0 B 1 C√

2 D√

3 E 2


SAMPLE

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Review


10 The equation of the normal to the curve with equation y = x2 at the point where x = a is:

A y = −1

2ax + 2 + a2 B y = −1

2ax + 1

2+ a2 C y = 2ax − a2

D y = 2ax + 3a2 E y = 1

2ax + 2 + a2

Short-answer questions (technology-free)

1 a Find the equation of the tangent to the curve y = x3 − 8x2 + 15x at the point with

coordinates (4, – 4).

b Find the coordinates of the point where the tangent meets the curve again.

2 Find the equation of the tangent to the curve y = 3x2 at the point where x = a. If this

tangent meets the y-axis at P, find in terms of a the y-coordinate of P.

3 Find the equation of the tangent to the curve with equation

y = x3 − 7x2 + 14x − 8

at the point where x = 1.

Find the x-coordinate of a second point at which the tangent is parallel to the tangent at

x = 1.

4 Use the formula A = �r2 for the area of a circle to find:

a the average rate at which the area of a circle changes with respect to the radius as the

radius increases from r = 2 to r = 3

b the instantaneous rate at which the area changes with r when r = 3

5 Find the stationary points of the graphs for each of the following and state their nature:

a f (x) = 4x3 − 3x4 b g(x) = x3 − 3x − 2 c h(x) = x3 − 9x + 1

6 Sketch the graph of y = x3 − 6x2 + 9x

7 The derivative of the function y = f (x) is:

dy

dx= (x − 1)2(x − 2)

Find the x-coordinate and state the nature of each stationary point.

8 Find the equation of the tangent to the curve y = x3 − 3x2 − 9x + 11 at x = 2.

9 Let f : R → R, where f (x) = 3 + 6x2 − 2x3

Determine the values of x for which the graph of y = f (x) has a positive gradient.

10 For what value(s) of x do the graphs of y = x3 and y = x3 + x2 + x − 2 have the same

gradient?

11 For the function with rule f (x) = (x − 1)45 :

a State the values for which the function is differentiable, and find the rule for f ′.b Find the equations of the tangent at the point (2, 1) and (0, 1).

c Find the coordinates of the point of intersection of the two tangents.


SAMPLE

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Rev

iew


12 A spherical bubble, initially of radius length 1 cm, expands steadily, its radius increases by

1 cm/s and it bursts after 5 seconds.

a Find the rate of increase of volume with respect to the change in radius when the radius

is 4 cm.

b Find the rate of increase of volume with respect to time when the radius is 4 cm.

Extended-response questions

1 The diagram shows a rectangle with sides 4 m and x m

and a square with side x m. The area of the shaded region

is y m2.y m2

4 m

x m

x ma Find an expression for y in terms of x.

b Find the set of possible values for x.

c Find the maximum value of y and the corresponding

value of x.

d Explain briefly why this value of y is a maximum.

e Sketch the graph of y against x.

f State the set of possible values for y.

2 A flower bed is to be L-shaped, as shown in the

figure, and its perimeter is 48 m.

y m

3y m

x m

x ma Write down an expression for the area, A m2,

in terms of y and x.

b Find y in terms of x.

c Write down an expression for A in terms of x.

d Find the values of x and y that give the maximum area.

e Find the maximum area.

3 It costs (12 + 0.008x) dollars per kilometre to operate a truck at x kilometres per hour. In

addition it costs $14.40 per hour to pay the driver.

a What is the total cost per kilometre if the truck is driven at:

i 40 km/h? ii 64 km/h?

b Write an expression for C, the total cost per kilometre, in terms of x.

c Sketch the graph of C against x for 0 < x < 120.

d At what speed should the truck be driven to minimise the total cost per kilometre?

4 A box is to be made from a 10 cm by 16 cm sheet of metal by cutting equal squares out of

the corners and bending up the flaps to form the box. Let the lengths of the sides of the

squares be x cm and V cm3 the volume of the box formed.

a Show that V = 4(x3 − 13x2 + 40x)

b State the set of x-values for which the expression for V in terms of x is valid.

c Find the values of x such thatdV

dx= 0.

d Find the dimensions of the box if the volume is to be a maximum.


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e Find the maximum volume of the box.

f Sketch the graph of V against x for the domain established in b.

5 A manufacturer finds that the daily profit, $P, from selling n articles is given by

P = 100n − 0.4n2 − 160

a i Find the value of n which maximises the daily profit.

ii Find the maximum daily profit.

b Sketch the graph of P against n. (Use a continuous graph.)

c State the allowable values for n for a profit to be made.

d Find the value of n which maximises the profit per article.

6 A rectangle has one vertex at the origin, another on

the positive x-axis, another on the positive

y-axis and a fourth on the line y = 8 − x

2

x

y

0

(x, y)

xy = 8 –12

What is the greatest area the rectangle can have?

7 At a factory the time, T seconds, spent in producing a certain size metal component is

related to its weight, w kg, by T = k + 2w2, where k is a constant.

a If a 5 kg component takes 75 seconds to produce, find k.

b Sketch the graph of T against w.

c Write down an expression for the average time A (in seconds per kilogram).

d i Find the weight that yields the minimum average machinery time.

ii State the minimum average machining time.

8 A manufacturer produces cardboard boxes that have a square

base. The top of each box consists of a double flap that opens

as shown. The bottom of the box has a double layer of

cardboard for strength. Each box must have a volume of

12 cubic metres.h

x

x

double

a Show that C, the area of cardboard required, = 3x2 + 4xh

b Express C as a function of x only.

c Sketch the graph of C against x for x > 0.

d i What dimensions of the box will minimise the amount of cardboard used?

ii What is the minimum area of cardboard used?

9 An open tank is to be constructed with a square base and vertical sides to contain 500 m3 of

water. What must be the dimensions of the area of sheet metal used in its construction if

this area is to be a minimum?

10 A piece of wire of length 1 m is bent into the shape of a sector of a circle of radius a cm

and sector angle �. Let the area of the sector be A cm2.

a Find A in terms of a and �. b Find A in terms of �.

c Find the value of � for which A is a maximum.

d Find the maximum area of the sector.


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11 A piece of wire of fixed length, L cm, is bent to form the boundary

OPQO of a sector of a circle. The circle has centre O and radius r cm.

The angle of the sector is � radians.

θ

P Q

O

a Show that the area, A cm2, of the sector is given by:

A = 1

2r L − r2

b i Find a relationship between r and L for whichdA

dr= 0

ii Find the corresponding value of �.

iii Determine the nature of the stationary points found in i.

c Show that, for the value of � found in b ii, the area of the triangle OPQ is

approximately 45.5% of the area of sector OPQ.

12 A Queensland resort has a large swimming pool as illustrated

with AB = 75 m and AD = 30 m. A boy can swim at 1 m/s

and run at 12

3m/s. He starts at A, swims to a point P on DC,

and runs from P to C. He takes 2 seconds to pull himself out

of the pool.

PD C

BA

a Let DP = x m and the total time be T s.

Show that T = √x2 + 900 + 3

5(75 − x) + 2

b FinddT

dxc i Find the value of x for which the time taken is a minimum.

ii Find the minimum time.

d Find the time taken if the boy runs from A to D and then D to C.

13 The point S is 8 km offshore from the point O which is located on the straight shore of a

lake, as shown in the diagram. The point F is on the shore, 20 km from O. Contestants race

from the start, S, to the finish, F, by rowing in a straight line to some point, L, on the shore

and then running along the shore to F. A certain contestant rows at 5 km per hour and runs

at 15 km per hour.

O L F

8 km

x km20 km

S

a Show that, if the distance OL is x km, the time

taken by this contestant to complete the course is

(in hours):

T (x) =√

64 + x2

5+ 20 − x

15

b Show that the time taken by this contestant to complete the course has

its minimum value when x = 2√

2. Find this time.


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14 At noon the captain of a ship sees two fishing boats approaching. One of them is 10 km due

east and travelling west at 8 km/h. The other is 6 km due north, travelling south at 6 km/h.

At what time will the fishing boats be closest together and how far apart will they be?

15 A rectangular beam is to be cut from a non-circular tree trunk

whose cross-sectional outline can be represented by the equation

y2 = 2 − 2x2

x

y

(x, y)

y2 = 2 – 2x2

0

a Show that the area of the cross-section of the beam is

given by A = 4x√

2 − 2x2 where x is the half-width of

the beam.

b State the possible values for x.

c Find the value of x for which the cross-sectional area of

the beam is a maximum and find the corresponding value of y.

d Find the maximum cross-sectional area of the beam.

16 An isosceles trapezoid is inscribed in the parabola y = 4 − x2 as illustrated.

x

y

(–2, 0) (2, 0)0

(x, y)

a Show that the area of the trapezoid is:

1

2(4 − x2) (2x + 4)

b Show that the trapezoid has its greatest area when

x = 2

3.

c Repeat with the parabola y = a2 − x2

i Show that the area, A, of the trapezoid = (a 2 − x2)(a + x)

ii Use the product rule to finddA

dx.

iii Show that a maximum occurs when x = a

3.

17 It is believed that, for some time after planting in ideal conditions, the area covered by a

particular species of ground-cover plant has a rate of increase of y cm2/week, given

by y = − t3 + bt2 + ct where t is the number of weeks after planting.

a Find b and c given the following table of observations:

t 1 2

y 10 24

b Assuming that the model is accurate for the first 8 weeks after planting, when, during

this period, is:

i the area covered by the plant a maximum?

ii the rate of increase in area a maximum?

c According to the model, if the plant covered 100 cm2 when planted, what area will it

cover after 4 weeks?


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d Discuss the implications for the future growth of the plant if the model remains accurate

for longer than the first 4 weeks.

18 Let f (x) = x3 − 3x2 + 6x − 10

a Find the coordinates of the point on the graph of f for which f ′(x) = 3.

b Express f ′(x) in the form a(x + p)2 + q

c Hence show that the gradient of f is greater than 3 for all points on the curve of f other

than that point found in a.

19 A curve with equation of the form y = ax3 + bx2 + cx + d has zero gradient at the point(1

3,

4

27

)and also touches, but does not cross, the x-axis at the point (1, 0).

a Find a, b, c and d.

b Find the values of x for which the curve has a negative gradient.

c Sketch the curve.

20 The volume of water, V m3, in a reservoir when the depth indicator shows y metres is given

by the formula:

V = �

3[( y + 630)3 − (630)3]

a Find the volume of water in the reservoir when y = 40.

b Find the rate of change of volume with respect to height, y.

c Sketch the graph of V against y for 0 ≤ y ≤ 60.

d If y = 60 m is the maximum depth of the reservoir, find the capacity (m3) of the

reservoir.

e IfdV

dt= 20 000 − 0.005� ( y + 630)2 , where t is the time in days from 1 January,

sketch the graph ofdV

dtagainst y for 0 ≤ y ≤ 60.

21 Water is being poured into a flask. The volume, V mL, of water in the flask at time,

t seconds, is given by:

V (t) = 3

4

(10t2 − t3

3

), 0 ≤ y ≤ 20

a Find the volume of water in the flask when:

i t = 0 ii t = 20

b Find the rate of flow of water,dV

dt, into the flask.

c Sketch the graph of V(t) against t for 0 ≤ t ≤ 20.

d Sketch the graph of V ′(t) against t for 0 ≤ t ≤ 20.

e At what time is the flow greatest and what is the flow at this time?

22 A cone is made by cutting out a sector with central angle � from a circular piece of

cardboard of radius 1 m and joining the two cut edges to form a cone of slant height 1 m as

shown in the following diagrams.


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θ 1 m

h m1 m

r m

The volume of a cone is given by the formula V = 1

3�r2h

a i Find r in terms of �. ii Find h in terms of �.

iii Show that V = 1

3�

(2� − �

2�

)2√

1 −(

2� − �

2�

)2

b Find the value of V when � = �

4c Find the value of � for which the volume of the cone is 0.3 m3.

d i Use a calculator to determine the value of � that maximises the volume of the cone.

ii Find the maximum volume.

e Determine the maximum volume using calculus.

23 a For the function with rule f (x) = x3 + ax2 + bx plot the graph of each of the following

using a calculator. (Give axes intercepts, coordinates of stationary points and the nature

of stationary points.)i a = 1, b = 1 ii a = −1, b = −1 iii a = 1, b = −1

iv a = −1, b = 1

b i Find f ′(x).

ii Solve the equation f ′(x) = 0 for x, giving your answer in terms of a and b.

c i Show that the graph of y = f (x) has exactly one stationary point if a2 − 3b = 0

ii If b = 3, find the corresponding value(s) of a which satisfy the condition

a2 − 3b = 0. Find the coordinates of the stationary points and state the nature of

each.

iii Plot the graph(s) of y = f (x) for these values of a and b using a graphics calculator.

iv Plot the graph of the corresponding derivative functions on the same set of axes.

d State the relationship between a and b if no stationary points exist for the graph of

v = f (x)

24 Consider the function with rule f (x) = 6x4 − x3 + ax2 − 6x + 8a i If x + 1 is a factor of f (x) find the value of a.

ii Using a calculator plot the graph of y = f (x) for this value of a.

b Let g(x) = 6x4 − x3+ 21x2 − 6x + 8

i Plot the graph of y = g(x)

ii Find the minimum value of g(x) and the value of x for which this occurs.

iii Find g′(x).

iv Using a calculator solve the equation g′(x) = 0 for x.


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v Find g′(0) and g′(10). vi Find the derivative of g′(x).

vii Show that the graph of y = g′(x) has no stationary points and thus deduce that

g′(x) = 0 has only one solution.

25 For the quartic function f, with rule f (x) = (x − a)2(x − b)2, a > 0 and b > 0:

a Show that f ′(x) = 2 (x − a) (x − b) [2x − (b + a)].

b i Solve the equation f ′(x) = 0 for x. ii Solve the equation f (x) = 0 for x.

c Hence find the coordinates of the stationary points of the graph of y = f (x)

d Plot the graph of y = f (x) on a calculator for several values of a and b.

e i If a = b, f (x) = (x − a)4. Sketch the graph of y = f (x)

ii If a = −b, find the coordinates of the stationary points.

iii Plot the graph of y = f (x) for several values of a, given that a = −b.

26 For the quartic function f with rule f (x) = (x − a)3 (x − b), a > 0 and b > 0:

a Show that f ′(x) = (x − a)2 [4x − (3b + a)]

b i Solve the equation f ′(x) = 0 ii Solve the equation f (x) = 0

c Find the coordinates of the stationary points of the graph of y = f (x) and state the

nature of the stationary points.

d Using a calculator plot the graph of y = f (x) for several values of a and b.

e If a = −b state the coordinates of the stationary points in terms of a.

f i State the relationship between b and a if there is a local minimum for x = 0.

ii Illustrate this for b = 1, a = −3 on a calculator.

g Show that if there is a turning point for x = a + b

2, then b = a and f (x) = (x − a)4

27 A cylinder is to be cut from a sphere. The cross-section through

the centre of the sphere is as shown. The radius of the sphere is

10 cm. Let r cm be the radius of the cylinder.

y cmr cm

10 cm

Oa i Find y in terms of r and hence the height, h cm, of the

cylinder.

ii The volume of a cylinder is given by V = �r2h. Find V

in terms of r.

b i Plot the graph of V against r using a calculator.

ii Find the maximum volume of the cylinder and the

corresponding values of r and h. (Use a calculator.)

iii Find the two possible values of r if the volume is 2000.

c i FinddV

dr.

ii Hence find the exact value of the maximum volume and the volume of r for which

this occurs.

d i Plot the graph of the derivative functiondV

dragainst r, using a calculator.

ii From the calculator, find the values of r for whichdV

dris positive.

iii From the calculator, find the values of r for whichdV

dris increasing.


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28 A wooden peg consists of a cylinder of length h cm

and a hemispherical cap of radius r cm so that the

volume, V cm3, of the peg is given by V = �r2h + 2

3�r3

r cm

h cm

a If the surface area of the peg is 100� cm:

i Find h in terms of r. ii Find V as a function of r.

iii Find the possible values of r (i.e. find the domain of the function defined in ii).

iv FinddV

drv Sketch the graph of V against r.

b If h = 6, V = 6�r2 + 2

3�r3. For r = 4:

i Show that a small increase of p cm in the radius results in an increase of 80�p in the

volume.

ii Show that a small increase of q% in the radius will cause an approximate increase of

2.3% q in the volume.

29 The diagram shows a container open at

the top, whose dimensions in cm are as

indicated. The cross- section profile AOB

is a parabola whose vertex is at a lowest

point O. ABCD is a horizontal rectangle. 16

O

C

BA

30

60

x

y

D

a Water is poured into the container at

a rate of 100 cm3/s. Find the rate at which the level of liquid is rising when the depth of

liquid in the container is 9 cm, given that the volume V cm3 of liquid when the depth of

water is y cm is given by V = 300y32 .

b Find the equation of the parabola.

c The water is being poured in at a rate of 100 cm3/s.

i Find the rate of change of x with respect to time when y = 9.

ii The surface of the water is a rectangle (the rectangle with dashed lines in the

diagram). Find the rate of change of the surface area when y = 9.

30 A triangular prism has dimensions as shown

in the diagram. All lengths are in centimetres.

The volume of the prism is 3000 cm3. C

E

y

F

D

12x

12x

5x

5x

13x

A B13x

a i Find y in terms of x.

ii Find S cm2, the surface area of the

prism, in terms of x.

b i FinddS

dx.

ii Find the minimum surface area, correct

to three decimal places.

c Given that x is increasing at 0.5 cm/s find the

rate at which the surface area is increasing when x = 10.


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P1: FXS/ABE P2: FXS CHAPTER 10downloads.cambridge.edu.au/education/extra/209... · P1: FXS/ABE P2: FXS 0521842344c10.xml CUAU030-EVANS August 26, 2008 6:26 348 Essential Mathematical