chapter 5_applications of differentiation

LFSE012 SciEngMathsA Applications of Differentiation

Page 1 of 23

CHAPTER 5 APPLICATIONS OF DIFFERENTIATION

5.1 INSTANTANEOUS RATES OF CHANGE: KINEMATICS

(a) Displacement

The position of a body moving in a straight line changes with time. This change in position is

called displacement. This displacement may be given by the expression

s f t where s is displacement (metres)

t is the time taken in seconds

(b)Velocity

We know that average velocity is calculated by using change in displacement

time

s

t

.

We can calculate instantaneous velocity by using the derivative.

0t

s dsv lim

t dt

where v is the velocity in metre per second (m/s)

Example 1

The displacement s (in metres)of a body is given by 3 22 3 2s t t t

(a) Find the velocity at any time t.

(b) Find the velocity when t = 5 seconds.

Solution:

3 2

2 2

(a) 2 3 2 (b) When t = 5 seconds:

3 4 3 3 5 4 5 3 m/s

s t t t

dsv t t v

dt

52 m/s

(c) Acceleration

When an object is accelerating, its velocity is changing. Average acceleration is calculated by

using change in velocity

time

v

t

.

We can calculate instantaneous acceleration by using the derivative.

0t

v dva lim

t dt

where a is the acceleration in metre per second per second (m/s2)

So acceleration is the second derivative of displacement

s f t v f t a f t


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Example 2

The velocity of a body is given by 2 5 7v t t

(a) Find the acceleration at any time.

(b) Find the acceleration when t = 3.6 seconds.

(c) At what time is the acceleration of the body zero?

Solution:

2

2 2

(a) 5 7

2 5

(b) When t = 3.6 seconds:

2 3 6 5 m/s 2.2 m/s

(c) When acceleration is 0 :

5 2 5 0 seconds = 2.5 seconds

2

.

v t t

dva t

dt

a

t t

Example 3

The displacement of an object is given by 3 22 5 4 15s t t t

(a) Find the velocity when t = 2.5 seconds.

(b) Find when the acceleration is zero.

(c) When is the object stationary?

Solution:

3 2

2

2

(a) 2 5 4 15

6 10 4

When t = 2.5 seconds :

6 2.5 10 2.5 4 m/s = 8.5 m/s

(b) 12 10

When acceleration is zero

s t t t

dsv t t

dt

v

dva t

dt

2

:

10 5 12 10 0 seconds = of a second

12 6

(c) When 0

6 10 4 0

1 2 3 1 2 0 , 2

3

But 0 So car is stationary when 2 secon

:

t t

v

t t

t t t

t t

ds


Page 3 of 23

Exercise 5.1

1. Find the velocity at any time t for the following displacement functions.

2

3 2

i 5 23 ii 5

iii 5 3 2 23 iv 27

-

s t s t t

s t t t s

2. Find the acceleration at any time t for the following velocity functions.

2 3

3 5 4 3 212

i 5 2 ii 5

iii 2 21 iv 3 3 5 5

-

v t v t t

v t t v t t t t t

3. The displacement s (in metres)of a body is given by 22 3 12s t t

(a) Find its displacement at t = 3 seconds.

(b) Find its velocity at t = 3 seconds.

(c) Find its acceleration at t = 3 seconds.

4. The velocity of a ball rolling down a slope is given by 22 4 9v t t . Find the

acceleration when t =2.5 seconds.

5. The displacement s (in metres)of a body is given by 3 22 4 12s t t t

(a) Find its displacement at t = 2 seconds.

(b) Find its velocity at t = 1 second.

(c) When is the body stationary?

(d) Find its acceleration at t = 3 seconds.

(e) Find the acceleration of the body when it is stationary.


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5.2 CURVE SKETCHING

Differentiation is used in sketching curves.

Stationary points are points on a curve where the tangent to the curve is parallel to the x-axis.

These stationary points may be either turning points or points of inflexion.

5.2.1 TURNING POINTS

Consider the cubic function 3 22 3 12y x x x . Its derivative is 26 6 12dy

x xdx

The graphs of both functions are drawn below.

3 22 3 12y x x x 26 6 12

dyx x

dx

y

(2,20)

+ +

+ +

+ +

+ +

+ +

3 2 x

+ +

(1,7)

y = 2x3+3x

2-12x

dy

dx

2 1 x

dy

dx = 6x

2+6x 12

Notes : The + and signs refer to the gradient of the graph

For 2 the gradient is positive,x For 2 is positive,dy

xdx

For 2 < 1 the gradient is negative,x For 2 < 1 is negative,

dyx

dx

For 1 the gradient is positive,x For 1 is positive,

dyx

dx

Turning point at 2x 0 at 2

dyx

dx

Turning point at 1x 0 at 1

dyx

dx

Point of inflexion at1

2x .(the graph changes

from concave up to concave down)

A turning point is either a local maximum or a local minimum.

The point (2, 20) is a local maximum. The y-value at (2, 20) is greater than the y-value of all neighbouring points.

The point (1, 7) is a local minimum. The y-value at (1, 7) is less than the y-value of all neighbouring points.


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We can use the second derivative to determine whether the turning point is a local maximum or a

local minimum.

For the function 3 22 3 12y x x x , the second derivative is:2

212 6 .

d yx

dx

2 2

2 2

2 2

2 2

At 2 12 2 6 18 0 (local maximum)

At 1 12 1 6 18 0 (local minimum)

: . .

: . .

d y d yx

dx dxd y d y

xdx dx

Definitions:

2

2

2

2

A function has a at if 0 and 0

dy d yx a

dx dx

dy d yx a

dx dx

local maximum

local minimum

Example 1

For the function 3 3 5y x x :

(a) Find any turning points and determine whether they are local maximums or minimums (b) Sketch the curve of the function, showing the turning points and the y-intercept.

Solution:

23 2

2

2

3

3

3 5 3 3 6

If 0 3 3 0

3( 1 1 0 1 or 1

If 1 1 3 1 5 7 1 7 is a stationary point

If 1 1 3 1 5 3

dy d yy x x x x

dx dxdy

xdx

x x x x

x y

x y

:

)

, ,

,

2

2

2

2

1 3 is a stationary point

If 1 6 1 6 0 1 7 is a local maximum

If 1 6 1 6 0 1 3 is a local minimum

d yx

dxd y

xdx

,

, ,

, ,

intercept : 0 5y x y

y

(1,7)

5

(1,3)

x

y = x3 3x+5


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5.2.2 STATIONARY POINTS OF INFLEXION

The graph of the function 3y x has been drawn at

right.

2

2

2

2

2

2

3

If 0

3 0 0 0 0 is a stationary point.

6

At 0 6 0 0

0 0 is neither a local maximum nor a local minimum

.

. ,

.

, .

,

dyx

dxdy

dxx x

d yx

dxd y

xdx

y

2 0 2 x

3y x

In fact (0,0) is a stationary point of inflexion

When 0 0 and when 0 0 , ,dy dy

x xdx dx

Definition:

2

2A function has a at if 0 and = 0

dy d yx a

dx dx stationary point of inflexion

Example 1

For the function 4 34 2y x x ,

(i) Find any local maximum and minimum points, stationary points of inflexion

(ii) Draw the graph, showing the intercept.y

Solution:

24 3 3 2 2

2

3 2 2

4 3

4 3

4 2 4 12 12 24

Stationary points :

0 4 12 0 4 3 0 0 or 3

0 0 4 0 2 2 0 2 is a stationary point

3 3 4 3 2

dy d yy x x x x x x

dx dx

dyx x x x x x

dxx y

x y

,

29 3 29 is a stationary point ,

22

2

22

2

At 0 12 0 24 0 0

So 0 2 is a stationary point of inflexion.

At 3 12 3 24 3 108 72 36 0

So 3 29 is a local minimum.

intercept : 0 2

,

,

,

,

d yx

dx

d yx

dx

y x y

y

x

(3,29)

4 34 2y x x

(0,2)


Page 7 of 23

Exercise 5.2 A

Sketch the graphs of the following functions, locating any local maximum and minimum points, stationary

points of inflexion, and y-intercepts.

2 2

3 2 3 2

4 2

1 2 2 2 3 6 1

3 4 2 4 1

5 2 5

. .

. .

.

y x x y x x

y x x x y x x x

y x x

4 3

3 4

1 6 2 10

2

7 4

.

.

y x x

y x x

5.2.3 RATIONAL FUNCTIONS

A rational function is a function which can be expressed in the form

p x

q x, where p x and

q x are polynomials. Examples of rational functions are :

2

2 3

3

5 2

2

3 1

1

,

,

,

y xx

xy x

x

f x xx

The graphs of rational functions show asymptotic behaviour. An asymptote is a straight line which

the graph approaches (but never touches). Asymptotes may be vertical, horizontal or oblique lines.

We examine the behaviour of the function as it approaches these asymptotes.

Vertical asymptotes occur where the denominator equals zero.

To find horizontal (or oblique)asymptotes, we examine the behaviour of the function as

x .

Example 1

Sketch the graph of the function 1

3y

x

, finding any asymptotes, stationary points and axial

intercepts.

Solution:

Vertical asymptote : 3 0 3x x

As 3 from the positive side ,

As 3 from the negative side ,

x y

x y

Horizontal asymptote As 0 from below

As 0 from above

So the horizontal asymptote is 0

: ,

,

.

x y

x y

y


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1

2

1Stationary points 3

31

33

0 So there are no turning points.

:

, .

.

y xx

dyx

dx x

dy

dx

1 1-intercept 0

0 3 3

-intercept 0 no -intercept

:

:

y x y

x y x

y

3 x

1

3y

x

3x

1

3-

Example 2


2

xy

x


intercepts.

Solution:

Vertical asymptote: 2 0 i.e. 2x x

As 2 ,

As 2 ,

x y

x y

5 3Horizontal asymptote =1+

2 2

As 1 from below

As 1 from above

So the horizontal asymptote is

:

,

,

xy

x x

x y

x y

1.y

1

2

2

5 3Stationary points or 1

2 22 1 5 1

(quotient rule) 1 3 22

3 0

2

xy y

x xx xdy

xdx x

x

:

2

3 0

2

There are no stationary points.

dy

dx x

0 5 5-intercept 0

0 2 2

5-intercept 0 0

2

5 0 5

:

:

y x y

xx y

x

x x

y

5 x

5

2

xy

x

2x

1y 5

2


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Example 3


3

1y

x


intercepts.

Solution:

2Vertical asymptote : 1 0

1 1 0 1 1,

x

x x x

As 1 ,

As 1 ,

As 1 ,

As 1 ,

x y

x y

x y

x y

2

3Horizontal asymptote

1

As 0 from above

As 0 from above

So the horizontal asymptote is 0

:

,

,

.

yx

x y

x y

y

2

2

2 22 2

22

3Stationary points

1

1 0 3 2 6 (quotient rule)

1 1

6 If 0 6 0 0

1

If 0 0 If 0 0

:

, ,

yx

x xdy x

dx x x

xx x

x

dy dyx x x

dx dx

0 is a local max.

3 0 3 0 3 is local max.

0 1,x y

2

-intercept 0 3

3-intercept 0 0

1

There are no -intercepts

y x y

x yx

x

:

:

y

1 1 x

x = 1 x=1

2

3

1y

x

0y

-3


Page 10 of 23

Exercise 5.2B

Sketch the graph of the following functions, finding any asymptotes, stationary points and axial intercepts.

2

2

2 2 41. 2. 3.

4 2 5 1

2 1 24. 5.

1 4

xy y y

x x x

x xy y

x x

21

6. 2

yx x

5.3 MAXIMA AND MINIMA (OPTIMISATION)

There are many situations in science where we are required to find the optimum value of a

variable. Just as the gradient of a function is equal to zero at maximum and minimum points,

similarly we can find maximum or minimum values of a function by equating the derivative of that

function to zero.

Example 1

The height, h (in metres), of a projectile above the ground t (seconds) after it has been fired is

given by h = 12t - t2 . What is the greatest height reached by the projectile?

Solution:

Step 1. Find an equation for what is to be maximised or minimised.

In this example this is easy, the equation is given to us. i.e. h = 12t - t2

Step 2. Make sure that the equation for what we want to maximise (or minimise) is written in

terms of just one other variable.

In this case it is. The height, h, is written in terms of t.

Step 3. Differentiate this equation, equate it to zero and solve.

h = 12t - t2

therefore dh

dt = 12 - 2t

maximum occurs when dh

dt = 0

therefore 0 = 12 - 2t

2t = 12

t = 6

So the projectile reaches its maximum height after 6 seconds.


Page 11 of 23

Step 4. Answer the question!

The question in this case asks not when does it reach maximum height, but what is the

maximum height. We have found that the projectile reaches its maximum height after 6

seconds.

Since h = 12t - t2

when t = 6 h = 12 6 - 62 = 36

So the maximum height reached by the projectile is 36 metres.

Example 2

A farmer wishes to fence off a rectangular paddock using 400 metres of wire. If he is going to use

a straight river as one boundary of the paddock (meaning that he only has to fence the remaining 3

sides), what dimensions should he make the paddock in order to fence off the maximum possible

area?

Solution:

Step 1. Find an equation for what is to be maximised or minimised.

In this example the thing to be maximised is the area of the paddock, so we start with the

equation for the area of the rectangular paddock.

A = L W , where L is the length of the paddock and W is the width. (Always define

your variables)

L

W W

River

Step 2. Make sure that the equation for what we want to maximise (or minimise) is written in

terms of just one other variable.

In this case it is written in terms of 2 variables, L and W.

Using the information given in the question (i.e. that there is only 400 metres of fencing

wire and that one side of the paddock is a river) we find that


Page 12 of 23

L + 2W = 400

therefore L = 400 - 2W

Substituting this L back into the equation A = L W we get

A = (400 - 2W) W

A = 400W - 2W2

Step 3. Differentiate this equation, equate it to zero and solve.

A = 400W - 2W2

therefore dA

dW = 400 - 4W

maximum occurs when dA

dW = 0

therefore 0 = 400 - 4W

4W = 400

W = 100

So the paddock has maximum area when it is 100 metres wide.

Step 4. Answer the question!

The question in this case asks for the dimensions of the paddock of maximum area. If it is

100 metres wide, the length will be 200 metres (to make up 400 metres of fencing)

So the paddock of maximum area will be 200 metres 100 metres.


Page 13 of 23

Exercise 5.3

1. A rectangular block of wood has the length of its base equal to twice its width. If the total

surface area of this block is 300 cm2 , find the dimensions of the block so that it is of

maximum volume.

h

x

2x

2. A cuboid has a total surface area of 150 cm 2 with a square base of side x cm.

( a ) Show that the height, h cm, of the cuboid is given by 275

2

xh

x

.

( b ) Express the volume of the cuboid in terms of x.

( c ) Determine the maximum volume of cuboid as x varies.

3. A soft drink company wants to produce cylindrical cans of volume 375 cm3.

What dimensions should these cans be if they are to use the least possible amount of material

(i.e. least surface area; assume uniform thickness).

4. A rectangular patch of ground is to be enclosed with 100 metres of fencing wire.

Find the dimensions of the ground so that the area enclosed will be a maximum.

5. A cone has a sloping edge of 5cm. For the cone to have maximum volume, what length should

the radius be?

*6. Find the volume of the largest circular cone that can be cut from a solid sphere of radius 4 cm.

*7. Find the dimensions of the cylinder of greatest volume that can be turned from a wooden cone

of height 8cm and base radius 4cm.


Page 14 of 23

5.4 RATES OF CHANGE

Often we need to find the rate of change of some quantity.

Example 1

If we know that a spherical balloon is being inflated with 5 cm3 of air every second then we may

want to know at what rate the radius of the balloon is increasing. This is referred to as a related

rates problem, because we know one rate (the rate at which the volume of the balloon is

increasing) and another related rate (the rate at which the radius of the balloon is increasing) has to

be found.

The method of solving these problems relies on the use of the chain rule,

i.e. dy

dx

dy

du

du

dx .

Returning to the problem of the balloon being inflated at a rate of 5 cm3/sec.

We wish to know the rate of change of the radius of the balloon, which can be written as dr

dt (r is

the radius and t stands for time).

We can say that the rate of change of volume is dV

dt = 5 (Note: it is often useful to look at the

units of the given information, volume is measured in cm3 and time is measured in seconds)

We now set up a chain rule : what we want = what we know what we can find

So we have dr

dt

dV

dt

dr

dV , where

dV

dt = 5

Also we know that, for a sphere V = 4

33r

therefore dV

dr = 4r2

dr

dV r

1

4 2

Now, returning to the chain rule : dr

dt

dV

dt

dr

dV

therefore dr

dt = 5

1

4 2r

= 5

4 2r


Page 15 of 23

So we have found that the rate at which the radius changes depends on what the radius is.

Obviously when the balloon is first being inflated an increase of 5 cm3 will greatly increase its

radius, when the balloon is a large sphere, an increase of 5 cm3

to the volume will mean only a

small increase in radius.

If we want to find the rate of increase of the radius when the radius is, say, 8 cm, then

we get dr

dt =

5

4 2r

= 5

4 82

= 5

256

0.00622

So when the radius is 8 cm, it is increasing at a rate of approximately 0.00622 cm/s

Example 2

A vessel containing water has the shape of an inverted right circular cone of base radius 2 metres

and height 4 metres. The water is flowing from the apex of the cone at a constant rate of 0.5

m3/min.

Find the rate at which the water level is dropping when the depth is 3 metres.

Solution:

Step 1. Draw a diagram and define variables.

2m

r 4m

h


Page 16 of 23

Step 2. Set up the chain rule.

We are asked to find dh

dt (i.e. the rate at which the height of the water is changing) , given

that dV

dt = - 0.5 (i.e. the volume of water in the vessel is decreasing at a rate of 0.5

cm3/min.)

So dh

dt = dV

dt dh

dV

Now for a cone V = 1

3 r2 h ; V is written in terms of two variables, r and h. We want

it in terms of just h in order to find dV

dh and therefore

dh

dV .

From the diagram above and by the use of similar triangles it can be shown that

r

h2

4

therefore r = h

2 ,

substituting this into the formula for volume we find

V = 1

3 2

2

hh

V = h3

12

dV

dh

h 2

4

dh

dV h42

So going back to the chain rule we have dh

dt = dV

dt dh

dV

therefore dh

dt = - 0.5

42h

= 22h

when the depth is 3 metres dh

dt =

2

9 - 0.0707

So the height of the water in the vessel is decreasing at a rate of 0.0707 m/min.


Page 17 of 23

Example 3

A ladder 5 m long is leaning against a vertical wall as shown in the diagram below. The bottom of

the ladder is pulled along the ground away from the wall at a constant rate of 0.4 ms-1

. How fast

will the top of the ladder be falling at the instant it is 4 m above the ground?

Solution:

1

2 2 2 2 2 2

Given 0.4 m s

By Pythagoras' theorem, OR By using implicit differentiation,

5 5

dx

dt

x y x y

d

2 2 2 2 2 25 5

2 2 0 2 2 0

2 2

d d d dx y x y

dx dx dt dt dt

dy dx dyx y x y

dx dt dt

dyy x

dx

2 2

0.4

dy dxy x

dt dt

dy x dy x dx

dx y dt y dt

dy dy dx

dt dx dt

x

y

2 2 2

2

When 4, 4 5

25 16

y x

x

9

3

x

dy x dx

dt y dt

1

3 0.4

4

0.3 m s

y m

x m

5 m


Page 18 of 23

Exercise 5.4

1. A stone is thrown into a pond causes a circular ripple to spread out so that the radius increases

at a rate of 18 cm/s. At what rate is the area of the region enclosed by the ripple increasing

when its radius is 60 cm?

2. A spherical balloon is being filled with air at the rate of 100 cm3/min. At what rate is the radius

of the sphere increasing when the radius is 5cm?

3. Sand falls onto a conical pile at a rate of 10 cm3/min. The diameter of the base of the pile is

always equal to its height. How fast is the height of the pile increasing when the pile is 5 cm.

deep?

4. An inverted circular cone whose height is equal to the radius of its base, is filled with water. If

water is flowing out of a small hole in the vertex of the cone at a constant rate of 5cm3/s, at

what rate is the surface of the water descending when the radius of the surface is 6cm?

5. A boat 2 metres below the level of a dock is being pulled towards the dock by a rope. If the

rope is pulled in at 0.1 m/s , how fast is the boat moving through the water when 5 metres of

rope remains between the boat and the dock ?

6. A kite, 50 m high, is being carried horizontally by the wind at a rate of 4m/s. How fast is the

string being let out when the length of string is 100 m?

7. A metal sphere is dissolving in acid. It remains spherical and the rate at which its volume

decreases is proportional to its surface area. Show that its radius is decreasing at a constant

rate.


Page 19 of 23

5.5 SMALL INCREMENTS

For the function ( )y f x , the derivative of the function is given by

0limx

dy y

dx x

Hence if x is very small,

y dy

x dx

dyy x

dx

This result is often used in examples involving a function of two measurements, where one is

changed by a small amount and you are required to find the corresponding change in the other.

Example 1

When a metal cube is heated, the length of the edges increases from 50 mm to 51 mm. Calculate:

(a) the approximate increase in the surface area of one of its faces

(b) the approximate increase in its volume.

Solution:

If the length of the cube is denoted by x , then 51 50 1x mm

(a) The area of one face is given by 2A x

and 2dA

xdx

dAA x

dx

Hence 2 1A x

2 50 1

2100 mm

(b) The volume of the cube is given by 3V x

and 23dV

xdx

dVV x

dx

Hence 23 1V x

23(50) 1

37500 mm


Page 20 of 23

Exercise 5.5

1. A cubic block of metal has each side unit 4 cm long. Due to a temperature change, each side

has increased to 4.02 cm long. Find the approximate change in volume.

2. When a circular metal disc is heated, the radius increases from 100 mm to 102 mm. Calculate

the approximate change in area.

3. For a certain vessel the power P required to give the vessel a speed v is given by the

formula 32.5 P v . Find approximately how many additional units of power are required to

increase to speed from 20 to 21 units.

4. Calculate the approximate change in the kinetic energy E of a 100 kg mass when its velocity

increases from 12 m/s to 12.1 m/s if 21

2E mv .

5. The rate at which water flows out of an orifice at the bottom of a tank is given by

364 10V h cubic metres per minute, where h is the depth of the water in metres. Calculate

the approximate change in the flow rate when the depth of water falls from 2.25 m to 2.22 m.

6. A cylinder has diameter of 80 mm and its depth is 90 mm. If the internal curved wall is worn

away uniformly all around by 0.1 mm, the depth remaining the same, calculate the

approximate change in the volume.

7. A hollow glass sphere has an internal radius of 10 cm and a glass thickness of 0.2 cm. Use

calculus to obtain an approximate value for the volume of the glass.

8. Bernoullis equation from fluid mechanics may be written as 2

2

vH p

where p is the

pressure, is the constant density of the fluid, v is the velocity and H is a constant.

Obtain an approximate formula for the change in pressure p due to the small change in

velocity v .

9. When a metal equilateral triangle is heated, the length of the sides increases from 100 mm to

101 mm. Calculate the approximate increase in area.


Page 21 of 23

ANSWERS

Exercise 5.1

21. i 5 (ii) 5 2t (iii) 15t 6t 2 iv 0 ( ) ( )

2 2 4 3 22. ( ) 2 ( )10 3 ( )6 21 ( )15 4 9 5i ii t t iii t iv t t t t

3. 2 2(a) 15 metres (b) 15 m/s (c) 4 m/s 4. 14 m/s

2 223

5. (a) 4 metres (b) 3 m/s (c) second (d) 22 m/s (e) 8 m/st

Exercise 5.2A All graphs are drawn below.

513 27

6723 27

1 Local min: 1 3 -intercept: 2

2 Local max: 1 4 -intercept: 1

3 Local max: 1 1 local min: - -intercept: 0

4 Local max: - local min: 2 7 -inter

y

y

y

y

. , ;

. , ;

. , ; , ;

. , ; , ;

72

cept: 1

5 Local max: 0 5 local min: 1 6 local min: 1 6 -intercept: 5

6 Local min: 3 Stationary point of inflexion (0,10); -intercept: 10

y

y

. , ; , ; , ;

. , ;

1.

y

x

2

(1,3)

y = x2 2x 2

2

.

(1,4) y

x

y = 3x2 6x +1

1

3.

y

(1,1)

513 27-, x

y = x3 +x2 x

4

.

y

6723 27- ,

x

(2, 7)

y = x3 2x2 4x+1

1


Page 22 of 23

5.

y

x

(0, 5)

(1, 6) (1, 6)

y = x4 2x2 5

6.

y

(0,10)

x

723, -

4 312

2 10y x x

Exercise 5.2B All graphs are drawn below 12

5 22 5

1. Asymptotes : 4 ; 0 ; -inter : ; No stat. points

2. Asymptotes : ; 0 ; -inter : - ; No stat. points

3. Asymptotes : 1 ; 1 ; -inter : 4

x y y

x y y

x y x

51 22 2 5

12

; -inter : 4 ; No stat. points

4. Asymptotes : ; ; -inter : - ; -inter : 2 ; No stat. points

5. Asymptotes : 2 ; 1 ; Local max : 0 - ; -interce

y

x y x y

x y y

,

1pt :

2

6. Asymptotes : 0 2 ; 0 ; Local max : 1 1 x x y

.

, ,

1.

y x=4

1

2

y=0 4 x

2

4y

x

2.

y

x= 52

y=0

52

x

2

2 5y

x

25

-

(3, 27)

0 4

3 44y x x

7.


Page 23 of 23

3.

y

y=1

1 4 x

4

x= 1

4

1

xy

x

4.

y

2y

12

- x

1x

2 1

1

xy

x

1

5.

-8 -6 -4 -2 2 4 6 8

y

(0,- 12

) x

x=2 x=2

2

2

2

4

xy

x

y=1

6.

Exercise 5.3

1. 5 cm 10 cm 6 23 cm 2. (b) V = 375

2

x x ( c ) 125 cm 3

3. r = 3.91 h = 7.82 4. 25 m 25 m

5. 4.08 cm 6. 79.43 cm3 7. r = 2

3

2 cm h = 2

3

2 cm

Exercise 5.4

1. 2160 67.85.84 cm2/s 2.

1 0.3183 cm/min

3. 8

/ min5

cm

4. 0.044 cm/s

5. 121

42ms 6. 23 m/s

Exercise 5.5

1. 0.96 cm3 2. 400 mm2 3. 3000 units

4. 120 J 5. 31.92 10 6. 2262 mm3

7. 80 cm3 8. p v v 9. 86.6 mm2

y

x

x=2

(1, 1)

2

1

2y

x x

chapter 5_applications of differentiation

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