chapter 2 geometrical applications of calculus

7/24/2019 Chapter 2 Geometrical Applications of Calculus

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Anti-differentiation: The process of nding a primitive(original) function from the derivative. It is the inverseoperation to differentiation

Concavity: The shape of a curve as it bends around (it canbe concave up or concave down)

Differentiation: The process of nding the gradient of atangent to a curve or the derivative

Gradient of a secant: The gradient (slope) of a linebetween two points that lie close together on a curve

Gradient of a tangent: The gradient (slope) of a line thatis a tangent to a curve at a point on a function. It is thederivative of the function

Horizontal point of inexion: A stationary point (wherethe rst derivative is zero) where the concavity of thecurve changes

Instantaneous rate of change: The derivative of a function

Maximum turning point: A local stationary point (where

the rst derivative is zero) and where the curve isconcave down. The gradient of the tangent is zero

Minimum turning point: A local stationary point (wherethe rst derivative is zero) and where the curve isconcave up. The gradient of the tangent is zero

Monotonic increasing or decreasing function: A function isalways increasing or decreasing

Point of inexion: A point at which the curve is neitherconcave upwards nor downwards, but where theconcavity changes

Primitive function: The original function found byworking backwards from the derivative. Found by anti-differentiation

Rate of change: The rate at which the dependent variablechanges as the independent variable changes

Stationary (turning) point: A local point at which thegradient of the tangent is zero and the tangent ishorizontal. The rst derivative is zero

TERMINOLOGY

GeometricalApplications

of Calculus

2

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51Chapter 2 Geometrical Applications of Calculus

DID YOU KNOW?

Although Newtonand Leibnizare said to have discovered calculus, elements of calculus were

around before then. It was Newton and Leibniz who perfectedthe processes and developed the

notation associated with calculus.

Pierre de Fermat(160165) used coordinate geometry to nd maximum and minimum

values of functions. His method is very close to calculus. He also worked out a way of nding the

tangent to a curve.

The 17th-century mathematicians who developed calculus knew that it worked, but it was

not fully understood. Limits were not introduced into calculus until the nineteenth century.

INTRODUCTION

YOU LEARNED ABOUTdifferentiation in the Preliminary Course. This is the

process of finding the gradient of a tangent to a curve. This chapter looks at

how the gradient of a tangent can be used to describe the shape of a curve.Knowing this will enable us to sketch various curves and find their maximum

and minimum values. The theory also allows us to solve various problems

involving maximum and minimum values.

Gradient of a Curve

To learn about the shape of a curve, we first need to revise what we know

about the gradient of a tangent. The gradient (slope) of a straight line measures

the rate of change of ywith respect to the change in x.

Since the gradient of a curve varies, we find the gradient of the tangent at

each point along the curve.

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52 Maths In Focus Mathematics Extension 1 HSC Course

EXAMPLES

1.

2.

3.

4.

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In the examples on the previous page, where the gradient is positive, the

curve is going up, or increasing (reading from left to right).

Where the gradient is negative, the curve is going downwards, or

decreasing.

The gradient is zeroat particular points of the curves. At these points

the curve isnt increasing or decreasing. We say the curve is stationaryat thesepoints.

If 0,f x >l] g the curve is increasingIf 0,f x l

] g for all x(monotonic increasing)or f x 0f x

l

] g for increasing curve. xx

x

2 4 0

2 4

2

i.e >

>

>

So the curve is increasing for x 2> .

This function is a

parabola.

CONTINUED

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2. Find the stationary point on the parabolay x x6 32= + .

Solution

dx

dy

x2 6=

For stationary points,dx

dy

x

x

x

0

2 6 0

2 6

3

i.e.

=

=

=

=

When ,x y3 3 6 3 3

6

2= = +

=

] g

So the stationary point is ,3 6^ h.

3. Find any stationary points on the curvey x x48 73= .

Solution

y x3 482= l

`

For stationary points,

y

x

x

x

x

0

3 48 0

3 48

16

4

i.e. 2

2

2

=

=

=

=

=

l

When , ( )x y4 4 48 4 7

135

3= =

=

When , ( )x y4 4 48 4 7

121

3= =

=

] g

So the stationary points are ,4 135^ hand ,4 121^ h.You will use stationary points

to sketch curves later in this

chapter.

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PROBLEM

What is wrong with this working out?

Find the stationary point on the curvey x x2 12= + .

Solution

y x4 1= +l

.

y

x

x

x

0

4 1 0

4 1

0 25


i.e.

=

+ =

=

=

l

When . , ( . )x y0 25 4 0 25 1

1 1

0

= = +

= +

=

So the stationary point is . ,0 25 0

^ h.

Can you nd the

correct answer?

1. Find the parts of each curve

where the gradient of the tangent

is positive, negative or zero. Label

each curve with +, or 0.

(a)

(b)

(c)

(d)

2. Find all values of xfor which the

curvey x x2 2= is decreasing.

3. Find the domain over which

the function f x x4 2= ] g isincreasing.

4. Find values of xfor which the

curvey x x3 42= is

decreasing(a)increasing(b)

stationary.(c)

5. Show that the function

f x x2 7= ] g is always(monotonic) decreasing.

2.1 Exercises

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6. Prove thaty x3= is monotonic

increasing for all 0x .

7. Find the stationary point on the

curve ( ) xf x 3= .

8. Find all x values for which the

curvey x x x2 3 36 93 2= + + is

stationary.

9. Find all stationary points on the

curve

(a) y x x2 32=

(b) f x x9 2= ] g (c) y x x x2 9 12 43 2= +

(d) .y x x2 14 2= +

10. Find any stationary points on thecurvey x 2 4= ] g .

11. Find all values of xfor which

the curve ( ) x xf x 3 43= + is

decreasing.

12. Find the domain over which the

curvey x x x12 45 303 2= + + is

increasing.

13. Find any values of xfor which the

curvey x x x2 21 60 3

3 2= +

isstationary(a)

decreasing(b)

increasing.(c)

14.The function f x x px2 72= + +] g has a stationary point at x 3= .

Evaluatep.

15. Evaluate aand bif

3y x ax bx3 2= + has stationary

points at x 1= and x 2= .

16. (a) Find the derivative of

.y x x x3 27 33 2= +

(b) Show that the curve is

monotonic increasing for all

values of x.

17. Sketch a function with 0f x >l] g for x< 2, 0f 2 =l] g and 0f x .

18. Draw a sketch showing a curve

withdxdy 0< for x 4< ,

dxdy 0= when

x 4= anddx

dy0> for x 4> .

19. Sketch a curve withdx

dy0> for all

1x anddx

dy0= when x 1= .

20. Draw a sketch of a function

that has 0f x >l] g for x 2< ,x 5

>, 0f x

=l

] g for ,x 2 5=

and0


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Types of Stationary Points

There are three types of stationary points.

Local minimum point

The curve is decreasing on the left and increasing on the right of the

minimum turning point.

x LHS Minimum RHSf xl] g < 0 0 > 0

Local maximum point

The curve is increasing on the left and decreasing on the right of the

maximum turning point.

x LHS Maximum RHS

f xl] g > 0 0 < 0

Local maximum and minimum points are also called turning points,

as the curve turns around at these points. They can also be called relative

maxima or minima.

Point of horizontal inflexion

The curve is either increasing on both sides of the inflexion or it is decreasing

on both sides. It is not called a turning point as the curve does not turn

around at this point.

The derivative has the

same sign on both sides

of the inexion.

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x LHS Inflexion RHS

f xl] g > 0 0 > 0or < 0 0 < 0

The stationary points are important to the shape of a curve. A reasonablyaccurate sketch of the curve can be made by finding these points, together

with the intercepts on the axes if possible.

EXAMPLES

1. Find the stationary point on the curvey x3= and determine which

type it is.

Solution

dx

dyx3 2=

dx

dy

x

x

0

3 0

0


i.e. 2

=

=

=

0, 0x y

0

When 3= =

=

So the stationary point is ,0 0^ h.To determine its type, check the curve on the LHS and the RHS.

x 1 0 1

dx

dy3 0 3

Since the curve is increasing on both sides, (0, 0) is a point of inflexion.

Substitute x =1 intodx

dy.

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2. Find any stationary points on the curve f x x x x2 15 24 73 2= + ] g anddistinguish between them.

Solution

6 30 24f x x x2= +l] g ( )

( )( )

f x

x x

x x

x x

x

0

6 30 24 0

5 4 0

1 4 0

1 4


i.e.

or

2

2

`

=

+ =

+ =

=

=

l

( )f 1 2 1 15 1 24 1 7

4

= +

=

3 2] ] ]g g g

So ,1 4^ his a stationary point.

x 0 1 2f xl] g 24 0 12

,1 4` ^ his a maximum stationary point( ) ( )f 4 2 4 15 4 24 4 7

23

= +

=

3 2] ]g g

So ,4 23^ his a stationary point.

x 2 4 5

f xl] g 12 0 24

,4 23` ^ his a minimum stationary point.

Substitutex 0= andx 2=

into (x)f . Take care that

there is no other stationary

point between the x values

you choose.

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curvey x 12= and show that it

is a minimum point by checkingthe derivative on both sides of it.


curvey x4= and determine its

type.


curvey x 23= + and determine its

nature.

4. The function f x 2x x7 4= ] g has one stationary point. Find itscoordinates and show that it is a

maximum turning point.

5. Find the turning point on

the curvey x x3 6 12= + + and

determine its nature.

6. For the curvey x4 2= ] g find theturning point and determine its

nature.

7. The curvey x x6 53 2

= + has 2turning points. Find them and

use the derivative to determine

their nature.

8. Show that the curve f x x 15= +] g has a point of inflexion at ,0 1^ h.

9. Find the turning points on

the curvey x x3 53 2= + and

determine their nature.

10. Find any stationary points on thecurve f x 4 2 .x x2 3= ] g Whattype of stationary points are they?

11. The curvey x x3 23= + has

2 stationary points. Find their

coordinates and determine

their type.

12. The curvey x mx x7 55 3= + +

has a stationary point at x 2= .

Find the value of m.

13. For a certain function,

.f x x3= +l] g For what valueof xdoes the function have a

stationary point? What type of

stationary point is it?

14. A curve has ( 1).f x x x= +l] g Forwhat xvalues does the curve have

stationary points? What type are

they?

15. For a certain curve,

( ) ( ) .dx

dyx x1 22= Find the x

values of its stationary points and


16. (a) DifferentiateP x x250

= + with

respect to x.

Find any stationary points(b)

on the curve and determine their

nature.

17. For the function ,h h

A82 52 +

=

find any stationary points and


18. Find any stationary points for

the function 40 V r r3= and

determine their nature (correct to

2 decimal places).

19. Find any stationary points on the

curve rS r120

2= + (correct to 2

decimal places) and determine

their nature.

20. (a) Differentiate .A x x3600 2=

Find any stationary points for(b)

A x x3600 2= (to 1 decimal

place) and determine their nature.

2.2 Exercises

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Higher Derivatives

A function can be differentiated several times:

differentiating f x] ggives f xl] gdifferentiating f xl] ggives f xm] gdifferentiating f xm] ggives ,f xn] g and so onthe other notation is ,

dx

d y

dx

d y

2

2

3

3

and so on

EXAMPLES

1. Find the first 4 derivatives of ( )f x x x x4 3 23 2= + .

Solution

( )x x x3 8 3= +

( )

( )

x x

x

6 8

0

=

=

( )x 6=

f

f

f

2

f

l

m

n

mm

2. Find the second derivative ofy x2 5 7= +] g .Solution

dx

dyx

x

dx

d yx

x

7 2 5 2

14 2 5

14 6 2 5 2

168 2 5

6

6

2

25

5

$

$ $

= +

= +

= +

= +

] ]]

]

g gg

g

3. Find f 1m] gif f x x 14= ] g Solution

( )

( )( )

f x x

f x xf

4

121 12 1

12

3

2

2`

=

=

=

=

l

m

m ] g

Only the rst 2 derivatives

are used in sketching

graphs of curves.

Differentiating further will

only give 0.

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2.3 Exercises

1. Find the first 4 derivatives of

x x x x2 37 5 4 + .

2. If f x x 59 =] g , find .f xm] g 3. Find f xl] gand f xm] gif

f x x x2 15 3 +=] g .4. Find f 1l] gand f 2m] g, given

f t t t t 3 2 5 44 3= + ] g .5. Find the first 3 derivatives of

x x x2 4 77 6 4 + .

6. Differentiatey x x2 3 32= +

twice.

7. If f x x x x x2 5 14 3 2= + ] g , findf 1l] gand f 2m] g.

8. Differentiate x 4 twice.

9. If ,g x x=] g find .g 4m] g 10. Given h t t t 5 2 53 2= + + , find

dt

d h2

2

when t 1= .

11. Find the value of xfor which

dxd y 3

2

2

= giveny x x x3 2 53 2= + .

12. Find all values of xfor

which f x 0>m] g given thatf x 3 2x x x 9= + +

] g.

13. Differentiate x4 3 5] g twice.14. Find f xl] gand f xm] gif

.f x x2= ] g 15. Find the first and second

derivatives of .f xxx3 1

5=

+] g

16. Finddt

d v2

2

if ( ) .v t t3 2 1 2= + ] g17. Find the value of bin

y bx x x2 5 43 2= + + ifdx

d y2

2

2

=

when .x21

=

18. Find the exact value of f 2m] gif( )f x x x3 4= .

19. Find f 1m] gif ( )f t t t 2 1 7= ] g .20. Find the value of bif

f x bx x5 42 3= ] g and .f 1 3 = m] g

Sign of the Second Derivative

The second derivative gives extra information about a curve that helps us to

find its shape.

Since f xm] gis the derivative of f xl] g, then f xm] gand f xl] ghave the samerelationship as f xl] gand f x] g.

( ) ( )

( ) ( )

( ) ( )

f x f x

f x f x

f x f x

0

0

0

That is if then is increasing

if then is decreasing

if then is stationary

>

m] g then f xl] gis increasing. This means that the gradient of thetangentis increasing, that is, the curve is becoming steeper.

Notice the upward shape of these curves. The curve lies above the

tangents. We say that the curve is concave upwards.

If f x 0


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Class Investigation

How would you check that concavity changes?

The sign must change for

concavity to change.

What type of point is it?

If ,f x 0>m] g the curve is concave upwards.If ,f x 0


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2. Find all values of xfor which the curve xf x x x2 7 5 43 2= +] g isconcave downwards.

Solution

( )

( )

f x x x

f x x

6 14 5

12 14

2=

=

l

m

For concave downwards, ( )f x

x

x

0

12 14 0

12 14

2

2

As the curve is decreasing, the number of unemployed people is(b)

decreasing.

Since the curve is concave upwards, the gradient is increasing. This(c)

means that the unemployment rate is increasing.

Remember that the gradient,

or rst derivative, measures

the rate of change.

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1. For each curve, describe the sign

ofdx

dyand

dx

d y

2

2

.

(a)

(b)

(c)

(d)

(e)

2. The curve below shows the

population of a colony of seals.

Describe the sign of the first(a)

and second derivatives.

How is the population rate(b)

changing?

3. Inflation is increasing, but the

rate of increase is slowing. Draw a

graph to show this trend.

4. Draw a sketch to show the shape

of each curve:

(a) andf x f x0 0< l m] ]g g (d) andf x f x0 0> >l m

] ]g g

5. The size of classes at a local TAFE

college is decreasing and the rate

at which this is happening is

decreasing. Draw a graph to show

this.

6. As an iceblock melts, the rate at

which it melts increases. Draw a

graph to show this information.

7. The graph shows the decay of a

radioactive substance.

Describe the sign ofdt

dMand .

dt

d M2

2

2.5 Exercises

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8. The populationPof fish in a

certain lake was studied over

time, and at the start the number

of fish was 2500.

During the study,(a) .dt

dP0<

What does this say about the

number of fish during the study?

If at the same time,(b) ,d

d P0>

2

2

what can you say about the

population rate?

Sketch the graph of the(c)

populationPagainst t.

9. The graph shows the level of

education of youths in a certain

rural area over the past 100 years.

Describe how the level of

education has changed over this

period of time. Include mention

of the rate of change.

10. The graph shows the number ofstudents in a high school over

several years.

Describe how the schoolpopulation is changing over time,

including the rate of change.

dx

dy0>

dx

dy0

2

2

dx

d y0m] g there is a minimumturning point.

If f x 0=l] g and ,f x 0=l m] ]g g Maximumturning point: , .f x f x0 0


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EXAMPLES

1. Find the stationary points on the curve f x x x x2 3 12 73 2= +] g anddistinguish between them.

Solution

f x x x6 6 122= l] g For stationary points, f x 0=l] g

( ) ( )

x x

x x

x x

6 6 12 0

2 0

2 1 0

i.e. 2

2

=

=

+ =

( ) ( )

x

f

2 1

2 2 2 3 2 12 2 7

13

or2

` =

= +

=

3] ]g g

So ( , )2 13 is a stationary point.

( ) ( )f 1 2 1 3 1 12 1 7

14

3 2 = +

=

] ]g g

So ,1 14^ his a stationary point.Now ( )

( ) ( )

(concave upwards)

f x x

f

12 6

2 12 2 6

18

0>

=

=

=

m

m

So ,2 13^ his a minimum turning point.( ) ( )

(concave downwards)

f 1 12 1 618

0

=

=

m

,3 26` ^ his a minimum turning point( ) ( )

0 (concave wards)

1 6 1 6

12

down

2= =

=

m ] g

So (1, 0) is a minimum turning point.

When ,

(concave upwards)

x y1 12 1 48

0>

= =

=

2m

] g

So (1, 0) is a minimum turning point.

When ,

(concave downwards)

x y0 12 0 4

4

02

2

=

=

=

=

=

x 3` = gives a minimum value

So 300 items manufactured each year will give the minimum expenses.

(c) When ,x E3 3 6 3 12

3

2= = +

=

] g

So the minimum expense per year is $30 000.

2. The formula for the volume of a pond is given by

,V h h h2 12 18 503 2= + + where his the depth of the pond in metres.

Find the maximum volume of the pond.

Solution

V h h6 24 182= +l

For stationary points, 0

i.e. 6 24 18 0

( )( )

V

h h

h h

h h

4 3 0

1 3 0

2

2

=

+ =

+ =

=

l

E is the expense in units of

ten thousand dollars.

This will give any maximum

or minimum values of E.

x is the number of items inhundreds.

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So h = 1 or 3

V h12 24= m

When , ( )h V1 12 1 24= = m

(concave downwards)12

0

=

So h= 3 gives a minimum V.

When ,h V1 2 1 12 1 18 1 50

85

= = + +

=

3 2] ] ]g g g

So the maximum volume is 58 m3

.

In the above examples, the equation is given as part of the question.

However, we often need to work out an equation before we can start

answering the question. Working out the equation is often the hardest part!

EXAMPLES

1. A rectangular prism has a base with length twice its breadth. The volume

is to be 300 cm3. Show that the surface area is given by .S x x49002= +

Solution

Volume:

2

( )

V lbh

x x h

x h

x h

300 2

2

3001

2

2`

=

=

=

=

Surface area:

( )

( )

( )

S bh lh

x xh xh

x xh

x xh

lb2

2 2 2

2 2 3

4 6

2

2

2

= + +

= + +

= +

= +

CONTINUED

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Now we substitute (1) into this equation.

S x xx

x

x

4 62

300

4900

22

2

$= +

= +

2. ABCDis a rectangle withAB= 10 cm andBC= 8 cm. LengthAE= xcm

and CF=ycm.

Show that triangles(a) AEBand CBFare similar.

Show that(b) xy= 80.

Show that triangle(c) EDFhas area given by .A x x80 5320

= + +

Solution

(a) (given)

(corresponding , )(similarly, )

EAB BCF

BFC EBABEA FBC

AB DCAD BC

90

s

+ +

+ +

+ +

+

chapter 2 geometrical applications of calculus

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