differential calculus - haese mathematics · 630 differential calculus (chapter 21) when a ball...

21Chapter

Differentialcalculus

Contents:

Syllabus reference: 7.1, 7.2, 7.3

A

B

C

D

E

F

Rates of change

Limits

The derivative function

Simple rules ofdifferentiation

Tangents to curves

The second derivative

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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_21\625IB_STSL-2_21.CDR Tuesday, 16 February 2010 2:07:45 PM PETER

OPENING PROBLEM

Valentino is riding his motorbike

around a racetrack. A computer

chip on his bike measures the

distance Valentino has travelled as

time goes on. This data is used to plot a graph

of Valentino’s progress.

Things to think about:

a What is meant by a rate?

b What do we call the rate at which Valentino

is travelling?

c What is the difference between an instantaneous rate and an average rate?

d How can we read a rate from a graph?

e Which is the fastest part of the racetrack?

A rate is a comparison between two quantities of different kinds.

Rates are used every day to measure performance.

For example, we measure:

² the speed at which a car is travelling in km h¡1 or m s¡1.

² the fuel efficiency of the car in km L¡1 or litres per 100 km travelled.

²

Josef typed 213 words in 3 minutes and made 6 errors, whereas Marie typed 260 words

in 4 minutes and made 7 errors. Compare their performance using rates.

Josef’s typing rate =213 words

3 minutes= 71 words per minute.

Josef’s error rate =6 errors

213 words¼ 0:0282 errors per word.

Marie’s typing rate =260 words

4 minutes= 65 words per minute.

Marie’s error rate =7 errors

260 words¼ 0:0269 errors per word.

) Josef typed at a faster rate but Marie typed with greater accuracy.

RATES OF CHANGEA

Example 1 Self Tutor

0 30 60 90

100

0

200

300

400

distance travelled (m)

time (s)

the of a basketballer in points per game.scoring rate

626 DIFFERENTIAL CALCULUS (Chapter 21)


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INVESTIGATION 1 CONSTANT AND VARIABLE

RATES OF CHANGE

DEMO

EXERCISE 21A.1

1 Karsten’s pulse rate was measured at 67 beats per minute.

a Explain exactly what this rate means.

b How many heart beats would Karsten expect to have each hour?

2 Jana typed a 14 page document and made eight errors. If an average page of typing has

380 words, find Jana’s error rate in:

a errors per word b errors per 100 words.

3 Niko worked 12 hours for $148:20 whereas Marita worked 13 hours for $157:95. Who

worked for the better hourly rate of pay?

4 New tyres have a tread depth of 8 mm. After driving for 32 178 km, the tread depth on

Joanne’s tyres was reduced to 2:3 mm. What was the wearing rate of the tyres in:

a mm per km travelled b mm per 10 000 km travelled?

5 We left Kuala Lumpur at 11:43 am and travelled to Penang, a distance of 350 km. We

arrived there at 3:39 pm. What was our average speed in: a km h¡1 b m s¡1?

When water is added at a constant rate to a cylindrical container, the depth

of water in the container is a linear function of time. This is because the

volume of water added is directly proportional to the time taken to add it.

If water was not added at a constant rate the direct proportionality would

not exist.

The depth-time graph for a

cylindrical container is shown

alongside.

In this investigation we explore

the changes in the graph for

different shaped containers such

as the conical vase.

What to do:

1 What features of the graph indicate a rate of change in water level that is:

a constant b variable?

water

depth

time

depth

time

depth

627DIFFERENTIAL CALCULUS (Chapter 21)


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2 For each of the following containers, draw a depth-time graph as water is added:

a b c d e

3 Use the water filling demonstration to check your answers to question 2.

4 Write a brief report on the connection between the shape of a vessel and the

corresponding shape of its depth-time graph. You may wish to discuss this in parts.

For example, first examine cylindrical containers, then conical, then other shapes.

Gradients of curves must be included in your report.

5 Draw possible containers as in question 2 which have the following depth-time

graphs:

a b c d

AVERAGE RATE OF CHANGE

If the graph which compares two quantities is a straight line, there is a constant rate of

change in one quantity with respect to the other. This constant rate is the gradient of the

straight line.

If the graph is a curve, we can find the average rate of change between two points by

finding the gradient of the chord or line segment between them. The average rate of change

will vary depending on which two points are chosen, so it makes sense to talk about the

average rate of change over a particular interval.

The number of mice in a colony was

recorded on a weekly basis.

a Estimate the average rate of increase in

population for:

i the period from week 3 to week 6

ii the seven week period.

b What is the overall trend with regard to

population increase over this period?


depth

time

depth

time

depth

time

depth

time

population size350

300

250

200

150

100

50weeks

654321 7



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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_21\628IB_STSL-2_21.CDR Friday, 26 March 2010 12:13:41 PM PETER

The average rate of

change between two

points on the graph is

the gradient of the

chord between them.

4

1

5

7

2

6

3

41 52 63 7 8

distance (m)

time (s)

40

2 4 6 8 10 12 14

10

20

30

beetles (per m )2

dose (g)

DIFFERENTIAL CALCULUS (Chapter 21) 629

a i population growth rate

=increase in population

increase in time

=(240¡ 110) mice

(6¡ 3) weeks

¼ 43 mice per week

ii population growth rate

=(315¡ 50) mice

(7¡ 0) weeks

¼ 38 mice per week

b The graph is increasing over the period by larger and larger amounts, so the

population is increasing at an ever increasing rate.

EXERCISE 21A.2

1 For the travel graph given alongside,

estimate the average speed:

a in the first 4 seconds

b in the last 4 seconds

c in the 8 second interval.

2 The numbers of lawn beetles per m2 of lawn

which are left surviving after various doses of

poison are shown in the graph alongside.

a Estimate the rate of beetle decrease when:

i the dose increases from 0 to 10 g

ii the dose increases from 4 to 8 g.

b Describe the effect on the rate of beetle

decline as the dose goes from 0 to 14 g.

INSTANTANEOUS RATES OF CHANGE

The speed of a moving object such as a motor car, an aeroplane, or a runner, will vary over

time. The speed of the object at a particular instant in time is called its instantaneous speed.

To examine this concept in greater detail, consider the following investigation.


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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_21\629IB_STSL-2_21.CDR Friday, 5 March 2010 9:53:18 AM PETER

INSTANTANEOUS SPEEDINVESTIGATION 2

DEMO

42

80

60

40

20F ,(2 20)�

M ,(4 80)�

curve

chord

6

D

t


When a ball bearing is dropped from the

top of a tall building, the distance it has

fallen after t seconds is recorded, and the

following graph of distance against time

obtained.

In this investigation we will try to measure the speed of

the ball at the instant when t = 2 seconds.

The average speed in the time interval

2 6 t 6 4 is

=distance travelled

time taken

=(80¡ 20)m

(4¡ 2) s

= 60

2m s¡1

= 30 m s¡1

What to do:

1 Click on the icon to start the demonstration. F is the point where t = 2 seconds, and

M is another point on the curve. To start with M is at t = 4 seconds.

The number in the box marked gradient is the gradient of the chord FM. This

is the average speed of the ball bearing in the interval from F to M. For M at

t = 4 seconds, you should see the average speed is 30 m s¡1.

t gradient of FM

3

2:5

2:1

2:01

2 Click on M and drag it slowly towards F. Copy and

complete the table alongside with the gradient of the

chord when M is at various times t.

3 Observe what happens as M reaches F. Explain why this

is so.

t gradient of FM

0

1:5

1:9

1:99

4 When t = 2 seconds, what do you suspect the

instantaneous speed of the ball bearing is?

5 Move M to the origin, and then slide it towards F from

the left. Copy and complete the table alongside with the

gradient of the chord when M is at various times t.

6 Do your results agree with those in 4?


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From the investigation you should have discovered that:

The instantaneous rate of change of a variable at a

particular instant is given by the gradient of the tangent

to the graph at that point.

For example, the graph alongside shows how a

cyclist accelerates away from an intersection.

The average speed over the first 8 seconds is

100 m

8 sec= 12:5 m s¡1.

Notice that the cyclist’s early speed is quite

small, but it increases as time goes by.

To find the instantaneous speed at any time instant, for example t = 4, we draw the tangent

at the point and find its gradient.

This tangent passes through (2, 0) and (7, 40).) the instantaneous speed at t = 4

= gradient of tangent

=(40¡ 0) m

(7¡ 2) s

= 40

5m s¡1

= 8 m s¡1

EXERCISE 21A.3

1

a b

c d

tangent

P

80

20

100

40

60

041 52 63 7 8

distance (m)

time ( sec)t

0

80

20

100

40

60

041 52 63 7 8

distance (m)

time ( sec)t0

tangentat t = 4

pointwheret = 4

PRINTABLE

GRAPHS

For each of the following graphs, estimate the rate of change at the

point shown by the arrow. Make sure your answer has the correct units.

t = 1

4321

4

3

2

1

distance (m)

time (s)

profit ($ thousands)�

number of items sold

x = 30

5040302010

4

3

2

1

1086421 3 5 7 9

120

100

80

60

40

20

population of bat colony

time (weeks)

t = 5

distance (km)

time (h)

t = 3

4321

8

6

4

2



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THE GRADIENT OF A TANGENTINVESTIGATION 3

2 Water is leaking from a tank. The

number of thousands of litres of water

left in the tank after x hours is given

in the graph alongside.

a How much water was in the tank

originally?

b How much water was in the tank

after 1 hour?

c How quickly was the tank losing

water initially?

d How quickly was the tank losing

water after 1 hour?

We have seen that the gradient of the chord AB measures

the average rate of change of the function for the given

change in x-values.

The gradient of the tangent at point A measures the

instantaneous rate of change of the function at point A.

We say that in the limit as B approaches A, the gradient

of the chord AB will be the gradient of the tangent at A.

The gradient of the tangent at x = a is defined as the gradient of the curve at the point

where x = a, and is the instantaneous rate of change in f(x) with respect to x at that

point.

LIMITSB

curve

tangent

chord(secant) B

A

DEMO A(1, 1)

y

x

f(x) = x 2

2

4

6

8

0.5 1.5 2.5 3.5

volume (thousands of litres)

time ( hours)x


Given a curve f(x), we wish to find the gradient of the tangent at the point

(a, f(a)).

For example, the point A(1, 1) lies

on the curve f(x) = x2.

What is the gradient of the tangent

at A?


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What to do:

1 Suppose B lies on f(x) = x2 and B

has coordinates (x, x2).

a Show that the chord AB has

gradient

f(x)¡ f(1)

x¡ 1or

x2 ¡ 1

x¡ 1.

x Point B gradient of AB

5 (5, 25) 6

3

2

1:5

1:1

1:01

1:001

b Copy and complete the table alongside:

2 Comment on the gradient of AB as x gets

closer to 1.

3 Repeat the process as x gets closer to 1, but

from the left of A.

4 Click on the icon to view a demonstration of

the process.

5 What do you suspect is the gradient of the

tangent at A?

Fortunately we do not have to use a graph and table of values each time we wish to find

the gradient of a tangent. Instead we can use an algebraic and geometric approach which

involves limits.

LIMIT ARGUMENT

From the investigation, the gradient of AB =x2 ¡ 1

x¡ 1

) the gradient of AB =(x+ 1)(x¡ 1)

x¡ 1= x+ 1 provided that x 6=1

In the limit as B approaches A, x ! 1 and the gradient

of AB ! the gradient of the tangent at A.

So, the gradient of the tangent at the point A is

mT= lim

x!1

x2 ¡ 1

x¡ 1

= limx!1

x+ 1, x 6= 1

= 2

A(1, 1)

B(x, x )2

y

x

f(x) = x 2

As B approaches A, the

gradient of AB approaches

or to .converges 2

A

B1

B2

B3

B4

y = x2

tangent at A


Limit arguments like that above form the foundation

of differential calculus. However, formal treatment

of limits is beyond the scope of this course.


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Evaluate: a limx!2

x2 b limx!0

5x+ x2

xc lim

x!3

x2 ¡ 9

x¡ 3

a x2 can be made as close as we like to 4 by making x sufficiently close to 2.

) limx!2

x2 = 4.

b limx!0

5x+ x2

x

= limx!0

x(5 + x)

x

= limx!0

5 + x provided x 6= 0

= 5

c limx!3

x2 ¡ 9

x¡ 3

= limx!3

(x+ 3)(x¡ 3)

x¡ 3

= limx!3

(x+ 3) provided x 6= 3

= 3 + 3

= 6

EXERCISE 21B

1 Evaluate:

a limx!3

(x+ 4) b limx!¡1

(5¡ 2x) c limx!4

(3x¡ 1)

d limx!2

5x2 ¡ 3x+ 2 e limh!0

h2(1¡ h) f limx!¡1

1¡ 2x

x2 + 1

g limx!0

(x2 + 5) h limx!¡2

4

x

2 Evaluate the following limits by looking for a common factor in the numerator and

denominator:

a limx!0

x2 ¡ 3x

xb lim

h!0

2h2 + 6h

hc lim

h!0

h3 ¡ 8h

h

We have seen how the rate of change of a function at a particular instant is given by the

gradient of the tangent at that point.

We can hence describe a gradient function which, for any given value of x, gives the gradient

of the tangent at that point. We call this gradient function the derived function or derivative

function of the curve.

If we are given y in terms of x, we represent the derivative function bydy

dx. We say this as

‘dee y by dee x’.

If we are given the function f(x), we represent the derivative function by f 0(x). We say

this as ‘f dashed x’.

THE DERIVATIVE FUNCTIONC


1



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THE DERIVATIVE OF = x2INVESTIGATION 4 y

DEMO

x

y

1 2

2

4

6

8

D

y = x2

-1-2

A

x

y

2

4

6

8y = x2

-1-2-3-4

B

x

y

2

4

6

8y = x2

-1-2-3-4

C

x

y

2

4

6

8y = x2

-1-2-3-4

x

y

1 2 3 4

2

4

6

8y = x2

Ex

y

1 2 3 4

2

4

6

8

F

y = x2

x

y

1 2 3 4

2

4

6

8 G

y = x2


The graphs below show y = x2 with tangents drawn at the points where

x = ¡3, ¡2, ...., 3.

What to do:

1

x-coordinate ¡3

gradient of tangent

2 Use your table to help complete:

“the gradient of the tangent to y = x2 at (x, y) is m = ::::”

3 Click on the icon to check the validity of your statement in 2.

Click on the bar at the top to drag the point of contact of the tangent along the curve.

You should have found that the gradient of the tangent to y = x2 at the point (x, y) is given

by 2x.

So, y = x2 has the derivative functiondy

dx= 2x,

or alternatively, if f(x) = x2 then f 0(x) = 2x.

Use the shaded triangles to find the gradients of

the tangents to at the seven different points.

Hence complete the following table:

y x= 2


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DISCUSSION

THE DERIVATIVE OF = xnINVESTIGATION 5 y

dy

dx

x

DERIVATIVE

DETERMINER


In this investigation we seek derivative functions for other

functions of the form y = xn where n 2 Z .

What to do:

1 Click on the icon to run the derivative determiner software.

2 Choose the function y = x.

a Use the software to complete the table:

x ¡3 ¡2 ¡1 0 1 2 3

dy

dx

b Graphdy

dxagainst x.

Hence predict a formula for the

derivative functiondy

dx.

3 Repeat 2 for the functions:

a y = x3 b y = x4 c y = x¡1 d y = x¡2

Hint: When x = 0, the derivatives of both y = x¡1 and y = x¡2 are undefined.

Function Derivative function

x

x2 2x

x3

x4

x¡1

x¡2

4 Use your results from 2 and 3 to complete:

5 Predict the form ofdy

dxwhere y = xn.

You should have discovered that: If y = xn thendy

dx= nxn¡1.

Does the rule “if y = xn thendy

dx= nxn¡1” work when n = 0?


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x

y

A

Bf(x + h)

x x + h

f(x)

y = f(x)

h


FINDING THE DERIVATIVE FUNCTION USING LIMITS

Consider a general function y = f(x) where A is (x, f(x)) and B is (x+ h, f(x+ h)).

The chord AB has gradient

=f(x+ h)¡ f(x)

x+ h¡ x

=f(x+ h)¡ f(x)

h:

So, the gradient of the tangent at the variable point (x, f(x)) is the limiting value of

f(x+ h)¡ f(x)

has h approaches 0, or lim

h!0

f(x+ h)¡ f(x)

h.

This formula gives the gradient of the tangent for any value of the variable x. Since there is

only one value of the gradient for each value of x, the formula is actually a function.

The gradient or derivative function is defined as f 0(x) = limh!0

f(x+ h)¡ f(x)

h.

When we evaluate this limit to find the derivative function, we call this the method of first

principles.

Use the definition of f 0(x) to find

the gradient function of f(x) = x2.

f 0(x) = limh!0

f(x+ h)¡ f(x)

h

= limh!0

(x+ h)2 ¡ x2

h

= limh!0

(2x+ h) fas h 6= 0g

= 2x


If we now let B approach A, then

the gradient of AB approaches the

gradient of the tangent at A.

= limh!0

x2 + 2hx+ h2 ¡ x2

h

= limh!0

h(2x+ h)

h


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SIMPLE RULES OF DIFFERENTIATIONINVESTIGATION 6


EXERCISE 21C

1 Find, from first principles, the gradient function of f(x) where f(x) is:

a x b 5 c x3 d x4

Hint: (a+ b)3 = a3 + 3a2b+ 3ab2 + b3

(a+ b)4 = a4 + 4a3b+ 6a2b2 + 4ab3 + b4

2 Find, from first principles, f 0(x) given that f(x) is:

a 2x+ 5 b ¡x+ 4 c x2 ¡ 3x d 2x2 + x¡ 1

Differentiation is the process of finding a derivative or gradient function.

There are a number of rules associated with differentiation. These rules can be used to

differentiate more complicated functions without having to resort to the tedious method of

first principles.

In this investigation we attempt to differentiate functions of the form cxn

where c is a constant, and functions which are a sum or difference of terms

of the form cxn.

What to do:

1 Find, from first principles, the derivatives of:

a x2 b 4x2 c x3 d 2x3

Hence copy and complete: “If f(x) = cxn, then f 0(x) = ::::”

2 Use first principles to find f 0(x) for:

a f(x) = x2 + 3x b f(x) = x3 ¡ 2x2

Hence copy and complete: “If f(x) = u(x) + v(x) then f 0(x) = ::::”

We have now determined the following rules for differentiating:

SIMPLE RULES OF DIFFERENTIATIOND

Function f(x) f 0(x)

a constant a 0

xn xn nxn¡1

a constant multiple of xn axn anxn¡1

multiple terms u(x) + v(x) u0(x) + v0(x)


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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_21\638IB_STSL-2_21.CDR Wednesday, 17 February 2010 3:26:37 PM PETER

For example, if f(x) = 3x4 + 2x3 ¡ 5x2 + 7x+ 6 then

f 0(x) = 3 + 2(3x2)¡ 5(2x) + 7(1) + 0

Remember that1

xn= x

¡n.

(4x3)


Using the rules we have now developed we can differentiate sums of powers of x.

Find f 0(x) for f(x) equal to:

a 5x3 + 6x2 ¡ 3x+ 2 b 7x¡4

x+

3

x3c

x2 + 4x¡ 5

x

a f(x) = 5x3 + 6x2 ¡ 3x+ 2

) f 0(x) = 5(3x2) + 6(2x)¡ 3(1)

= 15x2 + 12x¡ 3

b f(x) = 7x¡4

x+

3

x3

= 7x¡ 4x¡1 + 3x¡3

) f 0(x) = 7(1)¡ 4(¡1x¡2) + 3(¡3x¡4)

= 7 + 4x¡2 ¡ 9x¡4

= 7 +4

x2¡

9

x4

c f(x) =x2 + 4x¡ 5

x

=x2

x+ 4¡

5

x

= x+ 4¡ 5x¡1

) f 0(x) = 1 + 5x¡2

Find the gradient function of f(x) = x2 ¡4

x.

Hence find the gradient of the tangent to the function at the point where x = 2.

f(x) = x2 ¡4

x

= x2 ¡ 4x¡1

) f 0(x) = 2x¡ 4(¡1x¡2)

= 2x+ 4x¡2

= 2x+4

x2

Now f 0(2) = 4 + 1 = 5.

So, the tangent has gradient = 5.



= 12x3 + 6x2 ¡ 10x+ 7


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EXERCISE 21D

1 Find f 0(x) given that f(x) is:

a x3 b 2x3 c 7x2

d x2 + x e 4¡ 2x2 f x2 + 3x¡ 5

g x3 + 3x2 + 4x¡ 1 h 5x4 ¡ 6x2 i 3¡ 6x¡1

j2x¡ 3

x2k

x3 + 5

xl

x3 + x¡ 3

x

2 Suppose f(x) = 4x3 ¡ x. Find:

a f 0(x) b f 0(2) c f 0(0)

3 Suppose g(x) =x2 + 1

x. Find:

a g0(x) b g0(3) c g0(¡2)

4 Find the gradient of the tangent to:

a y = x2 at x = 2 b y =8

x2at x = 9

c y = 2x2 ¡ 3x+ 7 at x = ¡1 d y = 2x¡ 5x¡1 at x = 2

e y =x2 ¡ 4

x2at x = 4 f y =

x3 ¡ 4x¡ 8

x2at x = ¡1

5 Finddy

dxfor y = (3x+ 1)2.

Hint: Start by expanding the brackets.

If y = 3x2 ¡ 4x, finddy

dxand interpret its meaning.

As y = 3x2 ¡ 4x,dy

dx= 6x¡ 4.

dy

dxis: ² the gradient function or derivative function of y = 3x2 ¡ 4x, from

which the gradient at any point can be found

² the instantaneous rate of change in y as x changes.

6 a If y = 4x¡3

x, find

dy


b The position of a car moving along a straight road is given by S = 2t2+4t metres

where t is the time in seconds. FinddS

dtand interpret its meaning.

c The cost of producing and selling x toasters each week is given by

C = 1785 + 3x+ 0:002x2 dollars. FinddC




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Consider a curve y = f(x).

The equation of the tangent is

y ¡ f(a)

x¡ a= f 0(a) fequating gradientsg

or y ¡ f(a) = f 0(a)(x¡ a).

Alternatively, the equation of the tangent at the point A(a, b) is

y ¡ b

x¡ a= f 0(a) or y ¡ b = f 0(a)(x¡ a).

The equation can also be written in the form y = mx+ c:

For example: If f(x) = x2 then f 0(x) = 2x.

At x = 3, m = f 0(3) = 6, so the tangent has gradient 6.

Since f(3) = 9, the tangent has equation y ¡ 9 = 6(x¡ 3)

or y = 6x¡ 9.

You can also find the equations of tangents at a given point using your graphics calculator.

For assistance with this, consult the graphics calculator instructions at the start of the book.

Find the equation of the tangent to f(x) = x2 + 1 at the point where x = 1.

TANGENTS TO CURVESE


A(a, f(a))

x = a

y = f(x)

point ofcontact

tangentgradient m

(1, 2)

y

x

f(x) = x + 12

1

If A is the point with -coordinate , then

the gradient of the tangent at this point is

.

x a

f a m'( ) =

Since f(1) = 1 + 1 = 2,

the point of contact is (1, 2).

Now f 0(x) = 2x, so m = f 0(1) = 2

) the tangent has equation

y ¡ 2

x¡ 1= 2

which is y ¡ 2 = 2x¡ 2

or y = 2x.


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Find the equations of any horizontal tangents to y = x3 ¡ 12x+ 2.

Since y = x3 ¡ 12x+ 2,dy

dx= 3x2 ¡ 12

Horizontal tangents have gradient 0, so 3x2 ¡ 12 = 0

) 3(x2 ¡ 4) = 0

) 3(x+ 2)(x¡ 2) = 0

) x = ¡2 or 2

When x = 2, y = 8¡ 24 + 2 = ¡14

When x = ¡2, y = ¡8 + 24 + 2 = 18

) the points of contact are

(2, ¡14) and (¡2, 18)

) the tangents are y = ¡14 and y = 18.

EXERCISE 21E

1 Find the equation of the tangent to:

a y = x2 at x = 4 b y = x3 at x = ¡2

c y = 3x¡1 at x = ¡1 d y =4

x3at x = 2

e y = x2 + 5x¡ 4 at x = 1 f y = 2x2 + 5x+ 3 at x = ¡2

g y = x3 + 2x at x = 0 h y = x2 + x¡1 at x = 0

i y = x+ 2x¡1 at x = 2 j y =x2 + 4

xat x = ¡1

2 Find the coordinates of the point(s) on:

a f(x) = x2 + 3x+ 5 where the tangent is horizontal

b f(x) = x3 + x2 ¡ 1 where the tangent has gradient 1

c f(x) = x3 ¡ 3x+ 1 where the tangent has gradient 9

d f(x) = ax2 + bx+ c where the tangent has zero gradient.

Given a function f(x), the derivative f 0(x) is known as the first derivative.

The second derivative of f(x) is the derivative of f 0(x), or the derivative of the first

derivative.

We use f 00(x) or y00 ord2y

dx2to represent the second derivative.

THE SECOND DERIVATIVEF



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f 00(x) is read as “f double dashed of x”.

d2y

dx2=

d

dx

µdy

dx

¶is read as “dee two y by dee x squared ”.

THE SECOND DERIVATIVE IN CONTEXT

Nowds

dt¼ 3:54t2 ¡ 60:94t+284:52 metres per minute is the instantaneous rate of change

in displacement per unit of time, or instantaneous velocity.

The instantaneous rate of change in velocity at any point in time is Michael’s acceleration,

sod

dt

µds

dt

¶=

d2s

dt2is the instantaneous acceleration.

In this cased2s

dt2¼ 7:08t¡ 60:94 metres per minute per minute.

We see that when t = 12, s ¼ 1050 m,

ds

dt¼ 63 metres per minute,

andd2s

dt2¼ 24 metres per minute per minute.

Michael rides up a hill and down the other side to his friend’s house. The dots on the graph

show Michael’s position at various times t.

The distance Michael has travelled at various times is given in the following table:

Time (t min) 0 2:5 5 7:5 10 12:5 15 17 19

Distance travelled (s m) 0 498 782 908 989 1096 1350 1792 2500

The model s ¼ 1:18t3 ¡ 30:47t2 + 284:52t¡ 16:08 metres fits this data well, although the

model gives s(0) ¼ ¡16:08 m whereas the actual data gives s(0) = 0. This sort of problem

often occurs when modelling from data.

A graph of the data points and the model is given below:

DEMO

s2500

2000

1500

1000

500

15105 t

s = 1.18t - 30.47t + 284.52t - 16.083 2

t = 0 t = 5

t = 10

t = 17 t = 19t = 15

friend’s houseMichael’s place


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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_21\643IB_STSL-2_21.cdr Friday, 5 March 2010 9:54:23 AM PETER


Find f 00(x) given that f(x) = x3 ¡3

x.

Now f(x) = x3 ¡ 3x¡1

) f 0(x) = 3x2 + 3x¡2

) f 00(x) = 6x¡ 6x¡3

= 6x¡6

x3

EXERCISE 21F

1 Find f 00(x) given that:

a f(x) = 3x2 ¡ 6x+ 2 b f(x) = 2x3 ¡ 3x2 ¡ x+ 5

c f(x) =2¡ 3x

x2d f(x) =

2x¡ x3

x2

2 Findd2y

dx2given that:

a y = x¡ x3 b y = x2 ¡5

x2c y =

4¡ x

x


3 The position of a bicycle rider from his starting point is given by

s = 2t3 ¡ 15t2 + 100t metres, where t is the time in minutes.

a Find his position after:

i 5 minutes ii 10 minutes.

b For t = 5, find the cyclist’s instantaneous:

i velocity ii acceleration.

4 The position of Kelly from her home is given by

s = 10t3 ¡ 100t2 + 200t metres, where t is the

time in minutes, 0 6 t 6 7:24.

a When is Kelly at home?

b When is Kelly furthermost from her home?

How far is she from home at this time?

c For t = 4, find Kelly’s:

i position

ii instantaneous velocity

iii instantaneous acceleration.

Explain what these results mean.


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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_21\644IB_STSL-2_21.cdr Tuesday, 16 February 2010 2:47:12 PM PETER

REVIEW SET 21A

REVIEW SET 21B

5040302010

8000

6000

4000

2000

number oftelevisions sold

time (months)


1 The total number of televisions sold

over many months is shown on the

graph alongside.

Estimate the rate of sales:

a from 40 to 50 months

b from 0 to 50 months

c at 20 months.

2 Use the rules of differentiation to find f 0(x) for f(x) equal to:

a 7x3 b 3x2 ¡ x3 c (2x¡ 3)2 d7x3 + 2x4

x2

3 Consider f(x) = x4 ¡ 3x¡ 1. Find: a f 0(x) b f 0(2) c f 0(0).

4 Find the equation of the tangent to y = ¡2x2 at the point where x = ¡1.

5 Find f 00(x) for: a f(x) = 3x2 ¡1

xb f(x) = (x+ 4)2

6 Find the derivative of f(x) = 3x¡ 1 from first principles.

1 Consider the function f(x) = x2+2x, with

graph shown alongside.

a Find the gradient of the line through

(1, 3) and the point on f(x) with

x-coordinate:

i 2 ii 1:5 iii 1:1

b Find f 0(x).

c Find the gradient of the tangent to f(x)

at (1, 3). Compare this with your

answers to a.

2 Finddy

dxfor:

a y = 3x2 ¡ x4 b y =x3 ¡ x

x2c y = 2x+ x¡1 ¡ 3x¡2

3 Find the equation of the tangent to y = x3 ¡ 3x+ 5 at the point where x = 2.

x

321

-1

-2-3-4

8

6

4

2

-2

y

(1, 3)

y = x + 2x2


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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_21\645IB_STSL-2_21.cdr Friday, 5 March 2010 9:56:31 AM PETER

REVIEW SET 21C


4 Find all points on the curve y = 2x + x¡1 which have a tangent parallel to the

x-axis.

5 If f(x) = 7 + x¡ 3x2, find: a f(3) b f 0(3) c f 00(3).

6 Find, using the limit method, the derivative function of f(x) = x3 ¡ 2x.

1 Use the rules of differentiation to find f 0(x) for f(x) equal to:

a x4 + 2x3 + 3x2 ¡ 5 b 2x¡3 + x¡4 c1

x¡

4

x2

2 Find the gradient of f(x) at the given point for the following functions:

a f(x) = x2 ¡ 3x, at x = ¡1 b f(x) = ¡3x2 + 4, at x = 2

c f(x) = x+2

x, at x = 3 d f(x) = x3 ¡ x2 ¡ x¡ 2, at x = 0

3 Find the equation of the tangent to y =12

x2at the point (1, 12).

4 Sand is poured into a bucket for 30 seconds. After t seconds, the weight of sand is

Find and interpret S0(t).

5 Findd2y

dx2for: a 7x3 ¡ 4x b 2x2 +

5

x

6 Evaluate:

a limx!1

x2 + 3 b limx!¡4

3¡ x

xc lim

h!0

h2 ¡ 3h

h

7 Find, from first principles, the derivative of f(x) = x2 + 2x.

S(t) = 0:3t3 ¡ 18t2 + 550t grams.


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Y:\HAESE\IB_STSL-2ed\IB_STSL-2ed_21\646IB_STSL-2_21.cdr Monday, 1 March 2010 11:49:48 AM PETER

differential calculus - haese mathematics · 630 differential calculus (chapter 21) when a ball...

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