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Chapter

21

A B C D E F Rates of change Limits The derivative function Simple rules of differentiation Tangents to curves The second derivative

Differential calculus

Syllabus reference: 7.1, 7.2, 7.3 Contents:

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DIFFERENTIAL CALCULUS

(Chapter 21)

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.

distance travelled (m) 400 300 200

100 Things to think about: a What is meant by a rate? 0 0 30 60 90 b What do we call the rate at which Valentino time (s) 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

Rates are used every day to measure performance. For example, we measure:

RATES OF CHANGE

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

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 the scoring rate of a basketballer in points per game.

Example 1

Self Tutor

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. 213 words 3 minutes 6 errors Josef's error rate = 213 words 260 words Marie's typing rate = 4 minutes 7 errors Marie's error rate = 260 words Josef's typing rate = = 71 words per minute. ¼ 0:0282 errors per word. = 65 words per minute. ¼ 0:0269 errors per word.

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

DIFFERENTIAL CALCULUS

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627

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 ?

INVESTIGATION 1

CONSTANT AND VARIABLE RATES OF CHANGE

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.

DEMO

water

depth

depth time depth

time

What to do: 1 What features of the graph indicate a rate of change in water level that is: a constant b variable?

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DIFFERENTIAL CALCULUS

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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

depth

b

depth

c

depth

d

depth

time

time

time

time

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.

Example 2

Self Tutor

350 300 250 200 150 100 50 population size

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?

1

2

3

4

5

6

weeks 7

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 population growth rate = (315 ¡ 50) mice (7 ¡ 0) weeks

The average rate of change between two points on the graph is the gradient of the chord between them.

ii

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.

7 6 5 4 3 2 1 time (s) 1 2 3 4 5 6 7 8 distance (m)

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.

beetles (per m2 ) 40 30 20 10 2 4 6 8 10 12 14 dose (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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DIFFERENTIAL CALCULUS (Chapter 21)

INVESTIGATION 2

INSTANTANEOUS SPEED

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.

D 80 60 40 20 2 F(2,¡20) t 4 6 M(4,¡80)

chord curve

The average speed in the time interval 2 6 t 6 4 is distance travelled = time taken (80 ¡ 20) m = (4 ¡ 2) s = = 30 m s¡1

60 2

m s¡1

DEMO

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 . 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. 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? t 3 2:5 2:1 2:01 gradient of FM

t 0 1:5 1:9 1:99

gradient of FM

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631

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 = 12:5 m s¡1 . 8 sec Notice that the cyclist's early speed is quite small, but it increases as time goes by.

distance (m) tangent P

100 80 60 40 20 0 0

1

2

3

4

5

time (t sec) 6 7 8

To find the instantaneous speed at any time instant, for example t = 4, we draw the tangent at the point and find its gradient.

distance (m) 100 80 60 40 20 0 0 1 2 point where t=4

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

tangent at t = 4 5 6 7 8 time (t sec)

PRINTABLE GRAPHS

(40 ¡ 0) m (7 ¡ 2) s

40 5

=

m s¡1

= 8 m s¡1

3

4

EXERCISE 21A.3

1 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. a b

4 3 2 1 1 t=1 time (s) 2 3 4 distance (m) 8 6 4 2 1 2 3 distance (km)

t=3 time (h) 4

c

4 3 2 1

profit ($¡thousands) x = 30 number of items sold 10 20 30 40 50

d

120 100 80 60 40 20

population of bat colony

t=5 time (weeks) 1 2 3 4 5 6 7 8 9 10

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DIFFERENTIAL CALCULUS

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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?

volume (thousands of litres) 8 6 4 2

0.5

1.5

2.5

3.5 time (x hours)

B

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.

chord (secant)

LIMITS

curve B

A

tangent

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.

INVESTIGATION 3

THE GRADIENT OF A TANGENT

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?

DEMO

y

f(x) = x 2

A(1, 1) x

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633

What to do: 1

y f(x) = x 2 B(x, x 2) A(1, 1) x

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) x2 ¡ 1 or . x¡1 x¡1

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?

x 5 3 2 1:5 1:1 1:01 1:001

Point B (5, 25)

gradient of AB 6

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 = ) the gradient of AB =

y = x2 B1 B2 B3 B4

x2 ¡ 1 x¡1 (x + 1)(x ¡ 1) = x + 1 provided that x 6= 1 x¡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 x2 ¡ 1 x!1 x ¡ 1

x!1

= lim x + 1, x 6= 1 =2

As B approaches A, the gradient of AB approaches or converges to 2.

A

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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Example 3

Self Tutor

a

x!2

Evaluate:

lim x2

b

x!0

lim

5x + x x

2

c

x!3

lim

x ¡9 x¡3

2

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

x!2

b

5x + x2 x!0 x x(5 + x) = lim x!0 x = lim 5 + x provided x 6= 0 lim

x!0

c

x2 ¡ 9 x!3 x ¡ 3 (x + 3)(x ¡ 3) = lim x!3 x¡3 1 = lim (x + 3) provided x 6= 3 lim

x!3

=5

=3+3 =6

EXERCISE 21B

1 Evaluate: a lim (x + 4)

x!3 x!2 x!0

b

x!¡1

lim (5 ¡ 2x)

c lim (3x ¡ 1)

x!4

d lim 5x2 ¡ 3x + 2 g lim (x2 + 5)

e lim h2 (1 ¡ h)

h!0

f

x!¡1

lim

1 ¡ 2x x2 + 1

h

x!¡2

lim

4 x

2 Evaluate the following limits by looking for a common factor in the numerator and denominator: x2 ¡ 3x 2h2 + 6h h3 ¡ 8h b lim c lim a lim x!0 h!0 h!0 x h h

C

THE DERIVATIVE FUNCTION

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 by `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'. dy . We say this as dx

DIFFERENTIAL CALCULUS (Chapter 21)

635

INVESTIGATION 4

THE DERIVATIVE OF y = x2

The graphs below show y = x2 with tangents drawn at the points where x = ¡3, ¡2, ...., 3.

y 8 6 4 2 -4 -3 -2 -1 y 8 6 4 2 -2 -1 D 1 2 x y = x2 8 6 4 2 E 1 2 3 4 y 8 6 4 2 1 2 3 4 x G y = x2 x x -4 -3 -2 -1 y y = x2 8 6 4 2 1 2 3 4 B y 8 6 4 2 x C -4 -3 -2 -1 y y = x2 F x y 8 6 4 2 x

A

y = x2

y = x2

y = x2

What to do: 1 Use the shaded triangles to find the gradients of the tangents to y = x2 at the seven different points. Hence complete the following table: 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 = ::::"

DEMO

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. dy So, y = x2 has the derivative function = 2x, dx or alternatively, if f (x) = x2 then f 0 (x) = 2x.

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DIFFERENTIAL CALCULUS (Chapter 21)

INVESTIGATION 5

THE DERIVATIVE OF y = xn

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 dy dx b Graph ¡3 ¡2 ¡1 0 1 2 3

x dy dx

dy against x. dx Hence predict a formula for the dy . derivative function 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. 4 Use your results from 2 and 3 to complete: Function x x2 x

3

Derivative function 2x

x4 x¡1 dy 5 Predict the form of where y = xn . dx x¡2

You should have discovered that:

If y = xn then

dy = nxn¡1 . dx

DISCUSSION

Does the rule "if y = xn then dy = nxn¡1 " work when n = 0? dx

DIFFERENTIAL CALCULUS (Chapter 21)

637

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)).

y f(x + h) B y = f(x)

The chord AB has gradient = = f (x + h) ¡ f (x) x+h¡x f (x + h) ¡ f (x) : h

f(x)

A

h x x+h x

If we now let B approach A, then the gradient of AB approaches the gradient of the tangent at A.

So, the gradient of the tangent at the variable point (x, f (x)) is the limiting value of f (x + h) ¡ f (x) h as h approaches 0, or lim f (x + h) ¡ f (x) . h

h!0

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) = lim f (x + h) ¡ f (x) . h

h! 0

When we evaluate this limit to find the derivative function, we call this the method of first principles. Example 4 Use the definition of f 0 (x) to find the gradient function of f (x) = x2 . f 0 (x) = lim

h!0

Self Tutor

f (x + h) ¡ f (x) h (x + h)2 ¡ x2 h x2 + 2hx + h2 ¡ x2 h h(2x + h) h fas h 6= 0g

= lim = lim = lim

h!0

h!0

h!0 h!0

= lim (2x + h) = 2x

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DIFFERENTIAL CALCULUS (Chapter 21)

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 + 3a2 b + 3ab2 + b3 (a + b)4 = a4 + 4a3 b + 6a2 b2 + 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

D

SIMPLE RULES OF DIFFERENTIATION

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.

INVESTIGATION 6

SIMPLE RULES OF DIFFERENTIATION

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: a f (x) = x2 + 3x Hence copy and complete: "If f (x) = cxn , then f 0 (x) = ::::" b f (x) = x3 ¡ 2x2 "If f (x) = u(x) + v(x) then f 0 (x) = ::::"

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

We have now determined the following rules for differentiating: Function a constant x

n

f (x) a x

n

f 0 (x) 0 nxn¡1 anxn¡1 u0 (x) + v0 (x)

a constant multiple of xn multiple terms

axn u(x) + v(x)

DIFFERENTIAL CALCULUS (Chapter 21)

639

Using the rules we have now developed we can differentiate sums of powers of x. For example, if f(x) = 3x4 + 2x3 ¡ 5x2 + 7x + 6 then f 0 (x) = 3(4x3 ) + 2(3x2 ) ¡ 5(2x) + 7(1) + 0 = 12x3 + 6x2 ¡ 10x + 7 Example 5 Find f 0 (x) for f (x) equal to: a 5x3 + 6x2 ¡ 3x + 2 a ) b 7x ¡ 4 3 + 3 x x c x2 + 4x ¡ 5 x

Self Tutor

f (x) = 5x3 + 6x2 ¡ 3x + 2 f 0 (x) = 5(3x2 ) + 6(2x) ¡ 3(1) = 15x2 + 12x ¡ 3 4 3 + 3 x x = 7x ¡ 4x¡1 + 3x¡3

b

f (x) = 7x ¡ )

f 0 (x) = 7(1) ¡ 4(¡1x¡2 ) + 3(¡3x¡4 ) = 7 + 4x¡2 ¡ 9x¡4 4 9 =7+ 2 ¡ 4 x x x2 + 4x ¡ 5 f (x) = x x2 5 = +4¡ x x = x + 4 ¡ 5x¡1 f 0 (x) = 1 + 5x¡2

Remember that 1 = x¡n . xn

c

)

Example 6 Find the gradient function of f (x) = x2 ¡

Self Tutor

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 Now f 0 (2) = 4 + 1 = 5. So, the tangent has gradient = 5.

) f 0 (x) = 2x ¡ 4(¡1x¡2 ) = 2x + 4x¡2 4 = 2x + 2 x

640

DIFFERENTIAL CALCULUS (Chapter 21)

EXERCISE 21D

1 Find f 0 (x) given that f (x) is: a x3 d x2 + x g x3 + 3x2 + 4x ¡ 1 j 2x ¡ 3 x2 b 2x3 e 4 ¡ 2x2 h 5x4 ¡ 6x2 k x3 + 5 x c 7x2 f x2 + 3x ¡ 5 i 3 ¡ 6x¡1 l x3 + x ¡ 3 x

2 Suppose f (x) = 4x3 ¡ x. Find: a f 0 (x) 3 Suppose g(x) = a g 0 (x) b f 0 (2) x2 + 1 . Find: x b g0 (3) c f 0 (0)

c g0 (¡2) 8 at x = 9 x2 d y = 2x ¡ 5x¡1 at x = 2 b y= f y= x3 ¡ 4x ¡ 8 at x = ¡1 x2

4 Find the gradient of the tangent to: a y = x2 at x = 2 c y = 2x2 ¡ 3x + 7 at x = ¡1 e y= 5 Find x2 ¡ 4 at x = 4 x2

dy for y = (3x + 1)2 . dx Hint: Start by expanding the brackets.

Example 7 If y = 3x2 ¡ 4x, find As y = 3x2 ¡ 4x, dy is: dx dy and interpret its meaning. dx

Self Tutor

dy = 6x ¡ 4. dx

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

dy 3 , find and interpret its meaning. x dx b The position of a car moving along a straight road is given by S = 2t2 + 4t metres dS where t is the time in seconds. Find and interpret its meaning. dt c The cost of producing and selling x toasters each week is given by dC and interpret its meaning. C = 1785 + 3x + 0:002x2 dollars. Find dx a If y = 4x ¡

DIFFERENTIAL CALCULUS (Chapter 21)

641

E

y = f(x) tangent gradient m point of contact

TANGENTS TO CURVES

Consider a curve y = f (x).

If A is the point with x-coordinate a, then the gradient of the tangent at this point is f'(a) = m.

The equation of the tangent is y ¡ f (a) = f 0 (a) fequating gradientsg x¡a or y ¡ f (a) = f 0 (a)(x ¡ a).

A(a, f(a))

x=a

Alternatively, the equation of the tangent at the point A(a, b) is y¡b = f 0 (a) x¡a y ¡ b = f 0 (a)(x ¡ a).

or

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. Example 8

Self Tutor

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

y f(x) = x 2 + 1

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

1

(1, 2) x

y¡2 =2 x¡1 which is y ¡ 2 = 2x ¡ 2 or y = 2x.

642

DIFFERENTIAL CALCULUS (Chapter 21)

Example 9 Find the equations of any horizontal tangents to y = x3 ¡ 12x + 2. Since y = x3 ¡ 12x + 2, dy = 3x2 ¡ 12 dx so 3x2 ¡ 12 = 0 ) 3(x2 ¡ 4) = 0 3(x + 2)(x ¡ 2) = 0 ) x = ¡2 or 2

Self Tutor

Horizontal tangents have gradient 0, ) 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 c y = 3x¡1 at x = ¡1 e y = x2 + 5x ¡ 4 at x = 1 g y = x3 + 2x at x = 0 i y = x + 2x¡1 at x = 2 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. b y = x3 at x = ¡2 4 d y = 3 at x = 2 x f y = 2x2 + 5x + 3 at x = ¡2 h y = x2 + x¡1 at x = 0 j y= x2 + 4 at x = ¡1 x

F

THE SECOND DERIVATIVE

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 or d2 y to represent the second derivative. dx2

DIFFERENTIAL CALCULUS (Chapter 21)

643

f 00 (x) is read as "f double dashed of x". µ ¶ d2 y d dy is read as "dee two y by dee x squared ". = dx2 dx dx

THE SECOND DERIVATIVE IN CONTEXT

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.

t=0 Michael's place t=5 t = 15 t = 17 t = 19 friend's house

DEMO

t = 10

The distance Michael has travelled at various times is given in the following table: Time (t min) Distance travelled (s m) 0 0 2:5 5 7:5 10 12:5 15 17 19

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:

2500 2000 1500 1000 500 5 10 15 t s s = 1.18t 3 - 30.47t 2 + 284.52t - 16.08

Now

ds ¼ 3:54t2 ¡ 60:94t + 284:52 metres per minute is the instantaneous rate of change dt 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, µ ¶ d2 s d ds = 2 is the instantaneous acceleration. so dt dt dt In this case d2 s ¼ 7:08t ¡ 60:94 metres per minute per minute. dt2

We see that when t = 12, s ¼ 1050 m, ds ¼ 63 metres per minute, dt and d2 s ¼ 24 metres per minute per minute. dt2

644

DIFFERENTIAL CALCULUS (Chapter 21)

Example 10

Self Tutor

3 . x

Find f 00 (x) given that f (x) = x3 ¡ Now f (x) = x3 ¡ 3x¡1 ) ) f 0 (x) = 3x2 + 3x¡2 f 00 (x) = 6x ¡ 6x¡3 6 = 6x ¡ 3 x

EXERCISE 21F

1 Find f 00 (x) given that: a f (x) = 3x2 ¡ 6x + 2 c f (x) = 2 Find 2 ¡ 3x x2 b f (x) = 2x3 ¡ 3x2 ¡ x + 5 d f (x) = 2x ¡ x3 x2

d2 y given that: dx2 b y = x2 ¡

a y = x ¡ x3

5 x2

c 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.

DIFFERENTIAL CALCULUS (Chapter 21)

645

REVIEW SET 21A

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.

8000 6000 4000 2000 time (months) 10 20 30 40 50 number of televisions sold

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 a f 0 (x) 1 x d b f 0 (2) 7x3 + 2x4 x2 c f 0 (0).

3 Consider f(x) = x4 ¡ 3x ¡ 1. Find:

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 ¡ b f (x) = (x + 4)2

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

REVIEW SET 21B

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 0 b Find f (x). c Find the gradient of the tangent to f (x) at (1, 3). Compare this with your answers to a. 2 Find dy for: dx b y= x3 ¡ x x2 c y = 2x + x¡1 ¡ 3x¡2

y = x2 + 2x y 8 6 4 2 -1 -4 -3 -2 -2 1 2 3 (1, 3) x

a y = 3x2 ¡ x4

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

646

DIFFERENTIAL CALCULUS (Chapter 21)

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.

REVIEW SET 21C

1 Use the rules of differentiation to find f 0 (x) for f (x) equal to: 4 1 ¡ 2 b 2x¡3 + x¡4 c a x4 + 2x3 + 3x2 ¡ 5 x x 2 Find the gradient of f (x) at the given point for the following functions: a f (x) = x2 ¡ 3x, at x = ¡1 2 c f (x) = x + , at x = 3 x 3 Find the equation of the tangent to y = b f (x) = ¡3x2 + 4, at x = 2 d f (x) = x3 ¡ x2 ¡ x ¡ 2, at x = 0

12 at the point (1, 12). x2 4 Sand is poured into a bucket for 30 seconds. After t seconds, the weight of sand is S(t) = 0:3t3 ¡ 18t2 + 550t grams. Find and interpret S 0 (t). 5 Find d2 y for: dx2 6 Evaluate: a lim x2 + 3

x!1

a 7x3 ¡ 4x

b 2x2 + 3¡x x

5 x c lim h2 ¡ 3h h

b

x!¡4

lim

h!0

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

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