Index of /Couchman and Jones 3 Unit Book 2

2

Writing Team Stan Jones Ken Couchman

Pearson Australia

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Produced by Pearson Australia Printed in Malaysia, VP National Library of Australia Cataloguing-in-Publication data

510

Jones, S.B. (Stanley Bruce) 3 unit mathematics. Book 2. For secondary school students ISBN 978-0582-81054-9

1. Mathematics. I. Couchman, K.E. (Kenneth Edgar). II Title.

Pearson Australia Group Pty Ltd ABN 40 004 245 943

The graph and derivative of y

=

ax--

the exponential function y =

ex--

differentiation of exponential functions integration the logarithm function -differentiation and integration - - the derivative of y

=

ax--

derivative of log

ax--

a value for

e.

Radian measure of angles --length of circular arc--area of a sector--graphs of the trigonometric functions - - graphical solution of equations - - differentiation of the trigonometric functions-- integration of the trig-onometric functions.

Chord properties of circles-- angles within a circle--angles in a segment of a circle-- cyclic quadrilaterals and concylic points-- arcs and angles-- chords and arcs-- tangents to circles-- tangents and the alter-nate segment.

The derivative applied to velocity and acceleration the use of primitives with motion in a straight line -rates of change-- exponential growth and decay--further growth and decay.

Parametric equations of the parabola --equation of a chord - - tangent and normal to a parabola - - the chord of contact - - general geometrical properties of the parabola - - locus problems.

Ratios of sums and differences - - ratios of double angles-- integration of sin2

x

_{and cos}2

x - -

_{the angle} between two straight lines-- trigonometrical ratios in terms of tan

f

--solution of trigonometrical equations.

Motion in a straight line-- acceleration as a function of position-- equations of motion under constant accel-eration-- vertical motion under gravity-- the motion of a projectile-- simple harmonic motion.

Inverse algebraic functions-- an important relationship

dy dx 1 . . . I f .

dx ·

d

= - -

mverse tngonometnca unctrons

-the derivatives of -the inverse trigonometrical functions --integration of the inverse trigonometrical functions.

Definition-- graphs of polynomial functions-- divi-sion of polynomials-- the remainder theorem-- the factor theorem-- roots and coefficients of polynomial equations-- numerical estimation of roots; halving the interval and Newton's method of approximation.

Discovering patterns-- Pascal's triangle-- the ex-pansion of (a

+

x)n --the general binomial theorem

--proof of the Pascal triangle relations-- the formula for nck --relations between coefficients.

Englishman Charles Babbage's Difference Engine was based on the same principles as today's computers. This machine could generate logarithmic and astronomical tables to six decimal places.

I

•

IC

Photograph courtesy of I 8 M

CHAPTER 19

The word exponent is often used instead of the word index. Functions where the variable is in the index (e.g. 3x, 102_{x) are called exponential functions. One aim of this chapter is to learn how to use}

exponential functions and in particular to discover two very important functions.

Exponential graphs have a particular shape. Here is part of the graph of y

=

2x and the table of values for it.

X -2 -1 0 1 2 3

2x 1 1

1 2 4 8 4 2

You can learn more about exponential graphs by thinking about the questions in the following exercise.

1. (a) What is the value of 2x if x

=

-3 and if x

=

-4? (b) What is the value of2x if x = 4 and if x = 5? (c) What is the value of 2x if x = 10?

8

6

4

2

-3 2 3

(d) How would you describe the behaviour of the graph of y = Y as x becomes larger and larger?

(e) Can you think of any value of x which would make 2x negative? (f) If x approaches negative infinity, what value does 2x approach?

2. Try these similar questions about y = Y.

(a) Do you expect Y ever to be negative?

(b) What is the behaviour of 3x as x becomes smaller and smaller (x ---+ - oo)? (c) Where does the graph of y

=

3x cut they axis?

(d) If xis any positive value, say x

=

4, is Y larger or smaller than 2x? (e) If a is negative which is larger 2a or 3a7

3. Complete the following table of values for y = Y and draw a sketch graph of the curve. For this

graph the domain is limited to lxl ~ 3. If necessary use a calculator.

1 3:

=

I

-3

1

-2

1

-1

I

0

I 1

1 5

I

2

I

2 5

I

4. On a different set of axes because of the scale difference sketchy = lOX and verify that the same fundamental properties hold, namely:

(a) 10x is never negativt;: (b) 10x --+ 0 as x --+ - oo

(c) 10x increases r~pidly and without limit as x gets larger (d) the graph passes through the point (0, 1).

5. Without drawing up a table of values draw a rough sketch of y

=

4x merely showing vital features of shape. What is one important point the graph passes through?

6. Draw your own graph or copy the graph of y = 2x and draw a tangent at the point (0, 1). Check if the gradient of this tangent is about 0·7.

7. Find rough values for the gradients (slopes) of the graphs y

=

3X, y

=

4x andy

=

wx at the point (0, 1) on each.

8. Verify from your answers and from the shape of your curves that the gradient of the tangent at (0, 1) on y

=

3x is greater than for y

=

2x and similarly it increases for y

=

4x and y

=

10x.

Given that the gradients of y

=

2x andy

=

Y at (0, 1) are approximately 0·7 and 1·1 respectively, guess what the gradient of y

=

2·5x at (0, 1) might be.

9. Drawupatableofvaluesandsketchy

=

2·5xfromx

=

-1 tox

=

2andestimatethegradient at (0, 1).

Consider any function y = ax where a is a constant greater than 1. Since a0 _{= 1 this graph passes}

through (0, 1) for any value of a. Let us try to find the gradient at this point in the way we have used in the calculus.

Take the point P(O, 1) and let Q be a point . on the curve near P so that OM= fix and let

QR

=

fly. Q, being on the curve y

=

aX, must have coordinates (fix, a"x)

fly QR a"x- 1 Then = =

-fix PR fix

The limit of this expression as fix --+ 0 is the gradient at (0, 1).

We will call this gradient m and use the ideas above to differentiate ax.

3

'

Ill

II

.;.,

Let P(x, y) and Q(x

+

L!x, y

+

L!y) be two points on the curve y =ax.

Now and so

thus

and

Y =ax

y

+

L!y = ax+tlx

L!y = ax+Ax - ax =ax. aAx _ax

= ax(a6_{x- 1)}

L!y _ ax(a6_{x - 1)}

L!x - L!x

dy = lim L!y dx t.x-->0 L!x

. ax(a6_{x - 1)}

= ltm

---'~---'-t.x-->o L!x

=a .ltm A

t.x->0 uX

)(

x . (at.x-

1)

But we have previously found that the limit lim (at.x - 1) is the gradient of the curve at the point

t.x->O L!x (0, 1) and which for any curve we call m.

So dy = max where m is the gradient at (0, 1). dx

A number of important results follow:

1. We can differentiate y = ax if we know its gradient at (0, 1). The gradient of y = Y at (0, 1) is 0·69 approx. so for that curve

Jx

= 0·69 x 2x.

2. On any exponential curve the gradient at any point is proportional to the value of the function at

that point (i.e. : =

my)

so when the function is small it increases slowly but when the function

is large its gradient is also large.

3. Looking at the expression : = max it follows that if we could find a function ax for some a where the gradient m at (0, 1) was 1 then we would have found a function which is unchanged upon differentiation. Such a function can be found, it is called the exponential function and it has great importance in mathematics.

The gradient of y = 2x at (0, 1) is approx 0·69. The gradient of y = Y at (0, 1) is about 1·1.

Somewhere between 2 and 3 is a value e such that ex has a gradient of 1 at (0, 1).

e turns out to be 2·71828 ... approximately.

y = ex is called the exponential function.

It has the property that if y

=

ex then:

=

ex.

Example:

Using tables or calculator find:

(a) e2·63

Solution:

(a) 13·87377

1. Evaluate: (a) eo·z

(b) e1·67s

(b) 0·49659

(c) e-o·s

(d) e-3·7

(c) 0·01479

(e) 5e3 6

(f)~

6

2. Draw the graph of y

=

ex for -1 ~ x ~ 3 and check that the gradient of the tangent at (0, 1) is 1.

3. (a) Find approximately the value of the function y = ex where x = 1· 5.

(b) Find the gradient of the graph y

=

ex drawn above when x

=

1· 5. Is the gradient approx-imately equal to the value of the function at this point?

4. Draw a rough sketch of y

=

e2x for -2 ~ x ~ 2.

5. Earlier we said that the gradient of y

=

ax at (0, 1) could be found as the limit of

a

11

~~

1 as

~x

becomes small. Using the calculator or tables for values of e11

x complete this table:

~X= 0·1 0·01 0·001

ellx = _{1·1 0517}

e l l x - 1 = _0·10517

e l l x - 1

1·0517

~X

Example:

Sketch the graphs of y

=

2x and y

=

rx

and compare.

Solution:

X -2 -1 0 1 2

2x 1 1 _{1 2 4}

4 2 4

2-x _{4 2 1} 1 1

2 4

2

The graphs are reflections of each other in they axis.

-2 0 2

1. On the same axes draw the graphs of y = ex andy = e-x for -3 ~ x ~ 3.

2. By adding ordinates of the above graphs or otherwise draw graphs of y = ex

+

e-x and of

ex +e-x y=

2

3. Sketch the graphs y = Y andy=

rx

and compare them.

4. Draw a sketch of the graph y = e-3_{x for -2 ~ x ~ 2. Where does this graph cut they axis?}

On the same axes also draw y = 2e-3_x.

5. Draw graphs of y = !ex andy = ef on the same axes.

Remember

Examples:

Differentiate with respect to x.

(a) e6x+3

Solution:

(a) y

=

e6x+3

1.

Let y = e" where u = 6x

+

3

dy = dy x du dx du dx

=

e" x 6

= 6e6x+3

Differentiate with respect to x:

(a) e3x (d) e-4x

(b) esx-6 _(e) e2x

+

9

(b) y = e2x3

Let y = e" where u = 2x3

dy = dy x du dx du dx

= e" x 6x2

=

6x2e2x3

(g) 3x

+

_!

+

e-x

X

(c) e?x+s _{(f) ex+ 5x2}

+

₂

(h) !(ex - e-x)

(i) 2e ?-x

2. Differentiate with respect to x:

(a) 5e6x (c) x - e7_-x

(b) ex2 (d) ex3

+

2x

3. Find second derivatives of:

(a) e2x (b) e3x+2

4. If f(x)

=

~(e~

+

e-~), find f'(a).

(e) e-4x2

(f) e.fX

(c) el-x

5. Find the gradient of the tangent to the curve y = ex at the point (1, e).

6. Show that the tangent to the curve y = 2eX, at the point where x = 2, is parallel to the tangent to the curve y = e2

7. Show that the curves y

=

e3_{x andy}

=

_{3ex each have the same gradient where they cross they axis.}

Find the equation of the tangents to each graph at these points.

8. If y

=

ex

+

e-x, show that

~J

=

y.

9.

If y = 4e-x

+

5e-3X, show

that~;;.

+

4:

+

3y =

0.

10. Show that the tangent to the graph y

=

ex at the point where it crosses they axis is inclined to the x axis at an angle of 45°. Hence find the angle between the tangents toy

=

ex andy

=

e-x

at the point (0, 1) which is on both graphs.

11. Show that if y = e 7" -2, then

Jx

= 7y and also

that~~

= 49y.

12. Find the equation of the tangent to the curve y = e"

+

1 at the point (1, e

+

1).

We consider now the use of the product and quotient rules and the application of the calculus t~

maxima and minima.

Example (i):

( ) a

D

11erentmte ·r~

.

y = - - -ex

+

1 . h w1t respect to x.

2x

(b) Find first and second derivatives ofy = x2_ex.

Solution:

(a) y = - -ex

+

1

2x

Let y

=

I!_ where u

=

ex

+

1 v and v = 2x

du dv

v u

-dy dx dx

dx v2

2xe" - (ex

+

1) X 2

4x2

2xex- 2ex- 2

4x2

xe"- ex- 1

2x2

Example (ii):

(b)

Let

and

y

=

x2ex

y = u x v where u = x2

and v = ex

dy = vdu

+

udv

dx dx dx

=

ex X 2x

+

x2 ex = (x2

+

2x)ex d2

-{- =

ex(2x

+

2)

+

(x2

+

2x)ex

dx

=

ex(x2

+

_4x

+

₂₎

Find the maximum and minimum values of y

=

x2 _{ex. (Note: Results above may be used).}

Solution:

if y = x2_ex

then from above y'

=

(x2

+

2x)ex

as ex is never zero then y' = 0

if x2

+

2x = 0 if x(x

+

2) = 0 if x

=

0 or x

= -

2

Substituting x

=

0 and x

= -

2

the turning points are (0, 0) and (- 2, 4e-2 ).

7

Given the turning points found opposite we use the second derivative to examine their nature From above y"

=

ex(x2

+

4x

+

2)

if x = 0, y" = e0 _{x 2 (positive)}

if x = -2, y" = e-2₍₄

-8

+

2)

=

-2e-2 _(negative)

So the point (0, 0) is a minimum and the point ( -2, 4e-2

1. Differentiate with respect to x:

(a) xex (d) 3x2_{ex (g)~} ex

X

(b) 3xex (e) xe2_x 1

(h) ex

+

1

(c) e-x(x - 1) (f)~ _ex (i) ex+ 1

e x - 1

2. Find second derivatives of:

(a) ex2 (b) xex (c) 2xe-x

3. Rewrite 5

~

in such a way that the quotient rule is not needed and differentiate with respect to x.

e·

4. Iff(x) = (1

+

2x)e3

X, findf'(2) andf"(O).

5. Find the equations of the tangents to the curves y = e3_x

+

_{2, y = e}2_x

+

_{1, andy = ex at the}

points where they cross the y axis. Show that these tangents all pass through the same point on the x axis.

6. Show that the tangent to the curve y = ex - 2x at the point (1, e - 2) passes through the origin.

7. Show that, if y = Ae -ct where A and c are constants, then

~:;

- c2 y = 0.

8. (a) Find the stationary value for the curve y = t(ex

+

e-x) and show that it is a minimum. (b) What happens to the value of y when x becomes large?

(c) Show that the values of y when x = 10 and when x = - 10 are the same. Also show that for any value a the values of y when x = a and when x = -a are the same. What does this tell you about the shape of the curve?

(d) Draw a rough sketch of the curve.

N.B. This curve is called a catenary. It is the shape of a chain or rope hanging freely.

9. A sketch of the curve y = xe-x is drawn opposite. Check these features:

(a) Show that the curve has one stationary point which is a maximum. Find it. (See A)

(b) Show that the curve passes through the origin. (c) Find the point of inflexion where the curve

changes curvature. (See B) (d) Find the point C where x = -1.

(e) Show that the curve approaches the x axis as

x becomes large. (See point D)

10. Consider the curve y = xex. Its graph is similar to the above but with obvious differences due to the positive sign.

(a) Find the first and second derivatives. (b) Find any maximum or minimum values. (c) Find any point of inflexion.

(d) Check if the curve passes through the origin or not. (e) What happens toy if x becomes large?

(f) Is y large or small, positive or negative when

(a)x=-3 (b)x=-10?

(g) Draw a sketch of the curve.

A

c

We need to use a little ingenuity in integrating exponential functions. Study these examples.

Example (i):

Show that if y = e5_{x+t, then}

t

=

_5e5_x+1.

Hence find fesx+1_dx.

Solution:

If y

=

e5_{x+1, then dy}

=

_5e5_x+t

dx

Hence

J

Sesx+ldx

=

eSx+l

+

C

So

J

esx+ldx _ .!_esx+l -5

+

C .

Example (ii):

Find

r

xex2dx.

Solution:

d 2 2

Note that -(ex)

=

2xex dx

So

J

2xex dx 2

=

ex 2

+

c

Thus

r

xex2dx

=

t

J:

2xex2dx

=

t[ex2Ji

=

!(e4 -

e).

Example. (iii):

Calculate the area under the curve y

=

!(e2_x

+

_e-2_{x) and between the lines x}

=

_{-1 and x}

=

_1.

Solution:

The area is given by

f

l !(e2x

+

e-2x)dx -1

As the graph is symmetrical about the y-axis this is

2

f

t(e2x

+

e- 2x)dx or

f

(e 2x

+

e- 2x)dx.

Either this integral or the original one may be calculated.

Area =

f

(e 2x

+

e- 2x)dx

=

[te2x _

!e-2x]6

=

(te2 - !e-2) - (t- t)

=

te2 - te-2

=i= 3·627 if simplified answer is required.

9

1. Integrate with respect to x:

(a) e4x (d) e3x

+

e-x

(b) e-2x (e) e-f

(c) eax+b (f) 3x2

+

e2x

2. Integrate with respect to x:

(a) 4xex2 (c) xe7x2+1

(b) ex - xex2 (d) xzezx3+1

3. Evaluate these definite integrals to 3 decimal places:

(a)

f

e3xdx (b)

f

xe-x2dx

(g) 6e4-x

(h) e1-6x

(i) e-2x

4

(e) xex2

+

xe-x2

(f)

2:,

e

4. Find the area bounded by the curve y

=

eX, the x and y axes, and the line x

=

2.

5. Find the area under the curve y

=

i(ex

+

e-x) from x

=

-2 to x

=

2. 6. The diagram shows the curves y

=

e2x and y

=

e-x.

Calculate the area under each curve from x

=

0 to x

=

1 and hence find the area shaded between the curves.

7. Show that the area between the curves y

=

ex andy

=

e2x from x

=

0 to x

=

1 is given by (ie2 - e

+

i).

8. The diagram shows part of the graphs of y

=

ex and

y = xex2• Show that these graphs both pass through the

point (1, e) and that one crosses they axis at y

=

0 and the other at y

=

1. Find the area shaded from x

=

0 to

9. The shaded region in the diagram below the curve y

=

ex

is rotated about the x axis. Find the volume of the solid of revolution formed.

Hint: Remember V = n

r

y2dx.

(1, e)

0

10. The region under the graph y

=

!(ex

+

e-x) and between the lines x

=

-2 and x

=

2 is rotated about the x axis. Find the volume of the solid of revolution formed.

11. Find the area shaded which is bounded by the curve

y = ex, the line y = e and they axis.

12. Sketch the curve y

=

e-2_{x. Find the area between the curve, the x axis, and the lines x}

=

_{1 and}

x = 2. If this region is rotated about the x axis find the volume of the solid generated. 13. Differentiate xex and show that

d~

(xex) - ex = xex. Hence prove that

f

xex = xex - ex

+

c.

Connected with the exponential function is another very important function called the logarithm function. You have already noted that an index can be called an exponent. It can also be called a logarithm. More particularly, if y

=

ax then x is called the logarithm of y to the base a (written X= logay)

then x

=

loga y

then x

=

loge y

You may have met logarithms before, including the common logarithms to the base 10. Logarithms to base e are called natural logarithms or Napierian logarithms after a Scotsman John Napier who published an article on logarithms in 1614.

Follow these examples to investigate the theory behind logarithms. Remember, a logarithm is an index.

Example (i):

Given 26

=

64 and 34

=

_{81 write down}

(a) log2 64 (b) log3 81

Solution:

(a) log2 64 = 6 (b) log3 81 = 4

(c) log10 100 = 2

because 26 = 64 because 34 _{= 81} because 102

=

100

Example (ii):

Find (a) log2 16

Solution:

(a) Since 16 = 24 then 1og2 16

=

4

Example (iii):

Find (a) log3

J27

Solution:

(a)

J27

= 3.[3 = 3t

.'.1og3

J27

=

1·5

Example (iv):

Prove that: (a) 1oga xy = 1oga x + loga y

X

(b) loga- = loga X - loga y y

(c) loga x" = 11 loga X

Solution:

Let ak

=

x and a"'

=

y

(a) xy = ak x a"'

= ak+m

Hence loga xy

=

k + m

(b)

I.e.

(c)

Example (v):

= loga X + loga y.

X ak k-m

- = - = a

y a"'

X

loga-

=

k - m

=

loga X - loga y.

y

x"

=

(ak)"

=

ak"

loga x" = kn = n loga x.

(c) log10 100

(b) logs 125

(b) Since 125 = 53 then logs 125 = 3

(b) log~ 32

(b) 32 = 2S = (J'i)10

so logv'Z 32

=

10

Given loge 3

=

1·09861 and loge 2

=

0·69315, calculate loge 6 and loge 81 to 4 decimal places.

Solution:

(a) loge 6 = loge (3 X 2)

= loge 3 + loge 2

= 1·09861 +0·69315

=

1·79176

= 1·7918 (to 4 places)

(b) log3 81

=

loge 34

=

4loge 3

= 1·09861 X 4

=

4·39444

Example (vi):

What is the natural logarithm of: (a) e3·57? (b) 1·7?

Solution:

(a) By definition if y = e3

'5

1,

then loge y = 3·57.

Note: In general loge ea =a.

(b) From tables or calculator loge 1·7 ~ 0·5306.

(c) 170? (d) 0·17?

(c) From calculator loge 170 = 5·1358 direct. Tables usually only have logarithms of numbers from 1 to 10. Hence if using tables proceed as follows:

loge 170 = loge (1·7 X 100) = loge 1·7

+

loge 100 = loge 1·7

+

2 loge 10

~ 0·5306

+

2 X 2·3026

~ 5·1358

(d) loge0·17 ~ -1·77196 fromcalculatordirect 1·7

= loge

lO

if using tables =loge 1·7- loge 10

= 0·5306 - 2·3026

=

-1·772.

1. Use calculator or tables to find to 4 places: (a) loge 5 (c) loge 1·384 (b) loge 7 (d) loge 10

(e) loge 1

(f) loge 2·7 2. Given loge 10 = 2·3026 and loge 7 = 1·9459, without tables or calculator find:

(a) loge 1000 (c) loge 70 (e) loge 0·7 (b) loge 0·01 (d) loge 7000 (f) loge 0·007

3. Given loge 2 = 0·69315, and loge 3 = 1·09861, find without tables or calculator to four decimal places:

(a) loge 4 (b) loge 27

(c) loge 1·5 (d) loge 12

4. Use your knowledge of indices (powers of 2 and 3) to find: (a) log₂ 128 (b) log₃

9J3

5. Use the fact that e0

= 1 to find loge 1.

6. What is loga 1 for any positive value of a?

(e) loge 18

(f) loge 13-5

(c) log2

J2

7. Let a > 0. Can you think of any number c such that ac would be negative? (Think of the graph ofy=2x.)

8. Is the statement loga( -4) meaningful if a is greater than zero?

9. Can you think of any number z such that ez = 0? Can there be any value for loge 0?

10. Draw the graph of y = loge x for values of x from 0·1 to 6. Use the same scale on the x andy axes. From your graph estimate the value of x for which loge x = l.

11. On the same axes as for question 10 draw the graph of y = ex but for convenience of scale draw for -2 ~ x ~ 3. On your graph draw also the line y = x. What do you notice about your two graphs in relation to the line y = x?

N.B. y = ex andy = loge x are called inverse functions.

12. On the same set of axes in each case draw the following curves. (N.B. Unless otherwise stated log x means loge x).

(a) log x and log 3x (N.B. log 3x = log 3

+

log x).

(b) log x, log (x

+

2) and log x

+

log 2. (c) log x and log x2_•

13. What is the maximum domain which can be assumed for log x? Assuming this domain, what is the range?

14. Simplify log a3

- log a2•

15. Simplify loge (e2

+

e) - loge (e

+

1).

16. Solve: (a) ex= 3 (b) e-2x = 4.

17. (a) Whatislog₂64? (b) Whatis21

og264?

18. (a) What is log3 81?

19. What do you think e1_og,2 _is?

N.B. If y

=

ax then X =loga y So y = a x = alog,y.

20. Evaluate: (a) e1 og 3

(b) elog 7·5

21. Solve the equations

(b) What is 31_og, 81 _?

(c) elog 10

(d) elog1

(a) e1og x = log 2 (b) elog 3x = 6

22 .. Show that 3 log x

+

2log y -!log z = log

(x~{).

dy 1

We now prove that dx = dx

Taking limits as ~x -+ 0

dy

l . lml -'t ~y =

---,---~x-+0 ~X l' 't ~X

l m l

-~x-->0 ~Y

1 It m -. ~X ~y-->0 ~y

Thus dy

= -

1

-dx dx

dy

This result can also be arrived at geometrically.

(because if ~x -+ 0 then ~y -+ 0)

dy dx

or- x

=

1

Given

Then this means

from which

But

y

=

logex X= eY

dx _ Y

dy-e. dy 1

dx

=

dx dy

1

eY

X

This is a very special result.

(previous result)

dy 1

If y.

=

log x then -

=

-e ' dx X Hence also

f

~dx

= loge x

+

C

Notice the importance of these results. Up to date we have had integrals for all other powers of x 1

except-. X

Example (i):

Find the derivative of (a) loge (5x - 2)

Solution:

(a) y = loge (5x - 2)

Let y = loge u where u = (5x - 2)

dy dy du dx = du · dx

=!

0 5

u

5 5x - 2

N.B. Here is a quick method you may wish to use.

dy f'(x)

If y

=

loge f(x) then dx

=

f(x).

Check that this is so using the example above.

Notation:

15

(b) y = loge (7x2

+

3x)

Let y

=

loge u u

=

7x2

+

3x dy dy du

dx

=

du · dx

=!.

(l4x

+

3) u

14x

+

3

1. Differentiate with respect to x:

(a) log 3x (e) log (3x

+

1) (b) log x

+

2x (f) log (2x - 3) (c) log x2 _{(g) log (x}2

- 5)

(d) e2x

+

log 2x (h) 5x - log 2x3

(i) log x4 (j) 8 log 3x

(k) 2 log x

+

5 log 2x

(1) t log 3x

2. Show that log (x

+

2)(x - 3) is the same as log (x

+

2)

+

log (x - 3). Perform the differen-tiation of this function in two ways. Which is easier?

3. Differentiate y = log x

+

1 with respect to x by the way you consider easier. x+4

4. Show that log

JX+9

= t log (x

+

9). Hence or otherwise differentiate this function with respect to x.

5. Differentiate with respect to x:

(a) In (2x

+

l)(x - 5)

X (b) l n

-x

+

1

6. Differentiate with respect to x:

(a) lnl X

(c) In (x

+

6)3

(d) In

JX+4

1

(b) In (2x - 1)

7. Differentiate these products with respect to x: (a) x In x (b) 2x3 In (x - 2)

() 1 _{e n} (x

+

1)(x - 1) (x

+

2)

(f) In (3x

+

2)2

1

(c) In 2x3

(c) x In x - x

8. Don't confuse y

=

log x2 andy

=

(log x)2• Differentiate each with respect to x.

9. Now think of y

=

log (log x). This can be written as log u where u

=

log x. Differentiate with respect to x.

10. Find second derivatives of:

(a) log (3x

+

1) (b) xlogx- x

d2y (dy)2

11. If y

=

log x, show that dx2

+

dx

=

0.

12. The line x = a cuts the curves y = log x andy = log 2x at P and Q. Show that the tangents to the curves at P and Q are parallel. Also show that the distance PQ remains constant for all values of a.

13. Find the equation of the tangent to the curve y

=

In x

(a) when x

=

1 (b) when x

=

3.

14. At what point on the curve y = In 2x is the gradient t? Find the equation of the tangent at this point.

15. Show that the minimum value of y = x log xis_!. Sketch the graph from x = t to x = 5.

e

16. Find the maxima, minima, or points of inflexion for the following functions and sketch a graph for each.

(a) y = X ln X - X (b) y

=

In x X

1

(c) y =In x

17. Differentiate with respect to x:

( ) a In x x

e

(b) e-x

In x

(c) ex. In x

(d) In ex

+

1

ex- 1

18. Remember that if x

=

-3, then

I

xi

=

3 and draw the graph of log lxl.

19. Write ln (x2 - 9) as the sum of two logarithms and hence differentiate with respect to x.

20. What is the maximum assumed domain of y = log (x

+

2)? Where does the curve cross the

x axis? What is its gradient at this point? Sketch the curve.

21. For what value of x is the ratio of the natural logarithm of a number to the number itself the greatest.

Hint: This question really asks-when is the function log x a maximum?

X

Remember _dx d _{(log x) =}

x'

1

d f'(x)

dx log f(x) = f(x) '

hence

f

~dx

= log x

+

C

hence

f

j.~j

dx

=

log f(x)

+

C

The art of integrating some fractional expressions is to try to put them into the form

ff'(~))

as shown below.

Example (i):

Find f_!_dx. 2x

Solution:

-dx

= -

-dx

=

-log x

+

C

f

2x

l 1fl

2 x

1

2

Example (ii):

Find f x2

6 :

7dx.

Solution:

Notice that if f(x)

=

x2

+

_{7, then f'(x)}

=

_2x.

So f-P-ax

=f

32x

2

;dx

=

3f 2

2 x

7dx

=

3log(x

2

+ 7) +C.

x+7 x+ x+

Example (iii): Find

J

4x

+

3 dx.

2x

+

1

Solution:

. 4x

+

2 We notice that

=

2.

2x

+

1

Hence we break up the fraction into two parts as follows:

1.

f

4x

+

3 dx =

f

4x

+

2

+

1 dx

2x+l 2x+l

=

J

(

~~

:

~

+

2x

~

1)

dx

=

f

2dx

+

~

f

2x

~

1 dx

=

2x

+

-!- log (2x

+

1)

+

C.

Integrate the following with respect to x:

1 4

(e) 2 2 5

(a)- _{(c) 4x- 3} 3x

+

x

-4x X

(b) _1_ 1 _(f) 3

x+2 (d) 6x - 7 2x - 1

2. Integrate these expressions with respect to x:

(a) 2x x-2 (e) ex

x2

+

1 (c) x2 - 4x

+

1 ex+ 1

(b) X

(d)~

(f)~

x 2 - 3 1 - x 2 x4

+

3

3. Simplify, if necessary, and integrate with respect to x:

(a) 5x2

+

;x - 6

(b) (x

+

~

2

Y

(c)

x2e2x2 - 1

X X

4. Evaluate:

J

e-1 1

(a)

-1dx

1

X+

J

2 X

(c) -₂

-1dx

0 X

+

(b) 2t

+

-dt

J

1 e 2 t

J

3 1 (d) 1 (2t - 1) dt

X

+

7 8

Je+

1 X

+

7

5. Show that

-1 can be written as 1

+ - -

1 and hence show that - -1

=

e

+

7.

x - x

-2 x

-6. Show that 1

1

+

x

= -

1

+ -

1 2

and hence find

J

1 1

+

x dx.

- x - x - x

. J2x

+

5 7. Fmd x

+

4 dx.

1 1 2 .

J

3 1

8. Show that

-1 - - -1 = -2 - -1 . Hence or otherwise show that -2 - -₁ dx = -!-log 1· 5.

x - x+ x - ₂X

-9. Find the area under the curve y

=

i

(a) from x

=

1 to x

=

5 (b) from x

=

1 to x

=

e.

10. Check the area under the curve y

=

i

from x

=

1 to x

=

e found above, by an approximate numerical method.

11. Find the area under the curve y

=

~ from x

=

1 to x

=

2.

13. Sketch the curve y = 2 - ~. Show that it cuts the x-axis where x =

t

and find the area between the curve and the x axis and from x =

t

to the line x = 1.

14. The gradient at any point (x, y) on a curve y = f(x) is - 1

-1. The curve crosses they axis at the

x+

point (0, 4). Find the equation of the curve.

15. The region under the curve y =

~from

x = 1 to x = e is rotated about the x axis. Find the './X

volume of the solid of revolution

16. Find the area under the curve y =~from x = 1 to x = 3. X

+

2

17. The region under the curve y =

J

x₂ 2

~

1 and bounded by the lines x = 0 and x = 1 is rotated about the x axis. Find the volume of the solid of revolution.

18. The region between the curve y = 1 -

~and

the x axis

X

and from x = 1 to x = 2 is rotated about the x axis.

Show that the volume generated is n(3 - 4 log 2) cubic units.

19. Find the area of the region shaded between the curve

y = In x and they axis and between the lines y = 0 and

y = 1. Do it this way:

(a) Rewrite y = In x as x = eY.

(b) Integrate eY with respect toy, fromy = 0 toy = 1.

0

(e. 1)

I I

I

I I

I

I

20. In a similar way (i.e. by first rewriting y = In x as x = eY), find the volume of the solid of

revolu-tion formed when the region described in quesrevolu-tion 19 is rotated about they axis.

21. Find the volume of the solid of revolution when the area between the curve y = In x and the y axis and from y = 0 to y = 2 is rotated about the y axis.

22. Sketch the graph of y = In 2x. Rewrite this equation as x equal to some function of y. Find the volume formed by rotating the region between this curve and they axis and from y = 0 toy= 1, about they axis.

So far we have only differentiated exponential expressions such as ex or ef<x). We look now at general expressions such as ax or af<x) where a is some constant.

Remember that if a = eY then y = In a by definition.

This means that a = e1_{"a (which is an important result)} Now ax = (e1nay

= exlna d(ax)

So ~=Ina. e.dna (as In a is a constant) =In a. ax.

Example:

lfy = a2x, find

z.

Solution: y = a2x

Let y = a" where u = 2x dy dy du

dx

=

du. dx

=In a. a". 2 = 2ln a. a2

x.

1. Find the slope of the tangents to the curves y = 2x and y = Y when x = 0 and compare with the values of0·7 and 1·1 estimated from graphs drawn in exercise 19.1.

2. Find the slope of the curve y = 10x when x = 2.

3. Differentiate with respect to x. Leave logarithms unsimplified.

(a) a4_{x (c) 5x . 4x}

(b) (6x2

+

_{1)ax (d) y}

X

4. If y = sx find:. Hence find

I

sx dx.

5. Find

I

wxdx.

6. Find the area under the curve y = 2x from x = 0 to x = 3 to 2 decimal places.

Before the advent of calculators, logarithms to base 10 were very useful. Logarithms are useful because being indices they can be added or subtracted when numbers are being multiplied or divided and most people find the first two operations easier. But why logarithms to base 10?

It is impossible to print a table with logarithms of all numbers shown separately because of space. Lists in tables are restricted usually to logarithms of numbers from 1 to 10. Logarithms of other numbers can then be calculated as, for example:

If logarithms are to the base 10 these transpositions are easy because log₁₀ 10 = 1 and log₁₀ 100 = 2 etc.

This all raises two questions

(a) How can logarithms be changed from one base to another?

(b) How can logarithms to different bases be differentiated or integrated?

CHANGE OF BASE

Let y =log" x Then x = aY

Taking logarithms of both sides to base b 1ogb x = 1ogb (aY)

= y. 1ogb a

= log" x . 1ogb a (by substitution)

This is usually written

1ogb x X =

-logb a

and a special case is :

log x

logloX = _ e _ loge 10

Example (i):

DIFFERENTIATION OF y

=

loga x Let y =log" x

Before differentiating rewrite this as loge X

y = - - (see opposite) loge a

dy 1

.. dx = loge a . ~ (as loge a is a constant)

Remember

and a special case is :

Use the fact that loge 8·2 = 2·104 and that loge 10 = 2·3026 to find log10 8·2 by change of base.

Solution:

log 8·2 By change of base, loglO 8·2 = 1 e 1

oge 0

Hence _{loglo 8·2 = 2·3026} 2·104

= 0·914.

N.B. As a check log10 8·2 = 0·914 to 3 places from calculator direct or from common logarithm

tables.

Example (ii):

Differentiate log10 (4x - 1) with respect to x.

Solution:

y = log10 (4x- 1)

= log10 u where u = (4x - 1)

dy_dy du

dx- du · dx

= - 1 - . 1 . 4 loge 10 u

4

loge 10. (4x- 1)'

1. Use tables or calculator to find the following logarithms to base e: (a) log 1·95 (b) log 5·803 (c) log 9·61

2. Using the change of base rule, change each logarithm above to an equivalent logarithm with a base of 10.

3. Show that log10 x

=

0·4343 ... x loge x.

4. Find the following logarithms to base e and then in each case change them to base 10. Answers to 3 decimal places.

(a) log 84 (b) log 0·059 (c) log 10

5. Differentiate with respect to x (leave unsimplified). (a) log₁₀ 3x (c) log10 _{x (e) log}

10 (5x

2 _{- 7x)}

X (b) 2x2 _log

10 x (d) log10 (x2)

6 . Sh ow th t a

!!_

d (1 og10 x -) - 0·4343 ...

X X

7. Prove that loge 10

=

-1 -1

- . Look up both loge 10 and log10 eon a calculator or in tables and

oglo e

check that the result is correct.

Some interesting properties of the number e:

(a) The area under the curve y

=

_!from 1 toe is 1 square unit. X

(b) The base for natural logarithms is e.

(c) If y

=

ex then

k

=

ex (i.e. unchanged). Likewise

I

exdx

=

ex.

(d) At all points on the curve y = ex the gradient is equal to the value of the function at that point. . dy

t.e. dx

=

y.

(e) If y = ekx,

then;~

= ky. (i.e. the gradient is k times the value of the function at any point).

(f) The curve y = ex has a gradient of 1 where it crosses they axis.

(g) e

=

lim

(1

+

_!_)".

An approximate value can be found fore by using this result which is proved

u--+oo n

in the next section.

Take the point P(Ax, e"'x) on the curve y

=

ex and join

P to Q(O, 1).

PR

Slope of PQ

=

RQ

e"'x- 1

Ax

But because the slope of the tangent at Q is 1

t.x 1 l i m e - =1 t.x->0 Ax

1. Ax 1

or 1m =

t.x->0 e"'x - 1

(by reciprocal)

22

t.x

0

~

f:

p (f. X, e"X)

e"x-1

R

Now let

~

=

e11x - 1 from which e11x = (1

+

~)

and Ax

=

loge ( 1

+

~}

By substituting these results for e11_{x and Ax in the line above we have:}

loge(l

+

l)

lim 11 = 1

llx->0 1

n

or lim 11 loge (1

+

l)

=

1 (because 11

~

oo when Ax

~

0)

n---+co n

( 1)"

lim loge 1

+ -

= 1

IJ->00 11 So

( 1)"

lim 1

+ -

=

e1

=

_e.

n-+oo n or

( 1)"

use your calculator to find the value of 1

+

n

when

(a) 11

=

100 (b) n = 1000 (c) n

=

10 000

( 1)"

Can you see that for n very large the value of 1

+

n

approaches 2·7182818 ... which is the approximation given by the calculator for ex when x = 1?

1. Differentiate with respect to x: (a) e-2x

(b) loge (1

+

x) (d) (3x

+

2)ex

(e) loge (Sx

+

2)

(f) lnx X

2. Find the equation of the tangent to the curve y

=

e3_{x at the point where x}

=

_{1. Find where}

this tangent cuts they axis.

3. Find primitives for:

(a) e2x+3 1

(b) (Sx - 4)

4. Find the area under the curve y

=

e2

x from x

=

1 to x

=

3.

4x

(c) x2

+

1

5. Find

r

2

(ex - e-x)dx. What is the area under the curve y

=

ex - e-x from X

=

1 to X

=

2?

d

2

d

6. If y

=

A

+

Be-41 _{show that _____11_}

+

4~ _{= 0.}

, dt2 _dt

d2y dy

7. If y

=

e2x

+

e4x, show that dx_{2 -} 6 dx

+

8y

=

0.

8. Differentiate with respect to x:

(a) e.fX (c) 3x ln x

(b) ef (d) ln

(x;

l)

23

(e) ex'+1 X

(f) ln-2--1

9. Find these integrals (leave answers unsimplified)

(a)

tl

x2:

3dx (b)

r

2

(5x

+

~)dx

10. Show that the tangent to the curve y = ex at the point where x = l passes through the origin. Find the area under the curve from x = 0 to x = l and also the area between the curve, the tangent mentioned above, and they axis.

11. Find the area under the curve y =

f

from x = 1 to x = 5. Also show that the area under this curve from x = l to x = e is l unit of area.

12. Find the area shaded in the diagram between the curve

y = ln x and the y axis and bounded by the lines

y = 0 (the x-axis) andy = 0·5.

Then find the volume of the solid of revolution formed if this region is rotated about they axis.

Hint: Note that y = ln x can be rewritten in another

form.

0.5

-13. Find the area under the curve y = i(ex

+

e-x) from x = 0 to x = 3.

X

14. Calculate the volume of revolution if the region under the curve y = ex and between the lines

x = 0 and x = 2 is rotated about the x-axis.

1

15. Prove that 1ogb a =

-1 oga b. You may assume the change of base rule.

16. Sum the series log10 2

+

log10 4

+

log10 8

+ · · ·

to 10 terms.

17. Consider the function y = ~

1

. X

+

(a) Find any stationary values.

(b) By writing the curve as y = - 1- ₁, show that as x becomes large y becomes very small.

x+-x What happens when x ---+ - oo?

(c) Sketch the curve (it is called the serpentine curve). (d) Find the area under the curve from x = 0 to x = 2.

18. Sketch the graph of y = 1 . Find the area bounded by the curve, the x axis and the lines 2x - 1

x = land x = 3.

19. By noticing that_!__

2 - _!__2 =

~

4

find

J

6

~

4

dx.

(Leave answer without using

x - x+ x - ₃ x

-tables or calculator).

20. Show that

I~

:

~

dx = x

+

6 loge (x

+

1)

+

c.

21. Use numerical integration to find an approximate value for the area under the curve y =

f

from

x = l to x = 3 and check by formal integration.

22. Use Simpson's Rule to find the area under the curve y =_!_from x = 2 to x = 4 and check 2x

In designing the Sydney Opera House. the engineers made the surface of every shell part of a sphere.

I

I

CHAPTER 20

In earlier work involving trigonometry the degree was used as a unit of measurement. It was a con-venient unit for practical, numerical measurement but when dealing with trigonometric functions a new unit, called the radian is more appropriate. On your calculator there is a selector switch which may be set to radians or degrees.

In a unit circle, with centre 0, two radii OP, OQ are drawn so that: the arc length PQ = radius OP = radius OQ

=

1 unit.

The angle POQ subtended at the centre by the unit arc PQ is defined

to have a measure of one radian. P

: A radian is the measure of the angle which an arc of unit length subtends at the

of a unit circle.

If, in a circle of unit radius, an angle ROS has its vertex at the centre, its radian measure is equal to the length of arc on which it stands.

If arc length RS = x units then L ROS = x radians.

Examples:

(i) If the length of arc RS

=

1· 3 units, then L ROS

=

1· 3 radians. (ii) If the arc RS measures 0·73 units, then LROS = 0·73 radians.

As in earlier trigonometry, if the rotation is anticlockwise the angle is taken as positive; if clockwise as negative. Hence in the unit circles below:

Q

In this way, any arc length, which may be measured by a real number, may be interpreted as an angle measured in radians. When theoretical work is being done with the trigonometrical functions, the radian measure is always used. The degree measure is still of vital importance in practical situations such as carpentry and surveying.

It is important to be able to convert between radians and degrees. This may be done by means of tables or on some scientific calculators there is a key to automatically carry out the conversion.

However, equivalent values may be found by means of a simple relationship. The circumference of a circle is given by

C

=

2nr

h · h . . 1 b d f' d h . f circumference d . . 1 1

w ere n ts t e uratwna num er e me as t e ratiO o d. an IS approximate y equa to

3·14159.

In the case of the unit circle, the circumference is

C

=

2n (as r

=

1).

Thus, the length of the semi-circular arc PQR

=

n units. If follows that the straight angle PO R measures n radians.

mmeter

R

c=2n

n is the length of the semicircular arc in a unit circle and hence is the measure in radians of the straight angle PO R.

Now the number of degrees in a complete revolution is 360 degrees and this corresponds to an arc length of 2n units and so the radian measure of a complete revolution is 2n.

It follows that

180° 1 radian =

7t ~ 57·2958°

and 1 degree

=

1 ;O radians

~ 0·01745 radians.

- - - --~ ---~---~---~~~

Examples:

(i) Express 90° in radians. 180°

=

n radians

1 o

=

1

~O

radians

. ₉₀o

=

90 x n

. . 180

n d'

=

2

ra mns.

(iii) Find the equivalent of~ radians in degrees.

n radians = 180°

n . 180° - rad1ans =

-3 3

=

60°.

(ii) Express -450° in circular measure. 180°

=

n radians

1 o

=

1 ~O radians _ 4500

= _

450n

180

5n d'

= - - r a mns

2

(on dividing numerator and denominator by 90)

1. Express the following angles in radians, leaving your answers in terms of n.

(a) 30° (f) 210° (k) -60°

(b) 45° (g) 300° (1) 54°

(c) 60° (h) 420° (m) -270°

(d) 120° (i) 22!0 (n) 480°

(e) 135° (j) 18° (o) 262!0

2. State in degrees the following angles measured in radians. (a) ~ (d)

i"z

(g) 9"o (j) 5s" (b) 3n (e) 5

4

(h) i~ (k)

?

2"

(c) 43" (f) 4n (i) i~ (1)

i

(m)

29

(n) s3n

(o) - 7s"

3. Using a calculator, express in degrees (correct to four decimal places) the following angles which are given in radians.

(a) 1·5 (b) 2·0

(c) 1·2 (d) 0·05

(e) 2·75

(f) 0·3

(g) - 3·15 (h) 6·28

4. Give the circular measure of the following angles, correct to four decimal places. (a) 28° (c) 113° (e) 116·5° (g) 298°30' (b) 77° (d) 137° (f) 97·35° (h) -317·652°

5. For each of the following angles, give its radian measure in terms of n and write its sine and

cosine as surds. (a) 60°

(b) 225°

(c) 300° (d) 120°

(e) 135°

(f) 420°

6. For each of the following angles, give its measure in degrees and evaluate its sine and tangent ratios, leaving the answers as surds.

(a) 2

3" radians (c) 83" radians

(b) 3n radians (d) 3

2" radians

(e) 1 ~" radians (f) 5

4

radians

7. With your calculator set to radian mode, find correct to 4 decimal places :

(a) sin

1

(d) sin 0·6 (g) cos i~ (j) tan (5n - 1)

(b) cos£ (e) cos 5·8 (h) tan 3·7 (k) sec 3

4

(c) tan 1·2 (f) tan 7

8. (a) Two angles of a triangle are~ and 2

3n radians. What is the measure of the third angle in degrees?

(b) Three angles of a quadrilateral have as their measures

-f,!

and 56n radians. Express the size

of the remaining angle in degrees.

(c) What is the size of each interior angle of the following regular polygons? Give your answers in radians.

(i) hexagon (ii) octagon (iii) dodecagon (12 sides).

(d) The sum of two angles is 2

33

J'

radians and their difference is 133

J'

radians. Find the size of each

angle in degrees.

In a circle of radius r units, an angle measuring

e

radians stands on an arc of length l units.

Now, as the length l, of the circular arc increases so does the angle at the centre.

Finally, the arc becomes the circumference of the circle and

e

becomes a complete revolution (2n radians).

Hence, length of arc, I units

e

radians

circumference of the circle (in the same units) 2n radians

e

l and

2nr 2n

. I= 2mB

· · 2n

and I=

re.

l =

re.

From this formula it follows that the measure of an angle

e

in radians is given by

e

ra tans d. = arc length . length of radms

Now consider the sector AOB (shaded in the above diagram). As the angle

e

increases, so does the area of the sector. Finally, when

e

=

2n radians the area of the sector becomes the area of the circle

(nr2).

H ence, area of sector . AOB

e

radians area of the c1rcle 2n radians·

Now, let the area of the sector AOB be A square units.

Then, A

e

In a given circle of radius r units, a sector contains an angle of 8 radians at the centre. The area of the sector is given by

Example (i):

Find the length of the arc in a circle of radius 6 em, subtending an angle of 0·86 radians at the centre.

Find also the area of the corresponding sector.

Solution:

l

=

r8

=

6 X 0·86

=

5·16 length of arc PQ = 5·16 em

A=

tr

28

=

t

X 6 X 6 X 0·86

=

15-48

area of sector POQ

=

15·48 cm2•

Example (ii):

Find the angle, in radians and degrees, subtending an arc of length 9 em in a circle of radius 8 em.

Solution:

8={

r

_ _2_

- 8

= 1·125

:. angle 8 = 1 ·125 radians

1·125 X 180 d

egrees

n

~ 64·4578 degrees by calculator.

Example (iii):

A sector of a circle, which contains an angle measuring

i

radians at the centre has an area 4·7059 cm2•

Find the radius of the circle correct to four decimal places.

Solution:

A=

tr

28

4·7059 =

tr

2 .

i

4·7059 X 2 X 6 2

- - - = r n

:. r = J12 x ;·7059

:. radius ~ 4·2397 em by calculator.

Example (iv):

In a circle of radius r, show that the minor segment (shaded), subtending an angle of 8 radians at the centre, has an area given by

tr

2_{(8- sin 8). Find (a) the area of the segment and (b) the length} of the chord PQ when r

=

10 em and 8

=

45°.

Solution:

Area of minor segment

=

area of sector 0 PQ - area of triangle 0 PQ

=

tr2 _{8 -}

tr

2 _{sin 8 (area of triangle}

= tab sin C)

=

tr

2(8 - sin 8).

p

of minor segment =

!r

2(8- sin 8) (a) 8

=

45°

=

i

radians.

area of minor segment

=

!

x l02_{(!f- sin !f)}

=

50(!f - 0·707107)

~ 3·91456 cm2 _{by calculator.}

(b) Draw OR .1 PQ.

Then L ROQ

=

t

L POQ

=

i

radians sin ROQ =

~~

RQ = 10 sin

i

chord PQ = 20 sin

i

~ 7·6537 em by calculator.

1. A circle has a radius of 3 · 5 em. Find : (a) its area and circumference.

(b) the length of arc subtending an angle of 1·75 radians at the centre. (c) the area of the sector formed by the arms of this angle.

p

2. For given circles, the following values of the radius are given together with values of() for the angle subtended by an arc at the centre. In each case find the corresponding arc length.

(a) r

=

8 em 8

=

1 radian (d) r

=

10 em ()

=

3 radians

(b) r

=

5 em 8

=

2·5 radians (e) I'

=

5·8 em 8

=

135°

(c) r = 6 em 8 = 1·76 radians

3. Copy and complete the following table, giving answers correct to three decimal places:

RADIUS (r) ANGLE AT CENTRE (8) ARC LENGTH (/)

(a) 6·2cm 3 radians

1

-(b) 2 radians 3·6 em

1

-(c) 2·8 em 8·21 em

1

-(d) 60 degrees 4·82 em

1

-(e) 3

4'

_{radians 15·86 em}

1

-(f) 134mm

4. The circumference of a circle is 198 em. Calculate the length of arc which subtends an angle of 120 degrees at the centre.

5. (a) Taking the radius of the earth as 6341 km, what angle would be subtended at the earth's centre by an Equatorial arc of length 9000 km?

(b) What arc length would be subtended by an angle of 1 minute at the earth's centre?

6. Find the area of the sectors in a circle of radius 10 em where the angles at the centre are;

(a) 0·78 radians (d) 150 degrees (g) 3

4" radians

(b) 1·04 radians (e) -if radians (h) 37 degrees

(c) 60 degrees (f) 2

3" radians

7. A sector of a circle has an area of 28·86 cm2

and contains a central angle of -if radians. Find:

(a) the radius of the circle (b) the arc length of the sector.

8. The area of a sector of a circle of radius 12 em is 188·5 cm2

. Find the central angle contained in

the sector in :

(a) radians (b) degrees.

9. Complete the following table for a given sector of a circle:

RADIUS ANGLE AT CENTRE AREA

(a) 15 em 90 degrees

- - -

-(b) 5·6 em -!f radians

---

- - -

-(c) 6·4cm radians 47·9 cm2

t - - -

---

-(d) 10m radians 26·18 cm2

t - - -

-(e) 5

6" radians 61·14 cm2

1 - - -

-(f) 1·762 radians 88·10 cm2

1 - - -

---(g) 2·8m 2

3" radians

- - -

-(h) 63 degrees 48·06 cm2

- - -

---(i) 7·14 em degrees 100 cm2

10. A pendulum of length 1·2 m swings through an angle of

i

radians. What is: (a) the length of arc traced out by the end of the pendulum?

(b) the area of the sector formed by the swing of the pendulum?

11. What is the area of the sector of a circle of radius 5 em, if the length of arc bounding the sector is :

W5~? ~3~? ~9~?

12. Find the area of the minor segment in a circle with:

(a) radius 4 em and subtending an angle of 1·32 radians at the centre (b) radius 5 em and subtending an angle of -j radians at the centre (c) radius 6·2 em and subtending an angle of 135 degrees at the centre.

13. (a) What is the area of the major segment of a circle in which the minor segment subtends a central angle of 5

6" radians and the radius is 7 em?

(b) What is the area of the minor segment cut off a circle of radius 10 em by a chord of length 12 em?

Check your answer by the approximate formula

area of minor segment ~ Ubase)

x

height.

14. A circular piece of wire has a radius of 12 em. This is cut and bent to form an arc of a circle whose radius is 60 em. Find the angle subtended at the centre by this arc. Give your answer to the nearest degree.

15. A circular pond has an area of 615·75 m2

• It is divided into sectors, one of which contains a

central angle of 2

3'.

_Find:

(a) the length of the arc in this sector (b) its area.

16. Find the shaded areas in the following figures where the arcs are quadrants of circles.

(a)

l

(b)

17. In an equilateral triangle with each side 8 em, circular arcs, each of radius 2 em are drawn as shown below.

Calculat~ the total arc length of the curve.

l

~---4cm---~

If an angle is expressed in radian measure then the trigonometric ratios may be defined as functions of a real variable. For example, sin xis de