m115 Further Calculus for Students

8/10/2019 m115 Further Calculus for Students

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Further Calculus 2

1 Implicit differentiation

y = f (x) denes y explicitly as a function of x. Sometimes we may have an equation relatingx and y which may be difficult or impossible to solve explicitly for y.

For example, y = x2 + 3 x + 4 denes y explicitly as a function of x;

however, sin(xy) + y2 + 2 xy + x2 = 1 denes y implicitly as a function of x. As xvaries, so do the values of y which solve this equation, i.e. we think of y as being afunction of x , even though we cannot write this function down explicitly.

Assuming the equation does dene y as a differentiable function of x, we can useimplicit differentiation to obtain the derivatives.

For example, consider the equation x2 + y2 = 1.

This has two explicit solutions: y = 1 x2 , y = 1 x2 (upper and lower semi-circles, respectively).

For y = 1 x2 (1 < x < 1), we can show thatdydx

= 2x2 1 x2

= x 1 x2 =

xy

using y = 1 x2 .Similarly for y = 1 x2 (1 < x < 1),

dydx

= x 1 x2

= xy

using y = 1 x2 .Proceeding implicitly: (consider y as being a function of x)

x2 + y(x)2

= 1 (assume y = y(x)) .

Consider the LHS as a function of x. Differentiate both sides with respect to x, using theChain Rule for the second term:

2x + 2y dydx

= 0, ()

thereforedydx

= xy

, (y = 0) .

Note that, in general, implicit differentiation gives dydx

as a function of both x and y.

We can also go on to nd the second derivative in terms of x and y using implicit differen-tiation.


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Further Calculus 3

Examples 1A

(i) Find the equation for the tangent to the curve

x2

= y3

+ x sin y + 4 at the point (2 , 0).

[Note: the equation is satised when x = 2, y = 0.]

Differentiate implicitly: (think of y = y(x) )

2x = 3y2 dydx

+ sin y + x cosy dydx

dy

dx (3y2 + x cosy) = 2 x sin y

i.e. dy

dx =

2x sin y3y2 + x cosy

. ()

Gradient at (2 , 0) = 22 sin030 + 2 cos0

= 42

= 2.

So tangent is y 0 = 2(x 2), i.e. y = 2( x 2).

Note that implicit differentiation simply involves the use of the usual rules of differ-

entiation (linearity, product rule, chain rule and (occasionally) the quotient rule) andwhenever a function of the dependent variable, i.e. f (y), appears we apply the chainrule:

ddx

f (y) = ddy

f (y) dydx

= f (y) dydx

.

More Examples in Lectures


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Further Calculus 4

Exercises: Implicit differentiation

1. For each of the following equations

(i) nd dy

dx by implicit differentiation,

(ii) by solving the equation for y nd two functions dened by the equation,

(iii) nd the derivative of each of the functions found in (ii),

(iv) verify that the results obtained in (i) and (iii) are in agreement.

(a) y2 = 4 x 8 (b) x2 + y2 = 25 (c) x2 y2 = 16(d) x2 + y2 2x 4y 4 = 0.

2. Find dydx

as a function of x and y given that

(a) x3 + y3 = 1 (b) 2 xy + y2 xy3 = 1(c) xy + sin x + cos y = 0 (d) x = y

x2 + y2

(e) sin(xy) = cos( x) cos(y) (f) sin(x + y2 ) = y.

3. Find the tangent to the given curve at the given point:

(a) y2 = 2 x3 , (2, 4) (b) (x + y)3 = 2 x + y + 3, (3, 1)(c) xy3 x3 y = 30, (2, 3) (d) x = y cos y, (/ 2, / 2).

4. Find y as a function of x and y given that

(a) x + y = 1 (b) sin x + tan y = 1 (c) y2 + xy = 1.

5. Find the value of dydx

on the given curve at the given point:

(a) 3x2 + 4 y2 = 19 , (1, 2) (b) cos(x + y) = y sin x, (0, / 2).


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Further Calculus 6

Equations of the above type are called parametric equations the third variable is calledthe parameter .

Suppose that we have x = g(t), y = f (t) and that we can express y as a functionh(x). Then

y = f (t) = h g(t) and y = h(x),so that by the chain rule,

dydt

= f (t) = h g(t) g (t) = h (x) dxdt

i.e. dy

dt =

dydx

dxdt

First derivative using parametric differentiationIf x and y are given as functions of parameter t, then

dydx

=

dydtdxdt

if dx

dt = 0 .

Differentiate both x and y with respect to parameter t ; then divide the derivative of y by the

derivative of x. Note that the derivative dydx

is obtained as a function of the parameter t.

Having found dydx

= z (t) as a function of parameter t, we can apply the same method to nd

d2y

dx2:

d2y

dx2 =

ddx

dydx

= ddx

z (t) =

ddt

z (t)

dxdt

Note : It is usual to denote derivatives with respect to t by a , i.e. x = dxdt

, y = d 2 ydt2

.

In this notation we have:

y = dy

dx =

yx

= z (t)

y = ddx

dydx

= ddx

z (t) = z x


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Further Calculus 7

Examples 2A

(i) Find the equations of the tangent and normal at the point12

, 32

on the circle

x2

+ y2

= 1.Set x = cos t, y = sin t. The given point corresponds to t =

3

.

dydx

=

dydt

dxdt

= cost

sin t = cot t =

1 3 at t =

3

.

Tangent has equation y 32 =

1

3 x 1

2 .The normal is perpendicular to the tangent. So the gradient of normal is 3. (Productof gradients is 1.)The equation of the normal is y

32

= 3 x 12

, i.e. y = 3 x.The normal is a line through the origin as expected (although we could prove thiswithout calculus).

(ii) We saw that if x = cos t, y = sin t, then dx

dt = sin t, dy

dx = cot t,

d

2y

dx2 =

ddt

(cot t )(sin t)

= cosec2 t

sin t = cosec

3 t

( = 1y3

as before in Implicit Differentiation) .



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Further Calculus 8

Exercises: Parametric Differentiation

1. By eliminating the parameter t identify and sketch the curves described by the followingequations. Indicate the direction of motion on the curve for increasing t.

(a) x = 2t 1, y = t + 1 , 0 t 2(b) x = 2t3 + 1 , y = t3 + 2 , 1 t 1(c) x = cos t, y = sin t, 0 t 2(d) x = 3 + cos 2 t, y = 2 + sin 2 t, 0 t .

2. Find dydx

as a function of the parameter t when x and y are given by

(a) x = 4t2

1, y = 2t + 1 (b) x = 2 sec t, y = tan t

(c) x = 1t21 + t2

, y = 2t1 + t2

(d) x = t + t 1 , y = t t 1 .

3. The coordinates of a particle moving in the ( x, y)-plane are given as functions of timet by

x = 2 cos2 t, y = sin(2 t) (t 0).Find the x and y components of the velocity at any time t and hence nd the magnitudeand the direction of the resultant velocity at any time t. Deduce that the particle never

comes to rest.

4. Find dydx

and d2 ydx2

as functions of the parameter t when x and y are given by

(a) x = cos2 t, y = sin 2 t (b) x = cos(2t), y = sin t

(c) x = t2 , y = t3 (d) x = t2 t + 1 , y = t2 + 1t

.

5. Find equations for the tangent and the normal at the point (2 , 1) on the curve given

parametrically by x = t2 + 1t

, y = t2 t + 1 .


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Further Calculus 9

3 Applications of differentiation

3.1 Related rates

Consider a practical problem where two quantities x and y are related by a formula y = f (x),and suppose that x depends on a variable t, x = g(t) so that

y = f (x) = f (g(t)) = h(t)

i.e. y is also a function of t. (Often, t represents time).

Then, by the Chain Rule,

dydt

= dydx

dxdt

or y = f (x)x,

where, as usual, we use x and y to represent the derivatives (i.e. the rates of change) withrespect to t.

For example, as water is pumped into a tank or bath, the height of the water level increases.The rate at which the level increases is related to the rate at which the water is supplied (orthe rate at which the volume changes).

To solve related rate problems:

1. Draw a diagram (if possible) and introduce appropriate variables (e.g. xand y).

2. Decide which rate of change is asked for (e.g. y), and which rate of changeis given (e.g. x).

3. Find an equation relating the variables (e.g. y = f (x)).

4. Use the Chain Rule to relate y to x: dy

dt =

dy

dx

dx

dt = f (x)

dx

dt.

5. Finally, substitute appropriate values of variables (if required).


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Further Calculus 10

Examples 3A

(i) Air is blown into a spherical soap bubble at a rate of 10 cm 3 s 1 (cubic cm per second).When its radius is 1.5 cm, how fast is

(a) its radius and (b) its surface area changing with time?

Let the bubble have volume V , radius r and surface area S .The information in the question tells us V = 10cm3 s 1 .

(a) We seek r . Volume is related to the radius via

V = 4

3r 3

dV

dr = 4r 2 .

Using the formula for related rates:

V = dV

dr r = 4 r 2 r , and re-arranging r =

V 4 r 2

.

Given that we require the rate when V = 10 and r = 1.5,

r = 104(1.5)2

= 109 0.35cms

1 .

(b) We seek S . Surface area of a sphere is related to the radius via

S = 4 r 2 , dS

dr = 8 r.

We found in (a) that r = 109

when r = 1.5, therefore

S = 8 r r = 8 1.5 109

= 40

3 cm2 s 1

at that instant.

(ii) Water is poured at a rate of 3 m3 per minute into an inverted conical tank. If thetank has depth 10 m and top radius 2 .5 m, how quickly is the depth of water changingwhen it is 8 m?(A cone of height h and base radius r has volume

r 2 h3

.)

Let be the angle of the cone. Let h(t), r(t) and V (t) be the depth, top radiusand volume of water in the tank at time t, respectively. The question tells us that

V = 3 m3

per minute.


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Further Calculus 11

Thentan =

2.510

= rh r =

14

h.

Therefore V can be re-written in terms of height h only :

V = 13

r 2 h = 13

h

2

16 h = 48

h3.

h

10

2.5

r

Using the formula for related rates:

V = dV

dhh =

16

h2

h h = 16 V

h2 =

163h

2 = 48h

2 (since V = 3).

When h = 8, h = 4864

= 0.75 m per minute.



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Further Calculus 12

Exercises: Related Rates

1. Oil spills from a tanker to form a circular slick. If the radius of the slick increases at arate of 0.3 m s 1 , how fast is its area increasing when the radius is 100 m?

2. A trough is 3 m long. Its ends are equilateral triangles of side 1 m. If water ows in ata rate of 0.5 m3 per minute, how fast is the water level rising when it is 0 .3 m deep?

3. The pressure P and volume V of a gas satisfy Boyles law P V = C (constant). If V increases at a rate of 10 cm3 per minute, at what rate is P changing when V = 2 litresand P = 2 atmospheres.

4. A 10m ladder leans against a wall. The bottom slips out at a rate of 1 m s 1 . How fastis the top moving down the wall when it is 6 m from the ground? At what rate is the

angle of inclination of the ladder to the horizontal then changing?5. A police helicopter H hovers stationary at a height of 1 mile above a straight motorway.

The pilot sees a car C on the road and, with radar, determines that the distance fromH to C is 1.5 miles, decreasing at a rate 55 m.p.h. Is the car driver speeding (assuminga motorway speed limit of 70 m.p.h.)?

6. A lecturer of height 5 ft 6 ins walks at 3 ft per second away from a lamp of height16 ft 6 ins. At what rate is his shadow lengthening, and how fast is the tip of hisshadow moving?

7. One car travels north at 40 m.p.h. and another car travels east at 60 m.p.h., on straightroads that cross at right angles. How fast are the cars approaching each other whenthey are, respectively, 8 miles south and 6 miles west of the junction?

8. Sand falls onto a conical pile at a rate of 0.1 m3 s 1 . The radius of the base of the pileis always equal to half its height. How fast is the height increasing when the pile is(a) 1 m high? (b) 2 m high?

9. A spherical block of ice melts such that its radius r decreases at a constant rate. Showthat, at any instant, the rate of decrease of its volume is r/ 2 times the rate of decreaseof its surface area.

10 . A drive wheel, centre O, radius 0.5 m, rotates at 30 rpm as shown. The connecting rodQP has length 2 m. How fast is the piston P moving when OP = 2 m?

O P

Q

0.5 2

x

drive wheel piston


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Further Calculus 13

3.2 Small variations and Error Analysis

Let (x0 , y0 ) be a point on a curve y = f (x).

Suppose that when x0 increases to x1 = x0 + x,y0 increases to y1 = f (x1 ) = f (x0 + x) = y0 + y.

(x , y )00

1 1(x , y )

y=f(x)

O

y

y

x x

(x , y )00

(x , y )1 1

y=f(x)

O

dy dx

x

x

y

x

Then since the derivative dydx

at (x0 , y0 ) is dened as

dydx

= lim x 0

f (x0 + x) f (x0 ) x = lim x 0 (y0 + y) y0 x

= lim x 0

y x

,

we expect that dy

dx y x

for x small.

This gives rise to the following formula which allows us to calculate approximately the changein the variable y, denoted by y, corresponding to a small change , x, in the variable x:

Small variations

y dydx

x when x is small.

Some terminology:

The absolute change in y is y.

The relative change in y is y

y0.

The percentage change in y is y

y0 100%.


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Further Calculus 14

Examples 3B

(i) Estimate 6213 .

Let y = x 13 . When x = 64 we know that y = 4. These are equivalent to the values(x0 , y0 ) described above.

Without using a calculator , what happens to the value of y when x changes fromx0 = 64 to x1 = 62 = 64 + x, where x = 62 64 = 2 ?We know that y =

13

x 23 . At the point x0 = 64, y0 = 4, this derivative has

valuey (x0 ) =

13

116

= 148

.

Hence, from the formula the corresponding change in y is given approximately by

y dydx

x = y(x0 ) x = 148

(2) = 124

= 0.0417.(This represents the change in y as x changes from 64 down to 62.)

Our approximate value for 6213 is therefore given by

6213 = y1 = y0 + y 40.0417 = 3.9583.

(ii) The cross-sectional area of a circular bore hole is to be accurate to within 5%. To whataccuracy does the drill have to be manufactured, i.e. what % error is allowed in theradius of the drill?

Area of the bore hole produced by the drill: A = r 2 where r is the radius of thehole.

Using the formula, A dA

dr r = 2r r.

Therefore AA 2r rr 2 = 2 rr , so that AA 2 rr .

Since AA 5%, we must have 2

rr 5%, i.e.

rr 2.5%.

One More Example in Lectures


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Further Calculus 15

3.3 Linear approximation

It is possible to approximate the values of the function y = f (x) near to a xed point x = x0by a linear function in the following way:

we saw in the last section that at point x = x0 + x,

y = y0 + y where y0 = f (x0 ) and y dydx

x = f (x0 ) x.

So y f (x0 ) + f (x0 ) x, and since x = x0 + x we have

Linear approximation

y f (x0 ) + f (x0 ) (x x0 ) for x near x0 .

This is of the form y = y0 + m (x x0 ), i.e. a straight line.Therefore it is called the linear approximation to f (x) near x0 .

Close to (x0 , y0 ) the curve y = f (x) and straight line y = f (x0 ) + f

(x0 ) (x x0 ) appearvery similar.

The straight line is a good approximation forthe curve.

However, further away from ( x0 , y0 ) the straightline and the curve diverge, and their values arevery different.

y=f(x)

O

y

x

(x , y )00

0 0 0 y=f(x ) + f (x ) (x x )


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Further Calculus 18

3.4 Curve sketching

3.4.1 Stationary Points

On a curve y = f (x), y is increasing as x increases where dydx

> 0, and y is decreasing as x

increases where dydx

< 0.

So the curve is at where dydx

= 0. Thus

A point at which dydx

= 0 is called a stationary point (SP) .

A stationary point is

a maximum turning point (Max TP), or

a minimum turning point (Min TP), or

a horizontal point of inection (PI),

as illustrated in the following diagrams.

PI

min

max

PI


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Further Calculus 19

3.4.2 The Nature of Stationary Points

The criterion dydx

= 0 identies stationary points on the curve y = f (x).

There are two tests to identify the nature of these points.

A test for max/min stationary (or turning) points using rst derivatives:

Near a stationary point x = a (so that dydx

= 0 at x = a):

if dy

dx < 0 for x < a and

dydx

> 0 for x > a , then x = a gives a min TP

if dy

dx > 0 for x < a and dy

dx < 0 for x > a , then x = a gives a max TP

if dy

dx has the same sign on either side of x = a, then x = a gives a PI

MIN for y

x a

dydx 0 +

slope \ /

MAX for y

x a

dydx

+ 0 slope / \

x a

dydx 0

slope \ \

PIs

x a

dydx

+ 0 +

slope / /


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Further Calculus 20

The second test relies on the idea of concavity .

If y > 0 at x = a then the curve lies above its tangent at x = a, and the curve is saidto be concave up at x = a.

If y < 0 at x = a then the curve lies below its tangent at x = a and the curve is saidto be concave down at x = a.

If y = 0 at x = a then further investigation is required.

concave downconcave up

Hence, we can derive a test for the nature of a stationary point based on the sign of thesecond derivative.

A test using the second derivative

y = 0 and y > 0 at x = a min TP at x = ay = 0 and y < 0 at x = a max TP at x = a

y = 0 and y = 0 or is undened at x = a : further investigation is required.


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Further Calculus 21

Examples 3D

(i) Find the stationary points of y = x4 + 4 x3 2x2 12x and identify their nature.Here

dy

dx = 4x3 + 12 x2

4x

12 = 4(x + 3)( x + 1)( x

1).

There are SPs where dydx

= 0, i.e. at x = 3, 1, 1 where y = 9, 7, 9 respectively,and no points where y is not dened.

x 3 1 1dydx 0 + 0 0 +

slope

\ /

\ /

MIN MAX MIN

Alternatively, using the 2nd derivative test y = 12x2 + 24 x 4 :

x y y type

3 0 10872 4 > 0 min1 0 1224 4 < 0 max

1 0 12 + 244 > 0 min



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Further Calculus 22

3.4.3 Asymptotes

An asymptote to a curve is a straight line that the curve approaches far away from theorigin.

The line y = b is a horizontal asymptote (HA) to a curve y(x) if y b as x and/or y b as x .

The line x = a is a vertical asymptote (VA) to a curve y(x) if eitherlim

x a+y = or limx a y = .

In particular, if

y = p(x)q (x)

with q (a) = 0 , p(a) = 0 then x = a is a VA.

The line y = mx + c is an inclined asymptote (IA) to a curve y(x) if y(x) (mx + c) 0 as x or as x .

In this case the curve approaches (or looks like ) the straight line y = mx + c asx .

y

x

max

O

min

b

y=mx+c

HA

VA

IA

a

Note: the graph of a function y = f (x ) can never cross a VA!


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Further Calculus 23

Examples 3E

(i) y = 3x + 1 0

as x , so y = 0 is a HA.Also, y as x 1

, so x = 1 is a VA.

(ii) y = 2x + 13x + 4

= 23

53

3x + 4. So y

23

as x + , and y 23

+

as x .Therefore y =

23

is a HA.

Also, y as x 43

, so x = 43

is a VA.

(iii) y = 6x2 + 5 x + 33x2 + 7

= 6 + 5x + 3x2

3 + 7x2 63

= 2 as x , so y = 2 is a HA.

Note that 3 x2 + 7 cannot be zero, so there is no VA.

(iv) y = x2 + 1

x 1 =

x2 1 + 2x 1

= (x 1)(x + 1) + 2

x 1 = x + 1 +

2x 1

.

Denominator is zero when x = 1; this is a VA. As x 1 , y .As x

, y

(x + 1) =

2

x 1 0 and y behaves like y = x + 1.

So y = x + 1 is an IA.

(v) y = x2 + 4 x 9

x 2 = x + 6 +

3x 2

.

Therefore, y (x + 6) = 3x 2

0 as x , and so y = x + 6 is an IA.Also y as x 2 so x = 2 is a VA.


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Further Calculus 24

3.4.4 Sketching a Curve

You are familiar with the graphs of linear functions y = mx + c (straight lines) and graphsof quadratic functions y = ax2 + bx + c (parabolae). A reasonable sketch of the graph of a function allows you to see how the function behaves as the independent variable changes.The features that we have considered above (stationary points and asymptotes) are oftensufficient to enable us to obtain good sketches of the graphs of other functions.

To sketch the curve y = f (x):

1. Find points where the curve cuts the coordinate axes,i.e. nd y when x = 0, and also solve y = f (x) = 0.

2. Find stationary points (i.e. solve f

(x) = 0) and identify their nature as MAX,MIN or PIs.

3. Find any asymptotes and identify how the curve approaches the asymptotes.

4. Choose appropriate scales for the x and y axes to accommodate the featuresidentied in 13 and plot these features.

5. Given that the function f (x) is continuous and differentiable everywhere else,you should now be able to complete the sketch by joining up the features with

a smooth curve.

Examples 3F

(i) Sketch the curve

y = x3

x2

3.

When x = 0, y = 0 and vice versa. Therefore the curve cuts the axes only at (0, 0).

Now,

y = (x2 3)3x2 x3 2x

(x2 3)2 =

x2 (x2 9)(x2 3)2

,

so y = 0 when x = 0, 3 (and y = 0, 92

).


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Further Calculus 25

x 3 0 3dydx

+ 0 0 0 +slope /

\ \ /

MAX PI MIN

Also, by polynomial division, we have

y = x(x2 3) + 3x

x2 3 = x +

3xx2 3

soy x =

3xx2

3 0

as x ,giving y = x as an IA.

Denominator is zero when x2 3 = 0, so there are VAs at x = 3 and x = 3.The following sign table gives the behaviour of the function ( ) as the VA isapproached.

x 3 0 3y ND + 0 ND +

Now, having chosen suitable scales and plotted the above features, we can completethe graph:

5 4 3 2 1 0 1 2 3 4 510

8

6

4

2

0

2

4

6

8

10



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Further Calculus 26

Exercises: Curve Sketching

1. Find and classify the stationary points of the following:

(a) y = 2x2

7x + 4 (b) y = 3x

x2 + 7 (c) y = x3

x2

x + 1

(d) y = x3 3x2 + 3 x + 3 (e) y = 3x5 10x3 (f) y = x + 4x

(g) y = x x + 3 (h) y = x 1 x2 + 2 x + 4 (i) y = 4

cos (2x)(j) y = x2 5x + 8 (k) y = x3 3x2 24x 3 (l) y = 4x2 + x 1

(m) y = x2 + x 2 (n) y = x

(1 + x2 ) (o) y = x 3 2x

(p) y = e x2

.

2. Show that the function y = x + ln 2x 1(x + 1)

5 has a maximum turning point at x = 32

and a minimum turning point at x = 2.

3. Determine the nature of the stationary point at x = 0 of y = (sin x) ln(1 x).

4. Find the turning points between x = 0 and x = 2 of y = e x sin x.

5. Sketch the following curves:

(a) y2 = x (b) y2 = x + 2

(c) y2 = x2 (x + 2) (d) y2 = ( x + 2)( x 1)(x 2).

6. Find the limits as x and x of the following:(a) y =

2x

(b) y = 3x2 + 2

(c) y = x + 12x

1

(d) y = 3x x2 + 7 (e) y = x2 + 2

x (f) y = x2 + 2 x x.


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Further Calculus 27

7. For each of the following functions,

(i) nd the intercepts with the coordinate axes,

(ii) nd and classify the stationary points,

(iii) nd any asymptotes,

(iv) plot the features found in (i)-(iii) and hence complete a sketch of the graph of the function.

(a) y = x2 3x 4 (b) y = x3 x2 (c) y = 1x + 4

(d) y = 1x2 + 1

(e) y = xx2 + 1

(f) y = 1

(x 2)2

(g) y = x 3x 1 (h) y = 2x + 33x 1 (i) y = xx2 1(j) y =

x + 2x2 + 2 x 3

(k) y = 2x2 + x 3

x + 2 (l) y = x +

1x

(m) y = x2 + 1x2

(n) y = x x + 2 (o) y = x e x .


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Further Calculus 28

3.5 Optimisation

At a point x = a, a curve y = f (x) has

a local maximum if f (x)

f (a) for all x close to a

a local minimum if f (x) f (a) for all x close to a,and has

a global maximum if f (x) f (a) for all x in the domain of f

a global minimum if f (x) f (a) for all x in the domain of f .Maxima and minima are referred to as extrema .

Optimisation means nding the best (or most appropriate) value of some variable. Oftenthis is a maximum value (prot, output) or a minimum value (energy, cost). So optimisationusually involves identifying the global maximum or the global minimum of a function.

Maximum and minimum turning points are always local (and may be global) maxima andminima, respectively, but there are other ways that extrema can occur.

Examples 3G

(i) y = |x|, < x < , has a global minimum at x = 0, but this is not a TP sincedydx

is not dened at x = 0. This function does not have a global maximum.

xO

y

The following is a useful result giving us sufficient conditions for a function to have globalextrema and also tells us where extrema (both local and global) can occur.


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Further Calculus 29

If f is a continuous function dened on a closed, nite interval [ a, b], then globalmaxima/minima exist .

Points to be checked when looking for the global max/min of the

function are :

points where f (x) = 0

points where f (x) is not dened (e.g. corners)

end points of the domain, a and b.

Examples 3H

(i) The curve

y = 3x3 , 2 x < 1x2 6x + 8 , 1 x 4

has dydx

= 0 at x = 3 (where y = 1), and at x = 0 (where y = 0).

However , the function has a global max y = 3 (at x = 1, a corner, where dydx doesnot exist), and global min y = 24 (at x = 2, an end-point of the domain).The test for SPs,

dydx

= 0, has not detected the global max/min.

We must check stationary points AND any other critical values of x, i.e. the end pointsand points where f (x) is not dened (x = 1).

O

4

1

2 2

y

x


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Further Calculus 30

If f (x) is not continuous, or is dened on a non-closed or non-nite interval, then we mustwork harder to nd the global maximum or minimum (usually by having a good understand-ing of the shape of the graph).

To solve optimisation problems:

1. Identify and name variables: a sketch of the physical situation will often help.

2. Decide the variable to be optimised (e.g. y), and a variable that can becontrolled or chosen (e.g. x).

3. Form an equation relating x and y, e.g. y = f (x).

4. Determine the domain of the function f , i.e. the set of all possible values of x.5. Consider critical points: stationary points within domain; end points of

domain; points where function is not differentiable.

6. If the function f (x) is continuous on a closed nite interval, then identify theglobal maximum or minimum by evaluating the function at all critical points(i.e. SPs, end points and points where f (x) is not dened).

7. Otherwise, identify the nature of the stationary points and locate the global

maximum or minimum (if it exists) from a good understanding of the shape of the graph.


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Further Calculus 31

Examples 3I

(i) Find the global extrema of the function f (x) = 2x3 3x2 12x + 10 on the domain3 x 3.We have a continuous function on a closed domain.

f (x) = 6 x2 6x 12 = 6(x + 1)( x 2 ) = 0 x = 1 or 2.f (1) = 17, f (2) = 10.At the end points of the closed interval, f (3) = 35 and f (3) = 1 .There are no points in the domain where the function is not differentiable.Thus, the global maximum is the largest of these values : 17 at x = 1 ; this SP mustbe a Max TP.The global minimum is 35 at the end point x = 3.

(ii) A rectangular metal plate with dimensions 4m by 2m has squares cut from its corners,and is then folded up to make an open box. What size is the box that has maximumvolume?

x

x

4

Let the squares have sides x m, with 0 < x < 1. (The domain is not closed: If x = 0then we cannot construct a box from the rectangular plate; if x = 1 then, we are leftwith no ends to fold up.) The volume of the box V (x)m2 is given by

V (x) = ( 4

2x)(2

2x)x = 4( x3

3x2 + 2 x),

soV (x) = 4(3 x2 6x + 2) , V (x) = 24( x 1).

At a SP, V (x) = 0 so

x = 6 36 24

6 = 1

126

= 1 33

,

but only x = 1 33

satises 0 < x < 1.

At this point V < 0, so it gives a MAX TP .


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Further Calculus 32

Since V (x) is continuous and differentiable for 0 < x < 1 and V 0 as x 0 or 1,we can see that x = 1

33

gives the absolute max .

The required box has dimensions (in metres)

2 + 2 33 by 2 33 by 1 33and

V max = 8 3

9 m3 .

(iii) A farmer wishes to enclose a rectangular eld of 80,000 m2 , one side of which is a wall(and so does not require to be fenced). Find the dimensions for the eld to use theleast fencing.

Let sides of eld be x metres and y metres, asshown.

Then amount of fencing used, F metres, is givenby

F = 2x + y.

We know that the required area is 80000 m 2 , so

xy = 80000 y = 80000

x .

x x

y

Writing the amount of fencing in terms of x only,

F (x) = 2x + 80000

x , F (x) = 2

80000x2

, F (x) = 160000

x3 (x > 0).

At SP, F (x) = 0 x = 200 , y = 80000

200 = 400 .

Since F > 0, this SP is a MIN TP , and there are no other SPs.

Also, F (x) is continuous and differentiable on 0 < x < with F (x) + as x 0and as x , so it is clear that this minimum TP gives the absolute min . Thus wehave two sides of length 200 m and one of length 400 m.



33/55


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Further Calculus 34

10 . A woman in a rowing boat is 2 miles from shore. She wants to get home, 10 milesalong the shore, as soon as possible.

She can run at 10mph and row at 8mph.

Where should she choose landing point A so that herroute from her position B to her home H is quickest,and what is the minimum time?

O A H

river

y 2

B

x

10

11 . An open box with a square base is to be made so that its volume is 1 m 3 . What is theminimum surface area of the ve sides?

12 . What is the size of the cylinder of largest volumethat can be cut from a sphere of radius a?

r

h

a

13 . The strength of a certain rectangular beam is proportional to the breadth and to thesquare of its depth. What are the dimensions of the strongest beam whose cross-sectional diagonal is pmm?

14 . A street of length 60m has a lamp at each end. The lamps are identical except thatone lamp is 8 times brighter than the other. Find the position of the dullest point onthe street if the brightness of the lamp at some point is inversely proportional to thesquare of the distance from it.

15 . If a small object is placed at a distance u in front of a thin lens of focal length f (< u ),the distance of the image behind the lens is v, where 1/u + 1 /v = 1/f . Show that the

minimum distance between the image and the object is 4 f .16 . A page of print is to contain 16,000 mm2 of printed area, with a margin of 40 mm at

the top and bottom and a margin of 25 mm at each side. Find the dimensions of thepage of the smallest area that fulls these requirements.

17 . A wall 2 2 m high is 1 m from the vertical wall of a house. Find the length of theshortest ladder that will reach over the wall to touch the house.

18 . The cost per hour of running a train is proportional to 100 + v2 / 36, where vm.p.h. isthe average speed on a trip. Find the speed v that makes the trip Glasgow-Londoncheapest. (Take the distance Glasgow-London to be 400 miles.)


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Further Calculus 35

Summary and important points:

Derivatives as evaluations of rates of change and gradients.

The denition of a derivative:f (x) = lim

h 0

f (x + h) f (x)h

Use of the Table of Standard Derivatives

The Linearity Rule:d

dx [f (x) + g(x)] = f (x) + g (x)

The Product Rule:d

dx f (x) g(x) = f (x) g(x) + f (x) g(x) .

The Quotient Rule:d

dx f (x)g(x)

= f (x) g(x) f (x) g(x)

g(x)2 .

Derivative of composite functions: the Chain Ruled

dxf g(x) = f g(x) g (x).

Higher Derivatives.

Implicit Differentiation.

Parametric differentiation:

dy

dx =

y

x .

Derivatives of inverse, exponential and logarithmic functions.


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Further Calculus 36

Summary (contd)

Related rates.

Small variations and error analysis: y dydx x ;absolute, relative and percentage changes.

The linear approximation of function y = f (x) at a point x = x0 ,y f (x0 ) + f (x0 )(x x0 ).

Stationary points (SP) (points where dydx

= 0).

If dydx

= 0 and d2 ydx2

> 0 at x = a then there is a minimum SP at x = a.

If dydx

= 0 and d2 ydx2

< 0 at x = a then there is a maximum SP at x = a.

Horizontal, vertical and inclined asymptotes.

Curve sketching.

Local and Global Extrema: these occur where dydx

= 0 or at end points

of the domain or at points where dydx

is not dened.

Optimisation problems.


37/55


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Further Calculus 38

Find the area bounded by the x-axis, lines x = 1, x = 4 and curve y = x.

Area = 4

1

x dx

= 23

x32

4

1

= 2

3 (8 1) =

143

.O

x

y

1 4

We now generalise the technique used above to calculate the area between two curves y =f (x) and y = g(x) where f (x) g(x) for a x b.

00000000000000000000000000000000000000000000

11111111111111111111111111111111111111111111

x

y=f(x)

ba x r

( r)

y=g(x)

f ( x ) g r ( r ) x

We will simplify our discussion by describing a typical strip at xr with width x andheight f (xr ) g(xr ) (0). The area of the strip is

( A)r = ( f (xr )

g(xr )) x.

Note that this is the correct area even if f (xr ) < 0 and/or g(xr ) < 0 as long as f (xr ) g(xr ).Hence the whole area is given approximately by

A n

r =1( A)r =

n

r =1(f (xr ) g(xr )) x.

This again is a Riemann sum as in the denition of a denite integral. Taking more andmore strips, i.e. letting x 0, gives better and better approximations, and

A = limn

n

r =1(f (xr ) g(xr )) x =

b

a(f (x) g(x)) dx. (2)


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Further Calculus 39

Example 4B

Find the area A between the parabolae f (x) = x2

4x + 6 and g(x) = 4 x

x2 .

The curves intersect where

x2 4x + 6 = 4 x x2

2(x2

4x + 3) = 02(x 3)(x 1) = 0x = 1 or x = 3.

y

x1 3

Hence

A = 3

1[(4x x

2 ) (x2

4x + 6)] dx = 3

1(2x

2 + 8 x 6) dx = 83

.

The area between a curve y = f (x) when f (x) < 0 and the x-axis can be regarded as a

special case of the above. In this case the typical strip has an area

(0 f (xr )) x = f (xr ) xand so

A = limn

n

r =1f (xr ) x =

b

a f (x) dx = b

af (x) dx. (3)

This will give a positive answer: areas are positive quantities but integrals may benegative .

Examples 4C


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Further Calculus 40

Examples in Lectures

000000000000000000000000000000000000000000111111111111111111111111111111111111111111 x=f(y)

yd

c

r

r( ) A

x= f ( r ) y

Similarly, to nd the area of a region between a curve x = f (y) and the y-axis as shown, weconsider a typical strip of width y parallel to the x-axis. The area of strip is

( A)r = f (yr ) y

so an approximation of the area of the region is

A n

r =1( A)r =

n

r =1f (yr ) y.

Thus as y 0 we getA =

d

cf (y) dy. (4)

Example 4E

Find the area A shown.

y

x

1

4

A

y=x 2


41/55

Further Calculus 41

Here y = x2 , so that x = y = f (y). So

A = 4

1 y dy = 2

3y

32

4

1=

23

(8 1) = 14

3 .

Again we can generalise this technique to calculate the area between two curves x = f (y)and x = g(y) where f (y) g(y) for c y d. A typical strip of width y parallel to thex-axis has area

( A)r = ( f (yr ) g(yr )) yso an approximation of the area of the region is

A n

r =1( A)r =

n

r =1(f (yr ) g(yr )) y.

Thus as y 0 we getA =

d

c(f (y) g(y)) dy. (5)

Examples 4F

Examples in Lectures


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Further Calculus 42

Exercises: Plane areas

1. Find the area between the curve y = x2 and the x-axis, from x = 0 to x = 1.

2. Find the area between the curve y = (2 x 1)3

and the x-axis, from x = 0 to x = 1.

3. Find the areas of the nite regions bounded by the curves:

(a) y = x, x = 4, y = 0;(b) y = (2 x)(x + 1) , y = 0;(c) y = (4 x2 )

12 , x = 0, x = 1, y = 0;

(d) y = x2 , y = x3 ;

(e) y = 2 + x x2 , y = x + 1;(f) y = x2 , x = y2 ;

(g) y = x3 , x = 0, y = 1, y = 8;

(h) y2 = 4

x, x = 1;

(i) x = y2 6y + 5 , x = 3.


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Further Calculus 43

5 Volumes of revolution

5.1 Rotation about x-axis

Suppose that we want to nd the volume V x swept out when the area between a curve

y = f (x) 0 and the x-axis for a x b, is rotated fully about the x-axis.

a b x

y=f(x)

( x f

x r

)r x

The volume of the disc swept out by the typical strip of width x and height f (xr ) whenit is rotated about the x-axis is

( V x )r = [f (xr )]2 x

(volume of cylinder is r 2 h) so

V x n

r =1( V x )r =

n

r =1[f (xr )]2 x.

Thus as x 0 we getV x =

b

a[f (x)]2 dx. (1)

Examples 5A

(i) Find V x for the solid obtained by rotating the region between

g(x) = 0 , f (x) = x2 , x = 1, x = 2

about the x-axis.


44/55

Further Calculus 44

y y=x 2

x=1 x=2 x

V x = 2

1[f (x)]2 dx =

2

1x4 dx =

x5

5

2

1=

315

.

(ii) Derive the formula for the volume of a sphere of radius a by nding V x for the solidobtained by rotating the region between f (x) = a2 x2 and the x-axis betweenx = a and x = a.

x=a x=a

y

x

y = a x2 2

V x = a

a[f (x)]2 dx =

a

a(a2 x

2 ) dx

= a2 x x3

3

a

a=

43

a 3 .


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Further Calculus 45

We can also calculate the volume generated by rotation of the area between the curvesy = f (x) and y = g(x) (f (x) g(x)) around the x-axis.

00000000000000000000000000001111111111111111111111111111 y=f(x) y=g(x) y

x

f( x r ) ( x r ) g

r xa b

The typical strip as shown, rotated about the x-axis generates a circular washer or annulus(a disc with a hole) with volume

( V x )r = [f (xr )]2 x [g(xr )]2 x

= [f (xr )]2

[g(xr )]2 x.

Summing and letting x

0 we get

V x = b

a[f (x)]2 [g(x)]

2 dx. (2)

Example 5B

Find the volume of revolution V x for the nite region between the curves

g(x) = x2 4x + 6 and f (x) = 4 x x2 .

000000000000000000000000000000000000000000111111111111111111111111111111111111111111 y=g(x)

y=f(x)

x=1 x=3


46/55


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Further Calculus 47

5.2 Rotation about y-axis

Suppose that we want to nd the volume V y swept out when an area between a curve y = f (x)and the x-axis is rotated fully about the y-axis.

y

f (xr )

r x

x

When rotated, the strip of width x at xr generates a cylindrical shell of height y = f (xr ),inside radius xr and thickness x. The volume of the shell is

( V y)r = (xr + x)2 f (xr ) (xr )2 f (xr )

= ((xr )2 + 2 xr x + x

2 )f (xr ) (xr )2 f (xr )

= 2 x r f (xr ) x + f (xr ) x2 .

Therefore,( V y)r 2x r f (xr ) x for x small

and

V y n

r =1( V y)r

n

r =12x r f (xr ) x.

Thus as x 0 we getV y = 2

b

axf (x) dx. (3)


48/55

Further Calculus 48

Example 5C

Find V y for the solid obtained by rotating the region between

g(x) = 0 , f (x) = x2 , x = 1, x = 2

about the y-axis.

y y=x 2

x=1 x=2 x

V y = 2 2

1xf (x) dx = 2

2

1x3 dx = 2

x4

4

2

1=

152

.

Again we can generalise this technique to calculate the volume generated by rotation of thearea between the curves y = f (x) and y = g(x) (f (x) g(x)) around the y-axis. Thecylindrical shell of height f (xr ) g(xr ) has volume

( V y)r 2x r (f (xr ) g(xr )) xand hence as x 0 we get

V y = 2 b

ax(f (x) g(x)) dx. (4)


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Further Calculus 49

Exercises: Volumes of revolution

1. In each of the following, nd the volume generated when the nite region bounded bythe given curves is rotated fully about the x-axis:

(a) y = x4 , x = 2, y = 0;

(b) y = (4 x2 )12 , x = 0, x = 1, y = 0;

(c) y4 = 16x, x = 1, x = 4, y = 0;

(d) y = x2 , y = 4;

(e) y = x2 + 1 , y = x + 3;

(f) y = 4 x2

, y = 2 x;(g) y = x3 , x = 0, y = 1, y = 8.

2. In each of the following, nd the volume generated when the nite region bounded bythe given curves is rotated fully about the y-axis:

(a) y = x, x = 4, y = 0;(b) y = 1/x, x = 1, x = 2, y = 0;

(c) y = x2 5x, y = 0;(d) y = x3 , x = 0, y = 1, y = 8;

(e) y = x2 , x = y2 .

3. By considering the line y = ax/h (with a and h constants) derive the formula for thevolume of a cone of height h and base radius a.

4. Find the volumes of the solids formed by the rotation of the loop of the curve y2

=x(4 x)2 fully about (a) the x-axis and (b) the y-axis.5. Find the volumes obtained when the nite region bounded by the line x = a and the

parabola y2 = 4 ax is rotated about (a) the line x = a and (b) the line y = 2a.

6. A bowl (of depth > 2 cm) has the shape of part of a sphere of radius 10 cm. It containssoup to a depth 2 cm. What is the volume of soup?

7. A hole of radius 1 is drilled through the centre of a solid sphere of radius 4. What isthe volume of the resulting solid?


50/55

Further Calculus 50

6 Answers to exercises

Implicit differentiation

1. (a) (i), (ii), (iii) 2y

(y = 0). The two functions are y = 4x 8.(b) (i), (ii), (iii)

xy

(y = 0). The two functions are y = 25 x2 .(c) (i), (ii), (iii)

xy

(y = 0). The two functions are y = x2 16.(d) (i), (ii), (iii)

1xy 2

(y = 2). The two functions are y = 2 8 + 2x x2 .

2. (a) x2

y2 (y = 0) (b) y(y2

2)

2x + 2y 3xy2 (2x + 2y 3xy2

= 0)

(c) y + 2 xy cos x

2 xy sin y x (2 xy sin y x = 0) (d)

3x2 + y2

1 2xy (2xy = 1)

(e) y cos(xy) + sin x cos yx cos(xy) + cos x sin y

(x cos(xy) + cos x sin y = 0)

(f) cos(x + y2 )1 2y cos(x + y2 )

(1 2y cos(x + y2 ) = 0)

3. (a) y + 3 x 2 = 0 (b) 11y + 10x 19 = 0(c) 46y 9x 120 = 0 (d) 2y x =

2

4. (a) y x (b) cos x cos

2 y (c) y

2y + x

5. (a) 38

(b) 1 + 2

Parametric differentiation

1. (a) Straight line 2 y x = 3, from (0, 1) to (4, 3).(b) Straight line 2 y x = 3, from (1, 1) to (3, 3).(c) Circle, centre at the origin, radius 1. As t varies from 0 to 2, the circle is traced

out clockwise, starting from (1 , 0).

(d) Circle, centre (3, 2), radius 1. As t varies from 0 to , the circle is traced outanticlockwise, starting from (4 , 2).


51/55

Further Calculus 51

2. (a) 1

4t (b)

12

cosec t (c) t2 1

2t (d)

1 + t2

t2 13. 2sin2t, 2cos2t, 2

The speed is always 2, so the particle never comes to rest.

4. (a) 1, 0 (b) 14 cosec t, 116 cosec3 t (c) 3t

2 , 3

4t

(d) 2t3 1

t2 (2t 1),

2(t3 3t + 1)t3 (2t 1)3

5. y = x 1, y + x 3 = 0

Related rates

1. 60 m2 s 1 (188.5 m2 s 1 ).

2. 11.2 3 m s

1 (0.48ms 1 ).

3.

4. 43

m s 1 ; 16

rads 1 (9.5 s 1 ).5. 73.8 m.p.h.

6. 1.5fts 1 , 4.5 ft s 1 .

7. 68m.p.h.

8. 0.13ms 1 ; 0.032ms 1 .

10. 1.61ms 1 .

Small variations and linear approximation

1. (a) 9.0556 (b) 9.8 (c) 0.515 (d) 0.7754

2. 9.95, 3.317

3. 0.93m2

4. 8720mm3

5. 1 cm


52/55

Further Calculus 52

6. (a) 0.2485 m (b) 0.01 s

7. (a) 3% (b) 1.996

8. (a) y =

2 x (b) y = 4x

3 (c) y =

7

4

+ 3

16

x (d) y = ex + e + 1

Curve sketching

1. (a) Minimum at74

, 178

(b) Maximum at32

, 374

(c) Minimum at (1 , 0), maximum at 13, 3227(d) Point of inection at (1 , 4)

(e) Maximum at 2, 8 2 , point of inection at (0 , 0), minimum at 2, 8 2(f) Maximum at ( 2, 4), minimum at (2 , 4)(g) Minimum at ( 2, 2)(h) Minimum at

52

, 73(i) Maximum y = 4 at x = 0, , 2, . . .,

minimum y = 4 at x = 2

, 32

, . . .

(j) Minimum at52

, 74

(k) Minimum at (4 , 83), maximum at ( 2, 25)(l) Minimum at

12

, 3

(m) Minima at (

1, 2) and (1, 2)

(n) Maximum at (1 , 2), minimum at ( 1, 2)(o) Maximum at (1 , 1)

(p) Maximum at (0 , 1)

3. Maximum

4. Maximum at4

, 1 2 e

/ 4 , minimum at54

, 1 2 e

5/ 4

6. (a) 0

(b) 0+

(c) 1

2

as x (d) 3

as x , 3+

as x (e) as x (f) 2 as x , + as x


53/55

Further Calculus 53

7. (a) (i) Intercepts (0 , 4), (1, 0) and (4, 0)(ii) Minimum at

32

, 254

(iii) No asymptotes.

(b) (i) Intercepts (0, 0) and (1, 0)(ii) Maximum at (0,0), minimum at

23

, 427

(iii) No asymptotes

(c) (i) Intercepts 0, 14

(ii) No stationary points

(iii) HA y = 0, VA x = 4(d) (i) Intercept (0 , 1)

(ii) Maximum at (0 , 1)

(iii) HA y = 0

(e) (i) Intercept (0 , 0)

(ii) Maximum at 1, 12

, minimum at 1, 12

(iii) HA y = 0

(f) (i) No intercepts

(ii) No SP

(iii) HA y = 0, VA x = 2

(g) (i) Intercepts (0 , 3) and (3, 0)

(ii) No SP

(iii) HA y = 1, VA x = 1

(h) (i) Intercepts (0 , 3) and 32

, 0

(ii) No SP

(iii) HA y = 23, VA x =

13

(i) (i) Intercept (0 , 0)

(ii) No SP

(iii) HA y = 0, VA x = 1 and x = 1(j) (i) Intercepts 0,

23

and (2, 0)(ii) No SP

(iii) HA y = 0, VA x = 1 and x = 3(k) (i) Intercepts 0, 32 , 32, 0 and (1, 0)


54/55

Further Calculus 54

(ii) Minimum at 2 + 12 6, 7 + 2 6 , maximum at 2

12 6, 7 2 6

(iii) VA x = 2, IA y = 2x 3(l) (i) No intercepts

(ii) Minimum at (1 , 2), maximum at ( 1, 2)(iii) VA x = 0, IA y = x(m) (i) No intercepts

(ii) Minima at ( 1, 2) and (1, 2)(iii) VA x = 0

(n) (i) Intercepts ( 2, 0) and (0, 0)(ii) Minimum at

43

, 43

23

(iii) No asymptotes

(o) (i) Intercepts (0 , 0)

(ii) Maximum at 1, 1e

(iii) HA y = 0

Optimisation1.

4

, 40

2. 400

33. 10, 100k

4. Height 20 1 / 3

cm (13.7 cm), radius 51 / 3

cm (3.4cm)

5. (a) 12, 12 , (b) (0, 1) and (1, 0)

6. (1, 2) or (1, 2)

7. Radius3V 8 2

1 / 3

, height (4 1) V 9 2

1 / 3

8. 10

9. C is 1 15 miles (0.26 miles) from O (between O and B)

10. x = 83

miles, minimum time 6960

hours (= 1 hr 9min).


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m115 Further Calculus for Students

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