A p p l i c a t i o n s o f d i f f e r e n t i a t i o n M B 1 2 Q l d - 2 19

Exercise 2A — Sketching curves 1 a 28y x= −

d 2dy xx

= −

Stationary point at d 0dyx

=

2 0

8 08

xy

− == −=

Stationary point (0, 8)

x −1 0 1 ddyx

+ 0 −

Slope / − \ Therefore (0, 8) is a local maximum turning point. b 3( ) 3f x x x= −

2'( ) 3 3f x x= − Stationary point at '( ) 0f x =

2

2

2

3 3 0

3 3

11

(1) 1 3 2( 1) 1 3 2

x

x

xx

ff

− =

=

== ±= − = −

− = − + =

Stationary points at (1, 2) and ( 1, 2)− −

x −2 −1 0 1 2 '( )f x + 0 − 0 +

Slope / − \ − / Therefore ( 1, 2)− − is a local maximum turning point

and (1, 2)− is a local minimum turning point.

c 2( ) 2 8g x x x= − ( ) 4 8g x x′ = − Stationary point at '( ) 0g x =

4 8 04 8

2(2) 8 16 8

xxx

g

− ==== − = −

Stationary point at (2, 8)−

x 1 2 3 '( )g x − 0 +

Slope \ − / Therefore (2, 8)− is a local minimum turning point.

d 2 3( ) 4 2f x x x x= − −

2'( ) 4 4 3f x x x= − − Stationary point at '( ) 0f x =

24 4 3 0x x− − =

23

2 2 2 22 33 3 3 3

4027

1327

2 3

(3 2)( 2) 0

or 2

( ) 4( ) 2( ) ( )

1

( 2) 4( 2) 2( 2) ( 2)8

x x

x

f

f

− − + =

= −

= − −

=

=

− = − − − − −= −

Stationary points at 2 133 27

( ,1 ) and ( 2, 8)− −

x −3 −2 −1 23

1

'( )f x − 0 + 0 − Slope \ − / − \

Therefore ( 2, 8)− − is a local minimum turning point

and 2 133 27

( ,1 ) is a local maximum turning point.

e 3 4( ) 4 3g x x x= −

2 3( ) 12 12g x x x′ = − Stationary points at '( ) 0g x =

2 3

2

12 12 0

12 (1 ) 00 or 1

(0) 0(1) 1

x x

x xx x

gg

− =

− == ===

Stationary points at (0, 0) and (1, 0)

x −1 0 12

1 2

'( )f x + 0 + 0 − Slope / − / − \

Therefore (0, 0) is positive point of inflection and (1, 1)

is a local maximum turning point. f

2 ( 3)y x x= +

3 2

2

3d 3 6d

x xy x xx

= +

= +

Stationary points at d 0dyx

=

23 6 03 ( 2) 0

0 or 2at 0 0at 2 4

x xx x

xx yx y

+ =+ =

= −= == − =

Stationary points at (0, 0) and ( 2, 4)−

x −3 −2 −1 0 1 ddyx

+ 0 − 0 +

Slope / − \ − /

Chapter 2 — Applications of differentiation

M B 1 2 Q l d - 2 20 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n

Therefore ( 2, 4)− is a local maximum turning point and (0, 0) is a local minimum turning point.

g 25 6y x x= − +

d 6 2dy xx

= − +

Stationary point at d 0dyx

=

6 2 02 6

3at 3 5 18 9

4

xxxx yy

− + ==== = − += −

Stationary point at (3, 4)−

x 2 3 4 ddyx

− 0 +

Slope \ − / Therefore (3, 4)− is a local minimum turning point.

h 3( ) 8f x x= +

2( ) 3f x x′ = Stationary point at '( ) 0f x =

23 00

(0) 8

xx

f

===

Stationary point at (0, 8)

x −1 0 1 ( )f x′ + 0 +

Slope / − / Therefore (0, 8) is a positive point of inflection. i 2 6y x x= − − +

d 2 1dy xx

= − −

Stationary point at d 0dyx

=

121 1 12

2 2 2254

14

2 1 02 1

at ( ) ( ) 6

6

−

− − −

− − =− =

=

= = − − +

=

=

xx

x

x y

Stationary point at 1 12 4

( , 6 )−

x −1 12

− 0

ddyx

+ 0 −

Slope / − \ Therefore 1 1

2 4( , 6 )− is a local maximum turning point.

j 4 3 23 8 6 5y x x x= − + +

3 2d 12 24 12dy x x xx

= − +

Stationary points at d 0dyx

=

3 2

2

2

12 24 12 0

12 ( 2 1) 0

12 ( 1) 00 or 1

at 0 5at 1 3 8 6 5

6

x x x

x x x

x xxx yx y

− + =

− + =

− === == = − + +=

Stationary points at (0, 5) and (1, 6)

x −1 0 12

1 2

ddyx

− 0 + 0 +

Slope \ − / − / Therefore (0, 5) is a local minimum turning point and

(1, 6) is a positive point of inflection. k 2( ) ( 27)g x x x= −

3

2

27

'( ) 3 27

x x

g x x

= −

= −

Stationary points at '( ) 0g x =

2

2

3 27 0

3( 9) 03( 3)( 3) 0

3 or 3

x

xx x

x

− =

− =− + =

= −

(3) 54

( 3) =54g

g= −

−

Stationary points at ( )3, 54− and ( )3, 54−

x −4 −3 −2 2 3 4 '( )g x + 0 − − 0 +

Slope / − \ \ − / Therefore ( 3, 54)− is a local maximum turning point and

(3, 54)− is a local minimum turning point,

l 3 24 3 2y x x x= + − −

2d 3 8 3dy x xx

= + −

Stationary points at d 0dyx

=

2

13

3 8 3 0(3 1)( 3) 0

or 3

x xx x

x

+ − =− + =

= −

1 1 1 13 2

3 3 3 36827

1427

at ( ) 4( ) 3( ) 2

2

= = + − −

= −

= −

x y

3 2at 3 ( 3) +4( 3) 3( 3) 2

16x y= − = − − − − −

=

Stationary points at 1 143 27( , 2 ) and ( 3,16)− −

x −4 −3 −2 13

1

ddyx

+ 0 − 0 +

Slope / − \ − /

A p p l i c a t i o n s o f d i f f e r e n t i a t i o n M B 1 2 Q l d - 2 21

Therefore ( 3,16)− is a local maximum turning point and 1 143 27

( , 2 )− is a local minimum turning point.

m 3( ) 12h x x= −

2'( ) 3= −h x x Stationary point at '( ) 0h x =

23 00

(0) 12

xx

h

− ===

Stationary point at (0,12)

x −1 0 1

( )h x′ − 0 − Slope \ − \

Therefore (0,12) is a negative point of inflection

n 3( ) ( 4)g x x x= −

4 3

3 2

4

( ) 4 12

x x

g x x x

= −

′ = −

Stationary points at '( ) 0g x =

3 2

2

4 12 0

4 ( 3) 00 or 3

(0) 0(3) 27

x x

x xx

gg

− =

− ==== −

Stationary points at (0, 0) and (3, 27)−

x −1 0 1 3 4 ( )g x′ − 0 − 0 +

Slope \ − \ − /

Therefore (0,0) is a negative point of inflection and (3, 27)− is a local minimum turning point.

2 a

b

c

d

e

f

g

h

i

j

k

l

m

n

3 a 3 2( ) 2 7 4f x x x x= − − −

2'( ) 3 4 7f x x x= − − Stationary points at '( ) 0f x =

2

73

3 4 7 0(3 7)( 1) 0

and 1

− − =− + =

= = −

x xx x

x x

7 7 7 73 23 3 3 3

50027

1427

at , ( ) ( ) 2( ) 7( ) 4

18

x f x= = − − −

= −

= −

Stationary point at 1 143 27

(2 , 18 )−

3 2at 1 ( ) ( 1) 2( 1) 7( 1) 4x f x= − = − − − − − −

1 2 7 4

0= − − + −=

Stationary point at (−1, 0)

x −2 −1 0 73

3

'( )f x + 0 − 0 + Slope / − \ − /

Therefore (−1, 0) is a local maximum turning point and

1 143 27(2 , 18 )− is a local minimum turning point.

b

4 a 3 2 16 16y x x x= − − +

2d 3 2 16dy x xx

= − −

Stationary point when d 0dyx

=

M B 1 2 Q l d - 2 22 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n

2

83

3 2 16 0(3 8)( 2) 0

or 2

x xx x

x

− − =− + =

= −

8 8 83 23 3 3

40027

2227

8at ( ) ( ) 16( ) 163

14

x y= = − − +

= −

= −

3 2at 2 ( 2) ( 2) 16( 2) 16

36= − = − − − − − +

=x y

Stationary points at 8 223 27

( , 14 )− and (−2, 36)

x −3 −2 −1 23

2 3

ddyx

+ 0 − 0 +

Slope / − \ − / Therefore (−2, 36) is a local maximum turning point and

2 22,

3 27(2 14 )− is a local minimum turning point.

b

5 a 4 2( ) 4g x x x= −

3( ) 4 8= −g x x x Stationary points when '( ) 0g x =

3

2

4 8 0

4 ( 2) 0

x x

x x

− =

− =

0 or 2x = ±

at 0, (0) 0

at 2 ( 2) 4

at 2, ( 2) 4

= =

= = −

= − − = −

x g

x g

x g

Stationary points at (0, 0), ( 2, 4) and ( 2, 4)− −

x −2 2− −1 0 1 2 2 '( )g x − 0 + 0 − 0 +

Slope \ − / − \ − / Therefore ( 2, 4)− − and ( 2, 4)− are local minimum

turning points and (0, 0) is a local maximum turning point.

b

6 a 4 26 8 3y x x x= − + −

3d 4 12 8dy x xx

= − +

Stationary points when d 0dyx

=

34 12 8 0

(1) 4 12 8 0x x

P− + =

= − + =

Therefore ( 1)x − is a factor

2

3

3 2

2

2

4 4 81 4 0 12 8

4 44 12 84 4

8 88 8

x xx x x

x xx xx x

xx

+ −− + − +

− −− +

− −− +− +

2

2

( 1)(4 4 8)4( 1)( 2)( 1)

4( 2)( 1)2 and 1

− + −= − + −

= + −= = −

x x xx x x

x xx

4 2

4

at 2, ( 2) 6( 2) 8( 2) 327

at 1 (1) 6(1) 8(1) 30

x y

x y

= − = − − − + − −= −

= = − + −=

Stationary points at ( 2, 27) and (1, 0)− −

x −3 −2 −1 0 1 2 ddyx

− 0 + + 0 +

Slope \ − / / − / Therefore ( 2, 27)− − is a local minimum turning point

and (1, 0) is a positive point of inflection. b Show that (1, 0) lies on the curve

4 2at 1 (1) 6(1) 8(1) 3

0x y= = − + −

=

Therefore (1, 0) lies on the curve c

7 a 4 3 25 6y x x x= + − −

3 2

2

d 4 3 10d

(4 3 10)(4 5)( 2)

y x x xx

x x xx x x

= + −

= + −= − +

Stationary points when d 0dyx

=

54

(4 5)( 2) 0

0, , 2

x x x

x

− + =

= −

5 5 54 3 24 4 4

2411256107256

at 0, 65at , ( ) ( ) 5( ) 64

9

x y

x y

= = −

= = + + −

= −

= −

A p p l i c a t i o n s o f d i f f e r e n t i a t i o n M B 1 2 Q l d - 2 23

4 3 2at 2, ( 2) ( 2) 5( 2) 6

18= − = − + − − − −

= −x y

Stationary points at 5 1074 256

(0, 6), , 9− − and ( 2, 18)− −

x −3 −2 −1 0 1 1

41 2

ddyx

− 0 + 0 − 0 +

Slope \ − / − \ − / Therefore ( 2, 18)− − and 1 107

4 256(1 , 9 )− are local

minimum turning points and (0, 6)− is local maximum turning point.

b To find y-intercept, let x = 0 6y = − c

8 a

4 2

3

2

( )

'( ) 4 2

2 (2 1)

= −

= −

= −

f x x x

f x x x

x x

2

2

Stationary points when '( ) 0

2 (2 1) 010 or2

12

=

− =

= =

= ±

f x

x x

x x

x

at 0, (0) 01 1 1at ,

42 21 1 1at ,

42 2

= =

= = − = − − = −

x f

x f

x f

x −1

12

− 12

− 0 12

12

1

( )f x′ − 0 + 0 − 0 + Slope \ − / − \ − /

Therefore 1 1,

42 − −

and 1 1,42

−

are local

minimum turning points and (0, 0) is a local maximum turning point.

At 0 0x y= = y-intercept is 0

b

3 2( ) 3f x x x= −

2'( ) 3 6

3 ( 2)f x x x

x x= −= −

Stationary points when ( ) 0f x′ =

3 ( 2) 00 and 2

at 0, ( ) 0at 2, ( ) 8 12 4

x xx

x f xx f x

− ==

= == = − = −

Stationary points are (0, 0) and (2, −4)

x −1 0 1 2 3 ( )f x′ + 0 − 0 +

Slope \ − \ − /

Therefore, (0, 0) is a local maximum turning point and (2, −4) is a local minimum turning point.

at 0 0, -intercept 0= = =x y y

c

3 4( ) 3g x x x= +

2 3

2

( ) 3 12

3 (1 4 )

g x x x

x x

′ = +

= +

Stationary points at ( ) 0g x′ =

2

14

1 1 1 13 44 4 4 4

1256

3 (1 4 ) 0

0 and

at 0, (0) 0

at , ( ) ( ) 3( )

x x

x x

x g

x g

+ =

= = −

= =

= − − = − + −

= −

Stationary points at 1 1 14 4 256(0, )( , )− − −

x −1 14

− 18

− 0 1

( )g x′ − 0 + 0 + Slope \ − / − /

Therefore 1 14 256

( , )− − is a local minimum turning point

and (0, 0) is a local maximum turning point. at 0, ( ) 0, -intercept 0x g x y= = =

d

3 2( ) 4 4g x x x x= − +

2'( ) 3 8 4

(3 2)( 2)g x x x

x x= − += − −

M B 1 2 Q l d - 2 24 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n

Stationary point when ( ) 0g x′ =

23

2 32 53 27 27

(3 2)( 2) 0

, 2

at , ( ) 1

at 2, ( ) 0

− − =

=

= = =

= =

x x

x

x g x

x g x

Stationary points at 2 53 27( ,1 ) and (2, 0)

x 0 23

1 2 3

( )g x′ + 0 − 0 + Slope / − \ − /

Therefore 2 53 27

( ,1 ) is a local maximum turning point and

(2, 0) is a local minimum turning point. at 0, ( ) 0, -intercept 0x g x y= = = e

3 2( ) 4 11 30h x x x x= − − +

2( ) 3 8 11

(3 11)( 1)h x x x

x x′ = − −

= − +

Stationary point when ( ) 0h x′ = (3 11)( 1) 0x x− + =

11311 11 11 113 23 3 3 3

40027

2227

or 1

at , ( ) ( ) 4( ) 11( ) 30

14

at 1, ( ) 36

= = −

= = − − +

= −

= −

= − =

x x

x h x

x h x

Stationary points at 2 22,

3 27(3 14 )− and (−1, 36)

x −2 −1 0 23

3 4

( )h x′ + 0 − 0 + Slope / − \ − /

Therefore (−1, 36) is a local maximum turning point and

2 223 27

(3 , 14 )− is a local minimum turning point.

at 0, ( ) 30, -intercept 30= = =x h x y

f

( ) ( 3)( 5)h x x x x= + −

3 2

2

2 15

( ) 3 4 15(3 5)( 3)

x x x

h x x xx x

= − −

′ = − −= + −

Stationary points when ( ) 0h x′ =

53

5 223 27

(3 5)( 3) 0

or 3

at , ( ) 14

at 3, ( ) 36

+ − =

= −

= − =

= = −

x x

x

x h x

x h x

Stationary points at 5 223 27

(3, 36) and ( ,14 )− −

x −2 53

− 0 3 4

( )h x′ + 0 − 0 + Slope / − \ − /

Therefore 2 22

3 27( 1 ,14 )− is a local maximum turning point

and (3, 36)− is a local minimum turning point. at 0, ( ) 0, -intercept 0= = =x h x y g

4 2( ) 2 1f x x x= − +

3( ) 4 4f x x x′ = −

24 ( 1)

4 ( 1)( 1)x xx x x

= −= − +

Stationary point when ( ) 0f x′ =

4 ( 1)( 1) 0

0,1, 1− + =

= −x x x

x

at 0, ( ) 1at 1, ( ) 0at 1, ( ) 0

= == == − =

x f xx f xx f x

Stationary points at ( 1, 0), (0,1) and (1, 0)−

x − 2 − 1 12

− 0 12

− 1 2

'( )h x − 0 + 0 − 0 + Slope \ − / − \ − /

Therefore ( 1, 0)− and (1, 0) are local minimum turning points and (0, 1) is a local maximum turning point.

at 0, ( ) 1,x f x= = y-intercept = 1

h

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

3

2( ) 3 1

x x

f x x

= +

′ = +

Stationary points at ( ) 0f x′ =

2

2

12 3

3 1 0

3 1

no stationary points

x

x

x

+ =

= −

= −

let 0, ( ) 0 -intercept 0= = =x f x y

A p p l i c a t i o n s o f d i f f e r e n t i a t i o n M B 1 2 Q l d - 2 25

i

3 2( ) 9 24 20g x x x x= + + +

2( ) 3 18 24

3( 2)( 4)g x x x

x x′ = + +

= + +

Stationary points when ( ) 0g x′ =

3( 2)( 4) 02, 4

at 2, ( ) 0at 4, ( ) 4

+ + == − −

= − == − =

x xx

x g xx g x

Stationary points at (−2, 0) and (−4, 4)

x −5 −4 −3 −2 −1 ( )h x′ + 0 − 0 +

Slope / − \ − / Therefore (−4, 4) is a local maximum turning point and

(−2, 0) is a local minimum turning point at 0, ( ) 20, -intercept 20.x g x y= = =

j

2 3( ) ( 1)h x x= − 2 2( ) 6 ( 1)h x x x′ = − Stationary point when ( ) 0h x′ =

26 ( 1) 00,1, 1

at 0, ( ) 1at 1, ( ) 0at 1, ( ) 0

− == −

= = −= == − =

x xx

x h xx h xx h x

Stationary points at (0, −1) (1, 0) (−1, 0)

x −2 −1 12

− 0 12

1 2

( )h x′ − 0 − 0 + 0 + Slope \ − \ − / − /

Therefore (−1, 0) is a negative point of inflection (1, 0) is a positive point of inflection and (0, −1) is a local minimum turning point.

at 0, ( ) 1, -intercept 1= = − = −x h x y 9 ( ) 0 when 2f x x′ < >

( ) 0 when 2f x x′ > <

x 1 2 3 Slope / − \

at 2, '( ) 0, local maximum= =x f x Answer is B.

10 3 2( ) 8 3= + − −f x x x x

2( ) 3 2 8f x x x′ = + −

2stationarypoints when 3 2 8 0x x+ − =

4 3

(3 4)( 2) 0

or 2

x x

x

− + =

= −

Answer is A. 11 4 3y x x= +

3 2d 4 3dy x xx

= +

has stationary points when 3 24 3 0x x+ =

2

34

(4 3) 0

0 or

x x

x x

+ =

= = −

x −1 34

− 12

− 0 1

ddyx

− 0 + 0 +

Slope \ − / − /

local minimum at 34

x = −

positive point of inflection at x = 0 Answer is C.

12 Turning point (2, 1)

2

2

( 2) 1

9 (0 2) 18 42

y a x

aa

a

= − +

= − +==

2Equation 2( 2) 1y x= = − + Answer is E.

13 a 3 local minx = − 0 local max=x b 2 local max= −x

1 local min4 local max

==

xx

c 2x = − negative point of inflection 3x = local min d 5x = − local min 2x = positive point of inflection e 3x = − local max 0x = local min 2x = local max f 1x = local max 5x = local min

14 2( ) 4 3f x x x= − + ( ) 2 4f x x′ = −

Stationary point when 2 4 0x − =

2 4

2at 2, ( ) 1

xxx f x

=== = −

x 1 2 3 ( )f x′ − 0 +

Slope \ − /

at 2, '( ) 0< <x f x i.e. graph is decreasing at 2, '( ) 0x f x> > i.e. graph is increasing

15 a i 1 3 23

( ) 2 2f x x x= + +

2

2

( ) 4

( ) 0 means 4 0( 4) 0

0 or 4

f x x x

f x x xx x

x

′ = +

′ = + =+ =

= −

x-intercepts of ( ) are 0 and 4f x′ −

M B 1 2 Q l d - 2 26 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n

(0) 0

-intercept of ( ) is 0upright parabola

fy f x

′ =′

ii ( )f x is increasing (i.e. ( )f x′ is above the x-axis) when 4 and 0x x< − >

iii ( )f x is decreasing (i.e. ( )f x′ is below the x-axis) when 4 0x− < <

b i 3 2( ) 2 7 5g x x x x= + − −

2

2

73

( ) 3 4 7( ) 0 gives -intercepts of '( )

3 4 7 0(3 7)( 1) 0

,1

g x x xg x x g x

x xx x

x

′ = + −′ =

+ − =+ − =

= −

73-intercepts of ( ) are ( ,0) and (1,0)

-intercept of ( ) is (0, 7)upright parabola.

x g x

y g x

′ −

′ −

ii ( )g x is increasing

1 3(i.e. ( ) is above the -axis) when 2 and 1g x x x x′ < − >

iii ( ) is decreasing (i.e. ( ) is below the -axis) g x g x x′

13

when 2 1− < <x

c i 4 3 2( ) 4 4h x x x x= + +

3 2( ) 4 12 8h x x x x′ = + +

3 2

2

( ) 0 gives -intercepts of ( )

4 12 8 0

4 ( 3 2) 04 ( 2)( 1) 0

0, 1, 2

h x x h x

x x x

x x xx x x

x

′ ′=

+ + =

+ + =+ + =

= − −

-intercepts are (0, 0)( 1, 0)( 2, 0)-intercept is (0, 0)

Cubic graph

xy

− −

ii ( ) is increasing (i.e. ( ) is above the -axis)h x h x x′

when 2 1and 0− < < − >x x

iii ( ) is decreasing (i.e. ( ) is below the -axis)h x h x x′

when 2 and 1 0x x< − − < <

16 ( ) 0 if 2 and 3f x x x′ = = − =

i.e. stationary points at 2 and 3= − = −x x

( ) 0 if 2 3f x x′ < − < < i.e. gradient is negative between −2 and 3 ( ) 0 all other valuesf x x′ >

i.e. gradient is positive everywhere else.

2 and 3x x< − >

17 a 3 2( ) = + +f x x ax bx

2

2

( ) 3 2at 2, ( ) 0

(2) 3(2) 2 (2) 012 4 0

4 12.........(1)

f x x ax bx f x

f a ba ba b

′ = + +′= =

′ = + + =+ + =

+ = −

4,3

4 4 423 3 3

at ( ) 0

( ) 3( ) 2 ( ) 0

16 8 03 3

16 8 3 08 3 16.............(2)

x f x

f a b

a b

a ba b

′= − =

′ − = − + − + =

− + =

− + =− + = −

Solve simultaneously

4 12 ........(1)

8 3 16 ........(2)1 2

a ba b

+ = −− + = −×

8 2 248 3 16

5 408

a ba b

bb

+ = −− + = −

= −= −

4 124 8 12

4 411, 8

+ = −− = −

= −= −= − = −

a ba

aaa b

b 2'( ) 3 2 8f x x x= − −

x −2 11

3− 0 2 3

( )f x′ + 0 − 0 + Slope / − \ − /

1at 1 ,3

x = − there is a local maximum

at 2,x = there is a local minimum

A p p l i c a t i o n s o f d i f f e r e n t i a t i o n M B 1 2 Q l d - 2 27

18 a 4 2( )f x x ax b= + +

3

4 2

( ) 4 2at 1, (1) 4 2 0

2 42

at 1, (1) 1 2(1) 41 4

5

f x x axx f a

aa

x f bbb

′ = +′= = + =

= −= −

= = + − + =− + =

=

b 34 4 0x x− =

24 ( 1) 0

0, 1,1x x

x− =

= −

other stationary points are at (−1, 4) and (0, 5). c

x −2 −1 12

− 0 12

1 2

( )f x′ − 0 + 0 − 0 + Slope \ − / − \ − /

(−1, 4) is a local min (0, 5) is a local max (1, 4) is a local min

Exercise 2B — Equations of tangents and normals 1 2y x x= +

d 2 1d

dat 2 5d

y xx

yxx

= +

= =

gradient of tangent is 5. Equation of tangent at (2, 6)

1 1( )6 5( 2)

5 10 65 4

− = −− = −

= − += −

y y m x xy x

y xy x

2 2 5 6y x x= + − Curve crosses the x-axis when y = 0

2 5 6 0( 6)( 1) 0

6 or 1d 2 5d

x xx x

xy xx

+ − =+ − =

= −

= +

at 6 gradient of tangent 7at 1 gradient of tangent 7Equation of tangent at ( 6, 0)

xx

= − = −= =

−

1 1( )0 7( 6)

7 42 or 7 42 0

y y m x xy x

y x x y

− = −− = − − −

= − − + + =

1 1

Equation of tangent at (1, 0)( )

0 7( 1)7 7

y y m x xy x

y x

− = −− = −

= −

3 23 5 4y x x= − +

d 6 5dat 1 gradient of tangent 1

gradient of normal 1

y xxx

= −

= == −

1 1

Equation of normal( )y y m x x− = −

at 1, 3 5 4 2

2 1( 1)3 or 3

x yy x

y x x y

= = − + =− = − −

= − + + =

4 21 3 72

= + −y x x

Curve crosses the y-axis when x = 0, y = −7

13

d 3d

at 0 gradient of tangent 3

gradient of normal

= +

= =

= −

y xxx

1 1

13

13

Equation of normal( )

7 ( 0)

7

3 21 or 3 21 0

y y m x x

y x

y x

y x x y

− = −

− − = − −

+ = −

+ = − + + =

5 a 2 1y x= +

12

at 1, 2d 2d

gradient of tangent at 1 is 2

gradient of normal at 1 is

x yy xx

x

x

= =

=

=

= −

i Eq. of tangent

2 2( 1)2

y xy x

− = −=

ii

12

Eq. of normal

2 ( 1)

2 4 12 5

y x

y xx y

− = − −

− = − ++ =

b 3 6= −y x x

2

16

at 2, 4d 3 6dgradient of tangent at 2 is 6

gradient of normal at 2 is

Eq. of tangent4 6( 2)

6 16

x yy xx

x

x

y xy x

= − =

= −

= −

= − −

− = − −= +

i

16

Eq. of normal

4 ( 2)

6 24 26 22

y x

y xx y

− = − − −

− = − −+ =

ii

c 1x

y =

12

2

14

at 2,

d 1d

gradient of tangent at 2 is

gradient of normal at 2 is 4

x y

yx x

x

x

= =

= −

= −

=

1 12 4

Eq. of tangent

( 2)

4 2 24 4

y x

y xx y

− = − −

− = − ++ =

i

M B 1 2 Q l d - 2 28 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n

12

12

Eq. of normal

4( 2)

4 8

2 8 15

y x

y x

y x

− = −

= − +

= −

ii

d 2( 1)( 2)y x x= − +

3 2

2

at 1, 6

2 2d 3 2 2d

x y

y x x xy x xx

= − = −

= − + −

= − +

17

17

17

gradient of tangent at 1 is 7

gradient of normal at 1 is

Eq. of tangent6 7( 1)

6 7 77 1

Eq. of normal

6 ( 1)

6 ( 1)

7 42 17 43 0

x

x

y xy x

y x

y x

y x

y xx y

= −

= − −

− − = − −+ = +

= +

− − = − − −

+ = − +

+ = − −+ + =

i

ii

e 12= =y x x

12

at 4, 2

d 1 1d 2 2

−

= =

= =

x y

yx x x

14

gradient of tangent at 4 is

gradient of normal at 4 is 4Eq. of tangent

x

x

=

= −i

14

2 ( 4)

4 8 44 4

y x

y xy x

− = −

− = −= +

Eq. of normalii

2 4( 4)

4 184 18

y xy x

x y

− = − −= − +

+ =

f y = 2 3x +

12

12

(2 3)at 3, 3d 1(2 3)d 2 3

−

= += =

= + =+

xx yy xx x

13

gradient of tangent at 3 is x =

gradient of normal at 3 is 3x = − Eq. of tangenti

133 ( 3)

3 9 33 6

y x

y xy x

− = −

− = −= +

Eq. of normalii

3 3( 3)

3 123 12

y xy x

x y

− = − −= − +

+ =

g ( 2)( 1)y x x x= + −

3 2

2

at 1, 2

2d 3 2 2d

x y

y x x xy x xx

= − =

= + −

= + −

gradient of tangent at 1 is 1gradient of normal at 1 is 1

xx

= − −= −

Eq. of tangenti

2 1( 1)

11

y xy x

x y

− = − − −= − +

+ =

Eq. of normalii

2 1( 1)

3y x

y x− = − −

= +

h 3 23 4y x x x= − +

2

at 0, 0d 3 6 4d

= =

= − +

x yy x xx

14

gradient of tangent at 0 is 4

gradient of normal at 0 is

Eq. of tangent

x

x

=

= −

i

0 4( 0)

4− = −

=y x

y x

Eq. of normalii

14

0 ( 0)

44 0

− = − −

= −+ =

y x

y xx y

i 3 22 6 2y x x x= + − +

2

at 1, 1d 6 2 6d

x yy x xx

= = −

= + −

12

gradient of tangent at 1 is 2

gradient of normal at 1 is

Eq. of tangent

x

x

=

= −

i

1 2( 1)

2 3y x

y x− − = −

= −

12

Eq. of normal

1 ( 1)

2 2 12 1 0

y x

y xx y

− − = − −

+ = − ++ + =

ii

6 a 4(2 3)y x= +

3d 8(2 3)dy xx

= +

at 1,x = −

18

gradient of tangent 8

gradient of normal

=

= −

Eq. of normal

181 ( 1)

8 8 18 7 0

y x

y xx y

− = − − −

− = − −+ − =

Answer is B.

A p p l i c a t i o n s o f d i f f e r e n t i a t i o n M B 1 2 Q l d - 2 29

b Gradient parallel to x-axis means gradient = 0

3

32

d 8(2 3) 0d

2 3 0

= + =

+ =

= −

y xx

x

x

Answer is C. 7 2( ) 4 1f x x x= + +

Parallel to 2 3 means'( ) 2'( ) 2 4 22 2

1

y xf xf x x

xx

= +== + == −= −

at 1, ( ) 2x f x= − = − Eq. of tangent

2 2( 1)2 2 2

2

− −− = −+ = +

=

y xy x

y x

8 2

211

xyx

+=−

at 0, 1x y= = − Quotient Rule

2 2

2 2

3

d ( 1)(2 ) (2 )( 1)d ( 1)

2

y x x x xx x

x

− − +=−

=32 2x x− −

2 2

2 2

2( 1)

4( 1)

xx

xx

−−

−=−

at x = 0, gradient of tangent = 0 Eq. of tangent

1 0( 0)1 0

1

y xy

y

− − = −+ =

= −

9 Eq. of tangent is y = −2x − 2.75

Exercise 2C — Maximum and minimum problems when the function is known 1 a 25 2 10c x x= − +

d 5 4d

c xx

= −

b 3 22 5p n n n= + −

13 2 2

12 2

2

2 5

d 53 4d 2

53 42

n n n

p n n nn

n nn

−

= + −

= + −

= + −

c 4 31 24 3

v h h= −

3 2d 2d

v h hh

= −

2 3 23 5 0 5Q n n n= − + ≤ ≤ (e) from inspection, min value is at (2, 0) Answer is D.

3 From inspection, max value is at x = 5 Answer is C.

4 1 32 96 600P n n= − + +

a 2d 3 96 0 for max or mind 2p nn

= − + =

2

2

3 192 0

64 0( 8)( 8) 0

8 or 8disregard 8

n

nn n

nn

− + =

− =− + =

= −= −

b 3

8 workers needed for max profit/weekat 8

1(8) (8) 96(8) 6002

1112max profit/week $1112

n

P

⇒=

= − + +

==

5 1 25

8 100 per rope, 0c x x x= − + >

25

d 2 8 0 for local max or mind 5

8

20

c xxx

x

= − =

=

=

x 19 20 21 ddcx

− 0 +

Slope \ − /

At x = 20 a local min occurs 20 m is the length of the cheapest tow rope that can be

produced.

6 3 212 21 105, 0P x x x x= − + + ≥

2

2

2

'( ) 3 24 21for min or max

3 24 21 0

3( 8 7) 03( 7)( 1) 0

7 or 1

= − +

− + =

− + =− − =

=

P x x x

x x

x xx x

x

x 0 1 2 7 8 '( )p x + 0 − 0 +

Slope / − \ − / When x = 1 a local max occurs and when x = 7 a local

min occurs. at 1, ( ) 115, at 7, ( ) 7x p x x p x= = = =

at 0 '( ) 105x p x= = 1x h= for maximum number of people 7x h= for minimum number of people

7 1 3 23

5 75 500, 5N x x x x= − + + + ≥ −

a 2'( ) 10 75N x x x= − + + Min or max when '( ) 0N x =

2

2

10 75 0

10 75 0( 15)( 5) 0

15 or 5

− + + =

− − =− + =

= −

x x

x xx x

x

23

at 15, 1625

at 5, 291

x N

x N

= =

= − =

x −6 −5 −4 14 15 16 '( )N x − 0 + + 0 −

Slope \ − / / − \

5 Cx = − ° for min number of rabbits

M B 1 2 Q l d - 2 30 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n

15 Cx = ° for max number of rabbits. b Minimum rabbits is 291. Maximum rabbits is 1625.

8 212 6 0.01(20 ) , 0 20L t t t= + + − ≤ ≤

'( ) 6 0.02(20 )

6 .4 0.020.02 5.6

L t tt

t

= − −= − += +

for min or max, '( ) 0L t = a i at birth, 0t =

212 0 0.01(20)

16 cm= + +=

L

ii at 20 weeks, 20t =

212 6 20 0.01(20 20)12 120132 cm

= + × + −= +=

L

b Rate of growth = L'(t) R = L'(t) = 0.02t + 5.6 c max or min growth rate is when

0, 0.02(0) 5.6 5.6 cm wk20, 0.02 20 5.6 6 cm/wk

t Rt R

= = + == = × + =

6 cm/wk = max. growth rate 5.6 cm/wk = min. growth rate.

9 40 25 200 2P n n= + − − a For max. or min. profit ( ) 0P n′ =

20( ) 2 025

P nn

′ = − =+

20 22510 25

100 2575

nn

nn

=+

= += +=

n 74 75 76 ( )P n′ + 0 −

Slope / − \

n = 75 is a local max. b ( 75) 40 75 25 200 2 75P − = + − − ×

400 200 15050

= − −=

Maximum profit is $50 per item. c Total profit = 75 × 50 = $3750.

Exercise 2D — Maximum and minimum problems when the function is unknown 1 10x y+ =

2

10( )

(10 )

10

y xP x x y

x x

x x

= −= ×= × −

= −

'( ) 10 2P x x= − Stationary point when '( ) 0P x =

10 2 0

2 105

xxx

− ===

x 4 5 6 '( )P x + 0 −

Slope / − \

at 5, 5x y= =

2 8x y+ =

3 2

3 2

8

( )

(8 )

y x

S x x y

x x

= −

= +

= + −

2

2

'( ) 3 2(8 )

3 2 16

S x x x

x x

= − −

= + −

Stationary point when ( ) 0S x′ =

2

83

3 2 16 0(3 8)( 2) 0

or 2

x xx x

x

+ − =+ − =

= −

Disregard 83

= −x as the numbers must be positive.

2, 6x y= =

3 a let x = width of frame and let y = length of frame Perimeter = 2 2 120x y+ =

2 120 2

60y xy x

= −= −

b ( )A x xy=

2

(60 )

60

x x

x x

= −

= −

c for maximum area '( ) 0A x =

( ) 60 260 2 0

2 6030

30

A x xxxx

y

′ = −− =

==⇒ =

Length and width are each 30 cm d Area

2

30 30

900 cm

== ×

=

xy

4 let x = width of paddock and let y = length of paddock Perimeter = 2 2 400x y+ =

2 400 2

200y xy x

= −= −

2

( )(200 )

200

A x xyx x

x x

== −

= −

Max area when ( ) 0A x′ =

( ) 200 2

200 2 02 200

A x xxx

′ = −− =

=

100x = 100y⇒ = Largest area = 100 × 100 = 10 000 2m

5 let l = length of tube and let r = radius of tube length + circumference = 120 cm

2 120120 2

l rl r

ππ

+ == −

Volume 2r lπ=

2

2 2 3

(120 2 )

120 2

r r

r r

π ππ π

= −

= −

for max volume, ( ) 0V x′ =

A p p l i c a t i o n s o f d i f f e r e n t i a t i o n M B 1 2 Q l d - 2 31

2 2

2 2

( ) 240 6

240 6 06 (40 ) 00 or 40 0

4040

V x r r

r rr r

r rr

r

π ππ π

π πππ

π

′ = −

− =− =

= − ==

=

disregard 040if

40then 120 2

120 8040

Circumference = 2402

8040Radius = cm = 12.73 cm

Length = 40 cmCircumference = 80 cm

r

r

l

r

π

ππ

π

ππ

π

=

=

= − ×

= −=

= ×

=

6

a 4 2 4 4 18x x L× + × + × =

92

8 4 4 184 18 12

3

x x LL x

L x

+ + == −

= −

b 2( ) 2V x x L=

2

2 3

92 32

9 6

x x

x x

= −

= −

c for max volume, ( ) 0V x′ =

2

2

'( ) 18 18

18 18 018 (1 ) 0

0 or 1

V x x x

x xx x

x x

= −

− =− =

= =

disregard x = 0 for max volume x = 1 m

92923 12 2

2 2 m

3 1

3

1 m

=

= − ×

= −

= =

x

L

Edges are 1 m, 2 m and 121 m for max volume.

d Max volume = 1 2 1.5× × 33 m=

7 let length of basex = let height of cuboidh = a Surface area of cuboid = S

2

2

2

2 4 200

4 200 2

200 24

502

S x xh

xh x

xhxx

x

= + =

= −

−=

= −

b Volume 2x h=

2

3

502

502

xxx

xx

= −

= −

c for max volume, ( ) 0V x′ =

2

3 22

10023

3( ) 50 02

50

103

V x x

x

x

x

′ = − =

=

=

= ±

disregard 103

−

10 5.77353

10 5.77353

x

h

⇒ = ≈

⇒ = ≈

Max volume 10 10 10 1000 393 3 3

= × × =

3192 cm to the nearest unit≈ Verify max volume

x 5 5.7735 6 '( )V x + 0 −

Slope / − \

8

let x = length of frame let r = radius of semicircle Area of frame = 1 2

22rx rπ+

Framework is 12 2 2 2 300 cmr x rπ× + + =

12

2 2 3002 300 2

150

+ + == − −

= − −

r r xx r r

x r r

ππ

π

1 122 21 2 2 22

12 22

( ) 2 (150 )

300 2

300 2

A r r r r r

r r r r

r r r

π π

π π

π

= + − −

= + − −

= − −

M B 1 2 Q l d - 2 32 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n

'( ) 300 4A r r rπ= − − For max area '( ) 0=A r

300 4 0(4 ) 300

300 42 cm4

r rr

r

ππ

π

− − =+ =

= ≈+

115021150 (42) 422

42

x r rπ

π

⇒ = − −

= − −

≈

Area = 1 22

2 42 42 42π× × + ×

26298.9 cm≈ Verify area is max

x 40 42 44 '( )A f + 0 −

Slope / − \

9 let x be the length of the cut out corner

Therefore the length of the tray is 50 2 cmx− and the

width is 40 2 cmx−

2 3

( ) (40 2 )(50 2 )

2000 180 4

V x x x x

x x x

= − −

= − +

for max volume, ( ) 0V x′ =

2

2

2

( ) 2000 360 12

12 360 2000 0

4(3 90 500) 0

V x x x

x x

x x

′ = − +

− + =

− + =

Solve using quadratic formula

290 ( 90) 4(3)(500)6

90 21006

22.64 or 7.36

x± − −

=

±=

≈

x 7 7.36 8 22 22.64 23 '( )V x + 0 − − 0 +

Slope / − − \ − / x = 7.36 gives a max value disregard x = 22.64 Max volume 2 32000(7.36) 180(7.36) 4(7.36)= − +

36564 cm≈

10 Distance walked through clear land 3 kmx= − Let distance walked through bush land = y km. Using Pythagoras

2 2 2

2

2

4

y x

y x

= +

= +

distanceTotal time taken = through clear landspeed

distanceplus through bush landspeed

12

2

2

3( )5 3

3 45 3

3 1 (4 )5 5 3

x yT x

x x

x x

−= +

− += +

= − + +

for min time ( ) 0T x′ =

121 1 1 2

5 3 2

2

'( ) ( )(2 )(4 )

15 3 4

T x x x

x

x

−= − + +

= − ++

2

2

2

22

22

2

1 053 4153 4

5 4325 4

925 4

916 4

9

x

xx

xx x

x x

x x

x

− =+

=+

= +

= +

− =

=

36216

6432

x

x

=

= ±

= ±

disregard 32x = −

Verify min

x 1 121 2

( )T x′ − 0 + Slope \ − /

1

21x = gives min time

1.5 km.x =

11 2 dollars50

1000 hrvc = +

distanceTime =speed

a 800 hrs=Tv

b Cost = cost × timehr

2

1

800501000

40 000 810

40 000 0.8

vv

vv

C v v−

+ ×

= +

= +

c Economical = Min cost when '( ) 0C v =

240 000'( ) 0.8C v

v−= +

240 000 0.8 0

v− + =

A p p l i c a t i o n s o f d i f f e r e n t i a t i o n M B 1 2 Q l d - 2 33

2

2

40 0000.8

50 000

50 000

v

v

v

=

=

= ±

disregard 50 000v = −

verify 50 000 223.61 is a minv = ≈

v 220 223.61 230 ( )c v′ − 0 +

Slope \ − / The most economical speed is 223.61 km/hr

12

Max cube inside a sphere means diagonal corners of the

cube touch the sphere. Using Pythagoras

2 2 2

2 2

2

2

24 ( 2 )

576 2

576 3

192

19213.86 cm

x x

x x

x

x

x

= +

= +

=

=

= ±≈ ±

disregard 13.86 cmx = − Therefore 13.86 cmx =

13

Using Pythagoras

12 2 222

2

22

8 ( )

644

644

r h

hr

hr

= +

= +

= −

Volume of cylinder = 2r hπ

2

3

( ) 644

644

hv h h

hh

π

ππ

= −

= −

for max volume, v′(h) = 0

2

2

256 23

2

3'( ) 64 04

3644

85.39.238

hv h

h

h

hh

ππ

ππ

= − =

=

=

≈≈ ±

disregard 9.238h = − verify 9.238 is a maxh =

h 9 9.238 10 '( )v h + 0 −

Slope / − \

Max volume 364 9.238 9.2384ππ= × − ×

31238 cm≈ 14 Let (1,0)P = and

let ( , )Q x y= As Q is on the line 2 3y x= + then ( ,2 3)Q x x= +

2 22 1 2 1( ) ( ) ( )d x x x y y= − + −

12

2 2

2 2

2

2

( 1) (2 3 0)

2 1 4 12 9

5 10 10

(5 10 10)

x x

x x x x

x x

x x

= − + + −

= − + + + +

= + +

= + +

121 2

2'( ) (10 10)(5 10 10)d x x x x −= + + +

2

5 5

5 10 10

x

x x

+=+ +

for min or max, '( ) 0d x =

2

5 5 05 10 10

x

x x

+= =+ +

then 5 5 0x + = 1x = − Verify

x −2 −1 0 '( )d x − 0 +

Slope \ − /

1x = − gives min distance

2( 1) 5( 1) 10( 1) 10d − = − + − +

5 10 10= − + 5= units

15 Let (5,0)P = and Let ( , )Q x y=

As Q is on the curve 2,y x= then

2( , )Q x x=

2 22 1 2 1( ) ( ) ( )d x x x y y= − + −

2 2 2

2 4

14 2 2

( 5) ( 0)

10 25

( 10 25)

x x

x x x

x x x

= − + +

= − + +

= + − +

1

1 3 4 2 22'( ) (4 2 10)( 10 25)d x x x x x x

−= + − + − +

M B 1 2 Q l d - 2 34 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n

3

4 2

2 5

10 25

x x

x x x

+ −=+ − +

for Min or Max '( ) 0d x =

3

4 2

2 5 010 25

x x

x x x

+ −= =+ − +

32 5 0x x+ − = Use graphic calculator to solve graph 32 5y x x= + −

adjust window settings and find where the graph crosses the x axis

solution is 1.235x = Verify

x 1 1.235 2 ( )d x′ − 0 +

Slope \ − / 1.235x = gives a minimum

2 4(1.235) 1.235 10(1.235) 25 1.235d = − + +

16.50154.06 units

=≈

16 1 22

4y x= −

Let (0,0)P = ( , )Q x y= As Q is on the curve, then 1 2

2( , 4)Q x x= −

12 2 22

( ) ( 0) ( 4 0)d x x x= − + − −

12 4 24 4 16x x x= + − +

12

1 4 24

1 4 24

3 16

( 3 16)

x x

x x

= − +

= − +

121 13 4 2

2 4( ) ( 6 )( 3 16)d x x x x x −′ = − − +

3

1 4 24

6

2 3 16

x x

x x

−=− +

for max or min '( ) 0d x =

3

1 4 24

2 6 02 3 16

L x

x x

−= =− +

3

2

6 0

( 6) 0

x x

x x

− =

− =

2

2

0 or 6 0

6

6

x x

x

x

= − =

=

= ±

For Max or min

x −3 6− −1 0 1 6 3 ( )d x′ − 0 + 0 − 0 +

Slope \ − / − \ − /

6 or 6x = − give Min values

1 4 24( 6) ( 6) 3( 6) 16d = − +

7=

1 4 24( 6) ( 6) 3( 6) 16d − = − − − +

7= The closest distance the comet gets to Mars

is 7 million km.

Investigation — Cross-country run

1 let y PC=

2 2 2

2 2

2 2

2.7

2.7 (in km)

2700 (in m)

y x

y x

y x

= +

= +

= +

2 Time distancespeed

=

2 22700

2xT +=

3 Total distance 2 2(3800 ) 2700x x= − + +

4 Total time 2 23800 2700

5 2x x− += +

5 Min or Max means ( ) 0T x′ =

12

12

2

1 1 1 2 25 2 2

1( ) 760 ( 2700 )5 2

'( ) ( )(2 )( 2700 )

xT x x

T x x x

2

−

= − + +

= − + +

2 2

15 2 2700

x

x= − +

+

2 2

2 2

22 2

152 2700

5 27002

25 27004

x

xx x

x x

=+

= +

= +

22

22

2

21 27004

4 270021

1388571.431178.38

x

x

xx

=

×=

== ±

disregard 1178.38x = − Verify min

x 1100 1178.38 1200 ( )T x′ − 0 +

Slope \ − / Minimum time at 1178.38 mx =

6 121178.38 1 2 2

5 2(1178.38) 760 (1178.38 2700 )T = − + +

1997.30 sec.= Convert to mins ( 60)÷

1997.30 60 33.29 mins

= 33 min 17 mins÷ =

7 Peter’s time

2 2(3800 1500) 270015005 2

12 580 000300

22073.41 sec

− += +

= +

=

A p p l i c a t i o n s o f d i f f e r e n t i a t i o n M B 1 2 Q l d - 2 35

34.56 min= Time difference 1.27= mins longer

or 76 sec longer.

Exercise 2E — Rates of change

1 a ddVr

b ddSh

c ddAt

d ddCx

e ddIr

f ddvt

2 2A rπ=

a d 2dA rr

π=

b at 2d10 m, 20 m /mdArr

π= =

3 4 33

V rπ=

a 2d 4dV rr

π= deflating

2d 4dV rr

π=> = −

b at 5r = cm, 3d 100 cm /cmdVr

π= −

4 Surface area of cube 26S x= =

d 12d

s xx

= melting

d 12d

s xx

⇒ = −

at 0.5x = cm, 2d 6 cm /cmd

sx

= −

5 210 20 5h t t= + − a Initial height is when 0t = 10h = m b 2(3) 10 20(3) 5(3) 25 mh = + − = (0) 10 mh =

average rate of change 25 103 0

−−

=

153

=

= 5 m/s c '( ) 20 10h t t= −

d i '(1) 10h = m/s ii (3) 10h′ = − m/s

When 1t = , the projectile is rising but when 3t = , it is falling

6 41 3

100 4(30 ) 0 90tV t t= − ≤ ≤

a 1 2 3100

d (90 )dv t tt

= −

b Use graphics calculator to sketch 1 2 3

100 (90 ),y x x= − adjust window settings to suit domain [0, 90]

c Greatest rate of flow is when the graph is a max; find

max using graphics calculator, max at 60t = min

7 a Velocity ddxt

=

12

12

2

2

2

(3 4)

d 1 (6 )(3 4)d 2

3

3 4

x t

x t tt

tvt

−

= +

= +

=+

b Acceleration ddvt

=

122(3 )(3 4)v t t −= +

Product Rule

3 1

1 2 22 22

d 3 ( )(6 )(3 4) 3(3 4)dv t t t tt

− −= − + + +

( )

( ) ( )

( )

2

3 22

2 2

3 32 2

32

9 3

3 43 4

9 3(3 4)

3 4 3 4

12

3 4

t

tt

t t

t t

t

−= +++

− += ++ +

=+

c 2

3 2 6 3(2) 1.52163 2 4

V ×= = = =× +

( )3 32

12 12 12 3(2)64 1643 2 4

a = = = =× +

8 3 212 36x t t t= − +

a 2d 3 24 36dx V t tt

= = − +

b When 0V =

2

2

3 24 36 0

3( 8 12) 03( 6)( 2) 0

6 or 2

t t

t tt t

t

− + =

− + =− − =

=

3 2at 6, 6 12(6) 36(6)

0t x= = − +

=

3 2at 2, 2 12(2) 36(2)

32t x= = = − +

=

c d 6 24dva tt

= = −

at 6, 6(6) 24

12t a= = −

=

at 2, 6(2) 24

12t a= = −

= −

M B 1 2 Q l d - 2 36 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n

9 3 22 8V t t t= − +

a 2d 43 4

dv t tt t

= − +

at 2d 44, 3(4) 4(4)d 4vtt

= = − +

334 m hr= Answer is D. b

3 2(1) 1 2(1) 8 1V = − + = 7 3 2(4) 4 2(4) 8 4V = − + = 48

Average rate of change 48 74 1

−=−

341 213 m /hr3 3

= =

Answer is A.

10 201

Ct

=+

1

2

2

20( 1)d 20( 1)d

20( 1)

tc tt

t

−

−

= +

= − +

−=+

at 2d 209 d 10ctt

−= =

0.2mL/hr = −

Chapter review 1 3( 2)y x= + has 1 point of inflection

2d 3( 2) 0dy xx

= + =

2x = −

x −3 −2 −1 ddyx

+ 0 +

Slope / − /

Answer is C. 2 3 22 2y x x x= + + −

2d 3 4 1

d(3 1)( 1)

y x xx

x x

= + +

= + +

Start points at d 0dyx

=

13

and 1x −= −

x −2 −1 12

− 13

− 0

ddyx

+ 0 − 0 +

Slope / − \ − /

Max turning point at 1x = −

Min turning point at 13

x = −

Answer is E. 3 1 3 2

34 9 5y x x x= − − +

2d 8 9

d( 9)( 1)

y x xx

x x

= − −

= − +

Start points at d 0dyx

=

9 and 1x = −

x −2 −1 0 9 10 ddyx

+ 0 − 0 +

Slope / − \ − / Max turning point at 1x = − Min turning point at 9x =

at 1 23 3

1, 4 9 5 9x y= − = − − + + =

Max T. P. at 23

( 1, 9 )−

Answer is A. 4 ( ) 0 at 3, 1, 4 meansg x x′ = = − stationary points at

3, 1 and 4x = − ( ) 0 at <3 and 1 4 meansg x x x′ < < < negative gradient

when 3 and 1 4x x< < < ( ) 0g x′ > for all other x means positive gradient for all

other x values Answer is D.

5 3 2( ) 2 3 36 5f x x x= + − +

2

2

'( ) 6 6 36

6( 6)6( 3)( 2)

f x x x

x xx x

= + −

= + −= + −

Stationary points when ( ) 0f x′ =

6( 3)( 2) 03 and 2

x xx

+ − == −

at 3 23, ( ) 2( 3) 3(2) 36( 3) 5x f x= − = − + − − + = 86 at 3 22, ( ) 2( 2) 3(2) 36(2) 5x f x= − = − + − + 39= −

x −4 −3 0 2 3 ( )f x′ + 0 − 0 +

Slope / − \ − / Max T. P. at ( 3, 86)− Min T. P. at (2, 39)−

6 2 2(16 )y x x= −

2 4

3

16d 32 4d

x xy x xx

= −

= −

Stationary points when d 0dyx

=

3

2

32 4 0

4 (8 ) 0

0 or 8 2 2

x x

x x

x

− =

− =

= ± = ±

at 0, 0x y= =

at 2 2, 64x y= =

at 2 2, 64x y= − =

x −3 2 2− −1 0 1 2 2 3 ddyx

+ 0 − 0 + 0 −

Slope / − \ − / − \

A p p l i c a t i o n s o f d i f f e r e n t i a t i o n M B 1 2 Q l d - 2 37

Max T. P. at ( 2 2, 64) and (2 2, 64)− Min T. P. at (0, 0) y-intercept at 0 x-intercepts at 0, 4, 4−

7 2,m = point 1(1 3)= −

1 1( )3 2( 1)

2 5

y y m x xy x

y x

− = −− − = −

= −

Answer is D. 8 ( 2)( 1)y x x x= + −

3 2

2

2d 3 2 2d

x x xy x xx

= + −

= + −

at 1, ( 1)(1)( 2)x y= − = − − 2y =

at 2d1, 3( 1) 2( 1) 2dyxx

= − = − + − −

= −1 Equation to tangent

1 1( )2 1( 1)

y y m x xy x− = −− = − − −

1

1 0y xy x

= − ++ − =

Answer is D. 9 2 36 3 2y x x x= − + −

at 1, 6 3 2 1x y= = − + − = 4

2d 3 4 3dy x xx

= − + −

at d1, 3 4 3dyxx

= = − + −

= 2 Equation of tangent

1 1( )4 2( 1)

2 62 6

y y m x xy x

y xx y

− = −− = − −

= − ++ =

10 2

54xy = −

12

ddy xx

=

Parallel to 3 7 0y x− − = 3 7y x= + means the gradient = 3

12

d 3dy xx

= =

6x = at 6, 4x y= = Equation to tangent

1 1( )

4 3( 6)3 14

y y m x xy xy x

− = −− = −= −

11 1y xx

= +

1x x−= +

2d 11dyx x

= −

4 10

3 3

3 4 10 03 4 10

y xy x

y x−

+ − == − +

= +

tangent gradient is 34

normal gradient is 43

−

32 4

d 11dyx x

= − =

1

241

2x

x

=

= ±

at 12

2, 2x y= =

at 12

2, 2x y= − = −

Equation to normal at 12

(2, 2 )−

1 15 42 3

( )

( 2)

y y m x x

y x

− = −

− = − −

Multiply by 6 6y 15 8 16x− = − + 6y 8 31 0x+ − = 6y 8 31x+ =

Equation to normal at 12

( 2, 2 )− −

5 42 3

( 2)y x− −− = − −

6y 15 8 16x+ = − − 6y 8 31x+ = − 6y 8 31 0x+ + =

12 1( ) 0 at 4 and 2f x x= = −

1

1

( ) 0 at 4

( ) 0 at 4

f x x

f x x

> < −

< > −

at 4x = − , local Maximum ( , 0, )+ − at 2x = , negative point of inflection ( , 0, )− − Answer is C.

13 2 2 40x y+ =

2 40 2

20y xy x

= −= −

Answer is B. 14 2 2 2h y x= −

2 2

2 2(20 )

400 40

h y x

x x

x

= −

= − −

= −

Answer is A. 15 Area 1

2base × height= ×

12 2 400 40

400 40

x x

x x

= × × −

= −

Answer is D.

16 12

1 12 21

2

(400 40 )d ( )( 40)(400 40 ) (400 40 )d

A x xA x x xx

−

= −

= − − + −

M B 1 2 Q l d - 2 38 A p p l i c a t i o n s o f d i f f e r e n t i a t i o n

20 400 40400 40

x xx

−= + −−

Max or Min when d 0dAx

=

20 400 40400 40

20 400 4060 400

x xxx xx

= −−

= −=

40060203

236

x =

=

=

Verify Max or Min

x 6 23

6 7

ddAx

+ 0 −

Slope / − \

Max at 23

6x =

Answer is A. 17 if 2

36x =

2 23 3

20 40 203 3

20 1200 8003 3

20 4003 3

(6 ) 400 40 6

400

A

×

−

= − ×

= × −

= ×

= ×

20 203 3400 3

9

= ×

×=

Answer is E. 18 4 3( ) 3 5f x x x x= − +

(1) 1 3 5 3(3) 81 81 15 15

ff

= − + == − + =

Average rate of change 15 33 1

−−

=

122

= = 6 Answer is B

19 Surface area of cube 26l=

d 12ds ll

= −

at 2d m40, 480

d msll

= = −

Answer is D.

Modelling and problem solving 1 2 (50 ) 180N t t= − +

d 100 4dN tt

= −

Max or Min at d 0dNt

=

100 4 0

4 10025

ttt

− ===

at 25, 1250 180 1430t N= = + =

Verify max

t 20 25 30 ddNt

+ 0 −

Slope / − \

Max at 25t = at 25t = days, there are 1430 bees

2 22 , 6

3( 6)xA x

x= >

−

2

12 ( 6)3x x −= −

a Product Rule

2

2 1d 2 4( 1)( 6) ( 6)d 3 3A x xx xx

− −= − − + −

2

22 4

3( 6)3( 6)x x

xx−= +

−−

for Min or Max d 0dAx

=

2

22 4

3( 6)3( 6)x x

xx=

−−

2 23 2 ( 6) 4 3( 6)2 4( 6)2 4 2424 212

x x x xx xx x

xx

/ // × − / = × / −/= −= −==

at 22(12)12,

3(12 6)x A= =

−

= 16 b Verify Min or max

x 10 12 14 ddAx

− 0 +

Slope \ − / at 12x = , local min. (h) Minimum Area = 16 square units

3

a Pythagoras

2 2 2

2

10

100

h r

h r

= −

= −

b Volume of Cone 1 23

r hπ=

( )1 2 23

100V r rπ= −

c 121 12 2

3 2d ( 2 )(100 )dv r r rr

π −= × × − −

122 2

3(100 )r rπ+ −

3

22

2 10033 100

r r rr

π π−= + −−

A p p l i c a t i o n s o f d i f f e r e n t i a t i o n M B 1 2 Q l d - 2 39

for Max or Min d 0dvr

=

3

22

2 10033 100

r r rr

π π= −−

32

22

2

2

3 1003 2

1002

3 1002

2003

r rr

r r

r

r

ππ

/ × / = −/ × //

= −

=

=

2003

r = ±

disregard 2003

−

r 8 2003

9

ddvr

+ 0 −

Slope / − \

Maximum at 2003

r =

1 2 23

100V r rπ= −

( )( )( )

221 200 2003 3 3

2003

1003

100

200 1009

2009

20009 3

2000 327

π

π

π

π

π

= −

= −

=

=

=

4 a Volume of can 2 50r hπ= =

250hrπ

=

Area of tin = 22 2r rhπ π+

22

22

502 2

1002

r rr

rr

π ππ

π

= +

= +

b 2d 1004dA rr r

π= −

for minimum or maximum, d 0dAr

=

2

3

3

1004

4 100

10034

rr

r

r

π

π

π

=

=

=

= 2 Verify Max or min

r 1 2 3 ddAr

− 0 +

Slope \ − / Min at r = 2 cm c Area at 2r =

2

2

1002

8 50

75.1 cm

A rr

π

π

= +

= +

≈

d 240 /100 cmc

275.1cm /can 10 000 cans×

= 2751000 cm

2

2751000 cm100 cm /40c

= 7510 × 40c = 300 400c = $3004

5 a at A, m = 4 24y x=

12

d 8d8 4 at A

y xxx

x

=

=

=

at 12

1x y= =

A = (0.5, 1) B = (−0.5, 1) b (0.5, 1), m = 4 Equation of tangent at A.

1 1( )1 4( 0.5)

4 1

y y m x xy x

y x

− = −− = −

= −

c y-intercept of tangent at A is −1 C = (0, −1)

d 2 22 1 2 1( ) ( )d x x y y= − + −

2 2(0 0) ( 1 0)

1 cm= − + − −=

e normal at A

14

12

98

1 ( 0.5)

4 4

8 8 2 18 2 9

14

y x

y x

y xy x

xy

− = − −

− = − +

− = − += − +

−= +

y-intercept 98= = 1.125 cm