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New Way Additional Mathematics 3 — Solution Guide (Chapter 14) 7.

8.

x2 + xy + y2 = 11 dy dy 2x + x + y + 2y =0 dx dx 2x + y dy ∴ = − x + 2y dx

4x3 + 4 y 3 ∴ dy dx

x3 + 2xy + y4 = 13 dy dy 3x2 + 2x + 2y + 4y3 =0 dx dx 3x 2 + 2 y dy ∴ = − 2x + 4 y3 dx

9.

∴ 15.

x + y = 10 1

1

1 − 2 1 − 2 dy =0 x + y 2 2 dx dy ∴ = − dx 2

10.

y x

2

16.

x3 + y3 = 1

(x − 2)y3 = x + 2 2 dy 3(x − 2) y + y3 = 1 dx 1 − y3 dy ∴ = 3 y 2 ( x − 2) dx

xy2 + x2y + 6 = 0 dy dy x(2y) + y 2 + x2 + 2xy = 0 dx dx 2 xy + y 2 dy = − 2 x + 2 xy dx

14.

1 x −1 1 − 2 x2 y

17.

dy ) dx 5 dy = 4x5 ⋅ 6 y + 4y6 ⋅ 5x4 dx 10 x 4 y 6 − 3 x( x 2 + y 2 ) 2 dy = 3 y ( x 2 + y 2 ) 2 − 12 x 5 y 5 dx

1 =1 y dy ⋅ =0 dx

+

dy y2 = − 2 dx x



18.

y2 = 4x − 8 dy 2y =4 dx 2 dy = y dx dy =1 dx y = 2 ∴

Gradient of tangent = −

Section B

3 (x2 + y2)2 (2x + 2 y

13.

= −

x3 + 2xy + y2 = 34 dy dy 3x2 + 2 x + 2y + 2 y =0 dx dx 3x 2 + 2 y dy = − dx 2 y + 2x dy 3(−1) 2 + 2(−5) 7 = − = − dx ( −1, −5) 12 2(−5) + 2(−1) ∴

(x2 + y2)3 = 4x5y6



(1, −1)

13 =1 (−1) 3

2(−2)(3) + 32 3 = − 2 (−2) + 2(−2) ⋅ 3 8 (−2 , 3 ) 3 ∴ Gradient of tangent = − 8

1

12.

= −

Gradient of tangent = 1

dy dx

2 − 3 2 − 13 dy =0 x + y 3 3 dx dy y ∴ = −3 dx x 11.

dy =0 dx x3 dy = − 3 y dx



∴ 19.

Gradient of tangent = 1 x4 + y4 = 2 33

xy + 3 =3 xy − 3 xy + 3 = 3xy − 9 xy = 6 dy y+ x =0 dx dy y = − dx x xy + 2x =

y

7 12

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) 1 2 x

1

=

6.

dy +2 2 y dx

⋅ y+ x⋅



1 dy ⋅ 2 y dx

y 1 − x2 + x y = x

20. − xy

7.

y = sin2 x dy = 2 sin x cos x dx

8.

y = sin x2 dy = cos x2 ⋅ 2x dx = 2x cos x2

dy x dy + 1− x + y+ =1 2 dx 2 y dx 1− x 2

2(1 − y ) y (1 − x 2 ) + 2 xy y dy = dx 2(1 − x 2 ) y + x 1 − x 2



EXERCISE 14.7

9.

2.

= cos

1 dy − = 1 (cot 2 x) 2 (− csc 2 2 x ⋅ 2) dx 2 csc 2 2 x = − cot 2 x

Section 14.7 Derivatives of trigonometric functions (page 83)

10.

x 4

11.

dy 1 x = − sin dx 4 4 3.

4.

y = tan (3x + 1) dy = 3 sec2 (3x +1) dx y

= cot

πx 3

π πx dy = (−csc2 ) dx 3 3 π πx = − csc2 3 3 5.

cot 2 x 1

y = sin 2x dy = 2 cos 2x dx y

y=

= (cot 2 x) 2

Section A 1.

= sec x3

dy = sec x3 tan x3 ⋅ (3x2) dx = 3x2 sec x3 tan x3

y + 4 xy dy = dx x−x



y

y = csc (1 − 2x) dy = −csc (1 − 2x) cot (1 − 2x) ⋅ (−2) dx = 2 csc (1 − 2x) cot (1 − 2x)

34

y = sin 3x − 4 cos x dy = 3 cos 3x − 4 (−sin x) dx = 3 cos 3x + 4 sin x π πx ) + 3 sec 4 4 dy π 3π πx πx 2 = −csc (x + )+ sec tan dx 4 4 4 4 y = cot (x +

12.

y = x − sin x dy = 1 − cos x dx

13.

y = x2 − cot x dy = 2x − csc2 x dx

14.

y = x sin 2x dy = 2x cos 2x + sin 2x dx

15.

y = (2x + 3) sec x dy = (2x + 3) sec x tan x + 2 sec x dx

16.

y = x2 tan x dy = x2 sec2 x + 2x tan x dx

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) 17.

18.

19.

20.

cos x x2 dy x 2 (− sin x) − cos x(2 x) = dx x4 − x sin x − 2 cos x = x3 y

=

25.

26.

cos x 1 − sin x (1 − sin x)(− sin x) − cos x(− cos x) dy = (1 − sin x) 2 dx 1 = 1 − sin x y

=

27.

1 − sin x 1 + sin x ( 1 + sin x )( − cos x) − (1 − sin x)(cos x) dy = (1 + sin x) 2 dx 2 cos x = − (1 + sin x) 2 y

=

sin 3 x x2 − 8 ( x 2 − 8)(3 cos 3x) − (sin 3 x)(2 x) dy = ( x 2 − 8) 2 dx y

y

F(θ ) =

23.

y = sec (1 + 3x5) dy = sec (1 + 3x5) tan (1 + 3x5) (15x4) dx = 15 x4 sec (1 + 3x5) tan (1 + 3x5)

24.

y = sin 2x cos 3x dy = sin 2x (−3 sin 3x) + 2 cos 2x cos 3x dx 1 = (5 cos 5x − cosx) 2

5

1 + cot

θ 3 1

θ = (1 + cot ) 5 3 F′ (θ ) =

4 1 θ − θ 1 (1 + cot ) 5 (− csc 2 ) ( ) 5 3 3 3 4

= −

dy π π = 4 cos3 (2x + ) [−sin (2x + )] (2) dx 3 3 π π = −8 cos3 (2x + ) sin (2x + ) 3 3 y = (2 + csc x)2 dy = 2(2 + csc x) (−csc x cot x) dx = −2(2 + csc x) csc x cot x

u +1 ) u −1 u + 1 (u − 1)(1) − (u + 1)(1) h′ (u) = sec2 ( )⋅ (u − 1) 2 u −1 2 u +1 = − sec2 ( ) (u − 1) 2 u −1 h(u) = tan (

29.

π ) 3

22.

f (t) = sin (cos t) f′ (t) = cos (cos t) (−sin t) = −sin t cos (cos t)

g(θ ) = sin10 (θ − θ 2) g′ (θ ) = 10 sin9 (θ − θ 2) cos (θ − θ 2) (1 − 2θ) = 10 (1 − 2θ) sin9 (θ − θ 2) cos (θ − θ 2)

=

= cos4 (2x +

= [sin (

28.

3( x 2 − 8) cos 3x − 2 x sin 3x = ( x 2 − 8) 2 21.

π + x)]4 2 dy π π = 4 sin3 ( + x) cos ( + x) dx 2 2 = 4 cos3 x (−sin x) = − 4cos3 x sin x y

30.

35

1 θ θ − csc 2 (1 + cot ) 5 15 3 3

d d cos x cot x = ( ) dx dx sin x sin x(− sin x) − cos x(cos x) = sin 2 x 2 − sin x − cos 2 x = sin 2 x 1 = − sin 2 x = −csc2 x d d 1 csc x = ( ) dx dx sin x 0 − cos x = sin 2 x − cos x 1 ⋅ = sin x sin x = −csc x cot x

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) 37.

d d 1 sec x = ⋅ dx dx cos x 0 − (− sin x) = cos 2 x sin x 1 ⋅ = cos x cos x = sec x tan x 31.

32.

sec x tan x − csc2 y

sin 2t f (t) = 1 − cos t 2(1 − cos t ) cos 2t − sin 2t sin t f ′ (t) = (1 − cos t ) 2

f′(

36.

39.

x + y = x sin (x + y) dy dy 1+ = (1+ ) x cos (x + y) + sin (x + y) dx dx x cos ( x + y ) + sin ( x + y ) − 1 dy = 1 − x cos ( x + y ) dx

40.

x sin y − y =4 dy dy x cos y + sin y − =0 dx dx sin y dy = 1 − x cos y dx

41.

y sin x − x cos y = 0 dy dy sin x + y cos x − cos y + x sin y =0 dx dx cos y − y cos x dy = x sin y + sin x dx

42.

cos (x2 − y2)

1+ cos θ cosθ sin θ 1 + cos 2 θ

π )=0 2

u sin u u +1 u 2 cos u + u cos u + sin u f ′ (u) = (u + 1) 2 f (u) =

f ′ (π) = − 35.

y = tan (x + y) dy dy = (1 + ) sec2 (x + y) dx dx dy = −csc2 (x + y) dx

2

f ′ (θ ) = −

34.

38.

f (x) = tan4 x f ′ (x) = 4 tan3 x sec2 x π f′( )=8 4

f (θ ) =

dy =0 dx

sec x tan x dy = csc 2 y dx

f ′ (π) = 1 33.

sec x + cot y = 3

π π +1

sin y = cos x dy cos y = −sin x dx sin x dy = − cos y dx

= xy dy dy − (2x −2y ) sin (x2 − y2) = x +y dx dx y + 2 x sin( x 2 − y 2 ) dy = 2 y sin( x 2 − y 2 ) − x dx

Section B 43.

1 cos y = tan 2x 5 1 1 dy − sin y = 2 sec2 2x 5 5 dx 10 sec 2 2 x dy − = 1 sin y dx 5

y

= cos x −

dy = −sin x + cos2 x sin x dx = sin x (−1 + cos2 x) = −sin3 x dy ∴ + sin3 x = 0 dx

36

1 cos3 x 3

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) 44.

y=

1 tan3 x − tan x + x 3

2.

x =1−t dx = −1 dt y = 2 + 5t dy =5 dt dy 5 ∴ = = −5 dx −1

3.

x = 3t2 dx = 6t dt y = 1 + 4t dy =4 dt dy 4 2 ∴ = = dx 6t 3t

dy = tan2 x sec2 x − sec2 x + 1 dx = tan2 x (tan2 x + 1) − (tan2 x + 1) + 1 = tan4 x dy ∴ − tan4 x = 0 dx 45.

y = x sec2 x − tan x dy = x (2 sec x sec x tan x) + sec2 x dx − sec2 x = 2x sec2 x tan x 2 x sin x = cos 3 x

46.

y=

1 − tan 2 x 1 + tan 2 x sin 2 x 1− cos 2 x = sin 2 x 1+ cos 2 x 2 = cos x − sin2 x = cos 2x dy ∴ = −2 sin 2x dx dy i.e. + 2 sin 2x = 0 dx

EXERCISE 14.8

4.



= t2 = 2t

y

=

1 3 t 2

dy 3 = t2 dt 2 ∴ 5.

Section 14.8 Differentiation of parametric equations (page 86)

Section A 1.

x dx dt

x = 6t + 7 dx =6 dt y = 5t dy =5 dt dy dy = dt dx dx dt 5 = 6

6.

37

3 2 dy 3 t = 2 = t dx 4 2t

x dx dt y dy dt dy ∴ dx

= 2t − 1 =2 = 1 − 4t2 = −8t =

−8t = −4t 2

x = t2 + t dx = 2t + 1 dt y = t2 − t dy = 2t − 1 dt dy 2t − 1 ∴ = dx 2t + 1

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) 7.

x

=

t



1 dx − = 1t 2 dt 2 y = 6 − 7t dy = −7 dt −7 1 dy ∴ = 1 − 12 = −14 t 2 t dx 2

8.

9.

10.

11.

x = 2 sin t dx = 2 cos t dt y = 5 cos t dy = −5 sin t dt dy −5 sin t ∴ = dx 2 cos t 5 = − tan t 2 x = 2 sec t dx = 2 sec t tan t dt y = 4 tan t dy = 4 sec2 t dt dy 4 sec 2 t ∴ = dx 2 sec t tan t 2 sec t = tan t = 2 csc t x = 3 sin 7t dx = 21 cos 7t dt y = 5 cos 7t dy = −35 sin 7t dt dy −35 sin 7t ∴ = dx 21 cos 7t 5 = − tan 7t 3

dy −2 sin 2t = dx 2 sin t cos t −2 sin 2t = sin 2t = −2

12.

x = t − sin 2t dx = 1 − 2 cos 2t dt y = 1 − cos 2t dy = 2 sin 2t dt dy 2 sin 2t ∴ = dx 1 − 2 cos 2t

13.

x=t−1 dx =1 dt y = 2t + 3 dy =2 dt dy ∴ = 2 dx t = 2

14.

x = 1 − t3 dx = −3t2 dt y = t2 − 2t + 1 dy = 2t − 2 dt 2t − 2 dy = − 3t 2 dx 2(0) − 2 dy −2 = 2 = dx t =0 − 3(0) 0 This is undefined.

15.

x dx dt y

= t2 + 2t = 2t + 2 =

dy = 1 dt 2

t

1 − t 2 1

x = sin2 t dx = 2 sin t cos t dt y = cos 2t dy = −2 sin 2t dt



38

1 1 −2 dy t 1 = 2 = 2 dx 4 ( t + 1 ) t 2t + 2 1 dy 1 1 = = dx t = 4 2 40 4(4 + 1)4

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) 16.

19.

1

x = 2t 3 2 dx 2 − = t 3 dt 3 2

y = 3t 3 1 dy = 2t − 3 dt 2t



1 3

dy 2 = 2 −3 dx t 3

x = cos3 θ dx = −3 cos2 θ sin θ dθ y = sin3θ dy = 3 sin2 θ cos θ dθ dy 3 sin 2 θ cos θ = dx − 3 cos 2 θ sin θ = −tan θ dy π ∴ dx π = − tan θ= 3 3

1

= 3t 3 dy dx

17.

= − 3 1

t = −8

= 3(−8) 3

Section B

=−6

20.

x = 2 sin θ dx = 2 cos θ dθ y = 3 cos θ dy = −3 sin θ dθ −3 sin θ dy = dx 2 cos θ 3 = − tan θ 2 dy 3 π dx θ = π = − 2 tan 4

18.

3 2

x = t sin t dx = t cos t + sin t dt y = t cos t dy = − t sin t + cos t dt dy −t sin t + cos t = dx t cos t + sin t dy − π sin π + cos π ∴ = dx t = π π cos π + sin π =

x .............................................= 1 − t

dx = −1 dt y = 1 − t2.......................................(2) dy = −2t dt dy −2t = dx −1 = 2t Putting x = 1 into (1), 1=1−t t=0 dy dy ∴ = dx (1, 1) dx t = 0 = 2(0) =0 (b) From (1), t = 1 − x ........................(3) Putting (3) into (2), y = 1 − (1 − x)2 = 1 − (1 − 2x + x2) ∴ y = f (x) = 2x − x2 (c) f ′ (x) = 2 − 2x dy ∴ = f ′ (1) dx (1, 1)

4

= −

(a) (1)

= 2 − 2 (1) =0

1 π

39

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) 21.

(a)

x dx dt y dy dt dy dx

= t3............................................(1) = 3t

When t = 1, 3(1) 2 y= 1 + (1) 3 3 = 2 3 3 ∴ ( , ) is a point on the curve 2 2 that corresponds to t = 1. (b) From (1), 3 2 dx (1 + t )(3) − (3t )(3t ) = (1 + t 3 ) 2 dt

2

= t2 ...........................................(2) = 2t

2t 3t 2 2 = 3t Putting x = 8 into (1), 8 = t3 t=2 dy dy ∴ = dx ( 4, 8) dx t = 2 =

3 − 6t 3 (1 + t 3 ) 2 From (2), (1 + t 3 )(6t ) − 3t 2 (3t 2 ) dy = (1 + t 3 ) 2 dt =

2 3(2) 1 = 3 =

=

1

(b) From (1), x 3 = t ...........................(3) Putting (3) into (2),

dy 6 − 3t 4 = dx 3 − 6t 3 dy 2 − t4 = ..................................(3) dx 1 − 2t 3 dy dy dx ( 3 , 3 ) = dx



1

y = ( x 3 )2 2

∴ y = g(x) = x 3 (c) g′ (x) = ∴

dy dx

1

2 −3 x 3 ( 8, 4 )

t=1

2 2

=

= g′ (8)

2 − 14 1 − 2(1) 3

= −1

1

2 − = (8) 3 3 1 = 3 22.

6 − 3t 4 (1 + t 3 ) 2

(c)

3t ........................................(1) 1+ t3 3 When x = , 2 3 3t = 2 1+ t3 1 + t3 = 2t 3 t − 2t + 1 = 0 (t − 1) ( t2 + t − 1) = 0

(a) x =

t = 1 or − 1± 5 2 3t 2 y= .......................................(2) 1+ t3

40

( 2) y , = t ..................................(4) (1) x Putting (4) into (1), y 3( ) x x= y 1 + ( )3 x y 3( ) x x= 3 x + y3 x3 ∴ The equation of the curve is x3 + y3 = 3xy ..........................(5)

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) (d) Differentiating (5) with respect to x, dy dy 3x2 + 3y2 = 3y + 3x ⋅ dx dx 2 y−x dy ∴ = 2 y −x dx dy dx

23.

3 3 ( , ) 2 2

We can consider the curve passes through the point (1, 1) twice, one π 3π at θ = and one at θ = . 2 2 Hence, there are two tangents, with gradients 1 and −1 respectively.

3 3 2 −( ) = 2 2 = −1 3 3 ( )2 − 2 2

(e) From (2), y2 = sin2θ cos2θ = (1 − cos2θ) cos2θ Putting (1) into it, y2 = (1− x2)x2 i.e. The equation of the curve can be represented by y2 = x2 − x4 ....................(5)

(a) x = cos θ......................................... (1) As −1 ≤ cos θ ≤ 1 for 0 ≤ θ ≤ 2π ∴ The range of values of x = [−1,1] (b) y = sin θ cos θ ...............................(2) 1 = ⋅ 2 sin θ cos θ 2 1 ∴ y = sin 2θ...........................(3) 2 As −1 ≤ sin 2θ ≤ 1 for 0 ≤ θ ≤ 2π ∴ The range of values of y 1 1 = [− , ] 2 2 (c) Putting x = 0 into (1), 0 = cos θ π 3π θ = or ...............(4) 2 2 Putting (4) into (2), π π π When θ = , y = sin cos =0 2 2 2 3π 3π 3π When θ = , y = sin cos =0 2 2 2 ∴ The point (0, 0) corresponds to π 3π θ= or . 2 2 dx (d) From (1), = −sin θ dθ dy From (3), = cos 2θ dθ dy cos 2θ ∴ = dx − sin θ π cos 2( ) dy = 2 =1 dx θ = π π − sin 2 2 3π cos 2( ) dy 2 = −1 dx θ = 3π = 3π − sin 2 2

(f) Differentiating (4) with respect to x, dy 2y = 2x − 4x3 dx dy x − 2x3 = dx y dy is in indeterminate form at (0, 0). dx We have to take the limit. Equation (5) can be considered as a combination of two functions, y = x 1 − x 2 and y = −x 1 − x 2 When y = x 1 − x 2 ,

dy dx

x − 2x3 = lim x→0 ( 0, 0 ) x 1 − x2

1 − 2x2 = lim x→0 1 − x2 =1 When y = −x 1 − x 2 ,

dy dx

( 0, 0 )

= lim x→0

x − 2x3 − x 1 − x2

1 − 2x2 = lim x→0 − 1 − x2 = −1

EXERCISE 14.9 Section A 1.

41

y = x5 − 7x3 + 2x y′ = 5x4 − 21x2 + 2 y′′ = 20x3 − 42x

Section 14.9 Second derivatives (page 90)

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) 2.

3.

4.

5.

y = 1 − 2x−1 − 3x−2 y′ = 2x−2 + 6x−3 y′′ = −4x−3 − 18x−4

10.

1 1 ) (5 − ) x x 2 1 = 15 + − 2 x x −2 2 y′ = 2 + 3 x x 4 6 y′′ = 3 − 4 x x

1

1

12.

y = (x − 1)3 (x − 4)3 y′ = 3(x − 4)2 (x − 1)3 + 3(x − 1)2 (x − 4)3 = 3(x − 1)2 (x − 4)2 (2x − 5) y′′ = 6(x − 1)2(x − 4)2 + 6(x − 1)2 (2x − 5) (x − 4) + 6(x − 1) (x − 4)2 (2x − 5) = 6(x − 1) (x − 4) (5x 2 − 25x + 29)

2

1

3 −2 8 −3 x − x 2 3

y = x 2 x − 8x

y=

14.

3

3

y′′ = − 2[(5 + 2t 2 ) − 2

1

7 2 x − 12 x 2 2 3

5

− 3 + t (− )(5 + 2t 2 ) 2 (4t )] 2

1

− y′′ = 35 x 2 − 6 x 2 4

8.

9.

5 + 2t 2

= − 2t (5 + 2t 2 ) − 2

3

5

1 3

x (x3 − 8x) 7

y=

− 1 y′ = (− )(5 + 2t 2 ) 2 (4t ) 2

= x 2 − 8x 2 y′ =

1 2t + 1 −2 y′ = (2t + 1) 2 8 y′′ = (2t + 1) 3 y=

3

5 2 x −8 2 1 15 2 y′′ = x 4 y′ =

7.

y = x3 (2x + 3)2 = 4x5 + 12x4 + 9x3 y′ = 20x4 + 48x3 + 27x2 y′′ = 80x3 + 144x2 + 54x

13.

y′ = 3 x 2 − 8 x 3

6.

11.

y = 2x 2 − 6x 3

y′′ =

9 − 5x 2

y = (3 +

4

3

− 5 y′ = − (9 − 5 x) 3 3 5 − 50 y′′ = − (9 − 5 x) 3 9

y = (x2 − 1) (x2 +1) = x4 − 1 y′ = 4x3 y′′ = 12x2

3

y=

5

= 2(4t 2 − 5)(5 + 2t 2 ) − 2

y = (x + 3)100 y′ = 100(x + 3)99 y′′ = 9 900(x + 3)98 y = (2x − 5)−4 y′ = −8(2x − 5)−5 y′′ = 80(2x − 5)−6

15.

42

t +1 t −1 (t − 1) − (t + 1) −2 y′ = = (t − 1) 2 (t − 1) 2 4 y′′ = (t − 1) 3 y=

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) 16.

y′ = =

t−2 +

=

t 2 t−2

3t − 4 2 t−2

y′′ = 1 ( 2

17.

24.

y = t t−2

3 t − 2 − (3t − 4) 3t − 8

t−2

2 t−2 )

19.

y = sin x2 y′ = 2x cos x2 y′′ = 2 cos x2 − 4x2 sin x2 y = tan x y′ = sec2 x y′′ = 2 sec2 x tan x

21.

y = x2 sin x y′ = x2 cos x + 2x sin x y′′ = −x2 sin x + 2x cos x + 2x cos x + 2 sin x = (2 − x2) sin x + 4x cos x

22.

y = cos2 x y′ = −2 cos x sin x = −sin 2x y′′ = −2 cos 2x

y = sin

1 x2

2 1 cos 2 x3 x 4 1 6 1 y′′ = − 6 sin 2 + 4 cos 2 x x x x y′ = −

y = sin 3x y′ = 3 cos 3x y′′ = −9 sin 3x y = cos 5x y′ = −5 sin 5x y′′ = −25 cos 5x

23.

25.

4(t − 2) t − 2

18.

20.

1

y = sin 3x cos 4x 1 = (sin 7 x − sin x) 2 7 1 y′ = cos 7 x − cos x 2 2 49 1 sin 7 x + sin x y′′ = − 2 2

cos x x − x sin x − cos x y′ = x2 [ x 2 (− x cos x − sin x + sin x) y′′ = − (− x sin x − cos x)2 x] x4 (2 − x 2 ) cos x + 2 x sin x = x3

26.

y = tan 2x sec2 2x = tan 2x + tan3 2x y′ = 2 sec2 2x + 6 tan 2 2x sec2 2x = 2 (1 + 4 tan2 2x + 3 tan 4 2x) y′′ = 2 (16 tan 2x sec2 2x + 24 tan3 2x sec2 2x) = 16 sec2 2x tan 2x (2 sec2 2x + tan2 2x)

27.

y = sin (cos x) y′ = −sin x cos (cos x) y′′ = −[cos x cos (cos x) − sin x sin (cos x) (−sin x)] = −sin2 x sin (cos x) − cos x cos (cos x)

28.

x + 3xy + 2y =5 dy dy 1 + 3x + 3y + 2 =0 dx dx dy 1+ 3y ∴ =− dx 3x + 2 dy (3 x + 2)(3 ) − (1 + 3 y ) ⋅ 3 d2y =− dx 2 dx (3 x + 2) 2 6(1 + 3 y ) = (3 x + 2) 2

29.

y2 = 12x dy 2y = 12 dx 6 dy = y dx 2 6 dy 6 6 36 d y = − 2 ( )= − 2 ( )= − 3 2 y dx y y y dx

y=

43

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) 30.

x2 − y 2 = 10 dy 2x − 2 y =0 dx x dy = y dx dy 10 y−x d2y = dx = − 3 2 y dx y2

31.

x2 + xy + y2 = 16 dy dy 2x + x + y + 2y =0 dx dx 2x + y dy = − x + 2y dx

34.

y = sin (x + y) dy dy = (1+ ) cos (x + y) dx dx cos ( x + y ) dy = dx 1 − cos ( x + y ) d2y dx 2 dy [1 − cos ( x + y )][−(1 + ) sin ( x + y )] dx = − cos ( x + y )[(1 + dy ) sin ( x + y )] dx [1 − cos ( x + y )]2 dy − (1 + ) sin ( x + y ) = dx [1 − cos ( x + y )]2 sin ( x + y ) = − [1 − cos ( x + y )]3

35.

sin y + cos x = 5 dy cos y − sin x = 0 dx sin x dy = cos y dx

d2y dx 2 = −

( x + 2 y )(2 +

dy dy ) − (2 x + y )(1 + 2 ) dx dx ( x + 2 y) 2

96 = − ( x + 2 y)3 32.

y3 + 2y −1= 3x dy dy 3y2 +2 =3 dx dx 3 dy = 2 3y + 2 dx dy 54 y 6y d2y = = − dx 2 ( − 3 ) (3 y 2 + 2) 3 dx (3 y 2 + 2) 2

1 2 x

+

=

=

x+ y

33.

dy cos y cos x + sin x sin y d2y = dx dx 2 cos 2 y

=4

1

dy =0 2 y dx dy dx

= −

36. y x

dy −y 1 x d2y 1 y − 2 dx = − ( ) ( ) dx 2 2 2 x x = = = =

1 (1 + 2x

y ) x

x+ y 2x x 4 2x x 2 x x 44

sin 2 x sin y cos y 2 cos y

cos y cos x +

cos 2 y cos x + sin y sin 2 x cos 3 y

y = (2x + 3)n dy = n(2x + 3)n−1 (2) dx = 2n(2x + 3)n−1 d2y = 2n(n − 1) (2x+3)n−2 (2) dx 2 = 4n(n − 1) (2x+3)n−2 dy d2y (2x + 3)2 2 − 9 (2x + 3) + 10y = 0 dx dx ∴ (2x + 3)2 ⋅ 4n (n − 1) (2x +3)n−2 − 9 (2x + 3) ⋅ 2n (2x + 3)n−1 + 10 (2x + 3)n = 0

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) The equation holds for all x, ∴ 4n2 − 4n −18n + 10= 0 2n2 − 11n + 5 = 0 (n − 5) (2n −1) = 0 1 n = 5 or (rejected) 2 ∴ n=5 37.

39.

y = A cos 3x + B sin 3x dy = −3A sin 3x + 3B cos 3x dx d2y = −9A cos 3x − 9B sin 3x dx 2 d2y + 6y = 5 cos 3x − sin 3x dx 2 ∴ (− 9A cos 3x − 9B sin 3x) + 6(A cos 3x + B sin 3x) = 5 cos 3x − sin 3x − 3A cos 3x − 3B sin 3x = 5 cos 3x − sin 3x Comparing coefficients of cos 3x and sin 3x on both sides, −3A = 5 and −3B= −1 5 1 ∴ A= − and B= 3 3

Section B 38.

(a) x = f (t), y = g(t) dy dy = dt dx dx dt 2 d y dx 2 d dy = ( ) dx dx dx d dy d dx dy ⋅ ( )− ( )⋅ dx dt dt = dt dx dt dx 2 ( ) dt dx d dy d1 dy ⋅ ( )− ⋅ dt dx dt dt dt = dx 2 ( ) dt dx d dy ⋅ ( ) = dt dx dt dx ( )2 dt d dy ( ) = dx dt dx dt t2 , y = 1− t 2 dy dy dt ∴ = dx dx dt −1 = t d dy ( ) d2y dt dx = 2 dx dx dt 1 = t2 t 1 = 3 t

(b) x =

b2x2 + a2 y2 = a2b2 dy 2b2x + 2a 2 y =0 dx b2 x dy = − 2 a y dx dy y−x 2 d2y b = − dx ) ( dx 2 a2 y2 =



b2 [ a2

y − x(−

b2 x ) a2 y

y2

= −

b2 a2 y + b2 x2 ( ) a2 a2 y3

= −

b 2 a 2b 2 ( ) a2 a2 y3

= −

b4 a2 y3

]

45

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) 40.

(a) x =

1+ t2 , y = dy dy = dt dx dx dt −t =

(c) If y = A sin mx + kx sin 2x, then d2y + 4y dx 2 d 2 (A sin mx + kx sin 2 x) = dx 2 + 4(A sin mx +kx sin 2x) d 2 (A sin mx) = [ + 4A sin mx ] dx 2 d 2 (kx sin 2 x) +[ + 4kx sin 2 x ] dx 2 = 0 + 2 cos 2x (given) = 2 cos 2x i.e. y = A sin mx + kx sin 2x is a solution of the equation in (b).

1− t2

1− t 2 t 1+ t2 1+ t2

= − = − (b)

1− t2 x y

dy y−x d2y = − dx dx 2 y2 x y − x (− ) y = − 2 y x2 + y2 = − y3 2 = − 3 ( y

41.

REVISION EXERCISE 14

Revision exercise 14 (page 94)

Section A 1.

1

f (x) = (1 + 8 x) 4 3

3 − 1 − (1 + 8 x) 4 (8) = 2(1 + 8 x) 4 4 3 2 f ′ (10) = 2[1 + 8(10)]− 4 = 27 y = (3 + 2x) 1 − 5 x dy dx 1 − 1 = (3+2x) ⋅ (1 − 5 x) 2 (−5) + 1 − 5 x (2) 2 1 − 1 = (1 − 5 x) 2 (−15 − 10 x + 4 − 20 x) 2 1 − 1 = (1 − 5 x) 2 (−11 − 30 x) 2

f ′ (x) =

x2 + y2 = 2)

(a)

y = A sin mx dy = mA cos mx dx d2y = −m2A sin mx dx 2 d2y + 4y = 0 dx 2 ∴ −m2A sin mx + 4A sin mx = 0 −m2 + 4 = 0 m=±2 (b) y = kx sin 2x dy = k[x(2 cos 2x) + sin 2x] dx d2y = k[x(−4 sin 2x) + 2 cos 2x dx 2 + 2 cos 2x] = k(−4x sin 2x + 4 cos 2x) d2y + 4y = 2 cos 2x dx 2 ∴ k(−4x sin 2x + 4 cos 2x) + 4kx sin 2 x = 2 cos 2x 4k cos 2x = 2 cos 2x 1 ∴ k = 2

2.

3.

(a)

y = x3 sin 2x dy = x3 (2 cos 2x) + (sin 2x) (3x2) dx = 2x3 cos 2x + 3x2 sin 2x d2y (b) = 2x3 (−2 sin 2x) + (cos 2x)(6x2) dx 2 + 3x2 (2 cos 2x ) + (sin 2x) (6x) = (−4x3 + 6x)sin 2x + 12x2 cos 2x 1

4.

y=

(1 + 3 x) 3 1

(1 − 2 x) 2 46

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) 1

1

= (1 + 3 x) 3 (1 − 2 x) − 2 f′ (

dy dx

π ) = 2

π π 2 sin 2( ) 2 2 1 − cos 2 (

3

1 − 1 = (1 + 3x) 3 [ − (1 − 2 x) 2 (−2)] 2 2 1 1 − − + (1 − 2 x) 2 [ (1 + 3 x) 3 (3)] 3 1 3

= (1 + 3x) (1 − 2 x) = (1 + 3 x)

2 − 3

3 2



(1 − 2 x)

+ (1 − 2 x)

3 − 2



=

1 2

(1 + 3 x)



9.

2 3

=

[(1 + 3x) + (1 − 2x)]

2

3

= (2 − x) (1 + 3 x) − 3 (1 − 2 x) − 2 dy dx

2

f (u) = 3

= (2 − 0) (1 + 0) − 3 (1 − 0) − 2

x=0

f (u) =

f ′ (u) =

=2 5.

y = (1 − 3

= 1 x

)

dy 1 = 2( dx 1− 3 x

2

3

1

=

6.

7

8.

=

4 1 − )[−(− ) x 3 4

sin 2 θ cos θ dy cosθ (2 sin θ cos θ ) − sin 2 θ (− sin θ ) = dθ cos 2 θ sin θ (1 + cos 2 θ ) = cos 2 θ = sin θ (1 + sec 2 θ ) =

10.

=

f ′ (t)

= =

f′(

1 − cos 2 t 2 2 cos t 2 sin t 2 ⋅ (2t ) 2 1 − cos 2 t 2 t sin 2t

f (θ ) = f ′ (θ ) =

2 1 y = tan x + tan 3 x + tan 5 x 3 5 2 2 y′ = sec x + 2 tan x sec 2 x + tan 4 x sec 2 x = sec6x f (t)

π 2

cos 2 u 1 + sin 2 u 1 − sin 2 u 1 + sin 2 u 1 −θ where θ = sin2 u 1+θ (1 + θ )(−1) − (1 − θ ) dθ du (1 + θ ) 2 −2 (2 sin u cos u ) (1 + sin 2 u ) 2 −2 sin 2u (1 + sin 2 u ) 2

π cos 2 π 4 = 1 f( )= π 4 3 1 + sin 2 4 π − 2 sin 2 ( ) 4 π 8 2 = − f′( )=  4 9 2 π  1 + sin ( 4 )   π π 1 8 ∴ f ( ) − 3 f′ ( ) = − 3(− ) = 3 4 4 3 9

]

− − 2 (1 − x 3 ) x 3 3

y

π 2 ) 2

11.

2

1 − cos 2 t 2

47

a 2 + b 2 − 2ab cosθ 2ab sin θ 2 a 2 + b 2 − 2ab cosθ π 2ab sin 2

π )= π 2 2 a 2 + b 2 − 2ab cos 2 ab = a 2 + b2

f (x) f′ (x)

= cos3 x = 3 cos2 x (−sin x) = −3 cos2 x sin x f′′ (x) = −3 cos2 x (cos x) + sin x [(−6 cos x) (−sin x)] = −3 cos3 x + 6 sin2 x cos x

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) 12.

Differentiating (1) with respect to x, dy dy 2x + 3y + 3x + 8y =0 dx dx 2x + 3y dy ∴ = − 3x + 8 y dx

d [sin (1 − x2)] = [cos (1 − x2)] (−2x) dx = −2x cos (1 − x2) d sin(1 − x 2 ) (b) [ ] dx x2

(a)

=

x2 ⋅

d d [sin (1 − x 2 )] − sin (1 − x 2 ) ⋅ ( x 2 ) dx dx x4

(b)

x 2 [−2 x cos (1 − x 2 )] − 2 x sin (1 − x 2 ) x4 2 2 − 2 x cos (1 − x ) − 2 sin (1 − x 2 ) = x3 =

13.

(a) y = f (x) =

=

=

25 − x ......................(1) 2

1 − dy 1 = (25 − x 2 ) 2 (−2 x) dx 2

=

1

= − x(25 − x 2 ) − 2 dy dx

= − 3(25 − 3 ) 2

( 3, 4 )

= −



1 2

=

(c)

25 − y 2 .

16.

1

= −y (25 − y 2 ) − 2

( 3, 4 )

= −4 (25 − 4 2 ) = −

14.



1 2

(a)

2x + 3 y ) 3x + 8 y (3x + 8 y ) 2

− 7 y + 7 x (−

− 7(3xy + 8 y 2 + 2 x 2 + 3 xy) (3 x + 8 y ) 3

x dx dθ y dy dθ ∴

4 3

x3 + xy2 = 5 dy 3x2 + x (2y) + y2 = 0 dx dy 3x 2 + y 2 ∴ = − dx 2 xy 2 2 dy 3(1) + 2 7 =− =− dx (1, 2 ) 2(1)(2) 4 ∴ Gradient of tangent = −

15.

dy dx (3 x + 8 y ) 2

− 7 y + 7x

=

1 dx 1 − = (25 − y 2 ) 2 (−2 y ) dy 2

dx dy

dy dy ) + (2 x + 3 y )(3 + 8 ) dx dx (3x + 8 y ) 2

(3x + 8 y )(−2 − 3

− 14( x 2 + 3 xy + 4 y 2 ) (3x + 8 y )3 −14(8) = (from (1)) (3 x + 8 y )3 112 = − (3 x + 8 y ) 3

3 4

(b) From (1), y2 = 25 − x2 x2 = 25 − y2 x≥0 ∴ The inverse function of f(x) is x=

d2y dx 2 d 2x + 3 y (− ) = dx 3 x + 8 y

= tan θ ....................................(1) = sec2θ = sec θ ....................................(2) = sec θ tan θ dy secθ tan θ = dx sec 2 θ tan θ = secθ = sin θ dy π 2 dx θ = π = sin 4 = 2 4

(b) As sec2θ − tan2 θ = 1, from (1) and (2), the required equation is y2 − x2 = 1.

7 4

(a) x2 + 3xy + 4y2 = 8 …………(1) 48

New Way Additional Mathematics 3 — Solution Guide (Chapter 14)

(c)

d dy d2y ( ) = 2 dx dx dx d (sinθ ) = dx d sin θ dθ ⋅ = dθ dx 1 = cosθ ⋅ sec 2 θ = cos3 θ d2y π 2 ∴ dx 2 π = cos 3 ( ) = 4 θ = 4 4

22.

(a) x = (1 + t)2, y = (1 − t)2 dy −2(1 − t ) = 2(1 + t ) dx t −1 = 1+ t dy t − 1 (b) = =0 dx 1 + t ∴ t=1 When t = 1, x = 4, y = 0. ∴ The required point is (4, 0). (c) x =y (1 + t)2 = (1 − t)2 ∴ t= 0 dx 0 −1 = = −1 dy t =0 1+ 0

23.

(a)

Section B 20.

(a) (uv)′ = uv′+ vu′ (uv)′′ = (uv′+ vu′)′ = uv′′+ u′v′+ vu′′+ v′u′ = u′′v + 2u′v′+ uv′′ (b) Let u = x2 , v = sin x. Then u′ = 2x, u′′ = 2; v′ = cos x, v′′ = −sin x. ∴ (x2 sin x)′′ = 2(sin x) + 2(2x)(cos x) + x2(−sin x) = 2 sin x + 4x cos x − x2 sin x

21.

(a) x2 + y2 −12x − 6y + 25 = 0...............(1) x2 + y2 + 2x + y − 10 = 0...............(2) (1) − (2), −14x − 7y + 35 = 0 ∴ y = 5 − 2x................................(3) Putting (3) into (2) yields x2 − 4x + 4 = 0 (x − 2)2 = 0 x=2 From (3), y=1 ∴ The point of intersection is (2, 1). (b) From (1), dy dy 2x + 2 y − 12 − 6 =0 dx dx 6− x dy ∴ = y −3 dx dy 6−2 = = −2 dx ( 2, 1) 1 − 3

(b) y = x2 − 2........................................(1) y = −2x2 + 10..................................(2) Putting (2) into (1), −2x2 + 10 = x2 − 2 3x2 = 12 x=±2 ∴ The points of intersection are (2, 2) and (−2, 2).

Similarly, for equation (2), dy = −2 dx ( 2, 1) (c) Yes ,it is because they have the same gradient. 49

New Way Additional Mathematics 3 — Solution Guide (Chapter 14) (c) For equation (1), dy = 2x dx dy m1 = =4 dx ( 2,2 )

At (3, 0), there are 2 tangents. For t = − 3 , the gradient of the tangent dy = dx t = − 3

For equation (2), dy = −4x dx dy m2 = = −8 dx ( 2,2 )

=

For t = 3 , the gradient of the tangent dy = dx t = 3

m2 − m1 1 + m2 m1

=

−8− 4 = 1 + (−8)(−4)

=

(a) x = t2 ...............................................(1) y = t3 − 3t.........................................(2) Putting y = 0 into (2), 0 = t3 − 3t 2 t (t − 3) = 0 t = 0 or ± 3 ......................(3) Putting (3) into (1), when t = 0, x = 02 = 0 when t = ± 3 , x = ( ± 3 )2 = 3 ∴

3( 3 ) 2 − 3 2( 3 ) 3

(d) The tangent is horizontal when 3t2 − 3 = 0 i.e. t = − 1 or 1 Putting these values of t into (1) and (2), when t = −1, x = (−1)2 = 1 and y = (−1)3 −3 (−1) = 2 when t = 1, x = (1)2 = 1 and y = (1)3 −3 (1) = −2 ∴ The required points are (1, 2) and (1, −2).

12 31 θ = 21°(cor. to the nearest degree) =

24.

2(− 3 )

= − 3

(d) Let θ be the acute angle between the tangents. tan θ =

3(− 3 ) 2 − 3

(e) From (2), y = t (t2 − 3) ∴ y2 = t2 (t2 − 3)2 Putting (1) into it, the equation of the curve can be represented as y2 = x (x − 3)2.

The curve C meets the x-axis at (0, 0) and (3, 0).

dx = 2t dt dy From (2), = 3t2 − 3 dt dy 3t 2 − 3 ∴ = dx 2t (c) At (0, 0), the gradient of the tangent dy = dx t = 0 3(0) − 3 = 2(0) −3 = 0 which is undefined. (b) From (1),

50