Chapter 16 Applications of differentiation
CHAPTER 16 Exercise 16A (p.119) 1. 4 x 2 + y 2 = 8 dy =0 4( 2 x ) + 2 y dx dy 4x =− dx y dy 4(1) =− dx (1, 2 ) (2) = −2 2. y 2 = x dy 2y =1 dx dy 1 = dx 2 y dy 1 = dx ( 4, 2 ) 2(2) 1 = 4 3. x 2 y + xy 2 = 6 dy dy 2 xy + x 2 + 2 xy + y2 = 0 dx dx dy y( y + 2 x ) =− dx x ( x + 2 y) (2)(2 + 2) dy =− (1)(1 + 4) dx (1, 2 ) 8 =− 5 4. x + 2 xy = 3 y 2
2x + 2x
2
dy dy + 2y = 6y dx dx dy x+y = dx 3 y − x dy 1+1 = dx (1, 1) 3(1) − 1 =1
5. x 3 + 3 x 2 y − 6 xy 2 + 2 y 3 = 0 dy dy dy 3x 2 + 3x 2 + 6 xy − 12 xy − 6 y2 + 6 y2 =0 dx dx dx dy 2 y 2 − 2 xy − x 2 = dx x 2 − 4 xy + 2 y 2 2 − 2 −1 dy = dx (1, 1) 1 − 4 + 2 =1
6. y = cos 2 x dy = −2 sin 2 x dx π dy = −2 sin 2( ) dx ( 12π , 23 ) 12 π = −2 sin 6 = −1 7. y = x 2 + 5 x − 3 dy = 2x + 5 dx dy = 2(1) + 5 = 7 dx (1, 3) The equation of the tangent is y − 3 = 7 ( x − 1) 7x − y − 4 = 0 The equation of the normal is y−3= − x + 7 y − 22 = 0
1 ( x − 1) 7
8. y = x 3 − 2 x 2 + 7 x − 4 dy = 3x 2 − 4 x + 7 dx dy = 3(2)2 − 4(2) + 7 = 11 dx ( 2, 10 ) The equation of the tangent is y − 10 = 11( x − 2) 11x − y − 12 = 0 The equation of the normal is −1 y − 10 = ( x − 2) 11 x + 11y − 112 = 0 3
9. y = x x = x 2
dy 3 12 = x dx 2 1 dy 3 = ( 4) 2 = 3 dx ( 4, 8) 2 The equation of the tangent is y − 8 = 3 ( x − 4) 3x − y − 4 = 0 The equation of the normal is y−8= − x + 3 y − 28 = 0
1 ( x − 4) 3
77
78
Chapter 16 Applications of differentiation
10. y = ( x 2 − 1)2 dy = 2( x 2 − 1)(2 x ) = 4 x ( x 2 − 1) dx dy = 4(1)(12 − 1) = 0 dx (1, 0 ) The equation of the tangent is y = 0 . The equation of the normal is x − 1 = 0 . 1
11. y = x − 1 = ( x − 1) 2 1 dy 1 = ( x − 1) − 2 dx 2 1 dy 1 1 = (2 − 1) − 2 = dx ( 2, 1) 2 2 The equation of the tangent is 1 y − 1 = ( x − 2) 2 x − 2y = 0 The equation of the normal is y − 1 = − 2( x − 2 ) 2x + y − 5 = 0 1
12. y = x x 2 + 1 = x ( x 2 + 1) 2 1 1 1 dy = ( x 2 + 1) 2 + x ( )( x 2 + 1) − 2 (2 x ) 2 dx 1 1 = ( x 2 + 1) 2 + x 2 ( x 2 + 1) − 2
= ( x 2 + 1) − 2 ( x 2 + 1 + x 2 ) 1
= ( x 2 + 1) − 2 (2 x 2 + 1) 1
dy dx ( 0,
= (0 + 1) − 2 = 1 1
0)
The equation of the tangent is y − 0 = (1)( x − 0) x−y=0 The equation of the normal is y − 0 = ( −1)( x − 0) x+y=0 13. y =
2 = 2( 4 − x ) −1 4−x
dy = 2[( −1)( 4 − x ) −2 ( −1)] = 2( 4 − x ) −2 dx dy = 2( 4 − 3) −2 = 2 dx (3, 2 )
The equation of the tangent is y − 2 = 2 ( x − 3) 2x − y − 4 = 0 The equation of the normal is 1 y − 2 = − ( x − 3) 2 x + 2y − 7 = 0
14. y =
2 = 2(1 − x ) −2 (1 − x )2
dy = 2[( −2)(1 − x ) −3 ( −1)] = 4(1 − x ) −3 dx dy = 4(1 − 0) −3 = 4 dx ( 0, 2 ) The equation of the tangent is y − 2 = 4x 4x − y + 2 = 0 The equation of the normal is 1 y−2= − x 4 x + 4y − 8 = 0 15. x 4 − x 2 y = 1 1 y = x2 − 2 x When y = 0 , x 4 − 1 = 0 ( x 2 + 1)( x − 1)( x + 1) = 0 ∴ x = 1 or −1 dy = 2 x − ( −2 x −3 ) = 2 x + 2 x −3 dx At (1, 0), dy =2+2= 4 dx (1, 0 )
The equation of the tangent is y − 0 = 4 ( x − 1) 4x − y − 4 = 0 The equation of the normal is 1 y − 0 = − ( x − 1) 4 x + 4y − 1 = 0 At (−1, 0), dy = 2( −1) + 2( −1) −3 = −4 dx ( −1, 0 ) The equation of tangent is y − 0 = − 4( x + 1) 4x + y + 4 = 0 The equation of normal is 1 y − 0 = ( x + 1) 4 x − 4y + 1 = 0 16. x 2 + 5 xy + y 2 = 4 When x = 0 , y 2 = 4 ∴ y = 2 or −2 dy dy 2 x + 5x + 5y + 2 y =0 dx dx dy 5y + 2 x =− dx 5x + 2 y
Chapter 16 Applications of differentiation
At (0, 2),
dy 5(2) + 2(0) 5 =− =− dx ( 0, 2 ) 5(0) + 2(2) 2 The equation of tangent is 5 y − 2 = − (x − 0) 2 5x + 2 y − 4 = 0 The equation of normal is 2 y − 2 = ( x − 0) 5 2 x − 5 y + 10 = 0 At (0, −2), dy dx ( 0,
− 2)
=−
5( −2) + 2(0) 5 =− 5(0) + 2( −2) 2
The equation of tangent is 5 y + 2 = − ( x − 0) 2 5x + 2 y + 4 = 0 The equation of normal is 2 y + 2 = ( x − 0) 5 2 x − 5 y − 10 = 0 17. Let ( x1, y1 ) be the point of contact. Slope of the given line = tan135° = −1 dy = 2x + 7 dx dy At ( x1, y1 ) , slope of the tangent = dx ( x1 , y1 ) = 2 x1 + 7 ∴ 2 x1 + 7 = −1 x1 = −4 When x1 = −4 , y1 = ( −4) + 7( −4) + 2 = −10 ∴ The point of contact is ( −4, − 10) . The equation of the tangent is y + 10 = ( −1)( x + 4) x + y + 14 = 0 2
dy dx ( x1 , y1 ) 9x =− 1 16 y1
At ( x1, y1 ) , slope of the tangent =
∴
−
9 x1 9 = 16 y1 8 x1 = −2 y1
When x1 = −2 y1 , 9( −2 y1 )2 + 16 y12 = 52 y12 = 1 y1 = 1, − 1 ∴ The points of contact are (−2, 1) and (2, −1). The equation of the tangent at (−2, 1) is 9 y − 1 = ( x + 2) 8 9 x − 8 y + 26 = 0 The equation of the tangent at (2, −1) is 9 y + 1 = ( x − 2) 8 9 x − 8 y − 26 = 0 19. Let ( x1, y1 ) be the point of contact. 1 Q Slope of the given line = − 3 ∴ Slope of the tangent = 3 dy = 3x 2 dx dy Slope of the tangent = dx ( x1 , y1 ) ∴
3 x12 = 3 x1 = 1, − 1
= 3 x12
When x1 = 1 , y1 = 6 When x1 = −1 , y1 = 4 ∴ The points of contact are (1, 6) and (−1, 4). The equation of the tangent at (1, 6) is y − 6 = 3 ( x − 1) 3x − y + 3 = 0 The equation of the tangent at (−1, 4) is
18. Let ( x1, y1 ) be the point of contact. 9 Slope of the given line = 8
9 x 2 + 16 y 2 = 52 dy 18 x + 32 y =0 dx dy 9x =− dx 16 y
79
y − 4 = 3 ( x + 1) 3x − y + 7 = 0 20. Let ( x1, y1 ) be the point of contact. Since ( x1, y1 ) lies on the curve, 4 x12 + 9 y12 = 36 ......................(1) dy 4x =− 1 Slope of the tangent = dx ( x1 , y1 ) 9 y1
∴
80
Chapter 16 Applications of differentiation
The slope of the line joining ( x1, y1 ) and (3, −3) = ∴
y1 + 3 x1 − 3 y1 + 3 4x =− 1 x1 − 3 9 y1
4 x12 + 9 x12 − 12 x1 + 27 y1 = 0 ...........(2) (1) − (2), 12 x1 − 27 y1 = 36 4 x1 − 9 x1 = 12 1 x1 = (12 + 9 y1 ) ...............(3) 4 Substitute (3) into (1), 1 4[ (12 + 9 y1 )]2 + 9 y12 = 36 4 (12 + 9 y1 )2 + 36 y12 = 144 y1 (117 y1 + 216) = 0 24 y1 = 0 or − 13 When y1 = 0 , x1 = 3
24 15 , x1 = − 13 13 ∴ The points of contact are (3, 0) and 15 24 (− , − ) . 13 13 At (3, 0), the equation of the tangent is x − 3 = 0 . 15 24 At ( − , − ) , 13 13 ) 4( − 15 13 = − 5 Slope of the tangent = − 24 18 9( − 13 ) When y1 = −
The equation of the tangent is 24 5 15 = − (x + ) 13 18 13 5 x + 18 y + 39 = 0 y+
(2) − (1), x1 + 10 y1 = −7 x1 = −7 − 10 y1 ......................(3)
Substitute (3) into (1), ( −7 − 10 y1 )2 − 2 y12 = 7 98 y12 + 140 y1 + 42 = 0 7 y12 + 10 y1 + 3 = 0 ( y1 + 1)(7 y1 + 3) = 0 3 7 When y1 = −1 , x1 = 3 y1 = −1 or −
3 19 When y1 = − , x1 = − 7 7 ∴ The points of contact are (3, −1) and 19 3 (− , − ) . 7 7 3 At (3, −1), slope of the tangent = −2 The equation of the tangent is 3 y + 1 = − ( x − 3) 2 3x + 2 y − 7 = 0 − 19 19 3 7 , − ) , slope of the tangent = 2( − 37 ) 7 7 19 = 6 The equation of the tangent is At ( −
3 19 19 = (x + ) 7 6 7 19 x − 6 y + 49 = 0 y+
22. y = 4 x 3 − 3 x ...................................(1) 21. Let ( x1, y1 ) be the point of contact. Since ( x1, y1 ) lies on the curve, ∴
x12 − 2 y12 = 7 ..........................(1) dy x Slope of the tangent = = 1 dx ( x1 , y1 ) 2 y1 The slope of the line joining ( x1, y1 ) and ( −1, 5) =
∴
y1 − 5 x1 + 1 y1 − 5 x = 1 x1 + 1 2 y1
x12 − 2 y12 + x1 + 10 y1 = 0 ...............(2)
dy = 12 x 2 − 3 dx 1 dy = 12( − )2 − 3 = 0 2 dx ( − 12 , 1) ∴ The equation of the tangent is y − 1 = 0 .........................................(2) Solving (1) and (2), ( x − 1)( 4 x 2 + 4 x + 1) = 0 ( x − 1)(2 x + 1)2 = 0 x = 1 or −
1 2
Chapter 16 Applications of differentiation
When x = 1 , y = 1 1 When x = − , y = 1 2 ∴ The required point is (1, 1) .
y1 = ( 4)2 − 7( 4) + 14 = 2 ∴ The coordinates of C are ( 4, 2) . (ii) Slope of the tangent at C = 2( 4 ) − 7 =1 The equation of the tangent at C is
23. x + 4 xy + 5 y = 1 2
2
2x + 4x
dy dy + 4 y + 10 y =0 dx dx dy x + 2y =− dx 2 x + 5y
Let ( x1, y1 ) be the point of contact.
y−2= x −4 x−y−2=0 25. Substitute (0, 4) into y = ax 3 + bx 2 + c ,
4 = a( 0 ) 3 + b ( 0 ) 2 + c
1 2 x1 + 2 y1 Slope of the tangent = − 2 x1 + 5 y1
Slope of the given line = −
∴
−
∴
c=4
dy = 3ax 2 + 2 bx dx
x1 + 2 y1 1 =− 2 x1 + 5 y1 2 y1 = 0
dy dx (1,
= 3a + 2 b = −3 ..................(1) 2)
Substitute y1 = 0 into the equation of the curve,
Substitute (1, 2) into y = ax 3 + bx 2 + 4 ,
x12 = 1 x1 = 1, − 1
a+b+4=2 a + b = −2 .................................(2) Solving (1) and (2),
∴ The points of contact are (1, 0) and (−1, 0). The equation of the tangent at (1, 0) is 1 y − 0 = − ( x − 1) 2 x + 2y − 1 = 0 The equation of the tangent at (−1, 0) is 1 y − 0 = − ( x + 1) 2 x + 2y + 1 = 0
24. (a) Slope of AB =
8−4 =1 6−2
(b) (i) Let the coordinate of C be ( x1, y1 ) .
dy = 2x − 7 dx dy = 2 x1 − 7 dx ( x1 , y1 )
∴
a = 1 , b = −3
26 − 28. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons.
Exercise 16B (p.133) 1.
y = x3 − 4x2 + 4x − 8 dy = 3x 2 − 8 x + 4 dx = ( x − 2)(3 x − 2) d2y = 6x − 8 dx 2
dy d2y = 0, > 0 and y = −8 dx dx 2 (2, − 8) is a minimum point.
When x = 2 , ∴
d2y 2 dy 184 < 0 and y = − , = 0, 2 3 dx 27 dx 2 184 ( , − ) is a maximum point. 3 27
Since the slope of the tangent at C is parallel to the chord AB,
When x =
2 x1 − 7 = 1 x1 = 4
∴
Substitute x1 = 4 into
Also, when x = 0 , y = −8
y1 =
A sketch of the curve is shown below:
x12
− 7 x1 + 14 ,
81
82
Chapter 16 Applications of differentiation
y
3. y = x 3 − 4x 2 + 4x − 8 x
O
(
2 184 ,− ) 3 27
−8
y = 12 x − x 3 dy = 12 − 3 x 2 = 3(2 − x )(2 + x ) dx d2y = −6 x dx 2 d2y dy < 0 and y = 16 When x = 2 , = 0, dx dx 2 ∴ (2, 16) is a maximum point.
dy d2y = 0, > 0 and y = −16 dx dx 2 ( −2, − 16) is a minimum point.
When x = −2 ,
(2, −8)
∴
Also, when x = 0 , y = 0 2.
y = 2 x − 3 x − 12 x + 10 dy = 6 x 2 − 6 x − 12 = 6( x − 2)( x + 1) dx d2y = 12 x − 6 = 6(2 x − 1) dx 2 dy d2y When x = 2 , = 0, > 0 and y = −10 dx dx 2 ∴ (2, − 10) is a minimum point. 3
2
A sketch of the curve is shown below: y (2, 16)
dy d2y = 0, < 0 and y = 17 dx dx 2 ( −1, 17) is a maximum point.
When x = −1 , ∴
y = 12x − x 3
Also, when x = 0 , y = 10 A sketch of the curve is shown below: y
O
(−1, 17)
x
10
y = 2x 3 − 3x 2 − 12x + 10 (−2, −16)
x
O
4.
(2, −10)
y = x 3 − 3x 2 + 1 dy = 3 x 2 − 6 x = 3 x ( x − 2) dx d2y = 6 x − 6 = 6( x − 1) dx 2 dy d2y When x = 2 , = 0, > 0 and y = −3 dx dx 2 ∴ (2, − 3) is a minimum point.
Chapter 16 Applications of differentiation
d2y dy < 0 and y = 1 = 0, dx dx 2 (0, 1) is a maximum point.
When x = 0 , ∴
6.
y = ( x + 2)( x − 2)3 dy = ( x + 2)[3( x − 2)2 ] + ( x − 2)3 dx = 3( x − 2)2 ( x + 2) + ( x − 2)3
A sketch of the curve is shown below:
= 4( x + 1)( x − 2)2
y
d2y = 4( x + 1)[2( x − 2)] + 4( x − 2)2 dx 2 = 8( x + 1)( x − 2) + 4( x − 2)2 = 12 x ( x − 2)
(0, 1) y = x 3 − 3x 2 + 1
x
O
When x = −1 , ∴
(2, −3)
5.
d2y dy > 0 and y = −27 = 0, dx dx 2
( −1, − 27) is a minimum point.
d2y dy = 0 and y = 0 = 0, dx dx 2 dy When x < 2 (slightly), >0 dx dy >0 When x > 2 (slightly), dx
When x = 2 ,
y = ( x − 1)3 + 1 dy = 3( x − 1)2 dx d2y = 6( x − 1) dx 2 dy d2y = 0, = 0 and y = 1 dx dx 2 dy When x < 1 (slightly), >0 dx dy When x > 1 (slightly), >0 dx dy This shows that does not change sign as x dx increases through 1, i.e. (1, 1) is neither a maximum point nor a minimum point.
dy does not change sign as x dx increases through 2. ∴ (2, 0) is neither a maximum point nor a minimum point.
When x = 1 ,
This shows that
Also, when x = 0 , y = −16 As x → − ∞ , y → + ∞ ; x → + ∞, y → + ∞. A sketch of the curve is shown below: y
∴ There is no turning point for the function. O
When x = 0 , y = 0
x 2
Also, as x → − ∞ , y → − ∞ ; y = (x + 2)(x − 2)3
x → + ∞, y → + ∞. −16
A sketch of the curve is shown below: y
(−1, −27)
(1, 1)
O
y = (x − 1) 3 + 1
83
x
7.
y = x 6 (7 − 6 x ) dy = 42 x 5 − 42 x 6 dx = 42 x 5 (1 − x )
84
Chapter 16 Applications of differentiation
d2y = 210 x 4 − 252 x 5 = 42 x 4 ( −6 x + 5) 2 dx d2y dy < 0 and y = 1 When x = 1 , = 0, dx dx 2 ∴ (1, 1) is a maximum point.
9.
d2y = ( x 2 − 1)[ −2( x 2 + x + 1) −3 ](2 x + 1) dx 2 + ( x 2 + x + 1) −2 (2 x )
dy d2y When x = 0 , = 0, = 0 and y = 0 dx dx 2 dy When x < 0 (slightly), <0 dx dy When x > 0 (slightly), >0 dx ∴ (0, 0) is a minimum point.
= −2( x 2 − 1)(2 x + 1)( x 2 + x + 1) −3 + 2 x ( x 2 + x + 1) −2
= −2( x 2 + x + 1) −3 ( x 3 − 3 x − 1) When x = 1 ,
A sketch of the curve is shown below: (1, 1)
8.
A sketch of the curve is shown below:
x2 y= 2 x +4 dy ( x 2 + 4)(2 x ) − x 2 (2 x ) 8x = 2 = 2 2 dx ( x + 4) ( x + 4)2 d 2 y ( x 2 + 4)2 (8) − 8 x[2( x 2 + 4)(2 x )] = dx 2 ( x 2 + 4) 4 8( x 2 + 4)( 4 − 3 x 2 ) = ( x 2 + 4) 4 8( 4 − 3 x 2 ) = 2 ( x + 4 )3 When x = 0 , ∴
(1,
Also, when x = 0 , y = 1 As x → − ∞ , y → 1; x → + ∞ , y → 1.
x
O (0, 0)
2 dy d2y = 0, > 0 and y = 3 dx dx 2
2 ) is a minimum point. 3 d2y dy < 0 and y = 2 When x = −1 , = 0, dx dx 2 ∴ ( −1, 2) is a maximum point.
∴
y y = x 6(7 − 6x )
x2 + 1 x2 + x + 1 dy ( x 2 + x + 1)(2 x ) − ( x 2 + 1)(2 x + 1) = dx ( x 2 + x + 1)2 2 2 = ( x − 1)( x + x + 1) −2 y=
y (−1, 2) 1
2 (1, ) 3
y=
x2 + 1 x + x+ 1 2
x
O
10. y 2 = x 2 (2 x − x 2 ) The curve is undefined for x < 0 or x > 2 , and it is composed of the two curves
d2y dy > 0 and y = 0 = 0, dx dx 2
1
1
y = x (2 x − x 2 ) 2 and y = − x (2 x − x 2 ) 2 , which are symmetrical about the x-axis. Consider
(0, 0) is a minimum point.
1
y = x (2 x − x 2 ) 2 1 1 1 dy = x[ (2 x − x 2 ) − 2 ](2 − 2 x ) + (2 x − x 2 ) 2 2 dx 1 = ( 2 x − x 2 ) − 2 (3 x − 2 x 2 )
Also, as
x → − ∞ , y → 1; x → + ∞ , y → 1.
A sketch of the curve is shown below: y 1
y=
1 d2y = ( 2 x − x 2 ) − 2 (3 − 4 x ) 2 dx 3 1 + (3 x − 2 x 2 )[ − (2 x − x 2 ) − 2 ](2 − 2 x ) 2 1 = ( 2 x − x 2 ) − 2 (3 − 4 x )
2
x x2 + 4
− (3 x − 2 x 2 )(1 − x )(2 x − x 2 ) − 2
3
O (0, 0)
x
= x (2 x − x 2 ) − 2 (2 x 2 − 6 x + 3) 3
85
Chapter 16 Applications of differentiation
When x =
A sketch of the curve is shown below:
3 3 d2y 3 dy < 0 and y = , , = 0, 2 4 2 dx dx
y
3 3 is a relative maximum of y. 4
∴
By symmetry, −
π 1 ( , ) 4 2
3 3 is a relative minimum of y. 4
π 2
O
3 3 3 ( , ) is a maximum point, and 2 4
∴
( 1 y = sin 2x 2
5π 1 , ) 4 2
x
π (
2π
3π 2
3π 1 ,− ) 4 2
(
7π 1 ,− ) 4 2
3 3 3 ( , − ) is a minimum point. 2 4 A sketch of the curve is shown below: y
(
3 3 3 , ) 4 2
y 2 = x 2(2x −x 2)
x
O
12.
y = x − 2 sin x dy = 1 − 2 cos x dx d2y = 2 sin x dx 2
When x = ∴
∴ y=
1 sin 2 x 2
∴
∴
∴
5π 5π , + 3 3
3)
d2y π dy 1 < 0 and y = , = 0, 2 4 dx 2 dx O
2
d y 5π dy 1 < 0 and y = , = 0, 2 4 dx 2 dx 5π 1 ( , ) is a maximum point. 4 2
x π π ( , − 3 3
2π 3)
13. (a) The coordinates of A are (n, 0) . The coordinates of B are ( m, 0) .
2
1 d2y 7π dy > 0 and y = − , = 0, 2 2 4 dx dx 7π 1 ( , − ) is a minimum point. 4 2
When x =
(
y = x − 2 sin x
3π dy 1 d y , = 0, > 0 and y = − 2 4 dx 2 dx 3π 1 ( , − ) is a minimum point. 4 2
When x = ∴
y
π 1 ( , ) is a maximum point. 4 2
When x =
5π dy 5π d2y , + 3 = 0, < 0 and y = 2 3 dx 3 dx 5π 5π ( , + 3 ) is a maximum point. 3 3
A sketch of the curve is shown below:
dy = cos 2 x dx d2y = −2 sin 2 x dx 2 When x =
π π ( , − 3 ) is a minimum point. 3 3
When x =
3 3 3 ( ,− ) 4 2
11.
d2y π dy π > 0 and y = − 3 , =0, 2 3 dx 3 dx
(b)
y = −( x − m)( x − n) dy = −( x − m ) − ( x − n) dx = −2 x + m + n m+n = −2( x − ) 2 d2y = −2 dx 2
86
Chapter 16 Applications of differentiation
When x =
d2y m+n dy < 0 and , =0, 2 dx dx 2
( m − n) 2 , 4 m + n ( m − n) 2 , ) is a maximum point. ∴ ( 2 4 y=
y = x 3 − 4 x 2 − 16 x + 15 dy = 3 x 2 − 8 x − 16 dx = ( x − 4)(3 x + 4) d2y = 6x − 8 dx 2 = 2 (3 x − 4 ) d2y dy > 0 and y = −49 = 0, dx dx 2 ( 4, − 49) is a minimum point.
When x = 4 ,
f ( x ) = x 2 ( x − p) + q
14. (a) (i)
15. (a)
f ' ( x ) = x 2 + ( x − p)(2 x )
∴
= x 2 + 2 x 2 − 2 px = 3 x 2 − 2 px
d2y −4 dy 23 < 0 and y = 26 , = 0, 2 3 dx 27 dx 4 23 ( − , 26 ) is a maximum point. 3 27
When x =
(ii) f ' ( x ) = 3 x 2 − 2 px = x (3 x − 2 p )
∴
By referring to the figure , when x = 2 , f ' ( x) = 0
3(2) − 2 p = 0 p=3
(b) When x = 0 , y = 15 A sketch of the curve is shown below: y
(b) Substitute p = 3 , f ( x ) = 0 and x = 1 into f ( x ) = x 2 ( x − p) + q ,
(−
4 23 , 26 ) 27 3
(1)2 (1 − 3) + q = 0 q=2 15
The equation becomes f ( x ) = x 2 ( x − 3) + 2 . A sketch of the curve is shown below:
y = x 3 − 4x 2 − 16x + 15
y −3
y = x 2(x − 3) + 2
x
O
6
2
1−
3
O
x 1+
1
3
(4, −49) (2, −2)
87
Chapter 16 Applications of differentiation
16. (a) y = x 3 + ax 2 + bx + c dy = 3 x 2 + 2 ax + b dx
(d) When x = 0 , y = −5 A sketch of the curve is shown below: y
Since C passes (5, 0), 0 = 125 + 25a + 5b + c .................(1)
Since C touches x-axis at ( −1, 0) ,
0 = −1 + a − b + c ........................(2) 0 = 3 − 2 a + b .............................(3) (1) − (2), 0 = 126 + 24 a + 6b
0 = 21 + 4 a + b ...........................(4) (4) − (3), 0 = 18 + 6 a ∴
y = x 3 − 3x 2 − 9x − 5
a = −3
Substitute a = −3 into (3), ∴
0 = 3+6+ b b = −9
(−1, 0) −2
x
−1 O −5
5
6
Substitute b = −9 into (2), ∴
0 = −1 − 3 + 9 + c c = −5
(b) By (a), y = x 3 − 3x 2 − 9 x − 5 dy = 3x 2 − 6 x − 9 dx = ( x + 1)( x − 3) d2y = 6x − 6 dx 2 = 6( x − 1) d2y dy < 0 and y = 0 When x = −1 , = 0, dx dx 2 ∴ ( −1, 0) is a maximum point. d2y dy > 0 and y = −32 = 0, dx dx 2 (3, − 32) is a minimum point.
When x = 3 , ∴
(c) When x = 5 , slope of the tangent = 3(5) − 6(5) − 9 = 36 2
The equation of the tangent is y − 0 = 36( x − 5) 36 x − y − 180 = 0
(3, −32)
17. (a) Substitute x = 0 into y = f ( x ) = 3 sin 2 x + cos 2 x , y = 3 sin 2(0) + cos 2(0) = 1 The coordinates of the point of intersection are (0, 1) . Substitute y = 0 into y = f ( x ) = 3 sin 2 x + cos 2 x,
3 sin 2 x + cos 2 x = 0 1 3 π 2x = − , 6 π x=− , 12
tan 2 x = −
5π 6 5π 12
The coordinates of the points of intersection π 5π are ( − , 0), ( , 0) . 12 12
88
Chapter 16 Applications of differentiation
(b)
y = 3 sin 2 x + cos 2 x dy = 2 3 cos 2 x − 2 sin 2 x dx d2y = −4 3 sin 2 x − 4 cos 2 x dx 2 dy When = 0 , 2 3 cos 2 x − 2 sin 2 x = 0 dx
tan 2 x = 3 2π π , 2x = − 3 3 π π x=− , 3 6
∴
When
dy = 3 x 2 − 18 + 15 dx d2y = 6 x − 18 dx 2
dy = 0 , 3 x 2 − 18 x + 15 = 0 dx
x2 − 6x + 5 = 0 ( x − 1)( x − 5) = 0 x = 1 or x = 5
d2y < 0 and y = 16 dx 2 (1, 16) is a maximum point.
When x = 1 , ∴
π d2y > 0 and y = −2 When x = − , 3 dx 2 π ∴ ( − , − 2) is a minimum point. 3 When x =
(b) From (a),
d2y > 0 and y = −16 dx 2 (5, − 16) is a minimum point.
When x = 5 , ∴
π d2y < 0 and y = 2 , 6 dx 2
(c) A sketch of the curve y = x 3 − 9 x 2 + 15 x + 9 is shown below:
π ( , 2) is a maximum point. 6
y (1, 16)
(c) A sketch of the curve is shown below: y π ( , 2) 6
y = x 3 − 9x 2 + 15x + 9
1
−
π 2
−
π O 12
5π π 12 2
y=
x
3 sin 2x + cos 2x
O
x 1
3
5
π (− , −2) 3
18. y = x 3 − 9 x 2 + 15 x + 9, 0 ≤ x ≤ 7 (a)
dy = 3 x 2 − 18 x + 15 dx dy For ≤ 0 , 3 x 2 − 18 x + 15 ≤ 0 dx x2 − 6x + 5 ≤ 0 ( x − 1)( x − 5) ≤ 0 1≤ x ≤ 5
(5, −16)
Chapter 16 Applications of differentiation
A sketch of the curve y 2 = x 3 − 9 x 2 + 15 x + 9 is shown below: y
2. Let x cm be the length of side of square base, y cm be the height, and V cm 3 be the volume. Q
y=
∴ O
1
3
x
−4
19 − 22. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons.
Exercise 16C (p.142) 1. Let x cm, y cm be the length and width respectively and A cm 2 be the area. Q 2( x + y) = 100 x + y = 50 x = 50 − y ∴ A = xy = (50 − y) y
x 2 + 4 xy = 300 300 − x 2 4x 75 x = − x 4
y 2 = x 3 − 9x 2 + 15x + 9
4
89
V = x2y
75 x − ) x 4 x3 = 75 x − 4 1 dV = 75 − (3 x 2 ) 4 dx 3 d 2V =− x 2 dx 2 dV When = 0, dx = x2 (
75 −
3x 2 =0 4 x = 10 or −10 (rejected)
d 2V 3 = − (10) = −15 < 0 2 2 dx Hence the volume of the box is maximum when x = 10 . When x = 10 , y = 5 When x = 10 ,
∴ If the box has a maximum volume, its dimensions are 10 cm × 10 cm × 5 cm .
= − y 2 + 50 y dA = −2 y + 50 dy d2A = −2 dy 2 dA When =0, dy −2 y + 50 = 0 y = 25 d2A When y = 25, = −2 < 0 dy 2 Hence the area of the rectangle is maximum when y = 25.
3. Transpose a circle of radius 5 cm so that its centre is transposed to the origin. Let θ be the acute angle between a diagonal and the positive x-axis, A cm 2 be the area of the inscribed rectangle. y
5
cm
θ
O
When y = 25, A = −(25)2 + 50(25) = 625 ∴ The maximum area of the rectangle is
625 cm 2 .
A = 4(5 cos θ)(5 sin θ) = 100 cos θ sin θ = 50 sin 2θ
x
90
Chapter 16 Applications of differentiation
dA = 100 cos 2θ dθ d2A = −200 sin 2θ dθ 2 π dA When = 0, θ = 4 dθ
d2A = −200 < 0 dθ 2 θ = π 4
∴ The area of a rectangle is maximum when π θ= . 4 π Maximum area = 50 sin 2( ) cm 2 = 50 cm 2 4
1 1 A = x 2 + π[ (3 − 2 x )]2 = x 2 + (3 − 2 x )2 π π dA 1 4 = 2 x + (2)(3 − 2 x )( −2) = 2 x − (3 − 2 x ) dx π π 6 dA When = 0, x = π+4 dx 6 When x = , π+4 4 d2A = 2 − ( −2) 2 π dx 8 =2+ >0 π
6 , the area is minimum but π+4 not maximum.
∴ When x =
4. Let x m, y m be the length and width respectively, and A m 2 be the area.
x + 2 y = 120 A = xy ∴
A = y(120 − 2 y) = 120 y − 2 y 2 dA = 120 − 4 y dy
d2A = −4 dy 2 dA = 0 , y = 30, When dy d2A = −4 < 0 dy 2 Hence the area of enclosure is maximum when y = 30 . When y = 30, x = 60 ∴ Maximum area of enclosure is A = (60)(30) m 2 = 1 800 m
2
5. Let x m be the length of the side of the square, 4x m be part of the wire used to form the square, (6 − 4x) m be the remaining part of the wire to form the circle of radius r m, A m 2 be the sum of the areas of the circle and the square. Q 6 − 4x ≥ 0 3 ∴ x≤ 2 3 i.e. 0 ≤ x ≤ 2 2 πr = 6 − 4 x 1 r = (3 − 2 x ) π
9 π 3 9 When x = , A = 2 4
When x = 0 , A =
9 9 > π 4 ∴ If the total area is maximum, the length of the side of the square is 0 .
Q
x y + = 1. a b Substitute P(5, 3) into the equation,
6. (a) The equation of AB is
5 3 + =1 a b 3a + 5b = ab 3a b= a−5 1 ab square units 2 3a 2 square units = 2( a − 5)
(b) Area of ∆OAB =
(c) Let A square units be the area of ∆OAB. 3a 2 2( a − 5) dA 3 ( a − 5)(2 a) − a 2 = da 2 ( a − 5)2 2 3( a − 10 a) = 2( a − 5)2 A=
When
dA = 0 , a = 10 or 0 (rejected) da
Chapter 16 Applications of differentiation
d2A da 2 3 ( a − 5)2 (2 a − 10) − ( a 2 − 10 a)(2 a − 10) = 2 ( a − 5) 4
d2A 3 = >0 2 da a =10 5 Hence the area of ∆OAB is minimum when a = 10 . ∴ Least area of ∆OAB 3(10)2 square units 2(10 − 5) = 30 square units
=
When
91
dA = 0, r = 5 dr
1 000 π d2A = 4π + 2 dr r3 2 d A = 12 π > 0 dr 2 r = 5 Hence when the base radius of each can is 5 cm , the material required will be the least. 9. Let r cm be the radius of the cylindrical can, h cm be the height and V cm 3 be the volume. 2 πr 2 + 2 πrh = 600 π
7. Let x cm be the length of side of square base, h cm be the height and $C be the cost.
V = πr 2 h = π(300 r − r 3 ) dV = π(300 − 3r 2 ) dr
x 2 h = 64 C = 20 x 2 + 40 xh 2 560 = 20 x 2 + x dC 2 560 = 40 x − 2 dx x
When
dC = 0, x = 4 dx
5 120 d 2C = 40 + 3 2 dx x 2 d C = 120 > 0 dx 2 x = 4 Hence the cost of the container is minimum when x = 4. 2 560 ∴ C = 20( 4)2 + 4 = 960 The minimum cost of the container is $960 . 8. Let r cm be the radius of each can, h cm be the height, 2
When
dV = 0 , r = 10 dr
d 2V = π( −6r ) dr 2 d 2V = −60 π < 0 dr 2 r =10 Hence the can will have the largest volume when the base radius is 10 cm . 10. Let s km be the distance between A and B after t hours. Then s 2 = (15t )2 + (90 − 20t )2 ds 2s = 450t + 2(90 − 20t )( −20) dt ds 625t − 1 800 = dt s When
ds = 0 , t = 2.88 dt
When t > 2.88 (slightly),
A cm be the surface area. πr 2 h = 250 π
A = 2 πr 2 + 2 πrh ∴
500 π r dA 500 π = 4 πr − 2 dr r A = 2 πr 2 +
ds >0 dt
ds <0 dt When t = 2.88 , s attains its minimum. Hence two ships will be closest at When t < 2.88 (slightly),
2 : 52 : 48 p.m.
92
Chapter 16 Applications of differentiation
11. Let x km be BP, s km be the length of the road.
13. (a) πr 2 h = 48π 48 h= 2 r
s = x 2 + 4 + (10 − x )2 + 9 1
1
= ( x 2 + 4) 2 + (100 − 20 x + x 2 + 9) 2 1
1
= ( x 2 + 4) 2 + (109 − 20 x + x 2 ) 2 1 ds 1 2 = ( x + 4) − 2 (2 x ) dx 2 1 1 + (109 − 20 x + x 2 ) − 2 ( −20 + 2 x ) 2 1 1 = x ( x 2 + 4) − 2 + ( x − 10)(109 − 20 x + x 2 ) − 2
When
ds = 0, x = 4 dx
3 3 d 2s = 4( x 2 + 4) − 2 + 9(109 − 20 x + x 2 ) − 2 2 dx d 2s >0 dx 2 x = 4
Hence S = 2 πrh + πr 2 48 = 2 πr ( 2 ) + πr 2 r 96 π = + πr 2 r (b) By (a), 96 π S= + πr 2 r dS 96 π = − 2 + 2 πr dr r d 2 S 192 π = 3 + 2π dr 2 r dS Set = 0, dr −
Hence the length of the road is its smallest when the location is 4 km from B. 12. (a) CD = CE 2 + DE 2
1
r = ( 48) 3
= 36 x 2 + 64 x 2 = 10 x
d 2S 192 π = + 2π = 4π + 2π = 6π > 0 2 1 48 dr r = ( 48) 3
(b) 8 x + y + 6 x + y + 10 x = 144 2 y + 24 x = 144 2 y = 144 − 24 x y = 72 − 12 x (c) Let A cm 2 be the area of ABCD. 1 A = (6 x )(8 x + y + y) 2 = 3 x (8 x + 2 y) = 3 x (8 x + 144 − 24 x ) = 3 x (144 − 16 x )
= 48(9 x − x 2 ) dA = 48(9 − 2 x ) dx d2A = −96 dx 2 dA When = 0 , 48(9 − 2 x ) = 0 dx 9 x= 2 9 When x = , 2 d2A 81 81 = −96 < 0 A = 48( − ) = 972 and 2 4 dx 2 Hence the maximum area of ABCD is
972 cm 2 .
96 π + 2 πr = 0 r2 96 π 2 πr = 2 r r 3 = 48
1
Hence S attains a minimum when r = ( 48) 3 .
14. (a)
1 2 128 πr h = π 3 3 128 r2 = h
(b) The slant height of the cone = h 2 + r 2 cm 128 cm = h2 + h (c) Let l cm be the slant height. 128 h dl 1 2 128 − 12 ) (2 h − 128h −2 ) = (h + dh 2 h 128 − 12 ) (h − 64h −2 ) = (h 2 + h h 3 − 64 = 3 1 h 2 (h 3 + 128) 2 l = h2 +
Chapter 16 Applications of differentiation
When
16. (a)
dl = 0, h = 4 dh
When h > 4 (slightly),
93
B
dl >0 dh
16 m
dl <0 dh Hence the slant height is minimum when h = 4. When h < 4 (slightly),
θ
C 2m
128 l = 42 + 4 =4 3
15. (a) xy − (2)(2 y) − (2)( 4)( x − 4) = 200 8( x + 21) y= x−4 (b) The area of the board including the margins = xy m 2 8 x ( x + 21) 2 = m x−4
4m
Let φ be ∠ACD.
16 + 2 x tan θ + tan φ 18 = 1 − tan θ tan φ x tan θ + 2x 18 = 2 1 − (tan θ)( x ) x 16 x (x > 0) tan θ = 2 x + 36 (b) Let T = tan θ tan(θ + φ) =
dT ( x 2 + 36)(16) − 16 x (2 x ) = dx ( x 2 + 36)2 −16( x 2 − 36) = ( x 2 + 36)2
(c) Let A m 2 be the area of the board. 8 x ( x + 21) x−4 dA 8( x − 4)(2 x + 21) − 8( x 2 + 21x ) = dx ( x − 4)2 2 8 x − 64 x − 672 = ( x − 4)2 8( x − 14)( x + 6) = ( x − 4)2 A=
(c) When
dT >0 dx Thus when x = 6 , T attains its maximum value.
16(6) 6 2 + 36 θ = 0.93 (corr. to 2 d.p.) ∴ The required position is 6 m from the screen and the greatest view angle is 0.93 rad. tan θ =
dA <0 dx Hence the area of the board attains minimum when x = 14 . When x < 14 (slightly),
When x = 14 , y = 28
The minimum area of the board is 392 m 2 and the corresponding height is 14 m .
dT <0 dx
When x < 6 (slightly),
dA >0 dx
A = (14)(28) = 392
dT = 0 , x = 6 or −6 (rejected) dx
When x > 6 (slightly),
dA = 0 , x = 14 or −6 (rejected) dx
When x > 14 (slightly),
φ
xm
∴ The minimum slant height is 4 3 cm .
When
A D
17. (a)
y 2 = 2 mx dy 2y = 2m dx dy m = dx y
94
Chapter 16 Applications of differentiation
dy m = dx ( 2 mt 2 , 2 mt ) 2 mt 1 = 2t
Thus when t =
minimum value and so does L. 1 1 When t = or − , 2 2
2 ms − 2 mt 2 ms 2 − 2 mt 2 2 m( s − t ) = 2 m( s − t )( s + t ) 1 = s+t
Slope of AB =
m 2 (1 + 2)3 1 = 27m 2 L = 3 3m or −3 3m (rejected) L 2=
∴ The minimum value of L is 3 3m .
(b) Q Line AB is perpendicular to the tangent to C at A. ∴
(
18. (a) Base radius = R cot θ
1 1 )( ) = −1 2t s + t 1 = −2t ( s + t )
Height = ( R cot θ) tan 2θ 1 ∴ V = π( R cot θ)2 ( R cot θ tan 2θ) 3 1 3 = πR tan 2θ cot 3 θ 3
1 = −2ts − 2t 2 2ts = −(1 + 2t 2 ) s=−
1 + 2t 2 2t
(b)
L 2 = (2 mt 2 − 2 ms 2 )2 + (2 mt − 2 ms)2 = 4 m 2 [(t 2 − s 2 )2 + (t − s)2 ] = 4 m 2 [(t − s)2 (t + s)2 + (t − s)2 ] =
dV 1 3 d tan 2θ cot 3 θ = πR dθ 3 dθ 1 = πR3 ( −3 tan 2θ cot 2 θ csc 2 θ 3 + 2 cot 3 θ sec 2 2θ)
m 2 (1 + 4t 2 )3 4t 4
=
m 2 (1 + 4t 2 )3 4t 4 d( L 2) m2 = [( 4t 4 )3(1 + 4t 2 )2 (8t ) 8 dt 16t − (1 + 4t 2 )3 (16t 3 )]
(c) L 2 =
m (1 + 4t ) (2t − 1) t5 d( L 2) When = 0, dt m 2 (1 + 4t 2 )2 (2t 2 − 1) = 0 =
2
2 2
2
Q
1 + 4t 2 ≠ 0 and m 2 ≠ 0
∴
2t 2 − 1 = 0 1 1 t= or − 2 2
When
d (L ) dt 2 t = −
= 1 2
m 2 ( 4 + 9 + 5) 1 8
1 dV = 0 , cos 2θ = 3 dθ
When 2θ > cos −1
1 dV (slightly), >0 3 dθ
1 dV (slightly), <0 3 dθ ∴ The minimum volume of the cone occurs 1 at cos 2θ = . 3 (c) By (b), When 2θ < cos −1
1 3 1 2 2 cos θ − 1 = 3 2 1 2 cos θ = , sin 2 θ = 3 3 ∴
= 144 m 2 > 0
cot 2 θ = 2
Since 0 < θ <
2
2
1 3 cot 2 θ sin 2θ(1 − 3 cos 2θ) πR 3 cos 2 2θ sin 2 θ
cos 2θ =
d 2 ( L 2 ) m 2 (32t 6 + 18t 2 + 5) = dt 2 t6 2 2 2 d (L ) m ( 4 + 9 + 5) = = 144 m 2 > 0 2 1 dt 1 8 t= 2
1 1 or − , L 2 attains a 2 2
∴
π , 4
cot 3 θ = ( 2 )3 =2 2
Chapter 16 Applications of differentiation
sin 2θ = 1 − cos 2 2θ 1 = 1− 9 8 = 3 sin 2θ tan 2θ = cos 2θ =2 2 ∴ The minimum volume of the cone 1 = ( πR3 )(2 2 )(2 2 ) 3 8 3 = πR 3
∴ When r = 10 ,
3. Let x, y be OA, OB respectively. x 2 + y 2 = 132 dx dy 2x + 2y =0 dt dt (a) When x = 5 , y = 12 , ∴
5 unit/s towards the 6
5 dm −(5)( − 6 ) + (12)(2) = dt 52 169 = 150 The slope of AB is increasing at the rate of 169 per second. 150
When x = 5 ,
The acceleration of the particle is −4 m / s 2 .
S = 4 πr 2 dS dr = 8πr dt dt dr 3 = dt 8πr 4 V = πr 3 3 dV dr 3 3 = 4 πr 2 = 4 πr 2 ( )= r dt dt 8πr 2
dy =0 dt dy 5 =− dt 6
(b) Let m be the slope of AB. y m=− x dy dx dm − x dt + y dt = dt x2
∴ The velocity of the particle is 0 m / s .
2. Let r cm, S cm 2 and V cm 3 be the radius, surface area and volume of the balloon respectively.
dx =2 dt
origin.
When t = 1 , v = 0 and a = −4
0 ≤ t ≤ 3 is 29 m .
2(5)(2) + 2(12)
∴ B is moving at
Exercise 16D (p.153)
(b) When t = 0 , s = 20 m When t = 3 , s = 29 m When t = 1 , s = 21 m ∴ The maximum distance between the particle and A within the time internal
dV 3 = (10) = 15 dt 2
∴ The increase rate is 15 cm 3 / s .
19 − 22. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons.
1. (a) L e t v m / s , a m/s 2 b e t h e v e l o c i t y a n d acceleration of the particle respectively. ds v= = 4t 3 − 12t 2 + 8t dt dv a= = 12t 2 − 24t + 8 dt
95
4. Let r cm, h cm, A cm 2 , V cm 3 and l cm be the radius of surface, height area, volume and slant height of water inside the cone respectively. r 6 = h 12 h 2 1 1 V = πr 2 h = πh 3 3 12 dV 1 2 dh = πh dt dt 4 dh 4π 1 When h = 6 , = = dt π(6)2 9 ∴
h = 2r , r =
∴ The rate of the water level rising is
1 cm / s . 9
96
Chapter 16 Applications of differentiation
8. Let s km be the distance between two cars after t hours.
h 5 l = h 2 + ( )2 = h 2 2 h 5h 5π 2 )= A = πrl = π( )( h 2 2 4 dA 5π dh = h dt dt 2 When h = 6 ,
s 2 = 6 400t 2 + 3 600t 2 = 10 000t 2 When t = 3 , s = 300 2s
dA 5π 1 = (6) dt 2 9 5π = 3
ds 60 000 = dt 2(300) = 100 ∴ After 3 hours, two cars are separating at the rate 100 km / h .
When t = 3 ,
∴ The rate of change of the area of the wet part of the cone at this instant is
ds = 20 000t dt
5π cm 2 / s . 3
5. Let x cm and V cm 3 be the side and volume of the cube respectively. V = x3 dV dx = 3x 2 dt dt dV = 3(20)2 (0.01) dt = 12 ∴ The rate of change of the volume is When x = 20 ,
12 cm 3 / h . 6. Let x m be the distance between the man and the lamp post, y m be the distance between the lamp post and tip of the shadow. y − x 180 1 = = y 540 3 3x = 2 y Differentiate both sides with respect to time t: dx dy 3 =2 dt dt dy 3 = (2) = 3 dt 2 ∴ The tip of his shadow is moving at the rate 3 m / s. 7. Let x cm, y cm and A cm 2 be the length, width and area of the rectangle respectively. A = xy dA dy dx =x +y dt dt dt When x = 20 , y = 10 ,
dA −3 = 20( ) + 10(1) dt 2 = −20 ∴ The rate of change of the area is −20 cm 2 / s .
9. Let r cm, C cm and A cm 2 be the radius, circumference and area of the circle respectively. A = πr 2 dA dr = 2 πr dt dt −6 dr = dt 2 πr −3 = πr C = 2 πr dC dr = 2π dt dt −3 dC = 2 π( ) πr dt −6 = r dC 6 = − = −2 dt 3 ∴ The rate of change of the circumference is −2 cm / s . When r = 3 ,
10. Let s m be the distance between B and the balloon after t seconds. s 2 = 36t 2 + 2 500
2s
ds = 72t dt ds 72t 36t = = dt 2s s
When t = 20 , s = 130 , ds (36)(20) = dt 130 7 =5 13 ∴ The rate of change of the distance of the 7 balloon from place B is 5 m / s. 13
Chapter 16 Applications of differentiation
11.
s = 5t 2 ds = 10t dt
dh = dt
When t = 5 , ds = 10(5) = 50 dt ∴ The velocity of the stone is 50 m / s . Let x m be the distance between the stone and boat. x = 10 000 + (200 − 5t ) dx 2x = 2(200 − 5t 2 )( −10t ) dt dx −50t ( 40 − t 2 ) = dt x 2
2 2
When t = 5 , x = 125 dx 50(5)( 40 − 25) =− dt 125 = −30 ∴ The rate of change of the distance between the stone and the boat is −30 m / s .
12.
y = x2 + 8 1
= ( x 2 + 8) 2 1 dx dy 1 2 = ( x + 8) − 2 (2 x ) dt 2 dt x dx = x 2 + 8 dt
When x = 1 ,
dy 1 = (3) dt 3 =1
13. Let x m, h m and V m 3 be the side, height and volume of water in the trough respectively. h = x cos 30° 2 x= h 3 1 V = ( x 2 sin 60°)(2) 2 3 2 x = 2 2 3 2 h = 3 dV 4 3 dh h = dt dt 3 When h =
1 , 2
3 4 3 1 ( ) 3 2
=
97
3 2
∴ The rate of change of the water level rising is 3 m / min. 2 14. Let r cm, h cm and V cm 3 be the base radius, height and volume of the conical pile. 2 r 3 1 2 V = πr 2 h = πr 3 3 9 dV 2 2 dr = πr dt 3 dt h=
When r = 12 , dV 2 1 = π(12)2 ( ) = 12 π dt 3 8 ∴ The rate of change of the volume is
12π cm 3 / s . 15. Let r cm, h cm and V cm 3 be the radius, height and volume of water inside the circular cone respectively. 1 r V = πr 2 h , h = = 3r 3 tan 30° 3 3 1 3 V= πr = πh 3 9 dV 1 2 dh = πh dt 3 dt When h = 6 , dV 1 = π(6)2 (0.2) dt 3 = 2.4 π ∴ The rate of water leaking away is (10 − 2.4 π) cm 3 / s . 16. Let x m be the length of shadow, θ be the angle of elevation. ∴ x = 180 cot θ dx dθ = −180 csc 2 θ dt dt When θ = 30° , dx 0.25π = −180( 4)( ) dt 180 = −π ∴ The rate of change of the length of the shadow is −π m / min.
98
Chapter 16 Applications of differentiation
17. (a) Let r cm be the radius of the stain. After 4 seconds, the area of the stain = 4 cm . 2
∴
πr 2 = 4 4 1 r = ( )2 π ∴ The radius of the stain 4 seconds after the 4 1 ink first touched the shirt was ( ) 2 cm . π
(b) Let A cm 2 be the area of the stain. A = πr 2 dA dr = 2 πr dt dt 4 1 dA Q = 1, r = ( ) 2 (at this instant, from (a)) dt π 4 12 dr ∴ 1 = 2 π( ) π dt dr 1 = dt 4 π ∴ The rate of increase of the radius of the 1 cm / s . stain at this instant is 4 π 18. No solution is provided for the H.K.C.E.E. question because of the copyright reasons. 19. (a) Using cosine formula, we obtain PQ 2 = OP 2 + OQ 2 − 2(OP)(OQ)cosθ π When Q = , 3 5 2 π ( ) = ( 3 )2 + x 2 − 2 3 ( x ) cos 2 3 13 2 x − 3x − =0 4
x=
3+4 2
or
3−4 (rejected) 2
5 (b) ( )2 = ( 3 )2 + x 2 − 2 3 ( x ) cos θ 2 13 x 2 − 2 3 x cosθ − = 0 4
∴
−6 x dx = dt 2 x − 3 3 +4 2 ) 3 +4 2 )− 3
−6(
=
2(
−3( 3 + 4) 4 ∴ The rate of change of OQ with respect to =
time is
−3( 3 + 4) m / s. 4
20. (a) When the depth is h m, using the properties of similar figures, h upper base = 1 + 2( ) = 1 + h 2 ∴ V = 1 (1 + h + 1)h(3) 2 3 = (h 2 + 2 h) 2 3 2 (b) V = (h + 2 h) 2 dV 3 dh = (2 h + 2) dt 2 dt dh = 3(h + 1) dt 1 When h = , 2 dh 4 8 = = dt 3( 12 + 1) 9 ∴ The rate of change of the water level is 8 m / min. 9 21. (a) Let r cm, h cm and V cm 3 be the radius, height and volume of space left inside the circular conical vessel. r = h tan 45° = h 1 ∴ V = πh 3 3 dV dh = πh 2 dt dt
Differentiate both sides with respect to time t,
When h = 6 ,
dx dθ dx + 2 3 x sin θ − 2 3 cos θ =0 dt dt dt π dθ When Q = and = 2, 3 dt π π dx dx + 2 3 x (sin )(2) − 2 3 cos =0 2x dt 3 3 dt dx dx + 6x − 3 =0 2x dt dt
dh −12 = dt π(36) 1 =− 3π ∴ The rate of the water level rising is 1 cm / s . 3π
2x
Chapter 16 Applications of differentiation
(b) Let A cm 2 be the area of the surface of water in the vessel. A = πr 2 = πh 2 dA dh = 2 πh dt dt When h = 6 , dA 1 = 2 π(6)( − ) dt 3π = −4 ∴ The rate of decrease of the area of the surface of the water at that moment is 4 cm 2 / s . π (16 + 12 h − h 3 ) 3 dV π = (12 − 3h 2 ) dh 3 = π( 4 − h 2 )
22. (a) V =
∴
∴
dh dt
dV dh = π( 4 − h 2 ) dt dt dV When h = 1 and = −6π , dt dh −6 π = π[ 4 − (1)2 ] dt dh = −2 dt ∴ The rate of change of h is −2 m / hour.
(c) (i) Radius of the surface = 2 2 − h 2
A = π( 4 − h 2 ) 2 = π( 4 − h 2 )
(ii)
dA dh = π( −2 h) dt dt dA When h = 1 , = π( −2)( −2) dt = 4π ∴ The rate of change of A is 4π m 2 / hour.
23. (a)
A=−
=−
x = A sin 2t + B cos 2t dx = 2 A cos 2t − 2 B sin 2t dt d2x = −4 A sin 2t − 4 B cos 2t dt 2
2 3 , B=− 25 50
2 3 sin 2t − cos 2t 25 50
(b) x = −
1 4 3 ( sin 2t + cos 2t ) 10 5 5
1 4 cos(2t − θ) where θ = tan −1 ( ) 10 3 ∴ The maximum displacement =−
= − =
(b) From (a),
∴
− 4 A sin 2t − 4 B cos 2t + 6 A cos 2t − 6 B sin 2t − 4 A sin 2t − 4 B cos 2t = sin 2t
−8 A − 6 B = 1 6 A − 8B = 0
dV dV dh = ⋅ dt dh dt = π( 4 − h 2 )
99
1 10
1 10
(c) Let v be the speed of the particle. dx v= dt 4 3 = − cos 2t + sin 2t 25 25 2 A2 + B2 − x 2 1 4 9 2 3 = 2[ + − ( − sin 2t − cos 2t )2 ] 2 625 2 500 25 50 4 4 = 2[ − sin 2 2t 625 625 2 3 − 2( sin 2t )( cos 2t ) 25 50 1 9 9 + − cos 2 2t ] 2 2 500 2 500 4 = 2[ cos 2 2t 625 1 2 3 9 − 2( )( )(sin 2t cos 2t ) + sin 2 2t ] 2 25 50 2 500 16 4 3 =[ cos 2 2t − 2( )( )(sin 2t cos 2t ) 625 25 25 1 9 2 2 + sin 2t ] 625 4 3 = ( − cos 2t + sin 2t )2 25 25 4 3 = − cos 2t + sin 2t 25 25 ∴
v = 2 A2 + B2 − x 2
100
Chapter 16 Applications of differentiation
(d) v = 2 A2 + B2 − x 2 As x 2 ≥ 0 , it is obvious that v is maximum when x = 0. ∴ The maximum speed = 2 A2 + B2 1 = 5
24. (a) (i) Since the minute-hand and the hour-hand move in the same direction at constant rates, the required rate = the rate at which the hour-hand moves − the rate at which the minute-head moves π 2π =( − ) rad. / min. 360 60 −11π = rad. / min. 360 2π 3 If OA coincide with OB after t minutes, 11π 2π t= 360 3 9 t = 21 11 ∴ OA first coincides with OB at 4 : 21 : 49 p.m. (corr. to the nearest second).
s 2 = 2 2 + 12 − 2(2)(1) cos θ = 5 − 4 cos θ ∴
2s
ds dθ = 4 sin θ dt dt ds 2 sin θ dθ = dt s dt
At 4:00 p.m., 2π θ= , 3
s 2 = 5 − 4 cos s= 7 Q
dθ 11π =− dt 360 2π , 3 2 sin 23π 11π ds =( )( − ) 360 dt 7
∴ When θ =
(ii) At 4:00 p.m., θ =
(b) Let A cm 2 be the area of ∆OAB. 1 (2)(1)sin θ 2 = sin θ dA dθ = cos θ dt dt A=
When θ =
2π , 3
2 π 11π dA = cos (− ) 3 360 dt = −0.05 (corr. to the nearest 0.01) ∴ The rate of change of the area of ∆OAB with respect to time at 4:00 p.m. is − 0.05 cm 2 / min. (c) Let s cm be AB. Using cosine formula,
2π 3
=−
11 3π 360 7
= −0.06 (corr. to the nearest 0.01)
∴ The rate of change of AB with respect to time at 4:00 p.m. is −0.06 cm / min. 25 − 26. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons.
Revision Exercise 16 (p.158) 1. y + xy = 6 x 3
dy 1 + dx 2
y 1 + x 2 (1 +
1 2
x dy = 18 x 2 y dx 1 x dy = 18 x 2 − ) 2 y dx
y x y
dy 18 x − 2 x = y dx 1+ 1 2
1
2
x
The slope of the tangent to the curve at (1, 4) is 1 dy 18 − 2 4 = dx 1 + 1 1 2 4 68 = 5
Chapter 16 Applications of differentiation
When x = 0 , y = 3
2. y 2 + 10 x = 0
2y
A sketch of the curve is shown below :
dy + 10 = 0 dx dy 5 =− dx y
2 17 y (− , 4 ) 3 27
Slope of the line = − ∴
5 1 − =− y 2 y = 10
∴
x=
1 2
y = 2x 3 − x 2 − 4x + 3 3
− y2 = −10 10
∴ The equation of the tangent is 1 y − 10 = − ( x + 10) 2 x + 2 y − 10 = 0 O
3.
f ( x ) = x 3 − 3x + 4 5.
f ' ( x ) = 3x 2 − 3 When f ' ( x ) > 0 , the value of f ( x ) increases. 3x 2 − 3 > 0
Consider
x −1 > 0 ( x + 1)( x − 1) > 0 x < −1 or x > 1 2
4.
(1, 0)
y = 2 cos 2 x − cos 4 x dy = −4 sin 2 x + 4 sin 4 x dx
When
dy = 0, dx
−4 sin 2 x + 4 sin 4 x = 0 2 sin 2 x cos 2 x − sin 2 x = 0 sin 2 x (2 cos 2 x − 1) = 0 1 2 π 5π 2 π or 2 x = , 3 3 π 5π π or x = , 6 6
y = 2x3 − x2 − 4x + 3 dy = 6x2 − 2x − 4 dx dy When = 0, dx 6x2 − 2x − 4 = 0
∴
3x 2 − x − 2 = 0 (3 x + 2)( x − 1) = 0
d2y = −8 cos 2 x + 16 cos 4 x dx 2
x=− d2y = 12 x − 2 dx 2
2 or 1 3
2 d2y 17 When x = − , < 0 and y = 4 , 3 dx 2 27 2 17 ∴ ( − , 4 ) is a maximum point. 3 27 d2y > 0 and y = 0 , dx 2 (1, 0) is a minimum point.
When x = 1 , ∴
sin 2 x = 0 or cos 2 x =
2 x = 0, π, x = 0,
For x = 0 ,
π , 2
d2y > 0 , y = 1; dx 2
x=
π d2y 3 < 0, y = ; , 6 dx 2 2
x=
π d2y > 0 , y = −3 ; , 2 dx 2
x=
5π d 2 y 3 , < 0, y = ; 6 dx 2 2
x = π,
d2y > 0 , y = 1. dx 2
x
101
102
Chapter 16 Applications of differentiation
5π 3 π 3 ( , ) and ( , ) are maximum points; 6 2 6 2
∴
(0, 1), (
π , −3) and (π, 1) are minimum points. 2
A sketch of the curve is shown below: y
π 3 ( , ) 6 2
5π 3 ( , ) 6 2
(0, 1)
d2A πH =− <0 2 4 dr
∴ When the radius is 8 cm , the curved surface area is maximum. 7. Let x m be the distance between the man and the light, y m be the height of the man’s shadow on the wall.
(π, 1)
2m
x
π
O
ym
xm 10 m
y = 2 cos 2x − cos 4x
π ( , −3) 2
2 y = x 10 20 y= x 20 dx dy =− 2 dt x dt
dx 8 000 20 = = dt 3 600 9
Q
6. Let r cm, h cm and A cm 2 be the radius, height and surface area of the cylinder respectively. Let H cm be the height of the cone.
∴ When x = 4 , dy 20 20 25 =− ( )=− dt 16 9 9 ∴ The rate of change of the height of the man’s shadow on the wall is 25 1 − × 3 600 × km / h = −10 km / h 9 1 000
H cm
8. Let r cm, h cm and V cm 3 be the radius, depth and volume of water in the filter respectively.
h cm
6 cm
16 cm
r cm
Using the properties of similar triangles, 16 H = r H−h H h=H− r 16 H r) 16 dA H H H = 2 π[( H − r ) + r ( − )] = 2 π( H − r ) dr 16 16 8
r cm 12 cm
h cm
A = 2 πrh = 2 πr ( H −
When H−
dA = 0, dr
H r=0 8 r=8
6 r = h 12 1 r= h 2 1 2 1 1 1 πh 3 V = πr h = π( h 2 )h = 3 3 4 12 dV 1 2 dh = πh dt dt 4
Chapter 16 Applications of differentiation
dh = 0.5 dt ∴ The rate of change of the volume of water 1 = π(8)2 (0.5) 4 = 8π cm 3 / s
When x = π + tan −1 (2) ,
When h = 8 ,
Since the rate of change of the volume of water = the rate of water pouring into the filter − the rate of water flowing out of the filter ∴ The rate of water pouring into the filter is
d2y >0 dx 2
∴ When x = tan −1 (2) , y attains a maximum. The maximum value of y = 5 11 − 16. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons. 17. (a) When f ' ( x ) = 0 , x = 0 , 2 or 3 When x = 0 , f "( x ) < 0
(8π + 4) cm 3 / s .
When x = 2 , f "( x ) > 0 When x = 3 , f "( x ) < 0 7 ∴ (0, 10), (3, − ) are maximum points, 2
9. (a) Let $P be the daily profit.
P = x (800 − x ) − ( x 2 + 52 x + 38)
and (2, − 6) is a minimum point.
= 800 x − x 2 − x 2 − 52 x − 38 = −2 x 2 + 748 x − 38
(b) A sketch of the curve is shown below: y
dP = −4 x + 748 dx d2P = −4 dx 2 dP When = 0 , −4 x + 748 = 0 dx x = 187
10
(0, 10)
y = f (x)
2
d P = −4 < 0 dx 2 ∴ When the number of fans produced daily is 187 , the daily profit is a maximum.
When x = 187 ,
O
x 1 7 (3, − ) 2
(b) The maximum daily profit
= $[ −2(187)2 + 748(187) − 38] = $69 900
(2, −6)
18. (a) 10. (a)
dy = 2 cos x − sin x dx d2y = −2 sin x − cos x dx 2
(b) When
103
dy 3( 4t 2 )2 3 = = t dx ( 4t 2 , 4t 3 ) 2 8( 4t 3 )
dy = 0, dx
2 cos x − sin x = 0 tan x = 2 x = tan −1 (2) , π + tan −1 (2)
When x = tan −1 (2) ,
4 y2 = x 3 dy 8y = 3x 2 dx dy 3 x 2 = dx 8 y
d2y < 0 and y = 5 dx 2
The equation of the tangent is 3 y − 4t 3 = t ( x − 4t 2 ) 2 2 y − 8t 3 = 3tx − 12t 3 3tx − 2 y − 4t 3 = 0
104
Chapter 16 Applications of differentiation
(b) 4 y 2 = x 3 ...............................................(1) 3tx − 2 y − 4t 3 = 0 ..................................(2) 3 tx − 2t 3.........................(3) 2 Substitute (3) into (1), 3 4( tx − 2t 3 )2 = x 3 2 9t 2 x 2 − 24t 4 x + 16t 6 − x 3 = 0 x 3 − 9t 2 x 2 + 24t 4 x − 16t 6 = 0 ................(4)
Substitute x = 0 , ∴
From (2), y =
Let g( x ) = x 3 − 9t 2 x 2 + 24t 4 x − 16t 6 g(t 2 ) = t 6 − 9t 6 + 24t 6 − 16t 6 = 0 ∴
x = t 2 is a root of equation (3).
∴ The coordinates of B are (t 2 , −
t3 ). 2
(c) Let P(x, y) be the mid-point of AB. 4t 2 + t 2 2 5 2 = t ................................................(5) 2 3 4t 3 − t2 y= 2 7 3 = t ................................................(6) 4 x=
(5)3 ÷ (6)2 , 3
x = y2 x3 =
y 49 250 y 2 = 49 x 3
250 y = 49 x . 2
3
9 y = ax + bx + cx + d ......................(1) dy 9 = 3ax 2 + 2 bx + c dx dy 1 2 2 c = ax + bx + ........................(2) dx 3 9 9 2
Substitute (0, 0) into (1), ∴
0=0+d d=0
∴ ∴
0 = 27a + 9b + 3c + d 3a + b + 3 = 0 .................................(4)
(b) ∴ The equation of C is 9 y = x 3 − 6 x 2 + 9 x dy 1 2 4 = x − x +1 dx 3 3 dy When = 0, dx 1 2 4 x − +1 = 0 3 3 x2 − 4x + 3 = 0 ( x − 1)( x − 3) = 0 x = 1 or 3 4 When x = 1 , y = ; 9 when x = 3 , y = 0 d2y 2 4 = x− 2 3 3 dx
(3, 0) is a minimum point.
∴ The equation of the locus of P is
19. (a)
dy = 0 into (2), dx 1 2 c ∴ 0 = a(3)2 + b(3) + 3 9 9 2 ∴ 3a + b + 1 = 0 ...............................(3) 3 Substitute x = 3 , y = 0 into (1), Substitute x = 3 ,
d2y 2 d2y 2 ; = − < = >0 0 2 2 3 dx x =1 dx x =3 3 4 ∴ (1, ) is a maximum point, and 9
( 25 )3 t 6 ( 47 )2 t 6 250 2
3
1=
Solving (3) and (4), we obtain a = 1, b = −6 .
Substitute x = t 2 into (3), 3 t3 y = t (t 2 ) − 2 t 3 = − 2 2
c 9 c=9
dy = tan 45° = 1 into (2), dx
(c) When x = 0, y = 0; when x = −1, y = −
16 ; 9
4 9 A sketch of the curve is shown below:
when x = 4, y = y
3 (1, 4 ) y = x − 2 x 2 + x 9 9 3 (3, 0)
−1
O
x 3
4
Chapter 16 Applications of differentiation
20. (a) The length of the tangent is z = a 2 + b 2 + ( −2)a + ( −6)b + 5 z 2 = a 2 + b 2 − 2 a − 6b + 5 ...................(1) Since (a, b) lies on L, ∴ 3a + b + 4 = 0 b = −3a − 4 ........................(2) Put (2) in (1),
z = a + (3a + 4) − 2 a + 6(3a + 4) + 5 2
2
2
= a 2 + 9a 2 + 24 a + 16 − 2 a + 18a + 24 + 5
2r 2 h= 2 r −1 r 2 h − h = 2r 2
∴
r 2 (h − 2) = h r2 =
= 10 a + 40 a + 45
(b) Volume of cone
=
1 2 1 h πh 2 πr h = π( )h = 3 3 h−2 3(h − 2)
∴ 1
z = (10 a 2 + 40 a + 45) 2 1 dz 1 = (10 a 2 + 40 a + 45) − 2 (20 a + 40) da 2
When
dz = 0 , a = −2 da
dz > 0; da dz when a < −2 (slightly), <0 da ∴ z attains its minimum when a = −2 . when a > −2 (slightly),
Put a = −2 in (2), ∴ b=6−4=2 ∴ The coordinates of A are ( −2, 2) .
V=
πh 2 3(h − 2)
dV π 2 h(h − 2) − h 2 π(h 2 − 4h) = = dh 3 3(h − 2)2 (h − 2)2 Put
As 10 a 2 + 40 a + 45 = 10( a + 2)2 + 5 > 0 ,
dV = 0, dh
∴
h 2 − 4h = 0 h( h − 4 ) = 0 ∴ h = 4 or
h = 0 (rejected) dV When h > 4 (slightly), >0 dh dV When h < 4 (slightly), <0 dh ∴ When h = 4 , V attains a minimum. π( 4 2 ) 3( 4 − 2) 8 = π 3
∴ Minimum volume =
(c) Let O be the centre of C, ∴ the coordinates of O are (1, 3). 3−2 1 = 1+ 2 3 Slope of line L = −3 Slope of AO =
22. (a) Curved surface of one of the cones
= πr r 2 + h 2 1 ( −3) 3 = −1
(Slope of AO) × (Slope of L) = ∴ AO is perpendicular to L. 21. (a) Let ∠EBD = θ DE = r tan θ r tan θ = 1 AD = h = r tan 2θ r (2 tan θ) = 1 − tan 2 θ 2 r ( 1r ) = 1 − ( 1r )2
h h−2
(c) Let V be the volume of cone.
2
(b)
105
∴
1
S = 2 πr (r 2 + h 2 ) 2 + 2 πrh 1 S − h = (r 2 + h 2 ) 2 2 πr S ( − h)2 = r 2 + h 2 2 πr S2 hS − + h2 = r 2 + h2 4 π 2 r 2 πr hS S2 = − r2 πr 4 π 2 r 2 hS S 2 − 4 π 2 r 4 = πr 4π 2 r 2 S 2 − 4π 2 r 4 h= 4 πrS
106
Chapter 16 Applications of differentiation
2 2 πr h + πr 2 h 3 5 = πr 2 h 3 5 S 2 − 4π 2 r 4 = πr 2 ( ) 3 4 πrS 5r 2 = (S − 4π 2 r 4 ) 12 S 5 (c) From (b), V = (rS 2 − 4 π 2 r 5 ) 12 S dV 5 ( S 2 − 20 π 2 r 4 ) = dr 12 S dV When = 0 , S 2 − 20 π 2 r 4 = 0 dr
1 (1 + h) x 2 1 = (1 + h)( 4 cos θ − cot θ) 2 1 = (1 + 4 sin θ)( 4 cos θ − cot θ) 2 1 = (16 sin θ cos θ − cot θ) 2 1 = 4 sin 2θ − cot θ 2
(ii) A =
(b) V =
(b)
S2 20 π 2 S 12 r=( ) 2 5π
r4 =
d 2V 5 = ( −80 π 2 r 3 ) 2 12 S dr 3 −100 π 2 ( S ) 2 2 S 12 d V 2 5π = ) , When r = ( 3S dr 2 2 5π <0 S 12 ) , V attains a ∴ When r = ( 2 5π maximum, S 12 ) i.e. r0 = ( 2 5π
h0 =
4S 1 1 5 4 πS r0 S = 5πr0 =
h0 S = r0 5πr0 2 S 2 5π = 5π S 2 = 5
∴
1 1 ∴ When sin θ = ( ) 3 , x attains a maximum. 4
2 S 2 − 4π 2 ( S 2 )
20 π 1 S 4 πS( )2 2 5π 2
h0 : r0 = 2 : 5 1 OA OA = cot θ x = OB − OA = 4 cos θ − cot θ
23. (a) (i) tan θ =
dx = −4 sin θ + csc 2 θ dθ 1 − 4 sin 3 θ = sin 2 θ dx When = 0, dθ 1 − 4 sin 3 θ =0 sin 2 θ 1 − 4 sin 3 θ = 0 1 sin 3 θ = 4 1 1 sin θ = ( ) 3 4 1 1 dx When θ > sin −1[( ) 3 ] (slightly), <0 4 dθ 1 1 dx When θ < sin −1[( ) 3 ] (slightly), >0 4 dθ
(c)
1 dA = 8 cos 2θ + csc 2 θ 2 dθ 1 2 sin 2 θ 2 16 sin θ(1 − 2 sin 2 θ) + 1 = 2 sin 2 θ 4 −32 sin θ + 16 sin 2 θ + 1 = 2 sin 2 θ = 8(1 − 2 sin 2 θ) +
dA = 0, dθ −32 sin 4 θ + 16 sin 2 θ + 1 =0 2 sin 2 θ −32 sin 4 θ + 16 sin 2 θ + 1 = 0
When
sin 2 θ =
−16 ± 16 2 − 4( −32) 2( −32)
2+ 6 2− 6 or (rejected) 8 8 sin θ = 0.75 (corr. to the nearest 0.01) =
Chapter 16 Applications of differentiation
(b)
d2A = −16 sin 2θ − csc 2 θ cot θ dθ 2
0 0 −t 0
As θ is acute, −16 sin 2θ − csc 2 θ cot θ < 0 , i.e.
Area =
d2A < 0 when sin θ = 0.75 . dθ 2
∴ W h e n sin θ = 0.75 , t h e a r e a o f t h e trapezium attains a maximum.
24. (a) C1: x 2 + y 2 + 2tx = 0 ................(1) ∴ The coordinates of the centre of C1 are ( −t, 0) . 2y = 0 ...............(2) t ∴ The coordinates of the centre of C2 are 1 (0, − ) . t (1) − (2), C2 : x 2 + y 2 +
2y =0 t t2x − y = 0
2tx −
Put (3) into (1),
−2t 3 1 + t4 1 0 − t 0 0
1 2t 4 2 + ( ) 2 1 + t4 1 + t4 =1 ∴ The area of the quadrilateral is a constant. =
(c) Let L be the length of common chord. 2t 2 2t 3 2 L2 = (0 + 0 + + ) ( ) 1 + t4 1 + t4 4t 2 + 4t 6 = (1 + t 4 )2 4t 2 (1 + t 4 ) = (1 + t 4 )2 4t 2 = (1 + t 4 )
d ( L2 ) (1 + t 4 )8t − 4t 2 ( 4t 3 ) = dt (1 + t 4 )2 2L
x 2 + t 4 x 2 + 2tx = 0 (1 + t 4 ) x 2 + 2tx = 0 x[(1 + t 4 ) x + 2t ] = 0
dL 8t + 8t 5 − 16t 5 = dt (1 + t 4 )2 dL 4t − 4t 5 = dt L(1 + t 4 )2
When
−2t x = 0 or x = 1 + t4
dL = 0, dt
∴ 4t (1 − t 4 ) = 0 t = 0 or t = ±1 Put t = 0 into L,
Put x = 0 into (3), y=0
y=
1 −2t 2 1 + t4
When L > 0 ,
y = t 2 x .........................(3)
Put x =
107
L=0 Since the length of chord should be greater than zero,
−2t into (3), 1 + t4
−2t 3 1 + t4
∴
The coordinates of intersecting points are −2t −2t 3 , ). (0, 0) and ( 1 + t4 1 + t4 ∴ The coordinates of the vertices of the q u a d r i l a t e r a l a r e (0, 0) , ( −t, 0) , 1 −2t −2t (0, − ) and ( , ). 4 t 1+ t 1 + t4
L = 0 is rejected.
Put t = ±1 into L, 4 ∴ L2 = = 2 2 L = 2 or − 2 (rejected) ∴ The stationary value of the length of the common chord is
2.
3
25 − 34. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons.
108
Chapter 16 Applications of differentiation
Enrichment 16 (p.167)
Classwork 1 (P.116)
4x 1. (a) f ( x ) = 2 x − x +1
1.
1 3 Consider x 2 − x + 1 = ( x − )2 + 2 4 1 2 (x − ) ≥ 0 Q 2 1 2 3 (x − ) + > 0 2 4 x2 − x + 1 > 0 ∴
dy dx ( 2,
y − 4 = 4 ( x − 2) 4x − y − 4 = 0
f ( x ) is defined for all real values of x. ( x 2 − x + 1)( 4) − ( 4 x )(2 x − 1) ( x 2 − x + 1)2 −4( x 2 − 1) = 2 ( x − x + 1)2
The equation of the normal is 1 y − 4 = − ( x − 2) 4 x + 4 y − 18 = 0
( x 2 − x + 1)2 > 0
Consider ∴
x2 − 1 < 0 ( x + 1)( x − 1) < 0 −1 < x < 1
2.
(c) f "( x ) = [ −4( x 2 − x + 1)2 (2 x ) + 4( x 2 − 1) ( 4 x − 6 x + 6 x − 2)] ÷ ( x − x + 1) 3
2
2
4
−8 x 5 − 8 x 4 − 16 x 3 + 32 x 2 − 32 x + 8 ( x 2 − x + 1) 4 When x = 1 , f ' ( x ) = 0 , f "( x ) < 0 and y = 4 =
∴
= 3(2)2 − 4(2) 4)
=4 The equation of the tangent is
(b) f ' ( x ) =
Q
y = x3 − 2x2 + 4 dy = 3 x 2 − 2( 2 x ) dx = 3x 2 − 4 x
(1, 4) is a maximum point.
W h e n x = −1 , f ' ( x ) = 0 , f "( x ) > 0 a n d 4 y=− 3 4 ∴ ( −1, − ) is a minimum point. 3 (d) When x = 0 , y = 0 A sketch of the curve is shown below: y
x 2 − xy 2 + y 2 = 9 dy dy 2 x − y 2 − 2 xy + 2y =0 dx dx dy y2 − 2 x = dx 2 y − 2 xy dy (2)2 − 2( −1) = dx ( −1, 2 ) 2(2) − 2( −1)(2) 3 = 4 The equation of the tangent is 3 y − 2 = ( x + 1) 4 3 x − 4 y + 11 = 0
The equation of the normal is 4 y − 2 = − ( x + 1) 3 4 x + 3y − 2 = 0
(1, 4)
Classwork 2 (P.119) y=
O (−1, −
4x x2 − x + 1
1. Let ( x1, y1 ) be the point of contact of a required tangent. Slope of the given line = −9 x
4 ) 3
2 − 5. No solutions are provided for the H.K.C.E.E. questions because of the copyright reasons.
y = x 3 − 12 x + 4 dy = 3 x 2 − 12 dx At ( x1, y1 ) , slope of the tangent =
dy dx ( x1 , y1 )
= 3 x12 − 12
109
Chapter 16 Applications of differentiation
∴
Substitute x1 = 0 into (3),
3 x12 − 12 = −9 x1 = ±1
y1 = 3(0) − 5 = −5
When x1 = 1 ,
Substitute x1 = 3 into (3),
y1 = (1)3 − 12(1) + 4 = −7 ∴ (1, − 7) is a point of contact.
y1 = 3(3) − 5 =4
∴
( x1, y1 ) = (0, − 5) or (3, 4)
At (1, − 7) , the equation of the tangent is y + 7 = − 9( x − 1) 9x + y − 2 = 0
At (0, −5), slope of the tangent = 0. ∴ At (0, −5), the equation of the tangent is
When x1 = −1 , y1 = ( −1) − 12( −1) + 4 = 15
At (3, 4), slope of the tangent = −
y + 5 = 0.
3
∴
3 , 4 ∴ At (3, 4), the equation of the tangent is 3 y − 4 = − ( x − 3) 4 3 x + 4 y − 25 = 0
( −1, 15) is another point of contact.
At ( −1, 15) , the equation of the tangent is y − 15 = −9( x + 1) 9x + y − 6 = 0
2. Let ( x1, y1 ) be the point of contact of a required tangent. Since ( x1, y1 ) lies on the curve, ∴
x12 + y12 = 25 .........................(1) 2x + 2y
Classwork 3 (P.125) 1. A sketch of the curve is shown below: y
dy =0 dx dy x =− dx y
(0, 5)
dy dx ( x1 , y1 ) x =− 1 y1
Slope of the tangent =
y = f (x)
Slope of the line joining ( x1, y1 ) and (15, −5) y +5 = 1 x1 − 15 ∴
−
x1 y +5 = 1 y1 x1 − 15
(2, 1)
x12 + y12 − 15 x1 + 5 y1 = 0 ................(2)
x
O
(1) − (2),
15 x1 − 5 y1 = 25 3 x1 − y1 = 5 y1 = 3 x1 − 5 ........................(3) Substitute y1 = 3 x1 − 5 into (1), x12 + (3 x1 − 5) = 25 + 9 x12 − 30 x1 + 25 = 25 10 x1 ( x1 − 3) = 0 x1 = 0 or 3
f ( x ) = 2 x 3 + 3 x 2 − 12 x + 2
2.
f ' ( x ) = 6 x 2 + 6 x − 12 = 6( x − 1)( x + 2) ∴ When x = 1 or x = −2 , f ' ( x ) = 0
2
x12
x
x < −2 x = −2 −2 < x < 0 x = 0 0 < x < 1 x = 1 x > 1
f (x) f' (x)
22 +
0
−5
2 −
−
−
0
+
110
Chapter 16 Applications of differentiation
5π into f ( x ), 6 5π 5π 5π − 2 cos f( ) = 2 − 6 6 6 5π =2+ 3− 6 5π 5π f "( ) = 2 cos 6 6 =− 3 <0 5π 5π ∴ ( , 2 + 3 − ) is a maximum point. 6 6
A sketch of the curve is shown below:
Put x =
y (−2, 22)
y = 2x 3 + 3x 2 − 12x + 2
(c) When x = 0 , y = 0 A sketch of the curve is shown below: y (
2
5π ,2+ 6
5π ) 6
y = 2 − x − 2 cos x
x
O
3−
O
π ( ,2 − 6
(1, −5)
3−
x
π
π ) 6
Classwork 4 (P.127) (a)
y = 2 − x − 2 cos x dy = −1 + 2 sin x = 0 dx 1 sin x = 2 π 5π x= , 6 6
Classwork 5 (P.129) 2x 1 + x2 (1 + x 2 )(2) − (2 x )(2 x ) f ' ( x) = (1 + x 2 )2 2(1 − x 2 ) = (1 + x 2 )2 f ( x) =
(a)
(b) f ' ( x ) = −1 + 2 sin x Set f ' ( x ) = 0 for turning points. π 5π From (a), x = , 6 6 π Put x = into f ( x ), 6 π π π f ( ) = 2 − − 2 cos 6 6 6 π =2− 3− 6 f "( x ) = 2 cos x π When x = , 6 π π f "( ) = 2 cos 6 6 = 3>0 π π ∴ ( , 2 − 3 − ) is a minimum point. 6 6
Consider
∴
2(1 − x 2 ) >0 (1 + x 2 )2 2(1 − x 2 ) > 0 (1 + x )(1 − x ) > 0
−1 < x < 1
(b) When x = −1 or x = 1 , f ' ( x ) = 0 x
x < −1 x = −1 −1 < x < 0 x = 0 0 < x < 1 x = 1 x > 1 −1
f (x) −
f' (x)
0
1
0 +
+
+
0
−
The turning points of C are ( −1, − 1) and (1, 1) . dy changes sign from negative to positive dx as x increases through −1,
Since ∴
( −1, − 1) is a minimum point.
Chapter 16 Applications of differentiation
dy Since changes sign from positive to negative dx as x increases through 1, ∴ (1, 1) is a maximum point. (c) When x = 0 , y = 0 A sketch of the curve is shown below: y (1, 1)
y= O
1 or x = 1 , f ' ( x ) = 0 2 1 When x = − , f "( x ) = 18 > 0 and 2 27 f ( x) = − , 8 1 27 ∴ ( − , − ) is a minimum point. 2 8
(b) When x = −
When x = 1 , f "( x ) = 0 , f ( x ) = 0
2x 1 + x2 x
When x < 1 (slightly), f ' ( x ) > 0 When x > 1 (slightly), f ' ( x ) > 0 dy does not change sign as dx x increases through 1.
This shows that
(−1, −1)
∴ (1, 0) is neither a maximum point nor a minimum point.
Classwork 6 (P.130) (a)
f ( x ) = −( x + 1) 4
When x = 0 , f ( x ) = −2
f ' ( x ) = −4( x + 1)3
When x = −1 , f ( x ) = 0
f "( x ) = −12( x + 1)2
(b) When x = −1 , f ' ( x ) = 0
Also, as x → −∞ , y → +∞ ;
When x = −1 , f "( x ) = 0
x → +∞ , y → +∞ .
When x < −1 (slightly), f ' ( x ) > 0 When x > −1 (slightly), f ' ( x ) < 0 dy Since changes sign from positive to negative dx as x increases through −1, ∴
111
( −1, 0) is a maximum point.
(c) A sketch of the curve is shown below: y
−1
O
(1, 0)
x
(c) A sketch of the curve is shown below: y (−1, 0)
O
y = 2(x − 1)3(x + 1)
x −2
−1
y = −(x + 1)4 1 27 (− , − ) 2 8
Classwork 7 (P.131) 1. (a)
f ( x ) = 2( x − 1)3 ( x + 1) f ' ( x ) = 2( x + 1)[3( x − 1)2 ] + 2( x − 1)3
1. (a) y 2 = (1 − x )( x + 2)2
= 6( x + 1)( x − 1)2 + 2( x − 1)3 = 4(2 x + 1)( x − 1)2
The curve is composed of the two curves:
f "( x ) = 4(2 x + 1)[2( x − 1)] + 4( x − 1) (2) 2
= 8(2 x + 1)( x − 1) + 8( x − 1)2 = 24 x ( x − 1)
Classwork 8 (P.133)
1
y = (1 − x ) 2 ( x + 2) and 1
y = −(1 − x ) 2 ( x + 2) .
112
Chapter 16 Applications of differentiation
For each value of x, the values of y of the above two curves only differ from each other by an opposite sign. Therefore, the curve of the given equation is symmetrical about the x-axis. (b)
y 2 = (1 − x )( x + 2)2 dy 2y = (1 − x )[2( x + 2)] + ( x + 2)2 ( −1) dx = 2(1 − x )( x + 2) − ( x + 2)2 dy 2y = −3 x ( x + 2) dx −3 x ( x + 2) dy = 1 dx ±2(1 − x ) 2 ( x + 2) 1 3 = ± x (1 − x ) − 2 2 1 2
(c) Consider y = (1 − x ) ( x + 2) 1 dy 3 = − x (1 − x ) − 2 dx 2 3 1 d2y 3 1 3 = − x[ − (1 − x ) − 2 ] − (1 − x ) − 2 2 2 2 2 dx 1 3 3 − 23 = x (1 − x ) − (1 − x ) − 2 4 2 3 − 23 = (1 − x ) (3 x − 2) 4 dy d2y When x = 0 , = 0, < 0 and y = 2 dx dx 2 ∴ 2 is a relative maximum of y.
By symmetry, −2 is a relative minimum of y. ∴
(0, 2) and (0, − 2) are the turning points
where 0 < r < 15 (b)
dV 1 = π(30 r − 3r 2 ) dr 3 = πr (10 − r ) d 2V = π(10 − 2 r ) dr 2 = 2 π( 5 − r ) dV = 0 , r = 10 or r = 0 (rejected) dr d 2V When r = 10 , = −10 π < 0 dr 2 ∴ The volume of the cone is maximum when r = 10 . When
2. (a) The height of trapezium ABCD = (2 3 ) 2 − x 2 = 12 − x 2
The area of trapezium ABCD 1 = (10 + 10 + 2 x ) 12 − x 2 2 = (10 + x ) 12 − x 2 ∴
(d) When x = 1 , y = 0 ; when x = −2 , y = 0
12 − x 2 > 0 and x > 0 ,
i.e. 0 < x < 2 3
When x → −∞ ,
(c) Let A be the area of the trapezium.
1 2
y = (1 − x ) ( x + 2) → −∞ , 1 2
y = −(1 − x ) ( x + 2) → +∞ A sketch of the curve is shown below : y
A = (10 + x ) 12 − x 2 1 dA 1 = 12 − x 2 + (10 + x )[ (12 − x 2 ) − 2 ( −2 x )] dx 2 = 12 − x 2 − x (10 + x )(12 − x 2 ) − 2 1
= −2( x − 1)( x + 6)(12 − x 2 ) − 2 1
maximum point (0, 2)
When
y 2 = (1 − x)(x + 2)2
−2
1. h + r = 15 1 (a) V = πr 2 h 3 1 2 = πr (15 − r ) 3
(b) As the height and x are positive,
for the curve.
O
Classwork 9 (P.139)
x 1
(0, −2) minimum point
dA = 0 , x = 1 or x = −6 (rejected) dx or x = 2 3 (rejected)
dA <0 dx dA When x < 1 (sightly), >0 dx Hence the area of the trapezium is maximum when x = 1 . When x > 1 (sightly),
Chapter 16 Applications of differentiation
∴ The total rental and travelling cost of a factory is minimum when A is located 2.5 km from O, and the minimum total
Classwork 10 (P.142) 1. Let s km be the distance between A and B after t minutes. Then
1 1 s 2 = (5 + t )2 + ( 2 t )2 − 2(5 + t )( 2 t )cos 45° 2 2 1 2 1 2 = (5 + t ) + ( 2 t ) − 2t (5 + t ) 2 2 ds 1 1 1 2s = 2(5 + t )( ) + 2( 2 t )( 2 ) − t − 2(5 + t ) dt 2 2 2 ds 1 5t = ( − 5) dt 2 s 2 5 = (t − 2 ) 4s ds When = 0, t = 2 dt ds When t < 2 (slightly), <0 dt ds When t > 2 (slightly), >0 dt ∴ When t = 2 , s attains its minimum.
cost is $4 000 .
Classwork 11 (P.150) 15t +1 4 ds 15 = −3t 2 + 6t + (a) dt 4
1. s = −t 3 + 3t 2 +
(b) Let v and a be the velocity and acceleration respectively at time t. ds 15 = −3t 2 + 6t + dt 4 dv a= = −6t + 6 dt 15 When t = 0 , v = and a = 6 4 15 The required velocity is units per second. 4 The required acceleration is 6 units per square second. v=
When t = 2 , s = 2 5 Hence they will be closest to each other at 12: 02 p.m. with distance 2 5 km . 2. (a) Let T and R be the travelling cost and rental cost respectively. R x When x = 2 , C = Tx +
∴
(c) Since the particle is instantaneously at rest at ds point B, = 0 then. dt ds 15 From (a), = −3t 2 + 6t + =0 dt 4 12t 2 − 24t − 15 = 0
C = 1 600 + 2 500 5 000 = 800(2) + 2 5 000 C = 800 x + x
5 000 C = 800 x + (b) x dC = 800 − 5 000 x −2 dx d 2C = 10 000 x −3 dx 2 5 5 dC When = 0 , x = or − (rejected) 2 2 dx 2
5 d C , > 0 and C = 4 000 2 dx 2 5 Hence, when x = , C attains its minimum. 2
When x =
113
4t 2 − 8t − 5 = 0 (2t + 1)(2t − 5) = 0 t=−
5 1 (rejected) or t = 2 2
∴ It happens at 2.5 s .
2. Let θ, y km and x km be ∠ALP, AP and LP respectively. L θ
x km 3 km
A
y km
P
114
Chapter 16 Applications of differentiation
dy dθ = 3 sec 2 θ dt dt y dθ When = 4 π , y = 4 , x = 5 and tan θ = , 3 dt dy 5 2 = 3( ) ( 4 π) dt 3 100 π = 3 100 π km / min. ∴ The required velocity is 3
When h = 2 ,
dA 25π ⋅ 2 8 = (− ) dt 8 5π = −10 ∴ The rate of decrease of the area of water surface is 10 m 2 / hour. 2. (a) Using cosine formula, we obtain AB2 = OA2 + OB2 − 2(OA)(OB)cosθ
(2 2 )2 = (2 2 )2 + x 2 − 2(2 2 )( x ) cos θ
Classwork 12 (P.152) 1. (a) Let h m, r m and V m 3 be the depth, radius of the surface and volume of water in the tank respectively. Using the properties of similar triangles, r 5 = h 4 5 r= h 4 1 2 1 5 25 3 πh V = πr h = π( h)2 h = 3 3 4 48 dV 25 dh = π(3h 2 ) 48 dt dt 25 2 dh = πh 16 dt When h = 2 , 25 dh π( 2 ) 2 16 dt dh 8 = dt 5π 10 =
∴ The rate of increase of the depth of water 8 is m / hour. 5π (b) Let A m 2 be the area of the surface of water. A = πr 2 5 = π( h ) 2 4 25 2 πh = 16 dA 25 dh π( 2 h ) = dt 16 dt 25πh dh = 8 dt
0 = x 2 − 4 2 x cos θ 0 = x ( x − 4 2 cos θ) x = 0 (rejected) or x − 4 2 cos θ = 0 x = 4 2 cos θ
(b) Given that A is 2 revolutions per second, dθ = 4π dt From (a), x = 4 2 cos θ dx dθ = −4 2 sin θ dt dt π = −4 2 (sin )( 4 π) 4 = −16 π
∴ The rate of change of OB is −16π units per second.