Revision Paper 7: Differentiation II
Q1
Two variables are related by the equation y  x 2 
(a)
(b)
10
.
x
dy
.
dx
Find the rate of change of x at the instant when x = 1 given that y is increasing at a rate of 3
units per second.
Find
1
dy
when x = 1. If x is increasing at a rate of 1.3 units per
(5x  4)12 find the value of
4
dx
second when x = 1, find the corresponding rate of change of y.
Q2
Given that , y 
Q3
A container in the form of an inverted cone of height 50 cm and base radius 20 cm is filled with water
at a constant rate of 40 cm3s–1 to a depth of x cm. Find the volume of water in terms of x and hence the
rate of increase of x at the instant when x = 2.
Q4
A vessel is in the shape of an inverted right circular cone whose base radius is equal to the height and
whose axis is vertical. Liquid is poured into the vessel at a constant rate of 100 cm3s–1. The volume of
 x3
liquid in the vessel is
cm 3 when the depth of liquid is x cm. Calculate, at the instant when the
3
depth of liquid is 15 cm, the rate of increase of
(a)
(b)
the depth of the liquid,
the area of the horizontal surface of the liquid.
Q5
A ball is thrown into the air and the distance from the ground is given by y  kx4  7 x 2  5. When x  2 ,
dx 11
dy

 27.5 . Find the value of k .
and
dt 40
dt
Q6
Differentiate the following with respect to x.
(a)
5  6sin 4x
(b)
(d)
5 4  4cot 2 2x
(e)
3  2 cos2 2x
1
3  3sec 2 x
(c)
1  3tan 2 2x
(f)
cos sin  x 2  
y  sin  x  y  


[Answers for (g) and (h) may be given in terms of x and y.]
(g)*
sin x cos y  2
(h)*
Q7
Given that y  x  tan x  and that
Q8
The equation of a curve is y 
2
2
dy

 a  b 3 when x  , find the values of a and b.
dx
3
cos x

. Find the x-coordinate where 0 < x < , of the point at which the
5  sin x
2
tangent to the curve is parallel to the x-axis.
 d2 y 
 dy 
Given that y  ax  bx and  2   4    32 y , find the possible values of a and b.
 dx 
 dx 
2
Q9
2
Q10* Given that y  sin 1 x , show that


dy
1
for 1  x  1,   y  .

2
2
dx
1  x2
Q11
The base of a right circular cone is a section of a sphere, centre O, and of unit radius. The vertex of the
cone is on the sphere and the centre of the sphere is within the cone. If the distance of O from the base of

the cone is x, prove that the volume of the cone is 1  x  x 2  x 3  . Hence find the value of x which
3
makes the volume of the cone a maximum.
Q12
A right circular cylinder of radius x and height h rest on the base of a right circular cone (inside the cone)
and the other end is in contact with the curved surface. If H and R are the height and radius of the cone,
Hx 2 ( R  x)
show that the volume, V, of the cylinder is V 
. Find its stationary volume and determine if
R
the volume is a maximum or minimum.
Q13
In the figure, PQR is an isosceles triangle which fits inside the isosceles triangle ABC , where PQ  RQ
and Q is the midpoint of BC and PSB is a right angle. Given that AB  AC  10 cm , BC  16 cm ,
BS  x cm and PS  y cm .
3
3
(a)
Show that y  x and the area, A cm2, of the triangle PQR is given by A  6 x  x 2 .
4
4
A
(b)
Given that x can vary, find using calculus, the maximum value of .
Q14
Find the volume of the largest right circular cone that can be inscribed in a sphere of radius 9 cm.
Q15
Sketch the following curves, showing clearly the intercept(s) with the axes, turning points and
asymptote(s) if any.
(a) y  2 x  1 
(b) y 
5
2x 1
x2  3
x 1
(c) * y  x 2 
1
x
Q16
Given the curve f ( x) 
(i)
(ii)
xk
,k  0 ,
xk
dy
 0 for all values of x.
dx
Sketch the graph of y  f ( x) , showing clearly the intercept(s) with the axes, turning points and
asymptote(s) if any.
Show that
 xk 
Given the curve g( x)  
 ,k  0,
 xk 
(i)
Find the coordinate of the stationary point in terms of k.
(ii)
Sketch the graph of y  g( x) , showing clearly the intercept(s) with the axes, turning points and
asymptote(s) if any.
3
2
(iii) By adding a sketch to (ii), show that the equation m  x  k    x  k   0 has 3 distinct real
2
Q17
roots for some values of m.
Q18
Q19
4  ax 2
Given that the curve y 
has asymptotes at x  1 and y  1  x , find the values of a and b.
b x
dy
 0 for all values of x and hence sketch the curve.
Show that
dx
Find the coordinates of the stationary point(s) on the curve y  sin x  sin 2 x, 0  x  2 and state
whether it is a maximum or minimum point. Hence, find the greatest value of y in the same interval.
Q20
d2 y
1
Given that x  2t , y  , using chain rule, find
in terms of t.
t
dx 2
Q21
Given that x  2cos3 t , y  sin 3 t , 0  t 
Q22
Sketch the graph of y  f ( x) that is of degree greater than 2 and has the following properties:
2
54
64


normal to the curve at the point 
,

 125 125 
x
x2
2
2 x4
4
4 x6
6
x6
Q23

y
1
4
7
, using chain rule, find
dy
and hence the equation of the
dx
Derivatives
y' 0
y '  0, y ''  0
y' 0
y '  0, y ''  0
y' 0
y '  0, y ''  0
y' 0
Given that y  x3  x 2  kx , find the range of values of k for which y is an increasing function for all
values of x.
Revision Paper 7: Differentiation II
10
Q1 (a) y  x 2 
x
dy
10
 2x  2
dx
x
(b) y is increasing at a rate of 3 units per second 
dy
3
dt
dy
10
2
dx
1
= –8
When x = 1,
dx dx dy


dt dy dt
1
 3
8
 0.375
x is decreasing at a rate of 0.375 units per second.
1
(5x  4)12
4
dy 1
11
 12 5x  4  5 
dx 4
Q2 y 
 15 5x  4 
11
When x = 1,
dy
 15 .
dx
x is increasing at a rate of 1.3 units per second 
dx
 1.3
dt
dx dx dy


dt dy dt
1 dy
1.3 

15 dt
dy
 19.5
dt
Hence, y is increasing at a rate of 19.5 units per second.
Q3 (a) Using similar triangles,
r
x

20 50
2x
r
5
When the depth of the water is x, r 
2x
.
5
2
1  2x 
Volume of water when depth of water is x     x cm 3
3  5
4
  x 3 cm 3
75
dx
(b) Rate of increase of x 
dt
4
V
 x3
75
dV
4

 x2
dx 25
2
dV 4 2 
When x = 2,

dx
25
16

25
dV
 40
Container filled with water at a rate of 40 cm3s–1 
dt
dx dx dV


dt dV dt
25

 40
16
125

2
The rate of increase of x is
125
cms–1.
2
1
Q4 (a) V   x 3
3
dV
  x2
dx
Liquid is poured into the vessel at a constant rate of 100 cm3s–1 
Rate of increase of depth of liquid 
When x = 15,
dV
 225
dx
dx
dt
dV
 100
dt
dx dx dV


dt dV dt
1

 100
225
4

9
The rate of increase of the depth of liquid is
4
cms–1.
9
(b) Rate of increase of the area of the horizontal surface of the liquid 
Area of base, A =  x2
dA
 2 x
dx
When x = 15,
dA
dt
dA
 30
dx
dA dA dx


dt dx dt
4
 30 
9
40

3
The rate of increase of the area of horizontal surface of the liquid is
Q5
dy dy dt
 
dx dt dx
40
 27.5 
 100
11
y  kx 4  7 x 2  5
dy
 4kx 3  14 x
dx
dy
When x  2,
 4k (2)3  14(2)  100
dx
k 4
Q6 (a) Let y  5  6 sin 4x
1

dy 1
 (5  6 sin 4 x) 2 (24 cos 4 x)
dx 2
24 cos 4 x

5  6 sin 4 x
.
40
cm2s–1.
3
(b) Let y  3  2 cos2 2x .
1

dy 1
 (3  2 cos 2 2x) 2 4 cos 2x(2 sin 2x)
dx 2
4 sin 2x cos 2x

3  2 cos 2 2x
2 sin 4x

3  2 cos 2 2x
(c)
Let y  1  3tan 2 2 x .
1

dy 1
2
 (1  3tan 2x) 2  12 tan 2x sec 2 2x 
dx 2
6 tan 2x sec 2 2x

1  3tan 2 2x
6 sin 2x

3
cos 2x 1  3tan 2 2x
(d) Let y  5 4  4cot 2 2 x .
1

dy 5
 (4  4 cot 2 2 x) 2  16 cot 2 x cosec 2 2 x 
dx 2
40 cot 2 x cosec 2 2 x

4  4 cot 2 2 x
(e)
Let y 
1
.
3  3sec 2 x
dy
 (3  3sec 2 x) 2  6sec x sec x tan x 
dx
18sec 2 x tan x

2
 3  3sec2 x 
(f)
Let y  cos sin  x 2  .
dy
  sin sin  x 2   cos  x 2   2 x
dx
 2 x cos  x 2  sin sin  x 2  
(g) sin x cos y  2
dy
sin x   sin y   cos y cos x  0
dx
dy
 sin x sin y
 cos y cos x
dx
dy
cos x cos y

dx
sin x sin y
(h) y  sin  x  y  


dy
2
 dy 
 cos  x  y   2  x  y  1  


dx
 dx 
dy
1
1
dx

2
d
y


2  x  y  cos  x  y 


dx
1
dx

1
2
2  x  y  cos  x  y   dy


1
dx
1 
2
dy
2  x  y  cos  x  y  


2
1  2  x  y  cos  x  y   dx


2
dy
2  x  y  cos  x  y  


2
2  x  y  cos  x  y  
dy



dx 1  2  x  y  cos  x  y 2 


2
Q7
dy
 2( x  tan x)(1  sec 2 x)
dx
When x 

3
,
dy
 d 


 2   tan   1  sec 2 
3
dx
3  dx 
3
 2


 2 3  1  4 
 3


10
 10 3
3
Thus a 
10
, and b = 10.
3
cos x
5  sin x
dy 5  sin x  sin x   cos x  cos x 

dx
5  sin x 2
Q8 y 

1  5 sin x
5  sin x 2
Gradient of tangent = 
1  5sin x
5  sin x 2
1  5 sin x
5  sin x 2
0
5 sin x  1
sin x 
Since 0  x 

2
1
5
x = 0.201 rad, x =  – 0.201
= 2.94 rad
, x = 0.201 rad.
Q9
y  ax 2  bx
dy
 2ax  b
dx
d2 y
 2a
dx 2
2a  4  2ax  b   32  ax 2  bx 
2
2a  4  4a 2 x 2  4abx  b 2   32ax 2  32bx
By comparing coefficients,
 0  16a 2  32a  16a  a  2   0  a  0 or 2
 0  16ab  32b  16b  a  2   0  b  0
 2a  4b 2  b 2  4, b  2
a  0 or 2, b  0 or  2
Q10 y  sin 1 x
y  sin 1 x
sin y  x  sin 2 y  x 2
cos y
dy
1
dx
dy
1

dx cos y
1

1  sin 2 y

1
1  x2
Q11
1  x 2 (by Pythagoras thm)
Height of cone = 2  1  x   1  x
Base radius of cone =
Volume of cone
1
  r 2h
3
2
1
  1  x 2 1  x 
3
1
  1  x 2 1  x 
3





3

1  x  x
2
 x3

Q12
H
h
x
By similar triangles,
Rx R

h
H
H  R  x
h
R
R
Volume of Cylinder
  x2h
 H  R  x 
  x2 

R


2
 Hx ( R  x )

R
dV
3 Hx 2
 2 Hx 
0
dx
R
3 Hx 2
2 Hx 
R
x  0 (rej) or
x  2 H 

3 H
R
2R
3
d 2V
6 Hx
 2 H 
2
dx
R
 2R 
6 H 

2
dV
 3   0  Volume is max when x  2 R

2

H

dx 2 x  2 R
R
3
3
Q13
By Pythagoras Theorem,
AQ = 6 cm
Triangle BAQ is similar to triangle BPS (AAA)
PS
BS

AQ BQ
y x

6 8
3
y x
4
Area of triangle PQR,
1
A   PR  y
2
1
3
  (16  2 x)  x
2
4
3
 6 x  x 2 (shown)
4
dA
3
 6 x  0
dx
2
x4
d2 A
3
 0
2
dx
2
A is max.
3
Max A = 6(4)  (4) 2  12cm 2
4
Q14
1
Vcone   r 2 h
3
2
r  18h  h 2
1
Vcone   18h  h 2  h
3
dV 
  36h  3h 2   0
dh 3
 h  12, Vmax  288 cm3
Q15(a)
y  2x 1
5
2x 1
Q15(b)
y
x2  3
x 1
 x  1

 2  x  1  1  2  3
x 1
4
 x 1
x 1
2
Q15(c)
y  x2 
1
x
Q16
f ( x) 
f '( x) 

xk
xk
x kx  k
x k
2k
x k
2
2
 0 for k  0
Q17
 xk 
g( x)  
 ,k  0
 xk 
2
 x  k   2k 
g'( x)  2 


2
 x  k    x  k  

4k  x  k 
x k
3
0
x  k , g( x)  0   k , 0  is a stationary point
mx  k  x  k   0
3
2
mx  k   x  k 
3
2
 xk 
mx  k   

 xk 
Q18
b 1
ax  a
x  1 ax 2  4
 ax 2  ax
ax  4
ax  a
4a
y
ax 2  4
4a
 ax  a 
xb
x 1
 ax  a  a 1  x 
a 1
2
y  1 x 
3
x 1
3
dy
 1 
0
2
dx
 x  1
Hence,
dy
is strictly decreasing for all values of x.
dx
Q19
y  sin x  sin 2 x, 0  x  2
dy
 cos x  2 cos 2 x  0
dx
cos x  2  2 cos 2 x  1  0
4 cos 2 x  cos x  2  0
1  33
8
x  0.936, 5.35, 2.57, 3.71
cos x 
y  1.76,  1.76,  0.369, 0.369
The stationary points are:  0.936, 1.76  ,  5.35,  1.76  ,  2.57,  0.369  ,  3.71, 0.369 
Greatest y = 1.76
Q20
x  2t , y 
1
t
dx
dy
1
 2,   2
dt
dt
t
dy
1
 2
dx
2t
2
d y 1

dx 2 t 3
Q21
x  2 cos3 t , y  sin 3 t
dx
dy
 6 cos 2 t sin t ,  3sin 2 t cos t
dt
dt
2
dy 3sin t cos t

dx 6 cos 2 t sin t
sin t

2 cos t
1
  tan t
2
54
x
,
125
27
cos3 t 
125
3
cos t 
5
t  0.927...
dy
dx
tangent
1
  tan  0.927...
2

dy
dx

normal
2
3
3
2
Equation of normal:
64 3 
54 
y
 x

125 2 
125 
250 y  375 x  34
Q22
Q23
y  x 3  x 2  kx
dy
 3x 2  2 x  k  0
dx
 4  4(3)( k )  0
4  12k  0
k
1
3