Grade 10 ADD Math
28/01/24 Review Worksheet 12 – Differentiation & Integration
1)
tan 2 x dy
(i) Given that y = , find . [3]
x dx
tan 2 x
(ii) Hence find the equation of the normal to the curve y = at the point where
x
x= . [3]
8
2)
The diagram shows an empty container in the form of an open triangular prism. The triangular
faces are equilaeral with a side of x cm and the length of each rectangular face is y cm. The
container is made form thin sheet metal. When full, the container holds 200 3 cm 3 .
(i) Show that A cm 2 , the total area of the thin sheet metal used, is given by
3x 2 1600
A= + . [5]
2 x
(ii) Given that x and y can vary, find the stationary value of A and determine its nature. [4]
3)
(i) Differentiate sin x cos x with respect to x, giving your answer in terms of sin x. [3]
sin
2
(ii) Hence find xdx . [3]
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4)
4
The diagram shows part of the curve y = + 2 x and the line y = 4x.
(2 x + 1)2
(i) Find the coordinates of A, the stationary point of the curve. [5]
4
(ii) Verify that A is also the point of intersection of the curve y = + 2 x and the
(2 x + 1)2
line y = 4x. [1]
(iii) Without using a calculator, find the area of the shaded region enclosed by the line
y = 4x. , the curve and the y-axis. [6]
5)
(a) A particle P moves in a straight line. Starting from rest, P moves with constant
acceleration for 30 seconds after which it moves with constant velocity, kms −1 , for 90
seconds. P then moves with constant deceleration until it comes to rest; the magnitude of
the deceleration is twice the magnitude of the initial acceleration.
(i) Use the information to complete the velocity-time graph. [2]
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(ii) Given that the particle travels 450 metres while it is accelerating, find the value of k and
the acceleration of the particle. [4]
(b) A body Q moves in a straight line such that, t seconds after passing a fixed point, its
acceleration, a ms −2 , is given by a = 3t 2 + 6 . When t = 0, the velocity of the body is 5
ms −1 . Find the velocity when t = 3. [5]
6)
dy
(i) Given that y = e x , find
2
. [2]
dx
xe
x2
(ii) Use your answer to part (i) to find dx. [2]
2
2
(iii) Hence evaluate xe x dx. [2]
0
dy 1
7) A curve is such that = 4x + for x > 0. The curve passes through the point
dx (x + 1)2
1 5
, .
2 6
(i) Find the equation of the curve. [4]
(ii) Find the equation of the normal to the curve at the point where x = 1. [4]
8)
(a) The diagram shows the velocity-time graph of a particle P moving in a straight line with
velocity v ms −1 at time t s after leaving a fixed point.
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Find the distance travelled by the particle P. [2]
(b) The diagram shows the displacement-time graph of a particle Q moving in a straight line
with displcement s m from a fixed point at time t s.
On the axes below, plot the corresponding velocity-time graph for the particle Q. [3]
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(c) The displacement s m of a particle R, which is moving in a straight line, from a fixed
point at time t s is given by s = 4t − 16 ln (t + 1) + 13.
(i) Find the value of t for which the particle R is instantaneously at rest. [3]
(ii) Find the value of t for which the acceleration of the particle R is 0.25 ms −2 . [2]
9) The region enclosed by the curve y = 2 sin 3x, the x-axis and the line x = a, where
1
0 a 1 radian, lies entirely above the x-axis. Given that the area of this region is square
3
unit, find the value of a. [6]
10) A solid circular has a base radius of r cm and a volume of 4000 cm 3 .
8000
(i) Show that the total surface area, A cm 2 ,of the cylinder is given by A = + 2r 2 .
r
[3]
(ii) Given that r can vary, find the minimum total surface area of the cylinder, justifying that
this area is a minimum. [6]
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11) The diagram shows a thin square sheet of metal measuring 24 cm by 24 cm. A square of side
x cm is cut off from each corner. The remainder is then folded to form an open box, x cm
deep, whose square base is shown shaded in the diagram.
(i) Show that the volume, V cm 3 , of the box is given by V = 4 x 3 − 96 x 2 + 576 x. [2]
(ii) Given that x can vary, find the maximum volume of the box. [4]
( )
1
12) Find the equation of the normal to the curve y = x x 2 − 12 3 at the point on the curve where
x = 2. [6]
13) A particle moves in a straight line such that, t s after passing through a fixed point O, its
velocity, v ms −1 , is given by v = 5 − 4e −2t .
(i) Find the velocity of the particle at O. [1]
(ii) Find the value of t when the acceleration of the particle is 6 ms −2 . [3]
(iii) Find the distance of the particle from O when t = 1.5. [5]
(iv) Explain why the particle does not return to O. [1]
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14) The diagram shows the graph of y = cos3x + 3 sin 3x, which crosses the x-axis at A and
has a maximum point at B.
(i) Find the x-coordinate of A. [3]
dy
(ii) Find and hence find the x-coordinate of B. [4]
dx
(iii) Showing all your working, find the area of the shaded region bounded by the curve, the
x-axis and the line through B parallel to the y-axis. [5]
1
15) Given that a curve has equation y = + 2 x , where x > 0, find
x
dy
(i) , [2]
dx
d2y
(ii) . [2]
dx 2
Hence, or otherwise, find
(iii) the coordinates and nature of the stationary point of the curve. [4]
dy
16) Find when
dx
x
(i) y = cos 2 x sin , [4]
3
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tan x
(ii) y = , [4]
1 + ln x
17) A sector of a circle of radius r cm has an angle of radians, where . The perimeter of
the sector is 30 cm.
(a) Show that the area, A cm 2 , of the sector is given by A = 15r − r 2 . [3]
(b) Given that r can vary, Find the maximum area of the sector. [3]
18) Differentiate with respect to x
(i) x 4 e 3 x , [2]
(ii) ln (2 + cos x ), [3]
sin x
(iii) [3]
1+ x
= (2 x + 1) 2 . The curve passes through the point (4, 10).
dy 1
19) A curve is such that
dx
(i) Find the equation of the curve. [4]
1.5
(ii) Find y dx and hence evaluate 0
ydx. [5]
20) A curve has equation y = x 3 − 9 x 2 + 24 x.
dy
(i) Find the set of values of x for which 0. [4]
dx
The normal to the curve at the point on the curve where x = 3 cuts the y-axis at the point P.
(ii) Find the equation of the normal and the coordinates of P. [5]
dy
21) A curve is such that = 6 x 2 − 8x + 3 .
dx
(i) Show that the curve has no stationary points. [2]
Given that the curve passes through the point P(2, 10),
(ii) find the equation of the tangent to the curve at the point P, [2]
(iii) find the equation of the curve. [4]
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22)
2x dy k
(i) Given that y = , show that = , where k is a constant to be
x + 21
2 dx (x 2
+ 21 )
3
found. [5]
6 6
10
(ii) Hence find dx and evaluate dx. [3]
(x 2
+ 21 ) 3 2
(x 2
+ 21 )
3
23) The point A, where x = 0, lies on the curve y =
(
ln 4 x 2 + 3 )
. The normal to the curve at A
x −1
meets the x-axis at the point B.
(i) Find the equation of this normal. [7]
(ii) Find the area of the triangle AOB, where O is the origin. [2]
24)
The diagram shows parts of the line y = 3x + 10 and the curve y = x 3 − 5 x 2 + 3x + 10 .The line
and the curve both pass through the point A on the y-axis. The curve has a maximum at the point
B and a minimum at the point C. The line through C, parallel to the y-axis, intersects the line
y = 3x + 10 at the point D.
(i) Show that the line AD is a tangent to the curve at A. [2]
(ii) Find the x-coordinate of B and of C. [3]
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(iii) Find the area of the shaded region ABCD, showing all your working. [5]
25) A particle P is projected from the origin O so that it moves in a straight line. At time t
seconds after projection, the velocity of the particle, v ms −1 , is given by v = 2t 2 − 14t + 12.
(i) Find the time at which P first comes to instantaneous rest. [2]
(ii) Find an expression for the displacement of P from O at time t seconds.
(iii) Find the acceleration of P when t = 3. [2]
26)
The diagram shows a cuboid of height h units inside a right pyramid OPQRS of height 8 units and
with square base of side 4 units. The base of the cuboid sits on the square base PQRS of the
pyramid. The points A, B, C and D are corners of the cuboid and lie on the edges OP, OQ, OR
and OS, respectively, of the pyramid OPQRS. The pyramids OPQRS and OABCD are similar.
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(i) Find an expression for AD in terms of h and hence show that the volume V of the cuboid
h3
is given by V = − 4h2 + 16h units3. [4]
4
(ii) Given that h can vary, find the value of h for which V is a maximum. [4]
27) A curve, showing the relationship between two variables x and y, is such that
d2y 1
= 6cos3x . Given that the curve has a gradient of 4 3 at the point , − , find the
dx 2 9 3
equation of the curve. [6]
28)
(
(i) Variables x and y are such that y = ( x − 3) ln 2 x 2 + 1 )
dy
(a) Find the value of when x = 2. [4]
dx
(b) Hence find the approximate change in y when x changes from 2 to 2.03. [2]
5x − 1
(ii) Variables x and y are related by the equation y = .
3− x
dy
(a) Find ,simplifying your answer. [2]
dx
(b) Hence find the approximate change in x when y increases from 9 by the small amount
0.07. [3]
x2
(iii) Given that y = , find
cos 4 x
dy
(a) ,
dx
(b) the approximate change in y when x increases from to + p , where p is small.
4 4
29) The rate of change of a variable x with respect to time t is 4cos 2 t .
(i) Find the rate of change of x with respect to t when t = . [1]
6
The rate of change of a variable y with respect to time t is 3sin t .
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(ii) Using your result from part (i), find the rate of change of y with respect to x when t = .
6
[3]
30)
(i) The total surface area, A cm2, of a solid cylinder with radius r cm and height 5 cm is
0.2
given by A = 2 r 2 + 10 r .Given that r is increasing at a rate of cms-1, find the rate
of increase of A when r is 6. [4]
(ii)
The diagram shows a container in the shape of a cone of height 120 cm and radius 30 cm. Water
is poured into the container at a rate of 20 cm3 s −1
(a) At the instant when the depth of water in the cone is h cm the volume of water in the cone
h3
is Vcm3. Show that V = . [3]
48
(b) Find the rate at which h is increasing when h = 50. [3]
(c) Find the rate at which the circular area of the water’s surface is increasing when h = 50.
[4]
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Answers: 1)(i) =
( )
dy x 2 sec 2 2 x − tan 2 x
(ii) y = −0.27 x + 2.65
dx x2
2)(ii) x = 9.74, So A = 246; so the value is a minimum 3)(i) 1− 2 sin 2 x
− sin x cos x + c 4) (i) x = , y = 2 (ii) y = 4 = 2 (iii) Shaded area =
x 1 1 1 3
(ii)
2 2 2 2 4
5)(a)(i)
(ii) k = 30; a = 1 ms −2 (b) 50 [ms −1
]
dy 1 2 1 4 1 1
= 2 xe x (ii) e x (iii) e − = 26.8 7)(i) y = 2 x − +1
2
2
6)(i)
dx 2 2 2 x +1
(ii) 8x + 34y - 93 = 0 8)(a) Distance travelled = area under graph = 480 (b)
(c)(i) When v = 0, t = 3 (ii) t = 7 9) a =
9
2
d A
10)(ii) A (1395, 1390) Show that is positive 11)(ii) V =1024 12) 4 y = 3x − 22
dr 2
13)(i) 1 (ii) t = 0.144 (iii) s = 5.60 (iv) Velocity is always positive, so no change in direction
5
14)(i) x = (0.873) (allow 50 )
18
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dy 2
(ii) = 3 3 cos3x − 3 sin 3x ; x = ;(0.349) (allow 20 ) (iii) Area = or 0.667 or better
dx 9 3
1 1 2 1
15)(i) − 2 + (ii) 3 − (iii) x = 1 y = 3 minimum
x x 12 x 2x 32
1 x x
(sec 2 x )(1 + ln x ) − (tan x )
1
x
16) (i) cos 2 x cos − 2 sin 2 x sin (ii) 17)(ii) r = 7.5,
3
3 3 (1 + ln x )2
56.25 18)(i) x 4 (3e 3 x ) + 4 x 3 e 3 x (ii) (− sin x ) (iii)
1
2 + cos x
( ) 1 −
1
1 + x their cos x − their x 2 sin x
(2 x + 1) 3
2
2 19) (i) y = +1
(1+ x ) 2
2
3
2
1
(ii) (2 x + 1) 2 + 1dx = (2 x + 1) 2 + x(+ const ) ; 107
3 1 5
3 15 30
dy 1
20) (i) = 3x 2 − 18 x + 24, x 2, x 4 , (ii) y = their + x + c and c = their 17; P (0, 17)
dx 3
21) (ii) y = 11x − 12 (iii) y = 2 x − 4 x + 3x + 4 22) (i) k = 42
3 2
(x − 1) 82 x − ln (4 x 2 + 3)
(ii)
6 2x 8
or 0.145 23) (i)
dy
=
(4 x + 2)
k x 2 + 21 55 dx (x − 1)2
10 11 1
y = 0.910 x − 1.10, or y = x − (ii) Area 0.66 or 0.67 24) (ii) x = , x = 3
11 10 3
2t 3 14t 2 8−h
(iii) Area = 24.7 or 24.8 25) (i) t = 1 (ii) (s = )
8
− + 12t (iii) −2 26) (i) 4 (ii)
3 2 8 3
27) 28) (i)(a) (b)0.0393
(ii)(a) (b)0.005 (iii)(a) (b) -1.57 p
h
29) (i) 3 (ii) 0.5 30)(i) 6.8 (ii) (a)Radius is (b)0.128 (c) 2.51
4
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