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A math exam paper from may 2011, focusing on integration and differential equations. It includes various types of questions such as evaluating definite and indefinite integrals, solving differential equations, and using partial fractions. The exam covers topics like integration by parts, first and second order differential equations, and the motion of projectiles.
Typology: Exams
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Examiner: M.J. Forward, Extension 44011.
Time allowed: Three hours
ALL answers to Section A and the best THREE answers to Section B will be counted. Section A carries 55 % of the available marks. The marks shown against questions, or parts of questions, indicate their relative weight. Your attention is drawn to the Formulae Sheet which accompanies this exam paper.
(a)
∫ (4x−^3 ) dx , [2 marks]
(b)
∫ (2x + 5)^4 dx , [2 marks]
(c)
∫ (^) dx
(3x + 2)
, [2 marks]
(d)
∫ (xex (^2) + ) dx. [4 marks]
(a)
∫ (^2)
− 2
(3x^3 − x^2 ) dx , [2 marks]
(b)
∫ √ 10
1
2 x
x^2 − 1 dx , [2 marks]
(c)
∫ (^) π
− π 2
sin (4x) dx , [3 marks]
(d)
∫ (^9)
2
dx √ 2 x
. [3 marks]
(a)
d^2 y dx^2
dy dx
(b)
d^2 y dx^2
dy dx
(ii) For equation (b) of part (i), find the particular solution where
y(0) = 1 and
dy dx
[2 marks]
(i) Using the notation that the vertically upwards direction is denoted by y, show that the differential equation for the ball’s motion is given by
d^2 y dt^2
= −g
where t the time in seconds and g the acceleration due to gravity (g ≈ 10 ms−^2 ). [4 marks]
(ii) Find the general solution of this differential equation. [3 marks]
(iii) If the ball reaches a height of y = 15m after 1 second, show that its initial velocity is v 0 = 20ms−^2. Find the particular solution which describes the ball’s motion, assuming that the ball is thrown at time t = 0 from the ground, which corresponds to y = 0. [5 marks]
(iv) What is the maximal height the ball reaches? (Hint: compute the time the ball takes to reach its maximal height). [3 marks]
d^2 y dx^2
dy dx
and find the particular solution where
y(
π 6
) = 0 and
dy dx
π 6
[9 marks]
(b) Evaluate the following integral:
∫ √ 3 √ 6 2
dx √ 3 − x^2
Hint: Use the substitution x =
3 sin(u). [6 marks]
d^2 θ dt^2
where t is time and k^2 = 36.
(a) Find the general solution of this differential equation. [4 marks]
(b) Find the particular solution where θ = 2 and
dθ dt
= 0 when t = 0. [4 marks]
(c) Plot this particular solution on a graph, where the horizontal axis is t and the vertical axis is θ. The plot should display at least one full period of the function. Specify its period explicitly. [5 marks]
(d) Find the particular solution where θ = 0 and
dθ dt
= 0 at time t = 0. Explain
why this solution is different from any other particular solution. [2 marks]
Paper Code MATH 014 May 2011 Page 7 of 7 END
