2016 H2 Maths Paper 1 Q9 — Worked Solution

Part (i)(a)

1 mark

Since $y = \frac{dx}{dt}$ , differentiating with respect to $t$ gives $\frac{dy}{dt} = \frac{d^{2}x}{dt^{2}}$ . Substituting into the given differential equation:

\frac{dy}{dt} + 2y = 10

\frac{dy}{dt} = 10 - 2y\ \ \left( \text{shown} \right)

Part (i)(b)

6 marks

Approach

$\frac{dy}{dt} = 10 - 2y$ has $y$ and $t$ separable, so we integrate directly and use $y = 0$ at $t = 0$ (since $\frac{dx}{dt} = 0$ when $t = 0$ ) to fix the constant. Once $y$ is known in terms of $t$ , we integrate again — using $x = 0$ at $t = 0$ — to get $x$ .

Since $\frac{dy}{dt} = 10 - 2y$ , separating variables,

\int\frac{1}{10 - 2y}\, dy = \int 1\, dt

- \frac{1}{2}\ln|10 - 2y| = t + C

Using $y = 0$ when $t = 0$ :

- \frac{1}{2}\ln 10 = C

Substituting back,

- \frac{1}{2}\ln|10 - 2y| = t - \frac{1}{2}\ln 10

\ln|10 - 2y| = \ln 10 - 2t

|10 - 2y| = 10e^{- 2t}

Note: the stone starts at rest and speeds up towards a terminal value, so $y$ increases from $0$ towards $5$ ; $10 - 2y$ therefore stays positive throughout, and the modulus can be dropped.

10 - 2y = 10e^{- 2t}

y = 5 - 5e^{- 2t}

So $\frac{dx}{dt} = 5 - 5e^{- 2t}$ . Integrating with respect to $t$ :

x = \int\left( 5 - 5e^{- 2t} \right)dt

= 5t + \frac{5}{2}e^{- 2t} + D

Using $x = 0$ when $t = 0$ :

0 = \frac{5}{2} + D

D = - \frac{5}{2}

x = 5t + \frac{5}{2}e^{- 2t} - \frac{5}{2}

To check your answer: at

t = 0

,

x = 0 + \frac{5}{2} - \frac{5}{2} = 0

, matching the given initial condition. ✓

$\therefore$ $y = 5 - 5e^{- 2t}$ and $x = 5t + \frac{5}{2}e^{- 2t} - \frac{5}{2}$ .

Part (ii)

3 marks

Approach

here $\frac{d^{2}x}{dt^{2}}$ is given directly as a function of $t$ , so we integrate twice in succession, using $\frac{dx}{dt} = 0$ at $t = 0$ to fix the first constant and $x = 0$ at $t = 0$ to fix the second.

\frac{d^{2}x}{dt^{2}} = 10 - 5\sin\frac{1}{2}t

Integrating with respect to $t$ :

\frac{dx}{dt} = 10t + 10\cos\frac{1}{2}t + C_{1}

Using $\frac{dx}{dt} = 0$ when $t = 0$ :

0 = 10\cos 0 + C_{1}

C_{1} = - 10

\frac{dx}{dt} = 10t + 10\cos\frac{1}{2}t - 10

Integrating again:

x = 5t^{2} + 20\sin\frac{1}{2}t - 10t + C_{2}

Using $x = 0$ when $t = 0$ :

0 = 20\sin 0 + C_{2}

C_{2} = 0

x = 5t^{2} - 10t + 20\sin\frac{1}{2}t

$\therefore$ $x = 5t^{2} - 10t + 20\sin\frac{1}{2}t$ .

Part (iii)

2 marks

Approach

the pond is $5\ m$ deep, so for each model we set $x = 5$ and solve for $t$ . Neither equation can be solved for $t$ algebraically, so we treat each as a GC numerical-solution problem.

For Model (i), setting $x = 5$ :

5t + \frac{5}{2}e^{- 2t} - \frac{5}{2} = 5

By GC, $t = 1.47\ s$ (2 d.p.).

For Model (ii), setting $x = 5$ :

5t^{2} - 10t + 20\sin\frac{1}{2}t = 5

By GC, $t = 1.05\ s$ (2 d.p.).

$\therefore$ Model (i) predicts the stone reaches the bottom at $t = 1.47\ s$ ; Model (ii) predicts $t = 1.05\ s$ .

2016 H2 Maths Paper 1, Question 9

Part (i)(a)

Part (i)(b)

Part (ii)

Part (iii)

This is how we mark, and how we teach.