C4 January 2006 Q7
7. The volume of a spherical balloon of radius \(r\) cm is \(V\ \text{cm}^3\), where \(V = \frac{4}{3}\pi r^3\).
(a) Find \(\dfrac{\mathrm{d}V}{\mathrm{d}r}\). (1)
The volume of the balloon increases with time \(t\) seconds according to the formula
\[\frac{\mathrm{d}V}{\mathrm{d}t} = \frac{1000}{(2t + 1)^2}, \qquad t \geqslant 0.\]
(b) Using the chain rule, or otherwise, find an expression in terms of \(r\) and \(t\) for \(\dfrac{\mathrm{d}r}{\mathrm{d}t}\). (2)
(c) Given that \(V = 0\) when \(t = 0\), solve the differential equation \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = \dfrac{1000}{(2t + 1)^2}\), to obtain \(V\) in terms of \(t\). (4)
(d) Hence, at time \(t = 5\),
(i) find the radius of the balloon, giving your answer to 3 significant figures, (3)
(ii) show that the rate of increase of the radius of the balloon is approximately \(2.90 \times 10^{-2}\ \text{cm}\,\text{s}^{-1}\). (2)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2\) | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| Uses \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{\mathrm{d}V}{\mathrm{d}t}\cdot\dfrac{\mathrm{d}r}{\mathrm{d}V}\) in any form, \(= \dfrac{1000}{4\pi r^2(2t + 1)^2}\) | M1,A1 |
| (2) |
| Scheme | Marks |
|---|---|
| \(V = \displaystyle\int 1000(2t + 1)^{-2}\,\mathrm{d}t\) and integrate to \(p\,(2t + 1)^{-1}\), \(= -500(2t + 1)^{-1}\ (+c)\) | M1, A1 |
| Using \(V = 0\) when \(t = 0\) to find \(c\), (\(c = 500\), or equivalent) | M1 |
| \(\therefore V = 500\left(1 - \dfrac{1}{2t + 1}\right)\) (any form) | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| (i) Substitute \(t = 5\) to give \(V\), | M1, |
| then use \(r = \sqrt[3]{\left(\dfrac{3V}{4\pi}\right)}\) to give \(r\), \(= 4.77\) | M1, A1 |
| (3) | |
| (ii) Substitutes \(t = 5\) and \(r = \) ‘their value’ into ‘their’ part (b) | M1 |
| \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = 0.0289 \quad (\approx 2.90 \times 10^{-2})\) (cm/s) \(\ast\) AG | A1 |
| (2) | |
| (12 marks) |