C4 June 2014 (R) Q5
5. At time \(t\) seconds the radius of a sphere is \(r\) cm, its volume is \(V\) cm3 and its surface area is \(S\) cm2.
[You are given that \(V = \dfrac{4}{3}\pi r^3\) and that \(S = 4\pi r^2\)]
The volume of the sphere is increasing uniformly at a constant rate of 3 cm3 s−1.
| Scheme | Marks |
|---|---|
| From question, \(V = \dfrac{4}{3}\pi r^3,\ S = 4\pi r^2,\ \dfrac{\mathrm{d}V}{\mathrm{d}t} = 3\) | |
| \(\left\{V = \dfrac{4}{3}\pi r^3 \Rightarrow\right\}\ \dfrac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2\) \(\dfrac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2\) (Can be implied) | B1 oe |
| \(\left\{\dfrac{\mathrm{d}V}{\mathrm{d}r} \times \dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{\mathrm{d}V}{\mathrm{d}t} \Rightarrow\right\}\ \left(4\pi r^2\right)\dfrac{\mathrm{d}r}{\mathrm{d}t} = 3\) \(\left\{\dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{\mathrm{d}V}{\mathrm{d}t} \div \dfrac{\mathrm{d}V}{\mathrm{d}r} \Rightarrow\right\}\ \dfrac{\mathrm{d}r}{\mathrm{d}t} = (3)\dfrac{1}{4\pi r^2};\ \left\{= \dfrac{3}{4\pi r^2}\right\}\) \(\left(\text{Candidate's } \dfrac{\mathrm{d}V}{\mathrm{d}r}\right) \times \dfrac{\mathrm{d}r}{\mathrm{d}t} = 3\) or \(3 \div \text{Candidate's } \dfrac{\mathrm{d}V}{\mathrm{d}r}\); | M1 oe |
| When \(r = 4\) cm, \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = \dfrac{3}{4\pi(4)^2}\ \left\{= \dfrac{3}{64\pi}\right\}\) dependent on previous M1. see notes | dM1 |
| Hence, \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = 0.01492077591\ldots\) (cm s−1) (corrected from the printed mark scheme: cm2 s−1) anything that rounds to 0.0149 | A1 |
| (4) |
| Scheme | Marks |
|---|---|
| \(\left\{\dfrac{\mathrm{d}S}{\mathrm{d}t} = \dfrac{\mathrm{d}S}{\mathrm{d}r} \times \dfrac{\mathrm{d}r}{\mathrm{d}t} =\right\} \Rightarrow \dfrac{\mathrm{d}S}{\mathrm{d}t} = 8\pi r \times \dfrac{3}{4\pi r^2}\ \left\{\text{or } \dfrac{6}{r} \text{ or } 8\pi r \times 0.0149\ldots\right\}\) \(8\pi r \times \text{Candidate's } \dfrac{\mathrm{d}r}{\mathrm{d}t}\) | M1; oe |
| When \(r = 4\) cm, \(\dfrac{\mathrm{d}S}{\mathrm{d}t} = 8\pi(4) \times \dfrac{3}{4\pi(4)^2}\) or \(\dfrac{6}{4}\) or \(8\pi(4) \times 0.0149\ldots\) (corrected from the printed mark scheme: \(\dfrac{\mathrm{d}r}{\mathrm{d}t} =\)) | |
| Hence, \(\dfrac{\mathrm{d}S}{\mathrm{d}t} = 1.5\) (cm2 s−1) anything that rounds to 1.5 | A1 cso |
| (2) | |
| (6 marks) |
Notes
(a)
B1: \(\dfrac{\mathrm{d}V}{\mathrm{d}r} = 4\pi r^2\) Can be implied by later working.
M1: \(\left(\text{Candidate's } \dfrac{\mathrm{d}V}{\mathrm{d}r}\right) \times \dfrac{\mathrm{d}r}{\mathrm{d}t} = 3\) or \(3 \div \text{Candidate's } \dfrac{\mathrm{d}V}{\mathrm{d}r}\)
dM1: (dependent on the previous method mark)
Substitutes \(r = 4\) into an expression which is a result of a quotient of “3” and their \(\dfrac{\mathrm{d}V}{\mathrm{d}r}\).
A1: anything that rounds to 0.0149 (units are not required)
(b)
M1: \(8\pi r \times \text{Candidate's } \dfrac{\mathrm{d}r}{\mathrm{d}t}\)
A1: anything that rounds to 1.5 (units are not required). Correct solution only.
Note: Using \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = 0.0149\) gives \(\dfrac{\mathrm{d}S}{\mathrm{d}t} = 1.4979\ldots\) which is fine for A1.