A gourmet chef is renowned for her spherical shaped soufflé. Once it is put in the oven, its volume increases at a rate proportional to its radius.

(a)     Show that the radius r cm of the soufflé, at time t minutes after it has been put in the oven, satisfies the differential equation $\frac{{{\text{d}}r}}{{{\text{d}}t}} = \frac{k}{r}$, where k is a constant.

(b)     Given that the radius of the soufflé is 8 cm when it goes in the oven, and 12 cm when it’s cooked 30 minutes later, find, to the nearest cm, its radius after 15 minutes in the oven.

(a)     $\frac{{dV}}{{{\text{d}}t}} = cr$     A1

$V = \frac{4}{3}\pi {r^3}$

$\frac{{dV}}{{{\text{d}}t}} = 4\pi {r^2}\frac{{{\text{d}}r}}{{{\text{d}}t}}$     M1A1

$\Rightarrow 4\pi {r^2}\frac{{{\text{d}}r}}{{{\text{d}}t}} = cr$     M1

$\Rightarrow \frac{{{\text{d}}r}}{{{\text{d}}t}} = \frac{c}{{4\pi r}}$     A1

$= \frac{k}{r}$     AG

[5 marks]

(b)     $\frac{{{\text{d}}r}}{{{\text{d}}t}} = \frac{k}{r}$

$\Rightarrow \int {r{\text{d}}r = \int {k{\text{d}}t} }$     M1

$\frac{{{r^2}}}{2} = kt + d$     A1

An attempt to substitute either t = 0, r = 8 or t = 30, r = 12     M1

When t = 0, r = 8

$\Rightarrow d = 32$     A1

$\Rightarrow \frac{{{r^2}}}{2} = kt + 32$

When t = 30, r = 12

$\Rightarrow \frac{{{{12}^2}}}{2} = 30k + 32$

$\Rightarrow k = \frac{4}{3}$     A1

$\therefore \frac{{{r^2}}}{2} = \frac{4}{3}t + 32$

When t = 15 , $\frac{{{r^2}}}{2} = \frac{4}{3}15 + 32$     M1

$\Rightarrow {r^2} = 104$     A1

$r \approx 10{\text{ cm}}$     A1

Note: Award M0 to incorrect methods using proportionality which give solution r = 10 cm .

[8 marks]

Total [13 marks]

Candidates found this question quite difficult, with only the better students making appreciable progress on part (a). Relatively few candidates recognised that part (b) was asking them to solve a differential equation. Many students tried methods involving direct proportion, which did not lead anywhere.