Standard +0.3 This is a straightforward related rates problem requiring differentiation of the sphere volume formula with respect to time, then rearranging to show dr/dt ∝ 1/r². It's slightly easier than average as it's a 'show that' question with a clear target, uses a given formula, and involves only one differentiation step with basic algebraic manipulation.
The volume of a spherical bubble is increasing at a constant rate.
Show that the rate of increase of the radius, \(r\), of the bubble is inversely proportional to \(r^2\)
Volume of a sphere = \(\frac{4}{3}\pi r^3\)
[4 marks]
Question 10:
10 | Models the rate of change of
volume with a differential
equation of the correct form.
With respect to time, not
contradicted. | 3.3 | B1 | dv
=k
dt
dv
=4πr2
dr
dv dv dr
= ×
dt dr dt
dr
k =4πr2×
dt
dr k
=
dt 4πr2
dr 1
∴ ∝
dt r2
Obtains 4πr2by differentiation. | 1.1b | B1
Uses the chain rule to connect
rates of change substituting their
expressions for dv/dt and dv/dr.
Or
Integrates to obtain expression
for v=kt+c
Then differentiates wrt r
dv/dr=kdt/dr
Substitutes their expression for
dv/dr | 3.1b | M1
Completes argument, obtaining
dr
a correct expression for and
dt
dr 1
concluding that ∝
dt r2
OE statement | 2.1 | R1
Total | 4
Q | Marking instructions | AO | Mark | Typical solution
The volume of a spherical bubble is increasing at a constant rate.
Show that the rate of increase of the radius, $r$, of the bubble is inversely proportional to $r^2$
Volume of a sphere = $\frac{4}{3}\pi r^3$
[4 marks]
\hfill \mbox{\textit{AQA Paper 1 2019 Q10 [4]}}