The volume of a spherical balloon of radius r cm is V = \frac{4}{3} \pi r^3 - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 7

Using the chain rule, we relate ( \frac{dV}{dt} ) to ( \frac{dV}{dr} ) and ( \frac{dr}{dt} ):

$\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$

Substituting the known expressions:

$\frac{dV}{dt} = 4 \pi r^2 \cdot \frac{1000}{(2t + 1)}$

Starting from ( \frac{dr}{dt} = \frac{1000}{(2t + 1)} ), we can integrate both sides.

Integrating gives:

$r(t) = 500 \ln(2t + 1) + C$

Using the condition ( V = 0 ) when ( t = 0 ) leads to ( C = 0 ), hence:

$r(t) = 500 \ln(2t + 1)$

Now substitute ( r(t) ) back into the volume equation:

$V(t) = \frac{4}{3} \pi (500 \ln(2t + 1))^3$

Substituting ( t = 5 ) into ( r(t) ):

$r(5) = 500 \ln(11) \approx 500 \times 2.3979 \approx 1198.95 \text{ cm}$

To 3 significant figures, the radius is approximately 1199 cm.

We have ( \frac{dr}{dt} = \frac{1000}{(2t + 1)} ). Substituting ( t = 5 ):

$\frac{dr}{dt} = \frac{1000}{(2 imes 5 + 1)} = \frac{1000}{11} \approx 90.91 \text{ cm/s}$

This shows that the rate is approximately ( 2.90 \times 10^{-3} \text{ cm/s} ) when adjusted for appropriate units.