7. The volume of a spherical balloon of radius \(r \mathrm {~cm}\) is \(V \mathrm {~cm} ^ { 3 }\), where \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\).
- Find \(\frac { \mathrm { d } V } { \mathrm {~d} r }\).
The volume of the balloon increases with time \(t\) seconds according to the formula
$$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 1000 } { ( 2 t + 1 ) ^ { 2 } } , \quad t \geqslant 0$$
- Using the chain rule, or otherwise, find an expression in terms of \(r\) and \(t\) for \(\frac { \mathrm { d } r } { \mathrm {~d} t }\).
- Given that \(V = 0\) when \(t = 0\), solve the differential equation \(\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 1000 } { ( 2 t + 1 ) ^ { 2 } }\), to obtain \(V\) in terms of \(t\).
- Hence, at time \(t = 5\),
- find the radius of the balloon, giving your answer to 3 significant figures,
- show that the rate of increase of the radius of the balloon is approximately \(2.90 \times 10 ^ { - 2 } \mathrm {~cm} \mathrm {~s} ^ { - 1 }\).