10 In a model of the expansion of a sphere of radius \(r \mathrm {~cm}\), it is assumed that, at time \(t\) seconds after the start, the rate of increase of the surface area of the sphere is proportional to its volume. When \(t = 0\), \(r = 5\) and \(\frac { \mathrm { d } r } { \mathrm {~d} t } = 2\).
- Show that \(r\) satisfies the differential equation
$$\frac { \mathrm { d } r } { \mathrm {~d} t } = 0.08 r ^ { 2 }$$
[The surface area \(A\) and volume \(V\) of a sphere of radius \(r\) are given by the formulae \(A = 4 \pi r ^ { 2 }\), \(V = \frac { 4 } { 3 } \pi r ^ { 3 }\).]
- Solve this differential equation, obtaining an expression for \(r\) in terms of \(t\).
- Deduce from your answer to part (ii) the set of values that \(t\) can take, according to this model.