Question

Suppose that the volume \(V\) of a rolling snowball increases so that \(\frac{{dV}}{{dt}}\) is proportional to the surface area of the snowball at time \(t\). Show that the radius \(r\) increases at a constant rate, that is, \(\frac{{dr}}{{dt}}\) is constant.

Short answer

It is proved that the radius \(r\) increases at a constant rate, that is, \(\frac{{dr}}{{dt}}\) is constant.

Step-by-step solution

Write the formula of volume and area of sphere

Volume of sphere:\(V = \frac{4}{3}\pi {r^3}\)

Area of sphere:\(A = 4\pi {r^2}\)

Construct the diagram according to the given conditions

Step 3 Show that the radius \(r\) increases at a constant rate, that is, \(\frac{{dr}}{{dt}}\) is constant.

The volume \(V\) of a rolling snowball increases so that \(\frac{{dV}}{{dt}}\) is proportional to the surface of the snowball at time \(t\) then for some positive number \(K\) we have,

\(\begin{aligned}\frac{{dV}}{{dt}} &= K \cdot A\\\frac{{d\left( {\frac{4}{3}\pi {r^3}} \right)}}{{dt}} &= K \cdot 4\pi {r^2}\\\frac{4}{3}\pi \cdot \frac{{d{r^3}}}{{dt}} &= K \cdot 4\pi {r^2}\\\frac{4}{3}\pi \cdot 3{r^2}\frac{{dr}}{{dt}} &= K \cdot 4\pi {r^2}\\K \cdot 4\pi {r^2}\frac{{dr}}{{dt}} &= K \cdot 4\pi {r^2}\\\frac{{dr}}{{dt}} = K\end{aligned}\)

This implies that the rate of change of radius \(r\) with respect to \(t\) is constant.

Hence proved.