SCalcET$93\mathrm{.}9\mathrm{.}053.$

Suppose that the volume $V$ of a rolling snowball increases so that $\frac{dV}{di}$ 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.

The volume of the snowball is given by $V^{}=\frac{4}{3}{r}^3,$ so

Since the change in volume is proportional to the surface area $S,$ with $S=4r^2,$ we also have the following for some constant $k.$

$$

$\frac{dV}{dt}=$cdots $k^3$

$\frac{dV}{dt}=r4k^2$

$\frac{dV}{dt}=k*4r^2$

$\frac{dV}{dt}=k*r^3$

$\frac{dV}{dk}=k*2r$

$x$

Equating the timo expressions for $\frac{dv}{dt}$ oves $4r^2\frac{dr}{dt}=k*((1)-\frac{dr}{dt}=k_1$ that is$,\frac{d}{T}$ is constant.