Question

A spherical block of ice melts so that its surface area decreases at a constant rate: ds/dt = - 8 pi cm^2/s. calculate how fast the radius is decreasing when the radius is 3cm. (recall that s = 4 pi r^2.)

Answer 1

The rate a which the surface area, S, decreases is
$\frac{dS}{dt} =-8 \pi \, cm^{2}/s$

The surface area is
S = 4πr²
where r =  the radius at time t.

Therefore
$\frac{dS}{dt} = \frac{dS}{dr} \frac{dr}{dt} =8 \pi r \frac{dr}{dt} \\\\ -8 \pi = 8 \pi r \frac{dr}{dt} \\\\ \frac{dr}{dt} =- \frac{1}{r}$

When r = 3 cm, obtain
$\frac{dr}{dt}]_{r=3} = - \frac{1}{3} \, cm/s$

Answer:  -1/3  cm/s  (or -0.333 cm/s)