Question

If a snowball melts so that its surface area decreases at a rate of 6 cm2/min, find the rate at which the diameter decreases when the diameter is 8 cm.

Answer 1

Assuming a spherical snowball with radius \(r\), its surface area \(A\) is given by
\[A=2\pi r\]Differentiating with respect to time \(t\) yields
\[\frac{dA}{dt}=2\pi \frac{dr}{dt}\]\(A\) decreases at a rate of \(6\text{ cm}^2/\text{min}\), which means \(\dfrac{dA}{dt}=-6\). You have enough info to solve for \(\dfrac{dr}{dt}\).

The diameter is just twice the radius, i.e.
\[d=2r\]so if you're looking for \(\dfrac{dd}{dt}\), it's just a matter of differentiating with respect to \(t\):
\[\frac{dd}{dt}=2\frac{dr}{dt}\]