Question

It a camphor ball evaporates at a rate proportional to its surface area 4πr square . show that its radius decreases at a constant rate.

Answer 1

Let r be the radius of the sphere, V its volume, and A its surface area. We know V = (4/3)?r[sup:3pja7b27]3[/sup:3pja7b27] and A = 4?r[sup:3pja7b27]2[/sup:3pja7b27]. We are given dV/dt = c(dA/dt), where c is a nonzero constant. We want to show that dr/dt is a constant.

V = (4/3)?r[sup:3pja7b27]3[/sup:3pja7b27] ? dV/dt = (4/3)? ? d(r[sup:3pja7b27]3[/sup:3pja7b27])/dt = (4/3)? ? 3r[sup:3pja7b27]2[/sup:3pja7b27] ? dr/dt

A = 4?r[sup:3pja7b27]2[/sup:3pja7b27] ? dA/dt = 4? ? d(r[sup:3pja7b27]2[/sup:3pja7b27])/dt = 4? ? 2r ? dr/dt

dV/dt = c(dA/dt) ? (4/3)? ? 3r[sup:3pja7b27]2[/sup:3pja7b27] ? dr/dt = c[4? ? 2r ? dr/dt] ? r = 2c

I'm not sure how this result connects with dr/dt being a constant. Am I on the right track, and where do I go from here?