Related rates - Melting snowball Assume that a snowball melts in such a way that its volume decreases at a rate proportional to its surface area. If half the original snowball has melted away after 2 hours, how much longer will it take for the snowball to disappear completely? My try: $$\frac{dV}{dt}=K4\pi r^2$$ and $$V(0) = \frac{4}{3}\pi [r(0)]^3$$ $$V(2) = \frac{2}{3}\pi [r(0)]^3$$ but that's about it. I don't know how to find $t$ such that $V(t) = 0$ Also, it seems to me that this has some relationship to exponential decay, is that right or not?

$V = \frac {4}{3} \pi r^3\\\ \frac {dV}{dr} = 4\pi r^2\\\ \frac {dV}{dt} = \frac {dV}{dr}\frac {dr}{dt} = -4K\pi r^2\\\ \frac {dr}{dt} = -K$

$r(t) = r(0) - tK$

$V(2) = \frac 12 V(0)\\\ \frac 43 \pi (r(2))^3 = \frac 23 \pi (r(0))^3\\\ (r(2))^3 = \frac 12 (r(0))^3\\\ (r(0) -2K)^3 = \frac 12(r(0))^3\\\ r(0) - 2K = \frac {1}{2^{\frac 13}} r(0)\\\ K = \frac 12(1 - 2^{-\frac 13}) r(0)$

Find $t$ such that

$r(0) - Kt = 0\\\ r(0) - \frac 12(1 - 2^{-\frac 13}) r(0)t = 0\\\ 1 = \frac 12(1 - 2^{-\frac 13})t\\\ t = \frac {2}{2^{-\frac 13} - 1}\\\ t = \frac {4}{2 - 2^\frac 23}$

This $t$ is total time. As the question is "how much longer" we are not starting the clock until the 2 hour point.

$\frac {4}{2 - 2^\frac 23}-2$