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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?

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    $\begingroup$ Try $V=4\pi r^3/3$, differentiate with respect to $t$, and substitute into your DE above to get the DE in terms of $r$. Find out when $r=0$. $\endgroup$ – Adrian Keister Apr 5 '18 at 17:25
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$V = \frac {4}{3} \pi r^3\\ \frac {dV}{dr} = 4\pi r^2\\ \frac {dV}{dt} = \frac {dV}{dr}\frac {dr}{dt} = -4K\pi r^2\\ \frac {dr}{dt} = -K$

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

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

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.

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

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  • $\begingroup$ Thanks, very thorough. $\endgroup$ – pseudomarvin Apr 5 '18 at 17:55
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Note that $V=\frac 43 \pi r^3$ so then $\frac{dV}{dt}=4\pi r^2 \frac{dr}{dt}$

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  • $\begingroup$ Okay so I get $\frac{dr}{dt}=-K$, correct? Unfortunately I still do not see how to move from here. $\endgroup$ – pseudomarvin Apr 5 '18 at 17:30
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    $\begingroup$ Can you integrate this DE directly? $\endgroup$ – Adrian Keister Apr 5 '18 at 17:33
  • $\begingroup$ That's probably the problem, I'm following my first calculus textbook and at this point we haven't reached integration yet so I think they expect us to solve it without it. $\endgroup$ – pseudomarvin Apr 5 '18 at 17:34
  • $\begingroup$ But I gave it a shot: $r = -Kt + c$ $\endgroup$ – pseudomarvin Apr 5 '18 at 17:36

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