Half of a snowball melts in an hour, how long will it take for the rest to melt? assuming that it remains spherical and that its volume decreases at a rate proportional to its surface area.

Much appreciated(Headbang)

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- Jul 14th 2010, 01:47 AMzcus05Melting snowball at rate proportional to surface area?
Half of a snowball melts in an hour, how long will it take for the rest to melt? assuming that it remains spherical and that its volume decreases at a rate proportional to its surface area.

Much appreciated(Headbang) - Jul 14th 2010, 02:58 AMGrandad
Hello zcus05

Welcome to Math Help Forum!I am assuming the the phrase 'half of a snowball' refers to half the volume, not that the radius has been halved (which would reduce the volume to one-eighth of its original value).

The surface area and volume formulae are:

$\displaystyle S = 4\pi r^2$The rate of decrease of volume is proportional to the surface area is translated as:

$\displaystyle V = \frac43\pi r^3$

$\displaystyle \dfrac{dV}{dt}=-kS$If the initial volume is $\displaystyle V_0$, then:

$\displaystyle \Rightarrow \dfrac{dV}{dr}\cdot\dfrac{dr}{dt}= -kS$

$\displaystyle \Rightarrow 4\pi r^2 \dfrac{dr}{dt}=-kS$

$\displaystyle \Rightarrow \dfrac{dr}{dt}=-k$

$\displaystyle \Rightarrow r = -kt+r_0$, where $\displaystyle r_0$ is the initial radius.

$\displaystyle V_0 = \frac43\pi r_0^3$

So when $\displaystyle V = \frac12V_0 $:

$\displaystyle \frac43\pi r^3 = \frac23 \pi r_0^3$So, assuming we measure time in hours, when $\displaystyle \displaystyle t = 1$:

$\displaystyle \Rightarrow r = \dfrac{r_0}{\sqrt[3]{2}}$

$\displaystyle \dfrac{r_0}{\sqrt[3]{2}}=-k+r_0$When $\displaystyle r=0$:

$\displaystyle \Rightarrow k = \dfrac{r_0(\sqrt[3]2-1)}{\sqrt[3]2}$

$\displaystyle t=\dfrac{r_0}{k}$So the time taken for the remainder to melt is:

$\displaystyle =\dfrac{\sqrt[3]2}{\sqrt[3]2-1}$

$\displaystyle \dfrac{\sqrt[3]2}{\sqrt[3]2-1}-1=\dfrac{1}{\sqrt[3]2-1}$ hours.If, after all, the phrase 'half the snowball' meant that the*radius*has been halved, then, of course the additional time taken would be one hour, since the radius is decreasing at a constant rate.

Grandad

- Jul 15th 2010, 10:46 PMzcus05
Ive spent days trying to get my head around this problem, thanks heaps for the clear and comprehensive response (Rofl)