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Math Help - Melting snowball at rate proportional to surface area?

  1. #1
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    Melting 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
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  2. #2
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    Hello zcus05

    Welcome to Math Help Forum!
    Quote Originally Posted by zcus05 View Post
    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
    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:
    S = 4\pi r^2

    V = \frac43\pi r^3
    The rate of decrease of volume is proportional to the surface area is translated as:
    \dfrac{dV}{dt}=-kS

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

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

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

    \Rightarrow r = -kt+r_0, where r_0 is the initial radius.
    If the initial volume is V_0, then:
    V_0 = \frac43\pi r_0^3

    So when V = \frac12V_0 :
    \frac43\pi r^3 = \frac23 \pi r_0^3

    \Rightarrow r = \dfrac{r_0}{\sqrt[3]{2}}
    So, assuming we measure time in hours, when \displaystyle t = 1:
    \dfrac{r_0}{\sqrt[3]{2}}=-k+r_0

    \Rightarrow k = \dfrac{r_0(\sqrt[3]2-1)}{\sqrt[3]2}
    When r=0:
    t=\dfrac{r_0}{k}
    =\dfrac{\sqrt[3]2}{\sqrt[3]2-1}
    So the time taken for the remainder to melt is:
    \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
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  3. #3
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    Ive spent days trying to get my head around this problem, thanks heaps for the clear and comprehensive response
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