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A spherical raindrops evaporates at rate proportional to surface area?

  1. Jun 23, 2014 #1
    i want to find V(t)
    At first i found this problem was very simple but when i try to write differential equations i ended up with these

    V' = kA thats for sure

    then i confined the problem only to spherical shape and no other shapes of raindrops involved

    as i cant express A in term of V alone( surface area of sphere = 4∏r2, volume of sphere is 4/3∏r3 ) then i have to

    use chain rule, dV/dt= dVdrdrdt substitute dV/dt from


    i get

    4∏r2r'= K4∏r2


    r = kt+c
    r3 = (kt+c)3
    4∏r3/3 = (kt+c)34∏/r=V(t)
    im i correct? the answer to this problem is V'=kV2/3 im not sure how they transform Area to variable V alone
    Last edited: Jun 23, 2014
  2. jcsd
  3. Jun 23, 2014 #2


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    \left(\frac{3V}{4\pi}\right)^{1/3} = r.
    [/tex] Now substitute this into [itex]A = 4\pi r^2[/itex].
  4. Jun 24, 2014 #3
    then how can u solve for V(t) ?
  5. Jun 24, 2014 #4


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    The ODE [tex]
    V' = CV^{2/3}
    [/tex] for constant [itex]C[/itex] is separable. Divide both sides by [itex]V^{2/3}[/itex] and then integrate with respect to [itex]t[/itex].
  6. Jun 26, 2014 #5
    i mean how they derive for v'=cv^2/3
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