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Applied Diff. Eq. Problems

  1. Jan 26, 2008 #1

    I was given several applied differential equation problems (which are ungraded), and I have trouble solving 4 of them. This time I'm going to ask about 1 first though (and after that I'll try solving the other 3 again on my own). Help (even hint) is very much appreciated. Here it is:

    What I've come up with so far is just:

    [tex]V(t) = k_{1} A(t)^{3/2}[/tex]

    where V is the volume and A is the surface area of the raindrop, and

    [tex]\frac{dV}{dt} = -k_{2} A(t)[/tex]

    Please help.. thanks!
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Jan 26, 2008 #2


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    In our first equation, surely you meant
    [tex]\frac{dV}{dt}= -k_1A(t)^{3/2}[/tex]

    Since, for any sphere, the volume is given by [itex]V= (4/3)\pi r^3[/itex] and the surface area by [itex]A= 4\pi r^2[/itex], [itex]A^{3/2}= (4 \pi)^{3/2} r^3[/itex] and so "proportional to surface area to the 3/2 power" is the same as "proportional to the volume" which is, at least, "reasonable" since many things decay that way: for example, radioactive substance decay that way.

    Volume decreasing in proportion to surface area is also reasonable (perhaps more reasonable) since raindrops decrease by water evaporating into the air which can only happen at the surface.

    You have
    [tex]\frac{dV(t)}{dt}= -k_1 A(t)^{3/2}[/tex]
    [tex]\frac{d(t)}{dt}= -k_2 A(t)[/itex]
    Now replace V(t) by [itex](4/3)\pi r(t)^3[/itex] and A by [itex]4\pi r(t)^2[/itex] in each of those so you have equations for r as a function of t.
  4. Jan 26, 2008 #3
    @ HallsofIvy I think that the 1st equation pociteh wrote is correct since

    or am I missing something? :smile:

    Thus he can substitude [itex]V(t)[/itex] from the 1st equation to the 2nd and solve the resulting ODE for [itex]A(t)[/itex]
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