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Related rates problem

  1. Aug 29, 2005 #1
    This is the problem as it appears in the text. "As a spherical raindrop evaporates, its volume changes at a rate proportional to its surface area A. If the constant of proportionality is 3, find the rate of change of the radius r when r=2." My first question is does the constant of proportionality refer to [tex]\frac{dV}{dA}[/tex]? Secondly, am I being asked to find the rate of change of the radius with respect to time [tex]\frac{dr}{dt}[/tex] or another rate?
     
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  3. Aug 29, 2005 #2

    quasar987

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    Hi, I don't think the problem makes sense. But I will try to answer your questions.

    No, if you are told that A is proportional to B, it means there is a constant k such that A = kB. Here, it means that dV/dt = kA, for a certain constant k. Next, you are told that k = 3. But this is nonsense in my opinion because the raindrop is evaporating... and a positive dV/dt means V is augmenting. So I'd try doing the problem with k = -3 instead; it'd make more sense.

    Yes, the rate of change of radius wrt time is dr/dt.
     
  4. Aug 29, 2005 #3
    thanks quasar, with that information ill try the problem.
    [tex]\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}[/tex]

    Since [tex]\frac{dV}{dt}=-3A[/tex]

    Then [tex]-3(4{\pi}r^2)=\frac{dV}{dr}\frac{dr}{dt}[/tex]

    But now it seems if i differentiate the volume with respect to the radius ill get the equation for surface area and it will cancel out and leave
    [tex]\frac{dr}{dt}=-3[/tex]
    And that cant be right, where have i erred?
     
  5. Aug 30, 2005 #4

    quasar987

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    Why can't it be right? I'd say it can't be wrong. :smile:
     
  6. Aug 31, 2005 #5
    well then, thanks again quasar. so was it irrelevant to ask to find the rate of change when the radius was 2, meaning its a constant rate of change when the volume and surface area decrease proportionally?
     
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