1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?
     
  2. jcsd
  3. Aug 29, 2005 #2

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Related rates problem
Loading...