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Related rates and rain drops

  1. Jan 6, 2005 #1
    if a raindrop is a perfect sphere,and it accumulates moisture at a rate proportional to its surface area. Show that the radius increases at a constant rate.

    I know surface area of sphere = 4 * pi * r^2

    dV/ dr = 4 * pi * r^2

    I am not sure where to go from here

    Any help is appreciated

    Thanks :smile:
     
  2. jcsd
  3. Jan 6, 2005 #2

    quasar987

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    First thing to do is realise that the amount of moisture is nothing but the volume of the waterdrop. The rate at which it accumulate moisture is then dV/dt. We can calculate this derivative. You have done that but made a little mistake by forgetting the chain rule! :wink:

    [tex]\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=A\frac{dr}{dt}[/tex].

    Next, we are given the fact that it accumulates moisture at a rate proportional to its surface area. The keyword in this sentence is "proportionnal" because if means that dV/dt is of the form

    [tex]\frac{dV}{dt}=(a \ constant)A[/tex]

    Compare the two equations of dV/dt. What do you conclude?
     
  4. Jan 6, 2005 #3

    dextercioby

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    You didn't hat the idea of "rate".It usually implies 'variance in time'.The problem gives that the rate the volume is increasing is proportional to its area.
    Therefore
    [tex] \frac{dV(r(t))}{dt}=kS [/tex]
    ,where 'k' is a constant of proportionality.
    [tex] \frac{dV}{dr}\frac{dr}{dt}=kS [/tex]
    But for an infinitesimal increase of the radius 'dr',the volume increases with
    [tex] dV=Sdr [/tex]
    Therefore
    [tex] S\frac{dr}{dt}=kS [/tex]
    The surface is nonzero.U get:
    [tex] \frac{dr}{dt}=k [/tex]
    ,which,by integration leads to the desired result.

    Daniel.
     
    Last edited: Jan 6, 2005
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