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Hemispherical Related Rates

  1. Nov 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Water is dripping into a hemispherical bowl with a radius of 8 cm at a rate of 1 cubic cm per minute. At what rate is the depth increasing when it is 4 cm.

    A) 1/96π
    B) 1/48π
    C) 1/24π
    D) 1/16π
    E) 1/8π

    2. Relevant equations

    V = (4πr3)/3
    and any other relevant equations, but that's the only one I know of that will help.

    3. The attempt at a solution

    V = (4πr3)/3
    and....that's about it. B is the correct answer btw.
     
    Last edited: Nov 7, 2009
  2. jcsd
  3. Nov 7, 2009 #2
    dV/dt = 4πr2(dr/dt)
    -1/(4π) = r2(dr/dt)

    when h = 4, then r = 4

    -1/(4π) = 16(dr/dt)
    -1/(64π) = dr/dt

    But...that can't be correct.

    Since dr/dt is decreasing, dh/dt has to be increasing.
     
    Last edited: Nov 7, 2009
  4. Nov 7, 2009 #3
    Anyone?
     
  5. Nov 7, 2009 #4

    lanedance

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    hmmm... haven't checked your working, but as a short cut without driving the full volume of the filled region with depth, for the given depth, work out the radius of the cirlce at teh top at relate the total volume change to the volume change of a circluar prism, hieght dz....
     
  6. Nov 7, 2009 #5

    lanedance

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    so what i think you did here is differentiate the volume of a sphere, which doens;t work for this problem,

    it only applies if you are expanding shperically, (ie effectively addind or removing spherical shells of volume
     
  7. Nov 8, 2009 #6
    Ok, I tried a different approach to this, but still didnt get the right answer.

    pmvj8.jpg
     
  8. Nov 8, 2009 #7
    Sorry for the large image.
     
  9. Nov 8, 2009 #8
    Oh.....

    Ok....

    So if I look at it as being composed of tiny slices, each of Δh height and of ΔV volume I get:

    ΔV = (π)(r2)Δh

    1 = π(16h-h2)Δh
    1 = 48πΔh
    1/48π = Δh

    But, why doesn't the other method (the one in the image) work?
     
  10. Nov 8, 2009 #9
    I guess its because V doesnt equal (2π/3)(r2)h

    V = ∫ πr2 dh and then the derivative of that is just dV/dt = πr2(dh/dt)

    Any ideas?
     
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