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Related rates prob.

  1. Oct 19, 2008 #1
    Related rates prob. [solved]

    A hemispherical bowl of radius 8 in. is being filled with water at a constant rate. If the water level is rising at the rate of 1/3 in./s at the instant when the water is 6 in. deep, find how fast the water is flowing in by using the fact that if V is the volume of the water at time t, then dV/dt = (pi)(r)^2(dh/dt)

    I just need someone to check my work. From my picture, I have:

    dV/dt = (pi)(64-(8-h)^2) dh/dt.

    just plug in 6 for h and 1/3 for dh/dt and solve right? Thanks.
     
    Last edited: Oct 19, 2008
  2. jcsd
  3. Oct 19, 2008 #2

    Mark44

    Staff: Mentor

    The first formula you show looks fine, but can you tell us how you arrived at your second formula, namely dV/dt = pi (64 - (8 - h)^2) dh/dt?
    You mentioned that it comes from your picture, but since I can't see your picture, it's not clear to me how h is related to the radius of the disk you are using for your incremental volume. I'm not saying your formula is wrong, but you haven't provided enough information so that I can tell.
     
  4. Oct 19, 2008 #3
    [​IMG] [​IMG]

    I suck at drawing on paint but this is kind of how my pic looks. y + h is radius of the bowl which is 8. 8 - h = y and then i just used c^2 - a^2 = b^2. And thanks for your help.
     
  5. Oct 19, 2008 #4

    Mark44

    Staff: Mentor


    OK, thanks for including the drawing. Now I know how h and y are related. I don't see anything wrong in your second formula.
     
  6. Oct 19, 2008 #5
    Thanks again.
     
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