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Related Rates Involving a Cone

  1. Dec 9, 2009 #1
    1. The problem statement, all variables and given/known data

    Water is being drained from a container which has the shape of an inverted right circular cone. The container has a radius of 5.00 inches at the top and a height of 7.00 inches. At the instant when the water in the container is 4.00 inches deep, the surface level is falling at a rate of 0.7 in/sec. Find the rate at which water is being drained from the container.

    2. Relevant equations

    v=1/3 pi r^2 h

    3. The attempt at a solution

    The values I came up with are as follows.

    h=4 (of water level)
    r=2.86 (of water level found using equal triangles)
    dV/dt= what I am solving for
    dh/dt = -0.7

    Whenever I take the derivative of the volume equation I end up with a dr/dt that I have no idea what to do with. Am I just doing it wrong and getting the dr/dt when I shouldn't?

    I also noticed in a few google results they'd used similar triangles to get r in terms of h. But whenever I do that, I get an actual value for r and the h goes away. Perhaps this is where I am messing up?
    Last edited: Dec 9, 2009
  2. jcsd
  3. Dec 9, 2009 #2
    So I tried it without the r variable using r=5h/7 instead. That seemed to work because I got an answer that seems more reasonable, 17.955.

    Is there any reason to do that? I am guessing because in this case you can only deal with two variables because of that dr/dt you end up with??
  4. Dec 10, 2009 #3


    User Avatar
    Science Advisor

    Yes, you can do that. The water takes the shape of the cone so no matter what the height of the water in the cone, you will always have r= (5/7) h. Draw a picture and think about "similar triangles".
  5. Dec 10, 2009 #4
    So is the main reason why you use r= (5/7)h is to get rid of the r in the equation? Because I have the actual value for h. But if I plug it in, I end up with that dr/dh again that I don't know what to do wtih.
  6. Dec 10, 2009 #5


    Staff: Mentor

    Yes, you want to get rid of one variable so that you can write V as a function of a single variable, and then get dV/dh and dh/dt. After you have found those, then you can substitute the known values at the particular time.
  7. Dec 10, 2009 #6
    Sweet. Thank you both. It is much appreciated.
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