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Depth of a cone rate problem (question about the equation I'm using)

  1. Mar 31, 2008 #1
    1. The problem statement, all variables and given/known data

    A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

    2. Relevant equations

    Cone: V= (Ah/3)

    Right Circular Cone: V=[(pi)(r^2)(h)]/(3)

    3. The attempt at a solution

    So on this on I'm confused about the equation I need to derive to begin with..

    I looked up the volume of a cone and it had two choices (stated above)

    So I think I need to use just the cone one because the problem doesn't say its a right circular cone. Is that correct?

    Assuming I do in fact need to use V= (Ah/3)

    Do I then plug in A=(pi)(r^2) which is the area of a circle? the book says A equalls the area of the base, which appears to be a circle to me...
     
    Last edited: Mar 31, 2008
  2. jcsd
  3. Mar 31, 2008 #2
    ok the volume of the cone is

    [tex]V=\frac{1}{3}r^{2}\pi H[/tex]

    The info that you know are:

    [tex]\frac{dV}{dt}=10[/tex], [tex] r=5[/tex]

    We need to find [tex]\frac{dH}{dt}=?[/tex] when H=8(be carefule here, you are not to plug this value for H in, immediately on the equation for the volume, since H (depth) is changing). Now try to find a relation between the radius of the circle on the cone when H=8, using these information.
    Draw a picture it usually helps.
     
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