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Related Rates - Volume and Height

  1. Jan 7, 2008 #1
    1. The problem statement, all variables and given/known data
    "A man pours water into a conical glass (with radius 9cm and height 6cm) at a rate of 8cm^3.

    At what rate is the height of the water changing per second when the height of the glass is 2/3rd's full of water?

    NOTE: The answer is supposedly 2/9(pi) cm/sec. I am supposed to show that this is right.

    2. Relevant equations

    V= 1/3*(pi)*r^2*h

    h = (3V)/(2*(pi)^2)

    3. The attempt at a solution

    I tried to differentiate the h equation that I found above. Then, after differentiating, I subbed in the radius (6) and the Volume (48). Note that I found the radius to be 6, and volume to be 48 when the height is 2/3rds (and thus, 4). However, after I diff and then plug, I always come up with a negative answer.

    Thanks for any help.
  2. jcsd
  3. Jan 7, 2008 #2


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    I really don't see how you came up with the second equation, unless it's the result of making mistakes while trying to solve the first one for [itex]h[/itex]. (It should be clear to you that the volume should be proportional to the cube of the height -- and not linear with it.)

    In order to solve this problem, you will need to use a relationship between the radius of the cone and the height of the cone.
  4. Jan 7, 2008 #3
    Sorry, that second equation should be:


    What two equations dealing with the radius and height should I be comparing?
  5. Jan 7, 2008 #4
    V is proportional to h*r^2 and r and h are proportional to each other (since it is a cone), so essentially V is proportional to h^3.
  6. Jan 7, 2008 #5
    Use similar triangles to get rid of radius.

    [tex]\frac r h = \frac 9 6[/tex]

    Solve for radius and plug it back into your original equation.
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