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Math Help - Rate of Change: Depth of Water in a Tank

  1. #1
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    Rate of Change: Depth of Water in a Tank

    The base of a pyramid-shaped tank is a square with sides
    of length 12 feet, and the vertex of the pyramid is 10 feet
    above the base. The tank is filled to a depth of 4 feet, and
    water is flowing into the tank at the rate of 2 cubic feet
    per minute. Find the rate of change of the depth of water
    in the tank.



    I don't even know how to approach it.
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  2. #2
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    Help?
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  3. #3
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    Volume is changing at a constant rate: "water is flowing into the tank at the rate of 2 cubic feet per minute" so dV/dt= 2.

    Write the volume as a function of height of the water: if V= f(h) the dV/dt= f'(h) dh/dt= 2 and you can solve for dh/dt.

    To find the volume as a function of the height, volume of a pyramid of height h with base area A is \frac{1}{3}Ah. Here, of course, the base will be a square with side length depending on height. You can get the length of an edge of that square as a function of height by remembering that the pyramid of water will be similar to the entire pyramid. It has edge length 12 and height 10 so x/h= 12/10 or x= (6/5)h. The area of the base is [/tex]A= (36/25)h^2[/tex] so the volume is V= \frac{12}{25}h^3.
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    Quote Originally Posted by HallsofIvy View Post
    Volume is changing at a constant rate: "water is flowing into the tank at the rate of 2 cubic feet per minute" so dV/dt= 2.

    Write the volume as a function of height of the water: if V= f(h) the dV/dt= f'(h) dh/dt= 2 and you can solve for dh/dt.

    To find the volume as a function of the height, volume of a pyramid of height h with base area A is \frac{1}{3}Ah. Here, of course, the base will be a square with side length depending on height. You can get the length of an edge of that square as a function of height by remembering that the pyramid of water will be similar to the entire pyramid. It has edge length 12 and height 10 so x/h= 12/10 or x= (6/5)h. The area of the base is [/tex]A= (36/25)h^2[/tex] so the volume is V= \frac{12}{25}h^3.
    Is the pyramid standing on its base or is it inverted?
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