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Math Help - rate of change

  1. #1
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    rate of change

    The base of a cone-shaped tank is a circle of radius 5 feet, and the vertex of the cone is 12 below the base. The tank is filled to a depth of 7 feet, and water is flowing out of the tank at a rate of 3 cubic feet per minute.

    Find the rate of change of the depth of the water in the tank.
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  2. #2
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    Mr_green... do u go to scn high?

    This one has got me stumped too =[
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mr_Green View Post
    The base of a cone-shaped tank is a circle of radius 5 feet, and the vertex of the cone is 12 below the base. The tank is filled to a depth of 7 feet, and water is flowing out of the tank at a rate of 3 cubic feet per minute.

    Find the rate of change of the depth of the water in the tank.
    ...at the instant the height is 7 feet?...
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  4. #4
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    h= 7

    dV/dt= -3

    r= (5h)/12

    V= (1/3)*pi*(r)^2 * h = (pi*25h^3)/(432)
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by DINOCALC09 View Post
    h= 7

    dV/dt= -3

    r= (5h)/12

    V= (1/3)*pi*(r)^2 * h = (pi*25h^3)/(432)
    correct so far, now differentiate implicitly with respect to t
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  6. #6
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    dont understand..
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  7. #7
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  8. #8
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    there are a few final steps that it fails to explain.
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  9. #9
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    is the answer -0.11225 ft/sec
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  10. #10
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    Refer back. It's been updated.
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