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Math Help - Rates of Change

  1. #1
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    Exclamation Rates of Change

    Grain is ejected from a chute at the rate of 0.1 m3/min and is forming a heap on a flat horizontal floor. The heap is in the form of a circular cone of semi-vertical angle 45 degrees. Find the rate, in metres per minute, at which the height of the cone is increasing, at the instant 3 minutes after the opening of the chute.

    Setting out:
    dV/dt = 0.1 m3/min
    dh/dt
    = dV/dt * dh/dV
    = 0.1 * dh/dV
    V = 1/3 pi r^2 h
    = 1/3 pi h^3
    dV/dh = pi h^2
    dh/dV = 1/(pi h^2)
    dh/dt = 1/(10 pi h^2)

    BUT I am trying to substitute a value of t...HELP! where did i go wrong???
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  2. #2
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    You are correct that you can't substitute t directly. However, you can use t to find out what V is, then you can use V to find out what h is, then finally you can use that h to plug into your expression.

    Does this make sense?
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  3. #3
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    Ummm no ... how on earth wuld i use t...
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  4. #4
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    Since the volume is increasing at a rate of 0.1 m^3/min and we are looking at 3 minutes after the chute was opened, doesn't it make sense that the volume at that time is 0.3 m^3? I think this makes sense intuitively, but you can also write it in Calculus terms:

    V = \int_0^3 \frac{dV}{dt} \, dt =  \int_0^3 0.1 \, dt = [0.1t]_0^3 = 0.3

    Then, to find the height, use your formula relating volume and height.

    0.3 = \frac{1}{3} \pi h^3 ~\implies~ h=\sqrt[3]{\frac{0.9}{\pi}}
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