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Math Help - Quick question. related rates

  1. #1
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    Quick question. related rates

    I am doing the chain rule in calculus, and this question has got me stuck for a long time now. I am not sure how to approach or solve this, help would be appreciated


    The amount of cleaning fluid in a partially filled, spherical tank is given by V = (pi)R(h^2)-pi(h^3)
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    3

    Where R is the radius of the rank, and h is the depth of the center of the tank. If cleaning fluid is being pumped into the tank, of radius 12m, at a rare of 8 m^3/min, how fast is the fluid rising when the tank is 3/4 full?
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  2. #2
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    Hello, Sky12!

    This problem doesn't involve the Chain Rule.
    Exactly where is your difficulty?


    The amount of fluid in a partially-filled spherical tank
    . . is given by: .  V \:= \:\pi Rh^2 -\frac{\pi}{3}h^3
    where R is the radius of the tank, and h is the depth at the center of the tank.

    If fluid is pumped into the tank at a rate of 8 m/min, and R = 12 m,
    how fast is the fluid rising when the tank is 3/4 full?

    We have: .  V \:= \:\pi Rh^2 -\frac{\pi}{3}h^3

    Differentiate with respect to time: . \frac{dV}{dt}\:=\:24\pi h\!\left(\frac{dh}{dt}\right) - \pi h^2\!\left(\frac{dh}{dt}\right)

    Then: . \frac{dV}{dt}\:=\:\pi h(24-h)\frac{dh}{dt}\quad\Rightarrow\quad\frac{dh}{dt} \:=\:\frac{\frac{dV}{dt}}{\pi h(24 - h)}

    . . We are given \frac{dV}{dt} = 8 m/min . . . When the tank is \frac{3}{4} full:  h \,\approx\,16 m. **

    Therefore: . \frac{dh}{dt} \:=\:\frac{8}{\pi\!\cdot\!16(24-16)}\:=\:\frac{1}{16\pi} m/min

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    ** . No, I didn't make an error . . .

    If the tank is three-fourth full, then: . V \:=\:\pi\!\cdot\!12h^2 - \frac{\pi}{3}h^3\:=\:\frac{3}{4}\left(\frac{4}{3}\  pi\!\cdot\!12^3\right)
    . . and we have a cubic equation to solve: . h^3 - 36h^2 + 5184 \:=\:0

    which has a root near h = 16.17

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  3. #3
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    great help, thanks!
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