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Thread: Determining Rates

  1. #1
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    Determining Rates

    A conical reservoir has a depth of 24 feet and a circular top of radius 12 feet. It is being filled so that the depth of water i increasing at a constant rate of 4 feet per hour. Determine the rate in cubic feet per hour at which water is entering the reservoir when the depth is 5 feet.

    I know that A= 1/3(pi)hr^2 but i havent done rates for a while and I don't remember how to do this. Its calc ab
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  2. #2
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    $\displaystyle V = \frac{\pi}{3}r^2 h
    $

    $\displaystyle \frac{r}{h} = \frac{12}{24}$ ... $\displaystyle r = \frac{h}{2}$

    sub in h/2 for r, and get the volume formula strictly in terms of h ...

    $\displaystyle V = \frac{\pi}{3}\left(\frac{h}{2}\right)^2 h$

    $\displaystyle V = \frac{\pi}{12}h^3$

    now take the time derivative and determine $\displaystyle \frac{dV}{dt}$
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  3. #3
    Member OnMyWayToBeAMathProffesor's Avatar
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    so then $\displaystyle \frac{dv}{dt}=\frac{3\pi}{12}h^2=\frac{\pi}{4}h^2$. I get that, I would think you would just plug in 5 to get $\displaystyle \frac{dv}{dt}$ but then where does the 4 feet per hour come in? Thanks for everyones help.
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  4. #4
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    Quote Originally Posted by OnMyWayToBeAMathProffesor View Post
    so then $\displaystyle \frac{dv}{dt}=\frac{3\pi}{12}h^2=\frac{\pi}{4}h^2$. I get that, I would think you would just plug in 5 to get $\displaystyle \frac{dv}{dt}$ but then where does the 4 feet per hour come in? Thanks for everyones help.
    chain rule ...

    $\displaystyle \frac{dV}{dt} = \frac{\pi}{4} h^2 \cdot \frac{dh}{dt}$
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  5. #5
    Member OnMyWayToBeAMathProffesor's Avatar
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    aaaa, i see. Sorry, I totally forgot about that. But is my $\displaystyle \frac{dv}{dt}$ formula correct? My peers got different derivative equations. If it is, this would be the subsequent work.

    $\displaystyle \frac{dV}{dt} = \frac{\pi}{4} h^2 \cdot \frac{dh}{dt}$

    $\displaystyle \frac{dV}{dt}=\frac{\pi}{4} (5)^2 \cdot (4)$

    $\displaystyle \frac{dV}{dt}=25\pi$

    correct?
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  6. #6
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    units for $\displaystyle \frac{dV}{dt}$ ?
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  7. #7
    Member OnMyWayToBeAMathProffesor's Avatar
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    Thanks for catching that, so $\displaystyle \frac{dV}{dt}=25\pi $ feet per hour

    correct?
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