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Apostol Related Rates Problem

  1. Dec 18, 2011 #1
    1. The problem statement, all variables and given/known data

    Apostol, Vol 1: Section 4.12 Problem 26

    Water flows into a hemispherical tank of radius 10 feet (flat side up). At any instant, let h denote the depth of the water, measured from the bottom, r the radius of the surface of the water, and V the volume of the water in the tank. Compute dV/dh at the instant when h=5 feet. If the water flows in at a constant rate of 5√3 cubic feet per second, compute dr/dt, the rate at which r is changing, at the instant t when h=5 feet.


    2. Relevant equations

    Hemispherical Volume: [2π(r^3)]/3

    3. The attempt at a solution

    I tried to rewrite the volume in terms of h. If we consider the circle equation, its possible to say that the radius of the water volume is: r' = √[100 - (h^2)], where 100 is the radius of the tank squared and h is the height. So, we can rearrange the volume formula to: V = {2π{[100 - (h^2)]^3/2}}/3, if we derive this with respect to h, we can find: dV/dh = -2πh√[100 - (h^2)], if we change h for 5, it gives that dV/dh = -10π√75. However, according to the answers at the end, the correct answer would be 75. Sorry if I made any mistakes. I would really appreciate any tip, correction or solution. Thank you.
     
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  3. Dec 18, 2011 #2

    SammyS

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    ..."If we consider the circle equation, its possible to say that the radius of the water volume is: r' = √[100 - (h^2)], where 100 is the radius of the tank squared and h is the height."​
    This is not a hemisphere when r' < r .
     
  4. Dec 18, 2011 #3
    So I would have to consider the spherical cap volume formula instead of hemispherical one?
     
  5. Dec 18, 2011 #4

    SammyS

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    Yes.
     
  6. Dec 18, 2011 #5
    I was able to achieve the correct result in first part. Thank you very much. Also, I made a little mistake, and r and r' are the same thing. Although I got the right answer using spherical cap formulas.
    However, when I got into the second part, I've got another problems. It asks for dr/dt. It says that dV/dt equals 5√3. I take dV/dt and rewrite it as (dV/dr)*(dr/dt). V = (π/6)*h*(3(r^2) + h^2). So I try to rewrite it in terms of 'r', using the formula: 10 = (r^2 + h^2)/2h, where 10 is the radius of the tank. I solve this equation for 'h', using quadratic formula e substitute 'h' in the volume formula for 'r'. The derivative was very long and even solving it, the result was not correct.
     
  7. Dec 19, 2011 #6
    SammyS:
    After some searching, looking for spherical cap's formulas, I was able to derive the correct answer. Thank you very much again.
     
  8. Dec 19, 2011 #7

    SammyS

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    You're welcome.

    I'm glad you found the answer.
     
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