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Math Help - Water draining out of spherical bowl

  1. #1
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    Water draining out of spherical bowl

    Hi,

    Its been 3 years since I've posted here. And I find myself trying to review the basics of calc before moving on to bigger things.

    I found this problem in my old math book:

    Water draining out of spherical bowl-problem.jpg

    Sorry I couldn't rotate it I'm in a rush I tried to look online for some online image utility... oh well.

    So, water is flowing out of a half hemisphere blown (the bottom half, lol) at a constant rate of 6 m^3/min. Radius of sphere is R (13 meters), radius of water level is r. height of water is y Volume is V. formula of Volume at height y is:

    V = (pi/3)y^2(3R-y)

    Find an equation for r and find dr/dy at y=8 meters

    I assume, that r=R*sin[(pi/2)(y/R)], baseless assumption I admit, so that dr/dy=cos[(pi/2)(y/r)] using chain rule: f(x)=Rsin[g(x)] g(x)=y/R

    But that doesn't include the volume in any way. If correct it gives me a relationship between r and y (radius of water level vs depth)

    I'm also curious if I can derive an actual formula for r with what is given.

    This isn't homework I'm an adult returning to higher education and I'm doing a crash course math review on my own before some harder math classes the next semesters coming up. My algebra needs work.
    Last edited by AsSeenOnTV; August 28th 2013 at 10:37 AM.
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Water draining out of spherical bowl

    I don't see how you get r=R \sin(\frac {\pi} 2 \frac y R) - how did you get that? From Pythagoras its apparent that R^2 = r^2+ (R-y)^2, so  r = \sqrt {(R^2 - (R-y)^2}.

    Question C asks for the rate of change of r, or dr/dt, not dr/dy. You can get that using the chain rule:

     \frac {dr}{dt} = \frac {dr}{dy} \ \frac {dy} {dV} \ \frac {dV}{dt}
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  3. #3
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    Re: Water draining out of spherical bowl

    Wow that looks obvious. Thank you.
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