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Related Rates - Water poured in a bowl

  1. Jul 18, 2007 #1


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    The volume of a cap, of depth h cm, cut from a sphere of radius a cm (a>h) is given by [tex]V = {\textstyle{1 \over 3}}\pi h^2 (3a - h)[/tex]
    . A bowl is in the shape of a cap of depth a/2 cm cut from a spherical shell of radius a cm. Water is being poured into the bowl at a constant rate of [tex]\frac{{\pi a^3 }}{{24}}[/tex]
    . Show that, when the depth of the water in the bowl is x cm, [tex]24\frac{{dx}}{{dt}} = \frac{{a^3 }}{{x(2a - x)}}[/tex]
    . Hence show that, if the bowl is initially empty, it will take 5 seconds to fill up.

    I managed to do the first part of it:
    V = {\textstyle{1 \over 3}}\pi x^2 (3a - x) \\
    = \pi ax^2 - {\textstyle{1 \over 3}}\pi x^3 \\
    \frac{{dV}}{{dt}} = 2\pi ax\frac{{dx}}{{dt}} - \pi x^2 \frac{{dx}}{{dt}} \\
    = \frac{{dx}}{{dt}}(2\pi ax - \pi x^2 ) \\
    \frac{{dx}}{{dt}} = \frac{{\frac{{dV}}{{dt}}}}{{(2\pi ax - \pi x^2 )}} \\
    = \frac{{\pi a^3 }}{{24(2\pi ax - \pi x^2 )}} \\
    = \frac{{a^3 }}{{24(2ax - x^2 )}} \\
    24\frac{{dx}}{{dt}} = \frac{{a^3 }}{{x(2a - x)}} \\

    But with the second part, it says "Hence, show that if the bowl...". What im not getting is how to use the part above to show that it takes 5 seconds?

    I managed to do it like this: (i think it is correct, anyway)

    The maximum volume the bowl can hold is given by

    V = {\textstyle{1 \over 3}}\pi h^2 (3a - h) \\
    = {\textstyle{1 \over 3}}\pi ({\textstyle{a \over 2}})^2 (3a - {\textstyle{a \over 2}}) \\
    = \frac{{\pi a^2 (3a - 0.5a)}}{{12}} \\
    = \frac{{5\pi a^3 }}{{24}} \\

    The bowl is full at time T
    \int_0^T {\frac{{\pi a^3 }}{{24}}} dt = \frac{{5\pi a^3 }}{{24}} \\
    \left[ {\frac{{\pi a^3 }}{{24}}t} \right]_0^T = \frac{{5\pi a^3 }}{{24}} \\
    \frac{{\pi a^3 }}{{24}}T = \frac{{5\pi a^3 }}{{24}} \\
    T = 5 \\

    But i dont know how to do it using the fact that [tex]24\frac{{dx}}{{dt}} = \frac{{a^3 }}{{x(2a - x)}}[/tex].

    Any help greatly appreciated.

    Last edited: Jul 18, 2007
  2. jcsd
  3. Jul 18, 2007 #2
    [tex]24\frac{{dx}}{{dt}} = \frac{{a^3 }}{{x(2a - x)}}[/tex]

    after some simple manipulation, this will become:

    [tex] \frac{24x(2a-x)}{a^3} dx = dt [/tex]

    then integrate over both size...

    [tex] \int_0^{a/2}\frac{24x(2a-x)}{a^3} dx = \int_0^Tdt [/tex]

    solve for T and you will get the desired answer....
  4. Jul 18, 2007 #3


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    Ah ok, thanks for that :smile:
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