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Another Related Rates Problem

  1. May 6, 2005 #1

    [tex]V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (\frac{4}{5}h)^{2}h[/tex]

    [tex]\frac{dV}{dt}=\frac{16}{25}\pi h^{2}\frac{dh}{dt}[/tex]

    Now I replace [tex]\frac{dV}{dt}[/tex] with 0.1-0.001[tex]h^2[/tex]. This is where I am stuck. Any suggestions?

  2. jcsd
  3. May 6, 2005 #2
    This just gives a separable diff equ....

    It can be solved but why solve it ? :

    If the water reaches a highest point this means that h(t) has a maximum....

    dh/dt=0->h=10m....which curiously is exactly the height of the cone

    Is it a maximum ? It turns out the second derivative is zero, the third derivative is zero...aso..

    So does it overflow ?
    Last edited: May 6, 2005
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