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Rate of change

  1. May 9, 2004 #1
    Sorry to bore you all with this rate of change question, I really hate this area of maths and struggle to get my hea round it. Here it is, I'd be greatful for any help you have to offer:

    Fluid flows out of a cylindrical tank with constant corss section. At the time t minutes (t is greater of equal to 0), the volume of fluid remaining in the tank is V m^3. The rate at which the fluid flows, in m^3min^-1, is proportional to the square root of V. Show that the depth h metre's of fluid in the tank satisfies the differential equation:

    dh/dt = -kh^(1/2) where k is a +ve constant


  2. jcsd
  3. May 9, 2004 #2


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    dV/dt=-aV^(1/2), by assumption.

    V(t)=pi*R^(2)*h(t)=b*h(t) (where R is the radius)
    Rearranging a bit, and introducing k yields the desired result
  4. May 9, 2004 #3
    thanks v much, great help

  5. May 9, 2004 #4


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    In other words, the whole point of "rate of change problems" is to use a static equation (typically a geometry formula or something like that) that has no time variable, t, in it and then use the chain rule to differentiate both sides of the equation- giving a formula connecting the rates of change of the quantities.
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