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Related Rates

  1. Nov 22, 2005 #1
    Sand falls from a conveyor belt at the rate of 10 [itex] \frac{ft^{3}}{min} [/itex] onto a conical pile. The radius of the base of the pile is always equal to half the pile's height. How fast is the height growing when the pile is 5 ft high?

    So [tex] r = \frac{1}{2} h [/tex]. That means when [tex] h = 5 [/tex] , [tex] r = 2.5 [/tex]. We want to find [tex] \frac{dh}{dt} [/tex]. I know the volume of a cone is: [tex] \frac{1}{3}\pi r^{2}h [/tex].
    [tex] \frac{dV}{dt} = \frac{1}{3}\pi r^{2} h \frac{dh}{dt} [/tex]. So would I just do:

    [tex] 10 = \frac{1}{3}\pi (2.5)^{2}(5) \frac{dh}{dt} [/tex] and solve for [tex] \frac{dh}{dt} [/tex]?

    Last edited: Nov 22, 2005
  2. jcsd
  3. Nov 22, 2005 #2


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    Homework Helper

    Your general approach is correct. You must express the volume of the cone as a function of the height of the cone only. So that you have only 2 variables, something like:

    V = f(h), then differentiating you get dV/dt = df(h)/dt

    This is an incorrect differentiation. Because r and h are functions of t, you must use the product rule on the right hand side. It is easier however if you replace r by 0.5h as given in the question so that you only have 1 variable. You are differentiating with respect to time. All variables that change with respect to time must be treated accordingly. For reference, you should be getting

    [tex] \frac{dV}{dt} = \frac{1}{4}\pi h^2 \frac{dh}{dt}[/tex]
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