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Math Help - Related Rate question regarding application of the deriviative..

  1. #1
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    Related Rate question regarding application of the deriviative..

    I have tried this problem, but it's not working out.

    Sand poured on the ground at a rate of 3 meters cubed per minute forms a conical pile whose height is 1/3 the diameter of the base. How fast is the altitude increasing when the radius of the base is 2 m.

    First I replace r in the conical V=1/3 PIE r^2h
    I take h = 1/3 d
    h=1/3(2r)
    So I get r = 3h/2
    So then I put it in V=1/3 PIE (3h/2)^2h
    When I do the derivative I get
    dV/dt=3PIEh^2 dh/dt
    I plug in the numbers
    3=3PIE(2)^2 dh/dt
    and get .0795 m/min

    But the answer is .239m/min
    What am I doing wrong in this problem, if someone can help thanks.
    Joanne
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  2. #2
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    You have to apply the chain rule when differentiating.

    V(t)=\frac{9}{12}\pi h^3

    \frac{dV}{dt}=\frac{9}{12}\pi(3h^2)\frac{dh}{dt}

    Your h should be  h=\frac{1}{3}(2r)=\frac{4}{3} when r=2
    So just plug in your values. and solve for \frac{dh}{dt}
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  3. #3
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    I am a bit lost here. Why is it 9/12 PIE h^3?? I have tried to figure it out and it's not working.
    grrrrrrrrrr on not getting it
    Jo

    UPDATE here is what I got and finally got the answer at plugging at it for a while.
    h=1/3(2r)
    From this I got h=4/3 and r =3h/2
    Then V=1/3Pi(3h/2)^2h
    V=1/3 Pi (9h/4)^3
    dV/dt=1/3 Pi(27/4)h^2 dh/dt
    3=1/3 Pi(27/4)(4/3)^3
    dh/dt = 0.239 m/min

    I was shocked myself when I got it.
    I could not figure yours out above, the way you did it.
    Last edited by bradycat; February 12th 2011 at 10:06 AM.
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  4. #4
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    The volume of a cone is:

    V=\frac{1}{3}\pi r^2h

    and you're given the relationship between the height and diameter.

    h=\frac{1}{3}(2r)=\frac{2}{3}r

    so

    r=\frac{3}{2}h

    The formula for volume becomes:

    V=\frac{1}{3}\pi\left(\frac{9}{4}h^2\right)h=\frac  {9}{12}\pi h^3

    V=\frac{9}{12}\pi h^3

    \frac{dV}{dt}=\frac{9}{12}\pi (3h^2)\frac{dh}{dt}=\frac{27}{12}\pi h^2\frac{dh}{dt}

    So you want to find \frac{dh}{dt} where you're given \frac{dV}{dt}=3 and when r=2 you know that h \frac{4}{3}.

    3=\frac{27}{12}\pi\left(\frac{16}{9}\right)\frac{d  h}{dt}=\frac{432}{108}\pi\frac{dh}{dt}

    I'm calculating \frac{dh}{dt}=\frac{324}{432\pi}\approx .239
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