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Application of differntiation

  1. Sep 21, 2008 #1
    A hemisphere bowl of radius 12cm is initially full of water. Water runs out of a small hole at the bottom of the bowl at a rate of 48pi cm^3 s^-1. When the depth of the water is x cm , show that the depth is decreasing at a rate of 48/[x(24-x)] cm s^-1
    Also, find the rate at which the depth is decreasing when

    a) The bowl is full.
    b)The depth is 6cm.

    Another question is in this picture


    Thanks in advance! Really urgent :)
  2. jcsd
  3. Sep 21, 2008 #2


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    Hi Ose90! :smile:

    Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:
  4. Sep 21, 2008 #3
    Hi tiny-tim, thanks for replying.
    Glad to tell you that I have solved it.

    However, I will still write down my solution to see if it is correct.

    For the first question, I imagine the right side of the semi-circle displaced vertically, I get the equation (x-0)^2 + (y - r)^2 = r^2
    then, x^2 = r^2 - (y^2 + r^2 -2yr)
    x^2 = r^2 - y^2 -r^2 +2yr
    = 2yr - y^2
    x = [y(2r - y)]^(1/2)

    Hence, when h = x and r = 12 , dV/dx = pi(24x-x^2)

    from the question , dv/dt =48 pi

    dx/dt = dx/dv x dv/dt
    = 48 pi
    pi(24x - x^2)

    For the second question , I use similar triangle to solve it.

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