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Integrals: #1 Help with fraction #2 Moment of inertia

  1. Mar 8, 2013 #1
    1. The problem statement, all variables and given/known data
    25-2-EX9
    The time rate of change of the displacement (velocity) of a robot arm is ds/dt = 8t/(t^2 + 4)^2. Find the expression for the displacement as a function of time if s = -1 m when t = 0 s.
    26-5-9
    Find the moment of inertia of a plate covering the first-quadrant region bounded by [itex]y^2 = x, x = 9[/itex] and the x-axis with respect to x-axis.

    2. Relevant equations
    25-2-EX9 unknown
    26-5-9 [itex]Ix =k \int_c^d y^2(x_2 - x_1)dy[/itex]

    3. The attempt at a solution
    25-2-EX9
    This is a book example, so the solution is here. I am hung up on one step though.
    [itex]\int ds = \int \frac{8tdt}{(t^2 + 4)^2} = 4 \int (t^2 + 4)^{-2}(2t dt)[/itex]

    This next step throws me off. What happened to [itex](2t dt)[/itex]??

    it just disappears. Can someone explain? Thanks.

    [itex]s = 4(\frac{1}{-1})(t^2 + 4)^{-1} + C[/itex]

    26-5-9
    [itex]Ix =k \int_c^d y^2(x_2 - x_1)dy[/itex]

    [itex]9ky^2dx = 9kx = 9k\frac{1}{2}x^2[/itex]

    [itex]4.5x^2 = 4.5 * 3^2[/itex]

    [itex]4.5*9 = 40.5[/itex]


    However, the answer is [itex]\frac{162}{5}k[/itex]

    so where'd I go wrong? thanks.
     
  2. jcsd
  3. Mar 8, 2013 #2

    eumyang

    User Avatar
    Homework Helper

    U-substitution was used without really showing it. Try letting u = t2+4 and rewrite the integral in terms of u and see what happens.
     
  4. Mar 8, 2013 #3
    which formula would that be?
     
  5. Mar 9, 2013 #4
    any suggestions?
     
  6. Mar 9, 2013 #5
    Problem with derivative

    accidental post.
     
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