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Math Help - Differentiation Application

  1. #1
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    Differentiation Application

    This problem has been proving extremely tough for me. Thanks a lot for the help.

    The period of a pendulum is given T=2(Pi)Squareroot(L/G) where L is the length of the Pendulum in feet, g is the acceleration due to gravity and T is the time in seconds. Suppose that the pedulum has ben subjected ton an increase in temeprature such that the length has increased by 1/2%.
    a) Find the approximate percent change in the period.
    b) Using the result in part (a), find the approximate error in this pendulum clock in one day.
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  2. #2
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    Quote Originally Posted by aussiekid90 View Post
    This problem has been proving extremely tough for me. Thanks a lot for the help.

    The period of a pendulum is given T=2(Pi)Squareroot(L/G) where L is the length of the Pendulum in feet, g is the acceleration due to gravity and T is the time in seconds. Suppose that the pedulum has ben subjected ton an increase in temeprature such that the length has increased by 1/2%.
    a) Find the approximate percent change in the period.
    b) Using the result in part (a), find the approximate error in this pendulum clock in one day.
    Here is one way.

    T = 2pi*sqrt(L/g)
    Differentiate both sides,
    Since g is constant,
    T = [2pi / sqrt(g)]*sqrt(L)
    dT = [2pi / sqrt(g)][(1/2) / sqrt(L)]dL
    dT = [pi / sqrt(gL)]dL -------------------(1)

    When dL = (1/2)percent of L, or 0.005L, the L becomes (L +0.005L) = (1.005L).
    Plug those into (1), to get the change in T,
    dT = [pi / sqrt(g*1.005L)](0.005L)
    dT = [pi / sqrt(g)]*[(0.005L) / sqrt(1.005L)] ------(2)

    So the percent change in T is
    (dT / T) *100percent
    Where dT /T =
    = [pi / sqrt(g)]*[(0.005L) / sqrt(1.005L)] / [2pi*sqrt(L/g)]
    = [(pi)sqrt(g) / sqrt(g)*2pi]*[(0.005L) / sqrt(1.005L)*sqrt(L)]
    = [(1/2)]*[0.005 / sqrt(1.005)]
    = 0.0024938
    or,
    = 0.0025

    And that is 1/4)percent.
    Therefore, when the L was increased by (1/2) percent, the period T was increased by (1/4) percent ------------answer.

    ---------------------------------------------------------
    b) Using the result in part (a), find the approximate error in this pendulum clock in one day.

    T is in seconds.

    1 day = 24hrs*(60min/hr)(60sec/min) = 24*3600 = 86,400 sec
    (0.0025)(86,400) = 216 sec = 3min and 36sec.

    Therefore, in one day the pendulum clock is faster by 3 minutes and 36 seconds than the correct time. ----------answer.
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