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Thread: rates of change

  1. #1
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    rates of change

    With length, $\displaystyle l $, in meters, the period $\displaystyle T$, in seconds, of a pendulum is given by

    $\displaystyle T = 2 \pi \sqrt{ \frac{l}{9.8} }. $

    (a) How fast does the period increase as $\displaystyle l$ increases?
    (b) Does this rate of change increase or decrease as $\displaystyle l$ increases?



    I think I am correct for this one, just check please.

    (a)
    $\displaystyle \frac{dT}{dl} = \pi ( \frac{l}{9.8} )^{- \frac{1}{2} } * ( \frac {1}{9.8} ) $
    $\displaystyle \frac{dT}{dl} = ( \frac{ \pi \sqrt{9.8} }{9.8 \sqrt{l} } ) \frac{m}{s} $

    (b)
    $\displaystyle \frac{d^2T}{dl^2} = \frac{- \pi \sqrt{9.8} }{19.6l \sqrt{l}} $ <-- negative.
    Decreases
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  2. #2
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    Quote Originally Posted by Jacobpm64 View Post
    With length, $\displaystyle l $, in meters, the period $\displaystyle T$, in seconds, of a pendulum is given by

    $\displaystyle T = 2 \pi \sqrt{ \frac{l}{9.8} }. $

    (a) How fast does the period increase as $\displaystyle l$ increases?
    (b) Does this rate of change increase or decrease as $\displaystyle l$ increases?



    I think I am correct for this one, just check please.

    (a)
    $\displaystyle \frac{dT}{dl} = \pi ( \frac{l}{9.8} )^{- \frac{1}{2} } * ( \frac {1}{9.8} ) $
    $\displaystyle \frac{dT}{dl} = ( \frac{ \pi \sqrt{9.8} }{9.8 \sqrt{l} } ) \frac{m}{s} $

    (b)
    $\displaystyle \frac{d^2T}{dl^2} = \frac{- \pi \sqrt{9.8} }{19.6l \sqrt{l}} $ <-- negative.
    Decreases

    Yup. Ya got this one, too.

    -Dan
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