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Math Help - Rates!!

  1. #1
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    Rates!!

    The period T, in seconds, of a pendulum is given by the equation below with length, l, in meters.

    T = 2pi*sqrt(l/9.8)

    (a) How fast does the period increase?

    need some help please, these rate problems give me trouble...
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  2. #2
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    Quote Originally Posted by mathlete View Post
    The period T, in seconds, of a pendulum is given by the equation below with length, l, in meters.

    T = 2pi*sqrt(l/9.8)

    (a) How fast does the period increase?

    need some help please, these rate problems give me trouble...
    With respect to what?

    T(l)=2 \pi \sqrt{\frac{l}{9.8}}

    If you double the length 2l then


    T(2l)=2 \pi \sqrt{\frac{2l}{9.8}} =2 \pi \sqrt{2 \cdot \frac{l}{9.8}}=2 \pi \sqrt{2} \cdot \sqrt{\frac{l}{9.8}}

    T(2l)=\sqrt{2} \cdot \left( 2 \pi \sqrt{\frac{l}{9.8}} \right)=\sqrt{2} \cdot T

    So doubling the length increases the period by \sqrt{2} \approx 1.414
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  3. #3
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    no sorry, it wants us to find dT/dl, so we're supposed to differentiate with respect to l, sorry...
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by mathlete View Post
    The period T, in seconds, of a pendulum is given by the equation below with length, l, in meters.

    T = 2pi*sqrt(l/9.8)

    (a) How fast does the period increase?

    need some help please, these rate problems give me trouble...
    T=2\pi\sqrt{\frac{l}{9.8}}=\frac{2\pi}{\sqrt{9.8}}  \cdot \sqrt{l}= \frac{2 \pi}{\sqrt{9.8}} \cdot l^{1/2}

    Now taking the derivative we get

    \frac{dT}{dl}=\frac{2\pi}{\sqrt{9.8}}\cdot \left( \frac{1}{2}l^{-1/2}\right)=\frac{\pi}{\sqrt{9.8l}}
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