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Math Help - Find corresponding percentage increase

  1. #1
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    Find corresponding percentage increase

    If L cm is the length of a pendulum and t s the time of one complete swing, it is known that L = kt^2. If the length of the pendulum is increased by x%, x being small, find the corresponding percentage increase in time of swing.

    My workings:
    L = kt^2
    dL/dt = 2kt
    delta t = dt/dL x delta L
    delta t = 1/2kt x x% = x/2kt%

    The answer given is x/2%.

    Where where where have I gone wrong?!
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  2. #2
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    Hello darkvelvet
    Quote Originally Posted by darkvelvet View Post
    If L cm is the length of a pendulum and t s the time of one complete swing, it is known that L = kt^2. If the length of the pendulum is increased by x%, x being small, find the corresponding percentage increase in time of swing.

    My workings:
    L = kt^2
    dL/dt = 2kt
    delta t = dt/dL x delta L
    delta t = 1/2kt x x% = x/2kt%

    The answer given is x/2%.

    Where where where have I gone wrong?!
    If you've read my reply to your previous post, you'll probably see where you've gone wrong.

    What you mean is:
    \dfrac{\delta L}{L}=x%
    I'm sure you can complete it now.

    Grandad
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  3. #3
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    Alright, this is what I've got:

    L=kt^2
    \dfrac{dt}{dL}\approx\dfrac{\delta t}{\delta L}
    \dfrac{dt}{t} = \dfrac{1/2kt * \delta L}t
    \dfrac{dt}{t} = \dfrac{1/2kt * \delta L}{\sqrt(L/K)}

    Am I right so far?
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  4. #4
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    Hello darkvelvet

    Do it in the same way as the previous question:

    L = kt^2

    \Rightarrow \dfrac{dL}{dt}= 2kt

    \Rightarrow \dfrac{dL}{L}=\dfrac{2kt\,dt}{L}
    =\dfrac{2kt\,dt}{kt^2}
    Can you continue?

    Grandad
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  5. #5
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    Yeaps, I could continue easily with your method. Thanks so much once again!
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  6. #6
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    Hi Grandad,

    Just a quick question: does \frac{\delta L}{L} = x% mean the same as \frac{\delta L}{L} * 100% = x% ?

    Thanks!
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