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Math Help - Laplace transform

  1. #1
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    Laplace transform

    show that

    cost*(1/a*sinat) = sint*cosat
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by action259
    show that

    cost*(1/a*sinat) = sint*cosat
    I don't understand your notation.

    RonL
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by action259
    show that

    cost*(1/a*sinat) = sint*cosat
    Is this
    cos(t) \left ( \frac{1}{a}sin(at) \right ) = sin(t) \cdot cos(at)?

    or
    cos \left ( t \frac{1}{a}sin(at) \right ) = sin(t) \cdot cos(at)?

    or
    cos \left ( t \frac{1}{a}sin(at) \right ) = sin \left ( t \cdot cos(at) \right )?

    In any of the above cases, I have doubts about the statement's truth.

    -Dan
    Last edited by topsquark; August 27th 2006 at 01:12 PM. Reason: Addendum
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  4. #4
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    sorry about the confusion it should be

    <br />
cos(t) \left ( \frac{1}{a}sin(at) \right ) = sin(t) \cdot cos(at)<br />
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by action259
    sorry about the confusion it should be

    <br />
cos(t) \left ( \frac{1}{a}sin(at) \right ) = sin(t) \cdot cos(at)<br />
    When I meddle with this expression a bit I get that
    tan(at) = a \cdot tan(t)
    which simply isn't true.

    Is there a typo here, or is there possibly some restriction on the problem? (Or are we possibly solving for a?)

    -Dan
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  6. #6
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    Using a table of LaPlace transforms.

    cos(t)=\frac{p}{p^{2}+1}

    \frac{sin(at)}{a}=\frac{a}{ap^{2}+a^{3}}

    Multiply:

    \left(\frac{p}{p^{2}+1}\right)\left(\frac{a}{ap^{2  }+a^{3}}\right)

    = \frac{p}{(a^{2}+p^{2})(p^{2}+1)}.....[1]

    Now, sin(t)=\frac{1}{p^{2}+1} and cos(at)=\frac{p}{p^{2}+a^{2}} and the result follows from [1].
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  7. #7
    Forum Admin topsquark's Avatar
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    Hmmm...I appear to be missing something in the notation. Ah well.


    -Dan
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  8. #8
    Grand Panjandrum
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    Quote Originally Posted by topsquark
    Hmmm...I appear to be missing something in the notation. Ah well.


    -Dan
    Your not the only one. Looks as though there may be a forest of \mathcal{L}s missing here.

    RonL
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