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Math Help - amplitude?

  1. #1
    Member i_zz_y_ill's Avatar
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    amplitude?

    Given y=1/13(3sint-2cost)+e^-3t(2cost+3sint)
    Q describe behaviour of y for large values of t. Sketch with range.

    The amplitube is in actual fact 1/(13)^0.5

    So on the graph its between -1/(13)^0.5 and 1/(13)^0.5

    How can this be when the max andmin value is clearly going to be 1/13 do you think this is a typo?
    I think it has something to do with the technique givenx = A sinwt=Bcoswt you can wright it as
    x=(A^2+B^2)^0.5(A/<(sinwt) + B/<(cowt)) < indicates the (A^2+B^2)^0.5
    But im afraid i cant see the direct relation between this straight off tchnique and the rsin(w+/-x) method or rcos(w+/-x) method is it the same thing? I dont think this is the factor formulae either!
    Oh its ok i got it
    Last edited by i_zz_y_ill; June 10th 2008 at 08:51 AM. Reason: solved
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  2. #2
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    Quote Originally Posted by i_zz_y_ill View Post
    Given y=1/13(3sint-2cost)+e^-3t(2cost+3sint)
    Q describe behaviour of y for large values of t. Sketch with range.

    The amplitube is in actual fact 1/(13)^0.5

    So on the graph its between -1/(13)^0.5 and 1/(13)^0.5

    How can this be when the max andmin value is clearly going to be 1/13 do you think this is a typo?
    I think it has something to do with the technique givenx = A sinwt=Bcoswt you can wright it as
    x=(A^2+B^2)^0.5(A/<(sinwt) + B/<(cowt)) < indicates the (A^2+B^2)^0.5
    But im afraid i cant see the direct relation between this straight off tchnique and the rsin(w+/-x) method or rcos(w+/-x) method is it the same thing? I dont think this is the factor formulae either!
    Oh its ok i got it
    For large t: y=1/13(3sint-2cost)+e^-3t(2cost+3sint) ---> y=1/13(3sint-2cost) since e^-3t ----> 0 and (2cost+3sint) is bounded.
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