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Math Help - Need help solving sinusoidal functions

  1. #1
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    Need help solving sinusoidal functions

    I am having trouble solving the the "x" value of a sinusoidal function when given the y value.

    Here is a sample question:

    d(t) = 2 sin(30t) + 5, solve for t when given the d value of 3 m.

    Also another question, T(t) = - 18.9cos30t + 5.8 solve for t when given the T value of 2.

    Your help is deeply appreciated
    Last edited by oceanblack; January 4th 2010 at 03:29 PM.
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by oceanblack View Post
    I am having trouble solving the the "x" value of a sinusoidal function when given the y value.

    Here is a sample question:

    d(t) = 2 sin(30t) + 5, solve for t when given the d value of 3 m.

    Also another question, T(t) = - 18.9cos30t + 5.8 solve for t when given the T value of 38.

    Your help is deeply appreciated
    If d=3, then the first equation becomes 3=2\sin\!\left(30t\right)+5\implies 2\sin\!\left(30t\right)=-2\implies \sin\!\left(30t\right)=-1.

    To make it a little easier, let \omega=30t so \sin \omega=-1. Now what values of \omega cause this to be true? Take those value(s) and divide them by 30 to find t.

    Can you try to do something similar for the second problem?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    If d=3, then the first equation becomes 3=2\sin\!\left(30t\right)+5\implies 2\sin\!\left(30t\right)=-2\implies \sin\!\left(30t\right)=-1.

    To make it a little easier, let \omega=30t so \sin \omega=-1. Now what values of \omega cause this to be true? Take those value(s) and divide them by 30 to find t.

    Can you try to do something similar for the second problem?
    I'm unsure as to how to find what values cause this to be true.

    I tried to do the inverse of sine and I got -90, but dividing that by 30 got me the answer of -3 which is not the given answers of 60, 180 and 300.

    May you please give more detail as to how I find the appropriate values for w, is it guessing and checking? I am unsure.

    Thank you for helping me so far, I greatly appreciate it.
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  4. #4
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    Quote Originally Posted by oceanblack View Post
    I am having trouble solving the the "x" value of a sinusoidal function when given the y value.

    Here is a sample question:

    d(t) = 2 sin(30t) + 5, solve for t when given the d value of 3 m.
    d(t) = 2 \sin(30t) + 5

    3 = 2 \sin(30t) + 5

    -2 = 2 \sin(30t)

    -1 = \sin(30t)

    30t = \sin^{-1} (-1) = \frac{3\pi}{2} + 2k\pi, k = 0,1,2, \cdots

    solve for t.

    Also another question, T(t) = - 18.9cos30t + 5.8 solve for t when given the T value of 38.
    sure about this question?
    the max value of T = 18.9 + 5.8 = 24.7 so there is no t satisfy T(t) = 38
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  5. #5
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    Quote Originally Posted by dedust View Post
    d(t) = 2 \sin(30t) + 5

    3 = 2 \sin(30t) + 5

    -2 = 2 \sin(30t)

    -1 = \sin(30t)

    30t = \sin^{-1} (-1) = \frac{3\pi}{2} + 2k\pi, k = 0,1,2, \cdots

    solve for t.

    sure about this question?
    the max value of T = 18.9 + 5.8 = 24.7 so there is no t satisfy T(t) = 38
    I am sorry the appropriate value is 2, not 38, I will edit the original post.

    Thank yo so much for your reply, could you please elaborate on this specific aspect:

    30t = \sin^{-1} (-1) = \frac{3\pi}{2} + 2k\pi, k = 0,1,2, \cdots
    I am lost as to how you got \frac{3\pi}{2} + 2k\pi.

    I am also assuming (if you could please confirm) that k is referring to the cycles?
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  6. #6
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    Quote Originally Posted by Chris L T521 View Post
    If d=3, then the first equation becomes 3=2\sin\!\left(30t\right)+5\implies 2\sin\!\left(30t\right)=-2\implies \sin\!\left(30t\right)=-1.

    To make it a little easier, let \omega=30t so \sin \omega=-1. Now what values of \omega cause this to be true? Take those value(s) and divide them by 30 to find t.

    Can you try to do something similar for the second problem?
    I ended up looking back on your solution and figured out the problem.

    The answer given by the homework ended up being incorrect

    But thank you so much for your help.
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