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Math Help - quick question

  1. #1
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    quick question

    y = 15 sin(ωt + a), determine dy/dt.... i'm just confused by the ω on this one
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  2. #2
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    Quote Originally Posted by Ep!taph View Post
    y = 15 sin(ωt + a), determine dy/dt.... i'm just confused by the ω on this one
    It appears that \omega and a are arbitrary constants.

    So you can use the chain rule to find this derivative.

    Let u = \omega t + a.

    Then y = \sin{u}.


    So \frac{du}{dt} = \omega and \frac{dy}{du} = \cos{u} = \cos{(\omega t + a)}.


    Therefore

    \frac{dy}{dt} = \frac{du}{dt}\cdot\frac{dy}{du}

     = \omega \cos{(\omega t + a)}.
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  3. #3
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    ah thanks... just as i thought, what about this one?

    v = 10(1-e^(-t/5)), determine dv/dt?
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  4. #4
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    Quote Originally Posted by Ep!taph View Post
    ah thanks... just as i thought, what about this one?

    v = 10(1-e^(-t/5)), determine dv/dt?
    What is the problem you are having with this one?

    (multiply out the bracket, do you still have a problem, what is it? Consider using the chain rule)

    CB
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