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Math Help - Derivative of parametric equations?

  1. #1
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    Derivative of parametric equations?

    A particle is moving in the xy-plane and its position at time t is given by
    x=cos(\frac{\pi }{3}t)
    and y=2sin(\frac{\pi }{3}t). When t=3 the speed of the particle is?

    -Ok i know that i have to take the derivative. But i don't remember how to take the derivative of parametric equations...

    all help is appreciated..
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  2. #2
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    Quote Originally Posted by dandaman View Post
    A particle is moving in the xy-plane and its position at time t is given by
    x=cos(\frac{\pi }{3}t)
    and y=2sin(\frac{\pi }{3}t). When t=3 the speed of the particle is?

    -Ok i know that i have to take the derivative. But i don't remember how to take the derivative of parametric equations...

    all help is appreciated..
    Recall the Chain Rule:

    \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}
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  3. #3
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    Quote Originally Posted by dandaman View Post
    A particle is moving in the xy-plane and its position at time t is given by
    x=cos(\frac{\pi }{3}t)
    and y=2sin(\frac{\pi }{3}t). When t=3 the speed of the particle is?

    -Ok i know that i have to take the derivative. But i don't remember how to take the derivative of parametric equations...

    all help is appreciated..
    \frac{dx}{dt} = -\frac{\pi}{3}\sin\left(\frac{\pi t}{3}\right)

    \frac{dy}{dt} = \frac{2\pi}{3}\cos\left(\frac{\pi t}{3}\right)<br />

    speed = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}
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  4. #4
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    ok so i have
    \frac{dy}{dx}=\frac{dy}{dt}  \frac{dt}{dx}
    and
    \frac{dy}{dt}=\frac{2\pi }{3}cos(\frac{\pi }{3}t)
    and
    \frac{dx}{dt}=-\frac{\pi  }{3}sin(\frac{\pi }{3}t)

    but i have dx/dt not dt/dx.
    do i have to take the reciprocal so i can plug it into the equation?
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  5. #5
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    Quote Originally Posted by dandaman View Post
    \frac{dy}{dx}=\frac{dy}{dt}  \frac{dt}{dx}

    ignore this ...
    you do not need to find \frac{dy}{dx}
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