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Math Help - how to find 2nd derivative of parametric equation?

  1. #1
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    how to find 2nd derivative of parametric equation?

    here's the parametric equation:
    x= tcos(t)
    y= tsin(t)

    for first derivative i get
    dy/dx= \frac {tcos(t)+sin(t)}{-tsin(t)+cos(t)}

    using:
    d^2y/dx^2= \frac{(d/dt)(dy/dt)}{dx/dt}

    this is what i get for the second derivative:

    d^2y/dx^2= \frac{-tsin(t)+2cos(t)}{-tsin(t)+cos(t)}

    but the answer in the back of the book is:
    d^2y/dx^2= \frac{t^2+2}{(cos(t)-tsint)^3}

    what did i do incorrectly or what am i doing incorrectly?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by yoman360 View Post
    here's the parametric equation:
    x= tcos(t)
    y= tsin(t)

    for first derivative i get
    dy/dx= \frac {tcos(t)+sin(t)}{-tsin(t)+cos(t)}

    using:
    d^2y/dx^2= \frac{(d/dt)(dy/dt)}{dx/dt}

    this is what i get for the second derivative:

    d^2y/dx^2= \frac{-tsin(t)+2cos(t)}{-tsin(t)+cos(t)}

    but the answer in the back of the book is:
    d^2y/dx^2= \frac{t^2+2}{(cos(t)-tsint)^3}

    what did i do incorrectly or what am i doing incorrectly?
    If \frac{\,dy}{\,dx}=\frac{\,dy/\,dt}{\,dx/\,dt}, then

    \frac{\,d^2y}{\,dx^2}=\frac{\,d}{\,dx}\left(\frac{  \,dy}{\,dx}\right)=\frac{\,d}{\,dx}\left(\frac{t\c  os t+\sin t}{\cos t-t\sin t}\right)

    Now observe that \frac{\,d}{\,dx}=\frac{\,d}{\,dt}\frac{\,dt}{\,dx}.

    Therefore, \frac{\,d}{\,dx}\left(\frac{t\cos t+\sin t}{\cos t-t\sin t}\right)=\frac{\,d}{\,dt}\left(\frac{t\cos t+\sin t}{\cos t-t\sin t}\right)\cdot\frac{\,dt}{\,dx}.

    Now apply quotient rule, and keep in mind that \frac{\,dt}{\,dx}=\frac{1}{\,dx/\,dt}=\frac{1}{\cos t-t\sin t}; you will then get the desired result.

    Can you take it from here?
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