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Thread: Tangent Line for Parametric Curve

  1. #1
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    Tangent Line for Parametric Curve

    Maybe im doing this wrong, but this doesn't seem to exist, or the answer is undefined.

    Find an equation for the line tangent to the curve at the point defined by the given value of t. Also, find the value of $\displaystyle d^2y/dx^2$ at this point.

    $\displaystyle x= -cos t $
    $\displaystyle y= 7+sin t $
    $\displaystyle t= pi/2$

    Well, I did $\displaystyle -cos(pi/2) = 0$
    and $\displaystyle 7+sin(pi/2) = 8$

    $\displaystyle dx/dt = sin(pi/2) = 1$
    $\displaystyle dy/dt = cos(pi/2) = 0$
    $\displaystyle dy/dx= cos t / sin t $ or $\displaystyle dy/dx = cos t$

    so, I plugged back in the $\displaystyle pi/2$ to get $\displaystyle cot(pi/2) = UND$
    So I checked in my calculator, and the oval does not even go through the point $\displaystyle pi/2$, so it cant exist? Did I do something wrong?
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  2. #2
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    the tangent line is vertical ... the tangent line equation is x = a constant, specifically x = 0

    $\displaystyle \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\fra c{dx}{dt}}$
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  3. #3
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    Oh, I guess I didnt mention it has to be in slope intercept form, which should be $\displaystyle y-8=0(x-0)$
    or $\displaystyle y=8$

    OK, this makes more sense now, thank you
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  4. #4
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    so 2nd derivative of that parametric equation is: $\displaystyle -sint/sint$

    plug in $\displaystyle pi/2$
    and I get: -1
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