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Thread: Finding Slope of Tangent at point on polar curve

  1. #1
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    Finding Slope of Tangent at point on polar curve

    r=cos[theta/3], theta=pi

    So dy/dx is dy/d theta / dx/d theta and x=rcos theta, y=rsin theta and i found the derivatives of each which gave me:

    x=1/3(-sin 2th/3)-2sin(4th/3)
    y=1/3((cos(2th/3)+2cos(4th/3))

    I'm kinda stumped here. I have dy/dx as a big mess but I don't know what to do from here.
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  2. #2
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    Hello, sfgiants13!


    Find the slope of the tangent to $\displaystyle r\:=\;\cos\tfrac{\theta}{3}\:\text{ at }\,\theta=\pi$
    You're expected know (or be able to derive) that the slope formula is:

    . . . . . $\displaystyle \frac{dy}{dx} \:=\:\frac{r\cos\theta + r'\sin\theta}{\text{-}r\sin\theta + r'\cos\theta} $ .[1]


    We have:.$\displaystyle r \:=\:\cos\tfrac{\theta}{3},\quad r' \:=\:-\tfrac{1}{3}\sin\tfrac{\theta}{3} $

    Substitute into [1]: .$\displaystyle \frac{dy}{dx} \:=\:\frac{\cos\frac{\theta}{3}\cos\theta - \frac{1}{3}\sin\frac{\theta}{3}\sin\theta} {\text{-}\cos\frac{\theta}{3}\sin\theta - \frac{1}{3}sin\frac{\theta}{3}\cos\theta} $

    Let $\displaystyle \theta = \pi\!:\;\;\frac{dy}{dx} \:=\:\frac{\cos\frac{\pi}{3}\cos\pi = \frac{1}{3}\sin\frac{\pi}{3}\sin\pi}{\text{-}\cos\frac{\pi}{3}\sin\pi - \frac{1}{3}\sin\frac{\pi}{3}\cos\pi}$ .$\displaystyle \;=\;\frac{(\frac{1}{2})(-1) - (\frac{1}{3})(\frac{\sqrt{3}}{2})(0)} {\text{-}(\frac{1}{2})(0) - \frac{1}{3}(\frac{\sqrt{3}}{2})(-1)}
    $

    Therefore: .$\displaystyle \frac{dy}{dx} \;=\;\frac{\text{-}\frac{1}{2}}{\frac{\sqrt{3}}{6}} \;=\;-\sqrt{3}$

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