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Math Help - polar coordinates

  1. #1
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    polar coordinates

    if  r = e^{\theta} find points on curve where tangent is horizontal and vertical.

    I got  (e^{n \pi - \frac{\pi}{4}}, \pi n - \frac{\pi}{4}) for horizontal tangents.

    Is this correct?
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  2. #2
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    Hello, shilz222!

    If  r = e^{\theta} find points on curve where tangent is horizontal and vertical.

    I got  \left(e^{n \pi - \frac{\pi}{4}},\,\pi n - \frac{\pi}{4}\right) for horizontal tangents.

    Is this correct? . . . . Yes!

    First, we need \frac{dy}{dx}

    There is a formula for this, but we can derive it.

    We know the conversion formulas: . \begin{array}{c}x \:=\:r\cos\theta \\ y \:=\:r\sin\theta\end{array}

    Then: . \begin{array}{ccc}\frac{dx}{d\theta} & = & \text{-}r\sin\theta + r'\cos\theta \\ \frac{dy}{d\theta} & = & r\cos\theta + r'\sin\theta\end{array}

    . . Hence: . \frac{dy}{dx} \;=\;\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \;=\;\frac{4\cos\theta + r'\sin\theta}{\text{-}r\sin\theta + r'\cos\theta}


    Since r = e^{\theta} and r' = e^{\theta}, we have: . \frac{dy}{dx} \;=\;\frac{e^{\theta}\cos\theta + e^{\theta}\sin\theta}{\text{-}e^{\theta}\sin\theta + e^{\theta}\cos\theta}

    Divide top and bottom by e^{\theta}\!:\;\;\frac{dy}{dx}\;=\;\frac{\cos\thet  a + \sin\theta}{\text{-}\sin\theta + \cos\theta}


    We have horizontal tangents when the numerator equals zero.

    . . \cos\theta + \sin\theta \:=\:0\quad\Rightarrow\quad \sin\theta \:=\:-\cos\theta\quad\Rightarrow\quad\frac{\sin\theta}{\  cos\theta} \:=\:-1

    . . Therefore: . \tan\theta\:=\:-1\quad\Rightarrow\quad\boxed{\theta \:=\:-\frac{\pi}{4} + \pi n}


    We have vertical tangents when the denominator equals zero.

    . . -\sin\theta + \cos\theta\:=\:0\quad\Rightarrow\quad\sin\theta \:=\:\cos\theta\quad\Rightarrow\quad\frac{\sin\the  ta}{\cos\theta} \:=\:1

    . . Therefore: . \tan\theta \:=\:1\quad\Rightarrow\quad\boxed{\theta \:=\:\frac{\pi}{4} + \pi n}

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  3. #3
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    But  x = e^{\theta} \cos \theta and  y = e^{\theta} \sin \theta . Where the the  e^{\theta} go?
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