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Thread: polar coordinates

  1. #1
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    polar coordinates

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

    I got $\displaystyle (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 $\displaystyle r = e^{\theta} $ find points on curve where tangent is horizontal and vertical.

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

    Is this correct? . . . . Yes!

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

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

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

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

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


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

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


    We have horizontal tangents when the numerator equals zero.

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

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


    We have vertical tangents when the denominator equals zero.

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

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

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