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Thread: Polar Curve

  1. #1
    Senior Member slevvio's Avatar
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    Polar Curve

    A polar curve is defined by $\displaystyle r=1+cos\theta$.

    Show that $\displaystyle \frac{dy}{dx} = -cot\frac{3\theta}{2}$.

    I was wondering if anyone could help me with this. This is what I have.

    Let $\displaystyle r=f(\theta)$
    $\displaystyle tan\alpha=\frac{f(\theta)}{f'(\theta)}=\frac{1+cos \theta}{-sin\theta}=\frac{cos^2\frac{\theta}{2}}{sin\frac{\ theta}{2}cos\frac{\theta}{2}}$

    So $\displaystyle tan\alpha = cot\frac{\theta}{2}$

    I know that $\displaystyle tan(\alpha + \theta) = \frac{dy}{dx}$ but I don't know how to manipulate this. Thanks a lot.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by slevvio View Post
    A polar curve is defined by $\displaystyle r=1+cos\theta$.

    Show that $\displaystyle \frac{dy}{dx} = -cot\frac{3\theta}{2}$.

    I was wondering if anyone could help me with this. This is what I have.

    Let $\displaystyle r=f(\theta)$
    $\displaystyle tan\alpha=\frac{f(\theta)}{f'(\theta)}=\frac{1+cos \theta}{-sin\theta}=\frac{cos^2\frac{\theta}{2}}{sin\frac{\ theta}{2}cos\frac{\theta}{2}}$

    So $\displaystyle tan\alpha = cot\frac{\theta}{2}$

    I know that $\displaystyle tan(\alpha + \theta) = \frac{dy}{dx}$ but I don't know how to manipulate this. Thanks a lot.
    why is this in pre-calculus?

    note that we have $\displaystyle r = f( \theta )$

    now, $\displaystyle x = r \cos \theta = f ( \theta ) \cos \theta$ and $\displaystyle y = r \sin \theta = f ( \theta ) \sin \theta$

    now, $\displaystyle \frac {dy}{dx} = \frac {\frac {dy}{d \theta}}{\frac {dx}{d \theta}} = \frac {\frac {dr}{d \theta} \sin \theta + r \cos \theta}{\frac {dr}{d \theta} \cos \theta - r \sin \theta}$
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