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Math Help - polar coordinates slope of tangent to curve problem

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    polar coordinates slope of tangent to curve problem

    Find the slope of the tangent to the curve r=9+6cosθ at the value θ = π/2
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    Quote Originally Posted by muddyjch View Post
    Find the slope of the tangent to the curve r=9+6cosθ at the value θ = π/2
    Re-write the equation in parametric form:

    c : \left\{\begin{array}{l}x=(9+6 \cos(\theta)) \cdot \cos(\theta) \\ y = (9+6 \cos(\theta)) \cdot \sin(\theta)\end{array}\right.

    Then the slope is calculated by

    y'=\dfrac{\frac{dy}{d \theta}}{\frac{dx}{d \theta}}

    Expand the terms before differentiation. Use product rule and chain rule.

    I've got y' = \dfrac{12 \cos^2(\theta) + 9 \cos(\theta) - 6}{-12 \sin(\theta) \cos(\theta) - 9 \sin(\theta) }

    Plug in \theta = \frac \pi2. I've got as the final result y' = \frac23
    Last edited by earboth; July 21st 2010 at 09:57 PM.
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