# Thread: polar coordinates slope of tangent to curve problem

1. ## 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

2. Originally Posted by muddyjch
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 $y' = \frac23$