What is the slope of the polar curve f(theta) = theta - sec^2theta+costheta  at theta = (7pi)/12?

1 Answer
Mar 21, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 6.968 \mathmr{if} \theta = \frac{7 \pi}{12}$

Explanation:

The slope of the polar curve at $\theta = \frac{7 \pi}{12}$ is $\frac{\mathrm{dy}}{\mathrm{dx}}$ evaluated in terms of $\theta$ at $\theta = \frac{7 \pi}{12}$.

To find $\frac{\mathrm{dy}}{\mathrm{dx}}$, we can first find $\frac{\mathrm{dy}}{d \theta}$ and $\frac{\mathrm{dx}}{d \theta}$ and use the chain rule for derivatives to determine that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{d \theta} \times \frac{d \theta}{\mathrm{dx}} = \frac{\mathrm{dy}}{d \theta} \div \frac{\mathrm{dx}}{d \theta}$

$y = r \sin \theta = \left(\theta - {\sec}^{2} \theta + \cos \theta\right) \left(\sin \theta\right)$
$\frac{\mathrm{dy}}{d \theta} = \left(1 - 2 {\sec}^{2} \left(\theta\right) \tan \left(\theta\right) - \sin \theta\right) \left(\sin \theta\right) + \left(\theta - {\sec}^{2} \theta + \cos \theta\right) \left(\cos \theta\right)$

Evaluating this at $\theta = \frac{7 \pi}{12}$ gives that $\frac{\mathrm{dy}}{d \theta} = 111.12 \mathmr{if} \theta = \frac{7 \pi}{12}$

$x = r \cos \theta = \left(\theta - {\sec}^{2} \theta + \cos \theta\right) \left(\cos \theta\right)$
$\frac{\mathrm{dx}}{d \theta} = \left(1 - 2 {\sec}^{2} \left(\theta\right) \tan \left(\theta\right) - \sin \theta\right) \left(\cos \theta\right) + \left(\theta - {\sec}^{2} \theta + \cos \theta\right) \left(- \sin \theta\right)$

Evaluating this at $\theta = \frac{7 \pi}{12}$ gives that $\frac{\mathrm{dx}}{d \theta} = - 15.948 \mathmr{if} \theta = \frac{7 \pi}{12}$

$\frac{\mathrm{dy}}{d \theta} \div \frac{\mathrm{dx}}{d \theta} = \frac{111.12}{- 15.948} = - 6.968$

Therefore, the slope of the line tangent to $f \left(\theta\right)$ at the point where $\theta = \frac{7 \pi}{12}$ is $- 6.968$.