What is the equation of the tangent line of r=cos(theta-pi/2)/sintheta - sin(2theta-pi) at theta=(-3pi)/8?

Jan 12, 2016

Equation of tangent line:

$\frac{y + \frac{1}{2} \cdot \sqrt{\frac{2 - \sqrt{2}}{2}}}{x - {\left(2 - \sqrt{2}\right)}^{\frac{3}{2}} / 4} = - \sqrt{19 + 6 \sqrt{2}}$

Explanation:

The equation can be simplified to

$r = 1 + \sin \left(2 \theta\right)$

using trigonometric identities.

The polar plot is shown here http://www.wolframalpha.com/input/?i=r%3D1%2Bsin%282theta%29

To find the tangent line, you need to know a point it passes though and its gradient, $\frac{\mathrm{dy}}{\mathrm{dx}}$

We know that at $\theta = - \frac{3 \pi}{8}$,

$x = r \cos \theta$

$= \left(1 + \sin \left(2 \left(- \frac{3 \pi}{8}\right)\right)\right) \cos \left(- \frac{3 \pi}{8}\right)$

$= {\left(2 - \sqrt{2}\right)}^{\frac{3}{2}} / 4$

$\approx 0.112$

$y = r \sin \theta$

$= \left(1 + \sin \left(2 \left(- \frac{3 \pi}{8}\right)\right)\right) \sin \left(- \frac{3 \pi}{8}\right)$

$= - \frac{1}{2} \cdot \sqrt{\frac{2 - \sqrt{2}}{2}}$

$\approx - 0.271$

The tangent line passes through $\left(0.112 , - 0.271\right)$.

We also know that at $\theta = - \frac{3 \pi}{8}$,

$\frac{\mathrm{dx}}{d \theta} = \frac{d}{d \theta} \left(r \sin \theta\right)$

$= \frac{d}{d \theta} \left(\left(1 + \sin \left(2 \theta\right)\right) \cdot \cos \left(\theta\right)\right)$

$= \frac{- 2 \sin \left(\theta\right) + \cos \left(\theta\right) + 3 \cos \left(3 \theta\right)}{2}$

Substituting $\theta = - \frac{3 \pi}{8}$, we have

${\frac{\mathrm{dx}}{d \theta}}_{\theta = - \frac{3 \pi}{8}} = - \frac{1}{2} \cdot \sqrt{\frac{2 - \sqrt{2}}{2}}$

$\approx - 0.271$

Similarly,

$\frac{\mathrm{dy}}{d \theta} = \frac{d}{d \theta} \left(r \sin \theta\right)$

$= \frac{d}{d \theta} \left(\left(1 + \sin \left(2 \theta\right)\right) \cdot \sin \left(\theta\right)\right)$

$= 2 \sin \left(\theta\right) \cos \left(2 \theta\right) + \left(\sin \left(2 \theta\right) + 1\right) \cdot \cos \left(\theta\right)$

Substituting $\theta = - \frac{3 \pi}{8}$, we have

${\frac{\mathrm{dy}}{d \theta}}_{\theta = - \frac{3 \pi}{8}} = \frac{1}{2} \cdot \sqrt{\frac{26 - 7 \sqrt{2}}{2}}$

$\approx 1.42$

From the chain rule, we can calculate the gradient as

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

$= - \sqrt{19 + 6 \sqrt{2}}$

$\approx 5.24$

Equation of tangent line:

$\frac{y - \left(- \frac{1}{2} \cdot \sqrt{\frac{2 - \sqrt{2}}{2}}\right)}{x - {\left(2 - \sqrt{2}\right)}^{\frac{3}{2}} / 4} = - \sqrt{19 + 6 \sqrt{2}}$