# What is the equation of the tangent line to the polar curve  f(theta)=theta^2cos(3theta)-thetasin(2theta)+tan(theta/6)  at theta = pi?

Jun 14, 2016

Y = ((pi^2-1/sqrt[3]) (1/sqrt[3] - pi^2 + X))/( 4 pi-2/9)

#### Explanation:

The pass equations are

{ (x(theta) = r(theta)cos(theta)), (y(theta)=r(theta)sin(theta)) :}

The tangent space is obtained making

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

but

$\frac{\mathrm{dx}}{d \theta} = \frac{\mathrm{dr}}{d \theta} \cos \left(\theta\right) - r \left(\theta\right) \sin \left(\theta\right)$
$\frac{\mathrm{dy}}{d \theta} = \frac{\mathrm{dr}}{d \theta} \sin \left(\theta\right) + r \left(\theta\right) \cos \left(\theta\right)$

with

$r \left(\theta\right) = {\theta}^{2} C o s \left(3 \theta\right) - \theta S \in \left(2 \theta\right) + T a n \left(\frac{\theta}{6}\right)$

Deriving and calculating for $\theta = \pi$ we obtain

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\theta = \pi} = \frac{{\pi}^{2} - \frac{1}{\sqrt{3}}}{4 \pi - \frac{2}{9}}$

${x}_{0} = x \left(\pi\right) = - \left(\frac{1}{\sqrt{3}}\right) + {\pi}^{2}$
${y}_{0} = y \left(\pi\right) = 0$

so the tangent line at ${p}_{0} = \left\{{x}_{0} , {y}_{0}\right\}$ is

Y = ((pi^2-1/sqrt[3]) (1/sqrt[3] - pi^2 + X))/( 4 pi-2/9)