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

Dec 18, 2016

$y = - 0.48 \left(x - \frac{\sqrt{3}}{4} + \frac{2 \pi}{6}\right) + \frac{\sqrt{3} \pi}{3} - \frac{3}{4}$

Explanation:

We need the Cartesian coordinates of the point of tangency:

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

$x = r \cos \left(\theta\right) \mathmr{and} y = r \sin \left(\theta\right)$

$x = \left(\frac{2 \pi}{3} - \frac{\sqrt{3}}{2}\right) \cos \left(\frac{2 \pi}{3}\right) \mathmr{and} y = \left(\frac{2 \pi}{3} - \frac{\sqrt{3}}{2}\right) \sin \left(\frac{2 \pi}{3}\right)$

$x = \frac{\sqrt{3}}{4} - \frac{2 \pi}{6} \mathmr{and} y = \frac{\sqrt{3} \pi}{3} - \frac{3}{4}$

For the slope of the line we need $\frac{\mathrm{dy}}{\mathrm{dx}}$ in polar form evaluated at $\theta = \frac{2 \pi}{3}$. This reference Tangent Lines to polar equations gives us the equation:

dy/dx = ((dr)/(d""theta)sin(theta) + rcos(theta))/((dr)/(d""theta)cos(theta) - rsin(theta)

(dr)/(d""theta) = 1 + cos( -theta + pi/3)

Evaluated at $\theta = \frac{2 \pi}{3}$

(dr)/(d""theta) = 3/2

The slope, m, of the tangent line is:

m = ((3/2)sin((2pi)/3) + ((2pi)/3 - sqrt(3)/2)cos((2pi)/3))/((3/2)cos((2pi)/3) - ((2pi)/3 - sqrt(3)/2)sin((2pi)/3)

m = ((3/2)(sqrt3/2) + ((2pi)/3 - sqrt(3)/2)(-1/2))/((3/2)(-1/2) - ((2pi)/3 - sqrt(3)/2)(sqrt3/2)

m = -(9sqrt(3) + sqrt(3) - 2pi)/(9 + (2pi - sqrt(3))(sqrt3)

$m \approx - 0.48$

Use the point-slope form of the equation of a line:

$y = m \left(x - {x}_{0}\right) + {y}_{0}$

$y = - 0.48 \left(x - \frac{\sqrt{3}}{4} + \frac{2 \pi}{6}\right) + \frac{\sqrt{3} \pi}{3} - \frac{3}{4}$