Question

Equation of tangent line to the polar curve r=1-cos(theta) where theta=pi/3

Answer 1

first off, let us find the slope of it, at that point, so we can plug it in the point-slope form

thus $\bf r=1-cos(\theta)\qquad \begin{cases} \theta=\frac{\pi }{3}\\ --------------\\ r=1-cos\left( \frac{\pi }{3} \right)\to 1-\frac{1}{2}\to \frac{1}{2} \end{cases} \\\\ -----------------------------\\\\ \left. \cfrac{dr}{d\theta} \right|_{\theta=\frac{\pi }{3}}\implies \cfrac{dr}{d\theta} =0-[-sin(\theta)]\to sin(\theta)\qquad when\quad \theta=\frac{\pi }{3} \\\\\\ \cfrac{dr}{d\theta}=sin\left( \frac{\pi }{3} \right)\to \cfrac{\sqrt{3}}{2}\\\\ -----------------------------$

$\bf y-y_1=m(x-x_1)\textit{now, we know that } \begin{cases} \theta=\frac{\pi }{3}\\\\ r=\frac{1}{2}\\\\ m=\frac{\sqrt{3}}{2} \end{cases} \\\\\\ thus\implies y-\cfrac{1}{2}=\cfrac{\sqrt{3}}{2}\left( x-\cfrac{\pi }{3} \right) \\\\\\ y=\cfrac{\sqrt{3}}{2}x-\cfrac{\sqrt{3}\ \pi }{6}+\cfrac{1}{2}\implies y=\cfrac{\sqrt{3}}{2}x-\cfrac{\sqrt{3}\ \pi +3}{6}$

and I gather you could also write it as

$\bf y=\cfrac{\sqrt{3}}{2}x-\cfrac{\sqrt{3}\cdot \pi }{2\cdot 3}+\cfrac{1}{2}\implies y=\cfrac{\sqrt{3}}{2}x-\cfrac{\boxed{\sqrt{3}}\cdot \pi }{2\cdot \boxed{\sqrt{3^2}}}+\cfrac{1}{2} \\\\\\ y=\cfrac{\sqrt{3}}{2}x-\cfrac{\pi }{2\sqrt{3}}+\cfrac{1}{2}\implies y=\cfrac{\sqrt{3}}{2}x-\cfrac{\pi +\sqrt{3}}{2\sqrt{3}}$

and that's equivalent as well