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=(\pi )/(3)\\ --------------\\ r=1-cos\left( (\pi )/(3) \right)\to 1-(1)/(2)\to (1)/(2) \end{cases} \\\\ -----------------------------\\\\ \left. \cfrac{dr}{d\theta} \right|_{\theta=(\pi )/(3)}\implies \cfrac{dr}{d\theta} =0-[-sin(\theta)]\to sin(\theta)\qquad when\quad \theta=(\pi )/(3) \\\\\\ \cfrac{dr}{d\theta}=sin\left( (\pi )/(3) \right)\to \cfrac{√(3)}{2}\\\\ -----------------------------\\\\`


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

and I gather you could also write it as


`\bf y=\cfrac{√(3)}{2}x-\cfrac{√(3)\cdot \pi }{2\cdot 3}+\cfrac{1}{2}\implies y=\cfrac{√(3)}{2}x-\cfrac{\boxed{√(3)}\cdot \pi }{2\cdot \boxed{√(3^2)}}+\cfrac{1}{2} \\\\\\ y=\cfrac{√(3)}{2}x-\cfrac{\pi }{2√(3)}+\cfrac{1}{2}\implies y=\cfrac{√(3)}{2}x-\cfrac{\pi +√(3)}{2√(3)}`

and that's equivalent as well