Question

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

Answer 1

`x=r(\theta)\cos\theta\implies(\mathrm dx)/(\mathrm d\theta)=(\mathrm dr)/(\mathrm d\theta)\cos\theta-r\sin\theta`

`y=r(\theta)\sin\theta\implies(\mathrm dr)/(\mathrm d\theta)\sin\theta+r\cos\theta`

Since
`r=1-\cos\theta`, you have
`(\mathrm dr)/(\mathrm d\theta)=\sin\theta`. Now,


`(\mathrm dy)/(\mathrm dx)=((\mathrm dy)/(\mathrm d\theta))/((\mathrm dx)/(\mathrm d\theta))=(\sin^2\theta+(1-\cos\theta)\cos\theta)/(\sin\theta\cos\theta-(1-\cos\theta)\sin\theta)`

`(\mathrm dy)/(\mathrm dx)=(\cos\theta-\cos2\theta)/(\sin2\theta-1)`

When
`t=\frac\pi3`, you have


`(\mathrm dy)/(\mathrm dx)=\frac{\cos\frac\pi3-\cos\frac{2\pi}3}{\sin\frac{2\pi}3-1}=-2\sqrt3-4`

which will be the slope of the tangent line. At this point, you have
`r=1-\cos\frac\pi3=\frac12`, so now you need to find the corresponding x and y coordinates.


`x=r\cos\theta=\frac12\cos\frac\pi3=\frac14`

`y=r\sin\theta=\frac12\sin\frac\pi3=\frac{\sqrt3}4`

So the equation of the tangent line is


`y-\frac{\sqrt3}4=(-2\sqrt3-4)\left(x-\frac14\right)`

`y=-2(2+\sqrt3)x+1+\frac{3\sqrt3}4`