Question

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`?

Answer 1

#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

`(dy)/(dx) = ((dy)/(d theta))((d theta)/(dx))`

but

`(dx)/(d theta) = (dr)/(d theta) cos(theta) - r(theta)sin(theta)`
`(dy)/(d theta) = (dr)/(d theta) sin(theta) + r(theta)cos(theta)`

with

`r(theta) = theta^2 Cos(3 theta) - theta Sin(2 theta) + Tan(theta/6)`

Deriving and calculating for `theta = pi` we obtain

`((dy)/(dx))_{theta=pi} = ( pi^2-1/(sqrt[3]) )/( 4 pi-2/9 )`

`x_0 = x(pi) = -(1/sqrt[3]) + pi^2`
`y_0 = y(pi) = 0`

so the tangent line at `p_0 = {x_0,y_0}` is

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