Question

What is the slope of the tangent line of `r=thetacos(theta/4-(5pi)/3)` at `theta=(pi)/3`?

Answer 1

`r=r'(pi/3)(theta-pi/3)+pi/3cos(-19pi/12)`
with
`r'(pi/3)=cos(-19pi/12)-pi/12sin(-19pi/12)`

Explanation:

`(dr)/(d theta)= r'(theta)=cos(theta/4-5pi/3)+theta(-sin(theta/4-5pi/3)*1/4)` that

evaluated in `theta=pi/3` becomes

`r'(pi/3)=cos(pi/12-5pi/3)-pi/12sin(pi/12-5pi/3)`

`r'(pi/3)=cos(-19pi/12)-pi/12sin(-19pi/12)`

the point of tangence is `(pi/3, pi/3cos(-19pi/3))` and the wished line is

`r=r'(pi/3)(theta-pi/3)+pi/3cos(-19pi/12)`