Question

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

Answer 1

Find the change in radius with respect to theta,
`r=thetacos(theta/4-(5pi)/3)`
`r'=cos(theta/4−(5pi)/3)-theta/4sin(theta/4−(5pi)/3)`

Now, use the chain rule to derive a formula for gradient in polar coordinates:
`(dy)/(dx)=(dy/(d theta))/(dx/(d theta))=m`
We know the transformations for polar coordinates, `x=rcostheta` and `y=rsintheta`, and therefore this gradient formula becomes
`dy/(d theta)=r'sintheta+rcostheta`
`dx/(d theta)=r'costheta-rsintheta`

`m=(dy/(d theta))/(dx/(d theta))=(r'sintheta+rcostheta)/(r'costheta-rsintheta)`

now, let's find the values for r', r and theta at your point.
Theta is defined to be `-(5pi)/3`, so we use this to find the other values.

`r=thetacos(theta/4-(5pi)/3)=-(5pi)/3cos(-(5pi)/(3*4)-(5pi)/3)`
`r=-(5pi)/3cos(-(5pi)/(12)-(20pi)/12)=-(5pi)/3cos(-(25pi)/12)`
`r=-(5pi)/3(1+sqrt(3))/(2sqrt2)`

And now to find r',
`r'=cos(theta/4−(5pi)/3)-theta/4sin(theta/4−(5pi)/3)`
We already evaluated the first cosine expression in the last, so I will just skip through a couple of things
`r'=(1+sqrt(3))/(2sqrt2)-(5pi)/12sin((25pi)/12)=(1+sqrt(3))/(2sqrt2)-(5pi)/12(sqrt(3)-1)/(2sqrt2)`

Sorry, I don't have enough time to do the rest of it, but I think you can easily plug those values into the gradient formula derived and get your answer :)