Question

What is the slope of `r=tantheta-theta` at `theta=pi/8`?

Answer 1

Derivative with Polar Coordinates is `(\partial x) / (\partial y) = ((\partial r) / (\partial theta) sin(theta)+ rcos(theta))/((\partial r) / (\partial theta) cos(theta)- rsin(theta)`

Explanation:

`(\partial r) / (\partial theta) = 1/cos^2(theta)+1`
`r(theta = pi/8)=0.0 22`
`(\partial x) / (\partial y) = (tan(theta)/cos(theta)+sin(theta)+rcos(theta))/(1/cos(theta)+cos(theta)-rsin(theta))`
`(\partial x) / (\partial y)(theta=pi/8) =1.57`