Question

What is the slope of the polar curve `f(theta) = theta^2 - sec^3theta+tantheta` at `theta = (3pi)/4`?

Answer 1

`f'((3pi)/4)=-1.77`

Explanation:

Anytime a problem asks for the slope of a curve at a given point, it is equivalent to asking what the value of the derivative is at that point.

Step 1: Find the derivative of `f(theta)` with respect to `theta`
`f'(theta)=d/(d(theta))f(theta)`
`f'(theta)=d/(d(theta))(theta^2-sec^3theta+tantheta)`
`f'(theta)=2theta-3*tanthetasectheta*sec^2theta+sec^2theta`
`f'(theta)=2theta-3*tanthetasec^3theta+sec^2theta`

Step 2: Plug `theta=(3pi)/4` into `f'(theta)`

`f'((3pi)/4)=2*((3pi)/4)-3*tan((3pi)/4)sec^3((3pi)/4)+sec^2((3pi)/4)`

Finally,

`f'((3pi)/4)=-1.77`