Question

What is the slope of the polar curve `f(theta) = theta - sec^3theta+thetasin^3theta` at `theta = (5pi)/8`?

Answer 1

`dy/dx=-0.54`

Explanation:

For a polar function `f(theta)`, `dy/dx=(f'(theta)sintheta+f(theta)costheta)/(f'(theta)costheta-f(theta)sintheta)`

`f(theta)=theta-sec^3theta+thetasin^3theta`

`f'(theta)=1-3(sec^2theta)(d/dx[sectheta])-sin^3theta+3thetasin^2theta(d/dx[sintheta])`

`f'(theta)=1-3sec^3thetatantheta-sin^3theta+3thetasin^2thetacostheta`

`f'((5pi)/3)=1-3sec^3((5pi)/3)tan((5pi)/3)-sin^3((5pi)/3)+3((5pi)/3)sin^2((5pi)/3)cos((5pi)/3)~~-9.98`

`f((5pi)/3)=((5pi)/3)-sec^3((5pi)/3)+((5pi)/3)sin^3((5pi)/3)~~-6.16`

`dy/dx=(-9.98sin((5pi)/3)-6.16cos((5pi)/3))/(-9.98cos((5pi)/3)+6.16sin((5pi)/3))=-0.54`