Question

What is the slope of the tangent line of `r=(2theta+sin2theta)/cos^2theta` at `theta=(-3pi)/8`?

Answer 1

`"m"_"tan"=16+8sqrt2+(12+9sqrt2)pi`

Explanation:

`(dr)/(d theta)=((cos^2theta)(2+2cos2theta)+(2costhetasintheta)(2theta+sin2theta))/cos^4(theta)`

`r'(-3/8pi)=`
`((cos^2(-3/8pi))(2+2cos2(-3/8pi))+(2cos(-3/8pi)sin(-3/8pi))(2(-3/8pi)+sin2(-3/8pi)))/cos^4((-3/8pi))`

`=16+8sqrt2+(12+9sqrt2)pi`