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

1 Answer
Apr 16, 2017

"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