What is the slope of the polar curve #f(theta) = theta - sectheta+thetasin^3theta # at #theta = (7pi)/12#?

1 Answer
Mar 21, 2018

#f'((7pi)/12)=-13.8459#

Explanation:

Given:
#f(theta)=theta-sectheta+thetasin^3theta#
Differentiating wrt #theta#
#f'(theta)=1-secthetatantheta+theta(3sin^2theta)costheta+sin^3theta#
#=1-1/costhetasintheta/costheta+sin^3theta+3thetasin^2thetacostheta #
#theta=(7pi)/12=1.8326#

#(7pi)/12=pi-(5pi)/12#
#costheta=cos(pi-(5pi)/12)=-cos((5pi)/12)=-0.2588#
#sintheta=sin(pi-(5pi)/12)=sin((5pi)/12)=0.9659#
Substituting the values in #f'(theta)#

#f'((7pi)/12)=1-1/-0.2588xx0.9659/-0.2588+0.9659^3+3xx1.8326xx0.9659^2(-0.2588)#
#f'((7pi)/12)=-13.8459#