What is the slope of the polar curve #f(theta) = theta - sec^2theta+costheta # at #theta = (7pi)/12#?
1 Answer
Explanation:
The slope of the polar curve at
To find
#y=r sintheta = (theta - sec^2theta+costheta)(sintheta)#
#(dy)/(d theta) = (1 - 2sec^2(theta)tan(theta) - sintheta)(sintheta) + (theta-sec^2theta+costheta)(costheta)#
Evaluating this at
#x=r costheta = (theta - sec^2theta + costheta)(costheta)#
#(dx)/(d theta) = (1 - 2sec^2(theta)tan(theta) - sintheta)(costheta) + (theta-sec^2theta+costheta)(-sintheta)#
Evaluating this at
Therefore, the slope of the line tangent to