What is the slope of the polar curve #f(theta) = theta^2 - sec^3theta+tantheta # at #theta = (3pi)/4#?

1 Answer
Jun 21, 2016

#f'((3pi)/4)=-1.77#

Explanation:

Anytime a problem asks for the slope of a curve at a given point, it is equivalent to asking what the value of the derivative is at that point.

Step 1: Find the derivative of #f(theta)# with respect to #theta#
#f'(theta)=d/(d(theta))f(theta)#
#f'(theta)=d/(d(theta))(theta^2-sec^3theta+tantheta)#
#f'(theta)=2theta-3*tanthetasectheta*sec^2theta+sec^2theta#
#f'(theta)=2theta-3*tanthetasec^3theta+sec^2theta#

Step 2: Plug #theta=(3pi)/4# into #f'(theta)#

#f'((3pi)/4)=2*((3pi)/4)-3*tan((3pi)/4)sec^3((3pi)/4)+sec^2((3pi)/4)#

Finally,

#f'((3pi)/4)=-1.77#