What is the slope of #r=thetacostheta/2-theta^2sintheta# at #theta=(3pi)/8#?

1 Answer
Sep 16, 2017

#(dr)/(dθ)=-1/64 (9 π^2 - 32) sin(π/8) - 15/16 π cos(π/8)#

Explanation:

#r=theta cos(theta)/2 -theta^2 sintheta#
differentiate with respect to theta
#(dr)/(dθ)=costheta/2 +theta(-sintheta/2)-2 θ sintheta-theta^2costheta#
#(dr)/(dθ)=costheta/2 -5/2 θ sintheta-theta^2costheta#
#(dr)/(dθ)# is the slope
Hence Slope at #theta=(3pi)/8#
#(dr)/(dθ)=cos((3pi)/8)/2 -5/2 ((3pi)/8) sin((3pi)/8)-((3pi)/8)^2cos((3pi)/8)#
#(dr)/(dθ)=cos(pi/2-pi/8)/2 -5/2 (pi/2-pi/8) sin(pi/2-pi/8)-((3pi)/8)^2cos(pi/2-pi/8)#
#(dr)/(dθ)=sin(pi/8)/2 -5/2 (3pi/8) cos(pi/8)-((3pi)/8)^2sin(pi/8)#
#(dr)/(dθ)=(-15 pi Cos[pi/8])/16 + Sin[pi/8]/2 - (9 pi^2 Sin[pi/8])/64#
#(dr)/(dθ)=-1/64 (9 π^2 - 32) sin(π/8) - 15/16 π cos(π/8)#