Given: #r(theta)=theta^2cos(theta)-theta+sin(theta/3), theta=pi#
#r(pi)=pi^2cos(pi)-pi+sin(pi/3)#
#r(pi)=-pi^2-pi+sqrt(3)/2#
The polar point is #(-pi^2-pi+sqrt(3)/2,pi)#
To convert to rectangular coordinates, use:
#x = rcos(theta)# and y = #rsin(theta)#
x = #pi^2+pi-sqrt3/2#
#y = 0#
Using the point-slope form of the equation of a line:
#y = m(x - pi^2-pi+sqrt3/2) + 0#
To find the value of the slope, m, we must write #dy/dx# as a function of #theta# and then evaluate it at #theta = pi#.
From the reference Tangents with Polar Coordinates we obtain the following equation:
#dy/dx = ((dr)/(d theta)sin(theta)+rcos(theta))/((dr)/(d theta)cos(theta)-rsin(theta))" [1]"#
Because #theta = pi, sin(pi) = 0, and cos(pi)=-1#, this simplifies to:
#dy/dx = r/((dr)/(d theta))" [2]"#
Compute #(dr)/(d theta)#:
#(d(theta^2cos(theta)-theta+sin(theta/3)))/(d theta) = 2thetacos(theta)-theta^2sin(theta)-1-1/3cos(theta/3)#
Evaluate #2thetacos(theta)-theta^2sin(theta)-1+1/3cos(theta/3)# at #theta = pi#:
#-2pi-1-1/6 = -2pi-7/6#
#m = (-pi^2-pi+sqrt(3)/2)/(-2pi-7/6)#
The equation of the tangent line is:
#y = (-pi^2-pi+sqrt(3)/2)/(-2pi-7/6)(x - pi^2-pi+sqrt3/2)#
