Find #r# when #theta=pi/4#
#r=sin(2(pi/4)+pi) -theta=(-4-pi)/4#
Find #(dr) /(d theta):#
#r=sin(2 theta + pi) -theta#
#(dr) /(d theta)=cos(2 theta + pi) (2)-1=2cos(2 theta + pi) -1#
Substitute #theta=pi/4#:
#=2cos(2(pi/4)+pi)-1=-1#
Find the derivative:
#dy/(dx)=((-1)sin (pi/4) +((-4-pi)/4)cos (pi/4))/((-1)cos(pi/4) -((-4-pi)/4) sin (pi/4))=-8/pi-1#
Find the respective #x# and #y#.
#x=rcos(theta)=((-4-pi)/4) (sqrt(2)/2)=(-4sqrt(2)-pi sqrt(2))/8#
#y=rsin(theta)=((-4-pi)/4) (sqrt(2)/2)=(-4sqrt(2)-pi sqrt(2))/8#
The slope of a normal line is the negative reciprocal of the derivative.
#-1/(-8/pi-1)=pi/(8+pi)#
Putting it all together:
#(y-y_1)=m(x-x_1)#
#(y-(-4sqrt(2)-pi sqrt(2))/8)=(pi/(8+pi))(x-(-4sqrt(2)-pi sqrt(2))/8)#
