What is the slope of the tangent line of #r=3sin(theta/2-pi/4)# at #theta=(3pi)/8#?

1 Answer
Jan 14, 2017

Slope at #theta=(3pi)/8# is #(-2cos((3pi)/8)sin(pi/16)+sin((3pi)/8)cos(pi/16))/(2sin((3pi)/8)sin(pi/16)+cos((3pi)/8)cos(pi/16)#

Explanation:

#r=f(theta)=3sin(theta/2-pi/4)#

Relating this to Cartesian coordinates, we know that

#x=rcostheta=3costhetasin(theta/2-pi/4)#

Hence #(dx)/(d theta)=-3sinthetasin(theta/2-pi/4)+3/2costhetacos(theta/2-pi/4)#

#y=rsintheta=3sinthetasin(theta/2-pi/4)#

Hence #(dy)/(d theta)=3costhetasin(theta/2-pi/4)+3/2sinthetacos(theta/2-pi/4)#

and hence #(dy)/(dx)=(3costhetasin(theta/2-pi/4)+3/2sinthetacos(theta/2-pi/4))/(-3sinthetasin(theta/2-pi/4)+3/2costhetacos(theta/2-pi/4))#

= #(2costhetasin(theta/2-pi/4)+sinthetacos(theta/2-pi/4))/(-2sinthetasin(theta/2-pi/4)+costhetacos(theta/2-pi/4))#

and slope at #theta=(3pi)/8# is

#(2cos((3pi)/8)sin((3pi)/16-pi/4)+sin((3pi)/8)cos((3pi)/16-pi/4))/(-2sin((3pi)/8)sin((3pi)/16-pi/4)+cos((3pi)/8)cos((3pi)/8-pi/4))#

= #(-2cos((3pi)/8)sin(pi/16)+sin((3pi)/8)cos(pi/16))/(2sin((3pi)/8)sin(pi/16)+cos((3pi)/8)cos(pi/16)#