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

Question

Christopher Agresta · Accepted Answer

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)#

#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)#

Samantha Adkison · Answer

undefined To find the slope of the tangent line of the polar curve $ r = 3 \sin(\frac{\theta}{2} - \frac{\pi}{4}) $ at $ \theta = \frac{3\pi}{8} $, you need to take the derivative of $ r $ with respect to $ \theta $ and evaluate it at $ \theta = \frac{3\pi}{8} $. Then, use the formula for the slope of a tangent line in polar coordinates, which is given by $ \frac{dr}{d\theta} + r\frac{d\theta}{dr} $.

After finding the derivative of $ r $ with respect to $ \theta $ and evaluating it at $ \theta = \frac{3\pi}{8} $, and similarly, finding $ \frac{d\theta}{dr} $ and evaluating it at the same point, you can plug the values into the formula for the slope of the tangent line to get the final answer.