Question

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

Answer 1

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)`