To find the equation of the line, we need a color(red)(poi nt), and we need a color(blue)(slope).
First, let's find the color(red)(poi nt).
Plug in pi/4 into f(theta).
f(pi/4)=2sin(3(pi/4)+pi/3)-2(pi/4)(cos(pi/4))
f(pi/4)=(sqrt2(2-2sqrt3-pi))/4
We have the polar coordinate (pi/4, (sqrt2(2-2sqrt3-pi))/4).
We have to convert this to Cartesian form. Using the formulas:
x = rcostheta
y = rcostheta
We find that the Cartesian coordinate is color(red)(((2-2sqrt3-pi)/4,(2-2sqrt3-pi)/4).
Good! We have found the color(red)(poi nt).
Now, we need to find the color(blue)(slope).
The formula for the tangent line of a polar function is:
dy/dx = ((dr)/(d theta)sintheta+rcostheta)/((dr)/(d theta)costheta-rsintheta)
If you notice, sintheta and costheta will be the same in this case, because sin(pi/4) and cos(pi/4) are the same, so for this case, our dy/dx is simply:
dy/dx = ((dr)/(d theta)+r)/((dr)/(d theta)-r)
We solved r already previously, which was (sqrt2(2-2sqrt3-pi))/4.
Now, we need to find (dr)/(d theta).
f'(theta)=(dr)/(d theta)=6cos(3theta+pi/3)+2thetasintheta-2costheta
Now, we plug in pi/4 and get:
f'(pi/4)=(dr)/(d theta)=(sqrt2(-10-6sqrt3+pi))/4
Going back to the tangent line formula:
dy/dx = ((sqrt2(-10-6sqrt3+pi))/4+(sqrt2(2-2sqrt3-pi))/4)/((sqrt2(-10-6sqrt3+pi))/4-(sqrt2(2-2sqrt3-pi))/4)
After some tedious simplification, we find this reduces to color(blue)(dy/dx = (-8-8sqrt3)/(-12-4sqrt3+2pi)
We have our color(red)(poi nt):
color(red)(((2-2sqrt3-pi)/4,(2-2sqrt3-pi)/4)
We have our color(blue)(slope):
color(blue)(dy/dx = (-8-8sqrt3)/(-12-4sqrt3+2pi)
All that is left is to plug in these values into the point slope formula:
(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))
We get:
y - ((2-2sqrt3-pi)/4) = (-8-8sqrt3)/(-12-4sqrt3+2pi)(x-((2-2sqrt3-pi)/4))
So the equation is:
y = (-8-8sqrt3)/(-12-4sqrt3+2pi)(x-((2-2sqrt3-pi)/4)) +((2-2sqrt3-pi)/4)