What is the equation of the tangent line to the polar curve # f(theta)=2thetasin(theta)+thetacot^2(2theta) # at #theta = pi/8#?

Question

What is the equation of the tangent line to the polar curve #                   f(theta)=2thetasin(theta)+thetacot^2(2theta)   # at #theta = pi/8#?

William Abdalla · Accepted Answer

#yapprox-0.650x+0.949# #f(pi/8)=2(pi/8)sin(pi/8)+pi/8cot^2(2(pi/8)) =pi/4sin(pi/8)+pi/8cot^2(pi/4)# By the half angle formulae, #sin(x/2)=+-sqrt((1-cos(x))/2)# #sin(pi/8)=+-sqrt((1-cos(pi/4))/2)=+-sqrt((1-sqrt2/2)/2)=+-sqrt(2-sqrt2)/2# Using the CAST rule, we can deduce that #sin(pi/8)=+sqrt(2-sqrt2)/2# Hence #f(pi/8)=pi/4sqrt(2-sqrt2)/2+pi/8(1)^2=pi/8(sqrt(2-sqrt2)+1)# #f(theta)=2thetasin(theta)+thetacot^2(2theta)# #f'(theta)=2sintheta+2thetacostheta+cot^2(2theta)+theta(2cot(2theta)*(-csc^2(2theta))*2)# #therefore f'(pi/8)=2sin(pi/8)+pi/4cos(pi/8)+cot^2(pi/4)+pi/2cot(pi/4)*(-csc^2(pi/4))=2sin(pi/8)+pi/4cos(pi/8)+1-pi/2csc^2(pi/4)# By the half angle formulae, #cos(x/2)=+-sqrt((cos(x)+1)/2)# #cos(pi/8)=+-sqrt((cos(pi/4)+1)/2)=+-sqrt((sqrt2/2+1)/2)=+-sqrt(2+sqrt2)/2# Using the CAST rule, we can deduce that #cos(pi/8)=+sqrt(2+sqrt2)/2# #because csc(x) = 1/sin(x)# #therefore csc(pi/8)=1/(sqrt(2-sqrt2)/2)=2/(sqrt(2-sqrt2)# #therefore csc^2(pi/8)=(2/(sqrt(2-sqrt2)))^2=4/(2-sqrt2)# #therefore f'(pi/8)=2(sqrt(2-sqrt2)/2)+pi/4sqrt(2+sqrt2)/2+1-pi/2(4/(2-sqrt2))=sqrt(2-sqrt2)+(pisqrt(2+sqrt2))/8+1-(2pi)/(2-sqrt2)# #therefore (y-y_1)/(x-x_1)=y'# #(y-f(pi/8))/(x-pi/8)=f'(pi/8)# #y=f'(pi/8)x-pi/8f'(pi/8)+f(pi/8)# #therefore y=-0.65013x+0.94875approx-0.650x+0.949#

Luke Alvarez · Answer

undefined To find the equation of the tangent line to the polar curve $f(\theta) = 2\theta \sin(\theta) + \theta \cot^2(2\theta)$ at $\theta = \frac{\pi}{8}$, you first find the slope of the tangent line by taking the derivative of $f(\theta)$ with respect to $\theta$, then evaluate it at $\theta = \frac{\pi}{8}$. Finally, you use the point-slope form of a line to write the equation of the tangent line.