What is the slope of the tangent line of #r=thetacos(theta/4-(5pi)/3)# at #theta=(-5pi)/3#?

Question

Isaiah Allen · Accepted Answer

Find the change in radius with respect to theta,
#r=thetacos(theta/4-(5pi)/3)#
#r'=cos(theta/4−(5pi)/3)-theta/4sin(theta/4−(5pi)/3)# Now, use the chain rule to derive a formula for gradient in polar coordinates:
#(dy)/(dx)=(dy/(d theta))/(dx/(d theta))=m#
We know the transformations for polar coordinates, #x=rcostheta# and #y=rsintheta#, and therefore this gradient formula becomes
#dy/(d theta)=r'sintheta+rcostheta#
#dx/(d theta)=r'costheta-rsintheta# #m=(dy/(d theta))/(dx/(d theta))=(r'sintheta+rcostheta)/(r'costheta-rsintheta)# now, let's find the values for r', r and theta at your point.
Theta is defined to be #-(5pi)/3#, so we use this to find the other values. #r=thetacos(theta/4-(5pi)/3)=-(5pi)/3cos(-(5pi)/(3*4)-(5pi)/3)#
#r=-(5pi)/3cos(-(5pi)/(12)-(20pi)/12)=-(5pi)/3cos(-(25pi)/12)#
#r=-(5pi)/3(1+sqrt(3))/(2sqrt2)# And now to find r',
#r'=cos(theta/4−(5pi)/3)-theta/4sin(theta/4−(5pi)/3)#
We already evaluated the first cosine expression in the last, so I will just skip through a couple of things
#r'=(1+sqrt(3))/(2sqrt2)-(5pi)/12sin((25pi)/12)=(1+sqrt(3))/(2sqrt2)-(5pi)/12(sqrt(3)-1)/(2sqrt2)# Sorry, I don't have enough time to do the rest of it, but I think you can easily plug those values into the gradient formula derived and get your answer :)

Ezra Adams · Answer

undefined To find the slope of the tangent line of the curve represented by $ r = \theta \cos\left(\frac{\theta}{4} - \frac{5\pi}{3}\right) $ at the point where $ \theta = -\frac{5\pi}{3} $, we need to find the derivative of $ r $ with respect to $ \theta $, and then evaluate it at $ \theta = -\frac{5\pi}{3} $.

$$ \frac{dr}{d\theta} = \cos\left(\frac{\theta}{4} - \frac{5\pi}{3}\right) - \theta \sin\left(\frac{\theta}{4} - \frac{5\pi}{3}\right) \times \frac{1}{4} - \cos\left(\frac{\theta}{4} - \frac{5\pi}{3}\right) $$

Then, we substitute $ \theta = -\frac{5\pi}{3} $ into the derivative to find the slope of the tangent line at that point.