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

Nov 20, 2017

Find the change in radius with respect to theta,
$r = \theta \cos \left(\frac{\theta}{4} - \frac{5 \pi}{3}\right)$
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:
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{d \theta}}{\frac{\mathrm{dx}}{d \theta}} = m$
We know the transformations for polar coordinates, $x = r \cos \theta$ and $y = r \sin \theta$, and therefore this gradient formula becomes
$\frac{\mathrm{dy}}{d \theta} = r ' \sin \theta + r \cos \theta$
$\frac{\mathrm{dx}}{d \theta} = r ' \cos \theta - r \sin \theta$

$m = \frac{\frac{\mathrm{dy}}{d \theta}}{\frac{\mathrm{dx}}{d \theta}} = \frac{r ' \sin \theta + r \cos \theta}{r ' \cos \theta - r \sin \theta}$

now, let's find the values for r', r and theta at your point.
Theta is defined to be $- \frac{5 \pi}{3}$, so we use this to find the other values.

$r = \theta \cos \left(\frac{\theta}{4} - \frac{5 \pi}{3}\right) = - \frac{5 \pi}{3} \cos \left(- \frac{5 \pi}{3 \cdot 4} - \frac{5 \pi}{3}\right)$
$r = - \frac{5 \pi}{3} \cos \left(- \frac{5 \pi}{12} - \frac{20 \pi}{12}\right) = - \frac{5 \pi}{3} \cos \left(- \frac{25 \pi}{12}\right)$
$r = - \frac{5 \pi}{3} \frac{1 + \sqrt{3}}{2 \sqrt{2}}$

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
$r ' = \frac{1 + \sqrt{3}}{2 \sqrt{2}} - \frac{5 \pi}{12} \sin \left(\frac{25 \pi}{12}\right) = \frac{1 + \sqrt{3}}{2 \sqrt{2}} - \frac{5 \pi}{12} \frac{\sqrt{3} - 1}{2 \sqrt{2}}$

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 :)