Confusion about whether $3\pi/4$ can extend into the 4th quadrant in a differential calculus problem involving a polar curve.

Question

This problem came up in Khan Academy differential calculus. I don't remember the exact wording, but the basic problem was:

Given $r=2\cos(\theta)$, choose all $\theta$ where the tangent to the curve $r$ is horizontal. The problem stipulated that we were only to choose values between $0<\theta<\pi$.

I did the math:

$x = r \cos\theta$

$x'=-4\cos\theta \sin\theta$

and

$y = r \sin \theta$

$y' = 2\cos(2x)$

I found the angle where the $y'$ was zero and I answered $\theta = \pi/4$. This made sense when graphed, as the line from the origin at an angle of $pi/4$ clearly intersects the curve where there is a horizontal tangent.

But my answer was wrong because apparently $3\pi/4$ is also a solution. This green line is at the angle $3\pi/4$ and while it does intersect the curve $r$ at a horizontal tangent, I can't understand how it can be correct.

The question asks for values of $\theta$ between $0$ and $\pi$. I get that $3\pi/4$ is in this range, but the line only hits the tangent in the 4th quadrant. Does this actually count? It looks much more like $7\pi/4$ to me!

I'm not saying that Sal is wrong here - I am yet to find an incorrect solution. But I would like to understand how this fits the constraints of the question!

Answer 1

Here is an animation of how the circle is plotted in polar coordinates. The line represents the angle $\theta$ as it sweeps from $0$ to $\pi$. The signed distance of the red dot from the origin is given by $r = 2 \cos \theta$. So when $0 < \theta < \pi/2$, the ray from the origin to the red dot is a positive distance, depicted as a blue line segment.

However, when $\pi/2 < \theta < \pi$, the line continues to sweep out the light blue shaded area, but the signed distance along that line is now negative, and the ray from the origin to the red dot is now shown in orange because it is a negative distance along that line.

We can clearly see that the locus of the red dot is a circle, for $0 \le \theta < \pi$. This circle has exactly two points for which the tangent line is horizontal, namely at $(x,y) = (1,1)$, and $(x,y) = (1,-1)$. So there must be two solutions for $\theta$. The first solution you found, $\theta = \pi/4$, corresponds to the first horizontal tangent at $(1,1)$. The second can only be when $\theta = 3\pi/4$.