# Finding horizontal tangent line for polar graph - extraneous solns

The problem is to find the coordinates of a specific polar function at which there are horizontal tangent lines.

$$r=sin(\theta)cos^2(\theta)$$

So, I use for $$dy/dx= (rcos(\theta) + sin(\theta)dr/d\theta)/(-rsin(\theta)+ cos(\theta)dr/d\theta)$$ and since I am looking for horizontal tangent lines I set the numerator $$=0$$ and solve for $$\theta$$

My calculations for this yield the following:

$$(sin(\theta)cos(\theta))(2cos^2(\theta)-2sin^2(\theta))=0$$

I set each of these terms $$=0$$ and use the zero property rule to solve for $$\theta.$$

I get the following set of values for $$\theta$$

$$=0, \pi,$$ $$\pi/4, 3\pi/4, 5\pi/4, and, 7\pi/4$$

The text lists ONLY $$0 , \pi/4$$ and $$3\pi/4$$ as answers and I can see from the enclosed graph that those would be the answers.

So my question is how can I determine without graphing that those other answers are not valid? I did put them into the equation for $$dy/dx$$ to see if they caused the denominator to be zero (vertical tangent line) but they do not.

I don't know how I'm suppose to be able to rule out those other values. On the test I would have listed all 8 values.

The issue is that on the interval $$\theta \in [0, 2\pi)$$, the curve is traversed twice. This can be seen by observing that $$r(\theta + \pi) = -r(\theta).$$ Alternatively, converting the polar equation to a parametric one gives $$(x(\theta),y(\theta)) = (\cos^3 \theta, \sin^2 \theta \cos^2 \theta),$$ and we find that $$(x(2\pi-\theta), y(2\pi-\theta)) = (x(\theta),y(\theta)).$$
• I see that the curve is traversed twice. However, it's not clear to me why this means that $\theta =\pi$ isn't a solution algebraically for a horizontal tangent line. Same for $5\pi/4$ and $7\pi/4$ Commented Sep 26, 2018 at 18:17
• The values I mention above DO cause $dy/d\theta=0$ Commented Sep 26, 2018 at 18:25
• @user163862 I am not saying that your solution is invalid. I am stating that this is the apparent reason for the discrepancy between the stated solution and yours. There is no intrinsic reason to assume, for example, $\theta \in [0,2\pi)$ as opposed to $\theta \in (-\pi, \pi]$, or $\theta \in (-\infty, \infty)$, each of which gives a different solution set. That said, if the problem were to state that the solution set should cover a suitable range of $\theta \in [0, 2\pi)$ for which the curve is traversed once, then this gives the text solution. Commented Sep 26, 2018 at 19:21
• Now I completely understand! I did not know this polar formula of $r(\theta + \pi)= -r(\theta)$. Seems very useful as it changes the range of $\theta$ values to be considered. Even after you showed how the problem only goes from $0 to \pi$ I couldn't figure out why $\pi$ wasn't a good answer. But, when I put those values into the $r$ equation they yield the same point. So, now I understand. The 6 vs 8 came from and editing mismatch. Commented Sep 26, 2018 at 19:31