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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.

enter image description here

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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 $$(x(2\pi-\theta), y(2\pi-\theta)) = (x(\theta),y(\theta)).$$

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  • $\begingroup$ 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 $7\pi/4$ $\endgroup$
    – user163862
    Commented Sep 26, 2018 at 18:17
  • $\begingroup$ The values I mention above DO cause $dy/d\theta=0$ $\endgroup$
    – user163862
    Commented Sep 26, 2018 at 18:25
  • $\begingroup$ @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. $\endgroup$
    – heropup
    Commented Sep 26, 2018 at 19:21
  • $\begingroup$ @user163862 A very minor point, but I noticed that you stated that there are eight solutions at the end of your post, but as you listed them, there are of course only six. $\endgroup$
    – heropup
    Commented Sep 26, 2018 at 19:23
  • $\begingroup$ 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. $\endgroup$
    – user163862
    Commented Sep 26, 2018 at 19:31

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