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