You are given the polar curve r=cos(2θ). Find the points where the tangent line is horizontal and where the tangent line is vertical.

Question

Some answers are listed below that I have gotten right. Unfortunately I am not getting the right answers for the majority of them

a. (a) List all of the points $(r,\theta)$ where the tangent line is horizontal. In entering your answer, list the points starting with the smallest value of (r) and limit yourself to $r≥ \theta $ and $0 \leq \theta \leq 2\pi$ . If two or more points share the same value of $r$ , list those starting with the smallest value of $\theta$.

Point 1: $(r,\theta)=(?,?) $

Point 2: $ (r,\theta)=(?,?) $

Point 3: $ (r,\theta)=(2/3,?) $

Point 4: $ (r,\theta)=(2/3,?) $

Point 5: $(r,\theta)=(?,?) $

Point 6: $ (r,\theta)=(?,?) $

(b) List all of the points $(r,θ)$ where the tangent line is vertical. In entering your answer, list the points starting with the smallest value of $r$ and limit yourself to $r \geq 0$ and $0 \leq θ < 2π$. If two or more points share the same value of $r$, list those starting with the smallest value of θ.

Point 1: (?,?)

Point 2: (?,?)

Point 3: (?,?)

Point 4: (?,?)

Point 5: (?,0)

Point 6: (?,?)

Answer 1

If you use rectangular coordinates, then the curve is parameterized as $$\big(x(\theta),y(\theta)\big)=\big(r(\theta)\cos(\theta),r(\theta)\sin(\theta)\big)=$$ $$=\big(\cos(2\theta)\cos(\theta),\cos(2\theta)\sin(\theta)\big)=$$ $$=\Big(\big(\cos^2(\theta)-\sin^2(\theta)\big)\cos(\theta),\big(\cos^2(\theta)-\sin^2(\theta)\big)\sin(\theta)\Big)=$$ $$=\big(\cos^3(\theta)-\cos(\theta)\sin^2(\theta),\sin(\theta)\cos^2(\theta)-\sin^3(\theta)\big).$$

Now, for some $\theta$ the tangent is horizontal if $y'(\theta)=0$ and $x'(\theta)\neq 0$, and vertical if the opposite is true. If both are zero, then nothing can be said.

Now, when you get a $\theta_0$ such that, say, the tangent is vertical, your answer will be $$(r,\theta)=(\cos(2\theta_0),\theta_0).$$

For instance: $$x'(\theta)=-3\cos^2(\theta)\sin(\theta)+\sin^3(\theta)-2\sin(\theta)\cos^2(\theta)=$$ $$=\sin(\theta)\big(\sin^2(\theta)-5\cos^2(\theta)\big).$$ And these are zero when $$\sin(\theta)=0$$ and $$|\tan(\theta)|=\sqrt 5,$$ which gives six different values in the interval $[0,2\pi)$. Check that $y'(\theta)$ is not zero at each of those values and you'll get the six answers for vertical tangents: for instance, since one solution is $\theta=0$ the corresponding answer will be $(r,\theta)=\big(\cos(2\cdot 0),0\big)=(1,0)$.

Then, look the $\theta$ values such that $y'(\theta)=0$ to find the points with horizontal tangent line.