# 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.

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: (?,?)

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.
• You might want to add that the points listed with $r=2/3$ will throw you for a loop. The zero derivative condition gives instead $r=-2/3$. You have to render $(r,\theta)=(-r,\theta\pm\pi)$ to conform with the list even though it does not conform with the equation. – Oscar Lanzi Jan 27 '19 at 2:36