calculusderivativesreal-analysis
Where does:
$$r = 1 + 2\cos(\theta)$$
Have horizontal and vertical tangent lines?
$x = r\cos(\theta) = \cos(\theta) + 2\cos^2(\theta)$
$y = r\sin(\theta) = \sin(\theta) + 2\sin(\theta)\cos(\theta)$
$$dx/d\theta = -\sin(\theta) – 4\sin(\theta)\cos(\theta)$$
$$dy/d\theta = \cos(\theta) + 2\cos(2\theta)$$
By hand it is very hard to solve $dy/d\theta = 0, dx/d\theta = 0$
How can it be done?
$$\dfrac{dy}{d\theta}\cdot \frac{d\theta}{dx} = \dfrac{dy}{dx}$$ and we want to know when $\dfrac{dy}{dx} = 0$.
$$\frac{dx}{d\theta} = \frac{\cos3\theta}{r},\quad \frac{dy}{d\theta} = \frac{\sin 3\theta}{r} \implies \dfrac{dy}{dx} = \frac{\sin{3\theta}}{r}\cdot \dfrac{r}{\cos 3\theta}$$
Now solve for when $$\require{cancel} \dfrac{dy}{dx} = \frac{\sin{3\theta}}{\cancel{r}}\cdot \dfrac{\cancel{r}}{\cos 3\theta} = = \tan(3\theta)= 0$$
That's going to happen when $\sin 3\theta = 0$ (and note that when $\sin{3\theta} = 0 \iff 3\theta = k \pi \iff \theta = k\pi/3\;$ ($k$ any integer), we have that $\cos 3\theta = \pm 1\iff 3\theta = k\pi \iff \theta = k\pi/3$.)
The points at which the tangents are horizontal and/or vertical are given in polar coordinates, $(r, \theta)$.
hint: $\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}$
Best Answer
You have calculated $$ \frac{dx}{d\theta}=-\sin\theta(1+4\cos\theta) $$ and $$ \frac{dy}{d\theta}=\cos\theta+2\cos2\theta. $$ Since $\cos2\theta=2\cos^2\theta-1$, the $\frac{dy}{d\theta}$ simplifies to $$ \frac{dy}{d\theta}=\cos\theta+4\cos^2\theta-2. $$
Vertical and horizontal tangents
In fact, we do not need to find the values of $\theta$ if we want to find the points where we have horizontal/vertical tangents. It is enough to know $\cos\theta$ and $\sin\theta$ and then use the formulas $$ x=r\cos\theta=(1+2\cos\theta)\cos\theta,\quad\text{and}\quad y=r\sin\theta=(1+2\cos\theta)\sin\theta. $$ Anyways, I indicate how to solve the equations. First, we can assume that $0\leq \theta<2\pi$, since $r=1+2\cos\theta$ is $2\pi$ periodic.
Now, one should solve $\frac{dx}{d\theta}=0$ to find the vertical tangent lines. For the vertical ones, $dx/d\theta=0$. Since the product $\sin\theta(1+4\cos\theta)=0$ if and only if (at least) one of its factors is zero, we find that either $\sin\theta=0$ (this gives $\theta=0$ and $\theta=\pi$), or $\cos\theta=-1/4$ (this gives $\theta=\arccos(-1/4)$ or $\theta=2\pi-\arccos(-1/4)$. In total, we get four values of $\theta$.
To find the horizontal tangent lines we solve $dy/d\theta=0$. We note that $dy/d\theta$ is a second degree polynomial in $\cos\theta$. Thus, with $t=\cos\theta$, we should solve $4t^2+t-2=0$. This has solutions $t=\frac{1}{8}(-1\pm\sqrt{33})$. So, we find $\theta$ by solving $$ \cos\theta=\frac{1}{8}(-1+\sqrt{33})\approx 0.59,\quad\text{and},\quad \cos\theta=\frac{1}{8}(-1-\sqrt{33})\approx -0.84. $$ To calculate $\sin\theta$ you can use the fact that $\cos^2\theta+\sin^2\theta=1$. I leave it to you to fill in the rest of the details. If you succeed, you will probably find something like the picture below, where I have drawn the curve $r=1+2\cos\theta$ together with the points I got by doing these calculations (red for horizontal tangents, green for vertical ones).