# Horizontal and vertical asymptotes of polar curve $r = \theta/(\pi - \theta) \, , \, \in[0,\pi]$

I as supposed to find the vertical and horizontal asymptotes to the polar curve

$$r = \frac{\theta}{\pi - \theta} \quad \theta \in [0,\pi]$$

The usual method here is to multiply by $\cos$ and $\sin$ to obtain the parametric form of the curve, derive these to obtain the solutions. However I am not able to solve these equations.

$$x = r \cos \theta = \frac{\theta \cos \theta}{\pi - \theta}$$

$$y = r \sin \theta = \frac{\theta \sin \theta}{\pi - \theta}$$

The derivatives are given by

$$\frac{\text{d}x}{\text{d}\theta} = \frac{\cos(\theta)\pi-\theta\sin(\theta)\pi+\theta^2\sin(\theta)}{(\pi - \theta)^2}$$

$$\frac{\text{d}y}{\text{d}\theta} = \frac{\sin(\theta)\pi-\theta\cos(\theta)\pi+\theta^2\cos(\theta)}{(\pi - \theta)^2}$$

However I am not able to solve any of these two equations, how am I supposed to find the horizontal and vertical tangent to the polar curve?

(I am also supposed to find the asymptote, but I guess I have to save that question for later)

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Did you forget to add the homework tag? Also, you seem to confuse tangents and asymptotes. The question was about both. You were not asked to find the vertical tangent, just to show that there is exactly one such. –  Harald Hanche-Olsen Jan 20 '12 at 10:19
I wrote that post about 4AM, so excuse me for being ambigous. Also added the homework tag =) –  N3buchadnezzar Jan 20 '12 at 10:57

An asymptote is going to mean that $r\rightarrow\infty$, which happens when the denominator approaches zero. Vertical or horizontal means that $\theta$ is a multiple of $\pi/2$ (vertical for even, horizontal for odd multiples). The only asymptote would therefore be when $\theta\rightarrow\pi$, which would be a horizontal one above (and parallel to) the negative $x$ axis.

Now as $\theta\rightarrow\pi$, $$y = r\sin\theta = \frac{\theta\sin\theta}{\pi-\theta} = \frac{\theta\sin(\pi-\theta)}{\pi-\theta} \rightarrow \theta \rightarrow \pi$$ so the horizontal asymptote is $y=\pi$.

For visualization, we can reason that as $\theta$ increases from $0$ to $\pi$, $r$ increases from $0$ to $\infty$, crossing $1$ at the positive $y$ axis. Here are some plots for $\theta\in[0,.6\pi]$ and $[0,.99\pi]$ made with sage (online):

x_coords = [t/(pi-t)*cos(t) for t in srange(0, 0.6*pi, 0.02)]
y_coords = [t/(pi-t)*sin(t) for t in srange(0, 0.6*pi, 0.02)]
list_plot(zip(x_coords, y_coords))


x_coords = [t/(pi-t)*cos(t) for t in srange(0, 0.99*pi, 0.02)]
y_coords = [t/(pi-t)*sin(t) for t in srange(0, 0.99*pi, 0.02)]
list_plot(zip(x_coords, y_coords))


Horizontal and vertical tangents can be found by finding the roots of $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$ respectively (assuming that they are never simultaneously zero, in which case we'd have to resort to higher order derivatives). A simple way to find these roots is to observe that $$r = \left(\frac{\pi}{\theta}-1\right)^{-1} \implies \frac{dr}{d\theta} = -\left(\frac{\pi}{\theta}-1\right)^{-2} \cdot \left(-\pi\theta^{-2}\right) = \frac{\pi}{(\theta-\pi)^2}$$ so that $$0 = \frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta-r\sin\theta \iff \tan\theta = \frac{1}{r} \frac{dr}{d\theta} = \frac{\pi}{\theta(\pi-\theta)}$$ and $$0 = \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta+r\cos\theta \iff -\cot\theta = \frac{1}{r} \frac{dr}{d\theta} = \frac{\pi}{\theta(\pi-\theta)}$$ which show us that they never vanish simultaneously. From the shape of $\frac{\pi}{\theta(\pi-\theta)}$, which is symmetric and positive on $(0,\pi)$ with global minimum at $\frac{\pi}{2}$ and vertical asymptotes at the endpoints, we see that it will meet $\tan\theta$ at exactly one point $\theta_0\in(0,\frac{\pi}{2})$; thus, there is one unique vertical tangent to our curve. From the graph above, we might estimate $$\tan\theta_0\approx\frac{.4}{.25}=1.6 \implies \theta_0 \approx \frac{\pi}{2}-\sqrt{\frac{\pi}{2}\left(\frac{\pi}{2}-\frac{5}{4}\right)} \approx .8609$$ where we used the quadratic equation to deduce $\theta$ from $\tan\theta$. However, the actual solution seems to be closer to $0.97803904765198235$:

t=var('t'); find_root(t*(pi-t)*tan(t)==pi, 0, pi/2)


For horizontal tangents, the same function $\frac{\pi}{\theta(\pi-\theta)}$ must now meet $-\cot\theta$, which is equivalent to the equation $$\tan\theta=\theta\left(1-\frac{\pi}{\theta}\right)$$ which on $[0,\pi]$ has only solutions at the two endpoints, i.e., on our curve, at the origin and at the horizontal asymptote found above.

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Exellent! Now how do I show that the polar curve, has exactly one vertical tangent? I have tried a few various attempts, alas they have all been fruitless –  N3buchadnezzar Jan 20 '12 at 11:23
For $\theta$ in $(0,\frac{\pi}{2})$ & $(\frac{\pi}{2},\pi)$, the graph is in the first & second quadrant respectively (as shown & discussed). I don't think there is any vertical asymptote ($r\rightarrow\infty\iff\theta\rightarrow\pi$). –  bgins Jan 20 '12 at 11:31
Oh sorry, I meant to find the horizontal tangent. It certaintly have atleast one. =) –  N3buchadnezzar Jan 20 '12 at 11:33
The problem text already told you where the horizontal tangent is: At the origin. It also asked you to show that it is the only one, which is challenging, since we are looking at a transcendental equation with all the attendant difficulties. –  Harald Hanche-Olsen Jan 20 '12 at 13:11