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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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up vote 1 down vote accepted

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))

enter image description here

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))

enter image description here

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

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