Chapter H.1, Problem 46E

To determine

To show: the given curve has the line $x=1$ as a vertical asymptote and also prove that the curve lies entirely within the vertical strip $0\le x<1$ .

Explanation of Solution

Given:

Curve: $r=\sin \theta \tan \theta$

Line: $x=1$

Proof:

If $x=1$ is an asymptote we will want to rewrite the equation so that thereis only $randx$ left.

  $\begin{array}{l}r=\sin \theta \tan \theta =\sin \theta \frac{\sin \theta }{\cos \theta }=\frac{{\sin }^2\theta }{\cos \theta }=\frac{1-{\cos }^2\theta }{\cos \theta }\\ Sincex=r\cos \theta ,then\cos \theta =\frac{x}{r}\\ r=\frac{1-{\left(\frac{x}{r}\right)}^2}{\frac{x}{r}}=\left(1-\frac{x^2}{r^2}\right)\frac{r}{x}\\ \frac{x}{r}r=1-\frac{x^2}{r^2}\\ x+\frac{x^2}{r^2}=1\\ r^{2}x+x^2=r^2\\ x^2+r^{2}x-r^2=0\\ x=\frac{-r^2\pm \sqrt{{\left(r^2\right)}^2-4\left(1\right)\left(-r^2\right)}}{2\left(1\right)}=\frac{-r^2\pm \sqrt{r^4+4r^2}}{2}\end{array}$

Evaluate the limit. We can ignore the negative square root because thatwould make $x<0$ , which is would not be involved in an $x=1$ asymptote.Start by multiplying by the conjugate.

  $\begin{array}{l}\underset{r\to \pm \infty }{\lim }\frac{-r^2+\sqrt{r^4+4r^2}}{2}.\frac{-r^2-\sqrt{r^4+4r^2}}{\frac{-r^2-\sqrt{r^4+4r^2}}{2}}=\underset{r\to \pm \infty }{\lim }\frac{r^4-\left(r^4+4r^2\right)}{2\left(r^2+\sqrt{r^4+4r^2}\right)}\\ \underset{r\to \pm \infty }{\lim }\frac{-4r^2}{-2\left(r^2+\sqrt{r^4+4r^2}\right)}.\frac{\frac{1}{r^2}}{\frac{1}{r^2}}\\ =\underset{r\to \pm \infty }{\lim }\frac{2}{\left(2+\sqrt{\frac{1}{x^4}\left(r^2+4r^2\right)}\right)}\\ =\underset{r\to \pm \infty }{\lim }\frac{2}{\left(1+\sqrt{1+\frac{4}{r^2}}\right)}\\ =\frac{2}{\left(1+\sqrt{1+0}\right)}=\frac{2}{2}=1\end{array}$

  $\frac{-r^2+\sqrt{r^4+4r^2}}{2}$ and therefore $x$ approaches $1$ as $r\to \pm \infty$

This means $x=1$ is a vertical asymptote.

This may be a bit confusing to graph so we could look at the Cartesian graphfirst.

  

  $tan\theta$ is undefined at $\theta =\frac{\pi }{2}+n\pi$ ( $n$ is an integer} because cos at thosevalues equals 0.

For $0\le \theta <\frac{\pi }{2}$ , $r$ starts at $0$ and goes to $\infty$ . So the curve is in quadrant $1$ and between $0\le x<1$ for this interval and approaches $x=1$ from the left,similar to how it did in some of the problems just before this one.

For $\frac{\pi }{2}<\theta \le \pi$ , $r$ comes from $-\infty$ from the left of $x=1$ and goes to $0$ . $\theta$ is in quadrant $2$ in polar coordinates while $r$ is negative for this intervalso the curve is in quadrant $4$ , between $0\le x<1$

For $\pi \le x<\frac{3\pi }{2}$ , $r$ goes from $0$ back to $-\infty$ from the left of $x=1$ . $\theta$ is in quadrant $3$ in polar coordinates while $r$ is negative for this interval sothe curve is in quadrant $1$ , between $0\le x<1$ . This part of the curve wasdrawn earlier already so the period seems to be $\pi$ .

For $\frac{3\pi }{2}<\theta \le 2\pi$ , $r$ comes from $-\infty$ from the left of $x=1$ and goesto $0$ . $\theta$ is in quadrant $4$ in polar coordinates while $r$ is positive for thisinterval so the curve is in quadrant $4$ , between $0\le x<1$ . Again, this partof the curve was drawn earlier already.

The curve repeats for greater $\theta$ after this, and never goes outside of $0\le x<1$ .

  

Conclusion:

Therefore, the curve repeats for greater $\theta$ after this, and never goes outside of $0\le x<1$ .

The given curve has the line $x=1$ as a vertical