Consider a curve given in polar coordinates by $r(θ)=\frac{1}{1+e\cos\theta}$, where $e≥0$.

When $e>1$, show that the curve approaches two asymptotes, find them and sketch the curve. Hint: If the critical angles are $\pm\theta_0$, compute the vertical distance of the point of the curve at angle $\theta=θ_0+h$ to the line $θ=θ_0$, and take a limit using Taylor approximations.

Not really sure where to begin, I could get $y=\frac{\sin(\theta_0+h)}{1+e\cos(\theta_0+h)}$ and $y=(\tan\theta_0)x$ and then distance is $$\frac{\sin(\theta_0+h)}{1+e\cos(\theta_0+h)}-(\tan\theta_0)x$$

is this where I'm supposed to take limits?

  • $\begingroup$ I believe what it is asking you to do is make a linear approximation of the asymptote. It is technically a Taylor series but you could also just realize that $dr = \frac{dr}{d\theta}d\theta \rightarrow \Delta r \approx \frac{dr}{d\theta}\Delta \theta$. Integrating gives: $r(\theta + \Delta \theta ) \approx r(\theta) + \frac{dr}{d\theta} \Delta \theta$. In the linear approximation--i.e. the first non-constant term in a Taylor expansion. $\endgroup$ – Jared Apr 1 '15 at 4:09
  • $\begingroup$ Would this still work even though it's in polar coordinates? And shouldn't the asymptote depend on $e$. ie the slope is steeper if $e$ is bigger? $\endgroup$ – math_R Apr 1 '15 at 12:18
  • $\begingroup$ Frankly, I don't understand what calculus is involved here. You simply find the vertical asymptotes to $r$ which occur when $1 + e\cos(\theta_0) = 0 \rightarrow \cos\theta_0 = -\frac{1}{e}$. There are no solutions when $\big|e\big| < 1$. The angle tells you the slope: $\frac{\Delta y}{\Delta x} = \frac{r\sin(\theta_0)}{r\cos(\theta_0} = \tan(\theta_0)$. $\endgroup$ – Jared Apr 1 '15 at 19:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.