# hyperbolic spiral

• Feb 7th 2010, 12:26 AM
vuze88
hyperbolic spiral
The hyperbolic spiral is given by the formula (in polar form), $\displaystyle r\theta=a$ for $\displaystyle \theta>0$ and some postive constant $\displaystyle a>0$. Use the fact that $\displaystyle \lim_{\theta\rightarrow 0}\frac{\sin\theta}{\theta}=1$ to prove there is a horizontal asympotote at $\displaystyle y=a$ and hence sketch the curve.

Basically, all i've done is a sketch of $\displaystyle r=\frac{a}{\theta}$ which isnt of great help. Ive also tried multiplying both sides by $\displaystyle \sin\theta$ to give $\displaystyle y=r\sin\theta=\frac{a\sin\theta}{\theta}$ but im unsure how complete the question and the sketch. Can someone give me some help please.
• Feb 7th 2010, 01:33 AM
HallsofIvy
Quote:

Originally Posted by vuze88
The hyperbolic spiral is given by the formula (in polar form), $\displaystyle r\theta=a$ for $\displaystyle \theta>0$ and some postive constant $\displaystyle a>0$. Use the fact that $\displaystyle \lim_{\theta\rightarrow 0}\frac{\sin\theta}{\theta}=1$ to prove there is a horizontal asympotote at $\displaystyle y=a$ and hence sketch the curve.

Basically, all i've done is a sketch of $\displaystyle r=\frac{a}{\theta}$ which isnt of great help. Ive also tried multiplying both sides by $\displaystyle \sin\theta$ to give $\displaystyle y=r\sin\theta=\frac{a\sin\theta}{\theta}$ but im unsure how complete the question and the sketch. Can someone give me some help please.

Excellent! Now use the hint that they gave you: since the limit, as $\displaystyle \theta$ goes to 0, of $\displaystyle \frac{sin(\theta)}{\theta}$ is 1, $\displaystyle y= \frac{a sin(\theta)}{\theta}$ goes to a. y= a is a horizontal asymptote.
• Feb 7th 2010, 02:39 AM
vuze88
hm...i was thinking that, but dont you get asymptotes when x approaches infinite? if we use the limit they give you, we are making theta approach 0. Unless x approaches infinite as theta apporaches 0 but how do we know this?