1. 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.

2. 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.

3. 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?