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Math Help - hyperbolic spiral

  1. #1
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    hyperbolic spiral

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

    Basically, all i've done is a sketch of r=\frac{a}{\theta} which isnt of great help. Ive also tried multiplying both sides by \sin\theta to give 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.
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  2. #2
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    Quote Originally Posted by vuze88 View Post
    The hyperbolic spiral is given by the formula (in polar form), r\theta=a for \theta>0 and some postive constant a>0. Use the fact that \lim_{\theta\rightarrow 0}\frac{\sin\theta}{\theta}=1 to prove there is a horizontal asympotote at y=a and hence sketch the curve.

    Basically, all i've done is a sketch of r=\frac{a}{\theta} which isnt of great help. Ive also tried multiplying both sides by \sin\theta to give 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 \theta goes to 0, of \frac{sin(\theta)}{\theta} is 1, y= \frac{a sin(\theta)}{\theta} goes to a. y= a is a horizontal asymptote.
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  3. #3
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    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?
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