Hi guys, I was trying to sketch a polar curve but my curve was different from the curve on maple(I plotted the same curve on maple).(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

Here is the whole question, I am using t as theta.

The hyperbolic spiral is described by the equation rt=a whenever t>0,where a is a positive constant. Using the fact that lim(t->0)sint/t=1,show that the line y=a is a horizontal asymtote to the spiral. Sketch the spiral.

2. Relevant equations

3. The attempt at a solution

Since y=rsint, substituting r=y/sint into rt=a we get y=asint/t. By taking the limit of both sides as t->0 we get y=a. The thing I don't understand is, why is y=a a horizontal asymptote on the polar coordinates? Isn't y=a only an asymptote on the cartesian coordinates of the curve rt=a(when you convert it in terms of x and y)?

Ok so I plotted the polar curve. The difference between my curve and the one on maple is the behavior of the curve as t tends to 0. My curve is a bit like y=1/x as x tends to +infinity(I am talking about the SHAPE of my polar curve as t tends to 0). After that it is just like a spiral as t increases. However, the one shown on maple tends to y=a as t tends to 0(as in, it tends to a horizontal line y=a but my curve tends to the x-axis like y=1/x).

Obviously, both curves (mine and maple's) obey the fact that r tends to infinity as t tends to 0(because rt=a so r=a/t). But why the horizontal asymptote? Am I missing something in here?

Any help would be appreciated.

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# Homework Help: Asymptotes in polar curves?

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