# Homework Help: Asymptotes in polar curves?

1. Jun 8, 2010

### sakodo

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).

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.

2. Jun 8, 2010

### vela

Staff Emeritus
An asymptote is an asymptote. It doesn't depend on how you represent it. You could write it equivalently as y=a or r sin θ=a.
Without knowing what you did to plot it, it's hard to say anything other than you did it wrong somehow. Your own analysis clearly shows that as θ approaches 0, y should go to a.

You have two competing factors. As θ goes to 0, the point wants to move toward the x axis. However, as θ>0, increasing r will move you away from the x-axis, so you can always get to a point any distance away from the x axis if you make r big enough. With your particular function, r apparently grows fast enough so it overcomes the tendency of a decreasing θ pushing the graph toward the x-axis.

Last edited: Jun 8, 2010
3. Jun 8, 2010

### sakodo

Thanks, that cleared up my confusion. I was thinking that the same curve might be different in polar coordinates and cartesian coordinates lol. So when a curve is converted into polar form and plotted, the shape of the curve should still be the same right?

Sorry I don't quite understand this. What did you mean by " r apparently grows fast enough so it overcomes the tendency of a decreasing θ"?

Anyway, thanks for your help. I understand what I did wrong now, just a bit confused about your explanation.

4. Jun 8, 2010

### vela

Staff Emeritus
Right.
If you hold r constant, as θ goes to 0, you'd be moving toward the x axis. On the other hand, if you hold θ constant and increase r, you'd move away from the x axis. With the function r=a/θ, r increases, which wants to move you away from the x axis, while θ gets smaller, which wants to move you toward the x axis. What actually happens depends on how they change relative to each other. In this case, the effect of r increasing wins out, so the function approaches y=a instead of y=0.