Why is y=a a horizontal asymptote on the polar coordinates?

[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).

Homework Statement

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.

Homework Equations

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.

 

[vela]

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).

Homework Statement

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.

Homework Equations

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)?

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.

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?

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.

 

[sakodo]

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.

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?

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.

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.

 

[vela]

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?

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.

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.