Is the curve r = tan(t) defined at t = 0?

  • #1
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Homework Statement:
Consider the curve ##r=\tan t## for ##t\in[0,2\pi]##. Substituting ##t=0## into this equation implies that ##r=0##. We convert this into rectangular coordinates.

We get the curve ##\sqrt{x^2+y^2}=\frac{y}{x}##, which is oddly undefined for some values of ##(x,y)## satisfying the original equation, even though the polar parametrization was defined for all values of ##t##. Squaring both sides of the equation corrects this problem for the most part.

The line ##y=0## intersects the curve at ##x=0##. On the other hand, the line ##y=mx##, where ##m\neq0##, intersects the curve at exactly two non-zero points. This implies that the curve is not defined at the origin.

So the questions:
(1) Why is the rectangular representation of the curve not 1-1 w.r.t. the polar representation?
(2) Why does the algebra suggest that any line with non-zero slope does not intersect the curve at the origin?
Relevant Equations:
##x=r\cos t##
##y=r\sin t##
The curve ##\sqrt{x^2+y^2}=(\frac{y}{x})## is not defined for points ##(x,y)## in the second and fourth quadrants.

Consider the transformed curve ##x^2+y^2=(\frac{y}{x})^2##.

If ##y = 0##, then ##x^2+0^2=\frac{0^2}{x^2}=0##, which means that ##x=0##.

Along the line ##y=mx##, where ##m\neq0##, ##x^2(m^2+1)=m^2## is not defined if ##x=0##. This equation also suggests that the line intersects the curve at exactly two points.

I thought of something that suggests that the curve might not be defined at the origin. In the parametrization ##r=\tan t=\frac{r\sin t}{r\cos t}##, I think that it is implicitly assumed that ##r\neq0##. But plugging in ##t=0## contradicts this.

You can find a rough sketch of the curve attached to this post.
 

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Answers and Replies

  • #2
Delta2
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Cant answer your questions directly, however your curve (in polar form) is not defined for all values of ##t\in (0,2\pi)##, it is not defined for $$t=\frac{\pi}{2},\frac{3\pi}{2}$$.
 
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  • #3
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Whoops, I must have forgotten to mention that.

Also, I think I already figured out the answer to my second question: it's not defined at the origin. I assumed that it was because it seemed to be defined at the origin in polar coordinates, when in fact, it is not, for the reasons stated in the opening post. And the rectangular representation is not defined at the point ##(0,0)##.
 
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  • #4
andrewkirk
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We need to be precise if we want to resolve apparent inconsistencies.
First note that what you have described is not a curve, if we take the usual meaning of curve to be a map from a real interval to a topological space. The domain of the function you have given is ##D\triangleq [0,\pi/2)\cup(\pi/2,3\pi/2)\cup(3\pi/2,2\pi)##, which is a union of three intervals, not a single interval.

So we can talk about the function ##f:D\to\mathbb R^2## such that ##f(t)=(\tan(t), t)## in polar coordinates.

Or you could think about it as two curves ##g:(\pi/2,3\pi/2)\to\mathbb R^2## and ##h:(3\pi/2,5\pi/2)\to\mathbb R^2## whose images are respectively the lower and upper loops in your diagram.

Note that in set-theoretic terms ##f = g\cup h##.

Next, notice that your equation in ##x## and ##y## does not specify a curve either, as it is not parametrised. It could specify the image of a curve, otherwise known as a path (there is a many-to-one relationship between curves and paths). However, it doesn't specify the image of function f, because it omits the point (0,0). So a correct specification of the image of ##f## is:
$$\{(x,y)\in \mathbb R^2\ |\ x^2+y^2=(y/x)^2\vee x=y=0\}$$

The statement about the rectangular "curve" note being defined in the 2nd and 4th quadrants is incorrect, because we can take the negative square root to satisfy the equation. Better to make this explicit by squaring both sides to remove the square root sign. I did that in the definition of the set in the previous paragraph.

This equation also suggests that the line intersects the curve at exactly two points.
Again, this misses the fact that (0,0) belongs to each of the images of the two curves g and h. In fact a straight line through the origin intersects each of g and h twice, once at the origin and once elsewhere. So it intersects the image of f three times, one of which is at the origin.
 
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  • #6
Delta2
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Another problem I found with the polar form is that ##r## has to be positive and ##\tan\theta## is positive only for ##\theta \in(0,\frac{\pi}{2}),(\pi,\frac{3\pi}{2})## so the curve is only defined for those values of theta.
Your diagram actually depicts the curve $$r=|\tan\theta|$$
 
  • #7
haruspex
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Even wolfram bugs for this curve, it plots (in the cartesian form) quite different curve than your figure (which i agree is the correct representation of the curve).

https://www.wolframalpha.com/input/?i=plot+x^2+y^2=(y/x)^2
I don't understand why you think the Wolfram curve wrong. Looks fine to me.

I did worry that posing it in that Cartesian form might create extraneous solutions. It turns out not to alter the locus, but in an unexpected way.
I would interpret a negative r as, effectively, a rotation through 180°. I.e. (-1,pi/4) is the same point as (1,5pi/4).
Increasing theta steadily from 0 to 2pi, the curve runs from the origin up the branch in the first quadrant, returns to the origin via the branch in the fourth quadrant, off to infinity again in the third quadrant, then back to the origin in the second quadrant.

As regards whether the curve is defined at the origin, we first need agree what that means. It is not the same as saying some particular functional way to represent the curve is not defined there, and a curve is not a single point.
Is (0,0) a solution to ##r=\tan(\theta)##? Yes. Does it form part of the continuous curve of other solutions? Yes.
 
  • #8
Delta2
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I don't understand why you think the Wolfram curve wrong. Looks fine to me.
Because it goes at infinity at ##x=+-1## while ##r=\tan\theta## goes to infinity for ##\theta=\frac{\pi}{2},\frac{3\pi}{2}## or equivalently for x=0.
 
  • #9
haruspex
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Because it goes at infinity at ##x=+-1## while ##r=\tan\theta## goes to infinity for ##\theta=\frac{\pi}{2},\frac{3\pi}{2}## or equivalently for x=0.
I see no contradiction. ##x=r\cos(\theta), r=\tan(\theta), x=\sin(\theta)##. As ##\theta\rightarrow\pi/2, r\rightarrow\infty, x\rightarrow 1##.
 
  • #10
Delta2
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I see no contradiction. ##x=r\cos(\theta), r=\tan(\theta), x=\sin(\theta)##. As ##\theta\rightarrow\pi/2, r\rightarrow\infty, x\rightarrow 1##.
ehm but the point $$(r,\theta)=(\infty,\frac{\pi}{2})$$ is on y axis isn't it. It isn't at x=1.
 
  • #11
haruspex
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ehm but the point $$(r,\theta)=(\infty,\frac{\pi}{2})$$ is on y axis isn't it. It isn't at x=1.
I showed in post #9 that it is on x=1. Why should it be on the y axis?
 
  • #12
Delta2
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I showed in post #9 that it is on x=1. Why should it be on the y axis?
Because that's the basic interpretation of the point ##(r,\theta)##. r is the distance from the origin and ##\theta## is the angle with the x-axis.
 
  • #13
Delta2
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ok I think I understand whats going on here, because ##r=\infty## some trouble arises...

Ok, lets see. For any ##r## finite, the point ##(r,\frac{\pi}{2})## is on the y-axis, I think @haruspex you will agree with that.

However if ##r## is infinite the point ##(r,\frac{\pi}{2})## can be anywhere (in the first quadrant)...
 
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  • #14
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Homework Statement:: Consider the curve ##r=\tan t## for ##t\in[0,2\pi]##. Substituting ##t=0## into this equation implies that ##r=0##. We convert this into rectangular coordinates.

We get the curve ##\sqrt{x^2+y^2}=\frac{y}{x}##, which is oddly undefined for some values of ##(x,y)## satisfying the original equation, even though the polar parametrization was defined for all values of ##t##.
My two pennies about graphing the images of implicit equations: The graph of an implicit equation does not always correspond to the image of the parametric form, but it does sometimes

##\sqrt{x^2+y^2}=\frac{y}{x}\Rightarrow x\sqrt{x^2+y^2}=y\Rightarrow x^2(x^2+y^2)=y^2##

The righthand side equation represents the entire image of ##r=tan(t)## for ##t\in[0,2\pi]##


Another problem I found with the polar form is that ##r## has to be positive and ##\tan\theta## is positive only for ##\theta \in(0,\frac{\pi}{2}),(\pi,\frac{3\pi}{2})## so the curve is only defined for those values of theta.
Your diagram actually depicts the curve $$r=|\tan\theta|$$

I'm scared I could antagonize @Delta2 because he is eons better at doing mathematics than me
nailbiting.png


##(x^2+y^2)=\frac{y^2}{x^2}\Rightarrow (\pm\sqrt{x^2+y^2})=\frac{(\pm y)}{(\pm x)} \Rightarrow \text{ r can be negative.##
 
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  • #15
Delta2
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In polar coordinates ##r=+\sqrt{x^2+y^2}## (and ##\theta\in[0,2\pi]##), ##r## simply cannot be equal to ##-\sqrt{x^2+y^2}##
 
  • #16
haruspex
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Because that's the basic interpretation of the point ##(r,\theta)##. r is the distance from the origin and ##\theta## is the angle with the x-axis.
That still does not mean it converges to the y axis.
Consider any curve for which ##y/x\rightarrow+\infty## as ##x\rightarrow+\infty##. ##y=x^2## will do. In polar, that has ##r\rightarrow+\infty## as ##\theta\rightarrow+\pi/2##.
 
  • #17
andrewkirk
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In polar coordinates ##r=+\sqrt{x^2+y^2}## (and ##\theta\in[0,2\pi]##), ##r## simply cannot be equal to ##-\sqrt{x^2+y^2}##
You state that as if it is an axiom of the system we are using. It is not.

There is a context in which that statement is true, which is when we convert from rectangular to polar coordinates. In that case we define ##r## to equal the positive square root of ##x^2+y^2##.

The current context is different - conversion from polar to rectangular coordinates. And the OP implicitly allows negative values of ##r## by its choice of domain. Since we can make a perfectly logical interpretation of negative r values in polar coordinates*, that introduces no inconsistencies. There's no barrier to doing that. Whether one allows negative r values when using polar coordinates is purely a matter of choice and convention.

Also, there's nothing wrong with the Wolfram plot, as post #5 suggests. It's just a part of the plot that the OP drew. The part is small enough to not show the lines curving back towards the y axis. It's completely consistent with the OP's drawing. I just noticed that @haruspex has already pointed out that the wolfram plot is fine.

The mathematical objects involved are perfectly straightforward and consistent as long as we are careful with our definitions and terminology, and don't do things like include illegal points in our domain or call something a function or a curve when it is not.

* We just interpret r as the signed distance travelled in direction theta, to get from the origin to the point, or say that the location vector of the point is r times ##\hat{\mathbf{\theta}}## the unit vector in direction theta.
 
  • #18
haruspex
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there's nothing wrong with the Wolfram plot, as post #5 suggests. It's just a part of the plot that the OP drew. The part is small enough to not show the lines curving back towards the y axis.
No, the arms will not curve back towards the y axis. That is the same mistaken assumption @Delta2 is making. x=1 is an asymptote.
 
  • #19
andrewkirk
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No, the arms will not curve back towards the y axis. That is the same mistaken assumption @Delta2 is making. x=1 is an asymptote.
Ah yes, right you are. I had just accepted the drawing in the OP without checking, as it seemed plausible and wasn't the main point under discussion.

As ##\theta\to \pi/2## we have ##y=r\sin\theta = \tan\theta\sin\theta\to +\infty## and ##x=\tan\theta\cos\theta=\sin\theta\to 1## since ##\sin\frac\pi 2 = 1##.
 
  • #20
haruspex
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In polar coordinates ##r=+\sqrt{x^2+y^2}## (and ##\theta\in[0,2\pi]##), ##r## simply cannot be equal to ##-\sqrt{x^2+y^2}##
As noted above, it is reasonable to interpret ##(-r,\theta)## as equivalent to ##(r,\theta+\pi)##.
We get the same using the polar representation of the complex plane: ##re^{i(\theta+\pi)}=-re^{i\theta}##, and we are accustomed to ##(r,\theta)## being equivalent to ##(r,\theta+2\pi)##.
 
  • #21
Delta2
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Sorry @haruspex and @andrewkirk I cannot understand how r can be negative (and yes I am using it as some sort of axiom). I ll try to reread your posts.
 
  • #22
haruspex
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Sorry @haruspex and @andrewkirk I cannot understand how r can be negative (and yes I am using it as some sort of axiom). I ll try to reread your posts.
It is a matter of choice. In the given equation ##r=\tan(\theta)## you can choose to insist that r is nonnegative. In that case the excursions of the Wolfram curve into the second and fourth quadrants are invalid; they are spurious solutions resulting from conversion to Cartesian and eliminating the surd by squaring up.
Or you can choose to interpret a negative r in the way I described. No contradiction arises.

Edit: did a quick google of "negative r polar coordinate". All the authoritative-looking sites I saw allow negative r.

But to me this is the minor point here. The important one is that the curves do not swing back towards the y axis as r tends to infinity. The asymptotes are at ##x=\pm 1##.
 
  • #23
Delta2
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But to me this is the minor point here. The important one is that the curves do not swing back towards the y axis as r tends to infinity. The asymptotes are at x=±1.
Yes after careful consideration I agree to that as well.
 
  • #24
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@Eclair_XII

Hi

Click this link link to the Cartisian and Polar Grapher.

Type "tan(t)" into the Cartisian coordinates.

Move the slider to trace the image.

Enjoy
 

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