Shape of y = 3x +2 in r - \theta space

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Substituting x = $r \cos{ \theta }$ and y = $r \sin \theta$ into the equation y = 3x +2,

I got r = $\frac { -3 \cos{\theta} \pm \sqrt{ 4 – 31 ( \cos{\theta} )^2 } } { 10 ( \cos{\theta} ) ^2 – 1}$Plotting it in Mathematica,

 

[BvU]

And what is your question ?

 

[Ray Vickson]

Homework Statement

View attachment 222295

Homework Equations

The Attempt at a Solution

Substituting x = $r \cos{ \theta }$ and y = $r \sin \theta$ into the equation y = 3x +2,

I got r = $\frac { -3 \cos{\theta} \pm \sqrt{ 4 – 31 ( \cos{\theta} )^2 } } { 10 ( \cos{\theta} ) ^2 – 1}$Plotting it in Mathematica, View attachment 222296

Why don't you just leave the solution $r = r(\theta)$ in terms of both $\sin \theta$ and $\cos \theta$? It is much simpler that way.

 

[BvU]

Unless you want to solicit some comments, of course:

• This looks more like $\theta-r$ space, which is probably fine too.
• Your $r$ are all negative, not so nice...but that's because you took the negative of the denominator
• You plot over the range $[-2\pi,2\pi]$ which is everything twice: too much.
• With every $r$ there can only be either zero or two $\theta$ (a circle intersects a line in zero or two points). Your plot has four $\theta$ for some of the $r$
• From a plot of $y = 3x+2$ you can check that $r = {2\over 3}$ if $\theta = \pi$. Don't see that in your plot.
• Your attempt at solution doesn't show your work, just an outcome -- which may well be wrong

Look at the plot of $y=3x+2$. The half-lines leaving the origin with an angle $\theta$ wrt the positive x-axis don't intersect the line until $\tan \theta > 3$, and also not when $\tan \theta < -3$

 

[Mark44]

Why don't you just leave the solution $r = r(\theta)$ in terms of both $\sin \theta$ and $\cos \theta$? It is much simpler that way.

Yes indeed. Just replace x and y in the original equation, and then solve for r in terms of $\theta$.

This looks more like θ−r space, which is probably fine too

Or as it's usually called, in polar coordinates.

 

[BvU]

Or as it's usually called, in polar coordinates.

Yeah, I realized that usually we express $r$ in $\theta$, so a plot with $\theta$ on the horizontal axis seems a good candidate. But why don't we call it a $\theta, r$ space, then ? Must be a tongue issue (of which I know nothing).

 

[Mark44]

Yeah, I realized that usually we express $r$ in $\theta$, so a plot with $\theta$ on the horizontal axis seems a good candidate. But why don't we call it a $\theta, r$ space, then ? Must be a tongue issue (of which I know nothing).

The positive x- and y-axes of the Cartesian coordinate system correspond to the rays $\theta = 0$ and $\theta = \frac \pi 2$. But there is neither an r-axis nor a $\theta$ axis.

 

[BvU]

Does that mean one can not plot r as a function of theta ?

 

[Mark44]

One other thing, @Pushoam, the graph you show has nothing to do with the line y = 3x + 2. When this equation is converted to polar form, you should get the same shape as this line. In Mathematica, using PolarPlot with your polar equation should produce this line.

 

[BvU]

OP attempted to plot r as a function of theta !

 

[Mark44]

Does that mean one can not plot r as a function of theta ?

No, it doesn't mean that at all. It's just that in polar coordinates you go about things in a different way. In Cartesian (or rectangular) coordinates, you plot the point (1, 1) by going out 1 unit on the pos. x-axis, and then up 1 unit. In polar coordinates, the same point has coordinates $(\sqrt 2, \pi/4)$ (among others). You first look at the ray $\theta = \pi/4$ and then go out $\sqrt 2$ units. For a polar equation $r = f(\theta)$, you pick a value of $\theta$, and then calculate the r-value.

 

[Mark44]

OP attempted to plot r as a function of theta !

But using Cartesian form. I.e., by treating $\theta$ and r as aliases for x and y (using the Mathematica Plot function). Unless I'm completely misunderstanding the purpose of the exercise here, that isn't what should be done.

Here's an example of what I'm saying.
$r = 2\cos(\theta)$
If you graph this in Cartesian form, imagining that r really is y, and that $\theta$ really is x, the graph is a familiar cosine curve with amplitude 2, period $2\pi$, and crossing the vertical axis at (0, 2).
But as a polar equation, what you get is a circle of radius 1 whose center is at (1, 0). IOW, the circle whose Cartesian equation is $(x - 1)^2 + y^2 = 1$. It's relatively easy to convert the Cartesian equation to the corresponding polar equation, and vice versa. Both equations represent the same shape.

 

[BvU]

My interpretation of the exercise is what OP attempted ( but it needs some improvement )
Saying the line is the same in cartesian as in polar is a bit too easy a solution in my view.
@Pushoam, what's your impression ? Or @Ray Vickson ?

And: @Pushoam : show your working! you introduce extra solutions by squaring and miss the simple result Ray is hinting at.

 

[Mark44]

My interpretation of the exercise is what OP attempted ( but it needs some improvement )

It's possible your interpretation is correct, but IMO the problem is poorly worded. WTH is "$r - \theta$ space"? Everyone else in the world calls this a polar coordinate system.

Saying the line is the same in cartesian as in polar is a bit too easy a solution in my view.

Obviously we get a different equation in polar coordinates, and that's not a completely trivial exercise to convert from Cartesian to polar coordinates. Perhaps the question is trickly, and aims to get the student to recognize that even though you have very different equations, the shape of the two curves is the same.

 

[OmCheeto]

...Perhaps the question is trickly, and aims to get the student to recognize that even though you have very different equations, the shape of the two curves is the same.

I found the textbook where this exercise is taken from: "A Student's Guide to Lagrangians and Hamiltonians".
In the introduction, the author says; "Some of the exercises border on the trivial, and are included only to help you to focus on an equation or a concept."

I believe this is a trivial problem to show a concept. The whole chapter was only 1½ pages long. The exercise is on page 14. The context of his instruction may explain why he used the phrase "r -- θ space".

 

[Pushoam]

Substituting x = $r \cos{ \theta }$ and y = $r \sin \theta$ into the equation y = 3x +2,

I got r = $\frac { 2 } { \sin {\theta} - 3 \cos{\theta} }$Plotting it in Mathematica, taking r( $\theta$ ) along y - axis and $\theta$ along x - axis,

How does one get the straight line? Is there a different way to plot in polar space?

As r could be negative, so those values of $\theta$, for which r is negative, are not acceptable. So, I have,

I think I am supposed to do a polar plot with polar axes.

 

[Pushoam]

The question says only to determine the shape of eqn. y = 3x + 2 in polar space, not to plot it.

So, is there any way to determine the shape without plotting it?

 

[OmCheeto]

The question says only to determine the shape of eqn. y = 3x + 2 in polar space, not to plot it.

So, is there any way to determine the shape without plotting it?

I don't think so.

And I think you need to plot it on a graph that looks like the following:

I don't have access to "Mathematica", so I can't help you with that application.

 

[Pushoam]

OP attempted to plot r as a function of theta !

Yes, earlier I thought that the question asks to plot r as a function of $\theta$ . And I plotted it. I thought that taking y - axes as r and x- axes as theta creates the polar space. But, it is not so. Polar space is something different from what I thought.
Then, what is the space in which I plotted r as a function of $\theta$ in post 16?
Did I plot r as a function of $\theta$ in Cartesian space?

Polar space looks like what @OmCheeto has shown in post 18.

 

[Pushoam]

Btw, is it always true that a straight line in one space will remain a straight line in all different kind of spaces?

 

[OmCheeto]

...
Did I plot r as a function of $\theta$ in Cartesian space?
...

You set y = r and x = θ.
The problem is, y ≠ r, and x ≠ θ.
I'm not sure what it's called when you do plots like that. I'm guessing "wrong" would be a good description.

 

[OmCheeto]

Btw, is it always true that a straight line in one space will remain a straight line in all different kind of spaces?

Is that supposed to be a trick question, to see if I really have access to your textbook?

Exercise 1.13 A particle moves along a line of constant r from θ_1 to θ_2.
Show that this is a straight line in r--θ space but not in x--y space.​

 

[Pushoam]

Is that supposed to be a trick question, to see if I really have access to your textbook?

Exercise 1.13 A particle moves along a line of constant r from θ_1 to θ_2.
Show that this is a straight line in r--θ space but not in x--y space.​

Taking above as a polar space, the particle moving from $\theta_1$ to $\theta_2$ with constant r traces a curved path on the circumference of radius r. How is that the book tells it a straight line in polar space?

 

[Let'sthink]

If we take r and θ as polar coordinates and consider theta to have domain [0, infinity} and r too have domain [0, infinity}, then it is all right we can map the total cartesian space into corresponding (r, θ) space without any ambiguity.

 

[BvU]

I believe this is a trivial problem

Beg to differ. In exercise 1.13 the author wants the reader to show that the arc from $(r,\theta_1)$ to $(r,\theta_2)$ is a straight line in $r-\theta$ space. The OP has given the correct answer in #16

I don't have access to "Mathematica",

Wolframalpha is a good alternative; it shows both plots.

 

[BvU]

Btw, is it always true that a straight line in one space will remain a straight line in all different kind of spaces?

In the same way a specific dog is a dog, no matter in what language someone describes the animal.

 

[BvU]

If we take r and θ as polar coordinates and consider theta to have domain [0, infinity} and r too have domain [0, infinity}, then it is all right we can map the total cartesian space into corresponding (r, θ) space without any ambiguity.

There are some fine points to consider when dealing with polar coordinates

 

[BvU]

You set y = r and x = θ.

No. The picture in #16 clearly has theta and r as axis titles.

The problem is, y ≠ r, and x ≠ θ

ordinate, abscissa -- it's just naming. I grant you that confusion can be caused easily.

I'm not sure what it's called when you do plots like that. I'm guessing "wrong" would be a good description.

Nonsense.

I see you found exercise 1.13. Is that 'wrong' too ?

Apologies to Push for these nitpicking helpers -- but instructive it surely is

 

[Mark44]

You set y = r and x = θ.

No. The picture in #16 clearly has theta and r as axis titles.

Because Pushoam used the Mathematica Plot function. This function plots an equation using Cartesian coordinates. In this case it merely relabels y as r, and x as $\theta$, exactly as @OmCheeto said.

The problem is, y ≠ r, and x ≠ θ

ordinate, abscissa -- it's just naming. I grant you that confusion can be caused easily.

Especially as we don't see what the author means by "r - $\theta$ space" in the few pages of the linked-to textbook that we have access to. "r - $\theta$ space" is not standard mathematical terminology, in my experience. If the author of this book has not stated what he means by this terminology, that's a very sloppy oversight.

I'm not sure what it's called when you do plots like that. I'm guessing "wrong" would be a good description.

Nonsense.

I see you found exercise 1.13. Is that 'wrong' too ?

In an actual polar coordinate system, if r is fixed and $\theta$ is allowed to vary, you can an arc along a circle. In this so-called "$r - \theta$ space," where r and $\theta$ are merely aliases for x and y, then yes, you would get a straight line segment, one that is horizontal.

The whole exercise seems very flaky to me, and seems to boil down to this:
1. Start with the equation y = 3x + 2.
2. Convert to polar form: $r\sin(\theta) = 3r\cos(\theta) + 2$
3. Solve for r: $r = \frac 2 {\sin(\theta) - 3\cos(\theta)}$
4. Rename r to y and $\theta$ to x
5. Plot the resulting equation: $y = \frac 2 {\sin(x) - 3\cos(x)}$
Perhaps the author has a point in doing this, but without seeing more of the book than the few pages we have access to, it's not clear to me why we are doing this silly exercise.

 

[BvU]

I agree the exercise isn't up to high standards. I'm pretty convinced this 1.13 tells us what the composer meant and in that context the OP did well in post #16.
We are spending too much energy on it as it is .

BTW, I really like the transform from "author $r-\theta$" to polar (ahem) in the wikipedia picture:

taken from wikipedia (polar coordinates)