- #36
Mark44
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In the third plot, it's really plotting the equation ##y = \frac 2 {\sin(x) - 3\cos(x)}##, but with the axes labeled as r (for y) and ##\theta## (for x). I don't know what you did in the fourth graph.
It's x plotted as r, and y plotted as ##\theta##, on a polar graph.Mark44 said:...I don't know what you did in the fourth graph.
What was the equation you plotted? This doesn't look like the polar graph of ##r = 3\theta + 2##. That graph would be a spiral.OmCheeto said:It's x plotted as r, and y plotted as ##\theta##, on a polar graph.
It's just naming things. However, I only see black and blueOmCheeto said:This looks like plotting cats on the orange line, and dogs on the apples line
expect only one x=0 and one y=0OmCheeto said:It's x plotted as r, and y plotted as ##\theta##, on a polar graph
Can't remember now.Mark44 said:What was the equation you plotted? This doesn't look like the polar graph of ##r = 3\theta + 2##. That graph would be a spiral.
Mark44 said: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.
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.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.
Mark44 said:Sure, let's find out.OmCheeto said:I have his email address. I could ask him.
OmCheeto said:Hi [Om],
Thanks for your email...I never expected to ignite a firestorm when I wrote that exercise. I just meant for students to appreciate that in changing coordinate systems, "Straight lines may become curves and angles and distances may change...[but]...a point is still a point and a curve is still a curve."[Author]
The solution is obtained by writing x=r cos θ and y=r sin θ and obtaining r as a function of θ . I obtained r = 2/(sin (theta) - 3 cos(theta) )
Good enough for me!
Pretty much the objection I wrote earlier in this thread. With a change of variables similar to what LCKurtz described, you have a new coordinate system and new axes. Changing to polar coordinates, though, you don't have new axes, as there is no "r axis" and no "##\theta## axis."LCKurtz said:It's just that for a polar coordinate change of variable we don't normally look at the geometry in that way, and I don't personally think the problem is helpful given that fact.
What I understood:OmCheeto said:Hi [Om],
Thanks for your email...I never expected to ignite a firestorm when I wrote that exercise. I just meant for students to appreciate that in changing coordinate systems, "Straight lines may become curves and angles and distances may change...[but]...a point is still a point and a curve is still a curve."[Author]
The solution is obtained by writing x=r cos θ and y=r sin θ and obtaining r as a function of θ . I obtained r = 2/(sin (theta) - 3 cos(theta) )
Good enough for me!