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Polar Coordinates - Graphing the points of when theta<0

  1. Apr 2, 2014 #1
    Polar Coordinates --- Graphing the points of when theta<0

    Hi everyone,

    I'm working with an online graphing program desmos.com. It's great, and actually tons of fun.

    I'm currently working with polar coordinates but the only flaw of this grapher is that when working with polar coordinates, it does not include when theta<0

    ... this is a problem for me because well, it's really only half of the graph!

    the equation I have is r=arctan(theta), and I need to graph the entire thing INCLUDING when theta<0. I figured out how to do it using parametrics, but that seems a bit like taking the easy way out. there MUST be a way to express the points where theta<0 without actually using theta<0 in the domain, no? I tried things like r=arctan(arccos(cos(theta))), but of course this only graphs half a rotation. Anyone have any ideas?


    Other equations i'm having the same problem with:

  2. jcsd
  3. Apr 2, 2014 #2

    Simon Bridge

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    You could exploit: sin(-A)=-sin(A) and cos(-A)=cos(A) etc.
    Is it that it defines angle as only positive anticlockwise from the x axis?

    i.e. you want -180<A<180, then they use 0<A<360?
    (note: the second is standard for polar coords.)

    Then you just need to convert between the definitions.
    if A is negative, then feed the program 360+A perhaps?
  4. Apr 2, 2014 #3
    Hi, thanks for your reply.

    yes. I did try using the fact that cos(-theta) = cos(theta). But this limits the number of rotations I of theta because I have to introduce arccos, right? So I really only get a half rotation. I COULD just shift the graph over to see the negative values of theta, but that doesn't show me ALL the values where theta is negative.

    I'm literally looking to display a graph where the domain of theta appears to be (-infinity, infinity) of the graph r=arccos(theta), but the domain is actually [0, infinity).
    Is there a way I can rearrange/manipulate the equation r=arccos(theta) such that it creates a graph is identical to that of r=arccos(theta) with a domain (-infinity, infinity)?

    is it impossible?
  5. Apr 3, 2014 #4

    Simon Bridge

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  6. Apr 3, 2014 #5
    Unfortunately that won't work for arctan, since arctan(A) ≠ arctan(360 + A).
    However, for each value of θ < 0, we have arctan(θ) = -arctan(-θ) where we now have -θ > 0, since arctan is an odd function. Therefore, r = -arctan(θ) will have the same length as r = arctan(θ) when θ < 0, but we need its argument θ to be 2Pi - θ to agree with the fact that the radial vector is supposed to be at a negative angle when it receives that length. However, -arctan(2Pi-θ) is not the correct length, so we cannot use it for r.
    So unfortunately, I'm afraid you are stuck with the parametric solution of graphing (r, θ) = (-arctan(t), 2Pi - t) to get the part of the graph that corresponds to negative theta values. A similar argument would simplify the other functions you list as well.
    An alternative free Computer Algebra System that includes advanced functionality is Sage. You can even use it online without downloading a thing. The instructions for plotting a polar function are here and it does not restrict theta to only positive values.
  7. Apr 3, 2014 #6

    Simon Bridge

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    That's why I didn't suggest it for arctan ;)

    ... your use of theta suggests you are still thinking of the argument of the inverse trig function as an angle - it's not an angle: arctan(A) is an angle, A is the length of the opposite side in units of the adjacent side.

    r=arctan(A), and s=arctan(-A), then tan(s)=-A and tan(r)=A, so tan(s)=-tan(r).
    but careful: -pi/2 < r,s < \pi/2 ... but from there you can get the infinite set of solutions.

    I'm using octave and I don't get restricted to positive values for the argument of inverse trig functions.
  8. Apr 3, 2014 #7
    Hi Simon! Actually, I'm using θ as the standard notation for the second coordinate in the polar coordinate system, which is, in fact, an angle. It is not meant to imply that the intended argument of the the function f(x) = arctan(x) is an angle, any more than the equation r = |θ| might imply the standard argument of the absolute value function is meant to be an angle. It is merely an artifact of the chosen coordinate system. :)
    In particular, the standard domain of arctan, the value of which may be represented at each point by arctan(x), arctan(θ), arctan(ε) or so on remains the same, regardless of the interpretation of the dummy variable used to represent its input.
    That is to say, the equations r = arctan(θ) and y=arctan(x) represent the same sets of input-output pairs, albeit one being only by historical convention printed in the reverse order of (output, input): the particular symbols used for input and output merely used by us to interpret how we want to represent the data in a coordinate system.
    Last edited: Apr 3, 2014
  9. Apr 3, 2014 #8

    Simon Bridge

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    Assuming that the argument of the inverse trig function is the polar coordinate in question ... but is that the case?
    Under what circumstances would you take the arctan of the second polar coordinate?

    i.e. converting from cartesian: ##\theta = \arctan(y/x)## and ##r=\sqrt{x^2+y^2}##

    so ##\arctan\theta = \arctan\big( \arctan(y/x)\big)## ... what could this mean?

    The variable you choose is also, in context, a form of communication ... it has an implied/inferred meaning that may be important to the application and gives the reader clues to what is going on.

    It's actually important to the problem in hand - and is partly why I think adamjts (OP) should look again at what is to be achieved. I suspect adamjts would benefit from a different approach to whatever problem results in trying to get inverse trigs outside their defined range. It may even indicate a mistake or a misunderstanding of the problem but there is simply not enough information to be able to tell.

    But if everything is fine, then there is always the two-argument version atan2(y/x) here:
    ... aside from that I'm inclined to agree that parameterizing the curve is the way to go... the reason the arc-trig functions are the way they are is because, extending to all angles would mean they are no longer functions.

    It's just like how you have to parameterize a circle to plot all of it in one go.
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