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Polar Coordinates Problem

  1. Nov 17, 2011 #1
    1. The problem statement, all variables and given/known data

    "Consider the graph of [itex]r = e^{\theta}[/itex] in polar coordinates. Then consider the graph of [itex](\theta \cos{\theta}, \theta \sin{\theta})[/itex] where [itex]\theta \in \mathbb{R}[/itex] on the Cartesian plane (x - y axis). How are the two graphs related? What relationship (if any) can we define between [itex]e^{\theta}[/itex] and the trigonometric functions?

    3. The attempt at a solution

    What I considered was [itex]r^2 = x^2 + y^2[/itex] where [itex]x = \theta \cos{\theta}[/itex] and y = [itex]\theta \sin{\theta}[/itex]. Plugging this all in I get:

    [tex] e^{2\theta} = (\theta)^2 ((\cos{\theta})^2 + (\sin{\theta})^2) [/tex]

    which reduces to:

    [tex] e^{2 \theta} = (\theta)^2 [/tex]

    taking the ln of both sides, noting that [itex]\theta \ne 0[/itex]:

    [tex] 2 \theta = 2 ln \theta [/tex]

    [tex] \theta = ln \theta [/tex]

    So as it stands now, the above equation has no real solutions. So I thought maybe putting each side as a power of e would be the relation between the two graphs.

    [tex] e^{\theta} = e^{ln \theta} [/tex]

    [tex] e^{\theta} = {\theta} [/tex]

    which is kind of a circluar argument because I just rearranged the equation. They mention this has something to do with trigonometric functions, I'm not seeing the connection. I would apprectiate some help.
     
  2. jcsd
  3. Nov 17, 2011 #2
    What you have done is define two equations, [tex]r^2 = \theta^2[/tex] and [tex]r = e^\theta[/tex]and then solve them simultaneously. The value of θ satisfying [tex]e^\theta = \theta[/tex]is simply where the two graphs intersect.
     
  4. Nov 17, 2011 #3
    Yes, I realize this, but there are no real solutions to that equation. And also it doesn't tell me much about the relation between the two functions.
     
  5. Nov 17, 2011 #4

    Dick

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    I have no idea about the rest of the problem, but I can tell you one thing. The curves do intersect. Just plot them. Show there is a solution to exp(t)=t+2*pi. You can find the root numerically if you want. Call it c. Now put c into the exponential and c+2*pi into the other curve. You'll get the same x,y coordinates. There are an infinite number of other similar solutions.
     
    Last edited: Nov 17, 2011
  6. Nov 17, 2011 #5
    Oops, I omitted the negative sign! [tex]e^\theta = \theta[/tex] has no real roots, but [tex]e^\theta = -\theta[/tex] does!
     
  7. Nov 17, 2011 #6
    Ah right you are, I thought that seemed weird that there were real intersection points but no solution. One thought i had was that if you take successive derivatives of [itex]r = e^{\theta}[/itex] you will get a tighter and tigher spiral. Eventually you should get the parametric function: [itex] (\theta \cos{\theta}, \theta \sin{\theta})[/itex]. Any thoughts?
     
  8. Nov 17, 2011 #7
  9. Nov 17, 2011 #8
    I've seen that before, its very nice. But does it have an application here?
     
  10. Nov 17, 2011 #9

    Dick

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    Derivatives with respect to what? That doesn't sound right. I still don't quite get the point of this problem, I share that with you.
     
  11. Nov 18, 2011 #10
    Derivatives with respect to [itex]\theta[/itex]. I'm not sure, at this point i'm just grasping at answers. Any other thoughts?
     
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