# Polar Coordinates Problem

1. Nov 17, 2011

### gordonj005

1. The problem statement, all variables and given/known data

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

3. The attempt at a solution

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

$$e^{2\theta} = (\theta)^2 ((\cos{\theta})^2 + (\sin{\theta})^2)$$

which reduces to:

$$e^{2 \theta} = (\theta)^2$$

taking the ln of both sides, noting that $\theta \ne 0$:

$$2 \theta = 2 ln \theta$$

$$\theta = ln \theta$$

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.

$$e^{\theta} = e^{ln \theta}$$

$$e^{\theta} = {\theta}$$

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. Nov 17, 2011

### obafgkmrns

What you have done is define two equations, $$r^2 = \theta^2$$ and $$r = e^\theta$$and then solve them simultaneously. The value of θ satisfying $$e^\theta = \theta$$is simply where the two graphs intersect.

3. Nov 17, 2011

### gordonj005

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.

4. Nov 17, 2011

### Dick

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
5. Nov 17, 2011

### obafgkmrns

Oops, I omitted the negative sign! $$e^\theta = \theta$$ has no real roots, but $$e^\theta = -\theta$$ does!

6. Nov 17, 2011

### gordonj005

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 $r = e^{\theta}$ you will get a tighter and tigher spiral. Eventually you should get the parametric function: $(\theta \cos{\theta}, \theta \sin{\theta})$. Any thoughts?

7. Nov 17, 2011

### Biljo6985

8. Nov 17, 2011

### gordonj005

I've seen that before, its very nice. But does it have an application here?

9. Nov 17, 2011

### Dick

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.

10. Nov 18, 2011

### gordonj005

Derivatives with respect to $\theta$. I'm not sure, at this point i'm just grasping at answers. Any other thoughts?