How Are Polar and Cartesian Graphs of \( r = e^{\theta} \) Related?

[gordonj005]

Homework Statement

"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?

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.

 

[obafgkmrns]

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

 

[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.

 

[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.

 

[obafgkmrns]

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

 

[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?

 

[Biljo6985]

There is in fact a beautiful relationship between the trigonometric functions and the exponential function.

See: http://en.wikipedia.org/wiki/Euler's_formula

 

[gordonj005]

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

 

[Dick]

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?

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.

 

[gordonj005]

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