# Scalar fields

1. Mar 24, 2007

### Nibbler

Ok, I've been trying to work this out for a couple of hours now and I'm completely stumped. Not even Google was helping much.

The question is:
Find and Sketch the level curves of the scalar field T(x,y) = (x +y)/(x2 + y2) for T = -1, -0.5, 0, 0.5, 1

I know that I should equate the equation to the values of T given and I suspect that I'll be getting 4 circles but it doesn't seem to work.

Here's what I've done:

(x +y)/(x2 + y2) = -1
x + y = -(x2 + y2)

Where do I go from here?!

2. Mar 24, 2007

### interested_learner

0 is easy. 1 and .5 should be easy too. The negative values are a bit tricky. A brute force way would be to put the x's on one side and y's on the other and then set the $$x+x^2$$ to an arbitary value and then try to solve the quadratic equation for $$-y-y^2 = value$$. The only trick is that your teacher probably doesn't want you to wander into imaginary (i) numbers.

Last edited: Mar 24, 2007
3. Mar 24, 2007

### Dick

The correct way is to simply complete the squares to read off the properties of the circles. This has nothing to do with imaginary numbers.

4. Mar 24, 2007

### HallsofIvy

Staff Emeritus
You have
(x +y)/(x2 + y2) = -1
x + y = -(x2 + y[/sup]2[/sup])
Don't stop there!
x+ y= -x2- y2
x2+ x+ y2+ y= 0

Now, as Dick said, complete the square to get the equation of the circle.

5. Mar 24, 2007

### zenmaster99

Not circles

I'm surprised everyone thinks this field's level curves are circles. They can't be.

Switch to a polar coordinate system:

$$x=r\cos\theta$$
$$y=r\sin\theta$$
$$T(r,\theta)=(\cos\theta+\sin\theta)/r$$

For $$T=0$$, you get $$\cos\theta+\sin\theta=0$$ which occurs at $$\theta=-\pi/4$$

For the other values, you have

$$\cos\theta+\sin\theta=Tr$$ for $$T=-1,-0.5,0.5,1$$

and the only way these could be circles is if you could eliminate the $$\theta$$ dependence.

ZM

Last edited: Mar 24, 2007
6. Mar 24, 2007

### Dick

I'm really impressed! You never considered that the circles might not be centered on the origin. Did you?

7. Mar 24, 2007

### zenmaster99

Doh!

It's early in the morning. :)

I just gave the polar equation for a circle for $$T\ne0$$, didn't I? :-)

ZM---going back to bed for a few hours

Last edited: Mar 24, 2007
8. Mar 25, 2007

### Nibbler

Thanks guys! I'll have Complete The Square burned into my memory from now on.