# Analysis of a vector field

1. Apr 24, 2015

### Calpalned

Oops. I just realized that this is the physics homework forum... This is actually calculus homework...

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

2. Relevant equations
n/a

3. The attempt at a solution
I read, analyzed and reread the text, but I am still confused.

1) How was the position vector $<x,y>$ determined?

2) How is it related to the circle with centered at the origin?

3) We are trying to prove that $F(x,y)$ is tangent to a circle with center at the origin. Why then, are we taking a dot product? The dot product shows that $F(x.y)$ is perpendicular to the position vector $<x,y>$. But isn't being perpendicular the opposite of being tangent?

4) The magnitude of the vector $F(x,y)$ is equal to the radius of the circle. If the circle wasn't centered at the origin, I believe that it will still be true. What is the significance of the magnitude in other vector fields, ones that don"t involve circles?

Thank you all so much!

Last edited: Apr 24, 2015
2. Apr 24, 2015

### Orodruin

Staff Emeritus
The position vector is simply the vector from the origin to a given point. A circle centered on the origin fulfills $F(\vec x) = \vec x^2 = R^2$ and is therefore a level surface of $F$. The normal of a level surface is the gradient of the function and in this case it is proportional to $\vec x$.

A tangent to a surface is orthogonal to its normal, so a tangent to the circle is necessarily also orthogonal to $\vec x$.

3. Apr 24, 2015

### Calpalned

While your answer has helped me understand the answer to my first question, it has also raised new several new questions. For the circle centered in the origin, how was $F(\vec x) = \vec x^2 = R^2$ determined? Does $R^2$ mean two dimensional? A level surface is a projection (or shadow) of a 3D object onto the xy plane. The circle is not 3D. I agree that the normal to a level surface is the gradient, but I don't see how it's proportional to $\vec x$

Thank you Orodriun.

Last edited: Apr 24, 2015
4. Apr 24, 2015

### Orodruin

Staff Emeritus
No, $R$ is the radius of the circle.

No, a level surface is the set of points where a function has a given value. It has a dimension of one less than the full space. For this reason, a level surface in two dimensions is a one dimensional curve.

Did you try taking the gradient of the function $F(\vec x) = \vec x^2$?

5. Apr 24, 2015

### Calpalned

Is $F(\vec x) = \vec x^2 \hat i = R^2$ ?

6. Apr 24, 2015

### Orodruin

Staff Emeritus
No, $\vec x^2 = \vec x \cdot \vec x = x^2 + y^2$, which is a scalar quantity. A vector can never be equal to a scalar. This is just the equation for a circle, $x^2 + y^2 = R^2$.

7. Apr 24, 2015

### Calpalned

Is the $\vec x$ the same as the $x$ two original equations? The one found in $F(x,y) = -y \hat + x \hat j$ and/or $x = x \hat i + y \hat j$

8. Apr 24, 2015

### Orodruin

Staff Emeritus
$\vec x$ is the position vector $x\hat i + y \hat j$. The vector arrow is important.

9. Apr 24, 2015

### certainly

Dear @Calpalned,
I would recommend you re-read everything you have on vectors to clear up some basic concepts before proceeding in your studies in Calculus III.
Otherwise I fear you will face a lot of confusion.