# Unit vector field and gradient

1. May 2, 2009

### -EquinoX-

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

http://img245.imageshack.us/img245/2353/87006064.th.jpg [Broken]

I need to find the unit vector in the direction of $$\vec{F}$$ at the point (1, 2, -2).

2. Relevant equations

3. The attempt at a solution

well first of all I need to find what F is right, which is gradf.. how can I get that?? what is the level surface of f? is it a sphere?

Last edited by a moderator: May 4, 2017
2. May 2, 2009

### CFDFEAGURU

The grad (del operator) by itself is simply the derivative taken in 3 directions, x,y,z for your problem.

3. May 2, 2009

### dx

grad f for a function which depends only on r = √(x² + y² + z²) always points radially, and has a magnitude ∂f/∂r.

To see why, use the chain rule. For example ∂f/∂x = (∂f/∂r)(∂r/∂x) = (∂f/∂r)(x/r).

Also, √(1² + 2² + (-2)²) = 3, and you already know ∂f/∂r at r = 3.

4. May 2, 2009

### -EquinoX-

is the g'(4) there for a reason?

5. May 2, 2009

### dx

What was the question? I think you missed some words in your post. Unit vector in the direction of what?

6. May 2, 2009

### -EquinoX-

I just corrected the question

7. May 2, 2009

### dx

Ok, g'(4) is not required.

Hint: Read my first post. Pay attention to the fact that g'(3) is positive. The magnitude of g'(3) is irrelevant.

8. May 2, 2009

### -EquinoX-

I believe I just the derivative of r = √(x² + y² + z²) each with respect to x, y, and z to get the gradf?

9. May 2, 2009

### dx

The definition of grad f is (∂f/∂x, ∂f/∂y, ∂f/∂z). Which direction does this point in if f is a function of only √(x² + y² + z²)?

10. May 2, 2009

### -EquinoX-

hmm.. it points outwards as you said right?? can I say f(x,y,z) instead of f(rho)

11. May 2, 2009

### dx

No, I didn't say it points outward. I said it points radially. It could be radially outward, or radially inward. Just do what I suggested: evaluate grad f in detail, from the definition of grad, using my previous hints.

12. May 2, 2009

### -EquinoX-

gradf = ∂f/∂x √(x² + y² + z²)i + ∂f/∂y √(x² + y² + z²)j + ∂f/∂z √(x² + y² + z²)k

it's just that right?

Last edited: May 2, 2009
13. May 2, 2009

### dx

No..

What's the definition of grad f?

14. May 2, 2009

### -EquinoX-

as stated above:

The grad (del operator) by itself is simply the derivative taken in 3 directions, x,y,z for your problem.

15. May 2, 2009

### dx

grad f is a vector field. It as components. The x-component is ∂f/∂x. The y-component is ∂f/∂y. The z-component is ∂f/∂z.

Evaluate these derivatives using the chain rule, as I have done in post #3. What is the vector you get when you do this?

16. May 2, 2009

### -EquinoX-

yes I know it's a vector I just forgot to put my i,j,k above.. right?

17. May 2, 2009

### dx

Nope, still not right. Don't try to guess it. Do it in detail, as I've already shown you how to do #3.

18. May 2, 2009

### -EquinoX-

I really don't understand how post #3 could help in this case

19. May 2, 2009

### -EquinoX-

Last edited by a moderator: May 4, 2017
20. May 2, 2009

### dx

No. √(x² + y² + z²) is not the value of the function. Reread the question. It says that f is a function of √(x² + y² + z²), which means that the value of f depends only on √(x² + y² + z²). For example, if x = 1, y = 0, z = 0, then the value of f is g(√(x² + y² + z²)) = g(√(1² + 0² + 0²)) = g(1) which from the graph in the original question is approximately 1.4.

f(x, y, z) = g(√(x² + y² + z²))