# Unit vector field and gradient

-EquinoX-

## Homework Statement

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).

## 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:

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

Homework Helper
Gold Member
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.

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

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

-EquinoX-
I just corrected the question

Homework Helper
Gold Member
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.

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

Homework Helper
Gold Member
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)?

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

Homework Helper
Gold Member
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.

-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:
Homework Helper
Gold Member
No..

What's the definition of grad f?

-EquinoX-
as stated above:

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

Homework Helper
Gold Member
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?

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

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

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

Homework Helper
Gold Member
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))

-EquinoX-

the gradient is just then ∂f/∂r

is it just 1/3rho?

-EquinoX-
bump!

Homework Helper
Gold Member
The gradient cannot be ∂f/∂r, because ∂f/∂r is not a vector!

-EquinoX-
yes I know... but it seems like it's like that.. ∂f/∂r at point (1,2,-2) is 1/3.. so is it 1/3r

Homework Helper
Gold Member
Depeds, what is r in (1/3)r ?

-EquinoX-
it's x²i + y²j + z²k, right?

Homework Helper
Gold Member
Then how is (1/3)r a vector?

-EquinoX-
I think it's something like this 1/3x²i + 1/3y²j + 1/3z²k

Homework Helper
Gold Member
No. The gradient at (1, 2, -2) is (1/3)(1/3i + 2/3j - 2/3k).

-EquinoX-
why is r here 1/3i + 2/3j -2/3k?

Homework Helper
Gold Member
That's the answer you get if you evaluate the gradient using the definition ∇f = (∂f/∂x, ∂f/∂y, ∂f/∂z).

-EquinoX-
and so is (1/3)(1/3i + 2/3j - 2/3k) equal to $$\vec{F}$$

Homework Helper
Gold Member
Yes, at the point (1, 2, -2) .

-EquinoX-
The points (1, 2, -2) and (0, 3, 0) are both on the sphere x^2 + y^2 + z^2 = 9. How can I estimate $$\vec{F}$$ at (0, 3, 0)?

Homework Helper
Gold Member
You don't need to estimate it. The exact answer can be calculated from the definition. Please show your work/attempt if you have any more questions.

Last edited: