Unit vector field and gradient

[-EquinoX-]

Homework Statement

http://img245.imageshack.us/img245/2353/87006064.th.jpg

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

Homework Equations

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?

 

[CFDFEAGURU]

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

 

[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.

 

[-EquinoX-]

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

 

[dx]

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

 

[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.

 

[-EquinoX-]

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

 

[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)?

 

[-EquinoX-]

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

 

[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.

 

[-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?

 

[dx]

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.

 

[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?

 

[-EquinoX-]

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

 

[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.

 

[-EquinoX-]

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

 

[-EquinoX-]

http://img245.imageshack.us/img245/2892/testncs.th.jpg

taken from wikipedia, in our case:

f(x,y,z) = √(x² + y² + z²)

is this true?

 

[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))

 

[-EquinoX-]

hmm.. after reading your third post again:

the gradient is just then ∂f/∂r

is it just 1/3rho?

 

[-EquinoX-]

bump!

 

[dx]

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

 

[dx]

Depeds, what is r in (1/3)r ?

 

[-EquinoX-]

it's x²i + y²j + z²k, right?

 

[dx]

Then how is (1/3)r a vector?

 

[-EquinoX-]

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

 

[dx]

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?