# Directional derivative of vector valued functions?

1. Jan 13, 2007

### chy1013m1

Just to confirm, is the directional derivative of a vector valued function calculated as Lu ? where L is the Frechet derivative , and u is the unit vector in the direction.

There seem to be a lot of sources for a real valued function's directional derivative, but very little on vector valued function in Rm.

it says DuF(directional derivative) = lim (F(x + tu) - F(x)) / t as t->0
if F is diff'ble, then -] L , then just set h = tu, we get the desired result?..

2. Jan 13, 2007

### HallsofIvy

Staff Emeritus
There's a reason why you find "very little on vector valued function in Rm"!

If f is a differentiable function from Rm to Rn, then the derivative of f is a linear function from Rm to Rn which can be represented by a n by m matrix. In particular, the directional derivative of a n dimensional vector valued function on Rm in the direction of vector m dimensional vector v, is an n dimensional vector given by the product of the matrix representing the derivative and the vector v, represented as a column matrix.

3. Jan 13, 2007

### chy1013m1

so that means, if the Fretchet derivative is L, and the directional vector is u, it is just Lu.

4. Sep 16, 2007

### ragnarök

Sorry for bringing this thread back from the dead, but I would like to clarify a doubt, if possible.

In order to calculate the directional derivative of a scalar field, we must use a unit vector to define the direction.

What happens in the case of a vector field? Does the direction also have to be defined by a unit vector, or we can use any vector?

If anyone could help, I would appreciate.

5. Sep 16, 2007

### genneth

Consider the derivative operator $$D_v$$. We require it to be linear in the direction of integration, v. So $$D_{2v} = 2 D_v$$ and $$D_{u+v} = D_u + D_v$$. In fact, this correspondence is taken to be an isomorphism in differential geometry, where tangent vectors are defined to be directional derivative operators.

6. Sep 16, 2007

### ragnarök

I'll use 2 examples:

Scalar field

f(x,y)=x$$^{3}$$y$$^{2}$$

To find the directional derivative for this function in the point P(-1,2), in the direction of the vector a=4i-3j, we must start by finding a unit vector, u, with the same direction as a.

u=$$\frac{1}{\left\right\|a\|}$$a = $$\frac{1}{5}$$ (4i-3j) = $$\frac{4}{5}$$i - $$\frac{3}{5}$$j

The result is:

Du f(x,y) = 3x$$^{2}$$y$$^{2}$$ ($$\frac{4}{5}$$) + 2x$$^{3}$$y ($$\frac{-3}{5}$$)

For P(-1,2)

Du f(-1,2) = 12

Vector field

f(x,y)= (y logx, x$$^{3}$$ - 3y)

Suppose we want to find the directional derivative for this function in the point P(1,2), in the direction of the vector v=1i+4j.

My doubt is: do we need to find a unit vector or we simply use v=(1,4).

I solved it like this (but I'm not sure it's right):

Dv f(x,y) = [[y/x, logx][3x$$^{2}$$, -3]]*[[1][4]]
Dv f(1,2) = [[2][-9]]

Sorry for the matrix notation, but I couldn't get it right with LaTeX.

Last edited: Sep 16, 2007