Directional derivative of vector valued functions?

[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?..

 

[HallsofIvy]

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

If f is a differentiable function from R^m to Rⁿ, then the derivative of f is a linear function from R^m 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 R^m 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.

 

[chy1013m1]

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

 

[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.

 

[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.

 

[ragnarök]

Thanks for your reply, but my doubt remains.

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.