Directional derivative of vector valued functions?

  • Thread starter chy1013m1
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  • #1
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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?..
 

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  • #2
HallsofIvy
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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
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so that means, if the Fretchet derivative is L, and the directional vector is u, it is just Lu.
 
  • #4
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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
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Consider the derivative operator [tex]D_v[/tex]. We require it to be linear in the direction of integration, v. So [tex]D_{2v} = 2 D_v[/tex] and [tex]D_{u+v} = D_u + D_v[/tex]. In fact, this correspondence is taken to be an isomorphism in differential geometry, where tangent vectors are defined to be directional derivative operators.
 
  • #6
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Thanks for your reply, but my doubt remains. :blushing:

I'll use 2 examples:


Scalar field

f(x,y)=x[tex]^{3}[/tex]y[tex]^{2}[/tex]

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=[tex]\frac{1}{\left\right\|a\|}[/tex]a = [tex]\frac{1}{5}[/tex] (4i-3j) = [tex]\frac{4}{5}[/tex]i - [tex]\frac{3}{5}[/tex]j

The result is:

Du f(x,y) = 3x[tex]^{2}[/tex]y[tex]^{2}[/tex] ([tex]\frac{4}{5}[/tex]) + 2x[tex]^{3}[/tex]y ([tex]\frac{-3}{5}[/tex])

For P(-1,2)

Du f(-1,2) = 12


Vector field

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

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