Vector calculus chain rule

  • #1
302
0

Main Question or Discussion Point

Hi. I was looking for a chain rule in vector calculus for taking the gradient of a function such as f(A), where A is a vector and f is a scalar function. I found the following expression on wikipedia, but I don't understand it. It's taking the gradient of f, and applying that to A, and then writing nabla A ?? Can anyone tell me what's going on?

fcd0ce7679df0e7387af5f353182e420.png
 

Answers and Replies

  • #2
RUber
Homework Helper
1,687
344
Consider ##A = [a(x,y,z) \hat x + b(x,y,z) \hat y + c(x,y,z) \hat z ],## and ##f(A) = f(a,b,c),##
then ##\nabla f(A) = \hat x \frac{\partial f}{\partial x}\left[\frac{\partial a}{\partial x}+\frac{\partial b}{\partial x}+\frac{\partial c}{\partial x} \right] +
\hat y\frac{\partial f}{\partial y}\left[\frac{\partial a}{\partial y}+\frac{\partial b}{\partial y}+\frac{\partial c}{\partial y} \right]+
\hat z \frac{\partial f}{\partial z}\left[\frac{\partial a}{\partial z}+\frac{\partial b}{\partial z}+\frac{\partial c}{\partial z} \right]##
##\nabla f = \hat x \frac{\partial f}{\partial x}+ \hat y \frac{\partial f}{\partial y}+ \hat z \frac{\partial f}{\partial z}##
Also, ## \nabla A = \pmatrix{\frac{\partial a}{\partial x} &\frac{\partial b}{\partial x} & \frac{\partial c}{\partial x}\\
\frac{\partial a}{\partial y} &\frac{\partial b}{\partial y} & \frac{\partial c}{\partial y} \\
\frac{\partial a}{\partial z} &\frac{\partial b}{\partial z} & \frac{\partial c}{\partial z}}##
So if you carry out the matrix math and rearrange the terms,
fcd0ce7679df0e7387af5f353182e420.png

You will see that the equation is true.
 
  • #3
302
0
Thanks. How would that look in spherical coordinates ?
 
  • #4
Fredrik
Staff Emeritus
Science Advisor
Gold Member
10,851
406
I'm just going to answer the first question in a different notation. I like this version of the chain rule: ##(f\circ g)_{,i}(x) =f_{,j}(g(x)) g^j{}_{,i}(x)##. Here ##_{,i}## denotes partial differentiation with respect to the ##i##th variable, and ##g^i## denotes the real-valued function that takes ##x## to the ##i##th component of ##g(x)##. I'm using the convention to not write any summation sigmas, since the sum is always over the index that appears twice. For example, if I write ##X^i_k Y^k_j##, it means ##\sum_{k=1}^n X^i_k Y^k_j##.

Note that the ##i##th component of ##\nabla(f\circ A)(x)## is ##(f\circ A)_{,i}(x)##.
$$(f\circ A)_{,i}(x)= f_{,j}(A(x))A^j{}_{,i}(x) =\nabla f(A(x))\cdot A_{,i}(x) =(\nabla f\circ A)(x)\cdot A_{,i}(x).$$ I suppose we could also write this as
$$\nabla(f\circ A)(x) =(\nabla f\circ A)(x)\cdot\nabla A(x),$$ but I don't see why we'd want to.
 

Related Threads for: Vector calculus chain rule

Replies
2
Views
2K
Replies
2
Views
471
Replies
3
Views
560
Replies
7
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
1
Views
99
  • Last Post
Replies
10
Views
760
  • Last Post
Replies
2
Views
1K
Top