Understanding the Chain Rule in Vector Calculus for Gradient of Scalar Functions

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SUMMARY

This discussion focuses on the application of the chain rule in vector calculus for calculating the gradient of scalar functions, specifically in the context of a vector A defined as A = [a(x,y,z) hat;x + b(x,y,z) hat;y + c(x,y,z) hat;z]. The expression for the gradient of f(A) is derived using the chain rule, resulting in the formula: ∇ f(A) = hat;x (∂f/∂x)(∂a/∂x + ∂b/∂x + ∂c/∂x) + hat;y (∂f/∂y)(∂a/∂y + ∂b/∂y + ∂c/∂y) + hat;z (∂f/∂z)(∂a/∂z + ∂b/∂z + ∂c/∂z). The discussion also touches on the notation used for partial differentiation and the implications of expressing the gradient in different coordinate systems.

PREREQUISITES
  • Understanding of vector calculus concepts, particularly gradients.
  • Familiarity with scalar functions and their derivatives.
  • Knowledge of partial differentiation notation and operations.
  • Basic comprehension of coordinate transformations, including spherical coordinates.
NEXT STEPS
  • Study the application of the chain rule in different coordinate systems, focusing on spherical coordinates.
  • Explore the concept of gradient fields and their physical interpretations in vector calculus.
  • Learn about matrix representations of gradients and their applications in multivariable calculus.
  • Investigate advanced topics in vector calculus, such as divergence and curl, and their relationship to gradients.
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Students and professionals in mathematics, physics, and engineering who are looking to deepen their understanding of vector calculus, particularly in the context of gradients and scalar functions.

daudaudaudau
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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
 
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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.
 
Thanks. How would that look in spherical coordinates ?
 
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.
 

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