What Does the Notation in This Vector Field Equation Mean?

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Homework Help Overview

The discussion revolves around understanding the notation and simplification of a vector field equation involving the gradient of a scalar function. The original poster is particularly interested in the implications of the notation used in the vector field and how it relates to finding the flux across a sphere.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the meaning of the notation, particularly the gradient and the modulus of the vector. Questions arise about how to compute the gradient of a scalar function that does not explicitly include unit vectors.

Discussion Status

The discussion is active, with participants questioning the nature of the gradient and its application to the given function. There is a focus on clarifying the definition of the gradient and its relationship to scalar functions, but no consensus has been reached on the specific computations or simplifications.

Contextual Notes

Participants are navigating the complexities of vector calculus notation and the implications for solving the problem, with some uncertainty about the definitions and assumptions related to the gradient operation.

JaysFan31
Just a quick question about notation.

I was given the vector field

F = r + grad(1/bar(r)) where r= (x)i+(y)j+(z)k.
grad is just written as the upside down delta (gradient) and the bar I wrote in the above equation looks like an absolute value around just the r (although I don't know if it is absolute value). Basically I want to find the gradient of (1/bar(r)).

What would be a simplification of this vector field so that I can solve the rest of the problem?

I want to find its flux across the surface of a sphere.


I think F would be ((x^3)-1)/x^2+((y^3)-1)/y^2+((z^3-1)/z^2, but I'm not sure.
 
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[tex]|\vec{r}|[/tex] is the modulus, or magnitude, of vector [tex]\vec{r}[/tex].
 
If [itex]\vec{r}= x\vec{i}+ y\vec{j}+ z\vec{k}[/itex] then
[tex]\frac{1}{||\vec{r}||}= \frac{1}{\sqrt{x^2+y^2+ z^2}}= (x^2+y^2+ z^2)^{-\frac{1}{2}}[/tex]
What is the gradient of that function?
 
How can you take the gradient of the function if it doesn't have i, j, and k?

Is it 0?
 
JaysFan31 said:
How can you take the gradient of the function if it doesn't have i, j, and k?

Is it 0?

A better question would be, how could you take the grad of the function if it *did* have i, j, and k? Remember, the gradient acts on a scalar.
 
JaysFan31 said:
How can you take the gradient of the function if it doesn't have i, j, and k?

Is it 0?
Well, you would first have to know what "gradient" actually means!

Given a function f(x,y,z), how would YOU define
[itex]\nabla f[/itex]?
 

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