# Solving Operator Nabla Example Problem

• prehisto
In summary, the conversation discusses the use of the nabla operator on vectors and the importance of precise notation. The OP is struggling with a complex example involving the operation of nabla on vectors and is trying to determine the correct action to take. The conversation also touches on the use of spherical coordinates and the difference between taking the divergence and gradient of a vector.
prehisto

## Homework Statement

So I have this rather komplex example and I am looking for help.
∇(3(r*a)r)/R5 -a/R5)
r=xex+yey+zez
a-constant vector
R=r1/2

## The Attempt at a Solution

So the nabla " works" on every member individualy,and i have to careful here:(r*∇a),because of analogue with derivative rule,am I correct?

So 9(r*a)/R^5+9r/R^5+9r/R^5-15(r*a)r*r/R^5-∇a/R^5

1. Homework Statement [/

prehisto said:

## Homework Statement

So I have this rather komplex example and I am looking for help.
∇(3(r*a)r)/R5 -a/R5)

Hi again.
Your notation here is not clear: are r and a vectors? In which case ($\vec{r}$$\cdot$$\vec{a}$)$\vec{r}$ is a vector and you
cannot take its gradient (unless you're defining a matrix form, which i doubt you are)...
Now you can act on it with ∇$\cdot$ (divergence) or ∇$\times$ (curl), or you can have [∇($\vec{r}$$\cdot$$\vec{a}$/R5)]$\vec{r}$. So could you clarify your notation?

Yes,r and a are vectors.
$\vec{∇}$(3($\vec{r}$*$\vec{a}$)$\vec{r}$/R5-$\vec{a}$/R5)

Hmm,why can't I act on ($\vec{r}$*$\vec{a}$)$\vec{r}$ because its a vector..I can act on $\vec{r}$,which is a vector,so why not ?

No you can't (again, unless you're defining a matrix)!
The gradient turns a scalar function into a vector, the divergence does the opposite and the curl turns a vector into another vector. So you have to make sure exactly where you put your parenthesis and adopt a precise notation, here...

it would be better to use the nabla operator in spherical coordinates... I think it will simplify your problem :)

(plus I don't see any gradient or divergence in the OP's post)

ChrisVer said:
it would be better to use the nabla operator in spherical coordinates... I think it will simplify your problem :)

(plus I don't see any gradient or divergence in the OP's post)

Yes, using r is indeed an incentive to use spherical coordinates, for which you need the corresponding form for ∇. But the first issue here is to determine exactly what operation is asked, and on what type of object because there's something wrong in the problem as stated...

yes, because most of times you will not see div acting on scalars, as you won't see grad acting on vectors... So in that case intuitively you choose the correct action XD.
If someone has a vector, he'll use the div for the vector, and when they have a scalar they'll use the grad...

Ok, now I am starting to see that there is problem in essence of example.
But I have to solve it, in given cordinates.

I assume that
∇R=dR/dx*ex+dR/dy*ey+dR/dz*ez
where R is modul of vector r
and
∇r=dx/dx*ex+dy/dy*ey+dz/dz*ez

prehisto said:
Ok, now I am starting to see that there is problem in essence of example.
But I have to solve it, in given cordinates.

I assume that
∇R=dR/dx*ex+dR/dy*ey+dR/dz*ez
where R is modul of vector r

Yes

∇r=dx/dx*ex+dy/dy*ey+dz/dz*ez

No, if r is the position vector then:
∇$\cdot$$\vec{r}$=(dx/dx)(ex*ex)+(dy/dy)(ey*ey)+(dz/dz)(ez*ez)= dx/dx + dy/dy + dz/dz = 3
And ∇$\vec{r}$ is a matrix.

You still haven't told us whether you are taking the divergence of a vector, or the gradient of the vector. The gradient of the vector is a second order tensor, while the divergence of the vector is a scalar. So, which is it? (Irrespective of what coordinate system you are using)

Chet

## 1. What is the operator nabla?

The operator nabla is a mathematical symbol that represents the gradient in vector calculus. It is denoted by the symbol ∇ and is used to measure the rate and direction of change in a scalar or vector field.

## 2. What is the purpose of solving operator nabla example problems?

The purpose of solving operator nabla example problems is to gain a better understanding of how to use the operator in solving various mathematical and scientific problems. It also helps in developing critical thinking and problem-solving skills.

## 3. How do you solve an operator nabla example problem?

To solve an operator nabla example problem, you first need to identify the function or field that you are working with. Then, you can use the appropriate formulas and rules to compute the gradient and solve for the desired variables.

## 4. What are some common applications of the operator nabla?

The operator nabla is commonly used in many fields, including physics, engineering, and mathematics. Some of its applications include calculating electric and magnetic fields, determining fluid flow and heat transfer, and solving optimization problems.

## 5. Are there any other operators similar to nabla?

Yes, there are other operators similar to nabla, such as the Laplace operator (∆) and the divergence (∇·) and curl (∇×) operators. These operators are all used in vector calculus to describe different aspects of a vector field.

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