# Finding divF and curl F for F = r/r

• ineedmunchies
In summary: Sorry about that. In summary, the student attempted to solve a homework equation that was set incorrectly. They differentiated the equation and found that r² = x² + y² + z².
ineedmunchies

## Homework Statement

F = $$\frac{r}{r}$$

Find divF
and curl F

## Homework Equations

r = x$$\widehat{i}$$ + y$$\widehat{j}$$ + z$$\widehat{k}$$

r = $$\sqrt{(x^{2} + y^{2} + z^{2})}$$

## The Attempt at a Solution

F = $$\frac{x}{(\sqrt{x^{2} + y^{2} + z^{2}})}$$$$\widehat{i}$$ + $$\frac{y}{(\sqrt{x^{2} + y^{2} + z^{2}})}$$$$\widehat{j}$$ + $$\frac{z}{(\sqrt{x^{2} + y^{2} + z^{2}})}$$$$\widehat{k}$$

div F = $$\frac{\partial}{\partial x}$$ $$(x(x^{2} + y^{2} + z^{2})^{\frac{-1}{2}}$$ + $$\frac{\partial}{\partial y}$$$$(y(x^{2} + y^{2} + z^{2})^{\frac{-1}{2}}$$ + $$\frac{\partial}{\partial z}$$$$(z(x^{2} + y^{2} + z^{2})^{\frac{-1}{2}}$$

Take the partial derivative of x first.

let u = $$x^{2} + y^{2} + z^{2}$$
and a = $$xu^{2}$$

$$\frac{\partial u}{\partial x}$$ = 2x
$$\frac{\partial a}{\partial u}$$ = $$\frac{-1}{2}$$x$$u^{\frac{-3}{2}}$$

$$\frac{\partial a}{\partial x}$$ = -$$x^{2}$$$$u^{\frac{-3}{2}}$$
= -$$x^{2}$$($$\frac{1}{\sqrt{(x^{2} + y^{2} + z^{2})}}$$$$)^{3}$$

The other derivatives would give similar answers, and the final answer would be

-$$\frac{x^{2}}{r^{3}}$$-$$\frac{y^{2}}{r^{3}}$$-$$\frac{z^{2}}{r^{3}}$$

This is apparently the incorrect answer, can anybody help?

The three parts that didnt come up are partial derivatives; du/dx da/du and da/dx

And the r as the numerator in the first equation is meant to be in bold to denote that it is a vector.

Hi ineedmunchies!

(here's √ and ² and ∂ for you to copy-and-paste
oh … and tags like B don't work in LaTeX
hmm … nice LaTeX apart from that, though! )

ineedmunchies said:
let u = $$x^{2} + y^{2} + z^{2}$$
and a = $$xu^{2}$$

$$\frac{\partial u}{\partial x}$$ = 2x
$$\frac{\partial a}{\partial u}$$ = $$\frac{-1}{2}$$x$$u^{\frac{-3}{2}}$$

ooooh … this is horrible … can't look … need air

Just use r² = x² + y² + z², so 2r∂r/∂x = … ?

hmmm sorry where are you getting 2r∂r/∂x from?

ineedmunchies said:
hmmm sorry where are you getting 2r∂r/∂x from?

Differentiating r² with respect to x.

(Chain rule: ∂r²/∂x = dr²/dr ∂r/∂x = 2r ∂r/∂x.)

(Can you read r² on your computer? if not, it's supposed to be r^2. )

ah actually never mind, I've worked it out. I had left out part of the product rule in my differentiation. I would post the full worked solution but its quite long and the forums seem to be running slow for me today.

## 1. What is "divF" and "curl F" in the context of this equation?

"DivF" stands for divergence of F, which is a measure of the flow of a vector field away from or towards a point. "Curl F" stands for curl of F, which is a measure of the rotation of a vector field around a point. In this equation, F represents a vector field and r represents the position vector.

## 2. How do you find divF and curl F for F = r/r?

To find divF, you can use the formula divF = ∇ · F, where ∇ is the nabla operator and · represents the dot product. For F = r/r, this becomes divF = ∇ · (r/r) = 3. To find curl F, you can use the formula curl F = ∇ x F, where ∇ is the nabla operator and x represents the cross product. For F = r/r, this becomes curl F = ∇ x (r/r) = 0.

## 3. Is this equation commonly used in physics or mathematics?

Yes, this equation is commonly used in both physics and mathematics. In physics, it is often used in the study of electromagnetism and fluid dynamics. In mathematics, it is used in vector calculus and differential equations.

## 4. How is this equation related to the concept of a conservative vector field?

A conservative vector field is a vector field where the line integral from one point to another is independent of the path taken. This concept is closely related to the equation F = r/r, as the divergence and curl of a conservative vector field both equal 0. This means that the vector field is both irrotational (curl F = 0) and incompressible (divF = 0).

## 5. Can this equation be applied to real-world situations?

Yes, this equation can be applied to many real-world situations. For example, it can be used to model the electric field of a point charge or the flow of a fluid around a point. It can also be used to solve various mathematical problems involving vector fields and differential equations.

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