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In summary, the conversation discusses the differentiation of a function \phi = rk/r^{3} where r=xi + yJ + zk and r is the magnitude of r. The goal is to prove that \nabla\phi = (1/r^{}5)(r^{}2k-3(r.k)r) by using the quotient rule for grad, which states that grad(f/g)=(g*grad(f)-f*grad(g))/g^2. The conversation also mentions the use of the product rule, grad(z*r^(-3))=grad(z)*r^(-3)+z*grad(r^(-3)), in order to find the missing piece of the grad of z/r^3. After some confusion

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gtfitzpatrick

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## Homework Statement

if [tex]\phi[/tex] = **r****k**/r[tex]^{3}[/tex] where **r**=x**i** + y**J** + z**k** and r is the magnitude of **r**, prove that [tex]\nabla[/tex][tex]\phi[/tex] = (1/r[tex]^{}5[/tex])(r[tex]^{}2[/tex]**k**-3(**r**.**k**)**r**

so i differenciated wrt x then y then z and tried to tidy it all up but i got1/rClick to see the LaTeX code for this image(-3(r.k)r)

When i differenciated wrt x i got -3x/rClick to see the LaTeX code for this image and similar for y and z was this right?

then i just put these answers into Click to see the LaTeX code for this image = dx/dClick to see the LaTeX code for this image i + dy/dClick to see the LaTeX code for this image j + dz/dClick to see the LaTeX code for this image k

which gives

-3xz/rClick to see the LaTeX code for this image i - 3yz/rClick to see the LaTeX code for this image j - 3zz/rClick to see the LaTeX code for this image k

am i right so far? it just seemed to tidy up to 1/rClick to see the LaTeX code for this image(-3(r.k)r)

when it should be 1/rClick to see the LaTeX code for this image(rClick to see the LaTeX code for this imagek-3(r.k)r)

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- #2

tiny-tim

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Hi gtfitzpatrick!

This is too difficult to read … everything has "Click to see the LaTeX code for this image" in the middle.

Can you type it out again, perhaps using ² and ³ and ^4 and ^5?

- #3

gtfitzpatrick

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if [tex]\phi[/tex] = **r****k**/r[tex]^{3}[/tex] where **r**=x**i** + y**J** + z**k** and r is the magnitude of **r**, prove that [tex]\nabla[/tex][tex]\phi[/tex] = (1/r[tex]^{}5[/tex])(r[tex]^{}2[/tex]**k**-3(**r**.**k**)**r**

so i differenciated wrt x then y then z and tried to tidy it all up but i got 1/r[tex]^{5}[/tex](-3(**r**.**k**)**r**)

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gtfitzpatrick

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When i differenciated wrt x i got -3x/r[tex]^{5}[/tex] and similar for y and z was this right?

then i just put these answers into = dx/d[tex]\phi[/tex] **i** + dy/d[tex]\phi[/tex] **j** + dz/d[tex]\phi[/tex] **k**

which gives

-3xz/r[tex]^{5}[/tex] **i** - 3yz/r[tex]^{5}[/tex] **j** - 3zz/r[tex]^{5}[/tex] **k**

- #5

gtfitzpatrick

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am i right so far? it just seemed to tidy up to 1/r[tex]^{5}[/tex](-3(**r**.**k**)**r**)

when it should be 1/r[tex]^{5}[/tex](r[tex]^{2}[/tex]**k**-3(**r**.**k**)**r**)

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Dick

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gtfitzpatrick said:

am i right so far? it just seemed to tidy up to 1/r[tex]^{5}[/tex](-3(

r.k)r)when it should be 1/r[tex]^{5}[/tex](r[tex]^{2}[/tex]

k-3(r.k)r)

It's not terribly clear what you are doing, but some how you are winding up with only one part of a quotient rule answer. Your initial function is (r.k)/r^3. That's the same thing as z/r^3.

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Dick

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You know there is a 'quotient rule' for grad, yes? grad(f/g)=(g*grad(f)-f*grad(g))/g^2. Does that help?

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tiny-tim

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gtfitzpatrick said:

When i differenciated wrt x i got -3x/r[tex]^{5}[/tex] and similar for y and z was this right?

then i just put these answers into = dx/d[tex]\phi[/tex]

i+ dy/d[tex]\phi[/tex]j+ dz/d[tex]\phi[/tex]kwhich gives

-3xz/r[tex]^{5}[/tex]i- 3yz/r[tex]^{5}[/tex]j- 3zz/r[tex]^{5}[/tex]k

Hi gtfitzpatrick!

You're only differentiating the 1/r^5.

You need to differentiate the **r** also.

- #9

gtfitzpatrick

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i differenciated z(x^2+y^2+z^2)[tex]^{-3/2}[/tex] wrt x then y then z, and then filled it into the fomula was this not right?

"you know there is a 'quotient rule' for grad, yes? grad(f/g)=(g*grad(f)-f*grad(g))/g^2"

I didn't know this, does this mean I'm wrong?i haven't seen this formula before what are the f and g's?

Last edited:

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gtfitzpatrick

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so i can't use the product rule?

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Dick

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gtfitzpatrick said:

i differenciated z(x^2+y^2+z^2)[tex]^{-3/2}[/tex] wrt x then y then z, and then filled it into the fomula was this not right?

"you know there is a 'quotient rule' for grad, yes? grad(f/g)=(g*grad(f)-f*grad(g))/g^2"

I didn't know this, does this mean I'm wrong?i haven't seen this formula before what are the f and g's?

It means you are missing a piece of the grad of z/r^3. In terms of the quotient rule, f is z and r^3 is g. You are missing the grad(f) part. Yes, you could also use the product rule, grad(z*r^(-3))=grad(z)*r^(-3)+z*grad(r^(-3)). You are missing the grad(z)*r^(-3) part.

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gtfitzpatrick

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thanks for getting back to me, i figured it out i was missing it,god it took me a age to get there Thanks again though

## FAQ: Vector & Gradient: Proving \phi=rk/r^{3}

## 1. What is a vector in mathematics?

A vector in mathematics is a quantity that has both magnitude and direction. It is represented by an arrow pointing in the direction of the vector, with the length of the arrow representing the magnitude of the vector. Vectors are commonly used to represent physical quantities such as velocity, force, and displacement.

## 2. What is a gradient in mathematics?

A gradient in mathematics is a vector that represents the rate of change of a function with respect to its variables. In other words, it shows how much the value of a function changes in different directions. The gradient is often denoted by the symbol ∇ and is used in various fields such as physics, engineering, and economics.

## 3. How is the gradient of a scalar field related to its vector field?

The gradient of a scalar field is related to its vector field by the fact that the vector field is the gradient of the scalar field. This means that the vector field represents the direction and magnitude of the steepest increase of the scalar field at a given point. In other words, the vector field points in the direction of the maximum rate of change of the scalar field.

## 4. What does the equation \phi=rk/r^{3} represent?

The equation \phi=rk/r^{3} represents the potential of a point charge in a three-dimensional space. Here, \phi is the potential, r is the distance from the point charge, and k is a constant. This equation is commonly used in electrostatics to calculate the potential at a certain point due to a point charge.

## 5. How do you prove the equation \phi=rk/r^{3} using the vector and gradient concepts?

To prove the equation \phi=rk/r^{3}, we can use the fact that the potential at a point is equal to the negative of the gradient of the potential function. We can also use the fact that the potential function of a point charge is inversely proportional to the distance from the point charge. By combining these two concepts, we can show that the potential function for a point charge in a three-dimensional space is given by \phi=rk/r^{3}.

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