Vector & Gradient: Proving \phi=rk/r^{3} (2025)

Homework Statement

if [tex]\phi[/tex] = rk/r[tex]^{3}[/tex] where r=xi + yJ + zk 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)

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?

if [tex]\phi[/tex] = rk/r[tex]^{3}[/tex] where r=xi + yJ + zk 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)

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

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)

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.

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] k

which 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.

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?

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.