1. May 13, 2008

### gtfitzpatrick

1. The problem statement, all variables and given/known data

if $$\phi$$ = rk/r$$^{3}$$ where r=xi + yJ + zk and r is the magnitude of r, prove that $$\nabla$$$$\phi$$ = (1/r$$^{}5$$)(r$$^{}2$$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)

2. May 14, 2008

### tiny-tim

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. May 14, 2008

### gtfitzpatrick

if $$\phi$$ = rk/r$$^{3}$$ where r=xi + yJ + zk and r is the magnitude of r, prove that $$\nabla$$$$\phi$$ = (1/r$$^{}5$$)(r$$^{}2$$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$$^{5}$$(-3(r.k)r)

4. May 14, 2008

### gtfitzpatrick

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

then i just put these answers into = dx/d$$\phi$$ i + dy/d$$\phi$$ j + dz/d$$\phi$$ k

which gives
-3xz/r$$^{5}$$ i - 3yz/r$$^{5}$$ j - 3zz/r$$^{5}$$ k

5. May 14, 2008

### gtfitzpatrick

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

when it should be 1/r$$^{5}$$(r$$^{2}$$k-3(r.k)r)

6. May 14, 2008

### Dick

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.

7. May 14, 2008

### Dick

8. May 14, 2008

### tiny-tim

Hi gtfitzpatrick!

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

You need to differentiate the r also.

9. May 14, 2008

### gtfitzpatrick

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

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: May 14, 2008
10. May 14, 2008

### gtfitzpatrick

so i cant use the product rule?

11. May 14, 2008

### Dick

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.

12. May 14, 2008

### gtfitzpatrick

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