Another vector identity question

Benny
Messages
577
Reaction score
0
Hi, I'm stuck another vector identity question. It's of a different kind to the other one I asked about and looks so much easier but I just can't see what I need to do.

I am told to use standard identities to deduce the following result. The standard identities being referred to are listed in my notes and as I'm sure you can appreciate, it would be a PITA to type them all out. It's just the usual list comprised of various combinations of grad, div curl, and higher dimensional equivalents of product and quotient rules.

Anyway here is the identity I need to show.

<br /> \nabla \left( {r^n } \right) = nr^{n - 2} \mathop r\limits^ \to <br />

where r = \left\| {\mathop r\limits^ \to } \right\| and r = xi + yj + zk.

I haven't written this down but from a quick look at the relation, I would've thought that the RHS should have an (n-1) rather than an (n-2) as given. In any case, grad(r^n) looks so simple to compute that I don't see how I do it without just writing out an explicit expression for the gradient. I can't think of any identities which could help - I'm told to use identities for this question.

Any help would be good thanks.

Edit: The (n-2) is probably right now that I think about it. Anyway I'm off to sleep.
 
Last edited:
Physics news on Phys.org
If you write it with the unit vector \hat r it's:

\nabla (r^n)=nr^{n-1}\hat r

which ofcourse reminds you of d/dx (x^n)=nx^{n-1}, so this is, as you can guess, proved in a very similar way.
 
Thanks for the suggestion, I'll see what I can come up with.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top