Momentum Flux: Quick Question (Product Rule)

andrew.c
Messages
46
Reaction score
0

Homework Statement


Heya,
I can expand the term on the LHs and get the term on the RHS no problem :)
But,
I don't understand how the bottom line is arrived at; I think its basically just a backward engineering of the product rule, but I can't get it!

Any help would be useful
 

Attachments

  • momentum flux.JPG
    momentum flux.JPG
    15.1 KB · Views: 424
Physics news on Phys.org
\frac{\partial(\rho u u)}{\partial x} = \rho u \frac{\partial u}{\partial x} + u \frac{\partial (\rho u)}{\partial x}.
 
Thanks
That was easier than I thought!
 
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