Fluid Mechanics equations in Cartesian and Cylindrical coordinates?

[silentwf]

Homework Statement

Not really a homework question, but more of a concept question which I'm unfamiliar with. So as we know, equations can be in any coordinate, but how do you convert them from one to another?
For example, a few equations from fluid mechanics. the first equation is the vector form of equation 2 and 3. i don't get why there's an extra 1/r. the same goes for equations 4 and 5. 4 is in cartesian while 5 is in cylindrical.

(let u,v,w be velocities. u is a function of x, v is a function of y and w is a function of z; v(theta) is velocity, function of theta, v(r) velocity, function of r)
(typo in equation 4, it should be partial u and not partial mu)

http://img341.imageshack.us/img341/8350/fluid.png

Homework Equations

(none, i guess?)

The Attempt at a Solution

(dont know where to begin)

 

[hunt_mat]

How about the chain rule which is how co-ordinate transformations are usally calculated.

 

[HallsofIvy]

By the chain rule, as hunt mat suggests, for any function, u, of x and y,
\frac{\partial u}{\partial x}= \frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+ \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}

r= (x^2+ y^2)^{1/2} so
\frac{\partial r}{\partial x}= (1/2)(x^2+ y^2)^{1/2}(2x)= \frac{x}{(x^2+ y^2)^{1/2}}= \frac{r cos(\theta)}{r}= cos(\theta)

\theta= arctan(y/x) so
\frac{\partial \theta}{\partial x}= \frac{1}{1+ \frac{y^2}{x^2}}\left(-\frac{y}{x^2}\right)= -\frac{y}{x^2+ y^2}= -\frac{r sin(\theta)}{r^2}= -\frac{1}{r} sin(\theta)

So that
\frac{\partial u}{\partial x}= cos(\theta)\frac{\partial u}{\partial r}- \frac{1}{r}sin(\theta)\frac{\partial u}{\partial \theta}

 

[silentwf]

OH! ok, I was wondering how to do it for a while.
Sorry for the late response! and thanks guys!