# Finding 1st and 2nd total partial differential

1. Oct 23, 2007

### ScottO

I've been struggling a bit with the following problem...

Given:
$$r^2 = (x - a)^2 + (y - b)^2 + (z - c)^2$$

Find:
1. $$\frac{\partial r}{\partial x} + \frac{\partial r}{\partial y} + \frac{\partial r}{\partial z}$$

and

2. $$\frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2} + \frac{\partial^2 r}{\partial z^2}$$

I've found the following for 1.

$$2rdr = 2(x - a)dx + 2(y - b)dy + 2(z - c)dz$$

$$dr = \frac{(x - a) + (y - b) + (z - c)}{r}$$

This matches the answer in the book. So I think I'm OK to this point.

Figuring 2 is where I'm having trouble. How do I handle the r in the denominator? Do I substitute $$\sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2}$$ from the original given? Or start from here:

$$2rdr = 2(x - a)dx + 2(y - b)dy + 2(z - c)dz$$

$$2d^2r = 2dx^2 + 2dy^2 + 2dz^2$$

$$d^2r = 3$$

I don't think latter is correct, as it seems like it doesn't consider r. Plus, the answer in the book is

$$\frac{2}{r}$$

I've gotten close using the substitution, but things are a bit messy, and I'm missing some simplification or something.

Here's what I get for one partial using the substitution...

$$dr = \frac{(x - a) + (y - b) + (z - c)}{r}$$

$$\frac{\partial^2 r}{\partial x^2} = \frac{r - \frac{[(x - a) + (y - b) + (z - c)](x - a)}{r}}{r^2}$$

-Scott

2. Oct 23, 2007

### HallsofIvy

Staff Emeritus
I would consider the phrase "total partial differential" a contradiction in terms! Perhaps you mean what I would call just the "total differential".

This is neither a partial derivative nor a differential. In fact, since it would change with any change in coordinate system, I'm not sure it has any "real" meaning!

Okay, now that's a total differential.

What happened to the "dx", "dy", and "dz"?

This matches the answer in the book. So I think I'm OK to this point.[/quote]
If the problem really was to find
$$\frac{\partial r}{\partial x} + \frac{\partial r}{\partial y} + \frac{\partial r}{\partial z}$$
then there was no reason to find the total differential. Just the three partial derivatives would suffice.

If you are really asked to find just the sum of the second partials (which I would consider to have no real significance, as, unlike the gradient, etc., it depends on the coordinate system.) then just find the partial derivatives.
Since $r^2= (x-a)^2+ (y-b)^2+ (z-c)^2$ then $2r\partial r/\partial x= 2(x-a)$ and $r\partial r/\partial x= x-a. Now differentiate both sides again with respect to x: [itex]\left(\partial r/\partial x\right)^2+ 2r \partial^2 r/\partial x^2= 1$. Solve that for $\partial^2 r/\partial x^2$.