# Finding 1st and 2nd total partial differential

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

HallsofIvy
Homework Helper
I've been struggling a bit with the following problem...
I would consider the phrase "total partial differential" a contradiction in terms! Perhaps you mean what I would call just the "total differential".

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

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

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.

$$2rdr = 2(x - a)dx + 2(y - b)dy + 2(z - c)dz$$
Okay, now that's a total differential.

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

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
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 $\partial^2 r/\partial x^2$.