- #1

- 1

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Given:

[tex]r^2 = (x - a)^2 + (y - b)^2 + (z - c)^2[/tex]

Find:

1. [tex]\frac{\partial r}{\partial x} + \frac{\partial r}{\partial y} + \frac{\partial r}{\partial z}[/tex]

and

2. [tex]\frac{\partial^2 r}{\partial x^2} + \frac{\partial^2 r}{\partial y^2} + \frac{\partial^2 r}{\partial z^2}[/tex]

I've found the following for 1.

[tex]2rdr = 2(x - a)dx + 2(y - b)dy + 2(z - c)dz[/tex]

[tex]dr = \frac{(x - a) + (y - b) + (z - c)}{r}[/tex]

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 [tex]\sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2}[/tex] from the original given? Or start from here:

[tex]2rdr = 2(x - a)dx + 2(y - b)dy + 2(z - c)dz[/tex]

which leads to

[tex]2d^2r = 2dx^2 + 2dy^2 + 2dz^2[/tex]

[tex]d^2r = 3[/tex]

I don't think latter is correct, as it seems like it doesn't consider

**r**. Plus, the answer in the book is

[tex]\frac{2}{r}[/tex]

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...

[tex]dr = \frac{(x - a) + (y - b) + (z - c)}{r}[/tex]

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

-Scott