- #1
ScottO
- 1
- 0
I've been struggling a bit with the following problem...
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
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