1. ## partial differentiation

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

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

Got the following for $\frac{\partial r}{\partial x} + \frac{\partial r}{\partial y} + \frac{\partial r}{\partial z}$

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

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

Here, I go astray. The answer provided to me for $d^2r$ is $\frac{2}{r}$.

I get $\frac{\partial^2 r}{\partial x^2} = \frac{(1)r - ((x - a) + (y - b) + (z - c))(1)}{r^2}$

Likewise for the other partials leads to :
$d^2r = \frac{3(r - ((x - a) + (y - b) + (z - c))}{r^2}$

Obviously not doing something right.

-Scott

2. Originally Posted by ScottO
$r^2 = (x - a)^2 + (y - b)^2 + (z -c)^2$

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

Got the following for $\frac{\partial r}{\partial x} + \frac{\partial r}{\partial y} + \frac{\partial r}{\partial z}$

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

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

Here, I go astray. The answer provided to me for $d^2r$ is $\frac{2}{r}$.

I get $\frac{\partial^2 r}{\partial x^2} = \frac{(1)r - ((x - a) + (y - b) + (z - c))(1)}{r^2}$

Likewise for the other partials leads to :
$d^2r = \frac{3(r - ((x - a) + (y - b) + (z - c))}{r^2}$

Obviously not doing something right.

-Scott
yes, i don't know what you are doing.

when we find the partial derivative with respect to, say x, we treat all other variables as constants.

let me do $\frac {\partial^2 r}{\partial x^2}$ for you. you do the rest

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

$\Rightarrow 2r~\frac {\partial r}{\partial x} = 2(x - a)$

Now $\frac {\partial^2 r}{\partial x^2}$ means we find the second derivative of r with respect to x

$\Rightarrow 2 ~\frac {\partial^2 r}{\partial x^2} = 2$

based on what you said the answer was, you may be looking for it in a different form.

you may want to write: $r = \sqrt{(x - a)^2 + (y - b)^2 + (z - c)^2}$

and then use the chain rule

$\Rightarrow \frac {\partial^2 r}{\partial x^2} = 1$

3. Originally Posted by Jhevon
yes, i don't know what you are doing.

when we find the partial derivative with respect to, say x, we treat all other variables as constants.

let me do $\frac {\partial^2 r}{\partial x^2}$ for you. you do the rest

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

$\Rightarrow 2r~\frac {\partial r}{\partial x} = 2(x - a)$

Now $\frac {\partial^2 r}{\partial x^2}$ means we find the second derivative of r with respect to x

$\Rightarrow 2 ~\frac {\partial^2 r}{\partial x^2} = 2$

$\Rightarrow \frac {\partial^2 r}{\partial x^2} = 1$
OK. What I did was find the first total partial ...

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

Then I started from there on the second total partial, using the quotient rule since I had r in the denominator. What you're showing is I should find the individual partials, $\frac {\partial r}{\partial x}$, then $\frac {\partial^2 r}{\partial x^2}$, for each term, and then add them.

Which gives an answer of $d^2r = \frac {\partial^2 r}{\partial x^2} + \frac {\partial^2 r}{\partial y^2} + \frac {\partial^2 r}{\partial z^2} = 3$.

-Scott

4. Originally Posted by ScottO
OK. What I did was find the first total partial ...

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

Then I started from there on the second total partial, using the quotient rule since I had r in the denominator. What you're showing is I should find the individual partials, $\frac {\partial r}{\partial x}$, then $\frac {\partial^2 r}{\partial x^2}$, for each term, and then add them.

Which gives an answer of $d^2r = \frac {\partial^2 r}{\partial x^2} + \frac {\partial^2 r}{\partial y^2} + \frac {\partial^2 r}{\partial z^2} = 3$.

-Scott
yes. what you typed confused me at first. i don't think it was wise to do it like that. find each piece first and then add them, because the statement dr = ... which you had was not correct. for various notational and conceptual violations