Calculating $\frac{\partial x }{\partial r}$ if $r^2 = x^2 + y^2 +z^2$

Question

I am getting this $ \frac{\partial x}{\partial r} $ now im not so sure $\frac{\partial x}{\partial r}$ is zero . As r has affiliated $x $in it. Now if its zero then an physical interpretation would be great . If not then what is it ?

Answer 1

To determine $\partial x/\partial r$, you need to know what are the other variables (besides $r$) that $x$ depends on. One example is the "spherical coordinates" change of variables... \begin{align} r &= \sqrt{x^2+y^2+z^2}\\ \theta &= \arccos\frac{z}{\sqrt{x^2+y^2+z^2}}\\ \varphi &= \arctan\frac{y}{x} \end{align} and the inverse \begin{align} x &= r \sin \theta\cos \varphi\\ y &= r \sin \theta \sin \varphi\\ z &= r \cos \theta \end{align}

When it is written out that way, can you find $\partial x/\partial r$?

The tricky thing is: there are other choices of additional variables, compatible with $r = \sqrt{x^2+y^2+z^2}$, producing a different answer for $\partial x/\partial r$.

Answer 2

If your expression is correct and you want to differentiate with respect to $r$, then solve for $x$ :

$$r^2 = x^2 + y^2 + z^2 \implies x = \pm \sqrt{r^2-y^2-z^2}$$

Recall that $(\sqrt{x})' = \frac{1}{2\sqrt{x}}$ and that $f(g(x))' = f'(g(x))\cdot g'(x)$. Treat $-y^2-z^2$ as constants, if they are not depended by $r$ and then :

$$\frac{\partial x}{\partial r} = \frac{\partial}{\partial r}\Big(\sqrt{r^2-y^2-z^2}\Big) = \frac{1}{2\sqrt{r^2-y^2-z^2}}\cdot(r^2)' = \frac{r}{\sqrt{r^2-y^2-z^2}}$$

Can you find the derivative for the case of the minus sign ?

This derivative can never be zero, except if $$r=0 \implies x^2 + y^2 + z^2 = 0 \implies (x,y,z) = (0,0,0)$$ where you don't even have an $r$ parameter and that's just a $0=0$ equality.

Note : That's only for the case of $x,y,z$ being independed with respect to $r$ and $x$.