# Find grad *((bbr)/(r^2)) where bbr(x,y,z)=x bbi+y bbj+z bbk, r=sqrt(x^2+y^2+z^2) In this case, why can we not bring the scalar function r^(-2) outside the dot product?

Ayanna Jarvis 2022-10-23 Answered
Find $\mathrm{\nabla }\cdot \left(\frac{\mathbit{r}}{{r}^{2}}\right)$ where $\mathbit{r}\left(x,y,z\right)=x\mathbit{i}+y\mathbit{j}+z\mathbit{k}$, $r=\sqrt{{x}^{2}+{y}^{2}+{z}^{2}}$
In this case, why can we not bring the scalar function ${r}^{-2}$ outside the dot product?
When I leave the function in place, I calculate a result of $\frac{1}{{r}^{2}}$. When I pull it out, I find $\frac{3}{{r}^{2}}$
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cokeman206
It's not just a dot product, it's a divergence operator, and the variables x,y,z appear in ${r}^{-2}$, it's not a constant.
However, you may use the product rule:
$\mathrm{\nabla }\cdot \left(\psi \mathbit{F}\right)=\psi \mathrm{\nabla }\cdot \mathbit{F}+\mathrm{\nabla }\psi \cdot \mathbit{F}$
And with $\psi ={r}^{-2}$, it's not difficult to prove that
$\mathrm{\nabla }\psi =-\frac{2}{{r}^{4}}\mathbit{r}$
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faois3nh
$\mathrm{\nabla }\cdot \left(\frac{\stackrel{\to }{r}}{{r}^{2}}\right)=\mathrm{\nabla }\cdot \left(\frac{x}{{x}^{2}+{y}^{2}+{z}^{2}}\stackrel{\to }{i}+\frac{y}{{x}^{2}+{y}^{2}+{z}^{2}}\stackrel{\to }{j}+\frac{z}{{x}^{2}+{y}^{2}+{z}^{2}}\stackrel{\to }{k}\right)\phantom{\rule{0ex}{0ex}}=\frac{{y}^{2}+{z}^{2}-{x}^{2}}{\left({x}^{2}+{y}^{2}+{z}^{2}{\right)}^{2}}+\frac{{x}^{2}+{z}^{2}-{y}^{2}}{\left({x}^{2}+{y}^{2}+{z}^{2}{\right)}^{2}}+\frac{{x}^{2}+{y}^{2}-{z}^{2}}{\left({x}^{2}+{y}^{2}+{z}^{2}{\right)}^{2}}=\frac{{x}^{2}+{y}^{2}+{z}^{2}}{\left({x}^{2}+{y}^{2}+{z}^{2}{\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{1}{{x}^{2}+{y}^{2}+{z}^{2}}=\frac{1}{{r}^{2}}.$