Question Number 55816 by Rio Mike last updated on 04/Mar/19
$${find}\:{grad}\:{log}\:\mid\boldsymbol{{r}}\mid \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Mar/19
![grade ln∣r∣ ▽^→ =(i(∂/dx)+j(∂/∂y)+k(∂/∂z))(ln(√(x^2 +y^2 +z^2 )) ) =(1/2)i(∂/∂x)(ln(x^2 +y^2 +z^2 )+(1/2)j(∂/∂y)({ln(x^2 +y^2 +z^2 )}+ (1/2)k(∂/∂z){ln(x^2 +y^2 +z^2 )} =(1/2)[((2x)/(x^2 +y^2 +z^2 ))×i+((2yj)/(x^2 +y^2 +z^2 ))+((2zk)/(x^2 +y^2 +z^z ))] =((ix+jy+kz)/(x^2 +y^2 +z^2 )) =(r^→ /(∣r∣^2 ))](https://www.tinkutara.com/question/Q55817.png)
$${grade}\:{ln}\mid{r}\mid \\ $$$$\overset{\rightarrow} {\bigtriangledown}=\left({i}\frac{\partial}{{dx}}+{j}\frac{\partial}{\partial{y}}+{k}\frac{\partial}{\partial{z}}\right)\left({ln}\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }\:\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{i}\frac{\partial}{\partial{x}}\left({ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)+\frac{\mathrm{1}}{\mathrm{2}}{j}\frac{\partial}{\partial{y}}\left(\left\{{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\right\}+\right.\right. \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{k}\frac{\partial}{\partial{z}}\left\{{ln}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{2}{x}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }×{i}+\frac{\mathrm{2}{yj}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }+\frac{\mathrm{2}{zk}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{{z}} }\right] \\ $$$$=\frac{{ix}+{jy}+{kz}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} } \\ $$$$=\frac{\overset{\rightarrow} {{r}}}{\mid{r}\mid^{\mathrm{2}} } \\ $$
Commented by otchereabdullai@gmail.com last updated on 04/Mar/19
$$\mathrm{wow}!\:\mathrm{this}\:\mathrm{is}\:\mathrm{fantastic}\:\mathrm{Prof}\:\mathrm{chaudhury} \\ $$