Answer to Question #278380 in Vector Calculus for Mafin

Question

Let r=xi^+yj^+zk^ and r=r .Show that:∇(lnr)=rr2. and∇×(rnr)=0.

Answer 1

$\left(a\right)\ \ Let\ \ \ \overrightarrow{r}=x\overrightarrow{i}\ \ +\ \ y\overrightarrow{j}+z\overrightarrow{k}\ \ and\ \ \overline{r}=\left\|r\right\|=\sqrt{x^{\mathrm{2}}+y{}^{\mathrm{2}}_{\mathrm{\ }}{\mathrm{\ }}+z^{\mathrm{2}}} \\ Such\ \ \ that\ \ \mathrm{ln}\left(\overline{r}\right)=\mathrm{ln}\left(\sqrt{x^{\mathrm{2}}+y{}^{\mathrm{2}}_{\mathrm{\ }}{\mathrm{\ }}+z^{\mathrm{2}}}\right) \\ \mathrm{Re}call\ \ that\ \ \mathrm{\nabla }=\overrightarrow{i}\frac{\partial }{\partial x}\ \ +\ \ \overrightarrow{j}\frac{\partial }{\partial y}+\overrightarrow{k}\frac{\partial }{\partial z} \\ Then\ \ \mathrm{\nabla }\left(\mathrm{ln}\left(\overline{r}\right)\right)=\left(\overrightarrow{i}\frac{\partial }{\partial x}\ \ +\ \ \overrightarrow{j}\frac{\partial }{\partial y}+\overrightarrow{k}\frac{\partial }{\partial z}\right)\mathrm{ln}\left(\overline{r}\right) \\ \mathrm{\nabla }\left(\mathrm{ln}\left(\overline{r}\right)\right)=\left(\overrightarrow{i}\frac{\partial }{\partial x}\ \ +\ \ \overrightarrow{j}\frac{\partial }{\partial y}+\overrightarrow{k}\frac{\partial }{\partial z}\right)\mathrm{ln}\left(\sqrt{x^{\mathrm{2}}+y{}^{\mathrm{2}}_{\mathrm{\ }}{\mathrm{\ }}+z^{\mathrm{2}}}\right) \\ \mathrm{\nabla }\left(\mathrm{ln}\left(\overline{r}\right)\right)=\left(\overrightarrow{i}\frac{\partial \mathrm{ln}\left(\sqrt{x^{\mathrm{2}}+y{}^{\mathrm{2}}_{\mathrm{\ }}{\mathrm{\ }}+z^{\mathrm{2}}}\right)}{\partial x}\ \ +\ \ \overrightarrow{j}\frac{\partial \mathrm{ln}\left(\sqrt{x^{\mathrm{2}}+y{}^{\mathrm{2}}_{\mathrm{\ }}{\mathrm{\ }}+z^{\mathrm{2}}}\right)}{\partial y}+\overrightarrow{k}\frac{\partial \mathrm{ln}\left(\sqrt{x^{\mathrm{2}}+y{}^{\mathrm{2}}_{\mathrm{\ }}{\mathrm{\ }}+z^{\mathrm{2}}}\right)}{\partial z}\right) \\ \mathrm{\nabla }\left(\mathrm{ln}\left(\overline{r}\right)\right)=\left(\overrightarrow{i}\frac{\partial r}{\partial x}\frac{\mathrm{1}}{\sqrt{x^{\mathrm{2}}+y{}^{\mathrm{2}}_{\mathrm{\ }}{\mathrm{\ }}+z^{\mathrm{2}}}}\ \ +\ \ \overrightarrow{j}\frac{\partial r}{\partial y}\frac{\mathrm{1}}{\sqrt{x^{\mathrm{2}}+y{}^{\mathrm{2}}_{\mathrm{\ }}{\mathrm{\ }}+z^{\mathrm{2}}}}+\overrightarrow{k}\frac{\partial r}{\partial z}\frac{\mathrm{1}}{\sqrt{x^{\mathrm{2}}+y{}^{\mathrm{2}}_{\mathrm{\ }}{\mathrm{\ }}+z^{\mathrm{2}}}}\right) \\ \mathrm{\nabla }\left(\mathrm{ln}\left(\overline{r}\right)\right)=\frac{\mathrm{1}}{x^{\mathrm{2}}+y{}^{\mathrm{2}}_{\mathrm{\ }}{\mathrm{\ }}+z^{\mathrm{2}}}\left(\overrightarrow{i}x\ \ +\ \ \overrightarrow{j}y+\overrightarrow{k}z\right)\ \ =\ \ \frac{\ \overrightarrow{r}}{{\left(\overline{r}\right)}^{\mathrm{2}}} \\ \\ \\ \\ \\ \left(b\right)\ \ \ Let\ \ \ \ \overline{r}=\left\|r\right\|=\sqrt{x^{\mathrm{2}}+y{}^{\mathrm{2}}_{\mathrm{\ }}{\mathrm{\ }}+z^{\mathrm{2}}}\ \ \ and\ \ \ \ {\left(\overline{r}\right)}^n={\left(\sqrt{x^{\mathrm{2}}+y{}^{\mathrm{2}}_{\mathrm{\ }}{\mathrm{\ }}+z^{\mathrm{2}}}\right)}^n \\ \\ \ {\left(\overline{r}\right)}^n.\ \left(\overline{r}\right)\ \ =\ \ {\left(\sqrt{x^{\mathrm{2}}+y{}^{\mathrm{2}}_{\mathrm{\ }}{\mathrm{\ }}+z^{\mathrm{2}}}\right)}^n\left(\sqrt{x^{\mathrm{2}}+y{}^{\mathrm{2}}_{\mathrm{\ }}{\mathrm{\ }}+z^{\mathrm{2}}}\right) \\ Then\ \ \ \\ \mathrm{\nabla }\times \left(\ {\left(\overline{r}\right)}^n.\ \left(\overline{r}\right)\right)\ \ =\ \ \mathrm{??} \\ U\mathrm{sin}g\ \ the\ \ identity\ \ that\ \ \mathrm{\nabla }\times \left(\ A.\ B\right)\ \ =\ \ A\mathrm{\nabla }\times \left(B\ \right)\ \ -\ B\mathrm{\nabla }\times \left(\ A\right) \\ \ \ \ \ \\ Then\ \ \mathrm{\nabla }\times \left(\ {\left(\overline{r}\right)}^n.\ \left(\overline{r}\right)\right)\ \ =\ \ \ {\left(\overline{r}\right)}^n\mathrm{\nabla }\times \left(\left(\overline{r}\right)\ \right)\ \ -\ \left(\overline{r}\right)\mathrm{\nabla }\times \left(\ {\left(\overline{r}\right)}^n\right)\ \ \\ \\ \mathrm{\nabla }\times \left(\ {\left(\overline{r}\right)}^n.\ \left(\overline{r}\right)\right)\ \ =\ \ \ {\left(\overline{r}\right)}^n\mathrm{\nabla }\times \left(\left(\overline{r}\right)\ \right)\ \ -\ \left(\overline{r}\right)\mathrm{\nabla }\times \left(\ {\left(\overline{r}\right)}^n\right)\ \ \ \ \ =\ \ \ 0-0\ \ \ \ =\ \ 0\ \ \left(\mathrm{Sin}ce\ \ \left(\overline{r}\right)=cons\mathrm{tan}t\ \right) \\$