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# Answer to Question #278380 in Vector Calculus for Mafin

Question #278380

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

1
2021-12-13T13:51:08-0500

"\\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}}} \\\\ \nSuch\\ \\ \\ that\\ \\ \\mathrm{ln}\\left(\\overline{r}\\right)=\\mathrm{ln}\\left(\\sqrt{x^{\\mathrm{2}}+y{}^{\\mathrm{2}}_{\\mathrm{\\ }}{\\mathrm{\\ }}+z^{\\mathrm{2}}}\\right) \\\\ \n\\mathrm{Re}call\\ \\ that\\ \\ \\mathrm{\\nabla }=\\overrightarrow{i}\\frac{\\partial }{\\partial x}\\ \\ +\\ \\ \\overrightarrow{j}\\frac{\\partial }{\\partial y}+\\overrightarrow{k}\\frac{\\partial }{\\partial z} \\\\ \nThen\\ \\ \\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) \\\\ \n\\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) \\\\ \n\\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) \\\\ \n\\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) \\\\ \n\\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}}} \\\\ \n \\\\ \\\\ \\\\ \\\\\n\\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 \\\\ \n \\\\ \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) \\\\ \nThen\\ \\ \\ \\\\ \n\\mathrm{\\nabla }\\times \\left(\\ {\\left(\\overline{r}\\right)}^n.\\ \\left(\\overline{r}\\right)\\right)\\ \\ =\\ \\ \\mathrm{??} \\\\ \nU\\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) \\\\ \n\\ \\ \\ \\ \\\\ \nThen\\ \\ \\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)\\ \\ \\\\ \n \\\\ \n\\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) \\\\"

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