( a ) L e t r → = x i → + y j → + z k → a n d r ‾ = ∥ r ∥ = x 2 + y 2 + z 2 S u c h t h a t l n ( r ‾ ) = l n ( x 2 + y 2 + z 2 ) R e c a l l t h a t ∇ = i → ∂ ∂ x + j → ∂ ∂ y + k → ∂ ∂ z T h e n ∇ ( l n ( r ‾ ) ) = ( i → ∂ ∂ x + j → ∂ ∂ y + k → ∂ ∂ z ) l n ( r ‾ ) ∇ ( l n ( r ‾ ) ) = ( i → ∂ ∂ x + j → ∂ ∂ y + k → ∂ ∂ z ) l n ( x 2 + y 2 + z 2 ) ∇ ( l n ( r ‾ ) ) = ( i → ∂ l n ( x 2 + y 2 + z 2 ) ∂ x + j → ∂ l n ( x 2 + y 2 + z 2 ) ∂ y + k → ∂ l n ( x 2 + y 2 + z 2 ) ∂ z ) ∇ ( l n ( r ‾ ) ) = ( i → ∂ r ∂ x 1 x 2 + y 2 + z 2 + j → ∂ r ∂ y 1 x 2 + y 2 + z 2 + k → ∂ r ∂ z 1 x 2 + y 2 + z 2 ) ∇ ( l n ( r ‾ ) ) = 1 x 2 + y 2 + z 2 ( i → x + j → y + k → z ) = r → ( r ‾ ) 2 ( b ) L e t r ‾ = ∥ r ∥ = x 2 + y 2 + z 2 a n d ( r ‾ ) n = ( x 2 + y 2 + z 2 ) n ( r ‾ ) n . ( r ‾ ) = ( x 2 + y 2 + z 2 ) n ( x 2 + y 2 + z 2 ) T h e n ∇ × ( ( r ‾ ) n . ( r ‾ ) ) = ? ? U s i n g t h e i d e n t i t y t h a t ∇ × ( A . B ) = A ∇ × ( B ) − B ∇ × ( A ) T h e n ∇ × ( ( r ‾ ) n . ( r ‾ ) ) = ( r ‾ ) n ∇ × ( ( r ‾ ) ) − ( r ‾ ) ∇ × ( ( r ‾ ) n ) ∇ × ( ( r ‾ ) n . ( r ‾ ) ) = ( r ‾ ) n ∇ × ( ( r ‾ ) ) − ( r ‾ ) ∇ × ( ( r ‾ ) n ) = 0 − 0 = 0 ( S i n c e ( r ‾ ) = c o n s t a n t ) \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}}} \\
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\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 \\
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\ {\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) \\
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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)\ \ \\
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\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) \\ ( a ) L e t r = x i + y j + z k an d r = ∥ r ∥ = x 2 + y 2 + z 2 S u c h t ha t ln ( r ) = ln ( x 2 + y 2 + z 2 ) Re c a ll t ha t ∇ = i ∂ x ∂ + j ∂ y ∂ + k ∂ z ∂ T h e n ∇ ( ln ( r ) ) = ( i ∂ x ∂ + j ∂ y ∂ + k ∂ z ∂ ) ln ( r ) ∇ ( ln ( r ) ) = ( i ∂ x ∂ + j ∂ y ∂ + k ∂ z ∂ ) ln ( x 2 + y 2 + z 2 ) ∇ ( ln ( r ) ) = ( i ∂ x ∂ ln ( x 2 + y 2 + z 2 ) + j ∂ y ∂ ln ( x 2 + y 2 + z 2 ) + k ∂ z ∂ ln ( x 2 + y 2 + z 2 ) ) ∇ ( ln ( r ) ) = ( i ∂ x ∂ r x 2 + y 2 + z 2 1 + j ∂ y ∂ r x 2 + y 2 + z 2 1 + k ∂ z ∂ r x 2 + y 2 + z 2 1 ) ∇ ( ln ( r ) ) = x 2 + y 2 + z 2 1 ( i x + j y + k z ) = ( r ) 2 r ( b ) L e t r = ∥ r ∥ = x 2 + y 2 + z 2 an d ( r ) n = ( x 2 + y 2 + z 2 ) n ( r ) n . ( r ) = ( x 2 + y 2 + z 2 ) n ( x 2 + y 2 + z 2 ) T h e n ∇ × ( ( r ) n . ( r ) ) = ?? U sin g t h e i d e n t i t y t ha t ∇ × ( A . B ) = A ∇ × ( B ) − B ∇ × ( A ) T h e n ∇ × ( ( r ) n . ( r ) ) = ( r ) n ∇ × ( ( r ) ) − ( r ) ∇ × ( ( r ) n ) ∇ × ( ( r ) n . ( r ) ) = ( r ) n ∇ × ( ( r ) ) − ( r ) ∇ × ( ( r ) n ) = 0 − 0 = 0 ( Sin ce ( r ) = co n s tan t )
#332078 on Oct 2022
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