Question

Prove that div grad r^n = n(n+1) r^n-2​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO PROVE

$\sf{div \: grad \: ( {r}^{n} ) = n(n + 1) \: {r}^{n - 2} \: }$

CONCEPT TO BE IMPLEMENTED

$\sf{div \: grad \: ( {r}^{n} ) = { \nabla}^{2} \: ( {r}^{n} )}$

CALCULATION

$\sf{div \: grad \: ( {r}^{n} ) = { \nabla}^{2} \: ( {r}^{n} )}$

$= \displaystyle \sf{ \frac{ { \partial}^{2}( {r}^{n} )}{ \partial {x}^{2} } +\frac{ { \partial}^{2}( {r}^{n} )}{ \partial {y}^{2} } +\frac{ { \partial}^{2}( {r}^{n} )}{ \partial {z}^{2} } \: }$

Now

$\sf{ {r}^{2} = {x}^{2} + {y}^{2} + {z}^{2} \: }$

Differentiating both sides with respect to x partially we get

$\displaystyle {\sf \: 2r \frac{ \partial r}{ \partial x} = 2x}$

$\therefore \: \displaystyle {\sf \: \frac{ \partial r}{ \partial x} = \frac{x}{r} }$

Similarly

$\therefore \: \displaystyle {\sf \: \frac{ \partial r}{ \partial y} = \frac{y}{r} }$

$\therefore \: \displaystyle {\sf \: \frac{ \partial r}{ \partial z} = \frac{z}{r} }$

Now

$\displaystyle \sf{ \frac{ { \partial}^{}( {r}^{n} )}{ \partial {x}^{} } }$

$\displaystyle \sf{ = n {r}^{n - 1} \frac{ { \partial}^{}{r}}{ \partial {x}^{} } }$

$\displaystyle \sf{ = n {r}^{n - 1} \times \frac{x}{r} }$

$\displaystyle \sf{ = x \: n \: {r}^{n - 2} }$

Now

$\displaystyle \sf{ \frac{ { \partial}^{2}( {r}^{n} )}{ \partial {x}^{2} } }$

$\displaystyle \sf{ = n \bigg( {r}^{n - 2} + (n - 2) {r}^{n - 3}x \: \frac{ \partial r}{ \partial x} \bigg)}$

$\displaystyle \sf{ = n \bigg( {r}^{n - 2} + (n - 2) {r}^{n - 4} \: {x}^{2} \bigg)}$

Similarly

$\displaystyle \sf{ \frac{ { \partial}^{2}( {r}^{n} )}{ \partial {y}^{2} } } = \displaystyle \sf{ = n \bigg( {r}^{n - 2} + (n - 2) {r}^{n - 4} \: {y}^{2} \bigg)}$

$\displaystyle \sf{ \frac{ { \partial}^{2}( {r}^{n} )}{ \partial {z}^{2} } } = \displaystyle \sf{ = n \bigg( {r}^{n - 2} + (n - 2) {r}^{n - 4} \: {z}^{2} \bigg)}$

So

$\displaystyle \sf{ \frac{ { \partial}^{2}( {r}^{n} )}{ \partial {x}^{2} } +\frac{ { \partial}^{2}( {r}^{n} )}{ \partial {y}^{2} } +\frac{ { \partial}^{2}( {r}^{n} )}{ \partial {z}^{2} } \: }$

$\displaystyle \sf{ = n \bigg( 3{r}^{n - 2} + (n - 2) {r}^{n - 4} \:( {x}^{2} + {y}^{2} + {z}^{2}) \bigg)}$

$\displaystyle \sf{ = n \bigg( 3{r}^{n - 2} + (n - 2) {r}^{n - 4} \: {r}^{2} \bigg)}$

$\displaystyle \sf{ = n \bigg( 3{r}^{n - 2} + (n - 2) {r}^{n - 2} \: \bigg)}$

$\displaystyle \sf{ = n (n + 1) {r}^{n - 2} }$

Therefore

$\sf{div \: grad \: ( {r}^{n} ) }\displaystyle \sf{ = n (n + 1) {r}^{n - 2} }$

Hence proved

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LEARN MORE FROM BRAINLY

Divergence of r / r^3 is

(a) zero at the origin

(b) zero everywhere

(c) zero everywhere except the origin

(d) nonzero

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Answer 2

Answer:

Step-by-step explanation: