Question

If given that :-

$\quad \leadsto \quad { \bf q = t^{n} e^{\bigg(- \dfrac{r²}{4t} \bigg)} }$

Then for what value of 'n' . The following equation will be satisfied ?

$\quad \leadsto \quad { \bf \dfrac{1}{r²} \cdot \dfrac{\partial}{\partial r} \bigg\{ r² \bigg( \dfrac{\partial q}{\partial r} \bigg) \bigg\} = \dfrac{\partial q}{\partial t}}$
​

Answer 1

Step-by-step explanation:

Answer:

$\boxed{\huge{\mathbb\red{REFERR \:TO\: THE\:\: ATTACHMENT }}}$

Step-by-step explanation:

hope it help you.

thanks

Answer 2

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: q = t^{n} e^{\bigg(- \dfrac{r²}{4t} \bigg)} - - - - (1)$

On differentiating partially w. r. t. r, we get

$\rm \: \dfrac{\partial }{\partial r} q = t^{n}\dfrac{\partial }{\partial r} e^{\bigg(- \dfrac{r²}{4t} \bigg)}$

$\rm \: \dfrac{\partial q}{\partial r} = t^{n} e^{\bigg(- \dfrac{r²}{4t} \bigg)}\dfrac{\partial }{\partial r}\bigg( - \dfrac{ {r}^{2} }{4t} \bigg)$

$\rm \: \dfrac{\partial q}{\partial r} = q\bigg( - \dfrac{ 2r }{4t} \bigg)$

$\rm \: \dfrac{\partial q}{\partial r} = q\bigg( - \dfrac{ r }{2t} \bigg)$

$\rm \: \dfrac{\partial q}{\partial r} = - \dfrac{ qr }{2t} - - - - (i)$

So,

$\rm \: {r}^{2} \dfrac{\partial q}{\partial r} = - \dfrac{ q {r}^{3}}{2t}$

On differentiating both sides w. r. t. r, we get

$\rm \:\dfrac{\partial }{\partial r}\bigg( {r}^{2} \dfrac{\partial q}{\partial r} \bigg)= - \bigg(\dfrac{3 q {r}^{2}}{2t} + \dfrac{ {r}^{3} }{2t}\dfrac{\partial q}{\partial r} \bigg)$

$\rm\implies \:\rm \:\dfrac{1}{ {r}^{2} } \dfrac{\partial }{\partial r}\bigg( {r}^{2} \dfrac{\partial q}{\partial r} \bigg)= - \bigg(\dfrac{3 q }{2t} + \dfrac{ r }{2t}\dfrac{\partial q}{\partial r} \bigg) - - - (2)$

Now, Consider again

$\rm \: q = t^{n} e^{\bigg(- \dfrac{r²}{4t} \bigg)}$

On differentiating both sides w. r. t. t, we get

$\rm \: \dfrac{\partial }{\partial t}q =\dfrac{\partial }{\partial t} t^{n} e^{\bigg(- \dfrac{r²}{4t} \bigg)}$

$\rm \: \dfrac{\partial q}{\partial t} = t^{n}\dfrac{\partial }{\partial t} e^{\bigg(- \dfrac{r²}{4t} \bigg)} + e^{\bigg(- \dfrac{r²}{4t} \bigg)}\dfrac{\partial }{\partial t} {t}^{n}$

$\rm \: = t^{n} e^{\bigg(- \dfrac{r²}{4t} \bigg)}\dfrac{\partial }{\partial t}\bigg( - \dfrac{ {r}^{2} }{4t} \bigg) + e^{\bigg(- \dfrac{r²}{4t} \bigg)}(n {t}^{n - 1})$

$\rm \: = q\bigg(\dfrac{ {r}^{2} }{4 {t}^{2} } \bigg) + \dfrac{n {t}^{n} }{t} e^{\bigg(- \dfrac{r²}{4t} \bigg)}$

$\rm \: = q\bigg(\dfrac{ {r}^{2} }{4 {t}^{2} } \bigg) + \dfrac{nq}{t}$

$\rm \: = \dfrac{q {r}^{2} }{4 {t}^{2} } + \dfrac{nq}{t}$

$\rm\implies \:\rm \:\dfrac{\partial \: q }{\partial t} = \dfrac{q {r}^{2} }{4 {t}^{2} } + \dfrac{nq}{t} - - - (3)$

According to statement,

$\rm \: \quad { \bf \dfrac{1}{r²} \cdot \dfrac{\partial}{\partial r} \bigg\{ r² \bigg( \dfrac{\partial q}{\partial r} \bigg) \bigg\} = \dfrac{\partial q}{\partial t}}$

On substituting the values from equation (2) and (3), we get

$\rm \: - \bigg(\dfrac{3 q }{2t} + \dfrac{ r }{2t}\dfrac{\partial q}{\partial r} \bigg) = \dfrac{q {r}^{2} }{4 {t}^{2} } + \dfrac{nq}{t}$

On substituting the value from equation (i), we get

$\rm \: - \bigg(\dfrac{3 q }{2t} - \dfrac{q {r}^{2} }{ {4t}^{2} }\bigg) = \dfrac{q {r}^{2} }{4 {t}^{2} } + \dfrac{nq}{t}$

$\rm \: - \dfrac{3 q }{2t} + \dfrac{q {r}^{2} }{ {4t}^{2} } = \dfrac{q {r}^{2} }{4 {t}^{2} } + \dfrac{nq}{t}$

$\rm \: - \dfrac{3 q }{2t} = \dfrac{nq}{t}$

$\rm\implies \:n \: = \: - \: \dfrac{3}{2}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

FORMULAE USED

$\rm \: \dfrac{d}{dx} {e}^{x} = {e}^{x} \:$

$\rm \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \:$

$\rm \: \dfrac{d}{dx}uv = u\dfrac{d}{dx}v \: + \: v\dfrac{d}{dx}u$

$\rm \: \dfrac{d}{dx}k \: f(x) = k \: \dfrac{d}{dx}f(x)$