Question

Prove that the vector f(r)r vector is
always irrotational

​

Answer 1

SOLUTION

TO PROVE

The vector $f(r) \vec{r}$ is always irrotational if f(r) is differentiable

CONCEPT TO BE IMPLEMENTED

A vector $\vec{v}$ is called irrational if

$curl \: \vec{v} = \nabla \times \vec{v} = 0$

PROOF

Here

$\nabla \times ( \vec{r}f(r))$

$= \nabla \times \bigg( xf(r) \hat{i} + yf(r) \hat{j} +zf(r) \hat{k} \bigg)$

$= \displaystyle\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ \\ \frac{ \partial}{ \partial x} & \frac{ \partial}{ \partial y} & \frac{ \partial}{ \partial z} \\ \\ xf(r) & yf(r) & zf(r) \end{vmatrix}$

$= \displaystyle \hat{i} \bigg(z\frac{ \partial f}{ \partial y} -y\frac{ \partial f}{ \partial z} \bigg) + \hat{j} \bigg(x\frac{ \partial f}{ \partial z} -z\frac{ \partial f}{ \partial x} \bigg) + \hat{k} \bigg(y\frac{ \partial f}{ \partial x} -x\frac{ \partial f}{ \partial y} \bigg)$

Now

$\displaystyle \frac{ \partial f}{ \partial x}$

$= \displaystyle \bigg(\frac{ \partial f}{ \partial r} \bigg) \bigg(\frac{ \partial r}{ \partial x} \bigg)$

$= \displaystyle \frac{ \partial f}{ \partial r} \bigg(\frac{ \partial }{ \partial x} \sqrt{ {x}^{2} + {y}^{2} + {z}^{2} } \bigg)$

$= \displaystyle \frac{ x f'(r)}{\sqrt{ {x}^{2} + {y}^{2} + {z}^{2} }}$

$= \displaystyle \frac{ x f'(r)}{r }$

Thus

$\displaystyle \frac{ \partial f}{ \partial x} = \displaystyle \frac{ x f'(r)}{r }$

$\displaystyle \frac{ \partial f}{ \partial y} = \displaystyle \frac{ y f'(r)}{r }$

$\displaystyle \frac{ \partial f}{ \partial z} = \displaystyle \frac{ z f'(r)}{r }$

Therefore

$\nabla \times ( \vec{r}f(r))$

$= \displaystyle \hat{i} \bigg(z\frac{ y f'(r)}{r } -y\frac{ z f'(r)}{r } \bigg) + \hat{j} \bigg(x\frac{ z f'(r)}{r } -z\frac{ x f'(r)}{r } \bigg)+ \hat{k} \bigg(y\frac{ x f'(r)}{r } -x\frac{ y f'(r)}{r } \bigg)$

$= \hat{ 0 }$

Hence proved

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