Question

If $\overrightarrow{V}$ is a differentiable vector function and $f$ is a sufficient differentiable scalar function, then curl $\left(f\overrightarrow{V}\right)$ is equal to

• A. $\left(gradf\right)\times \left(\overrightarrow{V}\right)+f\left(curl\overrightarrow{V}\right)$
• B. $0$
• C. $fcurl\left(\overrightarrow{V}\right)$
• D. $\left(gradf\right)\times \left(\overrightarrow{V}\right)$

Answer 1

The correct option is A $\left(gradf\right)\times \left(\overrightarrow{V}\right)+f\left(curl\overrightarrow{V}\right)$

Let $\overrightarrow{V}=u\hat{i}+v\hat{j}+w\hat{k}$
$curl\left(f\overrightarrow{v}\right)=\left|\begin{array}{ccc}i& j& k\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ uf& vf& wf\end{array}\right|$
$=\left[\frac{\partial }{\partial y}\right(wf)-\frac{\partial }{\partial z}(vf\left)\right]\hat{i}+\left[\frac{\partial }{\partial z}\right(uf)-\frac{\partial }{\partial x}(wf\left)\right]\hat{j}+\left[\frac{\partial }{\partial x}\right(vf)-\frac{\partial }{\partial y}(uf\left)\right]\hat{k}$
$=f[(\frac{\partial w}{\partial y}-\frac{\partial v}{\partial z})\hat{i}+(\frac{\partial u}{\partial z}-\frac{\partial w}{\partial x})\hat{j}+(\frac{\partial v}{\partial x}-\frac{\partial u}{\partial y})\hat{k}]$
$+[(w\frac{\partial f}{\partial y}-\frac{\partial f}{\partial z})\hat{i}+(u\frac{\partial f}{\partial z}-\frac{\partial f}{\partial x})\hat{j}+(v\frac{\partial f}{\partial x}-\frac{\partial f}{\partial y})\hat{k}]$
$curl\left(f\overrightarrow{V}\right)=f\left(curl\overrightarrow{V}\right)+(\frac{\partial f}{\partial x}\hat{i}+\frac{\partial f}{\partial y}\hat{j}+\frac{\partial f}{\partial z}\hat{k})\times (u\hat{i}+v\hat{j}+w\hat{k})$
$curl\left(f\overrightarrow{V}\right)=f\left(curl\overrightarrow{V}\right)+\left(gradf\right)\times \overrightarrow{V}$

Alternative Solution:
$Curl\left(f\overrightarrow{V}\right)=▽\times \left(f\overrightarrow{V}\right)$
$=(▽f)\times \overrightarrow{V}+f(▽\times \overrightarrow{V})$
$f\left(curl\overrightarrow{V}\right)=\left(gradf\right)\times \overrightarrow{V}+f\left(curl\overrightarrow{V}\right)$
It is a vector identity just learn it without paying attention on its proof.