punniya.v

2022-01-19

please provide answer for this question Show that the vector field by ⃗= (x^2+4yz)̂+ (y^2+4zx)̂+ (z^2+4xy) ̂ is irrotational. Hence find the scalar potential.

nick1337

We are saked to show that the vector field
$\stackrel{\to }{f}=\left({x}^{2}+4yz\right)\stackrel{\to }{i}+\left({y}^{2}+4zx\right)\stackrel{\to }{j}+\left({z}^{2}+4xy\right)\stackrel{\to }{k}$
is irrational. Also we are asked to find scalar potential. Since we know that a vector filed $\stackrel{\to }{A}$ is irrational if
$curl\stackrel{\to }{A}=\stackrel{\to }{0}$ or $\stackrel{\to }{\mathrm{\nabla }}×\stackrel{\to }{A}=\stackrel{\to }{0}$
Also with respect to a irrational field $\stackrel{\to }{A}$ there exist a scalar potential $\oslash$ such that $\stackrel{\to }{A}=\stackrel{\to }{\mathrm{\nabla }}\oslash$
Now solving for proving $\stackrel{\to }{f}$ irrational we will find $curl\stackrel{\to }{f}$
$curl\stackrel{\to }{f}=\stackrel{\to }{\mathrm{\nabla }}×\stackrel{\to }{f}=\left[\begin{array}{ccc}\stackrel{\to }{i}& \stackrel{\to }{j}& \stackrel{\to }{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ {f}_{x}& {f}_{y}& {f}_{z}\end{array}\right]$
Where $\frac{\partial }{\partial x},\frac{\partial }{\partial y}$ and $\frac{\partial }{\partial z}$ are partial derivatives with respect to and $\stackrel{\to }{f}={f}_{x}\stackrel{\to }{i}+{f}_{y}\stackrel{\to }{i}+{f}_{z}\stackrel{\to }{k}$ x,y and z respectively.
Now $\stackrel{\to }{\mathrm{\nabla }}×\stackrel{\to }{f}=\left[\begin{array}{ccc}\stackrel{\to }{i}& \stackrel{\to }{j}& \stackrel{\to }{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ {x}^{2}+4yz& {y}^{2}+4zx& {z}^{2}+4xy\end{array}\right]$
$=\stackrel{\to }{i}\left[\frac{\partial }{\partial y}\left({z}^{2}+4xy\right)-\frac{\partial }{\partial z}\left({y}^{2}+4zx\right)\right]-\stackrel{\to }{j}\left[\frac{\partial }{\partial x}\left({z}^{2}+4xy\right)-\frac{\partial }{\partial z}\left({x}^{2}+4yz\right)\right]+\stackrel{\to }{k}\left[\frac{\partial }{\partial x}\left({y}^{2}+4zx\right)-\frac{\partial }{\partial y}\left({x}^{2}+4yz\right)\right]$
$=\stackrel{\to }{i}\left(4x-4x\right)-\stackrel{\to }{j}\left(4y-4y\right)+\stackrel{\to }{k}\left(4z-4z\right)$
$=\stackrel{\to }{i}\left(0\right)-\stackrel{\to }{j}\left(0\right)+\stackrel{\to }{k}\left(0\right)$
$\stackrel{\to }{\mathrm{\nabla }}×\stackrel{\to }{f}=\stackrel{\to }{0}$
Since $\stackrel{\to }{\mathrm{\nabla }}×\stackrel{\to }{f}=\stackrel{\to }{0}$
so $\stackrel{\to }{f}$ is irrational vector field.
Now for scalar potential $\oslash$
since $\stackrel{\to }{f}=\mathrm{\nabla }\oslash =\left(\stackrel{\to }{i}\frac{\partial }{\partial x}+\stackrel{\to }{j}\frac{\partial }{\partial y}+\stackrel{\to }{k}\frac{\partial }{\partial z}\right)\oslash$
$\stackrel{\to }{f}=\frac{\partial \oslash }{\partial x}\stackrel{\to }{i}+\frac{\partial \oslash }{\partial y}\stackrel{\to }{j}+\frac{\partial \oslash }{\partial z}\stackrel{\to }{k}$
$\left({x}^{2}+4yz\right)\stackrel{\to }{i}+\left({y}^{2}+4zx\right)\stackrel{\to }{j}+\left({z}^{2}+4xy\right)\stackrel{\to }{k}=\frac{\partial \oslash }{\partial x}\stackrel{\to }{i}+\frac{\partial \oslash }{\partial y}\stackrel{\to }{j}+\frac{\partial \oslash }{\partial z}<$

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