Question

A fluid motion is given by ~v = (y sin z - sin x)ˆi + (x sin z+ 2 yz)ˆj + (xy cos z + y
2
)
ˆk. Is
the motion irrotational? If so, find the potential function of ~v.

Answer 1

Potential of a function

We need to find the potential function of the given function if it is an irrational.

Solution:

The given function is,

$\bar {v} = (y sin z - sin x ) \hat {i} + ( x sin z + 2yz ) \hat {j} + (xy cos z +y^2 ) \hat {k }$

the motion vector is irrational if

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

So,

$$\vec \nabla \times \vec {v} = \begin{vmatrix} \hat {i} & \hat {j} & \hat {k}\\ \frac {\partial }{\partial x} & \frac {\partial }{\partial y} & \frac {\partial }{\partial z}\\ (y sin z - sin x ) & ( x sin z + 2yz) & (xy cos z + y^2) \end{vmatrix}$$

Now find the Determinant,

$$\vec \nabla \times \vec {v} = \hat {i} \space [ \frac {\partial }{\partial y} (x y cos z+ y^2) - \frac {\partial }{\partial z} (x sin z + 2yz ]$$

$$- \space \hat {j} \space [\frac {\partial }{\partial x} (xy cos z + y^2) - \frac {\partial }{\partial z} ( y sin z - sin x ]$$

$$+ \hat {k} \space [ \frac {\partial }{\partial x} (x sin z + 2yz ) - \frac {\partial }{\partial y} (y sin z- sin x)]$$

$=\hat {i} (x \space cos z + 2y - x cos z -2y) - \hat {j} [ y cos z y cos z] + \hat {k} [ sin z - sinz]$

= $= \hat {i} (0) - \hat {j} (0) + \hat {k} (0) = \vec {0}$

Therefore,

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

So,

$$\vec {v} \space is \space irrational$$

Now we can find the potential function

Let the potential function is $f(x, y, z)$

We know, the potential function must satisfy the below equation,

i.e.

$$\vec \nabla f = \vec v$$

We know,

$$df = \vec\nabla f . ( x \hat {i} + y \hat {j} + z \hat {k} )$$

$$df = \vec v . ( x \hat {i} + y \hat {j} + z \hat {k} )$$

(since , $\vec \nabla = v )$

$$df =( { (y sin z - sin x ) \hat {i} + ( x sin z + 2yz ) \hat {j} + (xy cos z +y^2 ) \hat {k } }) . \space( dx \hat {i} + dy \hat {j} + dz \hat {k} )$$

$df = y sin z dx - sinx dx + x sin z dy + 2 y z dy + x y cos z dz + y^2 dz$

$$df = (y sin z dx+ x sin z dy + xy cos z dz) + (2yz dy + y^2 dz) - sin x dx$$

$$df = d(xysinz ) + d (y^2 z) + d(cos x)$$

$$df = d( xy sin z + y^2 z + cos x)$$

Integrating, we get

$$f (x, y, z) = xy \space sin z + y^2 \space z + cos x$$

Answer:

potential function of the given function is

$$f (x, y, z) = xy \space sin z + y^2 \space z + cos x$$