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}\n \\hat {i} & \\hat {j} & \\hat {k}\\\\\n \\frac {\\partial }{\\partial x} & \\frac {\\partial }{\\partial y} & \\frac {\\partial }{\\partial z}\\\\\n (y sin z - sin x ) & ( x sin z + 2yz) & (xy cos z + y^2)\n\\end{vmatrix}"
Now find the Determinant,
"- \\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 = 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"
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