Answer to Question #98762 in Calculus for navya

Question #98762
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.
1
Expert's answer
2019-11-18T11:04:12-0500

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,




"\\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"


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