Answer to Question #225911 in Chemical Engineering for Lokika

Question #225911

If V = x²z²i-2y²z²j+ xy²zk then Curl V at (3,2,1) is


A) 20i + 14k


B)2i + 14k


C)14j+20k


D) 20i + 14j


1
Expert's answer
2021-08-18T08:23:02-0400

By definition, "curl(x^2z^2,2y^2z^2,xy^2z)=\u2207\u00d7(x^2z^2,2y^2z^2,xy^2z)" or, equivalently,

"\u2207\u00d7(x^2z^2,2y^2z^2,xy^2z)= \\begin{vmatrix}\n i & j & k\\\\\n \\frac{\\partial }{\\partial x} & \\frac{\\partial }{\\partial y} & \\frac{\\partial }{\\partial z}\\\\\nx^2z^2 & 2y^2z^2 & xy^2z\n\\end{vmatrix}"

"curl(x^2z^2,2y^2z^2,xy^2z) =(\\frac{\u2202}{\u2202y}(xy^2z)\u2212\\frac{\u2202}{\u2202z}(2y^2z^2),\\frac{\u2202}{\u2202z}(x^2z^2)\u2212\\frac{\u2202}{\u2202x}(xy^2z),\\frac{\u2202}{\u2202x}(2y^2z^2)\u2212\\frac{\u2202}{\u2202y}(x^2z^2))\\\\"

Now, just plug in the found partial derivatives to get the curl:

"curl(x^2z^2,2y^2z^2,xy^2z)=(2yz(x\u22122y),z(2x^2\u2212y^2),0)."

Finally, find the curl at the specific point.

"(curl(x^2z^2,2y^2z^2,xy^2z))|((x_0,y_0,z_0)=(3,2,1))=(\u22124,14,0)"

The answer should be -4i + 14j



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