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(x2z2,2y2z2,xy2z)=×(x2z2,2y2z2,xy2z)curl(x^2z^2,2y^2z^2,xy^2z)=∇×(x^2z^2,2y^2z^2,xy^2z) or, equivalently,

×(x2z2,2y2z2,xy2z)=ijkxyzx2z22y2z2xy2z∇×(x^2z^2,2y^2z^2,xy^2z)= \begin{vmatrix} i & j & k\\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ x^2z^2 & 2y^2z^2 & xy^2z \end{vmatrix}

curl(x2z2,2y2z2,xy2z)=(y(xy2z)z(2y2z2),z(x2z2)x(xy2z),x(2y2z2)y(x2z2))curl(x^2z^2,2y^2z^2,xy^2z) =(\frac{∂}{∂y}(xy^2z)−\frac{∂}{∂z}(2y^2z^2),\frac{∂}{∂z}(x^2z^2)−\frac{∂}{∂x}(xy^2z),\frac{∂}{∂x}(2y^2z^2)−\frac{∂}{∂y}(x^2z^2))\\

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

curl(x2z2,2y2z2,xy2z)=(2yz(x2y),z(2x2y2),0).curl(x^2z^2,2y^2z^2,xy^2z)=(2yz(x−2y),z(2x^2−y^2),0).

Finally, find the curl at the specific point.

(curl(x2z2,2y2z2,xy2z))((x0,y0,z0)=(3,2,1))=(4,14,0)(curl(x^2z^2,2y^2z^2,xy^2z))|((x_0,y_0,z_0)=(3,2,1))=(−4,14,0)

The answer should be -4i + 14j



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