Question #229231

If Ø(x, y, z) = x² - y²-2²-2, then ∇Ø at (2,1,-1) is ,,,choose the correct answer?


A) 4i +2j-2k

B)4i-2j-2k

C) 4i + 2j + 2k

D)4i - 2j +2k


1
Expert's answer
2021-08-31T23:57:14-0400

By definition, curl(x2y2z22)=×(x2y2z22)curl(x² - y² -z²-2)=∇×(x² - y²-z²-2) or, equivalently,

×(x2y2z22)=ijkxyzx2y2z22∇×(x² - y²-z²-2)= \begin{vmatrix} i & j & k\\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z}\\ x^2 & y^2 & z^2-2 \end{vmatrix}

curl(x2,y2,z22)=(y(y2)z(z22),z(z22)x(x2),x(x2)y(y2))curl(x^2,y^2,z^2-2) =(\frac{∂}{∂y}(y^2)−\frac{∂}{∂z}(z^2-2),\frac{∂}{∂z}(z^2-2)−\frac{∂}{∂x}(x^2),\frac{∂}{∂x}(x^2)−\frac{∂}{∂y}(y^2))\\

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

curl(x2,y2,z22)=(2y2z,2z2x,2x2y).curl(x^2,y^2,z^2-2)=(2y-2z,2z-2x,2x-2y).

Finally, find the curl at the specific point.

(curl(x2,y2,z22))((x0,y0,z0)=(2,1,1))=(4,6,2)(curl(x^2,y^2,z^2-2))|((x_0,y_0,z_0)=(2,1,-1))=(4,-6,2)

The answer should be 4i - 6j+2k


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