By definition, c u r l ( x 2 z 2 , 2 y 2 z 2 , x y 2 z ) = ∇ × ( x 2 z 2 , 2 y 2 z 2 , x y 2 z ) curl(x^2z^2,2y^2z^2,xy^2z)=∇×(x^2z^2,2y^2z^2,xy^2z) c u r l ( x 2 z 2 , 2 y 2 z 2 , x y 2 z ) = ∇ × ( x 2 z 2 , 2 y 2 z 2 , x y 2 z ) or, equivalently,
∇ × ( x 2 z 2 , 2 y 2 z 2 , x y 2 z ) = ∣ i j k ∂ ∂ x ∂ ∂ y ∂ ∂ z x 2 z 2 2 y 2 z 2 x y 2 z ∣ ∇×(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} ∇ × ( x 2 z 2 , 2 y 2 z 2 , x y 2 z ) = ∣ ∣ i ∂ x ∂ x 2 z 2 j ∂ y ∂ 2 y 2 z 2 k ∂ z ∂ x y 2 z ∣ ∣
c u r l ( x 2 z 2 , 2 y 2 z 2 , x y 2 z ) = ( ∂ ∂ y ( x y 2 z ) − ∂ ∂ z ( 2 y 2 z 2 ) , ∂ ∂ z ( x 2 z 2 ) − ∂ ∂ x ( x y 2 z ) , ∂ ∂ x ( 2 y 2 z 2 ) − ∂ ∂ y ( x 2 z 2 ) ) 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))\\ c u r l ( x 2 z 2 , 2 y 2 z 2 , x y 2 z ) = ( ∂ y ∂ ( x y 2 z ) − ∂ z ∂ ( 2 y 2 z 2 ) , ∂ z ∂ ( x 2 z 2 ) − ∂ x ∂ ( x y 2 z ) , ∂ x ∂ ( 2 y 2 z 2 ) − ∂ y ∂ ( x 2 z 2 ))
Now, just plug in the found partial derivatives to get the curl:
c u r l ( x 2 z 2 , 2 y 2 z 2 , x y 2 z ) = ( 2 y z ( x − 2 y ) , z ( 2 x 2 − y 2 ) , 0 ) . curl(x^2z^2,2y^2z^2,xy^2z)=(2yz(x−2y),z(2x^2−y^2),0). c u r l ( x 2 z 2 , 2 y 2 z 2 , x y 2 z ) = ( 2 yz ( x − 2 y ) , z ( 2 x 2 − y 2 ) , 0 ) .
Finally, find the curl at the specific point.
( c u r l ( x 2 z 2 , 2 y 2 z 2 , x y 2 z ) ) ∣ ( ( x 0 , y 0 , z 0 ) = ( 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) ( c u r l ( x 2 z 2 , 2 y 2 z 2 , x y 2 z )) ∣ (( x 0 , y 0 , z 0 ) = ( 3 , 2 , 1 )) = ( − 4 , 14 , 0 )
The answer should be -4i + 14j
Comments