Answer to Question #158034 in Calculus for Padma vasa

Question #158034

If a=(xz^(3))i-(2x^(2)yz)j+(2yz^(4))k find curl a at (1 -1 1)


1
Expert's answer
2021-01-26T04:38:31-0500
a=xz3i2x2yzj+2yz4k\vec a=xz^3\vec i-2x^2yz\vec j+2yz^4\vec k

curl a=×a=ijkxyzxz32x2yz2yz4\text{curl } \vec a=\nabla\times\vec a=\begin{vmatrix} \vec i & \vec j & \vec k \\ \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \\ xz^3& -2x^2yz & 2yz^4 \end{vmatrix}

=i(2z4(2x2y)j(03xz2)+k(4xyz0)=\vec i(2z^4-(-2x^2y)-\vec j(0-3xz^2)+\vec k(-4xyz-0)

=(2z4+2x2y)i+3xz2j4xyzk=(2z^4+2x^2y)\vec i+3xz^2\vec j-4xyz\vec k

curl a(1,1,1)=3j+4k\text{curl } \vec a\big|_{(1,-1,1)}=3\vec j+4\vec k




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