Mathematics : asked on AQUI

12.03.2022

If vector A=xz³î-2x²yj+2yz⁴k find CURL A at the point [1, -1, 1]

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09.07.2023, solved by verified expert

Mathematics, English, Physics. Research Scholar at Indian Institute of Technology, Roorkee

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Given that

its curl is

so that at the point (1, -1, 1), the curl is

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Mathematics

If vector A=xz³î-2x²yj+2yz⁴k find CURL A at the point [1, -1, 1]

Step-by-step answer

P Answered by Specialist

1

Given that

$\vec A = xz^3 \, \vec\imath - 2x^2 y \, \vec\jmath + 2yz^4 \, \vec k$

its curl is

$\displaystyle \nabla\times\vec A = \left(\frac{\partial\left(2yz^4\right)}{\partial y} - \frac{\partial\left(-2x^2y\right)}{\partial z}\right) \, \vec\imath - \left(\frac{\partial\left(2yz^4\right)}{\partial x} - \frac{\partial\left(xz^3\right)}{\partial z}\right) \, \vec\jmath \\ ~~~~~~~~~~~~ + \left(\frac{\partial\left(-2x^2y\right)}{\partial x} - \frac{\partial\left(xz^3\right)}{\partial y}\right) \, \vec k$

$\nabla\times\vec A = 2z^4 \, \vec\imath + 3xz^2 \, \vec\jmath - 4xy \, \vec k$

so that at the point (1, -1, 1), the curl is

$\nabla\times\vec A \bigg|_{(x,y,z)=(1,-1,1)} = \boxed{2 \, \vec\imath + 3 \, \vec\jmath + 4 \, \vec k}$

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P Answered by PhD

SI=(P*R*T)/100

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