Question #225909

If Ø(x, y, z) = x² - y²-2²-2, then ∇Ø at (2,1,-1) is


A) 4i +2j-2k

B)4i-2j-2k

C) 4i + 2j + 2k

D)4i - 2j +2k


1
Expert's answer
2021-08-16T02:19:54-0400

Ø(x,y,z)=x2y2z22Then Ø=xi,xj,xkØ=x(x2y2z22),x(x2y2z22),x(x2y2z22)Ø=2xi,2yj,2zkAt(2,1,1)Ø=4i+2j2kØ(x, y, z) = x² - y²-z²-2\\ Then \space ∇Ø= \frac{\partial}{\partial x}i, \frac{\partial}{\partial x}j, \frac{\partial}{\partial x}k\\ ∇Ø= \frac{\partial}{\partial x} (x² - y²-z²-2), \frac{\partial}{\partial x}(x² - y²-z²-2), \frac{\partial}{\partial x}(x² - y²-z²-2)\\ ∇Ø=2x i, 2yj,2zk\\ At (2,1,-1) \\ ∇Ø=4 i+ 2j-2k


Option A is correct.


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