Answer to Question #225909 in Chemical Engineering for Lokika

Question

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

Answer 1

$Ø(x, y, z) = x² - y²-z²-2\\ Then \space ∇Ø= \frac{\partial}{\partial x}i, \frac{\partial}{\partial x}j, \frac{\partial}{\partial x}k\\ ∇Ø= \frac{\partial}{\partial x} (x² - y²-z²-2), \frac{\partial}{\partial x}(x² - y²-z²-2), \frac{\partial}{\partial x}(x² - y²-z²-2)\\ ∇Ø=2x i, 2yj,2zk\\ At (2,1,-1) \\ ∇Ø=4 i+ 2j-2k$

Option A is correct.