Answer on Question #56125 – Math – Vector Calculus
Let φ = 2 x ∧ z ∧ 4 − x ∧ 2 y \varphi = 2x^{\wedge}z^{\wedge}4 - x^{\wedge}2y φ = 2 x ∧ z ∧ 4 − x ∧ 2 y , find
∣ ∇ φ ∣ |\nabla \varphi| ∣∇ φ ∣
5(v97)
3(v112)
(v105)
2(v93)
Solution
∇ φ = ( ∂ φ ∂ x , ∂ φ ∂ y , ∂ φ ∂ z ) = ( ∂ ( 2 x z 4 − x 2 y ) ∂ x , ∂ ( 2 x z 4 − x 2 y ) ∂ y , ∂ ( 2 x z 4 − x 2 y ) ∂ z ) = ( 2 z 4 − 2 x y , − x 2 , 8 x z 3 ) . \nabla \varphi = \left(\frac{\partial \varphi}{\partial x}, \frac{\partial \varphi}{\partial y}, \frac{\partial \varphi}{\partial z}\right) = \left(\frac{\partial (2xz^4 - x^2y)}{\partial x}, \frac{\partial (2xz^4 - x^2y)}{\partial y}, \frac{\partial (2xz^4 - x^2y)}{\partial z}\right) = (2z^4 - 2xy, -x^2, 8xz^3). ∇ φ = ( ∂ x ∂ φ , ∂ y ∂ φ , ∂ z ∂ φ ) = ( ∂ x ∂ ( 2 x z 4 − x 2 y ) , ∂ y ∂ ( 2 x z 4 − x 2 y ) , ∂ z ∂ ( 2 x z 4 − x 2 y ) ) = ( 2 z 4 − 2 x y , − x 2 , 8 x z 3 ) . ∣ ∇ φ ∣ = ( 2 z 4 − 2 x y ) 2 + ( − x 2 ) 2 + ( 8 x z 3 ) 2 . |\nabla \varphi| = \sqrt{(2z^4 - 2xy)^2 + (-x^2)^2 + (8xz^3)^2}. ∣∇ φ ∣ = ( 2 z 4 − 2 x y ) 2 + ( − x 2 ) 2 + ( 8 x z 3 ) 2 .
If we need the number in the answer we need to choose a point ( x 0 , y 0 , z 0 ) (x_0, y_0, z_0) ( x 0 , y 0 , z 0 ) where we calculate ∣ ∇ φ ∣ |\nabla \varphi| ∣∇ φ ∣ .
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