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vampirchik [111]
4 years ago
5

Find the directional derivative of the function at the given point in the direction of the vector v. f(x, y, z) = xey + yez + ze

x, (0, 0, 0), v = 4, 3, −1
Mathematics
1 answer:
attashe74 [19]4 years ago
8 0

Answer: 6 / √26

Step-by-step explanation:

Given that f(x, y, z) = xe^y + ye^z + ze^x

so first we compute the gradient vector at (0, 0, 0)

Δf ( x, y, z ) = [ e^y + ze^x,  xe^y + e^z,  ye^z + e^x ]

Δf ( 0, 0, 0 ) = [ e⁰ + 0(e)⁰, 0(e)⁰ + e⁰, 0(e)⁰ + e⁰ ] = [ 1+0 , 0+1, 0+1 ] = [ 1, 1, 1 ]

Now we were also given that  V = < 4, 3, -1 >

so ║v║ = √ ( 4² + 3² + (-1)² )

║v║ = √ ( 16 + 9 + 1 )

║v║ = √ 26

It must be noted that "v"  is not a unit vector but since ║v║ = √ 26, the unit vector in the direction of "V" is ⊆ = ( V / ║v║)

so

⊆ =  ( V / ║v║) = [ 4/√26, 3/√26, -1/√26 ]

therefore by equation   D⊆f ( x, y, z ) = Δf ( x, y, z ) × ⊆

D⊆f ( x, y, z ) = Δf ( 0, 0, 0 ) × ⊆ = [ 1, 1, 1 ] × [ 4/√26, 3/√26, -1/√26 ]

= ( 1×4 + 1×3 -1×1 ) / √26

= (4 + 3 - 1) / √26

= 6 / √26

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