Find the directional derivative of ∇ · (∇ϕ) at point (1, −2, 1) in the direction of the normal to the surface xy²z = 3x + z² where ϕ = 2x⁴yz³

Respuesta :

To find the directional derivative of ∇ · (∇ϕ) at the point (1, -2, 1) in the direction of the normal to the surface xy²z = 3x + z² where ϕ = 2x⁴yz³, we first need to find the gradient of ϕ and the normal to the surface.

The gradient of ϕ is given by:

∇ϕ = (∂ϕ/∂x, ∂ϕ/∂y, ∂ϕ/∂z)

= (8x³yz³, 6x⁴z², 6x⁴y²z²)

The normal to the surface is given by the gradient of the surface function xy²z - 3x - z²:

∇(xy²z - 3x - z²) = (y²z - 3 - 2z, 2xyz, xy² - 2x)

Now, to find the directional derivative of ∇ · (∇ϕ) in the direction of the normal to the surface at the given point, we compute the dot product of the gradient of ϕ and the normal to the surface at that point:

∇ · (∇ϕ) = ∇ϕ · ∇

= (8x³yz³, 6x⁴z², 6x⁴y²z²) · (y²z - 3 - 2z, 2xyz, xy² - 2x)

We evaluate this expression at the point (1, -2, 1) to find the directional derivative.