Question

For a scaler field ϕ(x, y, z) = x^n+y^n+z^n, show that (🔻ϕ ).r = nϕ , where n is a non-zero real constant.

Answer 1

[AI-generated answer — Explain Bot]

To show that (🔻ϕ).r = nϕ for the scalar field ϕ(x, y, z) = x^n + y^n + z^n, where n is a non-zero real constant, we need to find the Laplacian of ϕ and then compute the dot product of the gradient of ϕ with the position vector r.

First, let's find the Laplacian of ϕ:
Laplacian of ϕ, (🔻ϕ), is given by:

(🔻ϕ) = (∂²ϕ/∂x²) + (∂²ϕ/∂y²) + (∂²ϕ/∂z²)

To find the second partial derivatives of ϕ, let's differentiate ϕ with respect to each variable separately:

∂ϕ/∂x = n*x^(n-1)
∂²ϕ/∂x² = n(n-1)x^(n-2)

∂ϕ/∂y = n*y^(n-1)
∂²ϕ/∂y² = n(n-1)y^(n-2)

∂ϕ/∂z = n*z^(n-1)
∂²ϕ/∂z² = n(n-1)z^(n-2)

Now, let's add up the second partial derivatives:

(🔻ϕ) = n(n-1)(x^(n-2) + y^(n-2) + z^(n-2))

Next, we calculate the gradient of ϕ, (∇ϕ):

∇ϕ = (∂ϕ/∂x)i + (∂ϕ/∂y)j + (∂ϕ/∂z)k
= (n*x^(n-1))i + (n*y^(n-1))j + (n*z^(n-1))k

Finally, let's compute the dot product between (∇ϕ) and r:

(∇ϕ) . r = (n*x^(n-1))x + (n*y^(n-1))y + (n*z^(n-1))z
= n*x^n + n*y^n + n*z^n
= nϕ

So, we have shown that (🔻ϕ).r = nϕ for the given scalar field.