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, where ϕ(x, y, z) = xn + yn + zn and n is a non-zero real constant, we need to find the gradient of ϕ and evaluate (∇ϕ)·r.

First, let's find the gradient of ϕ, denoted as ∇ϕ. The gradient is a vector that contains the partial derivatives of the scalar field with respect to each variable (x, y, and z). Each component of the gradient will be a partial derivative of ϕ.

∇ϕ = (∂ϕ/∂x, ∂ϕ/∂y, ∂ϕ/∂z)

Finding the partial derivatives:

∂ϕ/∂x = n*x^(n-1), ∂ϕ/∂y = n*y^(n-1), ∂ϕ/∂z = n*z^(n-1).

Therefore, the gradient of ϕ is:

∇ϕ = (n*x^(n-1), n*y^(n-1), n*z^(n-1)).

Now, let's calculate the dot product of ∇ϕ and the position vector r = (x, y, z).

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

Simplifying this expression:

(∇ϕ)·r = n*x^n + n*y^n + n*z^n.

Now, recall the definition of ϕ(x, y, z) = xn + yn + zn. If we substitute this into the equation above, we get:

(∇ϕ)·r = nϕ.

Therefore, we have shown that (∇ϕ)·r = nϕ for the given scalar field ϕ(x, y, z) = xn + yn + zn and a non-zero real constant n.