Show that vector del(r^2) = 2vector r

To prove that ∇(r^2) = 2r, where ∇ is the vector del operator, we need to apply the del operator to the scalar function r^2.

The del operator, written as ∇, is defined in Cartesian coordinates as:

∇ = (∂/∂x)i + (∂/∂y)j + (∂/∂z)k

where i, j, and k are the unit vectors along the x, y, and z-axes, respectively.

Given the scalar function r^2, we can express it as:

r^2 = x^2 + y^2 + z^2

where x, y, and z are the Cartesian coordinates of a vector r.

Now, let's calculate ∇(r^2):

∇(r^2) = (∂/∂x)(x^2 + y^2 + z^2)i + (∂/∂y)(x^2 + y^2 + z^2)j + (∂/∂z)(x^2 + y^2 + z^2)k

To compute each partial derivative, we treat x, y, and z as independent variables while differentiating.

Taking the partial derivative of r^2 with respect to x, we get:

(∂/∂x)(x^2 + y^2 + z^2) = 2x

Similarly, taking the partial derivative with respect to y and z, we get:

(∂/∂y)(x^2 + y^2 + z^2) = 2y

(∂/∂z)(x^2 + y^2 + z^2) = 2z

Substituting these values back into the expression for ∇(r^2), we have:

∇(r^2) = 2x i + 2y j + 2z k

Recalling that the Cartesian coordinates x, y, and z correspond to the vector r, we can rewrite this as:

∇(r^2) = 2(xi + yj + zk)

Since xi + yj + zk represents the vector r, we have:

∇(r^2) = 2r

Thus, we have shown that ∇(r^2) = 2r.