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

To prove that the gradient of the scalar function f(r) = r^2 is equal to 2r where r is a vector representing position in three-dimensional space, we need to compute the gradient of the function and demonstrate its equality to 2r.

1. Start by expressing the scalar function f(r) = r^2 in terms of its components:
f(r) = x^2 + y^2 + z^2, where x, y, and z are the Cartesian components of the vector r.

2. Compute the gradient of f(r) using the del operator (∇), which is defined as:
∇ = (∂/∂x)î + (∂/∂y)ĵ + (∂/∂z)k̂

3. Apply the gradient operator to f(r):
∇(f(r)) = (∂/∂x)î(x^2 + y^2 + z^2) + (∂/∂y)ĵ(x^2 + y^2 + z^2) + (∂/∂z)k̂(x^2 + y^2 + z^2)

4. Compute each partial derivative:
(∂/∂x)î(x^2 + y^2 + z^2) = 2xî
(∂/∂y)ĵ(x^2 + y^2 + z^2) = 2yĵ
(∂/∂z)k̂(x^2 + y^2 + z^2) = 2zk̂

5. Combine the partial derivatives:
∇(f(r)) = 2xî + 2yĵ + 2zk̂

6. Notice that 2xî + 2yĵ + 2zk̂ is equivalent to 2(r) since each component (x, y, z) is multiplied by 2.

Therefore, we have shown that the gradient of the function f(r) = r^2 is equal to 2r, as desired.