Show that vector del(r^2) = 2vector r
To show that the vector del(r^2) is equal to 2 times vector r, we need to calculate the gradient of the scalar function r^2.
The gradient of a scalar function is defined as the vector whose components are the partial derivatives of the function with respect to each coordinate. In this case, we have a scalar function r^2, so we need to take the gradient of r^2.
The position vector r can be expressed as (x, y, z). Therefore, we have r^2 = x^2 + y^2 + z^2. To calculate the gradient of r^2, we need to take the partial derivatives of r^2 with respect to x, y, and z.
∂/∂x(r^2) = 2x
∂/∂y(r^2) = 2y
∂/∂z(r^2) = 2z
Therefore, the gradient of r^2 is (2x, 2y, 2z).
Now, let's compare this to 2 times the position vector r, which is 2(x, y, z) = (2x, 2y, 2z). We can see that the gradient of r^2 is indeed equal to 2 times the position vector r.
Hence, we have shown that vector del(r^2) = 2vector r.