How can we find a function f for the vector field f = x i + y j + z k, such that f is equal to itself and also satisfies the condition f(0, 0, 0) = 0?

To find a function f for the given vector field f = x i + y j + z k, we need to start by understanding what is meant by a function that satisfies f = f and that evaluates to zero at the origin, i.e., f(0, 0, 0) = 0.

The expression f = x i + y j + z k represents a vector field in three-dimensional space, where:

• i is the unit vector in the x-direction,
• j is the unit vector in the y-direction, and
• k is the unit vector in the z-direction.

Thus, the vector field describes a point in space based on the coordinates (x, y, z).

Next, we need to identify a scalar function F(x, y, z) such that F satisfies the given conditions. A common form to consider is:

F(x, y, z) = rac{1}{2}(x^2 + y^2 + z^2)

Here’s why this function works:

• This function is smooth and continuous across all points in space, ensuring that it behaves well.
• When we evaluate F at the origin (x = 0, y = 0, z = 0), we find:

F(0, 0, 0) = rac{1}{2}(0^2 + 0^2 + 0^2) = 0

• Moreover, if we compute the gradient of F, we get:

∇F = (F_x, F_y, F_z) = (x, y, z)

• This means that the gradient matches our original vector field f, confirming that this choice of F produces the resulting vector expression.

In conclusion, the function we are looking for is:

F(x, y, z) = rac{1}{2}(x^2 + y^2 + z^2)

It meets all the specified requirements: it is equal to itself in the context of deriving the vector field through its gradient, and it yields zero at the origin.