Gradient of x^2 + y^2: Vector, Steps, and Geometry

Differentiate once per variable, then stack the results

The gradient is not a new kind of derivative. It is the partial derivatives you already know, collected into one vector. Compute $\frac{\partial f}{\partial x}$ with $y$ frozen, then $\frac{\partial f}{\partial y}$ with $x$ frozen.

The $y^2$ term contributes nothing, because with $y$ held fixed it is just a number, and the derivative of a number is zero. Swapping roles gives the other component.

Now list them in variable order, $x$ first and $y$ second. That order is a convention, but it is the one every later formula assumes.

At $(3,4)$ this evaluates to $\nabla f(3,4) = \langle 6, 8\rangle$, which has length $\sqrt{6^2 + 8^2} = 10$.

The mistake: giving the gradient a third component

The graph of $f$ is a paraboloid sitting in three dimensions, so students often write the gradient as a three-component vector, something like $\langle 2x, 2y, -1\rangle$, and try to make it point up the hillside in space.

The gradient of a function of two variables has exactly two components, one per input variable. It lives in the $xy$-plane, the floor plan, not on the surface. Read it as a set of walking directions: stand at $(3,4)$ on the floor and head in the direction $\langle 6,8\rangle$, and the surface above you rises as fast as it possibly can.

• Number of components equals number of input variables, so two here.
• $\nabla f$ is a vector, never a scalar. Writing $\nabla f = 2x + 2y$ loses the direction.
• The vector $\langle 2x, 2y, -1\rangle$ is a normal to the surface $z = f(x,y)$, which is a different object with a different job.

What the vector says about this paraboloid

Notice that $\langle 2x, 2y\rangle = 2\langle x, y\rangle$, so the gradient at any point is a positive multiple of the position vector. It points radially outward from the origin, and it gets longer the further out you go.

That matches the level curves. Setting $f = c$ gives $x^2 + y^2 = c$, a circle of radius $\sqrt{c}$, and a radial vector is perpendicular to every circle centred at the origin. The gradient is always perpendicular to the level curve through the point.

The magnitude carries the steepness. Walking from $(3,4)$ in the unit direction $\mathbf{u} = \left\langle \tfrac{3}{5}, \tfrac{4}{5}\right\rangle$ gives the directional derivative

No other unit direction beats $10$, which is exactly what it means for the gradient to point along steepest ascent.