The Gradient and Directional Derivatives

≈ 25 min

The gradient collects the partial derivatives into a vector:

\nabla f = \left( \frac{\partial f}{\partial x}, \; \frac{\partial f}{\partial y} \right).

It has two headline properties: $\nabla f$ points in the direction of steepest ascent, and its length $|\nabla f|$ is the maximum rate of increase at that point.

The directional derivative measures the rate of change in an arbitrary direction given by a unit vector $\hat{u}$ :

D_{\hat{u}} f = \nabla f \cdot \hat{u}.

Worked example. For $f(x, y) = x^2 + xy$ , $\nabla f = (2x + y, \; x)$ . At $(1, 2)$ this is $(4, 1)$ . In the direction of $(3, 4)$ , the unit vector is $\hat{u} = \left(\tfrac{3}{5}, \tfrac{4}{5}\right)$ , so $D_{\hat u} f = 4\cdot\tfrac{3}{5} + 1\cdot\tfrac{4}{5} = \tfrac{16}{5} = 3.2.$

For $f(x, y) = x^2 + y^2$ , find the magnitude $|\nabla f|$ of the gradient at the point $(3, 4)$ .

What is the gradient $\nabla f$ of $f(x, y) = x^2 y$ ?

$(2xy, \; x^2)$

$(2xy, \; x^2 y)$

$(x^2, \; 2xy)$

$(2x, \; y)$

At a point where $\nabla f \neq 0$ , the gradient vector points in the direction of…

steepest ascent (fastest increase of $f$ )

steepest descent (fastest decrease of $f$ )

no change in $f$ (along a contour)

the origin

For $f(x, y) = x^2 + xy$ , find the directional derivative at $(1, 2)$ in the direction of the vector $(3, 4)$ .