Directional Derivatives | Tony's Intertwingled Memex

Core Definition

Let $f$ be a differentiable function of two or three variables at a point $P$ , and let $\vec{u}$ be a unit vector ( $\|\vec{u}\| = 1$ ). The directional derivative of $f$ at $P$ in the direction of $\vec{u}$ , denoted by $D_{\vec{u}} f(P)$ , is defined as the limit:

D_{\vec{u}} f(P) = \lim_{h \to 0} \frac{f(P + h\vec{u}) - f(P)}{h}

By the Gradient Theorem for Directional Derivatives, if $f$ is differentiable, this limit can be computed directly using the vector dot product of the gradient $\nabla f$ and the unit vector $\vec{u}$ :

Two Variables ( $f(x, y)$ with $\vec{u} = \langle u_1, u_2 \rangle$ ):
$D_{\vec{u}} f(x, y) = \nabla f(x, y) \cdot \vec{u} = f_x(x, y)u_1 + f_y(x, y)u_2$
Three Variables ( $f(x, y, z)$ with $\vec{u} = \langle u_1, u_2, u_3 \rangle$ ):
$D_{\vec{u}} f(x, y, z) = \nabla f(x, y, z) \cdot \vec{u} = f_x u_1 + f_y u_2 + f_z u_3$

The output $D_{\vec{u}} f$ is always a scalar quantity representing the instantaneous rate of change of $f$ per unit distance traveled in the specific direction of $\vec{u}$ .

Concept Explanation

Bridging from Partial Derivatives to Arbitrary Directions

Partial derivatives $f_x$ and $f_y$ measure the instantaneous rate of change of $z = f(x, y)$ strictly parallel to the coordinate axes:

$f_x = D_{\hat{i}} f = \nabla f \cdot \langle 1, 0 \rangle$ (moving due East)
$f_y = D_{\hat{j}} f = \nabla f \cdot \langle 0, 1 \rangle$ (moving due North)

In practical engineering applications, observers, vehicles, fluid streams, and structural loads rarely move parallel to grid axes. The directional derivative $D_{\vec{u}} f$ generalizes differentiation to any arbitrary compass heading or 3D spatial trajectory:

                  ^ y
                  |           ^ u = <cos(theta), sin(theta)>
                  |          /
                  |         /  (Heading direction)
                  |        /
                  |       * P(x_0, y_0)
                  |        \
                  |         \  grad f (Steepest ascent)
                  +----------> x

The Mandatory Unit Vector Normalization Rule

The dot product formula $D_{\vec{u}} f = \nabla f \cdot \vec{u}$ strictly requires that the direction vector is a unit vector ( $\|\vec{u}\| = 1$ ).

If a problem specifies direction via an arbitrary non-unit vector $\vec{v}$ (where $\|\vec{v}\| \neq 1$ ), you must normalize it first: $\vec{u} = \frac{\vec{v}}{\|\vec{v}\|} = \frac{\langle v_1, v_2 \rangle}{\sqrt{v_1^2 + v_2^2}}$
If direction is specified as moving from point $P$ toward a destination point $Q$ , compute the displacement vector $\overrightarrow{PQ} = Q - P$ and normalize it: $\vec{u} = \frac{\overrightarrow{PQ}}{\|\overrightarrow{PQ}\|}$

Pitfall Warning

Computing $\nabla f \cdot \vec{v}$ with an unnormalized vector scales your result by $\|\vec{v}\|$ . This yields an incorrect rate of change with distorted units.

Angle Parameterization in 2D

In two dimensions, any unit vector $\vec{u}$ making an angle $\theta$ measured counterclockwise from the positive $x$ -axis can be written as:

\vec{u} = \langle \cos\theta, \, \sin\theta \rangle

Substituting into the directional derivative formula gives:

D_{\vec{u}} f(x, y) = f_x(x, y)\cos\theta + f_y(x, y)\sin\theta

Geometric Interpretation via the Dot Product Angle Formula

From the geometric definition of the dot product:

D_{\vec{u}} f = \nabla f \cdot \vec{u} = \|\nabla f\| \|\vec{u}\| \cos\phi = \|\nabla f\|\cos\phi

where

\phi

is the angle between the gradient vector

\nabla f

and the direction vector

\vec{u}

. Because

-1 \le \cos\phi \le 1

Maximum Rate of Increase ( $\phi = 0$ ): When $\vec{u}$ points in the same direction as $\nabla f$ , $\cos(0) = 1$ , yielding $D_{\vec{u}} f = \|\nabla f\|$ .
Maximum Rate of Decrease ( $\phi = \pi$ ): When $\vec{u}$ points opposite to $\nabla f$ , $\cos(\pi) = -1$ , yielding $D_{\vec{u}} f = -\|\nabla f\|$ .
Zero Rate of Change ( $\phi = \pi/2$ or $3\pi/2$ ): When $\vec{u}$ is orthogonal to $\nabla f$ , $\cos(\pi/2) = 0$ , yielding $D_{\vec{u}} f = 0$ (moving tangent to a level curve).

Engineering Applications

Civil & Transportation Engineering: Roadway design engineers calculate ground grades along angled road cuts using directional derivatives. If a proposed highway alignment vector $\vec{u}$ yields a slope $D_{\vec{u}} h$ exceeding federal grade standards (e.g., $6\%$ ), the alignment must be modified.
Entertainment Engineering & Lighting Design: Lighting designers model stage illuminance fields $I(x, y)$ in lux. When a performer moves along a trajectory $\vec{u}$ , $D_{\vec{u}} I$ dictates the required dynamic dimmer response to prevent shadows and maintain consistent video capture.
Environmental & Atmospheric Modeling: Meteorology stations track atmospheric pressure $P(x, y)$ . The directional derivative $D_{\vec{u}} P$ in the direction of an aircraft's flight heading dictates localized altimeter drift and crosswind shear.

Example Problems

Example 1: Directional Derivative via Vector Normalization

Find the directional derivative of $f(x, y) = 3x^2 - 2y^2 + 5xy$ at the point $P(2, -1)$ in the direction of the vector $\vec{v} = \langle 3, 4 \rangle$ .

Solution

Compute and evaluate the gradient:
$f_x = 6x + 5y \implies f_x(2, -1) = 6(2) + 5(-1) = 12 - 5 = 7$ $f_y = -4y + 5x \implies f_y(2, -1) = -4(-1) + 5(2) = 4 + 10 = 14$ $\nabla f(2, -1) = \langle 7, 14 \rangle$
Normalize the direction vector:
$\|\vec{v}\| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ $\vec{u} = \frac{\langle 3, 4 \rangle}{5} = \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle$
Evaluate the dot product:
$D_{\vec{u}} f(2, -1) = \langle 7, 14 \rangle \cdot \left\langle \frac{3}{5}, \frac{4}{5} \right\rangle = 7\left(\frac{3}{5}\right) + 14\left(\frac{4}{5}\right) = \frac{21 + 56}{5} = \frac{77}{5} = 15.4$

Example 2: Three-Variable Directional Derivative Toward a Target Point

Find the directional derivative of $f(x, y, z) = x^2yz + 2z^2$ at the point $P(1, 3, 1)$ in the direction toward the point $Q(3, 1, 2)$ .

Solution

Compute partial derivatives and evaluate at $P(1, 3, 1)$ :
$f_x = 2xyz \implies f_x(1, 3, 1) = 2(1)(3)(1) = 6$ $f_y = x^2z \implies f_y(1, 3, 1) = (1)^2(1) = 1$ $f_z = x^2y + 4z \implies f_z(1, 3, 1) = (1)^2(3) + 4(1) = 7$ $\nabla f(1, 3, 1) = \langle 6, 1, 7 \rangle$
Find and normalize displacement vector $\overrightarrow{PQ}$ :
$\overrightarrow{PQ} = \langle 3 - 1, \, 1 - 3, \, 2 - 1 \rangle = \langle 2, -2, 1 \rangle$ $\|\overrightarrow{PQ}\| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$ $\vec{u} = \frac{\langle 2, -2, 1 \rangle}{3} = \left\langle \frac{2}{3}, -\frac{2}{3}, \frac{1}{3} \right\rangle$
Evaluate the directional derivative:
$D_{\vec{u}} f(P) = \langle 6, 1, 7 \rangle \cdot \left\langle \frac{2}{3}, -\frac{2}{3}, \frac{1}{3} \right\rangle = 6\left(\frac{2}{3}\right) + 1\left(-\frac{2}{3}\right) + 7\left(\frac{1}{3}\right) = \frac{12 - 2 + 7}{3} = \frac{17}{3} \approx 5.67$

Connections & References

Prerequisites:
- The Gradient Vector — Foundation for the vector differential operator $\nabla f$ .
- Vectors — Unit vectors, dot products, and displacement vectors $\overrightarrow{PQ}$ .
- Partial Derivatives — Coordinate-aligned rates of change $f_x$ and $f_y$ .
Direct Successors & Extensions:
- Tangent Planes to Level Surfaces — Normal vectors and surface geometry.
- Level Curves and Contour Maps — Orthogonal paths and contour intersections.
Course Sessions:
- Session 12 - Gradient and Directional Derivatives