Steepest Ascent Direction¶

Theorem¶

Let $f\colon A\subseteq\mathbb{R}^n\to\mathbb{R}$ be differentiable at $\boldsymbol{a}$ , and suppose $\boldsymbol{\nabla} f(\boldsymbol{a})\neq\boldsymbol{0}$ $0$ . Among all unit vectors $\boldsymbol{u}$ , the directional derivative

D_{\boldsymbol{u}} f(\boldsymbol{a})=\boldsymbol{\nabla} f(\boldsymbol{a})^{\top} \boldsymbol{u}

is maximized when

\boldsymbol{u}=\frac{\boldsymbol{\nabla} f(\boldsymbol{a})}{\|\boldsymbol{\nabla} f(\boldsymbol{a})\|}.

The maximum value is $\|\boldsymbol{\nabla} f(\boldsymbol{a})\|$ , and the minimum value is $-\|\boldsymbol{\nabla} f(\boldsymbol{a})\|$ in the opposite direction.

Interpretation¶

The gradient points in the direction in which the function rises fastest, and its length is that fastest rate. It answers the question: among all steps of the same small length from $\boldsymbol{a}$ , which direction gives the largest first-order increase of $f$ ?

Circular level sets of f around a point a, with three unit directions: steepest ascent along the gradient, steepest descent opposite to it, and zero change along the level set.

The grey circles are level curves of $f$ around a peak, and the dashed circle holds the unit directions at $\boldsymbol{a}$ . For a unit vector $\boldsymbol{u}$ at angle $\theta$ to the gradient, $D_{\boldsymbol{u}}f(\boldsymbol{a})=\|\boldsymbol{\nabla} f(\boldsymbol{a})\|\cos\theta$ $∥ \nabla f (a) ∥ cos θ$ : the rate is largest at $\theta=0$ (red, towards the peak), most negative at $\theta=\pi$ (green), and zero at $\theta=\pi/2$ (blue), along the level curve through $\boldsymbol{a}$ . So the gradient is perpendicular to that level curve.

The factor $\cos\theta$ is the Cauchy-Schwarz inequality at work: $|\boldsymbol{\nabla} f(\boldsymbol{a})^{\top}\boldsymbol{u}|\le\|\boldsymbol{\nabla} f(\boldsymbol{a})\|$ for every unit vector $\boldsymbol{u}$ , with equality exactly when $\boldsymbol{u}=\pm\boldsymbol{\nabla} f(\boldsymbol{a})/\|\boldsymbol{\nabla} f(\boldsymbol{a})\|$ $\pm \nabla f (a) /∥ \nabla f (a) ∥$ . Both hypotheses matter. Differentiability makes every directional derivative equal to $\boldsymbol{\nabla} f(\boldsymbol{a})^{\top}\boldsymbol{u}$ ; partial derivatives alone do not: for $f(x,y)=x^2y/(x^2+y^2)$ with $f(0,0)=0$ $0$ , both partial derivatives at the origin are $0$ , yet the rate along $(1,1)/\sqrt{2}$ is $1/(2\sqrt{2})$ . And if $\boldsymbol{\nabla} f(\boldsymbol{a})=\boldsymbol{0}$ $0$ , there is nothing to normalize and no preferred direction.

A common slip is to use the gradient itself as the direction: $\boldsymbol{\nabla} f(\boldsymbol{a})^{\top}\boldsymbol{\nabla} f(\boldsymbol{a})=\|\boldsymbol{\nabla} f(\boldsymbol{a})\|^2$ is not the steepest rate, because directional derivatives are taken along unit vectors. The same reasoning returns later in the course: at an interior local maximum no direction may give a positive rate, which forces $\boldsymbol{\nabla} f=\boldsymbol{0}$ (unconstrained extrema conditions).

Key Results¶

The result follows from the Cauchy-Schwarz inequality.
The direction of steepest descent is $-\boldsymbol{\nabla} f(\boldsymbol{a})/\|\boldsymbol{\nabla} f(\boldsymbol{a})\|$ .
If $\boldsymbol{\nabla} f(\boldsymbol{a})=\boldsymbol{0}$ $0$ , then every directional derivative is zero and there is no unique steepest direction from first-order information.

Examples¶

For $f(x,y)=x^2+y^2$ at $(1,2)$ , $\boldsymbol{\nabla} f=(2,4)$ $(2, 4)$ , so the steepest ascent direction is $(1,2)/\sqrt{5}$ .
For $f(x,y)=3x-y$ at any point, $\boldsymbol{\nabla} f=(3,-1)$ $(3, - 1)$ , so the steepest ascent direction is $(3,-1)/\sqrt{10}$ .

Prerequisites¶

Sources¶

Math IV: Multivariate Calculus · Week 1, Lecture 1

Aliases: direction of steepest ascent, steepest direction Keywords: maximum directional derivative Symbols: nabla gradient, pi, real numbers, square root, subset, theta

Last updated: 27 October 2026 11:52