The Gradient Vector
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 4.6: “Directional Derivatives and the Gradient” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/4-6-directional-derivatives-and-the-gradient |
| Textbook used in class | Stewart, Calculus, Section 14.6: “Directional Derivatives and the Gradient Vector” (Examples 4, 5, 6) |
Quick Reference
Gradient: $\nabla f(x,y) = \langle f_x, f_y\rangle$, also written $\text{grad}\,f$.
Properties:
- $\nabla f$ points in the direction of steepest increase of $f$.
- $|\nabla f|$ equals the maximum rate of change of $f$.
- $\nabla f$ is perpendicular to the level curves of $f$.
- $-\nabla f$ points in the direction of steepest decrease.
Normal to level surface $F(x,y,z) = c$: $\nabla F(a,b,c)$ is perpendicular to the surface at $(a,b,c)$.
Motivation
The gradient $\nabla f$ packages the partial derivatives into a single vector that encodes all directional information. It answers the question “in which direction does $f$ increase most rapidly?” instantly: the answer is the direction of $\nabla f$, with rate of increase $|\nabla f|$.
The perpendicularity of $\nabla f$ to level curves is equally important and leads directly to the method of Lagrange multipliers for constrained optimization (Section 14.8).
Key Concepts
1. Gradient Perpendicular to Level Curves
Suppose $f(x,y) = c$ is a level curve, parameterized by $\mathbf{r}(t) = \langle x(t), y(t)\rangle$. Differentiating $f(\mathbf{r}(t)) = c$ with respect to $t$: $$\nabla f \cdot \mathbf{r}'(t) = 0.$$
So $\nabla f$ is perpendicular to the tangent vector of every level curve. In 3D, $\nabla F$ is normal to the level surface $F(x,y,z) = c$.
2. Normal Lines and Tangent Planes to Level Surfaces
For a surface $F(x,y,z) = c$ and point $(a,b,d)$ on it:
- Normal line: $\mathbf{r}(t) = \langle a,b,d\rangle + t\nabla F(a,b,d)$.
- Tangent plane: $F_x(a,b,d)(x-a) + F_y(a,b,d)(y-b) + F_z(a,b,d)(z-d) = 0$.
This unifies the tangent plane formula from Section 14.4 (which covered $z = f(x,y)$, i.e., $F = f-z$) with the general surface case.
Worked Example
Find the equation of the tangent plane to $x^2 + 2y^2 + 3z^2 = 6$ at $(1,1,1)$. (Stewart 14.6, Example 5.)
$F(x,y,z) = x^2 + 2y^2 + 3z^2$.
$\nabla F = \langle 2x, 4y, 6z\rangle \implies \nabla F(1,1,1) = \langle 2, 4, 6\rangle$.
Tangent plane: $$2(x-1) + 4(y-1) + 6(z-1) = 0$$ $$2x + 4y + 6z = 12, \quad \text{or} \quad x + 2y + 3z = 6.$$
Example 2. For $f(x,y) = x^2 y - y^3$, find the direction of maximum increase at $(1,2)$ and the maximum rate of increase.
$\nabla f = \langle 2xy, x^2 - 3y^2\rangle \implies \nabla f(1,2) = \langle 4, 1-12\rangle = \langle 4, -11\rangle$.
Direction of maximum increase: $\mathbf{u} = \langle 4, -11\rangle / \sqrt{16+121} = \langle 4, -11\rangle / \sqrt{137}$.
Maximum rate: $|\nabla f(1,2)| = \sqrt{137}$.
$\nabla f$ is tangent to the level curves of $f$. The gradient is PERPENDICULAR (normal) to the level curves, not tangent to them. The tangent to the level curve $f = c$ lies in the direction perpendicular to $\nabla f$ in the $xy$-plane. A good memory check: if you stand on a hill where elevation is $f$, the gradient points UPHILL (perpendicular to the contour lines), not along a contour.
Leveled Practice
Problem 1. Find the gradient of $f(x,y,z) = x\sin(yz)$ and the unit vector in the direction of maximum increase at $(1, \pi/2, 1)$.
Show answer
$\nabla f = \langle \sin(yz),\; xz\cos(yz),\; xy\cos(yz)\rangle$.
At $(1, \pi/2, 1)$: $\sin(\pi/2) = 1$, $\cos(\pi/2) = 0$.
$\nabla f = \langle 1, 0, 0\rangle$.
The gradient has magnitude 1, so the unit vector in the direction of maximum increase is $\langle 1, 0, 0\rangle$ (the positive $x$-direction).