The Gradient Vector

Two partial derivatives, one vector

At each point of a surface z = f(x, y) there are two partial derivatives: $\frac{\partial f}{\partial x}$∂f/∂x, the slope in the x direction, and $\frac{\partial f}{\partial y}$∂f/∂y, the slope in the y direction. Written side by side, in that order, they form a vector, the gradient of f: $\nabla f=\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)$∇f = (∂f/∂x, ∂f/∂y). The symbol $\nabla$∇ is read “del”, and $\nabla f$∇f is read “grad f”.

For $f\left(x,y\right)=x^2+y^2$f(x, y) = x² + y², $\frac{\partial f}{\partial x}=2x$∂f/∂x = 2x and $\frac{\partial f}{\partial y}=2y$∂f/∂y = 2y, so $\nabla f=\left(2x,2y\right)$∇f = (2x, 2y). At (3, 4) the gradient is (6, 8), and at (1, 0) it is (2, 0). There is a gradient vector at every point of the plane, and it lies in the plane: it has two components, not three.

The level curve $x^2+y^2=25$x² + y² = 25 of $f=x^2+y^2$f = x² + y², dashed, and the gradient at (3, 4), drawn at a quarter of its length: (6, 8) shrunk to (1.5, 2). It points straight out from the origin, along the radius, so it crosses the circle at a right angle.

The steepest way up

Walk from (3, 4) in the direction of a unit vector u = (p, q). Over a short step, the height changes by about $\frac{\partial f}{\partial x}$∂f/∂x times the step in x plus $\frac{\partial f}{\partial y}$∂f/∂y times the step in y, so the slope in that direction is 6p + 8q. That is the dot product $\nabla f\cdot u$∇f · u.

Difference quotients at (3, 4) agree. Heading east, u = (1, 0), the slope is 6. Heading north, u = (0, 1), it is 8. Along u = (0.6, 0.8), the direction of the gradient itself, it is 6 × 0.6 + 8 × 0.8 = 10. Heading the opposite way, it is −10.

By the dot product, $\nabla f\cdot u=|\nabla f|\cos \theta$∇f · u = |∇f| cos θ, where $\theta$θ is the angle between $\nabla f$∇f and u. The cosine is largest, 1, when $\theta =0$θ = 0. So the slope is greatest in the direction of $\nabla f$∇f: the gradient points in the direction of steepest ascent, and $-\nabla f$−∇f in the direction of steepest descent.

Contours of the bowl $f=x^2+\frac{y^2}{2}$f = x² + y²/2. At P = (1, 1) the gradient is (2x, y) = (2, 1), the gold arrow, with length $\sqrt{5}\approx 2.24$√5 ≈ 2.24, and the dashed line is the tangent to the contour through P. The chalk arrow u is a unit direction at 70°, 43° from the gradient, and the slope that way is $\nabla f\cdot u=2\cos 70^{\circ }+\sin 70^{\circ }\approx 1.62$∇f · u = 2 cos 70° + sin 70° ≈ 1.62. Turned onto the gradient the slope reads 2.24, and turned along the dashed tangent it reads 0.

Square to the level curves

Along a level curve the height does not change, so the slope in the direction of the curve is 0. Then $\nabla f\cdot u=0$∇f · u = 0 for u along the curve, which means $\nabla f$∇f is perpendicular to the level curve.

At (3, 4) on the circle $x^2+y^2=25$x² + y² = 25, the tangent to the circle runs along (−0.8, 0.6), and $\nabla f\cdot u=6\times \left(-0.8\right)+8\times 0.6=-4.8+4.8=0$∇f · u = 6 × (−0.8) + 8 × 0.6 = −4.8 + 4.8 = 0. On a contour map, the gradient crosses every contour at a right angle, heading toward the higher ones.

Length is steepness

In the direction of $\nabla f$∇f, $\theta =0$θ = 0 and the slope is $|\nabla f|$|∇f| itself. So the length of the gradient is the steepest slope available at that point. For $x^2+y^2$x² + y² at (3, 4), $|\nabla f|=\sqrt{6^2+8^2}=10$|∇f| = √(6² + 8²) = 10, and no direction from (3, 4) climbs faster than 10 units of height per unit of distance.

Where the ground is steep the gradient is long, and where it is gentle the gradient is short. At the top of a hill, or the bottom of a bowl, both partial derivatives are 0 and $\nabla f=\left(0,0\right)$∇f = (0, 0): no direction is uphill.

A contour map of the dome $z=16\left(1-x^2-y^2\right)$z = 16(1 − x² − y²), cut at heights 4, 8 and 12, so $\nabla z=\left(-32x,-32y\right)$∇z = (−32x, −32y). At (0.22, 0.14), near the gentle top, the gradient is (−7.04, −4.48), with length 8.3. At (−0.3, −0.7), where the contours crowd, it is (9.6, 22.4), with length 24.4. Both arrows are drawn to the same scale and point toward the summit.

The usual mistakes

Swapping the components. The first component is always $\frac{\partial f}{\partial x}$∂f/∂x: for f = 3x + 5y, $\nabla f=\left(3,5\right)$∇f = (3, 5), not (5, 3).

Adding the partial derivatives into one number. For 3x + 5y, the gradient is the vector (3, 5), not 8.

Giving the unit vector as the steepest slope. The unit vector in the direction of $\nabla f$∇f gives only the direction; the slope that way is the length $|\nabla f|$|∇f|.

A hotplate

In the application below, the temperature of a hotplate is $T\left(x,y\right)=100-0.5x^2-y^2$T(x, y) = 100 − 0.5x² − y². At the probe’s position the gradient gives the direction in which the temperature rises fastest and the rate it rises at, and a robot that always follows the gradient traces a path found from it.