Directional Derivatives

The slope depends on the direction

Take the dome $f\left(x,y\right)=16\left(1-x^2-y^2\right)$f(x, y) = 16(1 − x² − y²) and stand at the point (0.5, 0.36), at height 9.93. The gradient there is $\nabla f=\left(-32x,-32y\right)=\left(-16,-11.52\right)$∇f = (−32x, −32y) = (−16, −11.52). It points toward the summit at the origin, and its length, $\sqrt{16^2+11.52^2}\approx 19.72$√(16² + 11.52²) ≈ 19.72, is the steepest slope available from that point.

Set off due south instead, along u = (0, −1). The summit lies partly south of you, so you still climb, but the slope that way is only 11.52. The slope of a surface in a chosen direction u is the directional derivative of f in that direction.

The dome $f=16\left(1-x^2-y^2\right)$f = 16(1 − x² − y²) seen from above, cut at heights 4, 8 and 12. At (0.5, 0.36) the gold arrow is $\nabla f$∇f, of length 19.72, and the plain arrow u points due south. The dashed line drops the tip of $\nabla f$∇f square onto u, and the gold bar along u is the part of $\nabla f$∇f in that direction: the slope due south, 11.52.

One part in each direction

Walk a distance t along the unit vector u = (a, b). Then $x=x_0+at$x = x₀ + a t and $y=y_0+bt$y = y₀ + b t, so x changes at the rate a and y at the rate b. By the chain rule for partial derivatives, the height changes at the rate $\frac{\partial f}{\partial x}\cdot a+\frac{\partial f}{\partial y}\cdot b$∂f/∂x · a + ∂f/∂y · b: each part of the step is multiplied by its own partial derivative, and the two add.

That sum is the dot product of $\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)$(∂f/∂x, ∂f/∂y) with (a, b). So the slope in the direction u is $\nabla f\cdot u$∇f · u. At the point on the dome, due south is a = 0 and b = −1, and the slope is (−16) × 0 + (−11.52) × (−1) = 11.52.

A unit vector first

The variable t above is the distance walked only because u has length 1. A direction is often given as a displacement instead, so divide it by its length first.

For $f\left(x,y\right)=x^{2}y$f(x, y) = x²y at (1, 2), $\frac{\partial f}{\partial x}=2xy=4$∂f/∂x = 2xy = 4 and $\frac{\partial f}{\partial y}=x^2=1$∂f/∂y = x² = 1, so $\nabla f=\left(4,1\right)$∇f = (4, 1). Head toward the point (4, 6): the displacement is (3, 4), its length is 5, and the unit vector is u = (0.6, 0.8). The slope is 4 × 0.6 + 1 × 0.8 = 3.2. Using the displacement itself gives $\nabla f\cdot \left(3,4\right)=4\times 3+1\times 4=16$∇f · (3, 4) = 4 × 3 + 1 × 4 = 16, five times too big, because (3, 4) is five units long.

A difference quotient agrees: f(1 + 0.6h, 2 + 0.8h) − f(1, 2), divided by h, comes to 3.2 for small h.

Steepest, gentlest, level

For a unit vector u, $\nabla f\cdot u=|\nabla f|\cos \theta$∇f · u = |∇f| cos θ, where $\theta$θ is the angle between $\nabla f$∇f and u. The slope is greatest, $|\nabla f|$|∇f|, when $\theta =0$θ = 0, along the gradient; it is least, $-|\nabla f|$−|∇f|, when $\theta =180^{\circ }$θ = 180°, straight downhill; and it is 0 when $\theta =90^{\circ }$θ = 90°.

For $x^{2}y$x²y at (1, 2), $|\nabla f|=\sqrt{17}\approx 4.12$|∇f| = √17 ≈ 4.12. Toward (4, 6) the slope 3.2 is $\cos \theta =3.2/4.12\approx 0.78$cos θ = 3.2 / 4.12 ≈ 0.78 of that, with $\theta \approx 39^{\circ }$θ ≈ 39°. Along $\left(-1,4\right)÷\sqrt{17}$(−1, 4) ÷ √17, at right angles to (4, 1), the slope is $\left(-4+4\right)÷\sqrt{17}=0$(−4 + 4) ÷ √17 = 0.

On the dome, due south makes an angle of 54.25° with $\nabla f$∇f, and 19.72 × cos 54.25° = 11.52. Turning to 125.75°, measured from east, puts u at right angles to $\nabla f$∇f, and the slope is 0.

The same point with u turned to 125.75°, at right angles to $\nabla f$∇f. The tip of $\nabla f$∇f drops onto u at the starting point itself, and a square corner marks the right angle: the part of $\nabla f$∇f along u is 0, and u is tangent to the contour through the point.

Why contours are level

A contour is a curve along which the height does not change, so the slope along it is 0. Where $\nabla f$∇f is not zero, the slope is 0 only in the two opposite directions at right angles to $\nabla f$∇f. So the contour through a point runs at right angles to $\nabla f$∇f there, and the gradient crosses every contour square on.

Contours of $f=x^2+\frac{y^2}{2}$f = x² + y²/2, with P at (1, 2), where $\nabla f=\left(2x,y\right)=\left(2,2\right)$∇f = (2x, y) = (2, 2), of length 2.83. The unit vector u points straight up, at 45° to $\nabla f$∇f, and the meter reads the slope $\nabla f\cdot u=2$∇f · u = 2. Turn u to 45° and it reads 2.83, the length of $\nabla f$∇f; turn it to 135°, along the dashed tangent to the contour, and it reads 0.

The usual mistakes

Leaving the direction unnormalized. For $x^{2}y$x²y at (1, 2) toward (4, 6), (3, 4) gives 16; the unit vector (0.6, 0.8) gives the slope, 3.2.

Adding the partial derivatives whole. With $\nabla f=\left(15,20\right)$∇f = (15, 20) and u = (0.6, 0.8), the slope is 15 × 0.6 + 20 × 0.8 = 25, not 35: each partial counts only for its own share of the step.

Swapping the shares. The 0.6 goes with $\frac{\partial f}{\partial x}$∂f/∂x, because 0.6 of the step lies along x: 15 × 0.8 + 20 × 0.6 = 24 is the slope in a different direction.

Calling the gradient the fall line. $\nabla f$∇f points uphill; straight downhill is $-\nabla f$−∇f, where the slope is $-|\nabla f|$−|∇f|.

A skier

In the application below, the height of a ski slope is a function of position. The gradient at the skier, dotted with the unit vector toward a lift, gives her rate of descent as she sets off, and setting $|\nabla h|\cos \theta$|∇h| cos θ equal to a gentler rate gives the angle of a traverse.