Multivariable calculus

Directional derivative of x^3 + y^3 - 3xy at (1, 2)

Q: What is the largest possible directional derivative at (1, 2)?

It is \lVert \langle -3, 9 \rangle \rVert = \sqrt{90} = 3\sqrt{10} \approx 9.4868, reached along \langle -3, 9 \rangle / \sqrt{90}. Our direction gets 9.48 of that.

Q: Why is the x component of the gradient negative here?

Because \frac{\partial f}{\partial x} = 3x^2 - 3y compares x^2 with y. At (1,2) we have x^2 = 1 and y = 2, so the cross term wins and increasing x decreases f.

At (1, 2) the gradient of f(x,y) = x^3 + y^3 - 3xy is (-3, 9). Dotting with the unit direction (-0.28, 0.96) gives 237/25, that is 9.48. The gradient has length 3 times the square root of 10, about 9.4868, so this direction is close to the steepest one but not equal to it.

f(x,y) = x^3 + y^3 - 3xy

D_{\mathbf{u}} f(1,2) = \frac{237}{25} = 9.48

At x = 1, y = 2, in the unit direction (-0.28, 0.96).

The cross term is where the care goes

The cubes are straightforward. The term $-3xy$ is the one to slow down on: with $y$ held fixed it differentiates to $-3y$ , and with $x$ held fixed it differentiates to $-3x$ .

\nabla f = \left\langle 3x^2 - 3y,\; 3y^2 - 3x \right\rangle, \qquad \nabla f(1,2) = \langle -3, 9 \rangle

At $(1,2)$ that is $3 - 6 = -3$ and $12 - 3 = 9$ . Note the $x$ partial is negative even though $x$ is positive, because the cross term dominates the small cube there.

D_{\mathbf{u}} f(1,2) = -3(-0.28) + 9(0.96) = 0.84 + 8.64 = \frac{237}{25}

The mistake: differentiating the cross term as a constant

Writing $\frac{\partial}{\partial x}(-3xy) = -3$ instead of $-3y$ is the standard error, and it survives unnoticed whenever the other variable happens to be 1. Here $y = 2$ , so it turns the $x$ partial from $-3$ into $0$ and the answer from 9.48 into 8.64.

The second mistake is interpretive. Since $9.48$ is so close to the maximum rate of about $9.4868$ , it is tempting to say the given direction is the direction of steepest ascent. It is not, it is about 2.2 degrees away, and the two are genuinely different vectors.

Given direction: $\langle -0.28, 0.96 \rangle$
Steepest ascent direction: $\langle -3, 9 \rangle / \sqrt{90} \approx \langle -0.3162, 0.9487 \rangle$
The gap is 2.17 degrees and its cosine is 0.99928, which is why the two rates look almost identical

The same surface has a point with no rate at all

This surface is the standard example for critical points, and comparing $(1,2)$ with those points sharpens what a directional derivative measures.

\nabla f = \langle 0, 0 \rangle \iff 3x^2 = 3y \text{ and } 3y^2 = 3x \iff (x,y) = (0,0) \text{ or } (1,1)

At $(1,1)$ every unit direction gives 0, because the gradient itself vanishes. One step away at $(1,2)$ , the same function is climbing at nearly 9.5 units per unit distance. Gradients can change fast, so evaluate at the point the question asks about and nowhere else.

The build re-checks that (-0.28, 0.96) has length 1, rebuilds the gradient of x^3 + y^3 - 3xy at (1, 2) by central differences on the function itself, and fails unless the dot product is still 9.48.

Frequently asked questions

What is the largest possible directional derivative at (1, 2)?

It is $\lVert \langle -3, 9 \rangle \rVert = \sqrt{90} = 3\sqrt{10} \approx 9.4868$ , reached along $\langle -3, 9 \rangle / \sqrt{90}$ . Our direction gets 9.48 of that.

Why is the x component of the gradient negative here?

Because $\frac{\partial f}{\partial x} = 3x^2 - 3y$ compares $x^2$ with $y$ . At $(1,2)$ we have $x^2 = 1$ and $y = 2$ , so the cross term wins and increasing $x$ decreases $f$ .