Tutorial 6 Q11–Q12, from Zero

The one idea behind both questions

Stand at a point on a hilly landscape, where f(x, y) is the height. Pick a direction and take a small step. How fast does the height change, per unit of distance walked? That rate is the directional derivative, written D_uf. The direction is a unit vector u = (u₁, u₂), meaning its length is 1. It has to be, because "per unit distance" only makes sense if the step you measure along is one unit long.

Two special directions you already know: walking along the x-axis, the rate is f_x; walking along the y-axis, it is f_y. A general direction is part x-step and part y-step: u₁ of the step goes along x and u₂ along y. For a smooth landscape the two changes simply add:

The pair ∇f = (f_x, f_y) is called the gradient. It's just the two partial derivatives written as a vector.

The word smooth is doing real work there. The formula needs f to be differentiable at the point: close up, the surface looks like a flat tilted plane. Q11 assumes that. Q12 is a function where it fails, and the formula gives the wrong answer.

Question 11 · Recover the gradient from two readings

The derivative of f(x, y) at P₀(1, 2) in the direction of i + j is 2√2 and in the direction of −2j is −3. What is the derivative of f in the direction of −i − 2j? Give reasons for your answer.

We are never told what f is. That's the point: two measurements are enough to work out the gradient, and the gradient gives the rate in every direction.

The gradient at P₀ is two numbers we don't know yet. Call them ∇f(P₀) = (a, b), where a = f_x(1, 2) and b = f_y(1, 2). Two unknowns need two equations, and the question gives us two measurements.

The formula needs u of length 1, and the given directions aren't. Divide each by its length:

• |(1, 1)| = √(1 + 1) = √2, so u₁ = (1/√2, 1/√2).
• |(0, −2)| = 2, so u₂ = (0, −1).

Measurement 1: ∇f · u₁ = 2√2, i.e.

(Multiply both sides by √2, and use √2 · √2 = 2.)

Measurement 2: ∇f · u₂ = −3, i.e.

Then a + 3 = 4, so a = 1. The gradient at P₀ is (1, 3): the slope is 1 in the x-direction and 3 in the y-direction.

The new direction is (−1, −2), with length √(1 + 4) = √5, so u = (−1/√5, −2/√5). Then

Why "not parallel" matters: if both readings were along the same line (say j and −2j), they would both only tell you b, and a could be anything.

Question 12 · When ∇f · u lies

Consider the function f : ℝ² → ℝ defined by f(x, y) = xy/(x² + y²) for (x, y) ≠ (0, 0), and 0 for (x, y) = (0, 0). Let u = ⟨u₁, u₂⟩ be a unit vector (u₁² + u₂² = 1). (a) Show that f_x(0, 0) and f_y(0, 0) exist and find their values. (b) Determine all unit vectors u for which D_uf(0, 0) exists using the limit definition. (c) Explain why the formula D_uf = ∇f · u cannot be used here.

It's a piecewise function: one formula everywhere except the origin, and a separate value, 0, at the origin. The formula can't be used at (0, 0) because it gives 0/0. So at the origin the quotient rule, the gradient formula and every shortcut are unavailable. The only tool is the definition, the same limit you used for derivatives in school.

Try some points. On the x-axis, f(x, 0) = 0/x² = 0. On the line y = x: f(t, t) = t²/(2t²) = ½ for every t ≠ 0, however close to the origin. On y = −x it is always −½. Each line through the origin carries its own constant value, and at the origin itself the value is 0.

f_x(0, 0) means: move only in x, a small amount h, and take the school limit.

The point (h, 0) is not the origin (since h ≠ 0), so use the formula: f(h, 0) = h · 0/(h² + 0) = 0. And f(0, 0) = 0 is given. So the quotient is (0 − 0)/h = 0 for every h, and its limit is 0. Exactly the same happens along the y-axis: f_y(0, 0) = 0. Both exist, and both equal 0, because f is 0 all along both axes.

Same idea as a partial derivative, but step along u: the point h units along u from the origin is (hu₁, hu₂).

First simplify the top. Put x = hu₁, y = hu₂ into the formula:

The h² cancels, and u₁² + u₂² = 1 because u is a unit vector. This is Step 1's picture as algebra: along the line in direction u, f is the constant u₁u₂. So the quotient is

Now let h → 0. There are two cases:

• If u₁u₂ ≠ 0, a fixed non-zero number is divided by something shrinking to 0. Take u = (1/√2, 1/√2), where u₁u₂ = ½: at h = 0.1 the quotient is 5, at 0.01 it's 50, at 0.001 it's 500, and at h = −0.01 it's −50. It blows up, with opposite signs on the two sides, so there is no limit.
• If u₁u₂ = 0, the quotient is 0/h = 0 for every h, so the limit is 0.

u₁u₂ = 0 means u₁ = 0 or u₂ = 0. With length 1, that leaves exactly four directions, the axis directions:

Drag the slider. At every angle except 0°, 90° and 180°, the right-hand graph is a hyperbola, shooting up on one side of h = 0 and down on the other, so no single value is approached. On an axis direction it flattens into the line 0, and the limit exists.

From part (a), ∇f(0, 0) = (f_x, f_y) = (0, 0). If the formula worked, it would say D_uf(0, 0) = (0, 0) · u = 0 in every direction. Part (b) showed the truth: in most directions the directional derivative doesn't even exist. So the formula must have a condition, and here it isn't met.

The condition is differentiability: close up, the surface must look like a flat plane. This f is nowhere near that at the origin. It isn't even continuous there, because along y = x the values stay at ½ right up to the origin, where the value is 0 (Step 1's jump). The chain of facts is:

The lesson the question is built around: having both partial derivatives is not enough. The partials only look along the two axes, and this function behaves perfectly along the axes (it's 0 there) while jumping in between.

The two questions in one line each

• Q11: when f is smooth, the gradient is two unknown numbers, each measured rate is one equation (after making the direction unit), and once you know ∇f you know the rate in every direction.
• Q12: at a piecewise point, use the definition. If the function jumps there, ∇f · u is not allowed, even though the partials exist.