4.2 Reading Critical Points from the Map

Where $\nabla f = \mathbf{0}$ , the terrain is momentarily flat: no direction climbs and none falls, to first order. Such critical points are where maxima and minima hide — but flatness alone does not say which, and a third possibility lurks: the saddle, uphill one way and downhill another. A contour map betrays the character of each critical point by a visual signature, and the discipline of this section is to read the map before touching the algebra.

A top-down view of a seventeen-level pink contour map of f of x and y equals x cubed over 3 minus x plus y cubed over 3 minus y: two families of nested oval contours sit in opposite corners of the square, and in the other two corners the contour lines cross themselves in X shapes. A translucent surface of the same function and four small colored marker points at its critical points are hidden in this view; they can be shown by opening the figure in XYZ 3D. — A straight-down view of a seventeen-level contour map of $f(x, y) = x^3/3 - x + y^3/3 - y$ , drawn in pink — just the map, like a hiker's chart. Two corners hold families of nested ovals, around $(1, 1)$ and $(-1, -1)$ ; near $(1, -1)$ and $(-1, 1)$ the level curves instead cross themselves in an X. The surface and four labeled marker points are hidden in this view; step 5 shows them in the XYZ 3D editor.

Explore

Zoom in on the corners near $(1, -1)$ and $(-1, 1)$ and describe what the contour lines do there. A curve that passes through itself in an X is something no drawn contour of Chapter 3's map ever did.
On a hiking map, what does a family of nested closed loops around a point mean about the terrain? What would a trail through the X-point feel like underfoot — and why is "flat exactly at the crossing, tilted all around it" the only consistent answer?
Classify all four points — $(1,1)$ , $(-1,-1)$ , $(1,-1)$ , $(-1,1)$ — as local max, local min, or saddle, using only the map. Write all four down before revealing anything.
One classification the unlabeled map cannot make: of the two nested-oval families, which is the peak and which the pit? Explain why the map alone is ambiguous — then commit to a guess anyway.
Press Open in XYZ 3D ↗ under the figure. In the editor's object list, turn on the four Point rows and then the Surface row with their eye toggles, and orbit down from the top view. Which oval family holds the blue marker (the pit)? Was your step 4 guess right — and your step 3 reads?
The saddle values here are exactly $f = 0$ , and the contour sheet starts at $z = 0$ . Still in the editor, step the Contours row's z offset box (steps of $0.5$ ) up to $0.5$ , down to $-0.5$ , and back to $0$ : only at $0$ does the sheet pass through both saddle points at once, because the X-shaped curve is the level curve at saddle height. Nested ovals never touch their center point this way. (The level count of 17 is deliberate — an odd count places one level exactly at $0$ ; move the row's Levels slider to an even count and the X dissolves into nearby hyperbola pairs. Set it back.)

The second derivative test

The signatures you just used are what the second derivative test formalizes. Near a critical point, $f$ is well approximated by a quadratic; the number

$D = f_{xx}\,f_{yy} - f_{xy}^2$

decides that quadratic's shape. If $D > 0$ the quadratic is a bowl — upright if $f_{xx} > 0$ (local minimum), inverted if $f_{xx} < 0$ (local maximum) — and its level curves are nested ovals. If $D < 0$ it is a saddle, and the level curve through the critical value crosses itself there: the X. If $D = 0$ the test is silent. Step 4's ambiguity is real mathematics: oval shape determines the sign of $D$ , but peak-versus-pit needs the sign of $f_{xx}$ (or labeled heights) — shapes alone cannot supply it.

Example 1 Four critical points, classified twice

Find all critical points of $f(x, y) = x^3/3 - x + y^3/3 - y$ and classify each with the second derivative test; then match the results to the map's signatures and markers.

Show solution

The partials $f_x = x^2 - 1$ and $f_y = y^2 - 1$ vanish when $x = \pm 1$ and $y = \pm 1$ : four critical points. The second derivatives are $f_{xx} = 2x$ , $f_{yy} = 2y$ , $f_{xy} = 0$ , so

$D = (2x)(2y) - 0 = 4xy.$

At $(1, 1)$ : $D = 4 > 0$ and $f_{xx} = 2 > 0$ — a local minimum, with $f = \tfrac{1}{3} - 1 + \tfrac{1}{3} - 1 = -\tfrac{4}{3}$ ; this is the blue marker in the pit. At $(-1, -1)$ : $D = 4 > 0$ but $f_{xx} = -2 < 0$ — a local maximum, $f = \tfrac{4}{3}$ , the red marker. At $(1, -1)$ and $(-1, 1)$ : $D = -4 < 0$ — saddles, and indeed $f = \tfrac{1}{3} - 1 - \tfrac{1}{3} + 1 = 0$ at both, which is why one contour level drawn at height $0$ produces both X's. Every algebraic verdict matches a signature you read in step 3 — and only the sign of $f_{xx}$ separated the two oval families.

4 \cdot x \cdot y

The discriminant, as the Hessian's determinant ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The second-derivative test's $D$ is not a recipe — it is the determinant of the matrix of second partials, $f_{xx}$ , $f_{xy}$ , $f_{yy}$ , arranged as stated in the solution above. The computer algebra system expands that determinant symbolically and lands on the same $4xy$ whose sign sorted all four critical points.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Aligned to OpenStax Calculus Volume 3 (Strang & Herman), © OpenStax (Rice University), licensed CC BY-NC-SA 4.0; no OpenStax content is reproduced, and this work is not affiliated with or endorsed by OpenStax or Rice University. License: CC-BY-NC-SA-4.0.