Difficulty: Intermediate | Prerequisites: Partial derivatives, second-order partial derivatives, critical points via the first derivative test, basic familiarity with 3D graphing (Desmos or similar).
This lab sits in the multivariable optimisation sequence of Calc III. You already know how to find critical points by setting the gradient to zero; now you need to classify them. The tool is the second derivative test, which hinges on the discriminant of the quadratic approximation at the critical point. The lab builds intuition by having you graph quadric surfaces of the form z = ax² + 2bxy + cy² and observe how the sign of the discriminant D = ac − b² and the sign of a determine whether the origin is a minimum, maximum, or saddle point.
For a quadric surface z = ax² + 2bxy + cy², the discriminant D = ac − b² tells you everything. When D > 0, the origin is a minimum (if a > 0) or maximum (if a < 0). When D < 0, the origin is a saddle. When D = 0, the test is inconclusive, and adding higher-order terms (like hxy²) does not change the classification when D ≠ 0 but can matter when D = 0.
Quadric surface (quadratic form in two variables)
A surface defined by z = ax² + 2bxy + cy², where a, b, and c are real constants. This is the second-order Taylor approximation of any smooth function near a critical point, so understanding it is equivalent to understanding the second derivative test.
In simple terms, it is the simplest curved surface you can build from x and y, and every smooth surface looks like one of these if you zoom in close enough to a critical point.
Discriminant (D)
D = ac − b². This single number, computed from the coefficients of the quadratic form, determines the shape of the surface at the origin. Positive D means the surface curves the same way in every direction; negative D means it curves up in some directions and down in others.
In simple terms, D tells you whether the surface is bowl-shaped (D > 0) or saddle-shaped (D < 0).
Paraboloid
A bowl-shaped surface. When D > 0 and a > 0, the surface opens upward (like a bowl sitting right-side up). When D > 0 and a < 0, it opens downward (like a bowl flipped over).
Saddle point
A critical point where the surface curves upward in some directions and downward in others, like a horse saddle or a mountain pass. Occurs when D < 0.
Second derivative test (for functions of two variables)
The classification rule: compute D = f_xx · f_yy − (f_xy)² at a critical point. If D > 0 and f_xx > 0, it is a local minimum. If D > 0 and f_xx < 0, it is a local maximum. If D < 0, it is a saddle point. If D = 0, the test is inconclusive.
In simple terms, this is the multivariable version of checking whether f″ is positive or negative at a critical point in single-variable calculus.
Higher-order perturbation (hxy² term)
Adding a small cubic term hxy² to the quadratic form to investigate whether it changes the classification of the origin. The lab shows that when D ≠ 0, small perturbations do not change the nature of the critical point, but when D = 0, they can.
The surface in question is z = ax² + 2bxy + cy². The coefficient of the cross-term is written as 2b (not just b) so that the discriminant formula comes out cleanly.
a controls the curvature in the x-direction.
c controls the curvature in the y-direction.
b controls how much the surface twists between the x and y directions.
D = ac − b².
This is really the determinant of the Hessian matrix divided by 4. For the general function f(x, y), the Hessian is:
| f_xx | f_xy | | f_yx | f_yy |
and D = f_xx · f_yy − (f_xy)². For the quadric z = ax² + 2bxy + cy², we get f_xx = 2a, f_yy = 2c, f_xy = 2b, so D = (2a)(2c) − (2b)² = 4(ac − b²). The sign is all that matters, so we work with D = ac − b².
When D > 0, the product ac exceeds b², meaning the "own-direction" curvatures dominate the cross-term. The surface curves the same way in all directions, producing a bowl (paraboloid).
When D < 0, the cross-term dominates, pulling the surface up in some directions and down in others, producing a saddle.
When D = 0, the quadratic form is degenerate: it factors into a product of linear terms, and the surface is a "trough" or "ridge" along a line rather than a proper bowl or saddle.
For z = ax² + 2bxy + cy² with D = ac − b²:
Sign of D | Sign of a | Shape | Origin is |
|---|---|---|---|
D > 0 | a > 0 | Paraboloid opening upward | Minimum |
D > 0 | a < 0 | Paraboloid opening downward | Maximum |
D < 0 | a > 0 | Saddle | Saddle |
D < 0 | a < 0 | Saddle | Saddle |
D = 0 | a > 0 | Parabolic trough opening upward | Neither (minimum along a line, not a point) |
D = 0 | a < 0 | Parabolic trough opening downward | Neither (maximum along a line, not a point) |
When D ≠ 0:
The origin is a minimum if a and c are both positive. (D > 0 guarantees they have the same sign, so checking a alone suffices.)
The origin is a maximum if a and c are both negative.
The origin is a saddle if a and c have opposite signs. (This forces D < 0.)
When D = 0:
The quadratic form degenerates. The graph is a parabolic trough or ridge. The origin is not a local min or max in the usual sense, but it can be part of a minimum or maximum line. The second derivative test is officially inconclusive here.
The lab's Exercise 2 asks what the graph can look like when D = 0. The answer: the graphs can only be troughs or saddles. The origin will never be a minimum or maximum point on its own, but it can sit on a minimum or maximum line.
This is the degenerate case. The quadratic form factors as a perfect square (times a constant), so the surface is flat in one direction.
The lab investigates z = ax² + 2bxy + cy² + hxy², where h is small. The research question: does the nature of the origin change when you add this higher-order perturbation, compared with h = 0?
The results from graphing:
If D > 0 and a > 0: the nature of the origin does not change. It remains a local minimum.
If D < 0 and a > 0: the nature of the origin does not change. It remains a local minimum (the saddle nature is set by the quadratic part and the cubic perturbation does not override it for small h).
If D = 0 and a > 0: the nature of the origin changes. It becomes "neither" (the second derivative test remains inconclusive, and the cubic term can tip the degenerate case).
When the discriminant is non-zero, small higher-order perturbations do not change the classification. The quadratic term dominates near the origin. This is a stability result: non-degenerate critical points are robust.
When D = 0, the critical point is degenerate and fragile. Even a tiny cubic perturbation can change its character. This is why the second derivative test says "inconclusive" rather than giving a definite answer.
Quadric surface: z = ax² + 2bxy + cy²
Discriminant: D = ac − b²
Hessian matrix (general f(x, y)): H = | f_xx f_xy | | f_yx f_yy |
Second derivative test: D = f_xx · f_yy − (f_xy)² at the critical point.
D > 0 and f_xx > 0 → local minimum
D > 0 and f_xx < 0 → local maximum
D < 0 → saddle point
D = 0 → inconclusive
Perturbed surface (Exercise 3): z = ax² + 2bxy + cy² + hxy²
Optimisation problems in engineering and economics come down to classifying critical points. When you design the shape of a satellite dish (a paraboloid), you need D > 0 and a > 0 so the dish focuses signals to a single point. The discriminant test is also how structural engineers check whether a load-bearing surface has a stable equilibrium (minimum) or an unstable one (saddle), which determines whether a structure stays put or buckles.
Students often think D < 0 means a maximum. It does not. D < 0 always means a saddle point, regardless of the sign of a.
Students sometimes check the sign of a alone without computing D first. The sign of a only determines min vs max after you have confirmed D > 0.
"D = 0 means it is a saddle" is wrong. D = 0 means the test gives no answer. The critical point might be a min, max, saddle, or something else entirely; you need other methods (like graphing or higher-order analysis) to decide.
Forgetting the factor of 2 in front of b. The surface is z = ax² + 2bxy + cy², so when the cross-term coefficient is, say, 1, the value of b is 0.5, not 1. Getting this wrong will give you the wrong D.
⚠️ The classification table (D > 0, a > 0 → min; D > 0, a < 0 → max; D < 0 → saddle) is a near-certainty on exams. Be able to apply it quickly to any given a, b, c.
⚠️ You will be asked to compute D for a general function f(x, y) using second partial derivatives, not just for the standard quadric. Make sure you can go from f(x, y) → compute f_xx, f_yy, f_xy → compute D → classify.
⚠️ The D = 0 case is a favourite trick question. If you get D = 0 on an exam, the correct answer is "the second derivative test is inconclusive," not "it is a saddle" or "it is a minimum."
⚠️ The stability result from Exercise 3 (non-degenerate critical points are robust to perturbation; degenerate ones are not) is conceptually important and may appear as a true/false or short-answer question.
True or false: If D > 0, the origin is always a minimum. (False: it is a minimum only if a > 0; if a < 0, it is a maximum.)
Fill in the blank: When D < 0, the origin is a ______. (Saddle point.)
True or false: When D = 0, the second derivative test tells us the origin is a saddle. (False: the test is inconclusive.)
Fill in the blank: The discriminant formula is D = ______. (ac − b², or equivalently f_xx · f_yy − (f_xy)².)
True or false: Adding a small hxy² term to a quadric with D > 0 changes the nature of the origin. (False: non-degenerate critical points are robust to small perturbations.)
Q: Given z = 2x² + 4xy + 3y², classify the critical point at the origin.
A: Here a = 2, the cross-term is 4xy so 2b = 4, meaning b = 2, and c = 3. Then D = ac − b² = (2)(3) − (2)² = 6 − 4 = 2 > 0. Since D > 0 and a = 2 > 0, the origin is a local minimum.
Q: Given z = −x² + 2xy − y², classify the critical point at the origin.
A: Here a = −1, 2b = 2 so b = 1, c = −1. Then D = (−1)(−1) − (1)² = 1 − 1 = 0. The second derivative test is inconclusive.
Q: Given z = x² − 6xy + y², classify the critical point at the origin.
A: Here a = 1, 2b = −6 so b = −3, c = 1. Then D = (1)(1) − (−3)² = 1 − 9 = −8 < 0. The origin is a saddle point.
Q: Suppose you add a small term 0.01xy² to a quadric with D > 0 and a > 0. Does the nature of the origin change?
A: No. When D ≠ 0, the critical point is non-degenerate and robust to small perturbations. The origin remains a local minimum.
Q: Why does the second derivative test fail when D = 0? Explain in one or two sentences.
A: When D = 0, the quadratic approximation is degenerate: it curves in only one direction and is flat in the other. The quadratic part alone cannot determine the local behaviour, so higher-order terms (or other methods) are needed.
This connects to the single-variable second derivative test (f″ > 0 → min, f″ < 0 → max) because the discriminant is the natural generalisation to two variables. Where single-variable calculus checks the sign of one number, multivariable calculus checks the sign of a 2×2 determinant.
This also connects to eigenvalues of the Hessian, which you may see later in the course or in linear algebra. The sign of D tells you whether the Hessian's eigenvalues have the same sign (D > 0) or opposite signs (D < 0). The full eigenvalue approach generalises to functions of three or more variables, where the 2×2 discriminant test no longer suffices.
The stability result (non-degenerate critical points are robust) connects to bifurcation theory and structural stability in differential equations, which are topics in more advanced courses.
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