MAST10006 · Calculus 2
Multivariable Calculus
A function z = f(x, y) takes a point in the plane and returns a height; its graph is a surface — a landscape of hills, valleys and passes. The whole chapter is the calculus of that landscape. You picture it with level curves / contours, test 2-D limits with the path test (a limit exists only if every path gives the same value), and slope the surface with partial derivatives (differentiate in one variable, holding the other constant; Clairaut gives fxy = fyx). The tangent plane is the first-order approximation, the chain rule sums over every path through the dependency tree, and the gradient ∇f points uphill, perpendicular to the level curve, with the directional derivative ∇f · û (unit vector). The flagship long-answer is classifying stationary points with the Hessian determinant D = fxxfyy − (fxy)², and the chapter closes with double integrals over rectangles (Fubini and the separable shortcut).
What this chapter covers
- 011-2 Surfaces, cross-sections and level-curve / contour maps
- 023 Limits and the path test
- 034-5 Continuity and partial derivatives (Clairaut)
- 046 The tangent plane (linear approximation)
- 057-8 The chain rule, gradient and directional derivative
- 069 Stationary points and the Hessian test
- 0710 Double integrals over rectangles (Fubini)
Worked example: classify stationary points with the Hessian
- +1Find stationary points: fx = 3x² − 3 = 0 gives x = ±1; fy = 2y = 0 gives y = 0. Points: (1, 0) and (−1, 0).
- +1Second partials: fxx = 6x, fyy = 2, fxy = 0, so the Hessian determinant is D = fxxfyy − (fxy)² = 12x.
- +1At (1, 0): D = 12 > 0 and fxx = 6 > 0 → local minimum.
- +1At (−1, 0): D = −12 < 0 → saddle (regardless of fxx).
Key terms
- Level curve / contour
- The set {(x, y) : f(x, y) = c} for a constant height c — exactly a topographic contour. Contours bunched together mark a steep region; far apart marks a flat one. The gradient always points across the contours, never along them.
- Path test
- The check that a 2-D limit lim(x,y)→(a,b) f exists: f must approach the same value along every path to (a, b). Finding two paths that give different values proves the limit does NOT exist; agreeing along every straight line is not enough to prove it does.
- Partial derivative
- fx = ∂f/∂x (hold y constant) and fy = ∂f/∂y (hold x constant), the slopes of the surface in the two axis directions. Clairaut's theorem says the mixed second partials agree, fxy = fyx, when they are continuous.
- Gradient
- ∇f = (fx, fy), the vector that points in the direction of steepest ascent; its length is the maximum rate of change and it is perpendicular to the level curve. The directional derivative in a direction û is ∇f · û, where û must be a UNIT vector.
- Hessian test
- At a stationary point (where ∇f = 0), the determinant D = fxxfyy − (fxy)² classifies it: D > 0 with fxx > 0 is a local minimum, D > 0 with fxx < 0 a local maximum, D < 0 a saddle, and D = 0 inconclusive.
Multivariable Calculus FAQ
How do I show a two-variable limit does not exist?
Use the path test: find two paths to the point that give different limiting values — common choices are y = 0, y = x, y = mx and y = x². For xy/(x²+y²) at the origin, the path y = 0 gives 0 while y = x gives ½, so the limit does not exist. To CONFIRM a limit exists, agreeing along straight lines is not enough; switch to polar coordinates and show the limit as r → 0 is independent of the angle.
How do I find a partial derivative?
Differentiate with respect to one variable while treating the other as a constant: fx holds y fixed, fy holds x fixed. For second-order partials, differentiate again; Clairaut's theorem guarantees the mixed partials fxy and fyx are equal when continuous, which is a useful check on your algebra.
Why do I have to normalise the direction in a directional derivative?
Because Dûf = ∇f · û only measures the rate of change per unit distance when û is a unit vector. If you are handed a direction like (3, 4), divide by its length |(3,4)| = 5 to get û = (3/5, 4/5) before taking the dot product, or your rate will be scaled wrongly.
How does the Hessian classify a stationary point?
Compute D = fxxfyy − (fxy)² at the point. D > 0 with fxx > 0 is a local minimum (a valley), D > 0 with fxx < 0 a local maximum (a hill), D < 0 a saddle (a mountain pass) regardless of fxx, and D = 0 is inconclusive. First solve ∇f = 0 to find all stationary points.
Exam move
Hold the topographic picture in mind — surface, contours, gradient across the contours — and it organises the whole chapter. For limits, default to the path test to disprove existence (two paths, different values) and reach for polar coordinates only when asked to confirm a limit. For partials, the discipline is holding the other variable constant, and Clairaut gives you a free consistency check. The flagship skill is stationary-point classification: solve fx = 0 and fy = 0 simultaneously and find them ALL, compute the second partials, evaluate D = fxxfyy − (fxy)² at each point, and read the sign table — remembering D < 0 is always a saddle and D = 0 is inconclusive. For directional derivatives always normalise the direction first, and for double integrals over rectangles iterate inner-then-outer and use the separable shortcut when f(x,y) = g(x)h(y). These are pure procedures; bank them by drilling past-exam questions by hand.
