Difficulty: Intermediate to Advanced | Prerequisites: Partial derivatives, chain rule for multivariable functions, single-variable Taylor series (MATH 124/125), double integrals.
This set of topics brings together the main tools for analysing multivariable functions: finding where they peak, dip, or saddle, approximating them locally with planes and polynomials, and computing weighted averages of regions. Local and absolute extrema are the multivariable version of max/min problems from Calc I, with a richer classification (saddle points have no single-variable analogue). Tangent planes generalise tangent lines. Taylor series extend polynomial approximation, and center of mass uses double integrals to find balance points. You need solid comfort with partial derivatives and double integrals before tackling these.
Critical points occur where both partial derivatives are zero (or undefined). The second derivative test classifies them using the discriminant D = fxx fyy - fxy². Tangent planes approximate a surface near a point using first-order partial derivatives. Taylor series generalise this to higher-order polynomial approximations. Center of mass finds the balance point of a 2D region using double integrals weighted by a density function.
Critical point
A point (a, b) where fx(a, b) = 0 and fy(a, b) = 0, or where one or both partial derivatives do not exist. Think of it as a candidate for a local maximum, local minimum, or saddle point.
Discriminant, D
D(a, b) = fxx(a, b) · fyy(a, b) - [fxy(a, b)]². This quantity determines the nature of a critical point. In simple terms, D tells you whether the surface curves the same way in all directions (bowl or dome) or differently (saddle).
Saddle point
A critical point where D < 0. The function increases in some directions and decreases in others. Think of it as the centre of a horse saddle or a mountain pass.
Tangent plane
The plane that best approximates a surface z = f(x, y) near a point (x₀, y₀, z₀). Its equation is z = f(x₀, y₀) + fx(x₀, y₀)(x - x₀) + fy(x₀, y₀)(y - y₀). In simple terms, the flat surface that "just touches" the graph at that point.
Linearisation
The function L(x, y) = f(x₀, y₀) + fx(x₀, y₀)(x - x₀) + fy(x₀, y₀)(y - y₀), which is the tangent plane expressed as an approximation to f near (x₀, y₀).
Taylor series (single variable)
The infinite polynomial Σ f⁽ᵏ⁾(b)/k! · (x - b)ᵏ that equals f(x) within the interval of convergence. Think of it as approximating a function with polynomials of increasing degree.
Taylor remainder, Rn(x)
The error in truncating the Taylor series at degree n: |Rn(x)| ≤ M/(n+1)! · |x - b|^(n+1), where M is a bound on |f⁽ⁿ⁺¹⁾| on the interval.
Center of mass
The balance point (x̄, ȳ) of a 2D region with density ρ(x, y), computed as x̄ = My/m and ȳ = Mx/m, where m = ∬ ρ dA, My = ∬ ρx dA, and Mx = ∬ ρy dA.
Finding critical points
Set fx = 0 and fy = 0 simultaneously and solve. Any point where both partials are zero (or where a partial does not exist) is a critical point.
Second derivative test
At each critical point (a, b), compute:
D = fxx(a, b) · fyy(a, b) - [fxy(a, b)]²
D > 0 and fxx > 0: local minimum
D > 0 and fxx < 0: local maximum
D < 0: saddle point
D = 0: the test is inconclusive (you need other methods)
The logic: D > 0 means the surface curves the same way in all directions. The sign of fxx then tells you whether it curves up (minimum) or down (maximum). D < 0 means the curvatures point in opposite directions, creating a saddle.
Tangent plane equation
For a surface z = f(x, y) at the point (x₀, y₀, z₀):
z = z₀ + fx(x₀, y₀)(x - x₀) + fy(x₀, y₀)(y - y₀)
This is the linear approximation L(x, y) to f near (x₀, y₀).
Normal vector to the tangent plane
The tangent plane has normal vector n = ⟨fx, fy, -1⟩, evaluated at (x₀, y₀). This comes from rewriting the surface as F(x, y, z) = f(x, y) - z = 0 and taking the gradient ∇F = ⟨fx, fy, -1⟩.
When to use linearisation
Linearisation is useful for approximating f(x, y) near a known point without recomputing the full function. It is the multivariable version of the tangent-line approximation f(x) ≈ f(a) + f'(a)(x - a) from Calc I.
To find the absolute maximum and minimum of f(x, y) on a closed, bounded region D:
Find all critical points inside D (set fx = 0 and fy = 0).
Find potential max/min values on the boundary of D. On each boundary curve, reduce f to a single-variable function and apply single-variable techniques.
Plug all candidates (interior critical points and boundary extrema, including corners and endpoints) into f. Compare values. The largest is the absolute maximum; the smallest is the absolute minimum.
This is the multivariable version of the closed interval method from Calc I. The key addition is that boundaries in 2D are curves, not just two endpoints, so step 2 involves more work.
Taylor series centred at b
Tn(x) = Σ (k=0 to n) f⁽ᵏ⁾(b)/k! · (x - b)ᵏ
As n → ∞, this approaches f(x) within the interval of convergence.
Remainder bound (Taylor's inequality)
|Rn(x)| ≤ M/(n+1)! · |x - b|^(n+1)
where M is an upper bound on |f⁽ⁿ⁺¹⁾(t)| for all t between b and x. This tells you the worst-case error when you truncate the series at degree n.
Operations on Taylor series
Multiplying by a constant, by x, or integrating/differentiating term by term: the result is still a Taylor series, with no change to the radius of convergence for multiplication or differentiation.
Adding two Taylor series: valid when both share the same centre and both converge. The sum converges on the intersection of their intervals.
Substitution: replace x with another expression and adjust the inequality describing the interval of convergence accordingly.
For a lamina (thin flat plate) occupying region D with density function ρ(x, y):
Total mass
m = ∬_D ρ(x, y) dA
Moments
My = ∬_D x · ρ(x, y) dA (moment about the y-axis)
Mx = ∬_D y · ρ(x, y) dA (moment about the x-axis)
Center of mass coordinates
x̄ = My / m, ȳ = Mx / m
Note the cross: x̄ uses My (the moment about the y-axis), and ȳ uses Mx (the moment about the x-axis). This trips students up. The logic is that My measures how the mass is distributed in the x-direction, so it gives you the x-coordinate of the balance point.
For uniform density (ρ = constant), the center of mass simplifies to the centroid, and ρ cancels out of the fractions.
Critical points: fx = 0 and fy = 0 (or DNE)
Discriminant: D = fxx fyy - fxy²
Tangent plane: z = f(x₀, y₀) + fx(x₀, y₀)(x - x₀) + fy(x₀, y₀)(y - y₀)
Normal to tangent plane: n = ⟨fx, fy, -1⟩
Taylor series: Tn(x) = Σ f⁽ᵏ⁾(b)/k! · (x - b)ᵏ
Taylor remainder: |Rn(x)| ≤ M/(n+1)! · |x - b|^(n+1)
Mass: m = ∬ ρ dA
Center of mass: x̄ = My/m = (∬ xρ dA) / (∬ ρ dA), ȳ = Mx/m = (∬ yρ dA) / (∬ ρ dA)
Optimisation of multivariable functions (finding extrema, classifying critical points) is the mathematical backbone of machine learning, engineering design, and economics. Every time a model is trained by minimising a loss function, these ideas are at work.
Center of mass calculations determine where to support a structure so it balances, and they appear in robotics (finding a robot arm's balance point), aerospace (locating an aircraft's centre of gravity), and structural engineering.
Students often forget that D > 0 is necessary but not sufficient on its own. You still need the sign of fxx to distinguish a local max from a local min. D > 0 only tells you the critical point is not a saddle.
The discriminant formula uses fxy squared, not fxx times fyy alone. Leaving out the fxy² term is a common algebraic slip that changes the classification entirely.
In center of mass, x̄ uses My (the moment about the y-axis), not Mx. The subscript on M refers to the axis the moment is taken about, not the coordinate it gives you. This cross-labelling is one of the most common errors on exams.
Students sometimes assume Taylor's remainder bound gives the exact error. It gives an upper bound on the error. The true error may be smaller.
⚠️ The second derivative test (computing D at critical points) is one of the most frequently tested procedures in MATH 126. Practise the full pipeline: find critical points, compute all four second partials, evaluate D, classify.
⚠️ Absolute extrema on a closed region is a multi-step problem that combines interior critical points with boundary analysis. Expect at least one exam question requiring you to parametrise each boundary segment.
⚠️ Tangent plane problems often ask you to find the equation of the tangent plane and then use it to approximate f at a nearby point. Know both steps.
⚠️ Taylor remainder questions typically ask: "How many terms are needed to approximate f(x) within a given error?" You solve M/(n+1)! · |x - b|^(n+1) < desired error for n.
⚠️ Center of mass problems require setting up the double integral correctly. Watch for non-uniform density functions, which change the integrand.
True or false: If D = 0 at a critical point, the point is a saddle.
Fill in the blank: The normal vector to the tangent plane of z = f(x, y) is ⟨fx, fy, ______⟩.
True or false: The Taylor remainder Rn(x) gives the exact error of the nth-degree approximation.
Fill in the blank: In center of mass, x̄ = ______/m.
True or false: To find absolute extrema on a closed region, you only need to check interior critical points.
Q: Find and classify the critical points of f(x, y) = x² + y² - 4x - 6y + 13.
A: fx = 2x - 4 = 0 gives x = 2. fy = 2y - 6 = 0 gives y = 3. One critical point: (2, 3). fxx = 2, fyy = 2, fxy = 0. D = (2)(2) - 0² = 4 > 0 and fxx > 0, so (2, 3) is a local minimum.
Q: Find the tangent plane to z = x² + y² at the point (1, 1, 2).
A: fx = 2x, fy = 2y. At (1, 1): fx = 2, fy = 2. Tangent plane: z = 2 + 2(x - 1) + 2(y - 1) = 2x + 2y - 2.
Q: Use the Taylor remainder to bound the error of the 3rd-degree Taylor polynomial of eˣ centred at 0, for x in [0, 1].
A: f⁽⁴⁾(x) = eˣ. On [0, 1], |f⁽⁴⁾(x)| ≤ e¹ ≈ 2.72. So |R₃(x)| ≤ e/(4!) · 1⁴ = e/24 ≈ 0.113.
Q: Find the center of mass of the region D = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} with density ρ(x, y) = 1 (uniform).
A: m = ∬ 1 dA = 1. My = ∬ x dA = ∫₀¹∫₀¹ x dy dx = 1/2. Mx = ∬ y dA = 1/2. So x̄ = 1/2, ȳ = 1/2. The center of mass is at (1/2, 1/2), as expected by symmetry.
Q: Classify the critical point of f(x, y) = x² - y² at (0, 0).
A: fx = 2x = 0 and fy = -2y = 0 at (0, 0). fxx = 2, fyy = -2, fxy = 0. D = (2)(-2) - 0 = -4 < 0. Saddle point.
The second derivative test connects to curvature: the discriminant D captures how the surface bends in two dimensions, much as curvature captures bending along a curve. Tangent planes are the foundation for the gradient and directional derivatives, which extend to gradient descent in optimisation. Taylor series reappear when deriving error bounds for numerical methods, and center of mass generalises to moments of inertia when you weight the integral by distance squared.
Critical points, second derivative test, discriminant, saddle point, local maximum, local minimum, absolute extrema, closed interval method, tangent plane, linearisation, linear approximation, normal vector to surface, gradient, Taylor series, Taylor polynomial, Maclaurin series, Taylor remainder, remainder bound, Taylor's inequality, center of mass, centroid, moment, double integral, density function, MATH 126, Calculus III, multivariable calculus, UW MATH 126, optimisation, extrema on a closed region