Difficulty: Intermediate to Advanced | Prerequisites: Partial derivatives, gradient, second-order partial derivatives, vector calculus basics
This set of topics covers two related areas: measuring how sharply a space curve bends (curvature), and finding the highs, lows, and saddle points of functions of two variables. These are among the most heavily tested topics on the MATH 126 final. You need to be comfortable taking partial derivatives and applying the second derivative test before tackling the optimisation problems.
Curvature measures how fast a curve changes direction. For multivariable functions, critical points are found by setting both partial derivatives to zero, then classified using the second derivative test (D = fₓₓfᵧᵧ − (fₓᵧ)²). Tangent planes give linear approximations to surfaces. Absolute extrema on closed regions require checking both the interior critical points and the boundary.
Curvature (κ)
A measure of how sharply a curve bends at a given point. Formally, κ = |r'(t) × r''(t)| / |r'(t)|³. Think of it as the reciprocal of the radius of the best-fitting circle at that point: tight curves have high curvature, gentle curves have low curvature.
Critical point
A point (a, b) where both partial derivatives are zero: fₓ(a, b) = 0 and fᵧ(a, b) = 0. In simple terms, the surface is flat (horizontal tangent plane) at that point. A critical point can be a local max, local min, or saddle point.
Saddle point
A critical point that is neither a local maximum nor a local minimum. The surface curves up in one direction and down in another. At a saddle point, the discriminant D < 0.
Second derivative test (discriminant D)
For a critical point (a, b): D = fₓₓ(a, b) · fᵧᵧ(a, b) − [fₓᵧ(a, b)]². If D > 0 and fₓₓ > 0, it is a local min. If D > 0 and fₓₓ < 0, it is a local max. If D < 0, it is a saddle point. If D = 0, the test is inconclusive.
Tangent plane
The plane that best approximates a surface z = f(x, y) at a point (a, b, f(a, b)). Equation: z − f(a, b) = fₓ(a, b)(x − a) + fᵧ(a, b)(y − b). Think of it as the 3D version of a tangent line.
Linear approximation
Using the tangent plane to estimate f(x, y) near the point of tangency. Plug the nearby (x, y) values into the tangent plane equation to get an approximation of f.
Absolute extrema on a closed region
The guaranteed maximum and minimum values of a continuous function on a closed, bounded region. You must check interior critical points and all boundary curves, then compare function values.
Given r(t) = ⟨x(t), y(t), z(t)⟩, compute r'(t) and r''(t).
Find the cross product r'(t) × r''(t) and take its magnitude.
Divide by |r'(t)|³.
If you need curvature at a specific point, first find the value of t that corresponds to that point.
Worked example: r(t) = ⟨t³, t² − 1, 3t + 7⟩ at the point (−8, 3, 1). First find t: t³ = −8 gives t = −2. Check: (−2)² − 1 = 3 ✓, 3(−2) + 7 = 1 ✓. Then r'(t) = ⟨3t², 2t, 3⟩, so r'(−2) = ⟨12, −4, 3⟩. r''(t) = ⟨6t, 2, 0⟩, so r''(−2) = ⟨−12, 2, 0⟩. Cross product: r' × r'' = ⟨(−4)(0) − (3)(2), (3)(−12) − (12)(0), (12)(2) − (−4)(−12)⟩ = ⟨−6, −36, −24⟩. Magnitude: √(36 + 1296 + 576) = √1908 = 6√53. |r'(−2)| = √(144 + 16 + 9) = √169 = 13. Curvature κ = 6√53 / 13³ = 6√53 / 2197.
Compute fₓ and fᵧ, set both to zero, and solve for all critical points.
Compute fₓₓ, fᵧᵧ, and fₓᵧ.
Evaluate D = fₓₓfᵧᵧ − (fₓᵧ)² at each critical point.
Saddle points are those where D < 0.
Worked example: f(x, y) = x²y² − 4x² − y². Partial derivatives: fₓ = 2xy² − 8x = 2x(y² − 4), fᵧ = 2x²y − 2y = 2y(x² − 1). Setting fₓ = 0: x = 0 or y = ±2. Setting fᵧ = 0: y = 0 or x = ±1. Combining: if x = 0 then y = 0 gives (0, 0). If y = 2 then x = ±1 gives (1, 2), (−1, 2). If y = −2 then x = ±1 gives (1, −2), (−1, −2). Second derivatives: fₓₓ = 2y² − 8, fᵧᵧ = 2x² − 2, fₓᵧ = 4xy. At (0, 0): D = (−8)(−2) − 0 = 16 > 0, fₓₓ = −8 < 0, so local max. At (±1, ±2): D = (0)(0) − (±4)² = −16 < 0, so all four are saddle points. Saddle points: (1, 2), (−1, 2), (1, −2), (−1, −2).
Evaluate f(a, b), fₓ(a, b), and fᵧ(a, b) at the given point.
Write the tangent plane: z = f(a, b) + fₓ(a, b)(x − a) + fᵧ(a, b)(y − b).
To approximate f at a nearby point, substitute the nearby x and y values into the tangent plane equation.
Worked example: f(x, y) = x²y² − 4x² − y² at (2, 1). f(2, 1) = 4 − 16 − 1 = −13. fₓ = 2xy² − 8x, so fₓ(2, 1) = 4 − 16 = −12. fᵧ = 2x²y − 2y, so fᵧ(2, 1) = 8 − 2 = 6. Tangent plane: z = −13 − 12(x − 2) + 6(y − 1). To approximate f(2.1, 0.8): z ≈ −13 − 12(0.1) + 6(−0.2) = −13 − 1.2 − 1.2 = −15.4.
Find interior critical points by setting fₓ = 0 and fᵧ = 0. Keep only those inside the region.
Parametrise each boundary curve and find the extrema of f along each (reduce to single-variable calculus).
Also check all corner points where boundary curves meet.
Compare all function values found. The largest is the absolute max, the smallest is the absolute min.
Worked example: f(x, y) = x² − y² + 2y on the region bounded by x = 0, y = 4, y = x². Interior: fₓ = 2x = 0 and fᵧ = −2y + 2 = 0 gives (0, 1), which is inside the region. f(0, 1) = 0 − 1 + 2 = 1. Boundary x = 0 (from y = 0 to y = 4): f = −y² + 2y. Critical point: y = 1, f = 1. Endpoints: f(0, 0) = 0, f(0, 4) = −8. Boundary y = 4 (from x = 0 to x = 2): f = x² − 8. Ranges from −8 to −4. Boundary y = x² (from x = 0 to x = 2): f = x² − x⁴ + 2x². Derivative: 2x − 4x³ + 4x = 6x − 4x³ = 2x(3 − 2x²) = 0 gives x = 0 or x = √(3/2). At x = √(3/2): y = 3/2, f = 3/2 − 9/4 + 2(3/2) = 3/2 − 9/4 + 3 = 9/4 = 2.25. Corners: (0, 0) gives 0, (2, 4) gives 4 − 16 + 8 = −4. Absolute max = 2.25 = 9/4 at (√(3/2), 3/2). Absolute min = −8 at (0, 4).
Curvature:
κ = |r'(t) × r''(t)| / |r'(t)|³
Second derivative test discriminant:
D = fₓₓ · fᵧᵧ − (fₓᵧ)²
Tangent plane to z = f(x, y) at (a, b):
z = f(a, b) + fₓ(a, b)(x − a) + fᵧ(a, b)(y − b)
Linear approximation:
f(x, y) ≈ f(a, b) + fₓ(a, b)(x − a) + fᵧ(a, b)(y − b) for (x, y) near (a, b)
Curvature determines how tightly a road must bank at a turn, and how lenses focus light (higher curvature = stronger lens). Optimisation of multivariable functions is the backbone of machine learning (gradient descent), engineering design, and economics (maximising profit subject to constraints).
Students often confuse the second derivative test for single-variable functions (f'' > 0 means min) with the multivariable version. In multivariable calculus, you need the discriminant D, not just fₓₓ.
A critical point where D = 0 is not a saddle point. The test is inconclusive and you need other methods.
When finding absolute extrema, students frequently forget to check the boundary. The extreme values are often on the boundary rather than at interior critical points.
For curvature, students sometimes evaluate r'(t) and r''(t) at the given point (x, y, z) directly instead of first finding the correct value of t.
⚠️ The second derivative test (computing D) is almost guaranteed on the MATH 126 final. Have the formula memorised.
⚠️ For absolute extrema problems, the boundary analysis is typically worth more marks than the interior check. Budget your time accordingly.
⚠️ Curvature problems require careful arithmetic with cross products. Show every step to earn partial credit even if the final number is wrong.
True or false: If D > 0 at a critical point, it must be a local minimum. (False. It could be a local max if fₓₓ < 0.)
Fill in the blank: The curvature formula involves the cross product of ______ and ______. (r'(t) and r''(t))
True or false: The tangent plane always passes through the point of tangency. (True.)
Fill in the blank: To find absolute extrema on a closed region, check interior critical points and the ______. (boundary)
Q: Find the curvature of r(t) = ⟨t³, t² − 1, 3t + 7⟩ at the point (−8, 3, 1).
A: At t = −2: r'(−2) = ⟨12, −4, 3⟩, r''(−2) = ⟨−12, 2, 0⟩. Cross product = ⟨−6, −36, −24⟩, magnitude = 6√53. |r'| = 13. Curvature κ = 6√53 / 2197.
Q: Find all saddle points of f(x, y) = x²y² − 4x² − y².
A: Critical points: (0, 0), (1, 2), (−1, 2), (1, −2), (−1, −2). D at (0, 0) = 16 > 0 (local max). D at each of the other four points = −16 < 0. Saddle points: (1, 2), (−1, 2), (1, −2), (−1, −2).
Q: Find the tangent plane to z = x²y² − 4x² − y² at (2, 1) and approximate f(2.1, 0.8).
A: f(2, 1) = −13. fₓ(2, 1) = −12. fᵧ(2, 1) = 6. Tangent plane: z = −13 − 12(x − 2) + 6(y − 1). Approximation: f(2.1, 0.8) ≈ −13 − 12(0.1) + 6(−0.2) = −15.4.
Q: Find the absolute maximum and minimum of f(x, y) = x² − y² + 2y on the region enclosed by x = 0, y = 4, y = x².
A: Interior critical point (0, 1) gives f = 1. Boundary analysis yields: f(0, 4) = −8 (minimum), f(√(3/2), 3/2) = 9/4 (maximum). Absolute max = 9/4, absolute min = −8.
The tangent plane is the multivariable extension of the tangent line from single-variable calculus. The second derivative test generalises the concavity test from Calculus I. These optimisation techniques connect to Lagrange multipliers (which may appear in MATH 324) and to the double integrals in the next set of notes, where the region of integration is often defined by the same kinds of boundary curves.
curvature, kappa, cross product curvature formula, saddle point, second derivative test, discriminant D, tangent plane, linear approximation, absolute extrema, closed bounded region, critical points multivariable, MATH 126, Calculus III, UW Math 126, optimisation two variables