SYSTEMATIC MATHEMATICS

Multivariable Calculus

This course contains 52 visible chapters. A temperature can depend on both east–west position and north–south position; a cost can depend on several choices at once. How do change, steepness, best choice, area, and volume work when there is more than one input? Begin with points, vectors, distances, quadric surfaces, curves, and contour pictures. Then use distance to define approach, boundaries, and continuity—this precise idea is called norm-based limits—explain a derivative as the best local linear prediction, and only then name its matrix the Jacobian and its direction of steepest rise the gradient. Build inverse maps, implicit systems, optimization, multiple constraints, double and triple integrals, graph surface area, parameter sensitivity, improper integrals, coordinate changes, joint, marginal, and conditional density from concrete regions. This course deliberately stops before general line integrals, surface integrals, Green’s theorem, Stokes’ theorem, and the Divergence Theorem; those form the separate Vector Calculus course.

Before this course: Completed Single-Variable Calculus I & II and Linear Algebra Units 1–3; students must be able to use vectors, matrices, determinants, systems, limits, derivatives, integrals, parametric curves, and trigonometric identities.

COURSE FACTSLevel, chapters, units, prerequisite, and outcome
Chapter 1

Points, displacement vectors, norms, and distance

Objective: Why is Q−P a vector even though P and Q are points?

A point names a location while a vector names a displacement; their coordinates may look alike but transform and combine for different reasons.

Start from the definition: For v=(v₁,…,vₙ), the Euclidean norm is ||v||=√(v₁²+⋯+vₙ²), and the distance from P to Q is ||Q−P||. The result to establish is: The Euclidean norm is nonnegative, homogeneous, and satisfies the triangle inequality ||u+v||≤||u||+||v||.

The exact worked question is: Find the displacement, distance, and midpoint from P=(1,−2,3) to Q=(5,1,−1). Compare it with this failure boundary: Coordinates without a declared origin cannot distinguish an absolute point from a displacement vector, and componentwise absolute values do not define Euclidean length.

The Euclidean norm is nonnegative, homogeneous, and satisfies the triangle inequality ||u+v||≤||u||+||v||.

Square both sides and expand ||u+v||²=||u||²+2u·v+||v||².

Use Cauchy–Schwarz u·v≤||u||||v|| to bound the cross term.

The right side is at most (||u||+||v||)²; both sides are nonnegative, so take square roots.

Find the displacement, distance, and midpoint from P=(1,−2,3) to Q=(5,1,−1).

  1. Q−P=(4,3,−4).
  2. Distance is √(4²+3²+(−4)²)=√41.
  3. Midpoint is (P+Q)/2=(3,−1/2,1).

Result: The displacement is (4,3,−4), distance √41, and midpoint (3,−1/2,1).