Multivariable and vector calculus

A hiker on the slopes of Kilimanjaro stands at a point $(x, y)$ of a map; her altitude is a function $h(x, y)$. Which way is steepest? How steep is it? A cooperative in Bangladesh wants the largest yield from fields of fixed total perimeter. A hydrologist in the Netherlands wants the total water flowing out through the surface of a region of soil. A physicist wants to know whether a force field can be described by a potential energy.

Each question involves a function of several variables — and each needs a new idea. With one variable, the derivative is a number and the integral is over an interval. With several, the derivative becomes a linear map (a matrix), optimisation must handle constraints, integrals run over regions, curves and surfaces, and the fundamental theorem of calculus grows into a family of theorems — Green's, Stokes's and Gauss's — relating the inside of a region to its boundary.

This lesson assumes calculus of one variable and linear algebra (matrices, determinants, linear maps). We write $|x|$ for the Euclidean length of $x \in \R^n$ and $x \cdot y$ for the dot product.

A function $f : \R^2 \to \R$ has a graph, a surface $z = f(x, y)$ in space, and level curves $f(x, y) = c$, like the contour lines on a map. Freeze $y$ and you get a function of $x$ alone, whose slope is the partial derivative $\partial f/\partial x$; freezing $x$ gives $\partial f/\partial y$. But slices along two directions do not see the whole surface. The right notion is that, near a point, the surface looks like a plane: the derivative is the best linear approximation,

where $Df(a)$ is a linear map — for $f : \R^n \to \R^m$, an $m \times n$ matrix of partial derivatives.

For integrals, the picture is "chop into small pieces and add". The new difficulty is that the natural pieces (polar rectangles, small patches on a curved surface) are distorted copies of squares, and the distortion factor — a Jacobian determinant — must be included. Finally, the one-variable theorem $\int_a^b F' = F(b) - F(a)$ says the integral of a derivative over an interval is determined by the boundary points $a, b$. In higher dimensions the same is true for regions, with "derivative" meaning curl or divergence and the boundary a curve or a surface.

Let $U \subseteq \R^n$ be open (every point of $U$ has a ball around it inside $U$), $f : U \to \R^m$ and $a \in U$. Limits and continuity are defined as in one variable, with $|\cdot|$ the Euclidean length.

$f$ is of class $C^1$ on $U$ if all partial derivatives exist and are continuous on $U$; $C^2$ if the second partials $\partial^2 f/\partial x_i \partial x_j$ are too. The Hessian of a $C^2$ function $f : U \to \R$ is the symmetric matrix $H_f(a) = \big(\partial^2 f/\partial x_i\partial x_j(a)\big)$, and $a$ is a critical point if $\nabla f(a) = 0$.

Multiple integrals $\iint_R f\,dA$ over rectangles are defined by Riemann sums exactly as in one variable, and over a bounded region $D$ by integrating $f \cdot \mathbf 1_D$ over a rectangle containing $D$; a continuous function on a closed bounded region with a reasonable boundary (for example, a piecewise $C^1$ curve) is integrable.

For $C^2$ functions, the order of differentiation does not matter: $\dfrac{\partial^2 f}{\partial x\,\partial y} = \dfrac{\partial^2 f}{\partial y\,\partial x}$ (the theorem of Schwarz, or Clairaut; the hypothesis is needed — the function $xy(x^2 - y^2)/(x^2 + y^2)$ has unequal mixed partials at the origin). Hence the Hessian is symmetric, and second-order Taylor expansion gives the second derivative test: at a critical point of a $C^2$ function, a positive definite Hessian means a strict local minimum, a negative definite one a strict local maximum, and one with eigenvalues of both signs a saddle point.

To find the extreme values of $f$ on a constraint set $\{g = c\}$, we cannot simply set $\nabla f = 0$: the extremum is taken along the constraint. The idea, due to Lagrange (1788), is that at a constrained extremum the level curve of $f$ must be tangent to the constraint curve.

With several constraints $g_1 = c_1, \dots, g_k = c_k$ whose gradients at $a$ are linearly independent, the same argument gives $\nabla f(a) = \sum_i \lambda_i \nabla g_i(a)$.

The multiplier itself has a meaning: if $M(c)$ is the optimal value as a function of the constraint level $c$, then (under mild hypotheses) $M'(c) = \lambda$ — the "shadow price" of the constraint in economics.

The proof (via uniform continuity and Riemann sums) belongs to real analysis; the general version for Lebesgue integrals is a cornerstone of measure theory. Without integrability the conclusion can fail: for $f(x, y) = \frac{x^2 - y^2}{(x^2 + y^2)^2}$ on $(0, 1]^2$ the two iterated integrals are $\frac\pi4$ and $-\frac\pi4$.

Why the determinant? A linear map $A$ multiplies all volumes by $|\det A|$ (from linear algebra: true for elementary matrices, hence for products). Near a point $u$, $T$ is approximately the affine map $u + h \mapsto T(u) + J_T(u)h$, so a tiny box of volume $\Delta V$ at $u$ is carried to a region of volume about $|\det J_T(u)|\,\Delta V$. Summing over a fine partition gives the formula; making "about" precise is the real work of the proof, done in analysis texts. Points where $\det J_T = 0$ or $T$ fails to be injective may be ignored if they form a set of volume zero — this is what allows polar coordinates, whose Jacobian vanishes at $r = 0$.

Polar coordinates. $T(r, \theta) = (r\cos\theta, r\sin\theta)$ has

so $dx\,dy = r\,dr\,d\theta$. Below, the widget shows $J_T$ at $(r, \theta) = (2, \pi/6)$ acting on the plane: the unit square becomes a parallelogram of area $2 = r$. A tiny polar rectangle of sides $\Delta r$ and $\Delta\theta$ is nearly a rectangle with sides $\Delta r$ and $r\,\Delta\theta$.

This is why the normal density $\frac{1}{\sqrt{2\pi}}e^{-x^2/2}$ has total probability $1$.

A test for conservative fields. If $F = (P, Q) = \nabla\varphi$ with $\varphi$ of class $C^2$, then $\partial P/\partial y = \partial^2\varphi/\partial y\,\partial x = \partial^2\varphi/\partial x\,\partial y = \partial Q/\partial x$ by Schwarz's theorem: a conservative $C^1$ field has zero curl. The converse holds on a disc, or any star-shaped open set (Poincaré's lemma), but not in general:

The vortex shows why the hypothesis "$C^1$ on all of $D$" matters: for the unit disc the left side is $2\pi$, the right side would be $0$, and indeed $F$ is undefined at the origin. Green's theorem with $P = -y/2$, $Q = x/2$ gives a formula for area, $\operatorname{Area}(D) = \tfrac12 \oint_{\partial D} x\,dy - y\,dx$ — the principle of the planimeter, a mechanical instrument that measures areas by tracing boundaries. For the ellipse $(a\cos t, b\sin t)$ it gives $\tfrac12\int_0^{2\pi} ab\,dt = \pi ab$.

Applied to the field $(-Q, P)$ rotated by a right angle, Green's theorem becomes its flux form: the outward flux $\oint_{\partial D} F \cdot \mathbf n\,ds$ of $F = (P, Q)$ equals $\iint_D \operatorname{div} F\,dA$. In three dimensions both forms have analogues.

For a box $V = [a_1, b_1] \times [a_2, b_2] \times [a_3, b_3]$ the divergence theorem is proved exactly like Green's theorem: $\iiint_V \partial F_3/\partial z\,dV$ equals, by Fubini and the fundamental theorem, the integral of $F_3$ over the top face minus that over the bottom face — the flux of $(0, 0, F_3)$ — and similarly for the other components. The general versions (and Stokes's theorem, which for a graph $z = h(x, y)$ reduces to Green's theorem by the chain rule) are proved in texts such as Hubbard and Hubbard, or Marsden and Tromba, and most cleanly as the single theorem $\int_{\partial M} \omega = \int_M d\omega$ of differential forms (see the geometry viewpoint).

Mathlib's derivative for maps between normed spaces is the Fréchet derivative: HasFDerivAt f f' a with f' : E →L[ℝ] F a continuous linear map — precisely the lesson's definition. Euclidean space $\R^n$ is one instance, but the same theorems hold in infinite-dimensional spaces, which is why statements are phrased with E and F rather than matrices.

The chain rule is cited from Mathlib (its proof is the $o(|h|)$ argument above). From it we derive directional derivatives: differentiate along the line $t \mapsto a + tv$. With an inner product, the derivative is $v \mapsto \langle \nabla f, v \rangle$, and Cauchy–Schwarz gives the steepest-ascent property:

/-- **The chain rule**: `D(g ∘ f)(a) = Dg(f(a)) ∘ Df(a)`, a composite of linear maps (in
coordinates, a product of Jacobian matrices). This applies Mathlib's `HasFDerivAt.comp`. -/
theorem chain_rule {E F G : Type*} [NormedAddCommGroup E] [NormedSpace ℝ E]
    [NormedAddCommGroup F] [NormedSpace ℝ F] [NormedAddCommGroup G] [NormedSpace ℝ G]
    {f : E → F} {g : F → G} {f' : E →L[ℝ] F} {g' : F →L[ℝ] G} {a : E}
    (hf : HasFDerivAt f f' a) (hg : HasFDerivAt g g' (f a)) :
    HasFDerivAt (g ∘ f) (g'.comp f') a :=
  hg.comp a hf