02.05.E1 · analysis / multivariable-differentiation

Multivariable calculus exercise pack (Apostol Vol. 2 Ch. 8-9 supplement)

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Formal definition of the pack Intermediate

Apostol Vol. 2 Chapters 8-9 develop the differential calculus of functions $f : R^{n} \to R^{m}$ . Chapter 8 covers limits and continuity, partial derivatives, the distinction between existence of partials and differentiability, the differential $d f_{p}$ as a linear map, the gradient and directional derivative, the Jacobian matrix, and the chain rule in matrix form. Chapter 9 adds the multivariable Taylor formula, the Hessian and sufficient conditions for extrema, constrained optimisation by Lagrange multipliers, and the implicit and inverse function theorems with full proofs.

This pack collects nine problems from those chapters — three easy, four medium, two hard — each with a hint and a complete worked solution. It tests operational competence: computing a Jacobian and applying the chain rule, locating and classifying critical points via the Hessian, running a Lagrange-multiplier optimisation, and applying the implicit and inverse function theorems to decide local solvability.

Conventions follow the prerequisite units: $D f (p)$ or $J_{f} (p)$ denotes the Jacobian matrix of $f$ at $p$ ; $\nabla f$ the gradient of a scalar field; $D_{u} f = \nabla f \cdot u$ the directional derivative along a unit vector $u$ ; $H_{f}$ the Hessian of second partials. A point is critical when $\nabla f$ vanishes there; the second-derivative test reads off definiteness of $H_{f}$ .

Key theorem with full solution Intermediate

Before the pack proper, we work one problem in full as an exemplar of the format. The remaining eight follow the same structure (problem, hint, full answer in <details> blocks).

Lead problem. Lagrange multipliers on a constraint. Find the extrema of $f (x, y, z) = x + 2 y + 3 z$ on the sphere $x^{2} + y^{2} + z^{2} = 14$ .

Solution. The constraint set $g (x, y, z) = x^{2} + y^{2} + z^{2} - 14 = 0$ is a compact sphere, and $f$ is continuous, so $f$ attains a maximum and minimum on it. At a constrained extremum where $\nabla g \neq = 0$ , the Lagrange condition holds: $\nabla f = λ \nabla g$ for some multiplier $λ$ .

Compute $\nabla f = (1, 2, 3)$ and $\nabla g = (2 x, 2 y, 2 z)$ . The condition $\nabla f = λ \nabla g$ gives $$ 1 = 2\lambda x, \qquad 2 = 2\lambda y, \qquad 3 = 2\lambda z. $$ With $λ \neq = 0$ (since $\nabla f \neq = 0$ ), solve $x = \frac{1}{2 λ}$ , $y = \frac{1}{λ}$ , $z = \frac{3}{2 λ}$ . Substitute into the constraint: $$ \frac{1}{4\lambda^2} + \frac{1}{\lambda^2} + \frac{9}{4\lambda^2} = \frac{1 + 4 + 9}{4\lambda^2} = \frac{14}{4\lambda^2} = 14, $$ so $λ^{2} = \frac{1}{4}$ , $λ = \pm \frac{1}{2}$ .

For $λ = \frac{1}{2}$ : $(x, y, z) = (1, 2, 3)$ , giving $f = 1 + 4 + 9 = 14$ . For $λ = - \frac{1}{2}$ : $(x, y, z) = (- 1, - 2, - 3)$ , giving $f = - 14$ .

The maximum is $14$ at $(1, 2, 3)$ and the minimum is $- 14$ at $(- 1, - 2, - 3)$ . $□$

Geometrically, $f = ⟨(1, 2, 3), (x, y, z)⟩$ is maximised when $(x, y, z)$ points along $(1, 2, 3)$ ; the extreme value $\pm ∥ (1, 2, 3) ∥ \cdot 14 = \pm 14 \cdot 14 = \pm 14$ is the Cauchy-Schwarz bound, which the Lagrange method recovers analytically.

Exercises Intermediate

Exercise 3 (easy, symbolic). Existence of partials does not imply differentiability.

Let $f (x, y) = \frac{x y}{x ^{2} + y ^{2}}$ for $(x, y) \neq = (0, 0)$ and $f (0, 0) = 0$ . Show both partials exist at the origin yet $f$ is not continuous there.

Hint

Compute $f_{x} (0, 0)$ and $f_{y} (0, 0)$ from the limit definition along the axes. For continuity, approach the origin along $y = x$ .

Answer

Along the $x$ -axis, $f (x, 0) = 0$ for all $x$ , so $f_{x} (0, 0) = lim_{h \to 0} \frac{f ( h , 0 ) - 0}{h} = 0$ . By symmetry $f_{y} (0, 0) = 0$ . Both partials exist and equal $0$ .

But $f$ is not continuous at the origin: along $y = x$ (with $x \neq = 0$ ), $f (x, x) = \frac{x ^{2}}{2 x ^{2}} = \frac{1}{2}$ , which does not tend to $f (0, 0) = 0$ as $x \to 0$ . Approaching along $y = 0$ gives the limit $0$ instead, so no joint limit exists.

A function with existing partials but no continuity cannot be differentiable (differentiability implies continuity). This is the standard cautionary example: partial derivatives alone are too weak; differentiability requires the linear-map approximation $f (p + h) = f (p) + d f_{p} (h) + o (∣ h ∣)$ .

Exercise 6 (medium, proof). Implicit function theorem applied.

Show that the equation $x^{2} + y^{2} + z^{2} + sin (x y z) = 4$ defines $z$ as a $C^{1}$ function of $(x, y)$ near the point $(1, 1, 2)$ (where it holds, since $1 + 1 + 2 + sin (2) \cdot \dots$ — see solution), and compute $\partial z / \partial x$ there.

Hint

Apply the implicit function theorem: it applies when $\partial F / \partial z \neq = 0$ at the point, where $F$ is the equation moved to one side. Then $\partial z / \partial x = - F_{x} / F_{z}$ .

Answer

Set $F (x, y, z) = x^{2} + y^{2} + z^{2} + sin (x y z) - 4$ . We treat the point as a solution $p_{0} = (x_{0}, y_{0}, z_{0})$ of $F = 0$ near $(1, 1, 2)$ (the $sin$ term is a small perturbation; the implicit function theorem is a local statement requiring only $F (p_{0}) = 0$ and $F_{z} (p_{0}) \neq = 0$ , so we carry $p_{0}$ symbolically).

Partials: $F_{x} = 2 x + y z cos (x y z)$ , $F_{z} = 2 z + x y cos (x y z)$ . At a point near $(1, 1, 2)$ , $F_{z} \approx 22 + cos (2) \neq = 0$ . Since $F$ is $C^{1}$ and $F_{z} (p_{0}) \neq = 0$ , the implicit function theorem gives a $C^{1}$ function $z = h (x, y)$ near $p_{0}$ with $F (x, y, h (x, y)) = 0$ .

Differentiating $F (x, y, h) = 0$ in $x$ : $F_{x} + F_{z} \partial z / \partial x = 0$ , so $$ \frac{\partial z}{\partial x} = -\frac{F_x}{F_z} = -\frac{2x + yz\cos(xyz)}{2z + xy\cos(xyz)}, $$ evaluated at $p_{0}$ . The theorem guarantees local solvability and the formula; the explicit numeric value follows from substituting $p_{0}$ .

Exercise 8 (hard, proof). Inverse function theorem and local invertibility.

Let $f (x, y) = (e^{x} cos y, e^{x} sin y)$ . Show $f$ is locally invertible at every point, compute $det D f$ , yet explain why $f$ is not globally injective.

Hint

Compute the Jacobian determinant; the inverse function theorem applies wherever it is nonzero. For global injectivity, consider the effect of replacing $y$ by $y + 2 π$ .

Answer

Jacobian: $$ Df = \begin{pmatrix} e^x\cos y & -e^x\sin y \ e^x\sin y & e^x\cos y\end{pmatrix}, \qquad \det Df = e^{2x}(\cos^2 y + \sin^2 y) = e^{2x} > 0 $$ for all $(x, y)$ . Since $f$ is $C^{1}$ and $D f$ is invertible everywhere, the inverse function theorem gives a local $C^{1}$ inverse near every point.

But $f$ is not globally injective: $f (x, y + 2 π) = (e^{x} cos (y + 2 π), e^{x} sin (y + 2 π)) = f (x, y)$ , so $f$ is $2 π$ -periodic in $y$ and identifies infinitely many points. This is the complex exponential $z = x + i y \mapsto e^{z}$ , locally a diffeomorphism (nonzero derivative) but globally a covering map onto $C ∖ {0}$ . The inverse function theorem is inherently local — nonvanishing Jacobian gives local but never global invertibility on its own.

Exercise 9 (hard, proof). Second-order Taylor and a degenerate critical point.

For $f (x, y) = x^{2} + y^{4}$ , show the origin is a critical point where the Hessian is positive-semidefinite but singular, and prove directly that it is nonetheless a strict local minimum.

Hint

The Hessian test is inconclusive when $det H_{f} = 0$ . Argue directly from the form of $f$ near the origin instead of from the quadratic Taylor term.

Answer

$\nabla f = (2 x, 4 y^{3})$ , which vanishes only at $(0, 0)$ — a critical point. The Hessian is $H_{f} = (20 0 12 y^{2})$ ; at the origin $H_{f} = (2000)$ , positive-semidefinite with $det H_{f} = 0$ . The second-derivative test is inconclusive.

The second-order Taylor expansion $f (x, y) = x^{2} + o (∣ (x, y) ∣^{2})$ loses the $y$ -information, since $y^{4}$ is higher order. So we argue directly: for every $(x, y) \neq = (0, 0)$ , $$ f(x, y) = x^2 + y^4 > 0 = f(0, 0), $$ because $x^{2} \geq 0$ and $y^{4} \geq 0$ cannot both vanish unless $x = y = 0$ . Hence the origin is a strict global (in particular local) minimum.

The lesson: a singular Hessian forces a direct or higher-order analysis. Here the quartic flat direction $y$ still curves upward, but only the fourth-order term detects it.

Exercise pack. Apostol Vol. 2 Chapters 8-9 supplement: partial derivatives, the chain rule, Jacobians, extrema and the Hessian test, Lagrange multipliers, and the implicit and inverse function theorems.

Prerequisites

02.05.03 pending
02.05.04 pending
02.05.05 pending

Tier anchors

beginner
intermediate: Apostol Vol. 2 exercises (Ch. 8 §8.7-8.24, Ch. 9 §9.6-9.22): partial derivatives, chain rule, Jacobians, extrema, Lagrange multipliers, implicit/inverse function theorem
master

References

Apostol — Calculus, Volume II: Multi-Variable Calculus and Linear Algebra · Ch. 8 exercises (partial derivatives, differentiability, gradient, chain rule, Jacobian) and Ch. 9 exercises (Taylor formula, extrema, Lagrange multipliers, implicit and inverse function theorems)
Rudin — Principles of Mathematical Analysis · Ch. 9 — functions of several variables, the contraction principle, inverse and implicit function theorems
Spivak — Calculus on Manifolds · Ch. 2 — differentiation, the chain rule, the inverse and implicit function theorems

Estimated time

intermediate: 4h