Learning goals and preparation

We assume scalar derivatives, trigonometric functions in radians, vector components and definite integration. You will distinguish a geometric curve from its traversal, derive first and second Cartesian slopes from a parameter, and relate vector velocity and acceleration to scalar speed.

6.1 A curve and a parameterization are different objects

A parametric curve is a function $r:I\to\mathbb R^n$, usually written $r(t)=(r_1(t),\ldots,r_n(t))^T$. Its image is the set of points visited; the parameterization additionally records order, timing and repeated visits. Eliminating $t$ may identify the image but can lose restrictions or direction.

Worked example 1: an ellipse traversed in time. For $r(t)=(a\cos t,b\sin t)^T$, $a,b>0$, eliminating $t$ gives $x^2/a^2+y^2/b^2=1$. On $[0,2\pi]$ it traverses the ellipse once counterclockwise, starting and ending at $(a,0)$. On $[0,\pi]$ it visits only the upper half. On $[0,4\pi]$ it visits the ellipse twice. The Cartesian equation alone does not distinguish these cases.

A reparameterization $t=\phi(s)$ gives $\tilde r(s)=r(\phi(s))$. If $\phi$ is a $C^1$ bijection $\phi:J\to I$ with $\phi'\ne0$, where $I,J$ are intervals and $r:I\to\mathbb R^n$ is a regular $C^1$ path, then $\tilde r$ traverses the same path regularly; the sign of $\phi'$ determines whether its orientation is preserved or reversed. A map that repeats parameter values can create repeated traversal and must not be assumed to preserve length counted once.

6.2 Vector limits and derivatives

A vector limit exists exactly when every component limit exists. This follows from bounding each component difference by the norm and bounding the norm by the sum of component magnitudes. Define

$$r'(t)=\lim_{h\to0}\frac{r(t+h)-r(t)}h.$$

Therefore $r'=(r_1',\ldots,r_n')^T$. Differentiate again componentwise for $r''$ where the derivatives exist. The parameter's domain still controls whether the derivative is two-sided or one-sided.

For differentiable vectors $u,v$ and scalar $f$,

$$(u+v)'=u'+v',\quad(fu)'=f'u+fu',\quad (u\cdot v)'=u'\cdot v+u\cdot v'.$$

The dot-product rule follows by writing the dot product as a finite sum and using scalar product rules. The chain rule is $d[r(\phi(s))]/ds=r'(\phi(s))\phi'(s)$. In three dimensions define $u\times v=(u_2v_3-u_3v_2,u_3v_1-u_1v_3,u_1v_2-u_2v_1)^T$. Differentiating each component gives $(u\times v)'=u'\times v+u\times v'$; order is retained because reversing a cross product changes its sign. No division by a vector is defined or needed.

Worked example 2: a three-coordinate path. For $r(t)=(t^2,e^t,\sin2t)^T$, derivative is $(2t,e^t,2\cos2t)^T$ and second derivative $(2,e^t,-4\sin2t)^T$. Each coordinate uses the usual scalar rule, including the inner factor two in the oscillatory coordinate.

If $r(t)\ne0$, differentiating $\|r\|^2=r\cdot r$ gives

$$\frac d{dt}\|r(t)\|=\frac{r(t)\cdot r'(t)}{\|r(t)\|}.$$

This is the rate of change of distance from the origin. It is generally not the speed $\|r'(t)\|$. The norm formula also requires a separate analysis at a zero vector.

6.3 Tangents and Cartesian derivatives

At a regular parameter value where $r'(t)\ne0$, the derivative is tangent to the path. Its unit direction is $T(t)=r'(t)/\|r'(t)\|$. For the local Cartesian formulas, let $x,y$ be $C^1$ on an open parameter interval about $t_0$, and suppose $x'(t_0)\ne0$. The first formula holds near $t_0$. For the second formula, assume additionally that $x,y$ are $C^2$ there:

$$\frac{dy}{dx}=\frac{y'(t)}{x'(t)},\qquad \frac{d^2y}{dx^2}=\frac1{x'(t)}\frac d{dt}\left(\frac{y'(t)}{x'(t)}\right).$$

The second formula is not $y''/x''$. It differentiates the first slope with respect to $x$, using the parameter chain rule again.

Why the local graph exists. Continuity of $x'$ lets us shrink the interval so that $x'$ has one fixed nonzero sign. The mean value theorem makes $x$ strictly monotone there. Continuity and the intermediate value theorem show that its image is an open interval: it contains every value between the images of two parameters, and every parameter has neighbors to its left and right in the open parameter interval. Its inverse is continuous: the images of any two parameters on opposite sides of a chosen parameter bound an open interval about its image, so sufficiently close image values have parameters between those two. Consequently, as a nearby Cartesian input tends to $x(t)$, its corresponding parameter is $t+h$ with $h\to0$. The graph difference quotient is

$$\frac{y(t+h)-y(t)}{x(t+h)-x(t)}=\frac{[y(t+h)-y(t)]/h}{[x(t+h)-x(t)]/h}\longrightarrow\frac{y'(t)}{x'(t)}.$$

If $x,y$ are $C^2$, the quotient $y'/x'$ is $C^1$ on that interval. Applying the first-derivative argument to this quotient yields $d/dx=(1/x')d/dt$ and therefore the second formula. A nonzero horizontal derivative at just one point, without the stated neighborhood regularity, is not used as an inverse theorem.

Worked example 3: a parametrized parabola. Let $x=t^2$, $y=t^4$ with $t>0$. Then $dy/dx=2t^2=2x$ and $d^2y/dx^2=2$. Although the eliminated curve is $y=x^2$, the parameter restriction visits only $x>0$. In the extension to the excluded parameter $t=0$, both first component derivatives vanish and the ratio formula cannot be used directly, even though the geometric parabola has a horizontal tangent there.

For the ellipse in Worked example 1, $dy/dx=-b\cos t/(a\sin t)$ when $\sin t\ne0$. At $t=0,\pi$, the horizontal component derivative is zero but the vertical derivative is nonzero, so the tangent is vertical. At an input where both vanish, no tangent conclusion follows from those two numbers alone; examine the curve or a higher-order expansion.

6.4 Velocity, speed and acceleration

For a position path, velocity is $v=r'$ and acceleration $a=r''$. Speed is $\|v\|\ge0$. At nonzero velocity,

$$\frac d{dt}\|v\|=\frac{v\cdot a}{\|v\|}.$$

A nonzero acceleration can change direction without changing speed.

Worked example 4: constant-speed circular motion. Let $r(t)=(R\cos\omega t,R\sin\omega t)^T$ with $R>0$, $\omega\ne0$. Then

$$v=R\omega(-\sin\omega t,\cos\omega t)^T,\qquad a=-R\omega^2(\cos\omega t,\sin\omega t)^T.$$

Speed is $R|\omega|$, constant, but acceleration has magnitude $R\omega^2$ and points toward the centre. The dot product $v\cdot a=0$ confirms zero speed derivative. Scalar speed alone does not determine the vector motion.

Worked example 5: changing parameter speed. Reparameterize a unit circle by $r(t)=(\cos(t^2),\sin(t^2))^T$, $t\ge0$. Its image is still a circle, but speed is $2t$ and the path repeatedly winds around it as time grows. The starting velocity is zero. A geometric shape does not prescribe a unique speed law.

6.5 Length and accumulated vector change

Why the real numbers need more than arithmetic

The rational numbers support the familiar ordered arithmetic, yet the set $\{q\in\mathbb Q:q>0,\ q^2<2\}$ has no rational least upper bound. The real number system adds a completeness property: bounded nonempty sets have least upper bounds. This is the source of many existence results later in analysis, not simply another rule for manipulating inequalities.

Definition

Upper bound and supremum

For a nonempty $A\subset\mathbb R$, $u$ is an upper bound if $a\le u$ for every $a\in A$. A number $s$ is $\sup A$ if it is an upper bound and $s\le u$ for every upper bound $u$. A maximum is an element of $A$ that is also an upper bound. The maximum, when it exists, is the supremum; a supremum need not belong to $A$. Lower bounds, infimum and minimum are defined by reversing inequalities.

Theorem

Completeness axiom

Every nonempty subset of $\mathbb R$ that is bounded above has a real supremum. Equivalently, every nonempty subset bounded below has a real infimum, via $\inf A=-\sup(-A)$.

For example, $\sup(0,1)=1$ although $1$ is not a maximum of that set. Conversely, $\sup[0,1]=1$ and it is a maximum. The empty set and an unbounded-above set do not have a real supremum under this definition.

Theorem

Supremum approximation criterion

An upper bound $s$ of a nonempty set $A$ is $\sup A$ if and only if, for every $\varepsilon>0$, there exists $a\in A$ such that $s-\varepsilon<a\le s$.

Proof

Proving both directions

If $s=\sup A$ and no element exceeds $s-\varepsilon$, then $s-\varepsilon$ is a smaller upper bound, contradiction. Conversely, suppose $s$ is an upper bound and the approximation property holds. If another upper bound $u$ were smaller than $s$, choose $\varepsilon=s-u>0$. The property gives $a>u$, contradicting that $u$ is an upper bound. Thus $s$ is the least upper bound.

Worked example

Use both halves of a supremum proof

Let $A=\{n/(n+1):n\in\mathbb N, n\ge1\}$. Every element is less than $1$, so $1$ is an upper bound. Given $\varepsilon>0$, choose $n>1/\varepsilon$ by the Archimedean property below. Then $1-n/(n+1)=1/(n+1)<\varepsilon$. Thus $\sup A=1$. No element equals $1$, so $A$ has no maximum. An upper-bound calculation alone would not prove leastness: $2$ is also an upper bound.

Archimedean property and density

Theorem

Archimedean property

For every real $x$ there exists a positive integer $n>x$. Consequently, for every $\varepsilon>0$ there is $n\in\mathbb N$ such that $1/n<\varepsilon$.

Proof

Completeness forces integers to be unbounded

If the positive integers were bounded above, completeness would give $s=\sup\mathbb N$. Since $s-1$ is not an upper bound, some $n\in\mathbb N$ satisfies $n>s-1$. But $n+1>s$ and $n+1\in\mathbb N$, contradicting that $s$ bounds the set. Apply this to $1/\varepsilon$ to obtain the second statement.

Between any real numbers $x<y$ there is a rational number. Indeed choose $n$ with $n(y-x)>1$; the Archimedean property and the integer-part property yield an integer $m$ with $nx<m<ny$, so $x<m/n<y$. The integer-part property follows from the unboundedness of integers in both directions and the well-ordering of the natural numbers. An irrational also lies between $x$ and $y$: choose a rational $q$ with $x-\sqrt2<q<y-\sqrt2$; then $q+\sqrt2$ is irrational and lies in the interval. Thus density does not say rationals and irrationals are equal sets.

Worked example

An infimum via reflection

For $B=\{2+1/n:n\ge1\}$, every element exceeds $2$. Given $\varepsilon>0$, choose $n>1/\varepsilon$ to obtain $2<2+1/n<2+\varepsilon$. Therefore $2$ is the infimum, but not a minimum. The maximum is $3$, attained at $n=1$.

Quantifiers in convergence

Definition

Convergence of a real sequence

A sequence $(x_n)$ converges to $L\in\mathbb R$ if for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that for every $n\ge N$, $|x_n-L|<\varepsilon$. Write $x_n\to L$. The index $N$ may depend on $\varepsilon$ but must work for all later indices.

To negate convergence to $L$, there must be some $\varepsilon_0>0$ such that for every $N$ there is an $n\ge N$ with $|x_n-L|\ge\varepsilon_0$. A few early terms do not affect convergence; a recurring failure arbitrarily late does.

Worked example

Construct an epsilon–N proof

For $x_n=(3n+1)/(n+2)$, guess $L=3$. Since

$$\left|\frac{3n+1}{n+2}-3\right|=\frac5{n+2}<\frac5n,$$

choose $N>5/\varepsilon$. Then every $n\ge N$ gives $|x_n-3|<\varepsilon$. The guess is motivation; the inequality and quantified choice establish the proof.

Theorem

Uniqueness, boundedness and limit laws

A convergent real sequence has a unique limit and is bounded. If $x_n\to x$ and $y_n\to y$, then $x_n+y_n\to x+y$, $x_ny_n\to xy$, and $x_n/y_n\to x/y$ when $y\ne0$ and the quotient is defined eventually. If $x_n\le z_n\le y_n$ eventually and both outer sequences converge to $L$, then $z_n\to L$.

Proof

Uniqueness and boundedness

If $x_n\to L$ and $x_n\to M$, then for any $\varepsilon>0$, sufficiently late terms satisfy both $|x_n-L|<\varepsilon/2$ and $|x_n-M|<\varepsilon/2$. The triangle inequality yields $|L-M|<\varepsilon$. Since this holds for every positive $\varepsilon$, $L=M$. For boundedness choose $N$ so that $|x_n-L|<1$ for $n\ge N$. The finite initial terms and the bound $|x_n|\le|L|+1$ for the tail together give a finite bound for the whole sequence.

The converse of boundedness is false: $(-1)^n$ is bounded but does not converge. If it converged to $L$, its even subsequence would converge to $1$ and its odd subsequence to $-1$, violating uniqueness.

Proof

Product law mechanism

A convergent sequence $(y_n)$ is bounded, say $|y_n|\le M$. Write

$$|x_ny_n-xy|\le|y_n||x_n-x|+|x||y_n-y|.$$

Both terms tend to zero. For the quotient, $y_n\to y\ne0$ implies $|y_n|\ge|y|/2$ eventually; this keeps the denominator away from zero and gives $1/y_n\to1/y$.

Subsequences

Definition

Subsequence

A subsequence of $(x_n)$ is $(x_{n_k})$ where $n_1<n_2<\cdots$ are positive integers. If $x_n\to L$, then every subsequence converges to $L$ because $n_k\ge k$ and therefore eventually lies beyond any prescribed index.

Theorem

Bolzano–Weierstrass

Every bounded real sequence has a convergent subsequence.

Proof

Nested-interval proof

Place every term in a closed bounded interval $I_1$. Bisect it. At least one half contains infinitely many terms; choose that half as $I_2$. Continue, obtaining nested closed intervals $I_k$ of lengths tending to zero, each containing infinitely many terms. Choose indices $n_k$ strictly increasing with $x_{n_k}\in I_k$; this is possible because each interval contains infinitely many terms. Completeness gives a common point $L$ in all intervals: the supremum of their left endpoints lies in each interval because every left endpoint is at most every right endpoint. Since $x_{n_k},L\in I_k$, their distance is at most the interval length and tends to zero.

The conclusion is about a subsequence, not necessarily the whole sequence. For example, $(-1)^n$ has constant subsequences converging to $1$ and $-1$, but does not converge.

Real closed sets: the sequence criterion and the open complement

Work on the real line in the usual classical set-theoretic setting, with countable witness selections available. A set $U$ is open when every $y$ in $U$ has some $r>0$ with $(y-r,y+r)$ contained in $U$. We use the Archimedean consequence that $1/n$ tends to zero. For every $K$ subset of the real line, the following conditions are equivalent:

1. Every sequence of points of $K$ that converges in the real line has its limit in $K$.
2. The complement of $K$ is open.

Proof of 2 implies 1. Suppose $x_n$ belongs to $K$ and $x_n$ tends to $x$. If $x$ were outside $K$, openness would give $r>0$ with $(x-r,x+r)$ disjoint from $K$. Convergence gives a threshold after which $|x_n-x|<r$, contradicting $x_n$ in $K$. Therefore $x$ belongs to $K$.

Proof of 1 implies 2. Fix $y$ outside $K$. If no positive-radius interval around $y$ avoided $K$, then for every positive integer $n$ there would be a point $x_n$ in $K$ with $|x_n-y|<1/n$. Choose these witnesses as a sequence. Given $\varepsilon>0$, choose $N$ with $1/N<\varepsilon$. For $n\ge N$, $|x_n-y|<1/n\le1/N<\varepsilon$, so $x_n$ tends to $y$. Condition 1 would force $y$ into $K$, a contradiction. Hence some interval around $y$ avoids $K$. This holds for every point of the complement, proving it open. The witness-sequence step uses the stated classical setting; this proof is not offered as a theorem about choice-free foundations. If $K$ is empty there is no sequence in $K$, and its complement is the whole open line. If $K$ is the whole line, its empty complement is open vacuously. Both edge cases also satisfy the equivalence.

Learning goals and definitions

A set $K\subseteq\mathbb R$ is compact if every family of open sets whose union contains $K$ has a finite subfamily still covering $K$. A set is closed if it contains all limits of convergent sequences of its points, equivalently if its complement is open. Sequential compactness means every sequence in $K$ has a subsequence converging to a member of $K$.

The empty set is compact, closed and bounded, and sequentially compact vacuously. In proofs using a selected point or an infinite sequence, treat that case separately. We may assume the real completeness property and its nested-interval and Bolzano–Weierstrass consequences from the earlier sequence units.

Theorem: Heine–Borel in the real line

For $K\subseteq\mathbb R$, the following are equivalent: compactness by open covers; closedness and boundedness; sequential compactness.

Compact implies closed and bounded. The intervals $(-n,n)$, $n\ge1$, cover $\mathbb R$, hence $K$. A finite subcover gives a bound. To show closedness, take $y\notin K$. For each $x\in K$, the interval $B(x,|x-y|/2)$ avoids $y$. A finite subfamily covers $K$. Let $\delta$ be the minimum of their positive half-distances to $y$. Every point in the chosen interval centered at $x$ has distance from $y$ strictly greater than $|x-y|/2\ge\delta$. Thus $B(y,\delta)$ misses $K$, showing its complement open. If $K$ is empty this is immediate.

Closed and bounded implies compact. Enclose nonempty $K$ in a closed bounded interval $I_0$. Suppose an open cover of $K$ has no finite subcover. Bisect $I_0$. At least one closed half meets $K$ in a set not finitely coverable; otherwise two finite collections would cover $K$. Choose such a half and repeat. We obtain nested closed intervals $I_n$ with lengths tending to zero, and $K\cap I_n$ not finitely coverable. Each intersection is nonempty; choose $x_n\in K\cap I_n$. The nested-interval property gives a unique common point $p$, and $x_n\to p$ because both lie in $I_n$. Closedness gives $p\in K$. Some member $O$ of the cover contains $p$, and openness gives a ball around $p$ inside $O$. For large $n$, the whole interval $I_n$ lies in that ball, so $O$ alone covers $K\cap I_n$, a contradiction.

Closed and bounded implies sequentially compact. Every sequence in $K$ is bounded, so Bolzano–Weierstrass supplies a convergent subsequence. Closedness retains its limit in $K$.

Sequentially compact implies closed and bounded. If unbounded, choose $x_n\in K$ with $|x_n|>n$; no subsequence can converge to a finite real number, a contradiction. If $p$ is a limit of points in $K$, choose $x_n\in K$ with $|x_n-p|<1/n$. A subsequence converges to a point of $K$ by sequential compactness, but it also converges to $p$; uniqueness of real limits gives $p\in K$. Thus $K$ is closed. This completes all implications.

Compact subsets of the real line

A set $K\subset\mathbb R$ is compact if every open cover has a finite subcover. In $\mathbb R$, Heine–Borel says this is equivalent to being closed and bounded; it is also equivalent to every sequence in $K$ having a subsequence converging to a point of $K$. The closedness requirement keeps the subsequential limit in the set; boundedness permits extraction. We use the sequential form in the proofs below.

Theorem

Extreme value theorem

If $K\subset\mathbb R$ is nonempty and compact and $f:K\to\mathbb R$ is continuous, then $f$ is bounded and attains both a maximum and a minimum on $K$.

Proof

Extract a maximizing subsequence

If $f$ were unbounded above, choose $x_n\in K$ with $f(x_n)>n$. A subsequence $x_{n_k}\to x\in K$ exists; continuity would give $f(x_{n_k})\to f(x)$, contradicting divergence to infinity. The same reasoning applied to $-f$ gives boundedness below. Let $s=\sup f(K)$. Choose $x_n\in K$ with $s-1/n<f(x_n)\le s$. A convergent subsequence tends to some $x\in K$, and continuity implies $f(x)=s$. Apply the argument to $-f$ for a minimum.

Worked example

Which hypothesis fails?

The continuous function $f(x)=x$ on $(0,1)$ is bounded but has no maximum or minimum: $0$ and $1$ are not in the domain. The set is bounded but not closed, hence not compact. On $[0,1]$, both endpoints are present and the extrema are attained. Compactness is a sufficient global hypothesis, not a condition required for every individual function to attain extrema.

Fermat's necessary condition. At a differentiable interior local maximum or minimum, $f'(a)=0$.

Proof. At a local maximum, sufficiently small positive increments give nonpositive difference quotients and negative increments give nonnegative ones. A two-sided derivative can satisfy both sign limits only by being zero. Negating the function treats a minimum. $\square$

The interior hypothesis matters: $f(x)=x$ on $[0,1]$ has extrema at endpoints despite derivative 1. The converse fails: $x^3$ has zero derivative at zero but no extremum there.

Rolle's theorem. If $a<b$, $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then $f'(c)=0$ for some $c\in(a,b)$.

Proof. The extreme-value theorem supplies maximum and minimum. If the function is constant, every interior point works. Otherwise a value differs from the equal endpoint values, so either a maximum or a minimum occurs in the interior. Fermat's condition applies there. $\square$

Mean value theorem. Under the same continuity and differentiability hypotheses but without equal endpoint values, some $c\in(a,b)$ satisfies $f(b)-f(a)=f'(c)(b-a)$.

Subtract the secant line $f(a)+[f(b)-f(a)](x-a)/(b-a)$ from $f$ and apply Rolle. This proves the theorem without assuming $f'$ is continuous. Consequences include: derivative zero on an interval makes the function constant; nonnegative derivative makes it nondecreasing; and $|f'|\le M$ on an interval gives $|f(y)-f(x)|\le M|y-x|$ for points in it. An interval-wide derivative bound cannot be inferred from one derivative value.

Uniform continuity

Definition

One delta for the entire domain

A function $f:A\to\mathbb R$ is uniformly continuous on $A$ if for every $\varepsilon>0$ there exists $\delta>0$ such that for all $x,y\in A$, $|x-y|<\delta$ implies $|f(x)-f(y)|<\varepsilon$. The chosen $\delta$ can depend on $\varepsilon$ but not on $x$ or $y$.

Theorem

Heine–Cantor theorem

Every continuous real-valued function on a compact set is uniformly continuous.

Proof

Contradiction using compactness

If uniform continuity failed, some $\varepsilon_0>0$ would permit, for every $n$, points $x_n,y_n\in K$ with $|x_n-y_n|<1/n$ but $|f(x_n)-f(y_n)|\ge\varepsilon_0$. Compactness supplies a subsequence $x_{n_k}\to x\in K$. Since the paired distances tend to zero, also $y_{n_k}\to x$. Continuity gives both $f(x_{n_k})\to f(x)$ and $f(y_{n_k})\to f(x)$, contradicting the output separation.

Worked example

Pointwise but not uniform on an unbounded domain

The function $f(x)=x^2$ is continuous on $\mathbb R$ but not uniformly continuous there. Take $x_n=n$ and $y_n=n+1/n$. Then $|x_n-y_n|=1/n\to0$, but $|f(y_n)-f(x_n)|=2+1/n^2\not\to0$. On any closed bounded interval, Heine–Cantor does give uniform continuity. It is also direct: if $|x|,|y|\le M$, then $|x^2-y^2|\le2M|x-y|$.

Learning goals and definitions

Develop the proof behind integrability statements and distinguish boundedness from integrability. Let $a<b$ and $f$ be bounded on $[a,b]$. For a partition $P:a=x_0<\cdots<x_n=b$, put $m_i=\inf_{[x_{i-1},x_i]}f$, $M_i=\sup_{[x_{i-1},x_i]}f$, and define $L(f,P)=\sum m_i\Delta x_i$, $U(f,P)=\sum M_i\Delta x_i$. The lower integral $\underline I$ is the supremum of all lower sums; the upper integral $\overline I$ is the infimum of all upper sums. Refining a partition makes lower sums nondecreasing and upper sums nonincreasing.

For any two partitions $P,Q$, their common refinement gives $L(f,P)\le U(f,Q)$, so $\underline I\le\overline I$. Boundedness ensures these quantities are finite. The function is Riemann integrable when they agree.

Theorem: Darboux criterion

The bounded function is integrable if and only if for every $\varepsilon>0$ there is a partition with $U(f,P)-L(f,P)<\varepsilon$.

Proof. If the two integrals equal $I$, choose a lower sum above $I-\varepsilon/2$ and an upper sum below $I+\varepsilon/2$, using supremum and infimum approximation. Their common refinement has gap less than $\varepsilon$. Conversely the inequalities $0\le\overline I-\underline I\le U(f,P)-L(f,P)$ force the integral gap to be zero when such partitions exist for every positive $\varepsilon$.

Theorem: finitely many discontinuities are permitted

If $f$ is bounded and continuous except possibly at finitely many points, it is integrable.

Proof. Let $|f|\le M$, taking $M>0$. Surround the exceptional points by finitely many small intervals, with endpoints included in the partition, having total length less than $\varepsilon/(4M)$. At endpoints of $[a,b]$ use one-sided intervals. Their contribution to the upper-lower gap is at most $2M$ times their total length, hence less than $\varepsilon/2$. On the remaining finitely many closed intervals, $f$ is continuous and therefore uniformly continuous. Subdivide each sufficiently finely that its oscillation on every resulting subinterval is less than $\varepsilon/[2(b-a)]$. Their total gap is less than $\varepsilon/2$. Combining all partition points gives gap below $\varepsilon$. If an exceptional interval endpoint would itself be exceptional, move it slightly so all actual exceptional points lie inside the selected intervals or at the domain endpoints. This keeps the remaining closed intervals continuous.

Continuous functions and all tagged meshes

Let $f:[a,b]\to\mathbb R$ be continuous, where $a<b$. The extreme value theorem makes $f$ bounded, and Heine–Cantor makes it uniformly continuous. For a partition $P$, write $L(f,P)$ and $U(f,P)$ for its lower and upper Darboux sums. The proved Darboux integrability theorem for continuous functions supplies the common integral $I$.

Fix $\varepsilon>0$. Uniform continuity supplies $\delta>0$ such that $|f(x)-f(y)|<\varepsilon/[2(b-a)]$ whenever $|x-y|<\delta$. If the mesh of $P$ is less than $\delta$, every subinterval has oscillation at most $\varepsilon/[2(b-a)]$. Therefore

$$U(f,P)-L(f,P)\le\frac{\varepsilon}{2}<\varepsilon.$$

For every selection of tags, its tagged sum $S(P)$ lies between $L(f,P)$ and $U(f,P)$; so does $I$. Thus $|S(P)-I|\le U(f,P)-L(f,P)<\varepsilon$. The same $\delta$ works for every partition of sufficiently small mesh and every choice of tags. This proves the tagged-mesh definition for continuous functions on the stated closed interval.

For $a<b$, take a partition $a=x_0<x_1<\cdots<x_n=b$, with $\Delta x_i=x_i-x_{i-1}$, tags $\xi_i\in[x_{i-1},x_i]$, and mesh $\|P\|=\max_i\Delta x_i$. A finite number $I$ is the definite integral $\int_a^bf(x)dx$ if, for every $\varepsilon>0$, there is $\delta>0$ such that every partition with $\|P\|<\delta$ and every choice of tags satisfy $|\sum_i f(\xi_i)\Delta x_i-I|<\varepsilon$. In that case $I$ is the common limit. Continuous functions on a closed interval are integrable.

Let $f$ be continuous on $[a,b]$ and $A(x)=\int_a^xf(t)dt$. For $x\in(a,b)$ and nonzero $h$ with $x+h\in[a,b]$, the integral estimate gives

$$\left|\frac{A(x+h)-A(x)}h-f(x)\right|\le\sup_{t\in[\min(x,x+h),\max(x,x+h)]}|f(t)-f(x)|\longrightarrow0\quad(h\to0).$$

Thus $A'(x)=f(x)$ in the interior; at the endpoints the same argument gives one-sided derivatives. If $F$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $F'=f$ there, the mean value theorem makes $F-A$ constant on $[a,b]$. Therefore $\int_a^bf=F(b)-F(a)$.

Componentwise integration and uniform continuity

Let $a<b$ and let $r=(r_1,\ldots,r_n):[a,b]\to\mathbb R^n$ be $C^1$, with $n\ge1$. Define the integral of a continuous vector function componentwise. For $a\le u<v\le b$, the scalar fundamental theorem applied to each coordinate gives

$$\int_u^v r'(t)\,dt=\bigl(r_1(v)-r_1(u),\ldots,r_n(v)-r_n(u)\bigr)=r(v)-r(u).$$

Each $r_i'$ is continuous on the compact interval and hence uniformly continuous. Given $\eta>0$, choose $\delta_i>0$ such that $|r_i'(t)-r_i'(s)|<\eta/\sqrt n$ whenever $t,s\in[a,b]$ and $|t-s|<\delta_i$. With $\delta=\min_i\delta_i$, every such pair with $|t-s|<\delta$ satisfies

$$\|r'(t)-r'(s)\|^2=\sum_{i=1}^n|r_i'(t)-r_i'(s)|^2<\eta^2.$$

Thus $r'$ is uniformly continuous in the Euclidean norm. These two facts apply even when $r'$ vanishes.

1. Define length before replacing it by an integral

For $r:[a,b]\to\mathbb R^n$, define its length as the supremum over finite partitions $a=t_0<\cdots<t_m=b$ of the polygonal sums $\sum_i\|r(t_i)-r(t_{i-1})\|$. The supremum allows arbitrarily fine polygonal approximations; one endpoint chord is generally only a lower bound.

Theorem: $C^1$ length formula

For a $C^1$ curve on a compact interval,

$$L(r)=\int_a^b\|r'(t)\|\,dt.$$

Upper bound. On a subinterval, the vector fundamental theorem gives displacement $d=\int r'(t)dt$. If $d\ne0$, dot with its unit direction to obtain $\|d\|\le\int\|r'(t)\|dt$; if $d=0$, that inequality is immediate. Summing over any partition gives polygonal length at most the speed integral.

Lower bound. Uniform continuity of $r'$ says that for any $\eta>0$, a sufficiently fine partition makes $\|r'(t)-r'(\xi_i)\|<\eta$ on every subinterval, where $\xi_i$ is any chosen tag there. Thus its displacement differs from $r'(\xi_i)\Delta t_i$ by a vector of norm at most $\eta\Delta t_i$. The reverse triangle inequality gives chord length at least $\|r'(\xi_i)\|\Delta t_i-\eta\Delta t_i$. The speed itself varies by at most $\eta$ there, so its integral differs from that tagged value by at most $\eta\Delta t_i$. Summing yields polygonal length at least the speed integral minus $2\eta(b-a)$. Since arbitrarily small $\eta$ are possible, the supremum reaches the integral. This proves both directions rather than only the familiar chord upper bound.

Regularity is not needed for this formula: a $C^1$ curve may stop. Regularity becomes essential when we invert arc length to use it as a coordinate.

Learning goals and setting

Prove local invertibility rather than only differentiating a formally assumed inverse. Use the Euclidean norm and its induced operator norm throughout. Let $U\subseteq\mathbb R^n$ be open, $F:U\to\mathbb R^n$ be $C^1$, $a\in U$, $b=F(a)$, and $A=DF(a)$ invertible. We will prove that some open neighborhoods $W$ of $a$ and $V$ of $b$ make $F:W\to V$ a $C^1$ bijection with derivative of its inverse equal to $[DF]^{-1}$.

Lemma: contraction on a closed ball

If a map $T$ takes a nonempty closed Euclidean ball $B$ into itself and $\|T(x)-T(y)\|\le q\|x-y\|$ for a fixed $0\le q<1$, then it has a unique fixed point in $B$.

Proof. Starting from any $x_0\in B$, define $x_{k+1}=T(x_k)$. If $q=0$, the inequality makes $T$ constant, and its value is the unique fixed point reached at the first iteration. For $0<q<1$, repeated contraction gives $\|x_{k+1}-x_k\|\le q^k\|x_1-x_0\|$. For $m>k$, summing consecutive differences bounds $\|x_m-x_k\|$ by $q^k\|x_1-x_0\|/(1-q)$, so the sequence is Cauchy. Coordinatewise real completeness gives a limit in $\mathbb R^n$, and closedness keeps it in $B$. Contraction implies continuity, so passing to the limit in $x_{k+1}=T(x_k)$ makes the limit fixed. Two fixed points would have distance at most $q$ times that same distance, forcing equality. This proof also gives the stated geometric error bound to the limit.

Existence and uniqueness of the local inverse

Continuity of $DF$ lets us choose $r>0$ with the closed ball $B=\overline B(a,r)\subset U$ and

$$\|I-A^{-1}DF(y)\|\le q=\frac12\quad\text{for all }y\in B.$$

For a target $x$ near $b$, set $T_x(y)=y-A^{-1}(F(y)-x)$. Its derivative has norm at most $q$ on the convex ball. Applying the one-variable fundamental theorem along a segment gives $\|T_x(y)-T_x(z)\|\le q\|y-z\|$. Also

$$\|T_x(y)-a\|\le q\|y-a\|+\|A^{-1}(x-b)\|.$$

Choose $V=B(b,\rho)$ with $\rho=(1-q)r/(2\|A^{-1}\|)$. For $x\in V$, the second term is less than $(1-q)r/2$, so $T_x$ maps $B$ into itself. The lemma supplies a unique $y=H(x)\in B$ satisfying $F(y)=x$. The fixed-point estimate at $a$ gives

$$\|H(x)-a\|\le\frac{\|A^{-1}(x-b)\|}{1-q}<r/2,$$

so all these roots are interior. Let $W=B(a,r)\cap F^{-1}(V)$. This is open, contains $a$, and $F:W\to V$ is bijective, because uniqueness holds even on the larger closed ball.

Continuity and differentiability of the inverse

For $x,z\in V$, compare the fixed-point equations:

$$\|H(x)-H(z)\|\le q\|H(x)-H(z)\|+\|A^{-1}\|\|x-z\|.$$

Thus $H$ is Lipschitz with constant $\|A^{-1}\|/(1-q)$. For $y\in B$, the matrix $A^{-1}DF(y)=I-E$ has $\|E\|\le q<1$. The convergent matrix series $\sum_{k=0}^{\infty}E^k$ is its inverse: multiplying finite partial sums gives $I-E^{N+1}$, which tends to $I$. Hence $DF(y)$ is invertible throughout $B$.

Fix $x\in V$, let $y=H(x)$, and take a small increment $k$ with $x+k\in V$. Set $h=H(x+k)-H(x)$. Differentiability of $F$ gives $k=DF(y)h+o(\|h\|)$. The Lipschitz estimate gives $\|h\|=O(\|k\|)$, so solving yields

$$h=[DF(y)]^{-1}k+o(\|k\|).$$

Therefore $DH(x)=[DF(H(x))]^{-1}$. The inverse matrices vary continuously: the adjugate-over-determinant formula is continuous where determinant is nonzero. Together with continuity of $DF$ and $H$, this proves $H$ is $C^1$. The theorem is now proved, including existence, uniqueness, neighborhoods and regularity.

Implicit functions as a corollary

Let $G:\mathbb R^{p+m}\supset U\to\mathbb R^m$ be $C^1$, $G(a,c)=0$, and $D_yG(a,c)$ invertible. Define $\Phi(x,y)=(x,G(x,y))$. Its block derivative is

$$D\Phi=\begin{pmatrix}I&0\\D_xG&D_yG\end{pmatrix},$$

which is invertible at $(a,c)$. The proved inverse theorem gives a local $C^1$ inverse. Shrink to product neighborhoods so that solving $\Phi(x,y)=(x,0)$ defines a unique $y=\varphi(x)$ and includes every zero in that product neighborhood. Differentiating the identity after existence is established gives

$$D\varphi(x)=-[D_yG(x,\varphi(x))]^{-1}D_xG(x,\varphi(x)).$$

The factor order is forced by solving the matrix equation and cannot be reversed.

2. Reparametrization and unit speed

Let $\phi:J\to I$ be a $C^1$ bijection with nonzero derivative. The new curve $q(u)=r(\phi(u))$ has velocity $q'=r'(\phi)\phi'$ and speed $\|q'\|=v(\phi)|\phi'|$. On a connected interval the derivative's sign cannot change without vanishing, by continuity. A positive sign preserves traversal direction; a negative sign reverses it. The absolute value is necessary for length in either direction.

For a regular curve, fix $t_0$ and define $s(t)=\int_{t_0}^t v(\tau)d\tau$. Then $s'=v>0$, so $s$ is strictly increasing and has an inverse $t=t(s)$ with derivative $1/v(t(s))$ on the interval interior, interpreted one-sidedly at included closed endpoints. The reparametrized curve $\gamma(s)=r(t(s))$ satisfies

$$\gamma'(s)=\frac{r'(t(s))}{v(t(s))},\qquad\|\gamma'(s)\|=1.$$

The value $s(t)$ may be negative before $t_0$; the nonnegative length between ordered parameters $t_1<t_2$ is $s(t_2)-s(t_1)$. The signed coordinate and an elapsed length are related but not identical notions.

The unit tangent $T=r'/v$ depends on traversal orientation. A direction-reversing reparametrization changes it to $-T$ at corresponding points, while the geometric tangent line is unchanged.

3. Curvature measures turning per unit length

For a regular $C^2$ curve, define curvature by $\kappa=\|dT/ds\|=\|T'(t)\|/v(t)$. Since $T\cdot T=1$, differentiation gives $T\cdot T'=0$. The turning of the unit tangent is perpendicular to that tangent.

In $\mathbb R^3$, differentiate $r'=vT$ to get $r''=v'T+vT'$. Therefore

$$\|r'\times r''\|=\|vT\times(v'T+vT')\|=v^2\|T'\|, \qquad \kappa=\frac{\|r'\times r''\|}{\|r'\|^3}.$$

The last equality uses $T\perp T'$ and $\|T\|=1$. If $T'=0$, both sides give zero; no normalizing division by $\|T'\|$ is allowed there. A principal unit normal $N=T'/\|T'\|$ is defined only where curvature is positive.

Curvature is invariant under the regular reparametrizations above: the new $T$ is the old $T$ times the fixed orientation sign, its derivative norm gains $|\phi'|$, and the new speed gains that same factor. The ratio is unchanged. Acceleration by itself has no such invariance because it also records changes of the parameter clock.

If a smooth curve is traced twice, this integral counts twice the once-traversed length.

For vector integration, integrate components. The fundamental theorem gives $r(b)-r(a)=\int_a^br'(t)dt$. This displacement vector can have zero norm even when traveled length is positive. For one full circular traversal displacement is zero while length is $2\pi R$.

Worked example 6: a helical path. For $r(t)=(\cos t,\sin t,2t)^T$, $0\le t\le2\pi$, speed is $\sqrt{\sin^2t+\cos^2t+4}=\sqrt5$. Length is $2\pi\sqrt5$, while displacement is $(0,0,4\pi)^T$. The path is longer than the straight displacement because it also circles horizontally.

6.6 Graded exercises

Foundation

1. Eliminate the parameter for $x=2t+1,y=t^2$, retaining the restriction if $t\ge0$.
2. Differentiate $r(t)=(t,e^{-t},t^3)^T$ twice.
3. For $r(t)=(3t,4t)^T$, calculate speed and the derivative of distance from the origin for $t>0$ and $t<0$.

Core

4. For $x=t,y=t^3-t$, find $dy/dx$ and $d^2y/dx^2$.
5. For $x=t^2,y=t^3$, $t>0$, derive both Cartesian derivatives and show why $y''/x''$ is not the second Cartesian derivative.
6. Find the tangent line to $r(t)=(\cos t,\sin t)^T$ at $t=\pi/4$.
7. For $r(t)=(t,t^2)^T$, find velocity, acceleration and the derivative of speed at $t=1$.
8. Find length of $r(t)=(2\cos t,2\sin t,3t)^T$ on $[0,\pi]$.

Synthesis

9. If $\|r(t)\|$ is constant and nonzero, prove $r\cdot r'=0$. Does this imply $r'=0$? Give a counterexample.
10. Compare $r_1(t)=(\cos t,\sin t)$ on $[0,2\pi]$ with $r_2(s)=(\cos2s,\sin2s)$ on $[0,2\pi]$. Determine image, speed, displacement and counted length for each.

6.7 Full solutions

1. $t=(x-1)/2$, so $y=(x-1)^2/4$. With $t\ge0$, restrict $x\ge1$; without that restriction the same eliminated parabola would include extra points.
2. $r'=(1,-e^{-t},3t^2)^T$, $r''=(0,e^{-t},6t)^T$.
3. Speed is five. Distance is $5|t|$, with derivative five for positive time and negative five for negative time. At zero the distance is nondifferentiable even though the vector path is differentiable.
4. Since $x'=1$, the derivatives are $3t^2-1$ and $6t$.
5. First slope is $3t^2/(2t)=3t/2$. Differentiate this with respect to $t$ and divide by $2t$ to get $3/(4t)$. In contrast $y''/x''=6t/2=3t$, usually different.
6. The point is $(1/\sqrt2,1/\sqrt2)$ and tangent direction $(-1/\sqrt2,1/\sqrt2)$. A tangent-line equation is $x+y=\sqrt2$.
7. $v=(1,2t)$, $a=(0,2)$, so at one the speed derivative is $(1,2)\cdot(0,2)/\sqrt5=4/\sqrt5$.
8. Speed is $\sqrt{4\sin^2t+4\cos^2t+9}=\sqrt{13}$, giving length $\pi\sqrt{13}$.
9. Differentiate the constant squared norm: $0=(r\cdot r)'=2r\cdot r'$. It does not imply a zero derivative; a unit-circle path has constant norm and nonzero tangent velocity.
10. Both images are the unit circle. The first speed is one and it traverses once, with zero displacement and length $2\pi$. The second speed is two and traverses twice, also with zero displacement but length $4\pi$. The interval and parameterization both matter.

Summary and onward map

A parameterization records traversal as well as image. Vector derivatives are componentwise and obey dot/product/chain rules. Cartesian slope formulas need a nonzero horizontal derivative; a stalled parameter does not automatically make the geometric curve singular. Velocity, speed, acceleration and distance-from-origin rates are different quantities. Length counts traversal while integrated velocity gives net displacement.