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Put $T={\alpha }^{\prime }(s)$. Since $s$ is arc length, $|T|=1$, so $T^{\prime }\perp T$. The curvature of a space curve isWhere $k(s)\ne 0$, define the principal normal and binormal byand define the torsion of a space curve byThus torsion requires nonzero curvature, so that $N$ and $B$ are defined.

Differentiating the orthonormality relations for the frame $(T,N,B)$ shows that its derivative matrix is skew-symmetric. Since $T^{\prime }=kN$, differentiating $B=T\times N$ and using the definition of $\tau$ gives the Frenet-Serret formulas

Under the Euclidean isometry $\tilde{\mathbf{r}}=\mathbf{a}+R\mathbf{r}$, where $R^{T}R=I$,ThereforeSince the integration parameter is unchanged,Both the induced metric and arc length are invariant.

Enforce the fixed arc length with a multiplier. Apart from an irrelevant common factor, the augmented integrand isThe Beltrami identity givesAfter absorbing constants into $z_0$ and a positive scale $B$, integration yields the catenaryThe endpoint conditions and prescribed length giveThe second equation determines $B>0$ implicitly because $L>2a$, and the first then determines $z_0$.

The periodic Wirtinger inequality says that if $f$ is continuously differentiable, $L$-periodic, and has mean zero, thenEquality holds exactly for a linear combination of $\cos (2\pi s/L)$ and $\sin (2\pi s/L)$.

First suppose $\partial \Omega$ is one positively oriented simple closed curve, parametrized by arc length as $\gamma (s)=(x(s),y(s))$ for $0\le s\le L$. Translating the origin, which changes neither length nor area, makes both $x$ and $y$ have mean zero. Green theorem and the Cauchy-Schwarz inequality giveApplying Wirtinger's inequality to both coordinate functions and using the unit-speed identity $(x^{\prime }{)}^2+(y^{\prime }{)}^2=1$ yieldsand therefore the planar isoperimetric inequalityFor several boundary components, apply the simple-curve result to the relevant enclosed regions and use ${\sum }_j{L}_j^2\le ({\sum }_j{L}_j{)}^2$; holes only decrease the area.

Equality in both inequalities forcesfor constant vectors $A,B$, with the unit-speed and Cauchy-Schwarz equality conditions making $A$ and $B$ perpendicular and equally long. Thus the boundary is a circle. Conversely, a circular domain has area $\pi R^2$ and perimeter $2\pi R$, so equality holds.

The planemeets the one-sheet hyperboloid in the two disjoint curvesThey are the two opposite meridians of the surface of revolution. Parametrize the generating profile by arc length. Its acceleration lies in the meridian plane and is orthogonal to the profile tangent; it is also orthogonal to the azimuthal tangent. It is therefore normal to the surface, so both meridians are geodesics. This is the disjoint case in plane-section geodesics of the unit one-sheet hyperboloid.

Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.

Let $\gamma$ be a $C^3$ regular curve parametrized by arc length, so $|{\gamma }^{\prime }(s)|=1$. Define the Frenet frameand define the torsion byThus curvature requires two derivatives, while $N,B,$ and torsion require $\kappa >0$; the displayed classical torsion requires three derivatives. At a point where $\kappa =0$, $N,B,$ and $\tau$ are not defined by this construction.

Since $(T,N,B)$ is an orthonormal frame, differentiating its inner products shows that its derivative matrix is skew-symmetric. The definition gives $T^{\prime }=\kappa N$. Next,so $N^{\prime }=-\kappa T+cB$. As $B^{\prime }=T\times N^{\prime }$ and $T\times B=-N$, we have $B^{\prime }=-cN$; hence $c=-B^{\prime }\cdot N=\tau$. The Frenet-Serret formulas are therefore

For a positively oriented local parametrization $X(u,v)$ with first fundamental formthe area element of a surface isUnder an orientation-preserving coordinate change, the Jacobian from $du\wedge dv$ cancels the inverse Jacobian in the square root of the metric determinant. Hence these local expressions agree and define a global two-form.

The Euler characteristic may be defined from any finite triangulation byor equivalently by the alternating sum of the dimensions of the rational homology groups. Subdivision leaves $V-E+F$ unchanged, and the homological formula shows that it is a topological invariant, so the definition does not depend on the triangulation.

Give the boundary its induced orientation and parametrize it by arc length. If $T$ is its unit tangent and $N$ is the chosen unit normal to the surface, its signed geodesic curvature iswhere $D_s$ is the surface covariant derivative. The Gauss-Bonnet theorem for a compact oriented surface with smooth boundary and no corners iswhere $K$ is the Gaussian curvature.

The velocity vector of the parametric curve isIts speed iswhere $\cosh t>0$ fixes the sign of the square root. Hence the arc length isThe coincidence of the endpoints does not duplicate any positive arc length.

Here $\varphi$ is a scalar field: the source's description as a vector field is a typographical slip, since the displayed formula is for its gradient. For cylindrical coordinates, write $x=r\cos \theta$, $y=r\sin \theta$. Away from $r=0$, a local choice of $\theta$ gives the partial derivativesThe chain rule therefore givesUse the Cartesian coordinates definition $\nabla \varphi ={\varphi }_x{\mathbf{e}}_x+{\varphi }_y{\mathbf{e}}_y+{\varphi }_z{\mathbf{e}}_z$ and the unit vectorsCollecting their coefficients provesThis is the gradient in cylindrical coordinates. The factor $1/r$ converts a change in angle into a change per unit arc length.

Use the convention that the Newtonian gravitational potential tends to zero at infinity and the gravitational field is $-\nabla \varphi$. A source element of mass $dm$ at position vector ${\mathbf{r}}^{\prime }$ contributes $-Gdm/|\mathbf{r}-{\mathbf{r}}^{\prime }|$, where $G$ is the gravitational constant. For the curve, $dm=\gamma d{s}^{\prime }$, soHere $d{s}^{\prime }$ is arc length along the source curve; the expression is finite away from it. This normalization fixes the otherwise arbitrary additive constant.

In Cartesian coordinates, a parametrization of the curve isDifferentiate using the product rule:Squaring and adding cancels the mixed terms, giving $(x^{\prime }{)}^2+(y^{\prime }{)}^2=(f^{\prime }{)}^2+f^2$. The arc length is the integral of the speed, soThis assumes the parametrization is piecewise continuously differentiable and counts length with multiplicity if the curve is retraced.

The cardioid has reflection symmetry about the $x$ axis because $1+\cos (-\theta )=1+\cos \theta$. It passes through $(2,0)$ at $\theta =0$, through $(0,\pm 1)$ at $\theta =\pm \pi /2$, and has a cusp at the origin when $\theta =\pi$. These points and the following sketch show its orientation.

Using the polar coordinates formula for arc length,The absolute value changes sign at $\theta =\pi$, so

Use the polar radius $r=at\in [0,1]$ as the parameter of the Archimedean spiral. Its angle in polar coordinates is $r/a$. The radial and angular contributions to its arc length are orthogonal, soThere is no need to evaluate this integral. The simple bounds $r\le \sqrt{a^2+r^2}\le r+a$ implyThe squeeze theorem therefore gives the scaled length of an Archimedean spiral:Geometrically, the radial contribution becomes negligible after multiplying the arc length by $a$; the remaining contribution is the area under the function graph $r\mapsto r$ on $[0,1]$.

Pair each point $\mathbf{x}$ of the square's boundary with $-\mathbf{x}$. The vector field obeys $\mathbf{f}(-\mathbf{x})=\mathbf{f}(\mathbf{x})$, whereas the outward unit normal vector changes sign: $\mathbf{n}(-\mathbf{x})=-\mathbf{n}(\mathbf{x})$. Corresponding arc length elements are equal. Thus the two contributions to the flux cancel, proving without evaluating any boundary integral thatThe same cancellation appears in the divergence, which is an odd function of $x$ over a region symmetric under $x\mapsto -x$. To justify using the planar divergence theorem despite the lack of differentiability at the origin, remove a disk of radius $\epsilon <1$. The vector field is continuously differentiable on the remaining region, its divergence has zero double integral by this symmetry, and the magnitude of the inner flux is at most $2\pi {\epsilon }^2$, since $\Vert \mathbf{f}\Vert \le \epsilon$ there. Letting $\epsilon \to 0$ recovers the result. This is flux around a removable point singularity.

On the unit circle, the positively oriented unit tangent vector is $\mathbf{u}=(-y,x,0)$, and $|\mathbf{u}|=1$. Hence $d\mathbf{r}=\mathbf{u}ds$ andSince the scalar field $\psi$ has no $z$ dependence, the curl product rule for a scalar field givesOn the unit upper hemisphere $S_1$, the outward unit normal vector is $(x,y,z)$, so its vertical component is $z=\cos \theta$. Applying the Stokes theorem to the vector field $\psi \mathbf{u}$ therefore provesThe smoothness needed here is that $\psi$ extend as a continuously differentiable scalar field to a neighbourhood of the projected unit disk. The left side is a scalar line integral with respect to arc length; the positive boundary orientation is used when identifying it with the vector line integral.

Use the parametric curveHere $|y(t)|\le \sqrt{5}<3$, so the ellipse stays strictly above the axis of rotation $y=-3$. Rotating in three dimensions gives a surface of revolution with parametrizationThe two parameter tangent vectors are orthogonal, with Euclidean norms $\sqrt{x^{\prime }(t{)}^2+y^{\prime }(t{)}^2}$ and $y(t)+3$. Hence the area element of a surface isThe generating ellipse is simple and has positive distance from the axis of rotation; its rotation sweeps the surface once, apart from the identified parameter endpoints. AlsoWriting $w(t)=\sqrt{8{\sin }^{2}t+2{\cos }^{2}t}$, the surface area is therefore $2\pi {\int }_0^{2\pi }(3+y(t))w(t)dt$. Since $y(t+\pi )=-y(t)$ and $w(t+\pi )=w(t)$,ConsequentlyThe cancellation is an instance of the surface area of revolution of a centrally symmetric curve: the surface area is the generating curve’s arc length times the distance $6\pi$ traveled by its center.

For the cardioid, $r^{\prime }=-a\sin \theta$. The expression under the square root simplifies toSince $0\le \theta \le {\theta }_2\le \pi$, the cosine is nonnegative. ThereforeThe upper half of the cardioid starts at $(2a,0)$, passes through $(0,a)$, bends to negative $x$, and ends at the cusp $(0,0)$ when ${\theta }_2=\pi$. Its full upper-half arc length is $4a$; reflecting gives $8a$ for the full cardioid.

The fundamental theorem of calculus givesThis decreases from $2a$ to zero, so the arc length increases fastest at ${\theta }_2=0$, interpreted as the right-hand rate at the initial endpoint. If only positive endpoints are allowed, the rate has supremum $2a$ as ${\theta }_2\downarrow 0$, rather than an attained interior maximum.

In polar coordinates, $x=r(\theta )\cos \theta$ and $y=r(\theta )\sin \theta$. The product rule givesSquaring and adding cancels the mixed terms and gives $(x^{\prime }{)}^2+(y^{\prime }{)}^2=(r^{\prime }{)}^2+r^2$. Hence the arc length is

For a piecewise continuously differentiable parametric curve, the arc length is the integral of speed:The Pythagorean theorem gives the length of an infinitesimal displacement as $\sqrt{d{x}^2+d{y}^2}$. Integration counts traversal length, including repeated portions if the parametrization retraces them. Piecewise continuous derivatives, or suitable absolute continuity, justify the formula; mere differentiability as stated in the source is not by itself the usual sufficient regularity hypothesis.