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For a regular parameter $t$, the curvature and torsion of a space curve areThe torsion of a space curve requires $\kappa \ne 0$. Reparameterize by arc length $s$ and set $T={\alpha }_s$, $N=T_s/\kappa$, $B=T\times N$. Differentiating the pairwise inner products of this orthonormal Frenet frame shows that its derivative matrix is skew symmetric. Since $T_s=\kappa N$, defining $\tau =N_s\cdot B$ gives the Frenet-Serret formulasThe fundamental theorem of regular space curves says that prescribed smooth $\kappa >0$ and smooth $\tau$ on an interval determine a unit-speed curve uniquely up to an orientation-preserving Euclidean rigid motion. Solve this frame's linear differential equation with an initial oriented orthonormal frame, then integrate ${\alpha }_s=T$; uniqueness proves the assertion.

Under the intended convention that the shared parameter $s$ is arc length, equality on an open interval gives equal position and equal Frenet frames at one point. Equal $\kappa ,\tau$ give identical differential equations, so uniqueness gives $\boxed{\tilde{\alpha }=\alpha \text{ on }I}$. The printed “regular smooth” hypothesis alone does not establish this convention. Without common arc length, an orientation-preserving smooth change of speed that is the identity on $J$ but not elsewhere reparameterizes a circle into a counterexample: both curves have $\kappa =1$, $\tau =0$ and agree on $J$, but differ elsewhere as parameterized curves.

For an oriented surface with unit normal $n$, define normal curvature ${\kappa }_n=T_s\cdot n$ and geodesic curvature ${\kappa }_g=T_s\cdot (n\times T)$. ThusA unit-speed curve is a geodesic exactly when ${\kappa }_g=0$. If also ${\kappa }_n\equiv 0$, then $T_s=0$, so it is a straight line. A vertical ruling of a circular cylinder is such a geodesic, although the cylinder contains no open piece of a plane. A nonconstant regular straight line cannot be a simple closed curve. Hence no planar patch is required, and such a geodesic cannot be simple and closed.

Finally let the principal curvatures be ${\kappa }_1,{\kappa }_2$. If the Gaussian curvature $K={\kappa }_1{\kappa }_2\ge 0$, they have the same sign or one vanishes. Euler formula for normal curvature makes ${\kappa }_n$ a weighted average of them. For a geodesic, $\kappa =|{\kappa }_n|$, soA horizontal circle on a cylinder of radius $R$ is a geodesic with $\kappa =1/R$, $K=0$, $|H|=1/(2R)$, attaining equality.

The length of a curve is $L(\alpha )={\int }_a^b\Vert {\alpha }^{\prime }(t)\Vert dt$. Its arc length parameter is $s(t)={\int }_a^t\Vert {\alpha }^{\prime }(u)\Vert du$. Regularity gives $s^{\prime }(t)>0$, so the inverse function theorem gives a smooth inverse and $\beta (s)=\alpha (t(s))$ has $\Vert {\beta }^{\prime }(s)\Vert =1$.

Reparametrize a length minimizer to constant speed on $[a,b]$ and use the energy functional $E=\frac{1}{2}{\int }_a^b\Vert {\alpha }^{\prime }{\Vert }^{2}dt$. By Cauchy-Schwarz inequality, every competitor has $E\ge L^2/(2(b-a))$, with equality for constant speed, so this parametrization also minimizes $E$. For a fixed-endpoint variation with tangential vector field $V$, differentiation and integration by parts giveThe boundary term is zero. Arbitrary smooth tangential $V$ of compact interior support are realizable by the stated variation fact. Taking $V$ to be a nonnegative cutoff times ${\nabla }_t{\alpha }^{\prime }$ proves ${\nabla }_t{\alpha }^{\prime }=0$. Hence a length minimizer is a geodesic after constant-speed reparametrization. The original parameter need not be affine: arbitrary varying-speed parametrizations preserve length but have nonzero covariant acceleration. This qualification is necessary for the literal wording.

For distinct nonantipodal $p,q$ on the punctured unit sphere, there is a unique shorter great circle arc, of length $d=\arccos (p\cdot q)<\pi$. A minimizing curve exists exactly when that shorter arc avoids the removed north pole. If it avoids the pole it attains the spherical lower bound. If it passes through the pole, arbitrarily small smooth detours have lengths tending to $d$, but equality would force the unique shorter arc, which is unavailable. Thus the infimum is not attained; the longer great circle arc is not a substitute minimizer.

For antipodal $p,q$ in the punctured sphere, one can choose a great circle semicircle avoiding the north pole; it attains length $\pi$. Thus every admissible antipodal pair has a minimizer. If identical endpoints are included despite the earlier distinctness stipulation, the infimum is zero, but no smooth regular curve attains it; only a nonregular constant curve does.

It is important to fix the mass flux convention. With literal mass units and the corresponding density-weighted buoyancy flux, the Boussinesq approximation and circular top-hat plume model giveReplacing $\rho$ by ${\rho }_0$ in inertial fluxes is the Boussinesq approximation; the density difference is retained in the buoyancy flux. Define the kinematic plume fluxesIf the symbols $Q,\mathbf{M},F$ instead denote the customary kinematic fluxes, simply omit the factors ${\rho }_0$ in all conversions below. Keeping the two conventions consistent is essential for the jet length and dimensional prefactors.

The Batchelor entrainment hypothesis prescribes inward edge velocityThe constant entrainment coefficient relates ambient inflow to the local axial velocity. A centreline segment of arc length $ds$ entrains volume $2\pi b{u}_{e}ds$. Hence, setting $E=2\alpha \sqrt{\pi }$,The buoyancy flux is conserved because the homogeneous entrained ambient has zero density deficit, and mixing has no buoyancy source or sink. The imposed zero source mass flux with nonzero source momentum flux is a singular point-source idealization, not a finite-radius nozzle with zero emitted fluid. Indeed $V=m/q$ and the plume reduced gravity $f/q$ diverge as $q\to 0$ with nonzero source fluxes. The Boussinesq approximation cannot remain valid arbitrarily close to that point. The top-hat plume model is applied outside the unresolved source region.

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.