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Spherical trigonometry calculates the sides and interior angles of spherical triangles using trigonometric functions. Its sides must be shorter great circle arcs, which are geodesics on a round sphere. A curve at constant nonzero latitude does not qualify. The spherical law of sines and spherical law of cosines play the roles of the familiar flat-triangle rules, but account for the sphere's curvature.

Use uppercase $A,B,C$ both for the vertices and for their interior angles, and lowercase $a,b,c$ for the opposite angular sides. For example, $a$ is the central angle subtended by the arc from $B$ to $C$. On a sphere of radius $R$, its physical arc length is ${\ell }_a=Ra$ when $a$ is in radians; similarly ${\ell }_b=Rb$ and ${\ell }_c=Rc$. Only on the unit sphere do the angular and physical side lengths have the same numerical values. See J. B. Tatum, Celestial Mechanics, chapter 3, section 3.5: great-circle sides and angular units.

One useful application of spherical trigonometry is finding great-circle distance from spherical coordinates. Write a unit vector as $u(\varphi ,\lambda )=(\cos \varphi \cos \lambda ,\cos \varphi \sin \lambda ,\sin \varphi )$, where $\varphi$ is latitude and $\lambda$ is the equatorial azimuth, or longitude, measured from a chosen reference direction. The dot product of two such unit vectors givesHere $\delta \in [0,\pi ]$ is the central angle and $d$ is the great-circle distance. The same identity follows from the spherical law of cosines by adding the north pole as the third vertex, with angular sides $\pi /2-{\varphi }_1$ and $\pi /2-{\varphi }_2$. This connects spherical geometry to position calculations in astronomy and to navigation on a spherical model of Earth. Choosing the surface model matters: Paula Neeley and Anthony Narkawicz, Map Projection Induced Variations in Locations of Polygon Geofence Edges, NASA/TM-2017-219675 (2017), sections 2-3 distinguishes spherical and ellipsoidal geodesics and gives this coordinate identity.

For a toy example, put two points at latitude $30^{\circ }$, with equatorial azimuths $-60^{\circ }$ and $60^{\circ }$. The great-circle distance and the arc length along constant latitude areThe great circle takes a shorter route by bending towards the pole in these spherical coordinates. To locate its highest latitude, the endpoint unit vectors are $p=(\sqrt{3}/4,-3/4,1/2)$ and $q=(\sqrt{3}/4,3/4,1/2)$. Their great circle lies in the plane through the origin with $z=2x/\sqrt{3}$. On its shorter arc, the maximum latitude occurs at $y=0$ and is $\arctan (2/\sqrt{3})\approx 49.106605^{\circ }$. Thus a geodesic can change latitude substantially even when its endpoints have equal latitude. These are lengths on an ideal sphere; the example illustrates the geometry rather than specifying an actual flight route.

Historically, spherical trigonometry was prominent in teaching as well as in astronomy. John Conway and Alex Ryba note its presence in early twentieth-century school textbooks in Remembering spherical trigonometry, The Mathematical Gazette 100(547), March 2016, pages 1-8, DOI 10.1017/mag.2016.3. Their introductory observation documents that teaching history; it does not identify a single inventor of the spherical law of sines or spherical law of cosines.

Parameterize a fixed yield surface by stress-space arc length $l$ with ${\xi }_{,l}=-\cos 2\varphi$ and ${\tau }_{,l}=-\sin 2\varphi$. For mean normal stress $\sigma$, force balance gives $\nabla \sigma =M\nabla l$, where $M=\left(\begin{array}{cc}\cos 2\varphi & \sin 2\varphi \\ \sin 2\varphi & -\cos 2\varphi \end{array}\right)$. Its eigenvectors along angles $\varphi$ and $\varphi +\pi /2$ have eigenvalues $+1$ and $-1$, proving the displayed slip line invariants.

The part of elastic energy associated with a change of material length. A simple quadratic constitutive equation assigns energy density $C(\alpha -1{)}^2/2$ per reference arc length, where $C>0$ is stretching stiffness and $\alpha$ is the stretch ratio. Such a small-strain model need not remain realistic as $\alpha \to 0$.

For the radial profile $(\rho (t),z(t))$, with $\rho >0$, use the parametrized surface $\mathbf{R}(t,\theta )=(\rho \cos \theta ,\rho \sin \theta ,z)$, $0\le \theta <2\pi$. The two tangent vectors are orthogonal and have Euclidean norms $\sqrt{{\rho }^{\prime 2}+z^{\prime 2}}$ and $\rho$. Their cross product gives the displayed surface area. Geometric surface area assumes a single covering of the radial profile. A general three-dimensional generating curve must first be replaced by its actual radius $\rho (t)=\sqrt{x(t{)}^2+y(t{)}^2}$; its spatial arc length cannot simply replace the radial profile's arc length.

Let a simple closed regular planar curve be invariant under point reflection through a center and remain strictly on one side of a coplanar axis of rotation, at distance $d$ from its center. If $L$ is its arc length, its surface of revolution has surface area $2\pi dL$. To prove this, choose coordinates with the center at the origin and the axis of rotation at $y=-d$. The surface area formula is $2\pi \oint (d+y)ds$. Pair each point with its reflected point: the arc length element $ds$ is unchanged and $y$ changes sign. Thus $\oint yds=0$, proving the result. The one-side condition prevents the surface from passing through the axis of rotation or requiring an absolute value with changing sign.

The total variation of a function $f:[a,b]\to \mathbb{R}$ measures its accumulated rises and falls, rather than just its change between the endpoints. For each partition of an interval $\Pi =(a=t_0<\cdots <t_n=b)$, add the absolute values of its sampled increments. Then take the supremum over all such partitions of an interval:The function has bounded variation when $V_a^b(f)<\infty$. This definition uses its value at every point of the interval, including both endpoints.^[ref]

For a nondecreasing function, all increments are nonnegative and the sum telescopes to $f(b)-f(a)$. More generally, the triangle inequality gives $|f(b)-f(a)|\le V_a^b(f)$. Equality can fail dramatically when the function changes direction: an object can travel a long distance and finish near its starting position. Here we measure movement of the function's value; arc length of its function graph also includes horizontal movement.

For a parametric surface $F(u,v)$ and a scalar implicit equation $f(P)=0$, set $g=f\circ F$. At a regular transversal intersection, $\nabla g\ne 0$, and $(-g_v,g_u)$ is a parameter tangent vector. Normalize its image under $DF$ to choose physical arc length steps. A predictor followed by Newton method correction of $g=0$ and a transverse plane equation gives a local continuation method. Certified patch bounds are useful for finding seeds; corner signs alone do not detect a closed intersection entirely inside a patch.

A wave's travel time along a elastic ray is the integral of local slowness against arc length. Its variation with source and receiver position describes the arrival branches, including transmitted and reflected paths. Different paths can have different travel times between the same endpoints.

For speed $v=k{r}^2$ in the punctured plane, the travel-time element is $ds/(k{r}^2)$. Under the inversion in a circle with radial coordinate $u=1/r$, this becomes $k^{-1}\sqrt{d{u}^2+u^{2}d{\theta }^2}$, the arc length in Euclidean space divided by $k$. Thus a shortest travel-time route corresponds to a straight line segment in the inverted plane, whenever that segment avoids the origin and lies in the allowed region. Between $(a,0)$ and $(0,a)$ the inverted segment has equation $u(\cos \theta +\sin \theta )=1/a$. The route is $r=a(\cos \theta +\sin \theta )$, and its minimum travel time is $\sqrt{2}/(ka)$. The triangle inequality for arc length proves global minimality.

For a regular curve $\mathbf{r}(t)$ with nonzero velocity vector, its unit tangent vector isThe Euclidean norm in the denominator removes the traversal speed, leaving a unit vector that records direction. A positive change of traversal speed leaves this direction unchanged; reversing the direction of travel reverses $T$. In arc length coordinates, $T=d\mathbf{r}/ds$.

For the parabola $\mathbf{r}(t)=(t,t^2)$, the velocity vector is $(1,2t)$ and the speed is $\sqrt{1+4t^2}$. At $t=1$, $T=(1,2{)}^T/\sqrt{5}$. The tangent vector $(1,2{)}^T$ and the unit tangent vector point in the same direction, but only the latter has Euclidean norm one. The principal normal vector is a further choice based on how this direction changes, where the curvature is nonzero. ^[ref]

For an oriented parametric curve $\mathbf{r}(t)$ and scalar field $V$, define the componentwise line integralIts value is a vector and reversing orientation changes its sign. It differs from the scalar arc length integral ${\int }_{C}Vds=\int V(\mathbf{r}(t))|{\mathbf{r}}^{\prime }(t)|dt$, which is independent of orientation. A request to integrate a scalar field along a curve should specify which differential is intended.

For the counterclockwise circle $\mathbf{r}(\varphi )=(R\cos \varphi ,R\sin \varphi ,z_0)$, suppose the restricted field has Fourier series $V(\varphi )=A_0+{\sum }_{n\ge 1}(A_n\cos n\varphi +B_n\sin n\varphi )$. Orthogonality givesThe oriented vector integral selects the first angular harmonics, whereas the scalar arc-length integral selects the mean.

For $-\pi \le t\le \pi$, this parametric curve closes with a corner on a unit circular cylinder. Its speed is $\cosh t$, so its arc length is $2\sinh \pi$. The cylindrical surface between the curve and $z=0$ has the same surface area, because unrolling gives the integral ${\int }_{-\pi }^{\pi }\cosh tdt$.