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Let $F=|\mathbf{F}|>0$ and choose an orientation of a regular curve following the vector field. Its unit tangent vector is $t=\sigma \mathbf{F}/F$, where $\sigma =\pm 1$ is constant along a connected regular segment. Arc length differentiation along that curve is $d/ds=(\sigma \mathbf{F}/F)\cdot \nabla$. ThereforeTaking the cross product with $t$ removes the term parallel to $\mathbf{F}$:Where the Frenet frame exists, $t\times t^{\prime }=\kappa (t\times n)=\kappa b$. ThusThe printed sign corresponds to orientation along or against the field. Taking magnitudes gives the orientation-independent curvature of an integral curve of a vector field, $\kappa =|\mathbf{F}\times (\mathbf{F}\cdot \nabla )\mathbf{F}|/F^3$. This scalar formula also applies at zero curvature, where $b$ itself may be undefined.

The original PDF measures $\theta$ at the rim point $O$, from the positive $x^{\prime }$-axis to $OP$; it is not the angle at the centre of the circle. The converted TeX omits this entire subpart. In rotating Cartesian coordinate system, the centre is $(a,0)$, and the nonzero intersection of the ray $\mathbf{r}=\rho (\cos \theta ,\sin \theta )$ with $(x^{\prime }-a{)}^2+y^{\prime 2}=a^2$ has $\rho =2a\cos \theta$. ThereforeThe second form stays regular even at the rim point $O$, where the polar radius is zero. Define the outward unit normal from the centre and the increasing-angle unit tangent vector byThe first two derivatives in the rotating reference frame are ${\mathbf{v}}^{\prime }=2a\dot{\theta }\mathbf{\tau }$ and ${\mathbf{a}}^{\prime }=2a\ddot{\theta }\mathbf{\tau }-4a{\dot{\theta }}^2\mathbf{n}$. These describe a bead on a horizontal hoop rotating about its rim.

The first term is the local viscous force per unit length from resistive-force theory. With unit tangent vector $\mathbf{t}={\mathbf{r}}_s$, the drag tensor gives $-c_{\parallel }{\mathbf{v}}_{\parallel }-c_{\perp }{\mathbf{v}}_{\perp }$, where $\mathbf{v}={\mathbf{r}}_t$ is velocity relative to the fluid.

The second term is the bending force of an inextensible filament. Varying its energy $E_b=(B/2)\int |{\mathbf{r}}_{ss}{|}^{2}ds$ gives bulk force density $-B{\mathbf{r}}_{ssss}$. The final term, ${\partial }_s(T{\mathbf{r}}_s)$, is the force density from filament tension; $T$ is the Lagrange multiplier enforcing $|{\mathbf{r}}_s|=1$. At low Reynolds number, these forces balance without a filament acceleration term.

With mass, length, and time dimensions $M,\mathsf{L},\mathsf{T}$, the coefficients haveThus the drag coefficients have the dimensions of dynamic viscosity, the filament bending modulus has dimensions force times length squared, and tension has dimensions force.

Use $x>0$ as the parameter of the smooth curve:Its velocity vector and speed areThe unit tangent vector is therefore

The parameter-independent formula for the curvature of a space curve givesDifferentiating,Hence $\kappa$ increases for $0<x<1/\sqrt{2}$ and decreases for $x>1/\sqrt{2}$. Its global maximum occurs at $x=1/\sqrt{2}$ and equalsThe corresponding point is

Use the spherical coordinates parametrizationThen ${\mathbf{R}}_{\varphi }\times {\mathbf{R}}_{\theta }=\sin \varphi \mathbf{R}$, so the outward unit normal vector is $\mathbf{n}=\mathbf{R}=\sin \varphi {\mathbf{e}}_r+\cos \varphi {\mathbf{e}}_z$ and the surface area element is $\sin \varphi d\varphi d\theta$. Since $\sin \varphi >0$ throughout this band, the vector field is smooth near it. The surface integral of the curl isFor the induced boundary orientation, the lower circle $\varphi =\beta$ is traversed with increasing $\theta$, and the upper circle $\varphi =\alpha$ with decreasing $\theta$. One way to check the signs is that the outward conormal is ${\mathbf{e}}_{\varphi }$ at the lower circle and $-{\mathbf{e}}_{\varphi }$ at the upper circle; the positive unit tangent vector is $\mathbf{n}\times \text{conormal}$.

Along a circle of fixed $\varphi$, $d\mathbf{r}=\sin \varphi {\mathbf{e}}_{\theta }d\theta$. Therefore the circulation iswhich equals the boxed surface integral and verifies Stokes theorem with the required orientation.

For a unit-speed curve with positive curvature of a space curve $\kappa$, the principal normal is $N=T^{\prime }/\kappa$, with unit tangent vector $T$. It is perpendicular to $T$ and points in the direction in which the tangent is turning. At zero curvature this formula does not define $N$.

Fix the standard orientation of a vector space on the plane. For an oriented regular curve in that plane, let $T=(T_x,T_y)$ be its unit tangent vector and let $N_L=(-T_y,T_x)$ be its left unit normal, obtained by an anticlockwise rotation through $\pi /2$. Its signed curvature $k$, also commonly written ${\kappa }_s$, is defined byHere $s$ increases with the chosen traversal and $\det$ is the determinant. Positive signed curvature means turning left; negative signed curvature means turning right. The left unit normal exists even when $k=0$; when $k<0$ it points opposite to the principal normal vector. See Ghomi's university lecture notes, section 1.10.

For an arbitrary regular parametric curve $r(t)=(x(t),y(t))$ traversed with increasing $t$, the signed curvature isFor the anticlockwise circle $r(t)=(R\cos t,R\sin t)$, the numerator is $R^2$ and the denominator is $R^3$, so $k=1/R$. For the clockwise traversal $q(t)=(R\cos t,-R\sin t)$, the numerator is $-R^2$, so $k=-1/R$, while unsigned curvature is $1/R$ in both cases. On this circle $N_L$ points inward for anticlockwise traversal and outward for clockwise traversal. For any simple closed regular curve traversed anticlockwise, $N_L$ is the inward unit normal, but signed curvature can still be negative along nonconvex portions.

To see why the formula uses the cube of the speed, write $r^{\prime }=vT$ and $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$. The term parallel to $T$ contributes zero to the determinant, giving $\det (r^{\prime },r^{\prime \prime })=v^3\det (T,dT/ds)=v^{3}k$. Under a reparametrization $q(u)=r(\varphi (u))$ with ${\varphi }^{\prime }\ne 0$, the numerator gains a factor $({\varphi }^{\prime }{)}^3$ and the denominator gains $|{\varphi }^{\prime }{|}^3$. Thus orientation-preserving reparametrization preserves signed curvature, while reversing traversal changes its sign at corresponding points. Unsigned curvature is preserved in either case.

Locally write the unit tangent vector as $T=(\cos \theta ,\sin \theta )$. Differentiating with respect to arc length gives $k=d\theta /ds$: signed curvature records the direction as well as the amount of turning. Its integral therefore measures net turning, as in the turning tangent theorem; the integral of $|k|$ is total curvature and counts both left and right turns positively.

A tangent vector at a point $p$ of a smooth manifold without boundary records an instantaneous velocity through $p$. Its base point is part of the data; measuring its length additionally requires a Riemannian metric. Intrinsically, a smooth curve $\gamma$ with $\gamma (0)=p$ defines its tangent vector by how every local smooth function changes along it:This is a derivation at a point. Two smooth curves through $p$ represent the same tangent vector if these derivatives agree for every such smooth function; equivalently, their coordinate velocity vectors agree in one manifold chart. The chain rule makes that agreement independent of the chosen manifold chart. ^[ref]

For an embedded submanifold of Euclidean space, the intrinsic tangent vector is represented by the ordinary derivative ${\gamma }^{\prime }(0)$, with its base point $p$ retained. A drawn arrow starting at $p$ represents that tangent vector; its endpoint $p+{\gamma }^{\prime }(0)$ need not lie on the embedded submanifold. Nor is a tangent vector required to be a unit tangent vector: a stationary smooth curve gives the zero vector, which belongs to the tangent space but has no direction to normalize.

Here are explicit calculations. In ${\mathbb{R}}^2$, $\gamma (t)=(t,t^2)$ and $\eta (t)=(t,2t^2)$ have different routes but the same tangent vector $(1,0)$ at $(0,0)$. For $f(x,y)=3x+y$, both give $v(f)=3$. For a circle of radius $R$, $c(t)=(R\cos (\omega t),R\sin (\omega t))$ has tangent vector $(0,R\omega )$ at $(R,0)$: changing $\omega$ changes the speed or reverses travel without changing the circle.

A reparametrization $\beta (s)=\gamma (h(s))$ with $h(0)=0$ gives, by the chain rule, ${\beta }^{\prime }(0)=h^{\prime }(0){\gamma }^{\prime }(0)$. A regular change of curve parameter has $h^{\prime }(0)\ne 0$. The singular choice $h(s)=s^3$ instead givesBoth describe the same parabola, whose tangent line at the origin is still the horizontal axis. The zero velocity diagnoses this singular change of curve parameter, whose inverse is not differentiable at zero; it does not show that the geometric tangent line is absent. The regular curve $\gamma$ supplies its nonzero direction.

The diagram is an analytical illustration of these exact parametric curves, with equal parameter increments marked on each route. It separates geometry from traversal: the tangent line depends on the route, while a velocity vector also depends on how the curve parameter advances.

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]

A wavefront connects points with the same wave phase at one specified time. It is a surface, rather than the trajectory of a particular piece of light. For a scalar monochromatic wave written as $u(\mathbf{r},t)=A(\mathbf{r})\cos \Phi (\mathbf{r},t)$, a wavefront is a connected level set of the phase function $\Phi$; different surfaces can share the same wave phase modulo $2\pi$. A plane wave has planar wavefronts. An outward wave from an ideal point source in a uniform medium has spheres as wavefronts, while an optical lens can turn nearly planar wavefronts into converging ones. The wavefront describes wave phase, not wave amplitude: drawing it does not say how bright the beam is. ^[ref]

The relation to light rays becomes precise in a stationary medium with isotropy, negligible loss and a real positive scalar refractive index $n(\mathbf{r},\omega )$. In the geometrical optics limit, spatial changes in the refractive index and wave amplitude are slow on the local wavelength scale. For a single smooth propagation branch, writeHere $\omega >0$ is the angular frequency, $c$ is the vacuum speed of light, $S$ has units of length, and the last relation is the optical eikonal equation. The local wavevector is $\mathbf{k}=\nabla \Phi =k_0\nabla S$. Its direction is normal to a smooth wavefront, since a gradient is normal to its level sets. A forward light ray has unit tangent vector $d\mathbf{r}/ds=\nabla S/n$, so it follows this normal under the stated assumptions. Along that light ray, $dS/ds=\nabla S\cdot (d\mathbf{r}/ds)=n$: the change in $S$ is the optical path length. This is the short geometric argument linking wavefronts, light rays and optical path length. ^[ref]

The diagram shows incident and transmitted branches separately, with the reflected branch omitted. Its stipulated refractive indices are $n_1=1$ and $n_2=3/2$, and its incident angle is $30^{\circ }$ to the interface normal. At a stationary interface, the frequency and tangential wavevector agree on the two sides: Snell's law gives $\sin {\theta }_2=1/3$, hence ${\theta }_2\approx 19.4712^{\circ }$. The wavefronts turn while their normal spacing decreases from ${\lambda }_0$ to $2{\lambda }_0/3$, where ${\lambda }_0$ is the vacuum wavelength. More explicitly, with $x$ normal to the interface and $y$ tangential, take $S_1=\sqrt{3}x/2+y/2$ for $x<0$ and $S_2=\sqrt{2}x+y/2$ for $x>0$. These functions agree at $x=0$; the drawn wavefronts are $S_j=m{\lambda }_0$ at $t=0$ with ${\varphi }_0=0$. This is an exact piecewise plane wave phase construction for the stipulated transparent media, not measured material data. ^[ref]

The normal-to-wavefront rule requires its assumptions. In a medium with anisotropy, the wavevector remains normal to a smooth wavefront, but the group velocity and energy flux need not follow it. ^[ref] Near a wave caustic, where multiple light rays meet, one smooth phase function may no longer describe the field; diffraction and the full wave amplitude matter. A wavefront is also not generally a surface of equal pulse arrival time. In a medium with optical dispersion, a monochromatic wave phase moves with the phase velocity, while a sufficiently narrow wave packet follows the group velocity. Even identical optical path lengths at one frequency can give different group delays. The usual equal-travel-time picture needs a common launch reference and a nondispersive ray model; an arbitrary broadband disturbance has no single monochromatic wavefront.^[ref]