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A curve parameter labels the points visited by a parametric curve $\gamma (t)$. Choosing time as the parameter determines its velocity; an arc-length parametrization instead measures accumulated arc length. A regular monotone reparametrization changes the labels and speed while preserving the route and its arc length.

Use the unit normal from part (c), and orient ${\gamma }_a$ by increasing $v$. Write $T={\gamma }_a^{\prime }/|{\gamma }_a^{\prime }|$ for its unit tangent vector. The signed geodesic curvature convention used below iswhere $s$ is arc length. The second equality follows by differentiating $T$: the term parallel to ${\gamma }_a^{\prime }$ disappears in the scalar triple product. It makes the correction for a parameter which is not arc-length parametrization explicit.

Put $C=\cos a$, $S=\sin a$, $D=1+S^2$, and $G=3+2C$. The speed is the constant $\sqrt{G}$. At $v=0$,soTaking its dot product with $N=(C,S,-S)/\sqrt{D}$ yields $-S(3+2C)/\sqrt{D}=-SG/\sqrt{D}$. ThereforeThe screw motion of part (b) makes this independent of $v$. Thus ${\gamma }_a$ is a geodesic precisely when $a$ is an integer multiple of $\pi$. Reversing the direction of traversal or the unit normal changes the sign of signed geodesic curvature; its magnitude is the absolute value of the displayed expression. The signed convention is needed in part (e).

A subdivision curve should support subdivision curve interrogation rather than merely supply successively larger control polygons. Its interface should provide the parameter interval, orientation and open/closed status; endpoint and limit-point evaluation to a prescribed accuracy; and tangent vector or derivative evaluation where defined. It should be possible to split a parameter interval, commonly at its midpoint, into equivalent restricted curve definitions while retaining all neighboring controls needed by the refinement rule.

For geometric algorithms, request a certified bounding volume for each restricted curve, such as a convex hull or an axis-aligned bounding box, and the minimum and maximum of a plane's linear function over that bound. Request also a certified flatness bound: the maximum distance of the restricted limit curve from its endpoint chord, together with its spatial extent. These allow safe rejection, adaptive drawing and accuracy control. A positive, constant-reproducing subdivision mask often makes a control-point convex hull a valid bound. With negative mask coefficients this must be proved separately or replaced by a padded bound; it cannot simply be assumed.

Further useful enquiries include roots of a scalar function along the curve, closest-point searches, arc length, and inversion of an arc-length parametrization. They can be implemented through evaluation, differentiation, certified bounds and recursive subdivision. Evaluation, restriction, enclosure and error bounds are the essential primitives needed for reliable clipping and rendering; an unqualified control-polygon approximation is insufficient.

The coordinate curves are assumed to be geodesics as geometric curves; use their arc-length parametrization in the geodesic equations. On $u=c$, choose increasing arc length $s$, so $\dot{u}=0$ and $\dot{v}=1/\sqrt{G}$. The transverse equation becomesThus $(F/\sqrt{G}{)}_v=G_u/(2\sqrt{G})=(\sqrt{G}{)}_u$. On $v=d$, similarly $\dot{v}=0$, $\dot{u}=1/\sqrt{E}$, and the transverse equation gives $(F/\sqrt{E}{)}_u=E_v/(2\sqrt{E})=(\sqrt{E}{)}_v$. Hence the geodesic coordinate grid identities are

For an arc-length parametrization, the unit tangent vector is $t=r^{\prime }(s)$, with $|t|=1$. Its derivative is perpendicular to $t$. The curvature of a space curve is $\kappa =|t^{\prime }|$; where $\kappa >0$, its principal normal vector is $n=t^{\prime }/\kappa$, so $t^{\prime }=\kappa n$. The binormal vector $b=t\times n$ completes the positively oriented Frenet frame.

Differentiate $b\cdot b=1$ to get $b^{\prime }\cdot b=0$. AlsoThus $b^{\prime }$ has only an $n$ component. Define the torsion of a space curve by $\tau =-b^{\prime }\cdot n$, giving $b^{\prime }=-\tau n$.

Next, $n^{\prime }\cdot n=0$, and differentiating the other inner products gives $n^{\prime }\cdot t=-\kappa$ and $n^{\prime }\cdot b=\tau$. Hence the Frenet-Serret formulas areThese statements require enough smoothness, for example a $C^3$ curve with positive curvature on the interval under discussion. If $\kappa =0$, the unit tangent vector is still defined, but the standard principal normal vector, binormal vector and torsion of a space curve need not be; their existence cannot be inferred just from smoothness of the curve.

Fix $t_0\in [a,b]$ and defineBecause $\gamma$ is a regular curve, $s^{\prime }(t)=\Vert {\gamma }^{\prime }(t)\Vert >0$. The inverse function theorem gives a smooth inverse $t=t(s)$ onto the corresponding interval. For $\tilde{\gamma }(s)=\gamma (t(s))$,Thus the arc-length parametrization exists.

For $R>r>0$, the closed curveis embedded on a torus and satisfiesAfter arc-length parametrization, it meets all the requirements of part (b). Its bending and twisting vary around the two waves, so neither curvature nor torsion is constant.

For a regular curve, the curvature and torsion definition givesIt measures the rate at which the unit tangent turns per unit arclength.

HereThus $|{\gamma }^{\prime }|=3$, $|{\gamma }^{\prime \prime }|=4$, and ${\gamma }^{\prime }\cdot {\gamma }^{\prime \prime }=0$. HenceThe arc-length parametrization from $t=0$ is $s=3t$, so the curvature remains the constant $\kappa (s)=4/9$ for $0\le s\le 3\pi$. Therefore the total curvature is

The curve $\tilde{\gamma }$ is a unit circle traversed once. Its curvature is one and its length is $2\pi$, so without further calculationThe helical curve $\gamma$ devotes part of its unit tangent to the constant vertical direction. Its tangent therefore turns more slowly on the unit sphere than the tangent to the planar circle, which explains the smaller total curvature.

A smooth curve $\alpha :I\to {\mathbb{R}}^3$ is regular when ${\alpha }^{\prime }(t)\ne 0$ for every $t\in I$. Its arc-length between parameters $t_0$ and $t_1$ is

Fix $t_0$ and defineRegularity gives $s^{\prime }(t)=|{\alpha }^{\prime }(t)|>0$, so $s$ is strictly increasing and has a smooth inverse $t=t(s)$ on its image. The reparametrized curve $\beta (s)=\alpha (t(s))$ satisfiesThus every regular curve has an arc-length parametrization.

For a unit-speed curve with nonzero curvature, letIts torsion isequivalentlyIf the curve lies in an affine plane, all three derivative vectors lie in the parallel two-dimensional vector plane, so their determinant is zero. Hence $\tau =0$ wherever torsion is defined.

Length functional. On the unit sphere, write the position vector in spherical coordinates asIts two coordinate tangent vectors satisfy $r_{\theta }\cdot r_{\varphi }=0$, $|r_{\theta }|=1$ and $|r_{\varphi }|=\sin \theta$. Consequently the induced Riemannian metric is $d{s}^2=d{\theta }^2+{\sin }^2\theta d{\varphi }^2$. On a segment for which $\theta$ increases and $\varphi =\varphi (\theta )$,Segments where $\theta$ is not monotone require another parameter or a subdivision. Endpoint turning points can be handled by a limit.

First integral. With $F=\sqrt{1+{\sin }^2\theta ({\varphi }^{\prime }{)}^2}$, there is no explicit $\varphi$ dependence. The Euler-Lagrange equation therefore says thatis constant. For $c\ne 0$, put $k=1/|c|$ and solve:The radicand must be nonnegative along the chosen segment. The printed expression with both limits fixed at ${\theta }_1,{\theta }_2$ describes a longitude difference between endpoints; a formula for $\varphi (\theta )$ needs the variable upper limit and the additive constant shown here. The case $c=0$ gives a meridian with constant longitude. The equator is also a stationary route, although it cannot be parameterized using changing $\theta$.

Integration and geometric meaning. Put $t=\cot \theta$, so $d\theta =-dt/(1+t^2)$ andFor a nonconstant segment $k>1$, and writing $K=\sqrt{k^2-1}$, the integral becomesRearranging, and absorbing phase and sign into two constants, givesMultiplying by $\sin \theta$ shows that the corresponding Cartesian coordinates satisfy $z=Ax+By$. Thus every such stationary route lies in a plane through the sphere's center: it is a great circle, the geodesic of the sphere. The meridians and equator are the limiting or separately parameterized cases of the same description.

Rome to Singapore. Use exactly the approximate coordinates supplied in the question: $(\theta ,\varphi )=(\pi /4,\pi /12)$ at departure and $(\pi /2,7\pi /12)$ at arrival. Inserting them into the preceding relation determinesIndeed the right side is one at Rome and zero at Singapore. This corresponds to $K=1$, $k=\sqrt{2}$. In latitude $\ell =\pi /2-\theta$, the route is $\ell =\arctan [\cos (\varphi -\pi /12)]$: it starts at $45^{\circ }$ north with an initially eastward heading and then curves southward to the equator. The first-integral expression in $\theta$ has an integrable endpoint singularity at departure, where $\theta$ has a turning point; the longitude parameter is regular there.

For an explicit regular arc-length parametrization, let $a=r(\pi /4,\pi /12)$ and $b=r(\pi /2,7\pi /12)$. Their dot product is $\sin (\pi /4)\cos (\pi /2)=0$. HenceBoth $|r(s)|$ and $|r^{\prime }(s)|$ equal one. The shorter great circle arc is the global shortest path: for any unit-speed curve from $a$, its central angle $\alpha =\arccos (a\cdot r)$ satisfies $|d\alpha /ds|\le 1$ by the Cauchy-Schwarz inequality, since the tangential projection of $a$ has length $\sin \alpha$. Integrating gives length at least the endpoint angle $\pi /2$, which this route attains. For a sphere of radius $R$, the corresponding shortest distance is $\pi R/2$.