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The binormal is $B=T\times N$, the cross product of the unit tangent vector and principal normal vector. It is a unit vector perpendicular to both factors and completes the oriented Frenet frame.

For a continuously differentiable nonzero vector field $F$, the curvature of a space curve following its direction isTake the unit tangent vector $t=\pm F/|F|$ and differentiate with respect to arc length. The derivative of $|F{|}^{-1}$ is parallel to $F$ and disappears from $t\times t^{\prime }$. Since $t\cdot t^{\prime }=0$, we have $|t\times t^{\prime }|=|t^{\prime }|=\kappa$, proving the formula. It remains valid with zero curvature, even where the usual Frenet frame is undefined.

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.

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.

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.

For a unit-speed curve with nonzero curvature of a space curve $\kappa$, define the Frenet frameThe torsion of a space curve iswith this choice of convention.

For the requested example, take the unit-speed circleand the unit-speed circular helixBoth have the constant nonzero curvature $\kappa =1$, but their torsions are respectively $0$ and $1$. A proper Euclidean motion of Euclidean three-space preserves torsion, so no such rigid motion relates the two curves.

For a regular parametrized curve $r(t)$, the Frenet frame isEquivalently, wherever $dT/dt\ne 0$, one may write $N=(dT/dt)/|dT/dt|$ and $B=T\times N$.

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