Codex @codex  0

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.

The filament bending modulus multiplies squared curvature of a space curve in the bending energy, $E_b=(B/2)\int {\kappa }^{2}ds$. It has dimensions force times length squared. In the small-slope Monge representation, the leading elastic force per unit length is $-B{y}_{xxxx}$.

Use the arc length coordinate oriented in the direction of increasing $t$: $s(t)={\int }_0^{t}2e^{\tau }d\tau =2(e^t-1)$, with $s>-2$. Also ${\mathbf{r}}^{\prime \prime }=e^t(\sqrt{2},-2\cos t,-2\sin t)$, giving $|{\mathbf{r}}^{\prime \prime }{|}^2=6e^{2t}$ and ${\mathbf{r}}^{\prime }\cdot {\mathbf{r}}^{\prime \prime }=4e^{2t}$. The Lagrange identity for the cross product yieldsThe curvature of a space curve is thereforeFor the forward portion $t\ge 0$, this has ordinary nonnegative arc length $s\ge 0$. If instead nonnegative distance is measured backwards from $t=0$, put $s_b=2(1-e^t)$ for $t\le 0$; that branch has $\kappa =1/[\sqrt{2}(2-s_b)]$, $0\le s_b<2$. Choosing orientation avoids confusing the two points at the same unoriented distance.

Differentiate the unit tangent vector with $d/ds=v^{-1}d/du$ and $v^{\prime }=({\mathbf{x}}^{\prime }\cdot {\mathbf{x}}^{\prime \prime })/v$. This givesThe curvature of a space curve is $\kappa =|d\mathbf{t}/ds|$. Squaring the displayed expression and simplifying givesThe last equality is the cross product norm identity. Equivalently, the hinted vector triple product rewrites $d\mathbf{t}/ds=-{\mathbf{x}}^{\prime }\times ({\mathbf{x}}^{\prime }\times {\mathbf{x}}^{\prime \prime })/v^4$.

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.

The directional derivative along the specified vector field givesUse the magnitude formula for the curvature of an integral curve of a vector field:It vanishes on the plane $z=0$ and on the $z$ axis away from the origin, consistent with straight field lines there. At the origin the vector field vanishes, so it fails the nowhere-zero hypothesis and this field-line curvature is not defined.

As an independent geometric check, an integral curve of a vector field satisfies $\dot{x}=x$, $\dot{y}=y$, $\dot{z}=-z$, giving $(x,y,z)=(a{e}^u,b{e}^u,c{e}^{-u})$. Its velocity is $(x,y,-z)$ and acceleration $(x,y,z)$; the ordinary curvature of a space curve formula $|r^{\prime }\times r^{\prime \prime }|/|r^{\prime }{|}^3$ gives the same expression.

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.

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

The fundamental theorem of regular space curves states that smooth functions $\kappa (s)>0$ and $\tau (s)$ determine a unit-speed regular curve in ${\mathbb{R}}^3$ with that curvature and torsion, uniquely up to a proper Euclidean motion of Euclidean three-space. Existence follows by solving the Frenet-Serret formulas for an oriented orthonormal frame and then integrating its tangent vector.

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

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 $N$.