# curvature determines the curve

Theorem.  If  $s\mapsto k(s)$  is a continuous real function, then there exists always plane curves satisfying the equation

 $\displaystyle\kappa\;=\;k(s)$ (1)

between their curvature $\kappa$ and the arc length $s$.  All these curves are congruent   (http://planetmath.org/Congruence).

Proof.  Suppose that a curve $C$ satisfies the condition (1).  Let the value  $s=0$  correspond to the point $P_{0}$ of this curve.  We choose $O$ as the origin of the plane.  The tangent  and the normal of $C$ in $O$ are chosen as the $x$-axis and the $y$-axis, with positive directions the directions of the tangent and normal vectors of $C$, respectively.  According to (1) and the definition of curvature, the equation

 $\frac{d\theta}{ds}\;=\;k(s)$

for the direction angle $\theta$ of the tangent of $C$ is valid in this coordinate system  ; the initial condition  is

 $\theta\;=\;0\quad\mbox{when}\quad s\;=\;0.$

Thus we get

 $\displaystyle\theta\;=\;\int_{0}^{s}k(t)d\!t\;:=\;\vartheta(s),$ (2)

which implies

 $\displaystyle\frac{dx}{ds}\;=\;\cos{\vartheta(s)},\qquad\frac{dy}{ds}\;=\;\sin% {\vartheta(s)}.$ (3)

Since  $x=y=0$  when  $s=0$, we obtain

 $\displaystyle x\;=\;\int_{0}^{s}\cos{\vartheta(t)}\,dt,\qquad y\;=\;\int_{0}^{% s}\sin{\vartheta(t)}\,dt.$ (4)

Thus the function  $s\mapsto k(s)$  determines uniquely these functions $x$ and $y$ of the parameter $s$, and (4) represents a curve with definite form and .

The above reasoning shows that every curve which satisfies (1) is congruent (http://planetmath.org/Congruence) with the curve (4).

We have still to show that the curve (4) satisfies the condition (1).  By differentiating (http://planetmath.org/HigherOrderDerivatives) the equations (4) we get the equations (3), which imply  $(\frac{dx}{ds})^{2}+(\frac{dy}{ds})^{2}=1$,  or  $ds^{2}=dx^{2}+dy^{2}$  which means that the parameter $s$ represents the arc length of the curve (4), counted from the origin.  Differentiating (3) we get, because  $\vartheta^{\prime}(s)=k(s)$  by (2),

 $\displaystyle\frac{d^{2}x}{ds^{2}}\;=\;-k(s)\sin{\vartheta(s)},\qquad\frac{d^{% 2}y}{ds^{2}}\;=\;k(s)\cos{\vartheta(s)}.$ (5)

The equations (3) and (5) then yield

 $\frac{dx}{ds}\frac{d^{2}y}{ds^{2}}-\frac{dy}{ds}\frac{d^{2}x}{ds^{2}}\;=\;k(s),$

i.e. the curvature of (4), according the parent entry (http://planetmath.org/CurvaturePlaneCurve), satisfies

 $\left|\begin{matrix}x^{\prime}&y^{\prime}\cr x^{\prime\prime}&y^{\prime\prime}% \end{matrix}\right|\;=\;k(s).$

Thus the proof is settled.

## References

• 1 Ernst Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset I.  WSOY. Helsinki (1950).
Title curvature determines the curve CurvatureDeterminesTheCurve 2016-02-22 16:14:25 2016-02-22 16:14:25 pahio (2872) pahio (2872) 11 pahio (2872) Theorem msc 53A04 fundamental theorem of plane curves  FundamentalTheoremOfSpaceCurves ErnstLindelof