curvature determines the curve

The curvature (http://planetmath.org/CurvaturePlaneCurve) of plane curve determines uniquely the form and of the curve, i.e. one has the

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

$\kappa =k(s)$ (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

$$\theta =\mathrm{\hspace{0.33em}0}\mathit{\hspace{1em}}\text{when}\mathit{\hspace{1em}}s=\mathrm{\hspace{0.33em}0}.$$

Thus we get

$\theta ={\displaystyle {\int }_0^s}k(t)𝑑t:=\vartheta (s),$ (2)

which implies

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

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

$x={\displaystyle {\int }_0^s}\cos \vartheta (t)𝑑t,y={\displaystyle {\int }_0^s}\sin \vartheta (t)𝑑t.$ (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  $d{s}^2=d{x}^2+d{y}^2$  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),

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

The equations (3) and (5) then yield

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

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

Thus the proof is settled.

Title	curvature determines the curve
Canonical name	CurvatureDeterminesTheCurve
Date of creation	2016-02-22 16:14:25
Last modified on	2016-02-22 16:14:25
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	11
Author	pahio (2872)
Entry type	Theorem
Classification	msc 53A04
Synonym	fundamental theorem of plane curves
Related topic	FundamentalTheoremOfSpaceCurves
Related topic	ErnstLindelof