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.