# 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

$\theta ={\displaystyle {\int}_{0}^{s}}k(t)\mathit{d}t:=\vartheta (s),$ | (2) |

which implies

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

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

$x={\displaystyle {\int}_{0}^{s}}\mathrm{cos}\vartheta (t)\mathit{d}t,y={\displaystyle {\int}_{0}^{s}}\mathrm{sin}\vartheta (t)\mathit{d}t.$ | (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),

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

$$\left|\begin{array}{cc}\hfill {x}^{\prime}\hfill & \hfill {y}^{\prime}\hfill \\ \hfill {x}^{\prime \prime}\hfill & \hfill {y}^{\prime \prime}\hfill \end{array}\right|=k(s).$$ |

Thus the proof is settled.

## References

- 1 Ernst Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset I. WSOY. Helsinki (1950).

Title | curvature determines the curve |
---|---|

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 |