curvature determines the curve
Theorem. If is a continuous real function, then there exists always plane curves satisfying the equation
Proof. Suppose that a curve satisfies the condition (1). Let the value correspond to the point of this curve. We choose as the origin of the plane. The tangent and the normal of in are chosen as the -axis and the -axis, with positive directions the directions of the tangent and normal vectors of , respectively. According to (1) and the definition of curvature, the equation
Thus we get
Since when , we obtain
Thus the function determines uniquely these functions and of the parameter , 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 , or which means that the parameter represents the arc length of the curve (4), counted from the origin. Differentiating (3) we get, because by (2),
The equations (3) and (5) then yield
i.e. the curvature of (4), according the parent entry (http://planetmath.org/CurvaturePlaneCurve), satisfies
Thus the proof is settled.
- 1 Ernst Lindelöf: Differentiali- ja integralilasku ja sen sovellutukset I. WSOY. Helsinki (1950).
|Title||curvature determines the curve|
|Date of creation||2016-02-22 16:14:25|
|Last modified on||2016-02-22 16:14:25|
|Last modified by||pahio (2872)|
|Synonym||fundamental theorem of plane curves|