PlanetMath: curvature determines the curve

(more info)

Math for the people, by the people.

Main Menu

sections Encyclopædia
Papers
Books
Expositions

meta Requests (200)
Orphanage
Unclass'd
Unproven (490)
Corrections (6)
Classification

talkback Polls
Forums
Feedback
Bug Reports

downloads Snapshots
PM Book

information News
Docs
Wiki
ChangeLog
TODO List
Legalese
About

Entry average rating: No information on entry rating

curvature determines the curve

(Theorem)

The curvature of plane curve determines uniquely the form and size 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

$\displaystyle \kappa = k(s)$

(1)

between their curvature $\kappa$ and the arc length

. All these curves are congruent.

Proof. Suppose that a curve satisfies the condition (1). Let the value correspond the point of this curve. We chose 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

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

for the direction angle $\theta$ of the tangent of

is valid in this coordinate system; the initial condition is

$\displaystyle \theta = 0\;\; \mathrm{when}\;\; s = 0.$

Thus we get

$\displaystyle \theta = \int_0^s k(t)\,dt := \vartheta(s),$

(2)

which implies

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

(3)

Since

when

, we obtain

$\displaystyle x = \int_0^s \cos{\vartheta(t)}\,dt,\,\,\,y = \int_0^s \sin{\vartheta(t)}\,dt.$

(4)

Thus the function $s\mapsto k(s)$ determines uniquely these functions

and

of the parameter

, and (4) represents a curve with definite form and size.

The above reasoning shows that every curve which satisfies (1) is congruent with the curve (4).

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

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

(5)

The equations (3) and (5) then yield

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

i.e. the curvature of (4), according the parent entry, satisfies

$\displaystyle \left\vert \begin{matrix}x' & y' \cr x'' & y'' \end{matrix} \right\vert = k(s).$

Thus the proof is settled.

Bibliography

1: ERNST LINDELÖF: Differentiali- ja integralilasku ja sen sovellutukset I. WSOY. Helsinki (1950).

"curvature determines the curve" is owned by pahio.

(view preamble)

Keywords:

plane curve

This object's parent.

Log in to rate this entry.
(view current ratings)

Cross-references: parameter, function, implies, initial condition, coordinate system, angle, normal vectors, positive, normal, tangent, plane, origin, point, arc length, equation, real function, continuous, curve, plane curve
There is 1 reference to this entry.

This is version 4 of curvature determines the curve, born on 2007-05-09, modified 2008-03-04.
Object id is 9353, canonical name is CurvatureDeterminesTheCurve.
Accessed 566 times total.

Classification:

AMS MSC:

53A04 (Differential geometry :: Classical differential geometry :: Curves in Euclidean space)

Pending Errata and Addenda

None.

Discussion

forum policy

No messages.

Interact