Showing that some curve is a circle

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SUMMARY

The discussion focuses on demonstrating that if a unit-speed curve \(\gamma(t)\) has all its normals passing through a specific point, then the trace of \(\gamma(t)\) is part of a circle. The user proposes that the relationship \(\gamma(t) - a = s(t) N(t)\) holds for all \(t\), where \(s(t)\) is a real-valued function and \(N(t)\) is the normal vector. The goal is to establish that \(|s(t)|\) remains constant across all \(t\), leading to a system of ordinary differential equations (ODEs) with a straightforward solution.

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  • Understanding of differential geometry concepts, particularly curves and normals.
  • Familiarity with unit-speed curves and their properties.
  • Knowledge of ordinary differential equations (ODEs) and their solutions.
  • Basic proficiency in vector calculus, especially regarding normal vectors.
NEXT STEPS
  • Study the properties of unit-speed curves in differential geometry.
  • Explore the derivation and solutions of ordinary differential equations (ODEs).
  • Learn about the geometric interpretation of normal vectors in curves.
  • Investigate the relationship between curvature and the shape of curves, particularly circles.
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Students and researchers in mathematics, particularly those focusing on differential geometry, as well as anyone interested in the geometric properties of curves and their applications in physics and engineering.

Werg22
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I'm trying out some exercises in differential geoemtry and came across this one:

If \gamma(t) is unit-speed, and that all its normals pass through a given point, show that the trace of \gamma(t) is part of a circle.

My solution so far:

Let a be a point where all the normals pass, then we know that \gamma(t) - a = s(t) N(t) \ \forall t, where s(t) is a real-valued function.

But where should I take it from there? All I know is that ultimately I want to show that |s(t)| is constant for all t.
 
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You might as well assume that you start at (1,0) with initial velocity (0,1), and that all your normals pass through zero. You get a system of ODEs with a very simple solution :)
 

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