Proving the Relationship Between Chord Length and Curve Type | Homework Question

  • Thread starter Thread starter akoska
  • Start date Start date
  • Tags Tags
    Curves Plane
Click For Summary
SUMMARY

The discussion focuses on proving that if the chord length ||f(s)-f(t)|| depends solely on |s-t|, then the function f must represent a line or a circle. Participants emphasize the need to demonstrate that |f'(t)| is a constant, which can be achieved by differentiating the expression and analyzing the curvature of f(t). The key approach involves defining a differentiable function a such that a(|s-t|) = ||f(s)-f(t)||, and exploring the implications of this definition on the behavior of f.

PREREQUISITES
  • Understanding of differential calculus and derivatives
  • Familiarity with the concepts of curvature and its mathematical implications
  • Knowledge of limits and continuity in functions
  • Basic understanding of geometric properties of lines and circles
NEXT STEPS
  • Study the properties of differentiable functions and their derivatives
  • Research the relationship between curvature and the shape of functions
  • Explore the concept of one-sided derivatives and their applications
  • Learn about the implications of constant derivatives in geometric contexts
USEFUL FOR

Mathematics students, particularly those studying calculus and geometry, as well as educators seeking to clarify concepts related to function behavior and curvature.

akoska
Messages
22
Reaction score
0

Homework Statement



I want to show that if the chhord length ||f(s)-f(t)|| depends only on |s-t| then the f is part of a line or circlle. f may not be regular or unit speed.

Homework Equations





The Attempt at a Solution



I'm trying to differentiete it and taking ||f'(t)||, but I really need help, because it's not working for me.
 
Physics news on Phys.org
You should be able to show |f'(t)| is a constant. For the rest of the problem I would think about trying to express the curvature of f(t) in terms of things like |f(s)-f(t)| as s->t and hence argue that it is also a constant.
 
How do I show |f'(t)| is a constant? I get to the part: |f'(t)| = lim(dt->0) |a(dt)/dt| where a is the function ||f(s)-f(t)|| =a(|s-t|)... and then I'm stuck
 
Divide both sides of your last expression by |s-t| and let s->t. I would conclude |f'(t)|=a'(0) for an appropriately defined a. Or this "|f'(t)| = lim(dt->0) |a(dt)/dt|". That looks to me like the definition of a'(0).
 
Last edited:
Yes, but I think the 'appropriately defined a' is the catch. a can be any differentiable function, adn although i can show that |f'(t)| is a constant for certain a, how can I show it for all a?
 
There's no 'catch'. You've defined a(|s-t|)=|f(s)-f(t)|. That defines a(x) for non-negative x. The only 'appropriately defined' case is whether you bother to define a(x) for negative x or keep referring to one sided derivatives. It's really nothing.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K