Proving the Relationship between Velocity and Curvature in a Vector Equation

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SUMMARY

The discussion focuses on proving the relationship between velocity and curvature in a vector equation, specifically demonstrating that V X A = k(ds/dt)³B, where B represents the binormal vector. The participants reference the equation a(t) = (d²s/dt²)T + K(ds/dt)²N as a foundational element in their proof. The conversation highlights the importance of using the acceleration vector a(t) rather than solely relying on the velocity vector v(t) for the proof. The clarification of V as (ds/dt)T is also emphasized as a critical component in understanding the relationship.

PREREQUISITES
  • Understanding of vector calculus and vector equations
  • Familiarity with curvature and its mathematical representation
  • Knowledge of the Frenet-Serret formulas, including T, N, and B vectors
  • Proficiency in taking derivatives and performing cross products in vector analysis
NEXT STEPS
  • Study the derivation of the Frenet-Serret formulas in detail
  • Learn about the geometric interpretation of curvature in vector functions
  • Explore advanced applications of cross products in physics and engineering
  • Investigate the relationship between acceleration, velocity, and curvature in motion
USEFUL FOR

This discussion is beneficial for students studying advanced calculus, physics majors focusing on mechanics, and anyone interested in the mathematical foundations of motion and curvature in vector fields.

Baumer8993
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Homework Statement


This is part of the proof for curvature.

Prove V X A = k(ds/dt)3B

B is the the binormal vector. I have seen something in the proof with a T X N which I know is B.

Homework Equations


I have already proven

a(t) = (d2s/dt2)T + K(ds/dt)2N


The Attempt at a Solution


I have seen some things where you start with v(t), and take the derivative then take a cross product, but I am not sure if this is how my professor wants us to do it. He wants us to use the a(t) part from above to prove it.
 
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Hi Baumer8993! :smile:

Isn't V = (ds/dt)T ? :confused:
 

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