Position vector perpendicular to tangent vector yields a sphere

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SUMMARY

The discussion centers on the mathematical proof that if a curve's position vector r(t) is always perpendicular to its tangent vector r'(t), then the curve must lie on a sphere centered at the origin. The key equations involved include the relationship between the position vector and its slope, specifically -1/r'(t), and the equation of a circle in two dimensions, x² + y² = 1. Participants emphasize the necessity of using vector equations to approach the problem effectively.

PREREQUISITES
  • Understanding of vector calculus, specifically vector equations.
  • Knowledge of the properties of perpendicular vectors.
  • Familiarity with the equation of a sphere in three-dimensional space.
  • Basic skills in solving mathematical proofs involving curves.
NEXT STEPS
  • Study the properties of vector functions and their derivatives.
  • Learn how to derive the equation of a sphere from vector equations.
  • Explore the implications of curves being confined to specific geometric shapes.
  • Investigate the relationship between curvature and the position vector in vector calculus.
USEFUL FOR

Students of mathematics, particularly those studying calculus and vector analysis, as well as educators looking to enhance their understanding of geometric properties of curves.

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Homework Statement


If a curve has the property that the position vector r(t) is always perpendicular to the tangent vector r'(t), show that the curve lies on a sphere with center the origin.


Homework Equations



-1/r'(t)= slope of position vector

x^{2}+y^{2}=1

The Attempt at a Solution


Not really sure where to begin for this one.
 
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hi physicsidiot1! :smile:
physicsidiot1 said:
… the position vector r(t) is always perpendicular to the tangent vector r'(t) … the curve lies on a sphere with center the origin.

this is a vector problem, so it needs a vector solution :wink:

start by writing the question out in two vector equations …

what do you get? :smile:
 

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