Motion on a Curve: Velocity Vector Always Tangent

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Discussion Overview

The discussion centers on understanding why the velocity vector is always tangent to a curve, exploring definitions and mathematical proofs related to this concept. It involves both theoretical and mathematical reasoning.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about proving that the velocity vector is tangent to the curve, despite feeling that it makes intuitive sense.
  • Another participant suggests comparing definitions, specifically mentioning the tangent vector derived from the parameterization of the curve.
  • A participant explains that for a curve parametrized by \(\vec{M}(t)=(x(t),y(t))\), the tangent vector \(\vec{T}(t)=(\dot{x}(t),\dot{y}(t))\) can be shown to be parallel to the position vector through calculations involving normal vectors and dot products.
  • One participant questions the parallelism of the position vector \(\vec{M}(t)\) and the tangent vector \(\vec{T}(t)\), seeking clarification on the definition of tangency.
  • Another participant proposes that the confusion may stem from a lack of a clear definition of what it means for a vector to be tangent to a curve, suggesting that tangency is defined by parallelism to the derivative at a point.
  • A later reply acknowledges the oversight regarding the definition of tangent vectors, indicating a realization about the foundational concepts involved.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the proof of the tangency of the velocity vector to the curve, and multiple viewpoints regarding the definitions and approaches to the problem are presented.

Contextual Notes

There are unresolved aspects regarding the definitions of tangency and the mathematical steps necessary to prove the claims made about the relationship between the velocity vector and the curve.

Swapnil
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Why is the velocity vector always tangent to the curve?


P.S.: I know it makes sense!
But I can't prove it and its driving me crazy!
 
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just compare the definitions
if your curve is given by r(t)
T=r'(t)/|r'(t)|
v=r'(t)
 
In the case your curve is parametrized by [itex]\vec{M}(t)=(x(t),y(t))[/itex], then the tangent vector is given by [itex]\vec{T}(t)=(\dot{x}(t),\dot{y}(t))[/itex]. Now, you can do two things to prove that [itex]\vec{M}[/itex] and [itex]\vec{T}[/itex] are parallel. One is to calculate the normal vector to the curve and then the dot product, proving that they are orthogonal. The second one is to calculate the cross product between the curve and the tangent.

This is really simple in [itex]\mathbb{R}^2[/itex], so i'll recommend you to prove it for [itex]\mathbb{R}^n[/itex] as well.

(If you are working with functions in [itex]\mathbb{R}[/itex], then [itex]\vec{M}(t)=(t,f(t))[/itex])
 
Last edited:
AiRAVATA said:
In the case your curve is parametrized by [itex]\vec{M}(t)=(x(t),y(t))[/itex], then the tangent vector is given by [itex]\vec{T}(t)=(\dot{x}(t),\dot{y}(t))[/itex]. Now, you can do two things to prove that [itex]\vec{M}[/itex] and [itex]\vec{T}[/itex] are parallel.

Why would [itex]\vec{M}(t)[/itex] and [itex]\vec{T}(t)[/itex] be parallel to each other? [itex]\vec{M}(t)[/itex] is the position vector correct?
 
It is my guess that you're trying to prove that the velocity vector is always tangent to the curve, but you don't have a clear idea (i.e. a definition!) of what it means for a vector to be tangent to a curve.

Like you noted, it makes sense that the velocity is tangent to the curve. So that is how we decide to define "tangency to the curve" (see matness). We'll say that some vector is tangent to the curve at some point if that vector is parallel to the derivative at that point.

With that definition, your problem is more than trivial.
 
OH Yes! I didn't even think about how tangent vectors were defined! How foolish of me... :smile:
 

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