Understanding Second Order PDEs: Explaining the Unit Tangent to a Curve

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Discussion Overview

The discussion revolves around the derivation and understanding of the unit tangent vector to a curve described by the function y=Y(X), particularly in the context of second order partial differential equations (PDEs). Participants seek clarification on the formula presented in lecture notes and its implications.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant expresses confusion about the origin of the unit tangent vector formula and requests an explanation.
  • Another participant notes that the tangent vector to a parameterized curve (x, y(x)) is (1, y'(x)) and discusses its length.
  • A different participant points out that the formula provides a tangent vector but does not specify the coordinates where the tangent touches the curve, emphasizing the need for the original function.
  • Further clarification is provided by another participant, who suggests parameterizing the curve and explains how the derivative Y'(X) relates to the tangent vector's formulation.
  • One participant draws an analogy with the gradient operator, questioning whether the discussion pertains to a one-dimensional case rather than a two-dimensional one, and relates it to a broader context of position vectors.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the clarity of the original formula or its implications, as multiple interpretations and clarifications are presented without resolution of the initial confusion.

Contextual Notes

Some assumptions about parameterization and the relationship between derivatives are not fully explored, and the discussion does not resolve the implications of the gradient analogy presented.

coverband
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Hi

In my lecturer's notes he describes the unit tangent to a curve y=Y(X) as

(i + Y'(X)j)/[(1+[Y'(X)]^2)^(0.5)]

in an introduction to second order PDEs

I'm a bit confused by this. Where did it come from?

Can anyone explain

Thanks
 
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A curve [itex](x, y(x))[/itex] parameterized by [itex]x[/itex] has tangent vectors [itex](1, y'(x))[/itex] with lengths [itex]\sqrt{1+(y')^2}[/itex].
 
Also, this function would give you a unit vector pointing in the positive x direction. Notice that the function will tell you only the tangent vector. It will not tell you at what x and y coordinates the actual tangent touches the curve. You need the original function for that.
 
coverband said:
In my lecturer's notes he describes the unit tangent to a curve y=Y(X) as

(i + Y'(X)j)/[(1+[Y'(X)]^2)^(0.5)]

Where did it come from?

Hi coverband! :smile:

It looks more logical if you parametrise the curve: X = x(t), Y = y(t).

Then your Y'(X) is dY/dX = y'/x', and the formula (after multiplying top and bottom by x') becomes …

(x',y')/√(x'2 + y'2) …

so the tangent is parallel to the gradient (x',y'),

and to make it a unit vector you divide by its magnitude. :smile:
 
Let's clarify the analogy with the gradient operator. This would actually be a one dimensional case of the gradient operator, not a 2D case correct?

y(x) = Grad[Y(x)] / Abs[Grad[Y(x)]]

Where in general, x is a position vector of arbitrary dimension. This of course goes much beyond the original question, but I want to check for myself.
 

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