Understanding Unit Tangent to a Curve in PDEs

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SUMMARY

The unit tangent to a curve defined by the equation y = Y(X) is expressed as (i + Y'(X)j) / [(1 + [Y'(X)]^2)^(0.5)]. This formulation arises from differentiating the position vector \(\vec{r} = x\vec{i} + Y(x)\vec{j}\) with respect to x, resulting in the tangent vector \(\vec{r}' = \vec{i} + Y'(x)\vec{j}\). The length of this tangent vector is calculated as |\(\vec{r}'\)| = \(\sqrt{1 + [Y'(x)]^2}\), which normalizes the tangent vector to yield the unit tangent vector.

PREREQUISITES
  • Understanding of vector calculus
  • Familiarity with derivatives and differentiation
  • Knowledge of parametric equations
  • Basic concepts of partial differential equations (PDEs)
NEXT STEPS
  • Study the derivation of unit tangent vectors in vector calculus
  • Explore the applications of unit tangent vectors in second order PDEs
  • Learn about parametric curves and their properties
  • Investigate the relationship between tangent vectors and curvature in differential geometry
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Students and professionals in mathematics, particularly those studying calculus, differential equations, and physics, will benefit from this discussion on unit tangents in curves.

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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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If y= Y(x), you can write the "position vector" of a point on the curve as [itex]\vec{r}= x\vec{i}+ y\vec{j}= x\vec{i}+ Y(x)\vec{j}[/itex]. Differentiating that vector with respect to x gives [itex]\vec{r}'= \vec{i}+ Y'(x)\vec{j}[/itex] as a vector tangent to the curve. It's length is, of course, [itex]|\vec{r}'|= \sqrt{1^2+ Y'(x)^2}= (1+ Y'(x)^2)^{1/2}[/itex]. Dividing [itex]\vec{i}+ Y'(x)\vec{j}[/itex] by that, which is the denominator in what you give, makes it a unit tangent vector.
 

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