Parametrizing a Hyperbola to Find Unit Tangent & Normal Vectors

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Discussion Overview

The discussion revolves around the parametrization of the hyperbola defined by the equation y² - x² = 1 (for y > 0). Participants explore methods to express the hyperbola in vector form, calculate the unit tangent vector, and derive the unit normal vector and curvature vector as functions of a parameter.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant suggests finding functions f(t) and g(t) such that f(t)² - g(t)² = 1 as a starting point for parametrization.
  • Another participant shares an attempt to set x = t and y = sqrt(1 + t²), expressing frustration over the complexity of the resulting expressions.
  • A participant references the unit circle equation x² + y² = 1 as a simpler case and questions whether a similar approach can be applied to the hyperbola's equation.
  • There is an inquiry about the deadline for the problem, indicating a potential urgency in the discussion.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the best method for parametrization, and multiple approaches are being explored without resolution.

Contextual Notes

Some assumptions about the parameterization methods and the complexity of the resulting equations remain unaddressed, and there are unresolved mathematical steps in the proposed approaches.

Who May Find This Useful

Students and educators interested in hyperbolic functions, vector calculus, and parametrization techniques may find this discussion relevant.

soccer*star
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Consider the hyperbola y^2-x^2=1 (y>0)
a.) Find a parameterization for the curve and write it in vector form, R(t)
(b) Calculate the unit tangent vector as a function of the parameter.
(c) Calculate the unit normal vector and the curvature vector as a function of the parameter.
 
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What did you try already??

You must find function f and g such that

[tex]f(t)^2-g(t)^2=1[/tex]
 


I tried to set x=t and y= sqrrt(1+t^2) ...it comes out nasty and ugly, so ugly that i didn't even finish it..i'm not sure if there's a better way to do it.
 


soccer*star, is this problem due tomorrow by any chance?
 


If you would have [itex]x^2+y^2=1[/itex], then there's an easy choice:

[tex]\sin^2(t)+\cos^2(t)=1[/tex]

But now you have [itex]x^2-y^2=1[/itex]. Can you do something similar?
 

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