Trouble with the unit normal vector for stokes theorem

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SUMMARY

The discussion centers on finding the unit normal vector for Stokes' Theorem using the equation x² + 2y² = 1. The user derives the minimum distance from the origin by substituting x² into the distance formula, resulting in the points (±1, 0) as the closest points. The calculations involve differentiating the distance squared and setting the derivative to zero, leading to the conclusion that the minimum distance is 1. The user later acknowledges that the post was mistakenly submitted.

PREREQUISITES
  • Understanding of Stokes' Theorem
  • Familiarity with distance formulas in Cartesian coordinates
  • Basic calculus, specifically differentiation
  • Knowledge of implicit functions and their properties
NEXT STEPS
  • Study the application of Stokes' Theorem in vector calculus
  • Learn about implicit differentiation techniques
  • Explore the geometric interpretation of distance in coordinate systems
  • Investigate the properties of unit normal vectors in multivariable calculus
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Students and professionals in mathematics, particularly those studying vector calculus, as well as educators looking for examples of applying Stokes' Theorem in practical scenarios.

hivesaeed4
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We're given x^2+2*y^2=1.
so x^2=1-2y^2

now using distance formula
d^2=x^2+y^2
since x^2=1-2y^2, substituting it in the distance formula we get:
d^2=1-2y^2+y^2=1-y^2;
differentiating and then setting the eq to 0 we get;
0=-4y
or y=0. now x^2=1-2y^2=1
so x=+-1
so point having min distance form origin is (+-1,0)

using the distance formula now
d^2=x^2+y^2
d=sqrt(1+0)=1
 
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Sorry. Wrong post. I intended to preview it but posted it by mistake.
 

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