How to Find a Tangent Plane on a Surface with Positive Z Values?

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SUMMARY

The discussion focuses on finding a parametrization of the surface defined by the equation x3 + 3xy + z2 = 2, with the constraint that z > 0. A tangent plane can exist at a point within the closure of the surface, which allows for the possibility of finding a tangent plane at the point (1, 1/3, 0). The participants discuss the challenges of plugging in values for x and y and the need for a proper parametrization to resolve these issues.

PREREQUISITES
  • Understanding of surface parametrization in multivariable calculus
  • Familiarity with the concept of tangent planes in differential geometry
  • Knowledge of the closure of a set in topology
  • Basic algebraic manipulation skills for working with equations
NEXT STEPS
  • Study the concept of surface parametrization in multivariable calculus
  • Learn how to derive tangent planes from implicit functions
  • Explore the closure of sets in topology and its implications for surfaces
  • Practice algebraic techniques for manipulating and solving equations involving multiple variables
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Students and professionals in mathematics, particularly those studying calculus and differential geometry, as well as educators looking for examples of surface parametrization and tangent plane derivation.

Tony11235
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The problem is find a parametrization of the surface x^3 + 3xy +z^2 = 2, z > 0, and use it to find the tangent plane at the point x=1, y=1/3, z=0. How is this possible when z > 0? I found a parametrization but when I plug the point in the x and the y places are undefined.
 
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(a) A tangent plane can exist at a point that is in the closure (it's been 30 years, since topology class, is that the right term?) of the surface, so it is possible.

(b) What is the parameterization you found? My inclination would be to define y in terms of x and z but sometimes I guess wrong.

Carl
 
Nevermind. I think I know what to do now. Thanks anyway.
 

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