What is the Surface of Revolution with Constant Curvature +1?

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SUMMARY

The discussion focuses on finding a surface of revolution with a constant Gauss curvature (K) of +1 that is not a sphere. The surface is parametrized using the equation \(\psi (t, \theta ) = ( x(t), y(t) \cos \theta , y(t) \sin \theta )\). The key equation for Gauss curvature is given as \(K = \frac{x' (x'' y' - x' y'')}{y(x'^2 + y'^2)^2}\). The challenge lies in the fact that surfaces of constant Gaussian curvature are locally isometric to spheres, leading to the conclusion that a non-unit-speed curve \(\alpha (t)\) may provide a solution.

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  • Understanding of Gaussian curvature and its implications in differential geometry.
  • Familiarity with surface parametrization techniques.
  • Knowledge of the properties of curves, specifically unit-speed curves.
  • Basic proficiency in calculus, particularly derivatives and their applications in geometry.
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  • Research the properties of surfaces of constant Gaussian curvature.
  • Explore the implications of non-unit-speed curves in differential geometry.
  • Study the relationship between curvature and surface isometry.
  • Investigate alternative parametrizations for surfaces of revolution.
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mooshasta
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Homework Statement



I'm trying to find a surface of revolution with Gauss curvature K of +1 at all points, which doesn't lie in a sphere.


Homework Equations



The surface is parametrized as \psi (t, \theta ) = ( x(t), y(t) cos \theta , y(t) sin \theta )

I have the equation
<br /> K = \frac{x&#039; (x&#039;&#039; y&#039; - x&#039; y&#039;&#039;)}{y(x&#039;^2 + y&#039;^2)^2}<br />


The Attempt at a Solution



I am thinking it has to do with the curve \alpha (t) = (x(t),y(t)) not having unit speed, but I am kind of stuck as to where to go from there.


Thanks!
 
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Is there one? The only way out I can think of is to make it disconnected. A surface of constant guassian curvature is locally isometric to a sphere.
 
I know that for a unit-speed \alpha (t), the equation reduces to K = \frac{-y&#039;&#039;}{y}, which does clearly represent a sphere.

The way the question is worded on my homework seems to point to the fact that that reduction only applies to unit-speed curves, which is why I think that perhaps an \alpha (t) that doesn't have unit-speed perhaps can give a surface of revolution with constant curvature +1 that isn't a sphere... but maybe there's another "gimmick" that I'm overlooking...

Anyways, thanks for your help!
 

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