Geodesic curvature

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Clairaut’s theorem. How to prove that geodesic curvature parallel Radius of the surface of revolution (i. e the value inversed to geodesic curvature) is equal to tangential line segment and meridian closed between the point of tangency and surface axis.enter image description here



I know that I should use the Clairaut's theorem.But I have no idea how to prove it. Can you help me to prove this?










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    1












    $begingroup$


    Clairaut’s theorem. How to prove that geodesic curvature parallel Radius of the surface of revolution (i. e the value inversed to geodesic curvature) is equal to tangential line segment and meridian closed between the point of tangency and surface axis.enter image description here



    I know that I should use the Clairaut's theorem.But I have no idea how to prove it. Can you help me to prove this?










    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      Clairaut’s theorem. How to prove that geodesic curvature parallel Radius of the surface of revolution (i. e the value inversed to geodesic curvature) is equal to tangential line segment and meridian closed between the point of tangency and surface axis.enter image description here



      I know that I should use the Clairaut's theorem.But I have no idea how to prove it. Can you help me to prove this?










      share|cite|improve this question









      $endgroup$




      Clairaut’s theorem. How to prove that geodesic curvature parallel Radius of the surface of revolution (i. e the value inversed to geodesic curvature) is equal to tangential line segment and meridian closed between the point of tangency and surface axis.enter image description here



      I know that I should use the Clairaut's theorem.But I have no idea how to prove it. Can you help me to prove this?







      differential-geometry






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      asked Jan 1 at 16:59









      Gera SlanovaGera Slanova

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