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I am studying differential geometry myself from Do carmo and i didn't understand the question :

show that if a surface is tangent to a plane along a curve , then the points of this curve are either parabolic or planar .

At the question i didn't understand the sentence ' surface is tangent to a plane along a curve ' Please firstly help me about it and later maybe hints for solution

bytrz
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Let $\pi$ be your plane, and $C$ some curve contained in $\pi$. The meaning of your problem would be that the surface $S$ is tangent to $\pi$ at all points $x\in C$. In other words, the surface touches the plane along the curve $C$ (in a tangential way).

For instance, the plane could be the $xy$-plane, and your surface could be a torus that rests on this plane. Then the point of contact between the surface and the plane is a circle.

Mankind
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  • thanks for explenation . On the other hand i cant see how i can use this ? i think the points on the curve already on the plane and so planar ? – bytrz Jun 02 '15 at 20:19
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    The points on the surface that touches the plane need not be planar. Take the torus example again. All points, where it touches the plane, has a non-zero principal curvature. – Mankind Jun 02 '15 at 20:48
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That a surface $S$ is tangent to a given plane $\pi$ along a curve $\gamma$ means that the tangent plane of $S$ at points of $\gamma$ is the given plane $\pi$. So the tangent plane of $S$ is the same along $\gamma$. Then the derivative of a unit normal vector field of $S$ along $\gamma$ is zero. So the shape operator of $S$ at points of $\gamma$ has $\gamma'$ in its kernel. Finally, keep in mind that parabolic or planar points are points of $S$ where the shape operator has kernel i.e. points where the Gauss curvature vanish.

Holonomia
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  • you said ' the derivative of a unit normal vector field of S along γ is zero ' because unit normal vector is constant along curve yes ? – bytrz Jun 03 '15 at 09:05
  • Yes, indeed. The unit normal is constant along $\gamma$ since it is the unit normal of the plane $\pi$ which is constant along the whole $\pi$. – Holonomia Jun 03 '15 at 11:40