What Are Tangent Planes and Vertical Curvature in Differential Geometry?

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SUMMARY

The discussion focuses on the concepts of tangent planes and vertical curvature in differential geometry, specifically for the surface defined by the function z = 2x² - y². The tangent plane at the point p = (-1, 2, -2) is expressed as Tp = {(-1, 2, -2) + t(1, 0, -4) + s(0, 1, -4)}. A basis for this tangent plane is confirmed to be the vectors (1, 0, -4) and (0, 1, -4). Additionally, the vertical curvature at point p is determined to be zero in the direction of a specific non-zero vector w, which needs to be identified.

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  • Learn about vertical curvature and its implications in geometry
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Consider the surface S defined as the graph of a function z = 2x ^ 2 - y ^ 2
i) find a basis of the tangent plane Tp surface S at the point p = (-1,2, -2)
ii) find a non-zero vector w in Tp with the property that the vertical curvature at point p in the direction of vector w is zero


for the i) question i found that the tangent plane has this form: Tp={(-1,2,-2)+t(1,0-4)+s(0,1,-4)}
but how i will find a basis of this tangent plane? maybe is <(1,0,4),(0,1,-4)> ?

and for the ii) how i will find this non zero vector?
 
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