Consider the surface z=(x/2)^2-(y/3)^2. Compute the tangent plane at each point. Find the point such that the normal vector is vertical at that point.

Deborah Proctor 2022-10-17 Answered
Consider the surface z = ( x 2 ) 2 ( y 3 ) 2 . Compute the tangent plane at each point. Find the point such that the normal vector is vertical at that point.
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Answers (1)

Plutbantonavv
Answered 2022-10-18 Author has 15 answers
Given surface
z = ( x 2 ) 2 ( y 3 ) 2
tangent plane at each point
Here f ( x , y , z ) = ( x 2 ) 2 + ( y 3 ) 2 z
let a point in given surface, whose posiion vector is
r 0 = a i + b j + c k
let p be any point of tangent plane
( r r 0 ) f = 0 ( ( x a ) u + ( y b ) j + ( z j k ) ) ( x a ) x 2 + 2 ( y b ) y 9 ( z c ) x ( x a ) 2 + 2 y 9 ( y b ) ( z c )
This is the equation of tangent plane at each point.
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