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Unformatted text preview: i . Solution : Think of the surface as deﬁned implicitly by F ( x,y,z ) = x 2 4 + y 2 9 + z 2 = 1. The gradient vector is ∇ F = h 1 2 x 2 , 2 9 y 2 , 2 z i , so if ( a,b,c ) is a point on the ellipsoid, then a normal vector to the tangent plane at ( a,b,c ) is ∇ F ( a,b,c ) = ² 1 2 a, 2 9 b, 2 c ³ . For the tangent plane to be normal to v , we need to solve the system ± 1 2 a, 2 9 b, 2 c ² = λ h 1 , 1 ,2 i , or 1 2 a = λ (1) 2 9 b = λ (2) 2 c =2 λ. (3) These equations imply a = 2 λ , b = 9 2 λ , and c =λ . Since ( a,b,c ) is a point on the ellipsoid, then c 2 = 1a 2 4b 2 9 . Substituting λ into this last equation gives λ 2 = 1(2 λ ) 2 4( 9 2 λ ) 2 9 or λ = ± 2 √ 17 . Therefore, the two points are ³ 4 √ 17 , 9 √ 17 ,2 √ 17 ´ and ³4 √ 17 ,9 √ 17 , 2 √ 17 ´ ....
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This note was uploaded on 03/28/2011 for the course MATH 20C taught by Professor Helton during the Winter '08 term at UCSD.
 Winter '08
 Helton
 Approximation, Linear Approximation

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