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Unformatted text preview: Math 192, Prelim 2  Solutions October 25, 2007. 7:309:00 1) a) 3 = 1 2 + 2 2 + 2 and 3 = 5 1 2 so (1 , , 3) is on both of those surfaces. b) Set f ( x, y, z ) = x 2 + 2 y 2 + 2 z . Then f ( x, y, z ) = 0 is a level surface. Now f ( x, y, z ) = 2 x i + 4 y j k and f (1 , , 3) = 2 i k is a normal to the tangent plane. Thus the equation of the tangent plane is 2( x 1) + 0( y 0) ( z 3) = 0 2 x z = 1 c) Set g ( x, y, z ) = 5 x y 2 z . Then g ( x, y, z ) = 0 is a level surface. Observe that g ( x, y, z ) = 5 i j k and g (1 , , 3) = 5 i j k is a normal to the tangent plane of S 2 at (1 , , 3). Since the line tan gent to the curve of intersection of S 1 and S 2 is in the tangent planes of both surfaces at (1 , , 3) then f (1 , , 3) g (1 , , 3) = i 3 j 2 k is a vector parallel to the line. Thus the parametric equation for the line is x = 1 t , y = 3 t and z = 3 2 t ....
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 Fall '06
 PANTANO
 Math, Multivariable Calculus

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