Prelim_2_Fall_2007 - Math 192, Prelim 2 - Solutions October...

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Unformatted text preview: Math 192, Prelim 2 - Solutions October 25, 2007. 7:30-9:00 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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Prelim_2_Fall_2007 - Math 192, Prelim 2 - Solutions October...

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