solutions to final spring 05

solutions to final spring 05 - Math 192 Final Exam...

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Math 192, Final Exam Solutions, Spring 2005 1) a) Proj ~ i +2 ~ j ( ~ i + ~ j + ~ k ) = ( ~ i + ~ j + ~ k ) · ( ~ i + 2 ~ j ) ( ~ i + 2 ~ j ) · ( ~ i + 2 ~ j ) ( ~ i + 2 ~ j ) = 3 5 ( ~ i + 2 ~ j ) so ~ i + ~ j + ~ k = ( 2 5 ~ i - 1 5 ~ j + ~ k ) + 3 5 ( ~ i + 2 ~ j ) is the desired decomposition. b) ( ~ i + ~ j + ~ k ) × ( ~ i - ~ j - ~ k ) = 2 ~ j - 2 ~ k so the unit vectors are ± 1 2 ( ~ j - ~ k ). c) The normal vector to the plane is ~ i + 2 ~ j + 3 ~ k . The point (0 , 0 , 0) is in the plane. The equation of the plane is 1( x - 0) + 2( y - 0) + 3( z - 0) = x + 2 y + 3 z = 0. d) We need an equation of the form Ax + By + Cz + D = 0 that reduces to 0 = 0 whenever t is plugged in for x , y , and z . The plane x - y = 0 has this property. 2) a) ~ f = 1 xy ~ i - ln ( x ) y 2 ~ j so ~ f ( e, 1 /e ) = ~ i - e 2 ~ j . A unit vector parallel to this is ~u = 1 1 + e 4 ~ i - e 2 1 + e 4 ~ j . The directions of maximal increase and maximal decrease are repsectively that of ~u and - ~u . b) Consider the the surface g ( x, y, z ) = 0 given by f ( x, y ) - z = 0. Then ~ g ( e, 1 /e, e ) = ~ i - e 2 ~ j - ~ k . The equations of the tangent plane and normal line at ( e, 1 /e, e ) are respectively 1( x - e ) - e 2 ( y - 1 e ) - 1( z - e ) = 0 and x = e + t , y = 1 e - e 2 t , z = e - t . c) Approaching (1 , 0) along the curve y = ln ( x ) we get lim x 1 ln ( x ) ln ( x ) = 1. Approaching along the curve y = ln ( x 2 ) = 2 ln ( x ) we get lim x 1 2 ln ( x ) ln ( x ) = 2. Thus lim ( x,y )
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