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exam1_sol - MA3160-08 Exam 1 Name(print No calculators...

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Feb 23, 2011 MA3160-08 Exam 1 No calculators! Justify all answers! Name (print): Solutions 1. (10 pts.) Find the equation of the plane whose graph contains the points P 1 = (0 , 0 , 2), P 2 = (1 , - 1 , 1), P 3 = (1 , 0 , 0). Solution The three points must satisfy the equation z = mx + ny + c P 1 = (0 , 0 , 2) : 2 = 0 + 0 + c P 3 = (1 , 0 , 0) : 0 = m + 0 + c P 2 = (1 , - 1 , 1) : 1 = m - n + c = 2 = c - m = c - m + n + 1 = c = c = 2 m = - 2 n = - 1 Therefore, the equation of the linear function is z = - 2 x - y + 2 2. Let f ( x , y ) = x 2 + xy + y 2 - 2. (a) (7 pts.) Find the di ff erential of f ( x , y ) at the point (2,1). Solution Note that f x ( x , y ) = 2 x + y and f y ( x , y ) = x + 2 y Therefore, d f = f x ( x , y ) dx + f y ( x , y ) dy = (2 x + y ) dx + ( x + 2 y ) dy and at the point (2,1) df = 5 dx + 4 dy (b) (5 pts.) Use part (a) to estimate f (1 . 9 , 1 . 2). Solution f (1 . 9 , 1 . 1) f (2 , 1) + f x (2 , 1) Δ x + f y (2 , 1) Δ y = 5 + 5( - 0 . 1) + 4(0 . 2) Therefore, f (1 . 9 , 1 . 1) 5 . 3
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Justify all answers! MA3160-08 Exam 1 page 2 3. (12 pts.) Find the directional derivative of f ( x , y , z ) = xz - y 2 at the point P = (1 , 0 , 1) in the direction of the point Q = (0 , 1 , 0). Solution The direction vector is ~ v = ~ PQ = (0 - 1) ~ i + (1 - 0) ~ j + (0 - 1) ~ k = - ~ i + ~ j - ~ k with length || ~ v || = p ( - 1) 2 + 1 2 + ( - 1) 2 = 3 . Therefore, the unit direction vector is ~ u = ~ v || ~ v || = - 1 3 ~ i + 1 3 ~ j - 1 3 ~ k .
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