HW05 - HW 5 Solutions Math 1920 Spring 2009 February 26,...

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HW 5 Solutions Math 1920 Spring 2009 February 26, 2009 Problem 14.6.6 f = (2 x - y ) i + ( - 2 y - x ) j - k f | (1 , 1 , - 1) = i - 3 j - k The tangent plane has normal f | (1 , 1 , - 1) and contains the point (1,1,-1) and hence has equation 1( x - 1) - 3( y - 1) - 1( z + 1) = 0 x - 3 y - z = - 1 The normal line has direction f | (1 , 1 , - 1) and contains the point (1,1,-1) and hence is given by x = 1 + t y = 1 - 3 t z = - 1 - t Problem 14.6.14 f = yz i + xz j + xy k f (1 , 1 , 1) = i + j + k g = 2 x i + 4 y j + 6 z k g (1 , 1 , 1) = 2 i + 4 j + 6 k . The tangent line to the curve of intersection is orthogonal to both f (1 , 1 , 1) and g (1 , 1 , 1), so we find its direction by taking the cross product of these two vectors. f (1 , 1 , 1) x g (1 , 1 , 1) = ± ± ± ± ± ± i j k 1 1 1 2 4 6 ± ± ± ± ± ± = 2 i - 4 j + 2 k Therefore the equation of the tangent line is given by x = 1 + 2 t y = 1 - 4 t z = 1 + 2 t Problem 14.6.28 a. f x = 3 x 2 y 4 f x (1 , 1) = 3 f y = 4 x 3 y 3 f x (1 , 1) = 4 f (1 , 1) = 1 Therefore L ( x, y ) = 1 + 3( x - 1) + 4( y - 1) = 3 x + 4 y - 6 1
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b. f x (0 , 0) = 0 f y (0 , 0) = 0 f (0 , 0) = 0 Therefore L ( x, y ) = 0 Problem 14.6.38 First we calculate the partials of f ( x, y, z ) f x = 2 x f y = 2 y f z = 2 z a. f x (1 , 1 , 1) = f y (1 , 1 , 1) = f z (1 , 1 , 1) = 2 and f (1 , 1 , 1) = 3. Therefore at (1,1,1) we have
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This note was uploaded on 12/25/2009 for the course MATH 1920 taught by Professor Pantano during the Spring '06 term at Cornell University (Engineering School).

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HW05 - HW 5 Solutions Math 1920 Spring 2009 February 26,...

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