Spring 2009 Prelim 1 Solutions

# Spring 2009 Prelim 1 Solutions - Solutions to Prelim Exam 1...

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Solutions to Prelim Exam 1 1. (a) lim ( x,y,z ) (1 , - 1 , - 1) 2 x - 3 y + z x 2 + z 2 lim x 1 sin( x - 1) ( x - 1) = 2(1) - 3( - 1) + ( - 1) 1 2 + ( - 1) 2 = 2 . (b) The limit does not exist. One can apply the Two Path Test using two paths of the form y = kx ( k 6 = 2), for example: k = 0 ,x 0 yields lim ( x,y ) (0 , 0) along y =0 p x 2 + y 2 2 x - y = lim ( x,y ) (0 , 0) along y =0 x 2 2 x = 1 2 . k = 1 ,x 0 yields lim ( x,y ) (0 , 0) along y = x p x 2 + y 2 2 x - y = lim ( x,y ) (0 , 0) along y = x 2 x 2 x = 2 . 2. Set f ( x,y,z ) = 2 x 2 - 3 y + z 2 - 5 and g ( x,y,z ) = x + y - 3. Then f = 4 x i - 3 j +2 z k and g = i + j . Hence, f | (1 , 2 , 3) = 4 i - 3 j + 6 k and g | (1 , 2 , 3) = i + j . The vector f × ∇ g is parallel to the tangent line we seek. f × ∇ g = ± ± ± ± ± ± i j k 4 - 3 6 1 1 0 ± ± ± ± ± ± = - 6 i + 6 j + 7 k . Using the point P (1 , 2 , 3), we ﬁnd parametric equations for the tangent line: x = 1 - 6 t y = 2 + 6 t z = 3 + 7 t. 3.

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Spring 2009 Prelim 1 Solutions - Solutions to Prelim Exam 1...

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