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Unformatted text preview: moseley (cmm3869) Exam03Review Gilbert (56380) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points Which of the following surfaces is the graph of 3 x + 6 y + 4 z = 12 in the first octant? 1. x y z correct 2. x y z 3. x y z 4. x y z 5. x y z 6. x y z Explanation: Since the equation is linear, its graph will be a plane. To determine which plane, we have only to compute the intercepts of 3 x + 6 y + 4 z = 12 . Now the xintercept occurs at y = z = 0, i.e. at x = 4; similarly, the yintercept is at y = 2, while the zintercept is at z = 3. By inspection, therefore, the graph is moseley (cmm3869) Exam03Review Gilbert (56380) 2 x y z 002 10.0 points Which one of the following equations has graph when the circular cylinder has radius 1. 1. z 2 + x 2 + 2 x = 0 correct 2. z 2 + x 2 + 4 x = 0 3. y 2 + z 2 + 4 z = 0 4. x 2 + z 2 4 z = 0 5. x 2 + z 2 2 z = 0 6. y 2 + z 2 + 2 z = 0 Explanation: The graph is a circular cylinder whose axis of symmetry is parallel to the yaxis, so it will be the graph of an equation containing no yterm. This already eliminates the equations y 2 + z 2 + 2 z = 0 , y 2 + z 2 + 4 z = 0 . On the other hand, the intersection of the graph with the xzplane, i.e. the y = 0 plane, is a circle centered on the xaxis and passing through the origin as shown in x z But this circle has radius 1 because the cylin der has radius 1, and so its equation is ( x + 1) 2 + z 2 = 1 as a circle in the xzplane. Consequently, after expansion we see that the cylinder is the graph of the equation z 2 + x 2 + 2 x = 0 . keywords: quadric surface, graph of equation, cylinder, Surfaces, SurfacesExam, 3D graph, circular cylinder, trace 003 10.0 points Reduce the equation x 2 y 2 + z 2 + 2 x 2 y 4 z + 2 = 0 to standard form and then classify the surface. 1. ( x + 1) 2 2 + ( y + 1) 2 2 + ( z 2) 2 2 = 1, ellipsoid 2. ( x + 1) 2 2 ( y + 1) 2 2 + ( z 2) 2 2 = 1, hyperboloid correct moseley (cmm3869) Exam03Review Gilbert (56380) 3 3. ( x + 1) 2 2 ( y 1) 2 2 + ( z 2) 2 2 = 1, hyperboloid 4. ( x 1) 2 2 + ( y + 1) 2 2 + ( z 2) 2 2 = 1, paraboloid 5. ( x + 1) 2 2 ( y 1) 2 2 + ( z + 2) 2 2 = 1, hyperboloid Explanation: Completing squares in all three variables gives ( x + 1) 2 ( y + 1) 2 + ( z 2) 2 = 2 or ( x + 1) 2 2 ( y + 1) 2 2 + ( z 2) 2 2 = 1, a hyper boloid. 004 10.0 points Find lim t + r ( t ) when r ( t ) = ( 8 cos t, 3 e t , 9 t ln t ) . 1. limit = ( , , ) 2. limit = ( 8 , , 9 ) 3. limit = ( 8 , 3 , ) correct 4. limit = ( , 3 , ) 5. limit = ( 8 , 3 , 9 ) 6. limit = ( 8 , , 9 ) Explanation: For a vector function r ( t ) = ( f ( t ) , g ( t ) , h ( t ) ) , the limit lim t + r ( t ) = (Big lim t + f ( t ) , lim t + g ( t ) , lim t + h ( t ) )Big ....
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This note was uploaded on 04/12/2011 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Sadler
 Multivariable Calculus

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