Gilbert_ReviewExam03sol

# Gilbert_ReviewExam03sol - moseley (cmm3869) Exam03Review...

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Unformatted text preview: moseley (cmm3869) Exam03Review Gilbert (56380) 1 This print-out should have 20 questions. Multiple-choice 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 x-intercept occurs at y = z = 0, i.e. at x = 4; similarly, the y-intercept is at y = 2, while the z-intercept 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 y-axis, so it will be the graph of an equation containing no y-term. 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 xz-plane, i.e. the y = 0 plane, is a circle centered on the x-axis 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 xz-plane. 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.

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Gilbert_ReviewExam03sol - moseley (cmm3869) Exam03Review...

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