Review Exam 3 - rhodes (ajr2283) Exam03Review Gilbert...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: rhodes (ajr2283) 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 6 x + 3 y + 4 z = 12 in the first octant? 1. x y z 2. x y z 3. x y z 4. x y z 5. x y z correct 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 6 x + 3 y + 4 z = 12 . Now the x-intercept occurs at y = z = 0, i.e. at x = 2; similarly, the y-intercept is at y = 4, while the z-intercept is at z = 3. By inspection, therefore, the graph is rhodes (ajr2283) 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 2. 1. z 2 + x 2 + 4 x = 0 correct 2. z 2 + x 2 + 2 x = 0 3. y 2 + z 2 + 2 z = 0 4. y 2 + z 2 + 4 z = 0 5. x 2 + z 2 2 z = 0 6. x 2 + z 2 4 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 2 because the cylin- der has radius 2, and so its equation is ( x + 2) 2 + z 2 = 4 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 + 4 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, hyperboloid 2. ( x 1) 2 2 ( y 1) 2 2 + ( z + 2) 2 2 = 1, hyperboloid 3. ( x 1) 2 2 + ( y + 1) 2 2 + ( z + 2) 2 2 = 1, ellipsoid rhodes (ajr2283) Exam03Review Gilbert (56380) 3 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 correct 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 ) = ( 7 cos t, 5 e t , 3 t ln t ) . 1. limit = ( , 5 , ) 2. limit = ( 7 , , 3 ) 3. limit = ( 7 , 5 , ) correct 4. limit = ( , , ) 5. limit = ( 7 , , 3 ) 6. limit = ( 7 , 5 , 3 ) 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 ....
View Full Document

This note was uploaded on 09/27/2010 for the course M 408 D taught by Professor Textbookanswers during the Spring '07 term at University of Texas at Austin.

Page1 / 9

Review Exam 3 - rhodes (ajr2283) Exam03Review Gilbert...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online