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Review Exam 3

# Review Exam 3 - rhodes(ajr2283 Exam03Review Gilbert(56380...

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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, it’s 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

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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 0 + r ( t ) when r ( t ) = ( 7 cos t, 5 e t , 3 t ln t ) . 1. limit = ( 0 , 5 , 0 ) 2. limit = ( 7 , 0 , 3 ) 3. limit = ( 7 , 5 , 0 ) correct 4. limit = ( 0 , 0 , 0 ) 5. limit = ( 7 , 0 , 3 ) 6. limit = ( 7 , 5 , 3 ) Explanation: For a vector function r ( t ) = ( f ( t ) , g ( t ) , h ( t ) ) , the limit lim t 0 + r ( t ) = (Big lim t 0 + f ( t ) , lim t 0 + g ( t ) , lim t 0 + h ( t ) )Big . But for the given vector function, lim t 0 + f ( t ) = lim t 0 + 7 cos t = 7 , while lim t 0 + g ( t ) = lim t 0 + 5 e t = 5 , and lim t 0 + h ( t ) = lim t 0 + t ln t = 0 , using L’Hospital’s Rule. Consequently, lim t 0 + r ( t ) = ( 7 , 5 , 0 ) . keywords: vector function, limit, trig function log function, exponential function 005 10.0 points Find a parameterization of the horizontal circle of radius 4 having center (3 , 5 , 2).

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