Gilbert_Hmwk12sol

# Gilbert_Hmwk12sol - moseley (cmm3869) – HW12 – Gilbert...

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Unformatted text preview: moseley (cmm3869) – HW12 – 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 Determine the vector function r ( y ) whose graph is the cross-section of the graph of z = f ( x, y ) = x 2 − 4 y 2 − 2 x − 3 y by the plane x = 3. 1. r ( y ) = ( y, 3 , 3 − 4 y 2 − 3 y ) 2. r ( y ) = ( 3 , y, 15 − 4 y 2 − 3 y ) 3. r ( y ) = ( 3 , y, 3 − 4 y 2 − 3 y ) correct 4. r ( y ) = ( y, 3 , 15 − 4 y 2 − 3 y ) 5. r ( y ) = ( y, 3 , 4 y 2 − 3 y ) 6. r ( y ) = ( 3 , y, 4 y 2 − 3 y ) Explanation: The graph of z = f ( x, y ) = x 2 − 4 y 2 − 2 x − 3 y is the set of all points ( x, y, f ( x, y )) as x, y vary in 3-space. So the intersection of this graph with the plane x = 3 is the set of all points (3 , y, f (3 , y )) , −∞ < y < ∞ . But f (3 , y ) = 3 − 4 y 2 − 3 y . Consequently, the cross-section is the graph of r ( y ) = ( 3 , y, 3 − 4 y 2 − 3 y ) . keywords: GraphsContoursMV, GraphsCon- toursMVExam, 002 10.0 points A space curve is shown in black on the surface x y z Which one of the following vector functions has this space curve as its graph? 1. r ( t ) = ( cos t, sin t, cos 2 t ) 2. r ( t ) = ( sin t, cos t, cos 2 t ) 3. r ( t ) = ( cos t, sin t, t ) 4. r ( t ) = ( sin t, cos t, − t ) 5. r ( t ) = ( cos t, sin t, cos 4 t ) 6. r ( t ) = ( cos t, sin t, sin 4 t ) correct Explanation: If we write r ( t ) = ( x ( t ) , y ( t ) , z ( t ) ) , then x ( t ) 2 + y ( t ) 2 = 1 for all the given vector functions, showing that their graph will always lie on the cylindrical cylinder x 2 + y 2 = 1 moseley (cmm3869) – HW12 – Gilbert – (56380) 2 To determine which particular vector function has the given graph, we have to look more closely at the graph itself. Notice that the graph oscillates with period 4, so r ( t ) is one of ( cos t, sin t, sin 4 t ) , ( cos t, sin t, cos 4 t ) . On the other hand, it passes it through (1 , , 0) and (0 , 1 , 0). Consequently, the space curve is the graph of r ( t ) = ( cos t, sin t, sin 4 t ) . keywords: 003 (part 1 of 2) 10.0 points The vector function r ( t ) = 4 i + (1 + 5 cos t ) j + (2 − 5 sin t ) k traces out a circle in 3-space as t varies. In which plane does this circle lie? 1. plane y = 4 2. plane z = − 4 3. plane x = 4 correct 4. plane z = 4 5. plane y = − 4 6. plane x = − 4 Explanation: Writing r ( t ) = x ( t ) i + y ( t ) j + z ( t ) k , we see that x ( t ) = 4 for all t . Consequently, r ( t ) traces out a curve in the plane x = 4 . 004 (part 2 of 2) 10.0 points Determine the radius and center of the cir- cle traced out by r ( t ). 1. radius = 5 , center = (2 , 1 , 4) 2. radius = 5 , center = (1 , 4 , 2) 3. radius = 1 , center = ( − 2 , 1 , 2) 4. radius = 1 , center = (4 , 1 , 2) 5. radius = 1 , center = (4 , 1 , − 2) 6. radius = 5 , center = (4 , 1 , 2) correct Explanation: Writing r ( t ) = x ( t ) i + y ( t ) j + z ( t ) k ,...
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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.

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Gilbert_Hmwk12sol - moseley (cmm3869) – HW12 – Gilbert...

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