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UT M 408D - Homework 12.6-solutions

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ha (lvh262) – Homework 12.6 – karakurt – (56295) 1This print-out should have 16 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.001 10.0 pointsIdentify the quadric surface1. 2x2+ y2+ 3z2= 12. x2+ y2− z2= 13. z = x2+ y24. x2+ y2+ z2= 15. z2− x2− y2= 1 correct6. z2= x2+ y2Explanation:The trace in any horizontal plane suffi-ciently far away from the xy-plane is alwaysa circle, while the trace on both the xz-planeand yz-plane is a hyperbola opening upwards.Consequently, the quadric surface is is thegraph of a two-sheeted hyperboloid, and theonly equation whose graph has these proper-ties isz2− x2− y2= 1 .keywords: Surfaces, SurfacesExam,002 10.0 pointsFor which of the following quadratic rela-tions is its g r aph a one-sheeted hyperboloid?1. 2x2+ y2+ 3z2= 12. z = x2+ y23. z2− x2− y2= 14. z2= x2+ y25. z = y2− x26. x2+ y2− z2= 1 correctExplanation:The graphs of each of the given quadraticrelations is a quadric surface in standard posi-tion. We have to decide which quadric surfacegoes wi th which equation - a good way of do-ing that is by taking plane slices parallel tothe coordinate planes, i.e., by setting respec-tively x = a, y = a and z = a in the equationsonce we’ve decided what the graphs of thoseplane slices should be.Now as slicing ofshows, slices of this one-sheeted hyperbolo idby z = a are circles, while the slices by x = 0and y = 0 are hyperbo las opening left andha (lvh262) – Homework 12.6 – karakurt – ( 56295) 2right. Only the graph ofx2+ y2− z2= 1has these properties.003 10.0 pointsWhich one of the fol lowing equations hasgraphwhen the circular cylinder has radius 2.1. x2+ y2− 2x = 02. y2+ z2− 4y = 03. x2+ y2+ 4y = 04. x2+ y2− 4x = 0 correct5. y2+ z2− 2y = 06. x2+ y2+ 2y = 0Explanation:The graph is a circular cylinder whose axisof symmetry is parallel to the z-axis, so itwill be the graph of an equation containing noz-term. This already eliminates the equationsy2+ z2− 2y = 0, y2+ z2− 4y = 0 .On the other hand, the intersection of thegraph with the xy-plane, i.e. the z = 0 plane,is a circle centered on the x-axi s and passingthrough the origin as shown inxyBut this circle has radius 2 because the cyl in-der has radius 2, and so its equation is(x − 2)2+ y2= 4as a circle in the xy-plane. Consequently,after expansion we see that the cylinder is thegraph of the equationx2+ y2− 4x = 0 .keywords: quadric surface, graph of equation,cylinder, 3D graph, circular cylinder, trace004 10.0 pointsWhich one of the following equations hasgraphwhen the circular cylinder has radius 2?1. x2+ y2− 2y = 02. y2+ z2+ 2y = 0ha (lvh262) – Homework 12.6 – karakurt – ( 56295) 33. x2+ y2− 4y = 0 correct4. y2+ z2− 2z = 05. y2+ z2− 4z = 06. y2+ z2+ 4y = 0Explanation:The graph is a circular cylinder whose axisof symmetry is parallel to the z-axis, so itwill be the graph of an equation containing noz-term. This already eliminates the equationsy2+ z2− 4z = 0, y2+ z2+ 2y = 0,y2+ z2+ 4y = 0, y2+ z2− 2z = 0 .On the other hand, the intersection of thegraph with the xy-plane, i.e. the z = 0 plane,is a circle centered on the y-axis and passingthrough the origin as shown inxyBut this circle has radius 2 because the cyl in-der has radius 2, a nd so its equation isx2+ (y − 2)2= 4as a ci r cl e in the xy-plane.Consequently, the graph is that of the equa-tionx2+ y2− 4y = 0 .keywords: quadric surface, graph of equation,cylinder, 3D graph, circular cylinder, trace005 10.0 pointsDetermine which one of the following equa-tions has gr aphxzy1. x2+ z = 12. y − x2= 13. y2= z − 14. x − z2= 15. z2= y − 1 correct6. x − y2= 1Explanation:The graph is a cylinder with axis parall elto the x-axis. In additio n its trace on theyz-plane is a parabola opening in the po si tivey-directi on and having posit ive y-intercept.Consequently, the graph is that of the equa-tionz2= y − 1 .keywords: quadric surface, graph of equation,cylinder, 3D graph, parabolic cylinder, Sur-faces, SurfacesExam,006 10.0 pointsha (lvh262) – Homework 12.6 – karakurt – ( 56295) 4Which one of the fol lowing equations hasgraphxzy1. y2− x2− z2= 12.x24+y29+ z2= 13.x24+ y2+z29= 1 correct4. x2− y2+ z2= 15. z2− x2− y2= 16. x2+ y2− z2= 1Explanation:The surface is an ellipsoid. As the graphshows, the trace on the y = 0 plane is anellipse centered at the origin, and so wi ll be ofthe formx2a2+z2b2= 1for some choice of a, b. But this ellipse willbe longer in the z-direction than in the x-direction. Thus b > a.Consequently, the graph is that of thequadric surfacex24+ y2+z29= 1.007 10.0 pointsFind and identify the trace of the quadricsurfacex2− 2y2− 2z2= 3in the plane x = 1 .1. hyperbola : y2−12z2= 12. ellipse : y2+12z2= 13. hyperbola : 2y2− 2z2= 14. circle : y2+ z2=125. no trace exists correct6. circle : y2+ z2= 17. ellipse :12y2+ z2= 18. hyperbola : 2z2− 2y2= 1Explanation:To determine the trace ofx2− 2y2− 2z2= 3in the plane x = 1 , we set x = 1 i nx2− 2y2− 2z2= 3 ,obtaining−2y2− 2z2= 2 .Since no (real) values of y and z satisy thisequation,no trace exists .008 10.0 pointsFind and identify the trace of the quadricsurfacex2− 2y2+ 2z2= 3in the plane x = −1.1. ellipse :12y2+ z2= 1ha (lvh262) – Homework 12.6 – karakurt – ( 56295) 52. hyperbola : 2z2− 2y2= 13. no trace exists4. hyperbola : y2−12z2= 15. ellipse : y2+12z2= 16. hyperbola : 2y2− 2z2= 17. hyperbola : z2− y2= 1 correct8. ellipse : 2y2+ z2= 1Explanation:To determine the trace ofx2− 2y2+ 2z2= 3in the plane x = −1, we set x = −1 inx2− 2y2+ 2z2= 3 ,obtaining2y2− 2z2= −2 .The graph of this equation is thehyperbola : z2− y2= 1 .009 10.0 pointsWhich one of the following functions hasthe surfacexyzyas its graph.1. y = 2 − x2− z22. z = 2 − x2− y23. y = x2+ z2− 24. z = x2+ y2− 25. x = y2+ z2− 26. x = 2 − y2− z2correctExplanation:The trace of the graph on the vertical planex = k i s a circle centered at the origin. Onthe other hand, the graph has tracexyzon the zx-plane, i.e., on the plane y = 0. Thisis a parabola opening downwards and inter-secting the x-axis at x = 2. Consequently, thegraph is t hat of the functionx = 2 − y2− z2.010 10.0 pointsReduce the equationsz2= −4x2+ 8y2+ 32to standard form and then classify the surface.1.x28+y24+z232= 1, ellipsoidha (lvh262) – Homework 12.6 – karakurt – ( 56295) 62. −x24−y28+z232= 1, ellipsoid3.x24−y28+z232= 1,


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