ha lvh262 Homework 12 6 karakurt 56295 This print out should have 16 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 10 0 points Identify the quadric surface 1 keywords Surfaces SurfacesExam 002 10 0 points For which of the following quadratic relations is its graph a one sheeted hyperboloid 1 2x2 y 2 3z 2 1 2 z x2 y 2 3 z 2 x2 y 2 1 4 z 2 x2 y 2 5 z y 2 x2 6 x2 y 2 z 2 1 correct 1 2x2 y 2 3z 2 1 2 x2 y 2 z 2 1 3 z x2 y 2 4 x2 y 2 z 2 1 Explanation The graphs of each of the given quadratic relations is a quadric surface in standard position We have to decide which quadric surface goes with which equation a good way of doing that is by taking plane slices parallel to the coordinate planes i e by setting respectively x a y a and z a in the equations once we ve decided what the graphs of those plane slices should be Now as slicing of 5 z 2 x2 y 2 1 correct 6 z 2 x2 y 2 Explanation The trace in any horizontal plane sufficiently far away from the xy plane is always a circle while the trace on both the xz plane and yz plane is a hyperbola opening upwards Consequently the quadric surface is is the graph of a two sheeted hyperboloid and the only equation whose graph has these properties is z 2 x2 y 2 1 shows slices of this one sheeted hyperboloid by z a are circles while the slices by x 0 and y 0 are hyperbolas opening left and ha lvh262 Homework 12 6 karakurt 56295 right Only the graph of 2 y x2 y 2 z 2 1 has these properties 003 x 10 0 points Which one of the following equations has graph But this circle has radius 2 because the cylinder has radius 2 and so its equation is x 2 2 y 2 4 as a circle in the xy plane Consequently after expansion we see that the cylinder is the graph of the equation x2 y 2 4x 0 when the circular cylinder has radius 2 keywords quadric surface graph of equation cylinder 3D graph circular cylinder trace 1 x2 y 2 2x 0 2 y 2 z 2 4y 0 004 10 0 points 3 x2 y 2 4y 0 2 2 4 x y 4x 0 correct Which one of the following equations has graph 5 y 2 z 2 2y 0 6 x2 y 2 2y 0 Explanation The graph is a circular cylinder whose axis of symmetry is parallel to the z axis so it will be the graph of an equation containing no z term This already eliminates the equations y 2 z 2 2y 0 y 2 z 2 4y 0 when the circular cylinder has radius 2 On the other hand the intersection of the graph with the xy plane i e the z 0 plane is a circle centered on the x axis and passing through the origin as shown in 1 x2 y 2 2y 0 2 y 2 z 2 2y 0 ha lvh262 Homework 12 6 karakurt 56295 3 x2 y 2 4y 0 correct 005 4 y 2 z 2 2z 0 3 10 0 points Determine which one of the following equations has graph z 5 y 2 z 2 4z 0 6 y 2 z 2 4y 0 Explanation The graph is a circular cylinder whose axis of symmetry is parallel to the z axis so it will be the graph of an equation containing no z term This already eliminates the equations y x y 2 z 2 4z 0 y 2 z 2 2y 0 y 2 z 2 4y 0 y 2 z 2 2z 0 On the other hand the intersection of the graph with the xy plane i e the z 0 plane is a circle centered on the y axis and passing through the origin as shown in y 1 x2 z 1 2 y x2 1 3 y 2 z 1 4 x z 2 1 5 z 2 y 1 correct x But this circle has radius 2 because the cylinder has radius 2 and so its equation is x2 y 2 2 4 as a circle in the xy plane Consequently the graph is that of the equation x2 y 2 4y 0 keywords quadric surface graph of equation cylinder 3D graph circular cylinder trace 6 x y 2 1 Explanation The graph is a cylinder with axis parallel to the x axis In addition its trace on the yz plane is a parabola opening in the positive y direction and having positive y intercept Consequently the graph is that of the equation z2 y 1 keywords quadric surface graph of equation cylinder 3D graph parabolic cylinder Surfaces SurfacesExam 006 10 0 points ha lvh262 Homework 12 6 karakurt 56295 Which one of the following equations has graph z 4 Find and identify the trace of the quadric surface x2 2y 2 2z 2 3 in the plane x 1 x y 1 1 hyperbola y 2 z 2 1 2 1 2 ellipse y 2 z 2 1 2 3 hyperbola 2y 2 2z 2 1 4 circle y 2 z 2 1 2 5 no trace exists correct 6 circle y 2 z 2 1 1 2 y z2 1 2 1 y 2 x2 z 2 1 7 ellipse x2 y 2 z2 1 4 9 z2 x2 y2 1 correct 3 4 9 8 hyperbola 2z 2 2y 2 1 2 Explanation To determine the trace of x2 2y 2 2z 2 3 4 x2 y 2 z 2 1 5 z 2 x2 y 2 1 2 2 6 x y z 2 in the plane x 1 we set x 1 in x2 2y 2 2z 2 3 1 Explanation The surface is an ellipsoid As the graph shows the trace on the y 0 plane is an ellipse centered at the origin and so will be of the form x2 z 2 2 1 a2 b for some choice of a b But this ellipse will be longer in the z direction than in the xdirection Thus b a Consequently the graph is that of the quadric surface x2 z2 y2 1 4 9 obtaining 2y 2 2z 2 2 Since no real values of y and z satisy this equation no trace exists 008 10 0 points Find and identify the trace of the quadric surface x2 2y 2 2z 2 3 in the plane x 1 007 10 0 points 1 ellipse 1 2 y z2 1 2 ha lvh262 Homework 12 6 karakurt 56295 5 as its graph 2 2 hyperbola 2z 2y 2 1 1 y 2 x2 z 2 3 no trace exists 2 z 2 x2 y 2 1 4 hyperbola y 2 z 2 …