Calculus-Sec-11-6-Solution


   11.6      Cylinders and Quadric Surfaces    by SGLee - HSKim - JHLee

 

 

1.(a) What does the equation  represent as a curve in ?

 

      Equation  represents a parabola of slope  passing through origin in .

 

   (b) What does it represent as a surface in ?

 

 

      The equation of the graph is , which doesn't involve  in .

      This means that any vertical plane with equation (parallel to the plane) intersects the graph in a curve with ,

      that is, a parabola. Below figure shows how the graph is formed by taking the parabola  in the plane

      and moving it in the direction of the axis.

      So the graph is a surface, called a parabolic cylinder, made up of infinitely many shifted copies of the same parabola.




   (c) What does the equation  represent?

       also represents a parabolic cylinder, this time with axis the axis in .

 




 2. (a) Sketch the graph of  as a curve in .

                 http://matrix.skku.ac.kr/cal-lab/cal-11-6-2.html 

 




 (b) Sketch the graph of  as a surface in .

 

 




 (c) Describe and sketch the surface .




3-5. Describe and sketch the surface.

 

 3.      

            http://matrix.skku.ac.kr/cal-lab/cal-11-6-3.html  

 

 




 4.       

            http://matrix.skku.ac.kr/cal-lab/cal-11-6-4.html 

 

 




 5. 

            http://matrix.skku.ac.kr/cal-lab/cal-11-6-5.html 

 

 




6-10. Find the traces of the given surface in . Then, identify the surface and sketch it.

 

 6. 

            http://matrix.skku.ac.kr/cal-lab/cal-11-6-5.html 

 




7. 

 

 

     The trace in  are ellipses of the form ,

      the trace in  are parabolas of the form ,

      and the trace in  are parabolas of the form .

      Combining these traces we form the graph.

                                         

 




 8. 

            http://matrix.skku.ac.kr/cal-lab/cal-11-6-8.html 

 

 




9. 

 

 

      The trace in  are hyperbolas of the form ,

      the trace in  are circles of the form ,

      and the trace in  are hyperbolas of the form .

      Combining these traces we form the graph.

 

                                      

 




 10. 

              http://matrix.skku.ac.kr/cal-lab/cal-11-6-10.html 

 




11-14. Reduce the equation to one of the standard forms, classify the surface, and sketch it.

 

11. 

 

      Dividing both sides by 15 gives ,

      an elliptic paraboloid with vertex  and axis the horizontal line .

                                                    




 12. 

              http://matrix.skku.ac.kr/cal-lab/cal-11-6-12.html 

 

 




13. 

 

      Completing squares in  and  gives  or

      , a hyperboloid of one sheet.

                                                   




 14. 

              http://matrix.skku.ac.kr/cal-lab/cal-11-6-14.html 

 

 




 15. 

              http://matrix.skku.ac.kr/cal-lab/cal-11-6-15.html 

 

 




 16. Sketch the region bounded by the surfaces  and  for .

              http://matrix.skku.ac.kr/cal-lab/cal-11-6-16.html 

 

 




17. Find an equation for the surface obtained by rotating the parabola  about the -axis.

     (Use revolution_plot3d to get the plot of this surface.)




18. Find an equation for the surface consisting of all points  for which

     the distance from  to the -axis is twice the distance from  to the -plane. Identify the surface.

 

 

      Let  be an arbitrary point whose distance from th axis is twice its distance from the plane.

      The distance from  to the axis is  and the distance from  to the plane() is .

      Thus         .

       So, the surface is a right circular cone with vertex the origin and axis the axis.

 




19. Find an equation for the surface consisting of all points that are equidistant

     from the point  and the plane .

                                                                      

 

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