Calculus Notes 13.6

# Calculus Notes 13.6 - Page 2 13.6 Quadric Surfaces A...

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Calculus- Stewart Dr. Berg Summer ‘09 Page 1 13.6 13.6 Cylinders and Quadric Surfaces Now we look at nonlinear objects in three dimensions: cylinders and quadric surfaces. To sketch the graph, it is useful to determine the curves of intersection with planes parallel to the coordinate planes. These curves are called traces (or cross-sections) of the surface. Cylinders A cylinder is a surface that consists of all lines (called rulings ) that are parallel to a given line and pass through a given plane curve. Example A Sketch the graph of the surface z = x 2 . Notice that the equation does not involve the variable y . This means that any plane of the form y = k (parallel to the xy -plane) will intersect the graph in a curve with the equation z = x 2 . Thus the traces are parabolas. We call this graph a parabolic cylinder . Example B The graph of the circular cylinder x 2 + y 2 = 4 is shown above. Remember, in R 2 this is the equation of a circle, but in R 3 , this is a circular cylinder.

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Calculus- Stewart Dr. Berg Summer ‘09
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Unformatted text preview: Page 2 13.6 Quadric Surfaces A quadric surface is the graph of a second-degree equation in three variables x , y , and z . The most general such equation is Ax 2 + By 2 + Cz 2 + Dxy + Eyz + Fxz + Gx + Hy + Iz + J = . By translating or rotating the graph, the equation can be brought into one of the two standard forms: Ax 2 + By 2 + Cz 2 + J = or Ax 2 + By 2 + Iz = . Quadric surfaces are the three dimensional counterparts to the conic sections in the plane. Example C We examine the traces of x 2 + y 2 9 + z 2 4 = 1 . For the traces parallel to the xy-plane, we set z = k for different values of k . These traces satisfy x 2 + y 2 9 = 1 − k 2 4 which describes an ellipse. Similarly, setting y = k gives the ellipses x 2 + z 2 4 = 1 − k 2 9 and likewise for setting x = k . Example D Above is the graph of x 2 4 + y 2 − z 2 4 = 1 a hyperboloid of one sheet. Exercise 1 Use traces to identify the surface given by 4 x 2 − 16 y 2 + z 2 = 16 ....
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