Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Section 12.6 - Cylinders and Quadric Surfaces - 12.6 Exercises - Page 840: 19

Answer

Hyperbolic Paraboloid

Work Step by Step

Rewrite the equation as: $y=z^{2}-x^{2}$ $ \dfrac{y}{1}=\dfrac{z^{2}}{1}-\dfrac{x^{2}}{1^1}$ On comparing the above form, we find that we have a: Hyperbolic Paraboloid $ \dfrac{z}{c}=\dfrac{x^{2}}{a^{2}}-\dfrac{y^{2}}{b^{2}}$ When $z=k$, traces are parabolas opening in the $-y$ direction and when $x=k$, traces are parabolas opening in the $+y$ direction.
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