Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 14 - Vector Calculus - 14.6 Surface Integrals - 14.6 Exercises - Page 1123: 11

Answer

$r=\lt u, v, \dfrac{16-2u+4 v}{3} \gt$; $ u, v \in (-\infty, + \infty)$

Work Step by Step

Let us consider that $r=\lt u, v \gt$ defines the parametric description of the plane. We are given the equation of a plane as: $2x -4y+3z=16 \implies z=\dfrac{16-2x+4y}{3} ~~~~(1)$ Here, $u=x $ and $v=y$, so we can write the equation (1) as: $z=\dfrac{16-2u+4 v}{3}$ and $ u, v \in (-\infty, + \infty)$ Thus, the parametric description of the plane can be expressed as: $r=\lt u, v, \dfrac{16-2u+4 v}{3} \gt$; $ u, v \in (-\infty, + \infty)$
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