Calculus: Early Transcendentals 8th Edition

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

Chapter 13 - Section 13.1 - Vector Functions and Space Curves - 13.1 Exercises - Page 854: 32

Answer

$(0.909,-0.416,2)$ and $(-0.909,-0.416,-2)$

Work Step by Step

The parametric equation from $r(t)= (\sin^2 t+\cos^2 t)+t^2=5$ is given as: $\sin^2 t+\cos^2 t=1$ We need to plug these parametric equations in to the given paraboloid equation as: $z=x^2+y^2$ $1+t^2=5$ or, $t^2=4$ or, $ t= \pm 2$ Now, the value of $t$ for the points of intersection are: $ \lt \sin 2, \cos 2, 2 \gt$ and $ \lt \sin (-2), \cos (-2), 2 \gt$ Thus, the the points of intersection are: $(0.909,-0.416,2)$ and $(-0.909,-0.416,-2)$
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