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: 31

Answer

$(0,0,0)$ and $(1,0,1)$

Work Step by Step

The parametric equation from $r(t)= ti +(2t-t^2) k$ is given as: $x=t; y=0, z=2t-t^2$ We need to plug these parametric equations in to the given paraboloid equation as: $z=x^2+y^2$ $2t-t^2=t^2+0^2$ or, $2t-2t^2=0$ or, $2t(1-t) =0 \implies t=0,1$ Now, the values of $t$ for the point of intersection are: $x=0; y=0$ and $z=2(0)-(0)^2=0$ and $x=1; y=0$ and $z=2(1)-(1)^2=1$ Thus, the the points of intersection are: $(0,0,0)$ and $(1,0,1)$
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