Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 13 - Vector Functions - 13.1 Vector Functions and Space Curves - 13.1 Exercises - Page 894: 31


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

Work Step by Step

General form of parametric equation for $r(t)= ti +(2t-t^2) k$ is: $x=t; y=0, z=2t-t^2$ We are given that the paraboloid equation is, $z=x^2+y^2$ This gives: $2t-t^2=(t)^2+(0)^2 \\2t-2t^2=0\\ 2t(1-t) =0 \\t=0,1$ Set the value of $t$ in order to get the point of intersection. This implies that we have $x=0; y=0$ and $z=2(0)-(0)^2=0\\ x=1; y=0$ and $z=2(1)-(1)^2=1$ So, the the point of intersection are: $(0,0, 0)$ and $(1,0,1)$
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