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 895: 49



Work Step by Step

Consider the first equation $t^2-4t+3=0...(1)$ This gives : $t^2--3t-t+3=0 \implies t= 3$ Consider the second equation $t^2-7t+12=0 ....(2)$ This gives: $t^2-6t-t+12=0 \implies t= 3$ Need to set the $z$ component of the two equations equal to each other first. By doing so, we have $t^2=5t-6$ ...(3) This gives: $t^2-5t+6=0 $ and $t^2-6t+t+6=0$ Then, we have $t=3$ Yes, all the three equations gives three equations.
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