University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Section 10.1 - Parametrizations of Plane Curves - Exercises - Page 563: 21

Answer

$x=t$; $y=\dfrac{4}{5}t-\dfrac{11}{5}$ and $-1\leq t \leq 4$ (Other answers are possible.)

Work Step by Step

The slope of a line between two points can be found as: $m=\dfrac{1-(-3)}{4-(-1)}=\dfrac{4}{5}$ Now, $y-y_0=m(x-x_0) \implies y-1=\dfrac{4}{5}(x-4)$ This implies that $y=\dfrac{4}{5}x-\dfrac{11}{5}$ Consider $x=t$ Then $y=\dfrac{4}{5}t-\dfrac{11}{5}$ and $-1\leq t \leq 4$
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