Calculus: Early Transcendentals 8th Edition

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

Chapter 10 - Section 10.1 - Curves Defined by Parametric Equations - 10.1 Exercises - Page 645: 9

Answer

$y=-x^{2}+1$, See image:
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Work Step by Step

a. Build a table of coordinates (x,y) $x=f(t)=\sqrt{t},\quad \qquad y=g(t)=1-t$ Plot the points and join with a smooth curve. Taking the initial t to be the first point in the table, track the direction in which the points "travel" as t increases. b. From $ x=x=\sqrt{t}, \quad x \geq 0, t\geq 0, t=x^{2}$ $x^{2}=\sin^{2}t$ Substituting t into the other equation, we get $y=1-t$ $y=-x^{2}+1$, a concave down parabola wing.
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