Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 10 - Section 10.1 - Curves Defined by Parametric Equations - 10.1 Exercises - Page 668: 1

Answer

$(x,y)=(2,\frac{1}{3})$ for $t=-2$ $(0,1)$ for $t=-1$ $(0,3)$ for $t=0$ $(2,9)$ for $t=1$ $(6,27)$ for $t=2$

Work Step by Step

$x=t^{2}+t$ $y=3^{t+1}$ When $t=-2$, $x=(-2)^{2}+(-2)=4-2=2$ $y=3^{(-2)+1}=3^{-1}=\frac{1}{3}$ $(x,y)=(2,\frac{1}{3})$ for $t=-2$ When $t=-1$, $x=(-1)^{2}+(-1)=1-1=0$ $y=3^{(-1)+1}=3^{0}=1$ $(x,y)=(0,1)$ for $t=-1$ When $t=0$, $x=(0)^{2}+(0)=0+0=0$ $y=3^{(0)+1}=3^{1}=3$ $(x,y)=(0,3)$ for $t=0$ When $t=1$, $x=(1)^{2}+(1)=1+1=2$ $y=3^{(1)+1}=3^{2}=9$ $(x,y)=(2,9)$ for $t=1$ When $t=2$, $x=(2)^{2}+(2)=4+2=6$ $y=3^{(2)+1}=3^{3}=27$ $(x,y)=(6,27)$ for $t=2$
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