Calculus: Early Transcendentals 8th Edition

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

Chapter 10 - Section 10.2 - Calculus with Parametric Curves - 10.2 Exercises - Page 655: 17

Answer

Horizontal tangent at $(0,-3)$ Vertical tangents at $(2,-2)$ and $(-2,-2)$ Graphed with desmos:
1559255231

Work Step by Step

$\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}$ The tangent is horizontal for $\displaystyle \frac{dy}{dt}=0$ $y=t^{2}-3$ $\displaystyle \quad \frac{dy}{dt}=2t,$ $2t=0\Rightarrow t=0$ When $t=0, (x,y)=(0,-3)$ The tangent is vertical for $\displaystyle \frac{dx}{dt}=0$ $x=t^{3}-3t$ $\displaystyle \quad \frac{dx}{dt}=3t^{2}-3=3(t+1)(t-1),$ $t=-1$ or $1$ When t=1, $(x,y)=(2,-2)$, and when t=$-1, (x,y)=(-2,-2).$
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