Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 14 - Calculus of Vector-Valued Functions - 14.1 Vector-Valued Functions - Exercises - Page 709: 1

Answer

The domain of $ r(t) $ is $$ D=\{t\in R; t\neq 0, -1\}.$$

Work Step by Step

Given $$ r(t)=e^ti+\frac{1}{t}j+(t+1)^{-3}k $$ We know that $1/t $ is not defined for $ t=0$ and $(t+1)^{-3}$ is not defined for $t=-1$ (because we can not divide by $0$). Thus, the domain of $ r(t) $ is $$ D=\{t\in R; t\neq 0, -1\}.$$
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