Calculus: Early Transcendentals 8th Edition

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

Chapter 13 - Section 13.2 - Derivatives and Integrals of Vector Functions - 13.2 Exercise - Page 860: 17



Work Step by Step

$\boldsymbol{r}(t)=\langle t^2-2t,1+3t,\frac{t^3}{3}+\frac{t^2}{2}\rangle$ $\boldsymbol{T}(t)=\frac{\boldsymbol{r'}(t)}{|\boldsymbol{r'}(t)|}=\frac{1}{\sqrt{(2t-2)^2+(3)^2+(t^2+t)^2}}\langle2t-2,3,t^2+t\rangle \Rightarrow \boldsymbol{T}(2)=\frac{1}{\sqrt{(2(2)-2)^2+(3)^2+((2)^2+(2))^2}}\langle2(2)-2,3,(2)^2+(2)\rangle=\frac{1}{7}\langle2,3,6\rangle=\langle\frac{2}{7},\frac{3}{7},\frac{6}{7}\rangle$
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