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 - 14.4 Curvature - Exercises - Page 734: 6


$$ r'(t) = \lt e^t, -e^{-t}, 2t \gt, \quad T(t)= \frac{ \lt e^t, -e^{-t}, 2t \gt}{\sqrt{e^{2t}+e^{-2t}+4t^2}}, \quad T(1)=\frac{ \lt e , -e^{-1}, 2 \gt}{\sqrt{e^{2}+e^{-2}+4}}.$$

Work Step by Step

Since $ r(t) = \lt e^t, e^{-t}, t^2 \gt$, then $ r'(t) = \lt e^t, -e^{-t}, 2t \gt$ and $\|r'(t)\|=\sqrt{ e^{2t}+e^{-2t}+4t^2} $ Hence, we have $$T(t)=\frac{r'(t)}{\|r'(t)\|}=\frac{ \lt e^t, -e^{-t}, 2t \gt}{\sqrt{e^{2t}+e^{-2t}+4t^2}}, \quad T(1)=\frac{ \lt e , -e^{-1}, 2 \gt}{\sqrt{e^{2}+e^{-2}+4}}.$$
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