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: 5


$$r'(t) = \lt - \pi\sin\pi t, \pi\cos\pi t, 1 \gt, \quad T(t)= \frac{ \lt - \pi\sin\pi t, \pi\cos\pi t, 1 \gt}{\sqrt{ 1+\pi^2}}, \quad T(1)=\frac{ \lt 0, -\pi, 1 \gt}{\sqrt{1+\pi^2}}.$$

Work Step by Step

Recall that $(\sin x)'=\cos x$. Recall that $(\cos x)'=-\sin x$. Since $ r(t) = \lt \cos\pi t, \sin\pi t, t \gt$, then $ r'(t) = \lt - \pi\sin\pi t, \pi\cos\pi t, 1 \gt$ and $\|r'(t)\|=\sqrt{ 1+\pi^2} $. Hence, we have $$T(t)=\frac{r'(t)}{\|r'(t)\|}=\frac{ \lt - \pi\sin\pi t, \pi\cos\pi t, 1 \gt}{\sqrt{ 1+\pi^2}}, \quad T(1)=\frac{ \lt 0, -\pi, 1 \gt}{\sqrt{1+\pi^2}}.$$
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