## Calculus (3rd Edition)

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