Chapter 14 - Calculus of Vector-Valued Functions - 14.4 Curvature - Exercises - Page 734: 9

$$\kappa(t)= 0 .$$

Since $ r(t) = \lt 4t + 1, 4t − 3, 2t\gt$, then $ r'(t) = \lt 4, 4, 2 \gt$ and hence $\|r'(t)\|=\sqrt{36 }=6$, $T(t)=\frac{r'(t)}{\|r'(t)\|}=\frac{ \lt 4, 4, 2\gt}{6}$. Now, the curvature is given by $$\kappa(t)=\frac{1}{\|r'(t)\|}\|\frac{dT}{dt}\|=\frac{1}{6}\|\frac{ \lt 0,0,0\gt}{ 6}\|=0 .$$