Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.3 Arc Length in Space - Exercises 13.3 - Page 759: 14

$$7$$

Since $$\mathbf{r}=(1+2 t) \mathbf{i}+(1+3 t) \mathbf{j}+(6-6 t) \mathbf{k} $$ Then \begin{align*} \mathbf{v}&=2 \mathbf{i}+3 \mathbf{j}-6 \mathbf{k} \\ |\mathbf{v}|&=\sqrt{2^{2}+3^{2}+(-6)^{2}}=7 \end{align*} Hence \begin{align*} s(t)&=\int_{0}^{t} 7 d \tau=7 t\\ \text{ Length }&=s(0)-s(-1)\\ &=0-(-7)=7 \end{align*}