## Thomas' Calculus 13th Edition

$$P(-\pi)=(12 \sin (-\pi),-12 \cos (-\pi),-5 \pi)=(0,12,-5 \pi)$$
Let $P\left(t_{0}\right)$ denote the point. Then $$\mathbf{v}=(12 \cos t) \mathbf{i}+(12 \sin t) \mathbf{j}+5 \mathbf{k}$$ and \begin{align*} -13 \pi&=\int_{0}^{t_{0}} \sqrt{144 \cos ^{2} t+144 \sin ^{2} t+25} d t\\ &=\int_{0}^{t_{0}} 13 d t\\ &=13 t_{0} \end{align*} Then $t_{0}=-\pi,$ and the point is $$P(-\pi)=(12 \sin (-\pi),-12 \cos (-\pi),-5 \pi)=(0,12,-5 \pi)$$