Answer
\[
-1=\frac{-1}{1}=\tau(t)
\]
Work Step by Step
The torsion for $\vec{r}(t)$ is given by
\[
\frac{[\overrightarrow{r^{\prime}(t)} \times \overrightarrow{r^{\prime \prime}(t)}] \cdot \overrightarrow{r^{\prime \prime \prime}(t)}}{\|\overrightarrow{r^{\prime}}(t) \times \overrightarrow{r^{\prime \prime}}(t)\|^{2}}=\tau(t)
\]
Differentiating $\vec{r}(t)=\langle -\sin t+t, -\cos t+1, t\rangle$
\[
\begin{array}{l}
\langle 1-\cos t, \sin t, 1\rangle =\overrightarrow{r^{\prime}}(t)\\
\langle\sin t, \cos t, 0\rangle=\overrightarrow{r^{\prime \prime}}(t) \\
\langle\cos t,-\sin t, 0\rangle=\overrightarrow{r^{\prime \prime \prime}}
\end{array}
\]
So:
\[
\begin{aligned}
\vec{r} \times \overrightarrow{r^{\prime \prime}}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1-\cos t & \sin t & 1 \\
\sin t & \cos t & 0
\end{array}\right|=-\cos t \hat{i}+\sin t \hat{j}+\left(-\cos ^{2} t-\sin ^{2} t+1\right) \hat{k} \\
&=-\cos t \hat{i}+\sin t \hat{j}
\end{aligned}
\]
We used that
\[
\cos ^{2} t+\sin ^{2} t=1 \forall t \in \mathcal{R}
\]
So
\[
\begin{aligned}
\tau(t)=\frac{(-\cos t \hat{i}+\sin t \hat{j}) \cdot(\cos t \hat{i}-\sin t \hat{j})}{\sqrt{(-\cos t)^{2}+\sin ^{2} t}} \\
&=\frac{-\cos ^{2} t-\sin ^{2} t}{\sqrt{1}}
\end{aligned}
\]
\[
-1=\frac{-1}{1}=\tau(t)
\]
