diff --git a/labs-set-2/template.pdf b/labs-set-2/template.pdf
index ff5cd98..099c6ae 100644
Binary files a/labs-set-2/template.pdf and b/labs-set-2/template.pdf differ
diff --git a/labs-set-2/template.tex b/labs-set-2/template.tex
index cae2a0b..3ede392 100644
--- a/labs-set-2/template.tex
+++ b/labs-set-2/template.tex
@@ -544,5 +544,59 @@ This corresponds with answer choice B.
   This corresponds with answer choice D.
 }
 
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} &= \langle 1, \frac{\cancel{\sec{t}}\tan{t}}{\cancel{\sec{t}}},0 \rangle \\
+    \Big\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\Big\| &=\sqrt{1+\tan^2{t}}=\sec{t} \\
+    \vec{T}(t) &= \langle \cos{t},\sin{t} ,0 \rangle \\
+    \frac{\mathrm{d}\vec{T}}{\mathrm{d}t} &= \langle -\sin{t},\cos{t},0\rangle \\
+    \Big\|\frac{\mathrm{d}\vec{T}}{\mathrm{d}t}\Big\| &= 1 \\
+    \vec{N}(t) &= \langle -\sin{t},\cos{t},0\rangle \\
+     \vec{B}(t) &= \vec{T}(t) \times \vec{N}(t) = \begin{vmatrix}
+       \hat{i} & \hat{j} & \hat{k} \\
+       \cos{t} & \sin{t} & 0 \\
+       -\sin{t} & \cos{t} & 0
+     \end{vmatrix} \\
+                &= \hat{k}\left[\cos^2{t}+\sin^2{t}\right ] \\
+                &= \hat{k}
+  \end{align*}
+  This corresponds with answer choice C.
+}
+
+\dfn{Torsion}{
+  \begin{align*}
+    \tau = \frac{(\vec{r}^\prime \times \vec{r}^{\prime\prime})\cdot \vec{r}^{\prime\prime\prime}}{
+      \left \| \vec{r}^\prime \times \vec{r}^{\prime\prime} \right \|^2
+    }
+  \end{align*}
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} &= \langle 4\cos{\tfrac{2}{5}t},3,-4\sin{\tfrac{2}{5}t} \rangle \\
+    \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} &= \langle -\tfrac{8}{5}\sin{\tfrac{2}{5}t},0,-\tfrac{8}{5}\cos{\tfrac{2}{5}t} \rangle \\
+    \frac{\mathrm{d}^3\vec{r}}{\mathrm{d}t^3} &= \langle -\tfrac{16}{25}\cos{\tfrac{2}{5}t},0,\tfrac{16}{25}\sin{\tfrac{2}{5}t} \rangle \\
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \times \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} &= \begin{vmatrix}
+      \hat{i} & \hat{j} & \hat{k} \\
+      4\cos{\tfrac{2}{5}t}&3&-4\sin{\tfrac{2}{5}t}\\
+      -\tfrac{8}{5}\sin{\tfrac{2}{5}t}& 0 & -\tfrac{8}{5}\cos{\tfrac{2}{5}t}
+    \end{vmatrix} \\
+                                                                                           &= \hat{i}\left [-\tfrac{24}{5}\cos{\tfrac{2}{5}t} \right]-\hat{j}\left [-\tfrac{32}{5}\cos^2{\tfrac{2}{5}t}-\tfrac{32}{5}\sin^2{\tfrac{2}{5}t}\right]+\hat{k}\left[\tfrac{24}{5}\sin{\tfrac{2}{5}t}] \\
+      \left \| \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \times \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} \right \| &= \sqrt{(\tfrac{24}{5})^2 +(\tfrac{32}{5})^2}=8 \\
+      \tau = \frac{(\vec{r}^\prime \times \vec{r}^{\prime\prime})\cdot \vec{r}^{\prime\prime\prime}}{
+      \left \| \vec{r}^\prime \times \vec{r}^{\prime\prime} \right \|^2
+    } &= \frac{\langle -\tfrac{24}{5}\cos{\tfrac{2}{5}t}, \frac{32}{5}, \tfrac{24}{5}\sin{\tfrac{2}{5}t} \rangle \cdot \langle -\tfrac{-16}{25}\cos{\tfrac{2}{5}t},0,\tfrac{16}{25}\sin{\tfrac{2}{5}t} \rangle  }{8^2}\\
+          &= \left(-\frac{24}{5}\cos{\tfrac{2}{5}t}\right)\left(-\frac{16}{25}\cos{\tfrac{2}{5}t}\right)
+         + \left(\frac{32}{5}\right)(0)
+         + \left(\frac{24}{5}\sin{\tfrac{2}{5}t}\right)\left(\frac{16}{25}\sin{\tfrac{2}{5}t}\right) \\
+      &= \frac{384}{125}\cos^2{\tfrac{2}{5}t} + \frac{384}{125}\sin^2{\tfrac{2}{5}t} \\
+      &= \frac{384}{125} \\
+    \tau &= \frac{(\vec{r}^\prime \times \vec{r}^{\prime\prime})\cdot \vec{r}^{\prime\prime\prime}}{\| \vec{r}^\prime \times \vec{r}^{\prime\prime} \|^2} 
+         = \frac{384/125}{8^2} 
+         = \frac{384}{10000} 
+         = \frac{6}{125}.
+  \end{align*} 
+  This corresponds with answer choice A.
+}
 
 \end{document}