diff --git a/labs-set-2/template.pdf b/labs-set-2/template.pdf
index 099c6ae..768ff36 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 3ede392..dce23e2 100644
--- a/labs-set-2/template.tex
+++ b/labs-set-2/template.tex
@@ -598,5 +598,45 @@ This corresponds with answer choice B.
   \end{align*} 
   This corresponds with answer choice A.
 }
+\qs{}{
+  \begin{align*}
+    \vec{r}(t) &= \langle 2t-7,\; t^2-2,\; 7\rangle \\
+    \vec{r}^{\prime}(t) &= \langle 2,\; 2t,\; 0\rangle \\
+    \vec{r}^{\prime\prime}(t) &= \langle 0,\; 2,\; 0\rangle \\
+    \vec{r}^{\prime\prime\prime}(t) &= \langle 0,\; 0,\; 0\rangle \\[4pt]
+    \vec{r}^{\prime}\times\vec{r}^{\prime\prime} &= 
+      \begin{vmatrix}
+        \hat{i} & \hat{j} & \hat{k} \\
+        2 & 2t & 0 \\
+        0 & 2 & 0
+      \end{vmatrix}
+      = \langle 0,0,4\rangle \\[4pt]
+    (\vec{r}^{\prime}\times\vec{r}^{\prime\prime})\cdot\vec{r}^{\prime\prime\prime} &= \langle 0,0,4\rangle\cdot\langle 0,0,0\rangle = 0 \\[4pt]
+    \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{0}{\| \langle0,0,4\rangle\|^2} = 0.
+  \end{align*}
+  This corresponds with answer choice A.
+}
+\qs{}{
+  \begin{align*}
+    \vec{r}(t) &= \langle 3\sin t,\; 3\cos t,\; -t\rangle \\
+    \vec{r}^{\prime}(t) &= \langle 3\cos t,\; -3\sin t,\; -1\rangle \\
+    \vec{r}^{\prime\prime}(t) &= \langle -3\sin t,\; -3\cos t,\; 0\rangle \\
+    \vec{r}^{\prime\prime\prime}(t) &= \langle -3\cos t,\; 3\sin t,\; 0\rangle \\[4pt]
+    \vec{r}^{\prime}\times\vec{r}^{\prime\prime} &=
+      \begin{vmatrix}
+        \hat{i} & \hat{j} & \hat{k} \\
+        3\cos t & -3\sin t & -1 \\
+        -3\sin t & -3\cos t & 0
+      \end{vmatrix}
+      = \langle -3\cos t,\; 3\sin t,\; -9\rangle \\[4pt]
+    \| \vec{r}^{\prime}\times\vec{r}^{\prime\prime}\|^2 &= (-3\cos t)^2+(3\sin t)^2+(-9)^2 = 9+81=90 \\[4pt]
+    (\vec{r}^{\prime}\times\vec{r}^{\prime\prime})\cdot\vec{r}^{\prime\prime\prime}
+      &= (-3\cos t)(-3\cos t) + (3\sin t)(3\sin t) + (-9)(0) = 9(\cos^2 t+\sin^2 t)=9 \\[4pt]
+    \tau &= \frac{9}{90} = \frac{1}{10}.
+  \end{align*}
+  This corresponds with answer choice B.
+}
 
 \end{document}