diff --git a/labs-set-2/template.pdf b/labs-set-2/template.pdf
index ec29ecd..ff5cd98 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 4bbb8b1..cae2a0b 100644
--- a/labs-set-2/template.tex
+++ b/labs-set-2/template.tex
@@ -426,7 +426,123 @@ This corresponds with answer choice B.
 \section{Work}
 
 \qs{}{
-
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle 1,0, \frac{-\sin{t}}{\cos{t}} \rangle = \langle 1,0,-\tan{t} \rangle \\
+    \|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| = \sqrt{1+\tan^2{t}}=\sec{t} \\
+    \vec{T}(t)=\cos{t}\cdot \langle 1,0,-\tan{t} \rangle = \langle \cos{t},0,-\sin{t} \rangle \\
+    \implies \vec{T}\prime(t) = \langle -\sin{t},0,-\cos{t} \rangle \\
+    \implies \|\vec{T}^\prime(t)\|=\sqrt{\sin^2{t}+\cos^2{t}}=1 \\
+    \vec{N}(t)=\langle -\sin{t},0,-\cos{t} \rangle 
+  \end{align*}
+  Corresponds with answer choice D.
 }
 
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} &= \langle 0,2t,2 \rangle \\
+    \Big\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\Big\| &= 2\sqrt{t^2+1} \\
+    \vec{T}(t) &= \frac{\langle 0,t,1 \rangle}{\sqrt{t^2+1}} \\[6pt]
+    \frac{\mathrm{d}\vec{T}}{\mathrm{d}t} &=
+      \left\langle 
+        0,\,
+        \frac{\mathrm{d}}{\mathrm{d}t}\Big(\frac{t}{\sqrt{t^2+1}}\Big),\,
+        \frac{\mathrm{d}}{\mathrm{d}t}\Big(\frac{1}{\sqrt{t^2+1}}\Big)
+      \right\rangle \\[6pt]
+    &=
+      \left\langle
+        0,\,
+        (t^2+1)^{-1/2} - \frac{t^2}{(t^2+1)^{3/2}},\,
+        -\frac{t}{(t^2+1)^{3/2}}
+      \right\rangle \\[6pt]
+    &=
+      (t^2+1)^{-3/2}\langle 0,1,-t\rangle \\[6pt]
+    \Big\|\frac{\mathrm{d}\vec{T}}{\mathrm{d}t}\Big\| &= (t^2+1)^{-1} \\[6pt]
+    \vec{N}(t) &=
+      \frac{\frac{\mathrm{d}\vec{T}}{\mathrm{d}t}}
+           {\Big\|\frac{\mathrm{d}\vec{T}}{\mathrm{d}t}\Big\|}
+      = \frac{(t^2+1)^{-3/2}\langle 0,1,-t\rangle}
+             {(t^2+1)^{-1}}
+      = \frac{\langle 0,1,-t\rangle}{\sqrt{t^2+1}}.
+  \end{align*}
+  This corresponds with answer choice B.
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} &= \langle 2t\cos{t}+\cancel{2\sin{t}-2\sin{t}},0,\cancel{2\cos{t}}-2t\sin{t}-\cancel{2\cos{t}} \rangle \\
+    \Big\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\Big\| &= 2t\sqrt{\cos^2{t}+\sin^2{t}}=2t \\
+    \vec{T}(t) &= \langle \cos{t},0,-\sin{t} \rangle \\
+    \frac{\mathrm{d}\vec{T}}{\mathrm{d}t} &= \langle -\sin{t},0,-\cos{t} \rangle \\
+    \Big\|\frac{\mathrm{d}\vec{T}}{\mathrm{d}t}\Big\| &= 1 \\
+    \vec{N}(t) &= \langle -\sin{t},0,-\cos{t} \rangle 
+  \end{align*}
+  This corresponds with answer choice C.
+}
+\newpage
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} &= \langle 4\cos{2t},-4\sin{2t},3 \rangle \\
+    \Big\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\Big\| &= \sqrt{4^2(\cos^2{2t}+\sin^2{2t})+3^2}=5 \\
+    \vec{T}(t) &= \langle \tfrac{4}{5}\cos{2t},-\tfrac{4}{5}\sin{2t},\tfrac{3}{5} \rangle \\
+    \frac{\mathrm{d}\vec{T}}{\mathrm{d}t} &= \langle -\tfrac{8}{5}\sin{2t},-\tfrac{8}{5}\cos{2t},0\rangle \\
+    \Big\|\frac{\mathrm{d}\vec{T}}{\mathrm{d}t}\Big\| &= \frac{8}{5} \\
+     \vec{N}(t) &= \langle -\sin{2t},-\cos{2t},0 \rangle 
+  \end{align*}
+  This corresponds with answer choice C.
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} &= \langle 6\cos{t},-6\sin{t},-1 \rangle \\
+    \Big\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\Big\| &= \sqrt{6^2+1}=\sqrt{37} \\
+    \vec{T}(t) &=\frac{1}{\sqrt{37}} \langle 6\cos{t},-6\sin{t},-1 \rangle \\
+    \frac{\mathrm{d}\vec{T}}{\mathrm{d}t} &= \frac{1}{\sqrt{37}}\langle -6\sin{t},-6\cos{t},0 \rangle \\
+    \Big\|\frac{\mathrm{d}\vec{T}}{\mathrm{d}t}\Big\| &= \sqrt{\frac{6^2}{\sqrt{37}^2}(\sin^2{t}+\cos^2{t})}
+      = \frac{6}{\sqrt{37}} \\
+    \vec{N}(t) &= \frac{1}{\sqrt{37}}\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} \\
+      \tfrac{6}{\sqrt{37}}\cos{t} & \tfrac{-6}{\sqrt{37}}\sin{t} & \tfrac{-1}{\sqrt{37}} \\
+      \tfrac{-1}{\sqrt{37}}\sin{t} & \tfrac{-1}{\sqrt{37}}\cos{t} & 0
+    \end{vmatrix} \\[6pt]
+    \vec{B}(t)&=\frac{-\langle \cos{t},-\sin{t},6 \rangle }{\sqrt{37}}
+  \end{align*}
+  This corresponds with answer choice D.
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} &= \langle t,-7,0 \rangle \\
+    \Big\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\Big\| &= \sqrt{t^2+49} \\
+    \vec{T}(t) &= \langle \frac{t}{\sqrt{t^2+49}},\frac{-7}{\sqrt{t^2+49}},0 \rangle \\
+    \frac{\mathrm{d}\vec{T}}{\mathrm{d}t} &= 
+      \left\langle 
+        \frac{\sqrt{t^2+49}-t\cdot \frac{t}{\sqrt{t^2+49}}}{t^2+49},\,
+        \frac{0 - (-7)\cdot \frac{t}{\sqrt{t^2+49}}}{t^2+49},\,
+        0
+      \right\rangle \\
+    &= \left\langle 
+        \frac{49}{(t^2+49)^{3/2}},\,
+        \frac{7t}{(t^2+49)^{3/2}},\,
+        0
+      \right\rangle \\
+    \Big\|\frac{\mathrm{d}\vec{T}}{\mathrm{d}t}\Big\| &= \frac{\sqrt{49^2 + (7t)^2}}{(t^2+49)^{3/2}} 
+      = \frac{7\sqrt{t^2+49}}{(t^2+49)^{3/2}} 
+      = \frac{7}{t^2+49} \\
+    \vec{N}(t) &= \frac{\frac{\mathrm{d}\vec{T}}{\mathrm{d}t}}{\Big\|\frac{\mathrm{d}\vec{T}}{\mathrm{d}t}\Big\|} 
+      = \frac{\langle 49,7t,0\rangle/(t^2+49)^{3/2}}{7/(t^2+49)} 
+      = \frac{\langle 7, t, 0 \rangle}{\sqrt{t^2+49}} \\
+    \vec{B}(t) &= \vec{T}(t) \times \vec{N}(t) 
+      = \begin{vmatrix}
+          \hat{i} & \hat{j} & \hat{k} \\
+          t/\sqrt{t^2+49} & -7/\sqrt{t^2+49} & 0 \\
+          7/\sqrt{t^2+49} & t/\sqrt{t^2+49} & 0
+        \end{vmatrix} \\
+      &= \frac{1}{t^2+49} \langle 0,0, t^2+49 \rangle 
+      = \langle 0,0,1 \rangle
+  \end{align*}
+  This corresponds with answer choice D.
+}
+
+
 \end{document}