halfway through the second lab

labs-set-2/template.tex

@@ -159,8 +159,82 @@
 
 \qs{}{
   \begin{align*}
-    \implies \int_0^3{\left \langle \frac{2}{\sqrt{1+t}},-9t^2,\frac{6}{(1+t^2)^2}\right \rangle \mathrm{d}t }\\
+    &\implies \int_0^3{\left \langle \frac{2}{\sqrt{1+t}},-9t^2,\frac{6t}{(1+t^2)^2}\right \rangle \mathrm{d}t }\\
-    \implies \int_0^3{\left \langle 2(1+t)^{-\tfrac{1}{2}},-9t^2,\frac{6}{(1+t^2)^2}\right \rangle \mathrm{d}t } 
+    &\implies \int_0^3{\left \langle 2(1+t)^{-\tfrac{1}{2}},-9t^2,\frac{6t}{(1+t^2)^2}\right \rangle \mathrm{d}t }\\
+&\implies \left [ \langle 4\sqrt{t+1},-3t^3,\frac{-3}{1+t^2} \right ]_0^3=\langle 8-4,-81, \tfrac{-3}{10}+3 \rangle = \langle 4,-81,\tfrac{27}{10} \rangle 
+  \end{align*}
+
+  This corresponds with answer choice C.
+}
+
+\qs{}{
+  \begin{align*}
+    u \coloneqq t^2+1\;\mathrm{du}=2t\mathrm{d}t \quad \implies \left \langle \left [8t^3-3t \right ]_0^1,\int_0^1{\frac{4}{u} \mathrm{d}u}, -\int_0^1{\frac{1}{2\sqrt{u}}\mathrm{d}u} \right \rangle \\
+    \left [ \left \langle 8t^3-3t, 4\ln{(t^2+1)},-\sqrt{t^2+1} \right \rangle \right ]_0^1 = \langle 8-3, 4\ln{2}-0,-\sqrt{2}+1\rangle   
+  \end{align*}
+
+  This corresponds with answer choice A.
+}
+
+\chapter{Lab 8}
+\section{Work}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} =\vec{v}=\langle -14t,\tfrac{1}{7}t^2 \rangle 
+  \end{align*}
+  This corresponds with answer choice B.
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} =\vec{v} = \langle -3\sin(3t),5\cos{t} \rangle \\
+    \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} =\vec{a} = \langle -9\cos(3t),-5\sin{t} \rangle
+  \end{align*}
+  This corresponds with answer choice D.
+}
+
+\qs{}{
+  \begin{align*}
+    \vec{v}(t)=\langle 18t+4, -15t^2,-2t \rangle \\
+    \vec{v}(3)=\langle 58,-135,-6\rangle 
+  \end{align*}
+  This correspon ds with answer choice B.
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} =\vec{v} = \langle -2\sin(2t),\frac{7}{t-3},-\tfrac{1}{3}t^2\rangle \\ 
+    \vec{v}(0)=\langle 0,\tfrac{-7}{3},0 \rangle
+  \end{align*}
+
+  This corresponds with answer choice D.
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} &= \vec{v} 
+      = \langle 8\cos(2t), \; 10\sin(2t), \; -6\csc(2t)\cot(2t) \rangle \\[6pt]
+    \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} &= \vec{a} 
+      = \langle -16\sin(2t), \; 20\cos(2t), \; 12(\csc(2t)\cot^2(2t) + \csc^3(2t)) \rangle \\[6pt]
+    \vec{a}\!\left(\tfrac{\pi}{4}\right) &= \langle -16\sin\!\big(\tfrac{\pi}{2}\big), \; 20\cos\!\big(\tfrac{\pi}{2}\big), \; 12(1\cdot 0^2 + 1^3) \rangle \\[6pt]
+    &= \langle -16,\; 0,\; 12 \rangle
+  \end{align*}
+This corresponds with answer choice B.
+}
+
+
+
+\qs{}{
+  \begin{align*}
+    \vec{v}(0) &= \langle \sqrt{2}, 0, \sqrt{2} \rangle, &
+    \vec{a}(0) &= \langle 0, 0, \tfrac{\pi}{2} \rangle, \\[2mm]
+    \vec{a}\cdot\vec{v} &= 0\cdot \sqrt{2} + 0\cdot 0 + \tfrac{\pi}{2}\cdot \sqrt{2} = \tfrac{\pi\sqrt{2}}{2}, \\[1mm]
+    \|\vec{a}\| &= \tfrac{\pi}{2}, &
+    \|\vec{v}\| &= \sqrt{(\sqrt{2})^2 + 0^2 + (\sqrt{2})^2} = \sqrt{4} = 2, \\[1mm]
+    \cos\theta &= \frac{\vec{a}\cdot\vec{v}}{\|\vec{a}\|\|\vec{v}\|} 
+      = \frac{\tfrac{\pi\sqrt{2}}{2}}{\tfrac{\pi}{2}\cdot 2} = \frac{\sqrt{2}}{2}, \\[1mm]
+    \theta &= \arccos\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4}.
   \end{align*}
 }
 

labs-set-2/template.toc

@@ -1,3 +1,5 @@
 \contentsline {chapter}{\numberline {1}Lab 7}{2}{chapter.1}%
 \contentsline {section}{\numberline {1.1}Work}{2}{section.1.1}%
+\contentsline {chapter}{\numberline {2}Lab 8}{6}{chapter.2}%
+\contentsline {section}{\numberline {2.1}Work}{6}{section.2.1}%
 \contentsfinish