started lab 10

labs-set-2/template.tex

@@ -303,4 +303,130 @@ This corresponds with answer choice B.
   \end{align*}
   This corresponds with answer choice C.
 }
+
+\chapter{Lab 9}
+\section{Work}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}= \langle -2,6,9 \rangle \quad \|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| = \sqrt{2^2+6^2+9^2}=\sqrt{4+36+81}=11 \\
+    L = \int_a^b{\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| \mathrm{d}t} = \int_{-10}^{-3}{11\mathrm{d}t}=11\left [ t \right]_{-10}^{-3}=11(-3+10)=77
+  \end{align*}
+  This corresponds with answer choice C.
+}
+
+\qs{}{
+  \begin{align*}
+        \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}= \langle 2\sin{t}+2t\cos{t}-2\sin{t},2\cos{t}-2t\sin{t}-2\cos{t} \rangle \\ 
+        \|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| = \sqrt{(2t)^2\cos^2{t}+(2t)^2\sin^2{t}}=2t \\
+        L = \int_a^b{\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| \mathrm{d}t} = \int_{-1}^7{2t\mathrm{dt}}=\left[ t^2 \right]_{-1}^7=[49-1]=48
+  \end{align*}
+  This corresponds with answer choice C.
+}
+
+\qs{}{
+  \begin{align*}
+     \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}= \langle 30t^2,33t^2,6t^2 \rangle \\
+     \|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| = \sqrt{30^2\cdot t^4 +33^2 \cdot t^4 + 6^2 \cdot t^4}=t^2\sqrt{30^2+33^2+6^2}= 45 t^2 \\
+     L = \int_a^b{\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| \mathrm{d}t} = \int_{-1}^{2}{45t^2}=15\left [t^2 \right]_{-1}^{2} =15(4-1)=45 
+  \end{align*} 
+  This corresponds with answer choice B.
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle 0, -18\cdot 7 \cos^2{7t}\sin{7t}, 18\cdot 7 \sin^2{7t}\cos{7t} \rangle = \langle 0, -126\cos^2{7t}\sin{7t},126\sin^2{7t}\cos{7t} \rangle \\
+    \|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| = \sqrt{126^2 \cos^4{7t}\sin^2{7t}+126^2\sin^4{7t}\cos^2{27}} \\
+    \implies 126\sqrt{(\sin^2{7t}\cos^2{7t})(\cos^2{7t}+\sin^2{7t})} =126\sqrt{\sin^2{7t}\cos^2{7t}}=126\sin{7t}\cos{7t} \\
+    L = \int_a^b{\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| \mathrm{d}t} = \int_{\tfrac{6\pi}{7}}^{\tfrac{13\pi}{14}}{126\sin{7t}\cos{7t}\mathrm{d}t} = 63\int_{\tfrac{6\pi}{7}}^{\tfrac{13\pi}{14}}{\sin{14t}\mathrm{d}t} \\
+    u=14t \quad \mathrm{d}u=14\mathrm{d}t \quad \implies L=\frac{-63}{14}\left [\cos{14t} \right ]_{\tfrac{6\pi}{7}}^{\tfrac{13\pi}{14}}=\tfrac{-9}{2} = \frac{-9}{2}(-1-1)=9
+  \end{align*}
+  This corresponds with answer choice B.
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}= \langle e^t\cos{t}-e^t\sin{t}, e^t\sin{t}+e^t\cos{t},5e^t \rangle \\ 
+    \|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| = \sqrt{e^{2t}(\cos{t}-\sin{t})^2+e^{2t}(\sin{t}+\cos{t})^2+25e^{2t}}=e^t\sqrt{1+\sin{2t}+1-\sin{2t}+25}=e^t\sqrt{27}=3\sqrt{3}e^t \\
+    L = \int_a^b{\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| \mathrm{d}t} = 3\sqrt{3}\int_{-\ln{2}}^{0}{e^t \mathrm{d}t} = 3\sqrt{3}(1-\tfrac{1}{2})=\frac{3\sqrt{3}}{2}
+  \end{align*}
+  This corresponds with answer choice D.
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}= \langle 6t^2, 6t^2, -3t^2 \rangle  \\
+    \|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| = \sqrt{36t^4+36t^4+9t^4}=t^2\sqrt{36+36+9}= 9t^2 \\
+    L = \int_a^b{\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| \mathrm{d}t} = 3\int_{-1}^{2}{3t^2 \mathrm{d} t} \\
+     \implies 3\left [ t^3 \right ]_{-1}^2=3(8+1)=27
+  \end{align*}
+  This corresponds with answer choice B.
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle 4\cos{2t},-4\sin{2t},3 \rangle \\
+    \|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| =  \sqrt{16\cos^2{2t}+16\sin^2{2t}+9} = \sqrt{16+9}=5 \\
+    \vec{T}(t)=\tfrac{1}{5}\langle 4\cos{2t},-4\sin{2t},3 \rangle 
+  \end{align*}
+  This corresponds with answer choice A. 
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle 84t^6,-28t^6,21t^6 \rangle \\
+    \|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| = t^6\sqrt{84^2+28^2+21^2}=91t^6 \\
+    \vec{T}(t)=\tfrac{1}{91}\langle 84t^6,-28t^6,21t^6 \rangle 
+  \end{align*}
+  This corresponds with answer choice B.
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle 0,6\cos{t}-6t\sin{t}-6\cos{t},6t\cos{t}+6\sin{t}-6\sin{t} \rangle \\
+    \|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| = \sqrt{36t^2\sin^2{t}+36t^2\cos^2{t}} = 6t \\
+    \vec{T}(t)=\frac{1}{6t}\langle 0,-6t\sin{t},6t\cos{t} \rangle =\langle 0,-\sin{t},\cos{t} \rangle 
+  \end{align*}
+  This corresponds with answer choice D.
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle -54\sin{9t},-54\cos{9t},0 \rangle \\
+    \|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| = 54\sqrt{\sin^2{9t}+\cos^2{9t}} = 54 \\ 
+    \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} = \langle -486\cos{9t},486\sin{9t},0 \rangle \\
+    \kappa=\frac{\| \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \times \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} \|}{\| \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \|^3} \\
+    \implies \kappa = \frac{\left \| \begin{vmatrix}
+        \hat{i} & \hat{j} & \hat{k} \\
+        -54\sin{9t} & -54\cos{9t} & 0 \\
+        -486\cos{9t} & 486\sin{9t} & 0 
+    \end{vmatrix} \right \|}{54^3} = -26244\cdot \tfrac{1}{157464}(\cos^2{9t}+\sin^2{9t}) \\
+    \implies \kappa = \frac{1}{6}
+  \end{align*}
+  This corresponds with answer choice C.
+}
+
+\qs{}{
+  \begin{align*}
+    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle 1, 0, \frac{\cancel{\sec{t}}\tan{t}}{\cancel{\sec{t}}} \rangle \\
+    \|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\| = \sqrt{1+\tan^2{t}}=\sqrt{\sec^2{t}}=\sec{t} \\
+    \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} = \langle 0,0,\sec^2{t} \rangle \\
+    \kappa=\frac{\| \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \times \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} \|}{\| \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \|^3} \\
+     \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} \\
+       1 & 0 & \tan{t} \\
+       0 & 0 & \sec^2{t}
+     \end{vmatrix} = -\hat{j}[\sec^2{t}] = \langle 0, -\sec^2{t},0 \rangle \\
+     \| \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \times \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} \| = \sec^2{t} \\
+     \implies \kappa = \frac{\sec^2{t}}{\sec^3{t}}=\cos{t}
+  \end{align*}
+}
+
+\chapter{Lab 10}
+\section{Work}
+
+\qs{}{
+
+}
+
 \end{document}

labs-set-2/template.toc

@@ -2,4 +2,8 @@
 \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}%
+\contentsline {chapter}{\numberline {3}Lab 9}{9}{chapter.3}%
+\contentsline {section}{\numberline {3.1}Work}{9}{section.3.1}%
+\contentsline {chapter}{\numberline {4}Lab 10}{13}{chapter.4}%
+\contentsline {section}{\numberline {4.1}Work}{13}{section.4.1}%
 \contentsfinish