\documentclass{report}

\input{preamble}
\input{macros}
\input{letterfonts}

\title{\Huge{Calculus III}}
\author{\huge{Krishna Ayyalasomayajula}}
\date{}

\begin{document}

\maketitle
\newpage% or \cleardoublepage
% \pdfbookmark[<level>]{<title>}{<dest>}
\pdfbookmark[section]{\contentsname}{toc}
\tableofcontents
\pagebreak

\chapter{Lab 7}
\section{Work}

\qs{}{
  \begin{align*}
    \vec{r} \in \mathbb{R} \therefore t \ge 0 \because e^{\sqrt{t}}\notin\mathbb{R} \; t|t<0 \\
    \text{This corresponds with answer choice D}
  \end{align*}
}

\qs{}{
  \begin{align*}
    \forall t>4, \quad \vec{r}\cdot\hat{j}\notin\mathbb{R} \\
    \ln(t-1) \text{ is not defined } \forall t\le1 \\
    \text{This corresponds with answer choice D} \\
  \end{align*}
}

\qs{}{
  \begin{align*}
    \vec{r}\cdot\hat{j}\in[-4,4] \quad \land \quad \vec{r}\cdot\hat{i}\in[-3,3] \\
    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}=\langle -3\sin t ,4\cos t \rangle \\
    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \right |_{t=\tfrac{\pi}{2}}= \langle -3,0 \rangle \quad \vec{r}(0)=\langle 3,0 \rangle \\ 
    \text{This corresponds with answer choice B}
  \end{align*}
}

\qs{}{
  \begin{align*}
    \text{This can be directly evaluated to: } \\
    \langle 28,-49 \rangle \\
    \text{This corresponds with answer choice D}
  \end{align*}
}

\qs{}{
    \begin{align*}
        \vec{r}\cdot\hat{j} \text{ is not defined for } t=1 \text{ but the limit as $t\to6$ can be evaluated without affecting the proceess} \\
    \vec{r}\cdot\hat{i} \text{ is not defined for } t=1 \text{ but the limit as $t\to6$ can be evaluated without affecting the proceess} \\
    \text{Using directevaluation: } \\
    \langle \frac{5}{35},-\frac{36+12-3}{5} \rangle  = \langle \tfrac{1}{7},-9 \rangle \\
    \text{This corresponds with answer choice B}
    \end{align*}
}

\qs{}{
  \begin{align*}
    \implies \langle 0,-6 \rangle \\
    \text{This corresponds with answer choice B}
  \end{align*}
}

\qs{}{
  \begin{align*}
    e^{-\ln 6}=6^{-1}=\tfrac{1}{6} \\
    \implies \lim_{t\to\ln 6}{\langle 6e^{-t}, 3e^{-t} \rangle}=\langle 1,\tfrac{1}{2} \rangle \\
    \text{This corresponds witha nswer choice B}
  \end{align*}
}

\qs{}{
  \begin{align*}
    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle -14t,\tfrac{1}{3}t^2 \rangle \\
    \text{This corresponds with answer choice C}
  \end{align*}
}

\qs{}{
  \begin{align*}
    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle -\csc^2 t, -\cot t\csc t \rangle \\
    \text{This corresponds with answer choice A}
  \end{align*}
}
\qs{}{
\begin{align*}
  \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle 8te^{t^2}, -3, 2t \rangle \\
  \text{This corresponds with answer choice B}
\end{align*}
}

\qs{}{
  \begin{align*}
    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle 18\frac{1}{6t},6t^2 \rangle \\
    \frac{\mathrm{d}^2\vec{r}}{\mathrm{d}t^2} = \langle \frac{-108}{36t^2}, 12t \rangle  \langle \frac{-3}{t^2}, 12t \rangle \\
    \text{This corresponds with answer choice C}
  \end{align*}
}

\qs{}{

  \begin{align*}
    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}  = \langle 15t^4,-60t^4,20t^4 \rangle \\
    \| \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} \| \\
    &= \sqrt{15^2\cdot t^8 +60^2\cdot t^8 +20^2\cdot t^8} \\
    &= t^4 \sqrt{15^2+60^2+20^2} = 65t^4 \\
    \hat{T}=\langle \tfrac{15}{65},\tfrac{-60}{65},\tfrac{20}{65} \rangle \\
    \text{This corresponds with answer choice B}
  \end{align*}

}

\qs{}{

  \begin{align*}
    \vec{r}(t) &= 6\sin^3(2t)\,\hat{i} + 6\cos^3(2t)\,\hat{j} \\[6pt]
    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t}
      &= \langle 36\sin^2(2t)\cos(2t),\; -36\cos^2(2t)\sin(2t) \rangle \\[6pt]
      &= 36\sin(2t)\cos(2t)\,(\sin(2t)\,\hat{i} - \cos(2t)\,\hat{j})
  \end{align*}

  \begin{align*}
    \left\|\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}\right\|
      &= \sqrt{(36\sin^2(2t)\cos(2t))^2 + (-36\cos^2(2t)\sin(2t))^2} \\[6pt]
      &= 36\sqrt{\sin^4(2t)\cos^2(2t) + \cos^4(2t)\sin^2(2t)} \\[6pt]
      &= 36\sqrt{\sin^2(2t)\cos^2(2t)} \\[6pt]
      &= 36\,|\sin(2t)\cos(2t)| \\[6pt]
      &= 18\,|\sin(4t)|
  \end{align*}

  \begin{align*}
    \hat{T}(t)
      &= \frac{\dfrac{d\vec r}{dt}}{\left\|\dfrac{d\vec r}{dt}\right\|}
       = \frac{36\sin(2t)\cos(2t)(\sin(2t)\,\hat{i} - \cos(2t)\,\hat{j})}{36|\sin(2t)\cos(2t)|} \\[6pt]
      &= \sin(2t)\,\hat{i} - \cos(2t)\,\hat{j}
  \end{align*}

  \text{This corresponds with answer choice B.}
}

\qs{}{
  Let $\vec{C}$ encode the components of the constant of integration such that $\vec{C}\in\mathbb{R}^3$  
  \begin{align*}
    \frac{\mathrm{d}\vec{r}}{\mathrm{d}t} = \langle t^4+5t^2,4t \rangle \\
    \int\mathrm{d}\vec{r}= \int \langle t^4+5t^2,4t \rangle \mathrm{d}t \\
    \vec{r}(t)&=\langle \tfrac{1}{5}t^5+\tfrac{5}{3}t^3,2t^2 \rangle + \vec{C}\\
    \vec{C}&=\vec{r}(0)=\langle 5,-6 \rangle 
  \end{align*}
  This corresponds with answer choice C.
}

\qs{}{
  \begin{align*}
    &\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{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*}
}

\qs{}{
  \begin{align*}
    \vec{v}_0=\langle 75\cos{\alpha},75\sin{\alpha} \rangle \quad \vec{a}=\langle 0,-32 \rangle 
  \end{align*}
  This corresponds with answer choice A.
}

\qs{}{
  \begin{align*}
    \vec{v}_0=\langle 800;34\degree \rangle \quad \vec{a}=\langle 0,-9.8 \rangle \\
    \vec{r}\cdot\hat{i}=800\cos{34\degree}t \; \implies \frac{20000}{800\cdot\cos{34\degree}}=t_h=30.1554487126\;\mathrm{s} \\
    \vec{r}(t_h)\cdot\hat{j}=800\sin{34\degree}30.1554487126-\tfrac{1}{2}9.8\cdot30.1554487126^2 \\
    \vec{r}(t_h)\cdot\hat{j}=9034.35001026\ge 0
  \end{align*}
  $\therfore$ The answer is B.
}

\qs{}{
  \begin{align*}
    \vec{v}(t)=\frac{\mathrm{d}\vec{r}}{\mathrm{d}t}=\langle 585\tfrac{\sqrt{2}}{2},585\tfrac{\sqrt{2}}{2}-9.8t \rangle \\
    \vec{v}(t)\cdot\hat{j}=0 \; \implies t_h=\frac{585\tfrac{\sqrt{2}}{2}}{9.8} \implies t_h=42.2099456116 \\
    \vec{r}(t_h)\cdot\hat{j}=585\tfrac{\sqrt{2}}{2}t_h-\tfrac{1}{2}9.8t_h^2=8730.22959184 \; \mathrm{m}
  \end{align*}
  This corresponds with answer choice D.
}

\qs{}{
  \begin{align*}
    12=\|\vec{v}\|\cos{33\degree}t \quad 0=\|\vec{v}\|\sin{33\degree}t -\tfrac{1}{2}9.8t^2 \\
    \text{Computationally solving this system of equations: } \; \|\vec{v}\| = 11.34589
  \end{align*}
    This corresponds with answewr choice D.
}

\qs{}{
  \begin{align*}
    \vec{r}(t)\cdot\hat{j}=-\tfrac{1}{2}32t^2+115\sin{44\degree}t+6.6 \\
    \vec{r}(4)=70.1428504111 \;\mathrm{ft}
  \end{align*} 
  This corresponds with answer choice C.
}

\qs{}{
  \begin{align*}
    \vec{r}(t)\cdot\hat{j}=-\tfrac{1}{2}32t^2+47\sin{39\degree}t+6.1 \\
    \vec{r}(t_f)\cdot \hat{j}=0 \; t_f=2.03589\;\mathrm{s} \\
    \vec{r}(t_f)\cdot\hat{i}=2.03589\cdot47\cos{39\degree}=74.3626334991\;\mathrm{ft}
  \end{align*} 
  This corresponds with answer choice C.
}
\qs{}{
  \begin{align*}
    0=30+39\sin{26\degree}t-\tfrac{1}{2}32t^2 \\
    \implies t=2.00411 \;\mathrm{s}
  \end{align*}
  This corresponds with answer choice B.
}

\qs{}{
  \begin{align*}
    x_{max}=\frac{\|\vec{v}_0\|^2\sin{2\alpha}}{g}=14000=\frac{380^2\sin{2\alpha}}{9.8} \\
  \implies \alpha = \tfrac{1}{2}\arcsin{\frac{14000\cdot9.8}{380^2}}
  \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{}{
  \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}
