multivar-work/labs-set-2/template.tex

\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{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{6}{(1+t^2)^2}\right \rangle \mathrm{d}t } 
  \end{align*}
}

\end{document}