115 lines
3.9 KiB
TeX
115 lines
3.9 KiB
TeX
\documentclass{report}
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\input{preamble}
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\input{macros}
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\input{letterfonts}
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\title{\Huge{Calculus III}\\Homework \# 1}
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\author{\huge{Krishna Ayyalasomayajula}}
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\date{}
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\begin{document}
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\maketitle
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\newpage% or \cleardoublepage
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% \pdfbookmark[<level>]{<title>}{<dest>}
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\pdfbookmark[section]{\contentsname}{toc}
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\tableofcontents
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\pagebreak
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\chapter{Lab 2 - 13.1 Apps}
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\section{Work}
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\qs{}{
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Let $\vec{T_1}$ represent the tension of the leftmost cable, while $\vec{T_2}$ encodes the tension force experienced by the rightmost cable. Our coordinate system will originate at the intersection of the two cables.
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\begin{align*}
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\vec{T_1} \coloneqq \langle \|\vec{T_1}\| ; 135\degree \rangle \\
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\vec{T_2} \coloneqq \langle \|\vec{T_2}|\| ; 15\degree \rangle \\
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500=\|\vec{T_1}\|\sin{135\degree} + \|\vec{T_2}\|\sin{15} \\
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0 = \|\vec{T_1}\|\cos{135\degree} + \|\vec{T_2}\|\cos{15} \\
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\implies 500=\|\vec{T_1}\|\tfrac{\sqrt{2}}{2} + \|\vec{T_2}\|\cdot0.258819045103 \\
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\implies0 = \|\vec{T_1}\|\tfrac{-\sqrt{2}}{2} + \|\vec{T_2}\|\cdot0.965925826289 \\
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\text{Solving the system numerically yields: }\\
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\|\vec{T_1}\| \Rightarrow 557.677\, \mathrm{lb}\\
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\|\vec{T_2}\| \Rightarrow 408.248\,\mathrm{lb} \\
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\text{This corresponds to answer choice A}
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\end{align*}
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}
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\qs{}{
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Let $\vec{T}$ represent the tension in the cable, and $\vec{C}$ represent the compression force experienced by the boom in neutralizing the horizontal component of $\vec{T}$.
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\begin{align*}
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\vec{T}\coloneqq \langle \|\vec{T}\| ; 142\degree \rangle \\
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\|\vec{T}\|\sin{142\degree} = 450 \, \mathrm{lb} \\
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\|\vec{T}\| = \frac{450 \, \mathrm{lb}}{\sin{142\degree}} \\
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\|\vec{T}\| \Rightarrow 730.921 \, \mathrm{lb} \\
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\|\vec{C}\| = |(\|\vec{T}\|\cos{142\degree})|\Rightarrow 575.973 \, \mathrm{lb} \\
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\text{ This corresponds to answer choice C}
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\end{align*}
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}
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\qs{}{
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Let the $x'$-axis of the new coordinate system be oriented along the unit vector $\hat{u}=\hat{T_1}$:
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\begin{align*}
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\vec{T_1} \coloneqq \langle 3600,0 \rangle \\
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\vec{T_2} \coloneqq \langle 1800;45\degree \rangle \\
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\|\vec{T_1}+\vec{T_2}\| = \langle 3600+1800\cos{45\degree},1800\sin{45\degree} \rangle = \langle 3600+1800\tfrac{\sqrt{2}}{2},1800\tfrac{\sqrt{2}}{2} \rangle \\
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\|\vec{T_1}+\vec{T_2}\| \Rightarrow 5036.278 \, \mathrm{lb} \\
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\text{This corresponds with answer choice D}
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\end{align*}
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}
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\qs{}{
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Let the force vector be $\vec{F}$.
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\begin{align*}
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\vec{F} \coloneqq 8\cdot\cos{30\degree}\hat i + 8\cdot\sin{30\degree}\hat j = 8\cdot\tfrac{\sqrt{3}}{2}\hat i + 8\cdot\tfrac{1}{2}\hat j = 4\cdot\sqrt{3} \hat i + 4 \hat j\\
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\text{This corresponds with answer choice A}
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\end{align*}
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}
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\qs{}{
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Let the velocity vector be $\vec{v_0}$.
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\begin{align*}
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\vec{v_0} \coloneqq 5\cdot\cos{56\degree}\hat i + 5\cdot\sin{56\degree}\hat j = 2.795 \hat i + 4.145 \hat j\\
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\text{This corresponds with answer choice A}
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\end{align*}
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}
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\qs{}{
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Let the velocity vector be $\vec{v_0}$.
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\begin{align*}
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\vec{v_0} \coloneqq \langle 817 ; 140\degree \rangle = \langle 817\cos{140\degree},817\sin{140\degree} \rangle \Rightarrow \langle -625.858,525.157 \rangle \\
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\text{This corresponds with answer choice A}
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\end{align*}
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}
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\qs{}{
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Let the wind's parameters be encoded in the vector $\vec{W}$.
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\begin{align*}
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\vec{W}\coloneqq\langle 5,14 \rangle \\
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\implies \theta = \arctan{\tfrac{14}{5}} \Rightarrow 70.346 \degree \\
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\text{This corresponds with answer choice C}
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\end{align*}
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}
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\pagebreak
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\qs{}{
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Let the boat's target velocity vector be $\vec{v}$.
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\begin{align*}
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\vec{v}\coloneqq \langle 0,31 \rangle - \langle -6,0 \rangle \\
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\vec{v} = \langle 6,31 \rangle \\
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\implies \theta = \arctan{\tfrac{31}{6}} \Rightarrow 79.04593\degree \\
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\text{This corresponds with answer choice C, when expressed as a bearing East of North}
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\end{align*}
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}
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\end{document}
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