297 lines
10 KiB
TeX
297 lines
10 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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\pagebreak
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\chapter{Lab 3 - 13.2}
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\section{Work}
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\qs{}{
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Let $P$ be the plane in question.
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\begin{align*}
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P = \{ (x,y,z) \in \mathbb{R}^3 : x = 1 \}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\mathrm{dist}(P_1,P_2) = \sqrt{(5-1)^2 + (-6+1)^2 + (-5+2)^2} = \sqrt{4^2 + 5^2 +3^2} = \sqrt{16+25+9}=\sqrt{50} = 5\sqrt{2}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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5^2 = (x+8)^2 + (y-10)^2+z^2 = x^2 + 8^2 + 16x +y^2 + 100 -20y +z^2 \\
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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 $C$ be the center point of the sphere, and $r$ be the radius.
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\begin{align*}
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x^2+y^2+z^2-18x-10y-6z=-15 \\
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\implies x^2 - 18x + (\tfrac{18}{2})^2 + y^2 - 10y + (\tfrac{10}{2})^2 + z^2 - 6z + (\tfrac{6}{2})^2 = -15 + (\tfrac{18}{2})^2 + (\tfrac{10}{2})^2 + (\tfrac{6}{2})^2 \\
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\implies (x-9)^2 + (y-5)^2 + (z-3)^2 = -15+81+25+9=10+81+9=100=10^2 \\
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\therefore C = (9,5,3) \quad r=10
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\end{align*}
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}
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\qs{}{
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The set $\{ (x,y,z) \in \mathbb{R}^3 : x^2 +y^2 +z^2 >1 \}$ can be described as the set of all real points outside of a sphere with radius one centered at the origin, non-inclusive of the boundary where $x^2+y^2+z^2=1$. This corresponds with answer choice C.
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}
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\qs{}{
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\begin{align*}
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\vec{v}=\vec{PQ}=Q-P \quad Q = (4,3,-3) \quad P = (-1,-3,0) \\
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\vec{v}=\langle 4+1,3+3,-3 \rangle = \langle 5,6,-3 \rangle = 5\hat{i} + 6\hat{j} -3\hat{k} \\
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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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\begin{align*}
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\vec{v}=\vec{AB}=B-A \quad B = (-2,-13,-2) \quad A = (-7,-6,-5) \\
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\vec{v}=\langle -2+7,-13+6,-2+5 \rangle = \langle 5,-7,3 \rangle = 5\hat{i} - 7\hat{j} +3\hat{k} \\
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\text{This corresponds with answer choice B}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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M \coloneqq \mathrm{midpoint}(A,B)=(\frac{3+5}{2},\frac{5+2}{2},\frac{5+4}{2})=(4,3.5,4.5) \\
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\vec{v} = M-C = \langle 4-1,3.5-1,4.5-1 \rangle = \langle 3,2.5,3.5 \rangle \\
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\text{This corresponds with answer choice B}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\vec{v} \coloneqq 2\vec{u}-6\vec{v} \quad \vec{u}= \langle 1,1,0 \rangle \quad \vec{v}=\langle 3,0,1 \rangle \\
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\vec{v} = \langle 2,2,0 \rangle - \langle 18,0,6 \rangle = \langle -16,2,-6 \rangle = -16\hat{i}+2\hat{j}-6\hat{k} \\
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\text{This corresponds with answer choice B}
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\end{align*}
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}
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\qs{}{
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The provided vector corresponds with answer choice D since:
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\begin{align*}
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5\hat{i}+10\hat{j}+10\hat{k}=15( \tfrac{1}{3}\hat{i}+\tfrac{2}{3}\hat{j}+\tfrac{2}{3}\hat{k}) \\
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\implies 5\hat{i}+10\hat{j}+10\hat{k}= \tfrac{15}{3}\hat{i}+\tfrac{30}{3}\hat{j}+\tfrac{30}{3}\hat{k} \\
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\implies \cancel{ 5\hat{i}+10\hat{j}+10\hat{k}}= \cancel{5\hat{i}+10\hat{j}+10\hat{k}} \\
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\implies 0=0
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\end{align*}
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}
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\qs{}{
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Let $\vec{v}=\langle -1,6,0 \rangle$
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\begin{align*}
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\|\vec{v}\|=\sqrt{1^2+6^2+0^2}=\sqrt{1+36}=\sqrt{37}
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\end{align*}
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}
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\chapter{Lab 4 - 13.3, 13.4}
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\section{Work}
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\qs{}{
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\begin{align*}
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\vec{u}\coloneqq \langle -5,7 \rangle \quad \vec{v} \coloneqq \langle 1,6 \rangle \quad \vec{w} \coloneqq
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\langle -11,2 \rangle \\
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\vec{u}\cdot(\vec{v}+\vec{w}) = \vec{u}\cdot(\langle1-11,6+2\rangle)=\vec{u}\cdot\langle-10,8\rangle \\
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\implies \vec{u}\cdot(\vec{v}+\vec{w}) = -5(-10)+7(8)=50+56=106
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\vec{r}\coloneqq \langle 7,-1,-3 \rangle \quad \vec{v} \coloneqq \langle 2,6,6 \rangle \quad \vec{w} \coloneqq
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\langle 7,4,7 \rangle \\
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\vec{w}\cdot(\vec{v}+\vec{r}) = \vec{w}\cdot\langle9,5,3\rangle=63+20+21=104
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\end{align*}
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}
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\qs{}{
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Let $o$ be the work done by the system.
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\begin{align*}
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o=\langle 158;180\degree-15\degree \rangle \cdot \langle -6,0 \rangle = \langle 158\cos{180\degree-15\degree},158\sin{180\degree-15\degree} \rangle \cdot \langle -6,0 \rangle \\
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o=\langle -152.616280554, 40.8934091262 \rangle \cdot \langle -6,0 \rangle = 915.697683324 \\
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\text{This corresponds with answer choice B}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\vec{u}\cdot\vec{v}=\|\vec{u}\|\cdot\|\vec{v}\|\cos{\theta} \\
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\implies \langle 1,-1 \rangle \cdot \langle 4,5 \rangle = \sqrt{2}\sqrt{4^2+5^2}\cos{\theta} = 4-5=-1=\sqrt{2}\sqrt{25+16}\cos{\theta} \\
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\implies \theta = \arccos{\frac{-1}{\sqrt{2}\sqrt{41}}}= 96.3401917459\degree \\
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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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\begin{align*}
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\langle 4,3 \rangle \cdot \langle 20,-8 \rangle = 80-24 \ne 0 \\
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\langle 15,-6 \rangle \cdot \langle 20,-8 \rangle = 300+48 \ne 0 \\
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\langle 20,4 \rangle \cdot \langle 20,-8 \rangle = 400-32 \ne 0 \\
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\because \langle -10,-25 \rangle \cdot \langle 20,-8 \rangle = -200+200 = 0 \\
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\implies \text{Answer choice D is correct.}
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\end{align*}
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}
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\qs{}{
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If $\vec{v}\cdot\vec{w}=0$:
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\begin{align*}
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\langle 1,-x \rangle \cdot \langle -4,-3 \rangle = -4+3x=0 \\
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\implies 3x=4 \quad \therefore x=\tfrac{4}{3}
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\end{align*}
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}
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\dfn{Scalar component of a projection}{
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\begin{align*}
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\mathrm{scal}_{\vec{b}}(\vec{a}) = \frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|}
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\end{align*}
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}
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\dfn{Vector projection}{
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\begin{align*}
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\mathrm{proj}_{\vec{b}}(\vec{a}) = \left( \frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|} \right)\frac{\vec{b}}{\|\vec{b}\|}
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\end{align*}
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}
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\qs{}{
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\begin{align*}
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\vec{v}=\langle 2,3 \rangle \quad \vec{w}=\langle 8,-6 \rangle \\
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\mathrm{proj}_{\vec{w}}{\vec{v}} = \frac{\vec{v}\cdot\vec{w}}{\|\vec{w}\|} \frac{\vec{w}}{\|\vec{w}\|} = \frac{16-18}{\sqrt{8^2+6^2}}\frac{\langle 8,-6 \rangle}{\sqrt{6^2+8^2}} \\
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\implies \mathrm{proj}_{\vec{w}}{\vec{v}} = \frac{-2}{10}\cdot\langle \tfrac{8}{10}, \tfrac{-6}{10} \rangle = \tfrac{-2}{10}\cdot\tfrac{2}{10}\cdot\langle 4,-3 \rangle = \tfrac{-4}{100}\langle 4,-3 \rangle=\tfrac{-1}{25}\langle 4,-3 \rangle \\
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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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\qs{}{
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\begin{align*}
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\vec{u}=\langle 4,5 \rangle \quad \vec{v} = \langle 1,1 \rangle \\
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\mathrm{scal}_{\vec{v}}{\vec{u}}=\frac{\langle 4,5 \rangle \cdot \langle 1,1 \rangle }{\sqrt{2}} \\
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\implies \mathrm{scal}_{\vec{v}}{\vec{u}} = \frac{4+5}{\sqrt{2}} = \tfrac{9}{\sqrt{2}} \\
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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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\begin{align*}
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\vec{u}=\langle 0,7,-2 \rangle \quad \vec{v}=\langle 4,-6,-7 \rangle \\
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\vec{u}\cdot\vec{v} = \|\vec{u}\|\cdot\|\vec{v}\|\cos{\theta} = 0-42+14 =-28=\sqrt{7^2+2^2}\sqrt{4^2+6^2+7^2}\cos{\theta} \\
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\theta = \arccos{\frac{-28}{\sqrt{53}\sqrt{101}}}= 1.9635142449\;\mathrm{rad} \\
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\text{This corresponds with answer choice B}
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\end{align*}
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}
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\end{document}
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