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.gitignore
vendored
2
.gitignore
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@@ -7,3 +7,5 @@
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**/*.bcf
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**/*.run.xml
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**/*.bbl
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**/**.dat
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**/**.script
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@@ -33,6 +33,12 @@
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\usepackage{tikz}
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\usepackage{titletoc}
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% add this to the preamble (near your other \usepackage lines)
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\usepackage{pgfplots}
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\pgfplotsset{compat=1.18} % or compat=newest if you prefer
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\usepgfplotslibrary{colormaps} % enables 'viridis' etc.
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\newcommand\mycommfont[1]{\footnotesize\ttfamily\textcolor{blue}{#1}}
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\SetCommentSty{mycommfont}
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\newcommand{\incfig}[1]{%
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@@ -40,6 +46,8 @@
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\import{./figures/}{#1.pdf_tex}
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}
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\usepackage{tikzsymbols}
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\tikzset{
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symbol/.style={
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@@ -32,6 +32,12 @@
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\usepackage{tikz}
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\usepackage{titletoc}
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% add this to the preamble (near your other \usepackage lines)
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\usepackage{pgfplots}
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\pgfplotsset{compat=1.18} % or compat=newest if you prefer
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\usepgfplotslibrary{colormaps} % enables 'viridis' etc.
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\newcommand\mycommfont[1]{\footnotesize\ttfamily\textcolor{blue}{#1}}
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\SetCommentSty{mycommfont}
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\newcommand{\incfig}[1]{%
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Binary file not shown.
@@ -331,4 +331,216 @@
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\end{align*}
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}
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\qs{}{
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In order to aptly encode the direction and magnitude of the side of the triangle let $\vec{s}_1, \vec{s}_2$ originate from $P$.
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\begin{align*}
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\vec{s}_1 = \langle 6-1,6-1,-3-1 \rangle = \langle 5,5,-4 \rangle \quad \vec{s}_2 = \langle 10-1,4-1,2-1 \rangle =\langle 9,3,1 \rangle \\
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A=\tfrac{1}{2}\|\vec{s}_1\times\vec{s}_2\| = \begin{vmatrix}
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\hat{i} & \hat{j} & \hat{k} \\
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5 & 5 & -4 \\
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9 & 3 & 1
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\end{vmatrix} \\
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\implies \begin{vmatrix}
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\hat{i} & \hat{j} & \hat{k} \\
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5 & 5 & -4 \\
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9 & 3 & 1
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\end{vmatrix} = \hat{i}(5+12)-\hat{j}(5+36)+\hat{k}(15-45)=\langle 17,-41,-30\rangle \\
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\implies \tfrac{1}{2}\|\vec{s}_1\times\vec{s}_2\| = \tfrac{1}{2}\cdot\sqrt{17^2+41^2+30^2} = \tfrac{1}{2}\sqrt{2870} \\
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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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\|\vec{F}\times\vec{PQ}\| =\|\vec{\tau}\| = \|\vec{PQ}\|\cdot\|\vec{F}\|\sin{45\degree}=\tfrac{9}{12}\cdot5\frac{\sqrt{2}}{2}=\frac{45\sqrt{2}}{24} = \tfrac{15}{8}\sqrt{2}\;\mathrm{ft}\cdot\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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\chapter{Lab 5 - 13.5}
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\section{Work}
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\qs{}{
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\begin{align*}
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\vec{r}=\langle 0,1,0 \rangle + \langle 3,0,-1 \rangle t \\
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\text{breaking it down into component form:}\\
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x=3t,\quad y=1,\quad z=-t \\
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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{r}=\langle -6,5,-5 \rangle t + \langle 5,-1,-5 \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{v}=P_2-P_1=\langle 3,0,4 \rangle \quad \vec{r}=\langle 3,0,4 \rangle t + \langle -3,7,3 \rangle \lor\vec{r} = \langle 3,0,4 \rangle t + \langle 0,7,7 \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{u}=\langle 6,4,4 \rangle \quad \vec{v}= \langle -7,-6,-4 \rangle \\
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\vec{v}_1\coloneqq \begin{vmatrix}
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\hat{i} & \hat{j} & \hat{k} \\
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6 &4 & 4 \\
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-7 & -6 & -4
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\end{vmatrix} = \hat{i}(-16+24)-\hat{j}(-24+28)+\hat{k}(-36+28)=\langle 8,-4,-8\rangle \\
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\vec{r}=\vec{v}_1 t+\langle -4,-7,4 \rangle \\
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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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\vec{v}_1=\langle 3,-3,-1 \rangle \quad \vec{v}_2 = \langle 4,-2,-5 \rangle \\ \nexists k | k\in \mathbb{R} \land k\cdot\vec{v}_1=\vec{v}_2\\ \therefore l_1 \nparallel l_2 \\
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1+3t &= 3+4s \\
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3-3t &= -1-2s \\
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z=-t &= 3-5s \\[6pt]
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s &= \tfrac{1}{4}(3t-2) \\
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3-3t &= -1-2\cdot\tfrac{1}{4}(3t-2) \\
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3-3t &= -1-\tfrac{1}{2}(3t-2) \\
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3-3t &= -\tfrac{3}{2}t \\
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3 &= \tfrac{3}{2}t \\
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t &= 2 \\
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s &= \tfrac{1}{4}(3\cdot 2-2)=1 \\[6pt]
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x &= 1+3(2)=7 \\
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y &= 3-3(2)=-3 \\
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z &= -2 \\[6pt]
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x &= 3+4(1)=7 \\
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y &= -1-2(1)=-3 \\
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z &= 3-5(1)=-2 \\[6pt]
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\therefore (x,y,z) &= (7,-3,-2) \\
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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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\exists k | k\in \mathbb{R} \land k\cdot\vec{v}_1=\vec{v}_2\\ \therefore l_1 \parallel l_2 \\
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3-t&=-6+5s \\
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9-t&=5s \\
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9-5s&=t \\
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1+6(9-5s)&=55-30s=1+54-30s \\
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55-30s&=55-30s \forall s\in\mathbb{R} \\
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4-2(9-5s)&=-14+10s=4-18+10s \\
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-14+10s&=-14+10s \forall s\in\mathbb{R} \\
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\therefore \text{$l_1$ and $l_2$ are identical. This corresponds to 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{n}\cdot(\vec{r}-\vec{r}_0)=0 \quad \langle 2,7,6 \rangle \cdot(\vec{r}-\langle 4,-3,2 \rangle) = 0 \\
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\vec{r}|\vec{r}=\langle x,y,z \rangle \implies \langle 2,7,6 \rangle \cdot \langle x-4,y+3,z-2 \rangle \\
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2(x-4)+7(y+3)+6(z-2)=2x-8+7y+21+6z-12=0 \\
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2x+7y+6z-8+21-12=0 \quad 2x+7y+6z=-1 \\
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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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\text{Considering answer choice A: } \\
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5(-1)-7(8)+58=-5-56+58=-3\ne 3 \therefore \cancel{\mathrm{A}} \\
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5(-1)+7(8)-58=-5+56-58=51-58=-7\ne3\therefore\cancel{\mathrm{B}} \\
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5(-1)+y(8)-58\ne-3 \therefore \cancel{\mathrm{C}} \\
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5(-1)-7(8)+58=-5-56+58=-3 \; \therefore \; \text{The answer is 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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-2x+2&=-2y \\
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-2x=2+5z&=2 \\
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z_0\coloneqq0 \quad -2y+5(0)=2 \; \implies y=-2 \\
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-2x+2(-2)&=-2=-2x-4 \\
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-2x&=2 \\
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\implies x&=-1 \\
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r_0=(-1,-2,0) \\
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\text{By parameterizing $r_0$, only answer choice D has the necessary constant terms.}
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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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x+y&=7-z \\
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7-z&=12 \\
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z&=-5 \\
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x&=t \\
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t+y&=12 \\
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y&=12-t \\
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\therefore x=t \quad y=12-t \quad z=-5
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\end{align*}
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This corresponds with answer choice B. Answer Choice A is the same line with different parametriazation to $x=-t$ instead of $x=t$
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}
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\chapter{Lab 6 - 13.6}
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\section{Work}
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\qs{}{
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\begin{tikzpicture}
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\begin{axis}[
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legend pos=outer north east,
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title=Q1,
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axis lines = box,
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xlabel = $x$,
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ylabel = $y$,
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zlabel = $z$,
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view={45}{30},
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]
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\addplot3[
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mesh,
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samples=50,
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color=blue,
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]
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{sqrt(10*x - 2*y^2/5)};
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\addplot3[
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mesh,
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samples=50,
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color=red,
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]
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{-sqrt(10*x - 2*y^2/5)};
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\addlegendentry{$z = \pm\sqrt{10x - \frac{2y^2}{5}}$}
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\end{axis}
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\end{tikzpicture}
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This image corresponds to that of an elliptical paraboloid, corresponding with answer choice A.
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}
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\qs{}{
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Since all terms are raised to the second degree, all behave linearly with respect to each other. Therefore, this is not a curved surface. Furthermore, the coefficient of the term containing $x$ is negative. Therefore, the equation represents an elliptical cone along the $x$-axis. This corresponds with answer choice D.
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}
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\qs{}{
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This surface is linear along the $y$-axis, and has a negative coefficient on the $x$ term. Therefore, the surface is a Hyperbolic paraboloid, corresponding with answer choice A.
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}
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\newpage
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\qs{}{
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The surface is linear along the $z$-axis, guaranteeing it to be a paraboloid. The $xz$-trace occurs when $y=0$. The $yz$-trace occurs when $x=0$.
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\begin{align*}
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\text{$xz$-trace: }\quad 16x^2-16(0)^2=z=16x^2 \\
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\text{$yz$-trace: }\quad 16(0)^2-16y^2-z=0=16y^2+z
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\end{align*}
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This corresponds with answer choice D.
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}
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\qs{}{
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Since two terms have a negative coefficient, and all terms are quadratic, the surface must be a hyperboloid of two sheets, leaving only answer choices B and D. A hyperboloid of two sheets only has hyperbolic cross-sections, so B must be the answer.
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}
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\qs{}{
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Since the equation only yields circular cross sections, and varies quadratically over $\hat{i},\hat{j},\hat{k}$, The answer is Figure 1, A.
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}
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\qs{}{
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The equation yield a circular cross section for any plane parallel to the $yz$-plane, and only discontinuous sets for the $xy$-plane and the $xz$-plane, the surface must be elliptical $\forall (y,z)$. Only Figure 3 satisfies the conditions, validating answer choice C.
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}
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\end{document}
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@@ -4,4 +4,8 @@
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\contentsline {section}{\numberline {2.1}Work}{5}{section.2.1}%
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\contentsline {chapter}{\numberline {3}Lab 4 - 13.3, 13.4}{7}{chapter.3}%
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\contentsline {section}{\numberline {3.1}Work}{7}{section.3.1}%
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\contentsline {chapter}{\numberline {4}Lab 5 - 13.5}{11}{chapter.4}%
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\contentsline {section}{\numberline {4.1}Work}{11}{section.4.1}%
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\contentsline {chapter}{\numberline {5}Lab 6 - 13.6}{14}{chapter.5}%
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\contentsline {section}{\numberline {5.1}Work}{14}{section.5.1}%
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\contentsfinish
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