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\documentclass{report}
\input{preamble}
\input{macros}
\input{letterfonts}
\title{\Huge{Calculus III}\\Homework \# 1}
\author{\huge{Krishna Ayyalasomayajula}}
\date{}
\begin{document}
\maketitle
\newpage% or \cleardoublepage
% \pdfbookmark[<level>]{<title>}{<dest>}
\pdfbookmark[section]{\contentsname}{toc}
\tableofcontents
\pagebreak
\chapter{Lab 2 - 13.1 Apps}
\section{Work}
\qs{}{
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.
\begin{align*}
\vec{T_1} \coloneqq \langle \|\vec{T_1}\| ; 135\degree \rangle \\
\vec{T_2} \coloneqq \langle \|\vec{T_2}|\| ; 15\degree \rangle \\
500=\|\vec{T_1}\|\sin{135\degree} + \|\vec{T_2}\|\sin{15} \\
0 = \|\vec{T_1}\|\cos{135\degree} + \|\vec{T_2}\|\cos{15} \\
\implies 500=\|\vec{T_1}\|\tfrac{\sqrt{2}}{2} + \|\vec{T_2}\|\cdot0.258819045103 \\
\implies0 = \|\vec{T_1}\|\tfrac{-\sqrt{2}}{2} + \|\vec{T_2}\|\cdot0.965925826289 \\
\text{Solving the system numerically yields: }\\
\|\vec{T_1}\| \Rightarrow 557.677\, \mathrm{lb}\\
\|\vec{T_2}\| \Rightarrow 408.248\,\mathrm{lb} \\
\text{This corresponds to answer choice A}
\end{align*}
}
\qs{}{
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 $\vec{T}$.
\begin{align*}
\vec{T}\coloneqq \langle \|\vec{T}\| ; 142\degree \rangle \\
\|\vec{T}\|\sin{142\degree} = 450 \, \mathrm{lb} \\
\|\vec{T}\| = \frac{450 \, \mathrm{lb}}{\sin{142\degree}} \\
\|\vec{T}\| \Rightarrow 730.921 \, \mathrm{lb} \\
\|\vec{C}\| = |(\|\vec{T}\|\cos{142\degree})|\Rightarrow 575.973 \, \mathrm{lb} \\
\text{ This corresponds to answer choice C}
\end{align*}
}
\qs{}{
Let the $x'$-axis of the new coordinate system be oriented along the unit vector $\hat{u}=\hat{T_1}$:
\begin{align*}
\vec{T_1} \coloneqq \langle 3600,0 \rangle \\
\vec{T_2} \coloneqq \langle 1800;45\degree \rangle \\
\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 \\
\|\vec{T_1}+\vec{T_2}\| \Rightarrow 5036.278 \, \mathrm{lb} \\
\text{This corresponds with answer choice D}
\end{align*}
}
\qs{}{
Let the force vector be $\vec{F}$.
\begin{align*}
\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\\
\text{This corresponds with answer choice A}
\end{align*}
}
\qs{}{
Let the velocity vector be $\vec{v_0}$.
\begin{align*}
\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\\
\text{This corresponds with answer choice A}
\end{align*}
}
\qs{}{
Let the velocity vector be $\vec{v_0}$.
\begin{align*}
\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 \\
\text{This corresponds with answer choice A}
\end{align*}
}
\qs{}{
Let the wind's parameters be encoded in the vector $\vec{W}$.
\begin{align*}
\vec{W}\coloneqq\langle 5,14 \rangle \\
\implies \theta = \arctan{\tfrac{14}{5}} \Rightarrow 70.346 \degree \\
\text{This corresponds with answer choice C}
\end{align*}
}
\pagebreak
\qs{}{
Let the boat's target velocity vector be $\vec{v}$.
\begin{align*}
\vec{v}\coloneqq \langle 0,31 \rangle - \langle -6,0 \rangle \\
\vec{v} = \langle 6,31 \rangle \\
\implies \theta = \arctan{\tfrac{31}{6}} \Rightarrow 79.04593\degree \\
\text{This corresponds with answer choice C, when expressed as a bearing East of North}
\end{align*}
}
\pagebreak
\chapter{Lab 3 - 13.2}
\section{Work}
\qs{}{
Let $P$ be the plane in question.
\begin{align*}
P = \{ (x,y,z) \in \mathbb{R}^3 : x = 1 \}
\end{align*}
}
\qs{}{
\begin{align*}
\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}
\end{align*}
}
\qs{}{
\begin{align*}
5^2 = (x+8)^2 + (y-10)^2+z^2 = x^2 + 8^2 + 16x +y^2 + 100 -20y +z^2 \\
\text{This corresponds with answer choice A}
\end{align*}
}
\qs{}{
Let $C$ be the center point of the sphere, and $r$ be the radius.
\begin{align*}
x^2+y^2+z^2-18x-10y-6z=-15 \\
\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 \\
\implies (x-9)^2 + (y-5)^2 + (z-3)^2 = -15+81+25+9=10+81+9=100=10^2 \\
\therefore C = (9,5,3) \quad r=10
\end{align*}
}
\qs{}{
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.
}
\qs{}{
\begin{align*}
\vec{v}=\vec{PQ}=Q-P \quad Q = (4,3,-3) \quad P = (-1,-3,0) \\
\vec{v}=\langle 4+1,3+3,-3 \rangle = \langle 5,6,-3 \rangle = 5\hat{i} + 6\hat{j} -3\hat{k} \\
\text{This corresponds with answer choice A}
\end{align*}
}
\qs{}{
\begin{align*}
\vec{v}=\vec{AB}=B-A \quad B = (-2,-13,-2) \quad A = (-7,-6,-5) \\
\vec{v}=\langle -2+7,-13+6,-2+5 \rangle = \langle 5,-7,3 \rangle = 5\hat{i} - 7\hat{j} +3\hat{k} \\
\text{This corresponds with answer choice B}
\end{align*}
}
\qs{}{
\begin{align*}
M \coloneqq \mathrm{midpoint}(A,B)=(\frac{3+5}{2},\frac{5+2}{2},\frac{5+4}{2})=(4,3.5,4.5) \\
\vec{v} = M-C = \langle 4-1,3.5-1,4.5-1 \rangle = \langle 3,2.5,3.5 \rangle \\
\text{This corresponds with answer choice B}
\end{align*}
}
\qs{}{
\begin{align*}
\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 \\
\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} \\
\text{This corresponds with answer choice B}
\end{align*}
}
\qs{}{
The provided vector corresponds with answer choice D since:
\begin{align*}
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}) \\
\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} \\
\implies \cancel{ 5\hat{i}+10\hat{j}+10\hat{k}}= \cancel{5\hat{i}+10\hat{j}+10\hat{k}} \\
\implies 0=0
\end{align*}
}
\qs{}{
Let $\vec{v}=\langle -1,6,0 \rangle$
\begin{align*}
\|\vec{v}\|=\sqrt{1^2+6^2+0^2}=\sqrt{1+36}=\sqrt{37}
\end{align*}
}
\chapter{Lab 4 - 13.3, 13.4}
\section{Work}
\qs{}{
\begin{align*}
\vec{u}\coloneqq \langle -5,7 \rangle \quad \vec{v} \coloneqq \langle 1,6 \rangle \quad \vec{w} \coloneqq
\langle -11,2 \rangle \\
\vec{u}\cdot(\vec{v}+\vec{w}) = \vec{u}\cdot(\langle1-11,6+2\rangle)=\vec{u}\cdot\langle-10,8\rangle \\
\implies \vec{u}\cdot(\vec{v}+\vec{w}) = -5(-10)+7(8)=50+56=106
\end{align*}
}
\qs{}{
\begin{align*}
\vec{r}\coloneqq \langle 7,-1,-3 \rangle \quad \vec{v} \coloneqq \langle 2,6,6 \rangle \quad \vec{w} \coloneqq
\langle 7,4,7 \rangle \\
\vec{w}\cdot(\vec{v}+\vec{r}) = \vec{w}\cdot\langle9,5,3\rangle=63+20+21=104
\end{align*}
}
\qs{}{
Let $o$ be the work done by the system.
\begin{align*}
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 \\
o=\langle -152.616280554, 40.8934091262 \rangle \cdot \langle -6,0 \rangle = 915.697683324 \\
\text{This corresponds with answer choice B}
\end{align*}
}
\qs{}{
\begin{align*}
\vec{u}\cdot\vec{v}=\|\vec{u}\|\cdot\|\vec{v}\|\cos{\theta} \\
\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} \\
\implies \theta = \arccos{\frac{-1}{\sqrt{2}\sqrt{41}}}= 96.3401917459\degree \\
\text{This corresponds with answer choice D}
\end{align*}
}
\qs{}{
\begin{align*}
\langle 4,3 \rangle \cdot \langle 20,-8 \rangle = 80-24 \ne 0 \\
\langle 15,-6 \rangle \cdot \langle 20,-8 \rangle = 300+48 \ne 0 \\
\langle 20,4 \rangle \cdot \langle 20,-8 \rangle = 400-32 \ne 0 \\
\because \langle -10,-25 \rangle \cdot \langle 20,-8 \rangle = -200+200 = 0 \\
\implies \text{Answer choice D is correct.}
\end{align*}
}
\qs{}{
If $\vec{v}\cdot\vec{w}=0$:
\begin{align*}
\langle 1,-x \rangle \cdot \langle -4,-3 \rangle = -4+3x=0 \\
\implies 3x=4 \quad \therefore x=\tfrac{4}{3}
\end{align*}
}
\dfn{Scalar component of a projection}{
\begin{align*}
\mathrm{scal}_{\vec{b}}(\vec{a}) = \frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|}
\end{align*}
}
\dfn{Vector projection}{
\begin{align*}
\mathrm{proj}_{\vec{b}}(\vec{a}) = \left( \frac{\vec{a} \cdot \vec{b}}{\|\vec{b}\|} \right)\frac{\vec{b}}{\|\vec{b}\|}
\end{align*}
}
\qs{}{
\begin{align*}
\vec{v}=\langle 2,3 \rangle \quad \vec{w}=\langle 8,-6 \rangle \\
\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}} \\
\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 \\
\text{This corresponds with answer choice C}
\end{align*}
}
\qs{}{
\begin{align*}
\vec{u}=\langle 4,5 \rangle \quad \vec{v} = \langle 1,1 \rangle \\
\mathrm{scal}_{\vec{v}}{\vec{u}}=\frac{\langle 4,5 \rangle \cdot \langle 1,1 \rangle }{\sqrt{2}} \\
\implies \mathrm{scal}_{\vec{v}}{\vec{u}} = \frac{4+5}{\sqrt{2}} = \tfrac{9}{\sqrt{2}} \\
\text{This corresponds with answer choice A}
\end{align*}
}
\qs{}{
\begin{align*}
\vec{u}=\langle 0,7,-2 \rangle \quad \vec{v}=\langle 4,-6,-7 \rangle \\
\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} \\
\theta = \arccos{\frac{-28}{\sqrt{53}\sqrt{101}}}= 1.9635142449\;\mathrm{rad} \\
\text{This corresponds with answer choice B}
\end{align*}
}
\qs{}{
\begin{align*}
\vec{u} &= \langle -5, -3, 4 \rangle, \quad
\vec{v} = \langle 5, -4, 2 \rangle, \quad
\vec{w} = \langle 10, -5, -8 \rangle \\[2mm]
\vec{u} \times \vec{v} &=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-5 & -3 & 4 \\
5 & -4 & 2
\end{vmatrix} \\[1mm]
&= \hat{i}((-3)(2) - (4)(-4))
- \hat{j}((-5)(2) - (4)(5))
+ \hat{k}((-5)(-4) - (-3)(5)) \\[1mm]
&= \hat{i}( -6 + 16 ) - \hat{j}( -10 - 20 ) + \hat{k}( 20 + 15 ) \\[1mm]
&= \langle 10, 30, 35 \rangle \\[1mm]
\vec{w} \cdot (\vec{u} \times \vec{v}) &=
\langle 10, -5, -8 \rangle \cdot \langle 10, 30, 35 \rangle \\[1mm]
&= 10\cdot10 + (-5)\cdot30 + (-8)\cdot35 \\[1mm]
&= 100 - 150 - 280 \\[1mm]
&= -330 \\
\text{This corresponds with answer choice B}
\end{align*}
}
\dfn{Area of a Triangle using a Cross Product}{
\begin{align*}
\text{Area}_{\triangle ABC} = \frac{1}{2} \|\vec{B} - \vec{A} \times \vec{C} - \vec{A}\|
\end{align*}
}
\dfn{Area of a Parallelogram using a Cross Product}{
\begin{align*}
\text{Area}_{\text{parallelogram}} = \|\vec{u} \times \vec{v}\|
\end{align*}
}
\qs{}{
In order to aptly encode the direction and magnitude of the side of the triangle let $\vec{s}_1, \vec{s}_2$ originate from $P$.
\begin{align*}
\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 \\
A=\tfrac{1}{2}\|\vec{s}_1\times\vec{s}_2\| = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
5 & 5 & -4 \\
9 & 3 & 1
\end{vmatrix} \\
\implies \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
5 & 5 & -4 \\
9 & 3 & 1
\end{vmatrix} = \hat{i}(5+12)-\hat{j}(5+36)+\hat{k}(15-45)=\langle 17,-41,-30\rangle \\
\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} \\
\text{This corresponds with answer choice D}
\end{align*}
}
\qs{}{
\begin{align*}
\|\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} \\
\text{This corresponds with answer choice D}
\end{align*}
}
\chapter{Lab 5 - 13.5}
\section{Work}
\qs{}{
\begin{align*}
\vec{r}=\langle 0,1,0 \rangle + \langle 3,0,-1 \rangle t \\
\text{breaking it down into component form:}\\
x=3t,\quad y=1,\quad z=-t \\
\text{This corresponds with answer choice C}
\end{align*}
}
\qs{}{
\begin{align*}
\vec{r}=\langle -6,5,-5 \rangle t + \langle 5,-1,-5 \rangle \\
\text{This corresponds with answer choice C}
\end{align*}
}
\qs{}{
\begin{align*}
\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 \\
\text{This corresponds with answer choice B}
\end{align*}
}
\qs{}{
\begin{align*}
\vec{u}=\langle 6,4,4 \rangle \quad \vec{v}= \langle -7,-6,-4 \rangle \\
\vec{v}_1\coloneqq \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
6 &4 & 4 \\
-7 & -6 & -4
\end{vmatrix} = \hat{i}(-16+24)-\hat{j}(-24+28)+\hat{k}(-36+28)=\langle 8,-4,-8\rangle \\
\vec{r}=\vec{v}_1 t+\langle -4,-7,4 \rangle \\
\text{This corresponds with answer choice D}
\end{align*}
}
\qs{}{
\begin{align*}
\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 \\
1+3t &= 3+4s \\
3-3t &= -1-2s \\
z=-t &= 3-5s \\[6pt]
s &= \tfrac{1}{4}(3t-2) \\
3-3t &= -1-2\cdot\tfrac{1}{4}(3t-2) \\
3-3t &= -1-\tfrac{1}{2}(3t-2) \\
3-3t &= -\tfrac{3}{2}t \\
3 &= \tfrac{3}{2}t \\
t &= 2 \\
s &= \tfrac{1}{4}(3\cdot 2-2)=1 \\[6pt]
x &= 1+3(2)=7 \\
y &= 3-3(2)=-3 \\
z &= -2 \\[6pt]
x &= 3+4(1)=7 \\
y &= -1-2(1)=-3 \\
z &= 3-5(1)=-2 \\[6pt]
\therefore (x,y,z) &= (7,-3,-2) \\
\text{This corresponds with answer choice D}
\end{align*}
}
\qs{}{
\begin{align*}
\exists k | k\in \mathbb{R} \land k\cdot\vec{v}_1=\vec{v}_2\\ \therefore l_1 \parallel l_2 \\
3-t&=-6+5s \\
9-t&=5s \\
9-5s&=t \\
1+6(9-5s)&=55-30s=1+54-30s \\
55-30s&=55-30s \forall s\in\mathbb{R} \\
4-2(9-5s)&=-14+10s=4-18+10s \\
-14+10s&=-14+10s \forall s\in\mathbb{R} \\
\therefore \text{$l_1$ and $l_2$ are identical. This corresponds to answer choice B}
\end{align*}
}
\qs{}{
\begin{align*}
\vec{n}\cdot(\vec{r}-\vec{r}_0)=0 \quad \langle 2,7,6 \rangle \cdot(\vec{r}-\langle 4,-3,2 \rangle) = 0 \\
\vec{r}|\vec{r}=\langle x,y,z \rangle \implies \langle 2,7,6 \rangle \cdot \langle x-4,y+3,z-2 \rangle \\
2(x-4)+7(y+3)+6(z-2)=2x-8+7y+21+6z-12=0 \\
2x+7y+6z-8+21-12=0 \quad 2x+7y+6z=-1 \\
\text{This corresponds with answer choice C}
\end{align*}
}
\qs{}{
\begin{align*}
\text{Considering answer choice A: } \\
5(-1)-7(8)+58=-5-56+58=-3\ne 3 \therefore \cancel{\mathrm{A}} \\
5(-1)+7(8)-58=-5+56-58=51-58=-7\ne3\therefore\cancel{\mathrm{B}} \\
5(-1)+y(8)-58\ne-3 \therefore \cancel{\mathrm{C}} \\
5(-1)-7(8)+58=-5-56+58=-3 \; \therefore \; \text{The answer is D}
\end{align*}
}
\qs{}{
\begin{align*}
-2x+2&=-2y \\
-2x=2+5z&=2 \\
z_0\coloneqq0 \quad -2y+5(0)=2 \; \implies y=-2 \\
-2x+2(-2)&=-2=-2x-4 \\
-2x&=2 \\
\implies x&=-1 \\
r_0=(-1,-2,0) \\
\text{By parameterizing $r_0$, only answer choice D has the necessary constant terms.}
\end{align*}
}
\qs{}{
\begin{align*}
x+y&=7-z \\
7-z&=12 \\
z&=-5 \\
x&=t \\
t+y&=12 \\
y&=12-t \\
\therefore x=t \quad y=12-t \quad z=-5
\end{align*}
This corresponds with answer choice B. Answer Choice A is the same line with different parametriazation to $x=-t$ instead of $x=t$
}
\chapter{Lab 6 - 13.6}
\section{Work}
\qs{}{
\begin{tikzpicture}
\begin{axis}[
legend pos=outer north east,
title=Q1,
axis lines = box,
xlabel = $x$,
ylabel = $y$,
zlabel = $z$,
view={45}{30},
]
\addplot3[
mesh,
samples=50,
color=blue,
]
{sqrt(10*x - 2*y^2/5)};
\addplot3[
mesh,
samples=50,
color=red,
]
{-sqrt(10*x - 2*y^2/5)};
\addlegendentry{$z = \pm\sqrt{10x - \frac{2y^2}{5}}$}
\end{axis}
\end{tikzpicture}
This image corresponds to that of an elliptical paraboloid, corresponding with answer choice A.
}
\qs{}{
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.
}
\qs{}{
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.
}
\newpage
\qs{}{
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 $x=0$.
\begin{align*}
\text{$xz$-trace: }\quad 16x^2-16(0)^2=z=16x^2 \\
\text{$yz$-trace: }\quad 16(0)^2-16y^2-z=0=16y^2+z
\end{align*}
This corresponds with answer choice D.
}
\qs{}{
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.
}
\qs{}{
Since the equation only yields circular cross sections, and varies quadratically over $\hat{i},\hat{j},\hat{k}$, The answer is Figure 1, A.
}
\qs{}{
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.
}
\end{document}
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