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\title{\Huge{Calculus III}\\Homework \# 1}
\author{\huge{Krishna Ayyalasomayajula}}
\date{}

\begin{document}

\maketitle
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\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*}
}

\end{document}
