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\begin{document}
\section*{Sample Quiz 5}
\rule{\textwidth}{0.5ex}
{\Large\sf Sample Quiz5 Problem 1}. {\bf Balancing Chemical Equations}: Tin oxide heated with hydrogen gas forms tin metal and water vapor.


The object of this problem is the derivation of the balanced chemical equation
\begin{center} SnO2(s) + 2 H2(g) $\to$ Sn(s) + 2 H2O(g). \end{center}
The chemical symbols are defined as follows:

\begin{tabular}{llll}
SnO2(s) &   Tin dioxide & (s)=solid state & Ref: \href{http://en.wikipedia.org/wiki/Tin_dioxide}{Tin dioxide}
\\
H2(g) & Hydrogen & (g)=gaseous state & Ref: \href{http://en.wikipedia.org/wiki/Hydrogen}{Hydrogen}
\\
Sn(s) & Tin & (s)=solid state & Ref: \href{http://en.wikipedia.org/wiki/Tin}{Tin}
\\
H2O(g) & Water & (g)=gaseous state & Ref: \href{http://en.wikipedia.org/wiki/Water}{Water vapor}
\end{tabular}

{\bf Chemical balance} means that matching elements on each side of a chemical equation have the same total amount of atoms. Notation 2 H2O documents 2x2=4 hydrogen atoms and 2x1=2 oxygen atoms. The given chemical equation is balanced, because the atom amounts for elements Sn, H, O total on each side of the equation to 1, 4, 2, respectively.

{\bf Chemical Balance by Linear Algebra}.
We begin with the unbalanced chemical equation
\begin{center} SnO2(s) +  H2(g) $\to$ Sn(s) +  H2O(g), \end{center}
which means tin dioxide and hydrogen gas chemically combine into tin and water vapor.

Let's write the balanced equation as
\begin{center} $x_1$ SnO2(s) +  $x_2$ H2(g) $\to$ $x_3$ Sn(s) +  $x_4$ H2O(g), \end{center}
where integers $x_1$ to $x_4$ are to be determined. Chemical balance means that the following algebraic equations hold:
$$
\begin{array}{rcrl}
x_1 &=& x_3 & \mbox{~~~Tin Sn} \\
2x_2 &=& 2x_4 & \mbox{~~~Hydrogen H} \\
2x_1 &=& x_4 & \mbox{~~~Oxygen O}
\end{array}
$$

(a) Write the equations in standard matrix form $A\vec{x}=\vec{0}$, where $A$ is a $3\times 4$ matrix and $\vec{x}$ has components $x_1$ to $x_4$.

(b) Show details for computing the reduced row echelon form $\rref(A)=\left(
\begin{array}{cccc} 1&0&0&-\frac12 \\ 0&1&0&-1 \\ 0&0&1&-\frac12
\end{array}\right)$
%A:=Matrix([[1,0,-1,0],[0,2,0,-2],[2,0,0,-1]]);
%linalg[linsolve](A,<0,0,0>);
%linalg[rref](A);

(c) There are more variables than equations. Therefore, there is at least one free variable. Identify the lead variables and the free variables.

(d) Solve the system $A\vec{x}=\vec{0}$. The answer is $x_1=\frac12 t_1$, $x_2=t_1$, $x_3=\frac12 t_1$, $x_4=t_1$, in terms of invented symbol $t_1$.

{\bf Balanced Equation}. The smallest positive integer values for $x_1$ to $x_4$ are obtained from $t_1=2$. Then $x_1=1$, $x_2=2$, $x_3=1$, $x_4=2$. Substitution gives the balanced chemical equation
\begin{center} (1)SnO2(s) +  (2)H2(g) $\to$ (1)Sn(s) +  (2)H2O(g) \end{center}

{\bf References}. Edwards-Penney section 5.1. Especially, this background theorem:
Theorem 3: {\small\em If variable $\vec{x}$ has more entries than there are equations in
            system $A\vec{x}=\vec{0}$, then the system has infinitely many solutions}.
Course document
\href{http://www.math.utah.edu/~gustafso/s2015/2280/linearequDRAFT.pdf}{Linear Algebraic Equations, No Matrices}.
How to balance chemical equations, a video by \href{http://video.about.com/chemistry/How-to-Balance-Chemistry-Equations.htm}{Anne Marie Helmenstine}
with English transcript.

\vfill
\newpage
\rule{\textwidth}{0.5ex}
{\Large\sf Sample Quiz5 Problem 2}. {\bf Solving Higher Order Initial Value Problems with Linear Algebra}.

The objective of this problem is to learn how to solve for the constants in a general solution, using given initial conditions. The differential equation and its known general solution are
$$
\begin{array}{l}
y''' + 4y'' = 0, \\
y(x) = c_1 + c_2 x + c_3 e^{-4x}.
\end{array}
$$
The initial conditions are
$$y(0)=1, y'(0)=2, y''(0)=-1.
$$

(a) Derive the general solution formula, as follows. Make the substitution $v=y''$ to obtain $v'+4v=0$. Solve for $v$ and then integrate twice to find $y$.

(b) Substitute the general solution into the three initial conditions to obtain the system of equations
$$
\left(\begin{array}{rrr}
1 & 0 & 1 \\
0 & 1 & -4 \\
0 & 0 & 16
\end{array}\right)
\colvector{c_1 \\ c_2 \\ c_3} = \colvector{1 \\ 2 \\ -1}
$$

(c) Solve the system of equations and then report the particular solution $y(x)$ so found.
\begin{quote}
 {\bf Remark}.\small\sf Picard's existence-uniqueness theorem says that the initial value problem has a unique solution, meaning that the linear algebra problem is in the unique solution case: no free variable.
\end{quote}

{\bf References}. Edwards-Penney sections 3.3, 3.4, 5.3.
Course manuscript
\href{http://www.math.utah.edu/~gustafso/s2015/2280/linearequDRAFT.pdf}{Linear Algebraic Equations, No Matrices}.

\rule{\textwidth}{0.5ex}
{\Large\sf Sample Quiz5 Problem 3}. {\bf An $RL$-Circuit with DC Voltage Source}.
\begin{center}
\includegraphics[scale=0.8]{seriesRL.png}
\end{center}
The RL-circuit in the figure has model $LI' + RI = V_s$ where symbols $L,R,V_s$ are constants. If the charge is $Q(t)$, then $I=Q'$ and $LQ''+RQ'=V_s$. Differentiate this equation to obtain $LQ'''+RQ''=0$.
Assume the current and charge are initially zero, before the switch is closed.

(a) Explain why $Q$ has initial conditions $Q(0)=Q'(0)=0$, $Q''(0)=V_s/L$.

(b) Solve for the charge $Q(t)$ and current $I(t)$ from the higher order equation $LQ'''+RQ''=0$ with initial conditions $Q(0)=Q'(0)=0$, $Q''(0)=V_s/L$, guided by Problem 2, above. The system of linear equations to be solved is
 $$\left( \begin {array}{ccc} 1&0&1\\ \noalign{\medskip}0&1&-{\frac {R}{
L}}\\ \noalign{\medskip}0&0&{\frac {R^2}{L^2}}\end {array}
 \right)\colvector{c_1\\[2ex] c_2\\[2ex] c_3} = \colvectorC{0\\[1.5ex] 0\\[1.5ex] \frac{Vs}{L}}.$$



(c) The steady-state current $\lim_{t\to\infty} I(t)$ computed directly from your solution formula should equal the answer from the equilibrium solution of $LI' + RI = V_s$. Does it?

{\bf References}. Edwards-Penney sections 3.7, 4.1, 5.1.
Course manuscript
\href{http://www.math.utah.edu/~gustafso/s2015/2280/linearequDRAFT.pdf}{Linear Algebraic Equations, No Matrices}. Edwards-Penney BVP 3.7, electric circuit supplement. All About Circuits Volume I -- DC
by \href{http://www.allaboutcircuits.com/}{T. Kuphaldt}. Course slides on
\href{http://www.math.utah.edu/~gustafso/s2015/2280/electricalCircuits.pdf}{Electric Circuits}.

%de:=L*diff(u(t),t)+R*u(t)=Vs; dsolve({de,u(0)=0},u(t));
%de1:=L*diff(u(t),t,t,t)+R*diff(u(t),t,t)=0; dsolve({de1,u(0)=0,D(u)(0)=0,D(D(u))(0)=Vs/L},u(t));
%y:=unapply(a+b*t+c*exp(-R*t/L),(t,a,b,c));
%y(0,a,b,c)=0;
%subs(t=0,diff(y(t,a,b,c),t))=0;
%subs(t=0,diff(y(t,a,b,c),t,t))=Vs/L;
%A:=Matrix([[1,0,1],[0,1,-R/L],[0,0,R^2/L^2]]); latex(%);
%LinearAlgebra[LinearSolve](A,<0,0,Vs/L>);

\end{document}
