\begin{problem*}{1.60}
The consumption of natural gas by a company satisfies the empirical 
equation
\begin{equation}
 V = 1.50 t + 0.00800 t^2,
\end{equation}
where $V$ is the volume in millions of cubic feet and $t$ is the time in 
months.  Express this equation in units of cubic feet and seconds.  
Assign proper units to the coefficients.  Assume that a month is $30.0$ 
days.
\end{problem*} % Probem 1.60

\begin{solution}
Adding units to the equation coefficients:
\begin{equation}
 V = 1.50 \p[{\frac{\text{million ft}^3}{\text{month}}}] t
     + 8.00\E{-3} \cdot\p[{\frac{\text{million ft}^3}{\text{month}^2}}] t^2
\end{equation}

We prepare some conversions:
\begin{align}
1 &= \p[{\frac{10^6 \text{ ft}^3}{\text{million ft}^3}}] \\
1 &= \p[{\frac{1 \text{ month}}{30 \text{ days}}
	\cdot \frac{1 \text{ day}}{24 \text{ hours}}
	\cdot \frac{1 \text{ hour}}{60 \text{ minutes}}
	\cdot \frac{1 \text{ minute}}{60 \text{ s}}}] \\
 &= \p[{\frac{ 1 \text{ month}}{ 2.592 \cdot 10^6 \text{ s}}}]
\end{align}

So converting the units in our equation to ft$^3$ and s:
\begin{align}
 V &= 1.50 \p[{\frac{\text{million ft}^3}{\text{month}}
	\cdot \frac{10^6\U{ft}^3}{\text{million ft}^3}
	\cdot \frac{1\U{month}}{2.592\E{6}\U{s}}}]
	t
     + 8.00\E{-3} \p[{\frac{\text{million ft}^3}{\text{month}^2}
	\cdot \frac{10^6\U{ft}^3}{\text{million ft}^3}
	\cdot \p{(\frac{1\U{month}}{2.592\E{6}\U{s}}})^2 }]
	t^2 \\
 V &= 0.579 \p[{\frac{\text{ft}^3}{\text{s}}}] t
	+ 1.19\E{-9} \cdot\p[{\frac{\text{ft}^3}{\text{s}^2}}] t^2
\end{align}

\end{solution}
