\magnification=\magstep1
\centerline{MONT 104N -- Modeling the Environment}
\centerline{Solutions for Mathematical Review Questions from Midterm Exam}
\centerline{October 28, 2019}
\bigskip
\noindent
I.  The following table gives amounts of carbon (in units of $10^{15}$ kilograms)
contained in the major ``reservoirs'' of this element on planet Earth:
$$\matrix{ {\bf Reservoir} & {\bf Carbon\ Content }\cr
           {\rm Atmosphere} & .59 \cr
           {\rm Crust\ (as\ fossil\ fuels)} & 3.7 \cr
           {\rm Vegetation} & 2.3 \cr
           {\rm Shallow\ Ocean} & .9  \cr
           {\rm Deep\ Ocean} & 37.3\cr}$$  
\item{A.}   What percent of the total carbon present on 
Earth is contained in each of these reservoirs?
\bigskip
\noindent
{\it Solution:}  The sum of these amounts is $44.79 \times 10^{15}$ kilograms.  
So, the percentage of the carbon in the deep ocean reservoir, for instance, is
$${37.3\over 44.79} \times 100\% \doteq 83.3\%$$
The other values are computed similarly.  Rounding to 1 decimal place, 

$$\matrix{ {\bf Reservoir} & {\bf Percent\ of\ Total\ Carbon}\cr
& \cr
           {\rm Atmosphere} & 1.3\% \cr
&\cr
           {\rm Crust\ (as\ fossil\ fuels)} & 8.3\%\cr
&\cr
           {\rm Vegetation} & 5.1\% \cr
&\cr
           {\rm Shallow\ Ocean} &  2.0\% \cr
&\cr
           {\rm Deep\ Ocean} & 83.3\% \cr}$$  

\item{B.}   Construct and draw a chart (your choice of type) showing how
the total carbon breaks down into these categories. 
\bigskip  
\noindent
{\it Solution:}  Either a pie chart or a bar chart is OK here.  If you 
use a pie chart, the percents above correspond to the fractions of the pie.
If you use a bar chart, it would be best to make the vertical axis
scale correspond to the actual amounts of carbon (in units of $10^{15}$ kg), so 
use something like a vertical scale of $0$ to $40$.  However, a scale where 
the vertical axis shows the percents is also acceptable.
\bigskip
\noindent
II.  Burning coal provides between 9500 and 14000 BTUs of heat energy per
pound.   Using the table below, answer questions A and B.
\item{$\bullet$}  $1\ {\rm lb} \doteq .454\ {\rm kg}$
\item{$\bullet$}  $1\ {\rm BTU} \doteq .252\ {\rm Kilocalorie}$
\bigskip
\item{A.} (8)  Express the heat energy from burning one pound of coal as a range of values
in calories.
\bigskip
\noindent
{\it Solution:}  There are $.252$ Kilocalories in one BTU.  Hence, 
burning one pound of coal yields between
$$9500\ {\rm BTU} \times .252\ {\rm Kcal\over BTU} \doteq 2.39 \times 10^3\ {\rm Kcal}$$
and 
$$14000\ {\rm BTU} \times .252\ {\rm Kcal \over BTU} \doteq 3.53 \times 10^3\ {\rm Kcal}.$$
\item{B.} (8)  What is the range of heat energy values provided by burning 1 kilogram of coal, 
expressed in BTUs?
\bigskip
\noindent
{\it Solution:}  From given information, one pound is $.454$ kg, so $1$ kg $= {1\over .454}\doteq 2.20$ lb.
Hence burning 1 kg of coal provides between
$$9500\ {\rm BTU/lb} \times 2.20\ {\rm lb} = 2.09 \times 10^4\ {\rm BTU}$$
and 
$$14000\ {\rm BTU/lb} \times 2.20\ {\rm lb} = 3.08 \times 10^4\ {\rm BTU}.$$  
\item{C.} (4)  If the 9500 BTU figure comes from bituminous coal and the 14000 BTU comes
from anthracite coal, what is the percentage difference in heat energy between anthracite and bituminous?  
\bigskip
\noindent
{\it Solution:}  The percentage difference is 
$${14000 - 9500 \over 9500} \times 100\% \doteq 47.3\%$$
(the anthracite coal generates 47.3\% more heat than the bituminous).   
\bigskip
\noindent
III.  (15)  The Honda Civic comes in a standard 
(gasoline engine) version and a hybrid (gasoline-electric)
version.  The standard version has a fuel efficiency of about 34 
miles per gallon, while the hybrid version gets about 44 miles per gallon.  
The hybrid version has a price about $\$ 4000$ more than the 
standard.  If you bought the hybrid and drive about 10000 miles per year, 
about how many years will it take for your savings in gasoline costs to make 
up for the difference in price?  Assume that gasoline will average $\$ 2.00 $ per 
gallon over the life of the vehicle. 
\bigskip
\noindent
{\it Solution:}  If you drive $10000$ miles with the hybrid version, then you use 
$10000 \times {1\over 44} \doteq 227.3$ gallons of gas in one year and you spend
$227.3 \times \$3.00 \doteq \$681.90$ on gasoline.  
With the standard version you use $10000 \times {1\over 34} \doteq 294.1$
gallons of gasoline and you spend $\$882.30$.  The difference is $\$200.40$.  To make up 
the $\$4000$ greater cost of the hybrid model, you would need to keep the hybrid for about 
${4000\over 200.40} \doteq 20$ years.  (Comment:  These are pretty realistic numbers, and 
this is one reason why hybrids are still seen as less attractive options by many people.)
\bigskip
\noindent
IV.  Consider the Excel scatter plot given in the exam. 
Place a check next to the best responses to the questions below.
\bigskip
\noindent
A.  (5) The least squares regression line for this data set would
have a slope $m$ about:

\leftline{ $m = -.1$ $\underline{\qquad X \qquad}$}

\leftline{ $m = .3$ $\underline{\qquad\qquad\qquad}$}

\leftline{ $m = 1$ $\underline{\qquad\qquad\qquad}$}
\bigskip
\noindent
The best answer is about $m = -.1$ since the slope definitely should be negative
and the velocity is decreasing by roughly 1.4 feet per second over the roughly
11 feet change in depth.  The actual slope is $m \doteq -0.12$.
\bigskip
\noindent
B.  (5)  Based on the scatter plot, the $R^2$ statistic is
probably:

\leftline{ between $.8$ and $1$ $\underline{\qquad\ X \qquad}$} 

\leftline{ between $.4$ and $.6$ $\underline{\qquad\qquad\qquad}$} 

\leftline{ between $.2$ and $.4$ $\underline{\qquad\qquad\qquad}$} 

\leftline{ between $.0$ and $.2$ $\underline{\qquad\qquad\qquad}$}

\bigskip
\noindent
The best answer is $.8 < R^2 < 1$ since the data is relatively linear (so
$R^2$ will be close to 1).  The actual value is $R^2 \doteq .88$.
 
\bigskip\noindent
V.  Wind power has emerged as one of the faster-growing methods
of electricity generation in recent years.  In 2016, the generating power 
of wind turbines installed around the world was about
$301$ gigawatts and it was increasing at about $33.2\%$ per
year.
\bigskip
\item{A.} The typical English unit of power is the horsepower.
$1 {\rm\ horsepower} = 7.457 \times 10^{-7}$ gigawatts.  Convert $301$ gigawatts
to the equivalent number of horsepower.
\item{A.} {\it Answer:}  
$$301\ {\rm Gw} = 301 {\rm Gw} \cdot {1\over 7.457\times 10^{-7}}\ {\rm horsepower}/{\rm Gw} \doteq
4.036 \times 10^8\ {\rm horsepower}.$$
\item{B.} Construct an exponential model for $WP = $
wind power generation as a function of
$t = $ years after 2016.  {\it Use units of $10^2$ gigawatts for WP} -- see
the entry for 2016 in the table below.
\item{} {\it Answer:}  The model would be given by the function 
$${\rm WP} = 301\cdot (1.332)^t$$
where WP is in gigawatts and $t$ is in years after 2016.    
\item{C.}  Compute the values of $WP$ predicted by
your model for the years $2017 - 2022$.   Round to 2 decimal places.
About how many years will it take for $WP$ to reach approximately 
double the 2016 level?
\item{} {\it Answer:}  The projected values are 
$$\matrix{{\rm Year} & {\rm Wind\ Power\ (in\ gigawatts)}\cr
&\cr
2017 & 400.93\cr
2018 & 534.04\cr
2019 & 711.34\cr
2020 & 947.51\cr
2021 & 1262.08\cr
2022 & 1681.09\cr}$$
You can think of these as the values at the start of each year.  
The amount of electricity generated by wind power will reach 
twice the 2016 level when 
$$602 = 301\cdot (1.332)^t,\quad {\rm so}\quad t = {\log(2)\over \log(1.332)} \doteq 2.4\ {\rm years}$$
(about $4/10$ of the way through the year 2018).  
\item{D.} How many years will it take for wind power generation
to reach $2.0 \times 10^3$ gigawatts according to your model?
\item{} {\it Answer:} We solve
$$2000 = 301\cdot (1.332)^t$$
and find 
$$t \doteq {\log(6.6445)\over \log(1.332)} \doteq 6.6\ {\rm years}$$
(that is, about $6/10$ of the way through the year 2022).  
\end
