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MONT 100N -- Modeling The Environment\\
Solutions for Final Examination -- December 15, 2017\\
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\noindent
I.  In 1990, forests covered $4.047 \times 10^9$ hectares of the Earth's surface.
By 2000, forest area had decreased to $4.038\times 10^9$ hectares.  Assuming
that the decrease in forest area is
linear, and that it will continue at the same rate into the future,  in
this problem you will develop a linear model for the forest area
remaining as a function of $t = $ years since 1990.
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\begin{enumerate}
\item[A.]  Determine the slope for the linear model of the forest area.
\item[] {\it Solution:}  The slope is
$$m = \frac{4.038 - 4.047}{2000 - 1990} = -.0009$$
(in units of $10^9$ hectares per year; this is equivalent to a net loss of 
900,000 hectares, or 9000 square kilometers per year).
\item[B.]  What is the linear equation modeling the forest
area as a function of $t = $ years since 1990.
\item[] {\it Solution:}  Writing $FA$ for the forest area, the model would be
$$FA = 4.047 - .0009 t$$
\item[C.]  (10)  According to your model, in what year will
the forest area reach $4.0 \times 10^6$ hectares?
\item[] {\it Solution: } We solve for $t$ from the equation:
$$4.0 = 4.047 - .0009 t,$$
so
$$t = \frac{4.047 - 4.0}{.0009} \doteq 52$$
This corresponds to the year $2042$.  
\item[D.]  (5) According to the United Nations Food and Agriculture
Organization, the actual forest area remaining in 2010 was $4.033 \times 10^9$
hectares.  How close is the prediction your model from part B gives for
the forest area in 2010?
\item[] {\it Solution:} Using the model equation 
from part B to predict the amount of forest area
that will remain in 2010: $2010$ is $t = 20$ years after 1990, so the 
model predicts
$$FA = 4.047 - (.0009)(20) = 4.029\ (\times 10^9 {\rm\ hectares}) $$ 
The rate of deforestation was slower between 2000
and 2010 than the rate between 1990 and 2010.  So the model predicted a
lower forest area remaining than the actual figure. 
\end{enumerate}
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\noindent
II.  Wind power has emerged as a fast growing source of
energy for electrical power 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.
\begin{enumerate}
\item[A.]  The typical English unit of power is the horsepower.
$1 {\rm\ horsepower} = 7.457 \times 10^{-7}$ gigawatts.  Convert $301$ megawatts
to the equivalent number of horsepower.
\item[] {\it Solution:}  Thinking about the unit conversion, we see
$${\rm horsepower} = {\rm megawatt} \times \frac{{\rm horsepower}}{{\rm megawatt}}$$
That is, there are 
$$\frac{1}{7.457 \times 10^{-7}} \doteq 1.341 \times 10^6$$
horsepower in one gigawatt.  Hence $301$ gigawatts is 
$$301 \times 1.341 \times 10^6 = 403.641 \times 10^6 \doteq 4.03 \times 10^8 {\rm\ horsepower} $$
\item[B.]  Construct an exponential model for $WP = $
wind power generation as a function of $t = $ years since 
2016.   \emph{Use units of $10^2$ gigawatts for WP} -- see
the entry for 2016 in the table below.
\item[] {\it Solution:}  In units of $10^2$ gigawatts, the 
model equation is:  
$$WP = (3.01)(1.332)^t$$
\item[C.] Fill in the table of values for $WP$ below with
values predicted by
your model for the years $2017 - 2021$.   Round to 2 decimal places.
About how many years will it take for $WP$ to reach approximately double the 2016 level?
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$$
\begin{array}{c|c|c|c|c|c|c|c|}
Year & 2016 & 2017 & 2018 & 2019 & 2020 & 2021 \\
\hline
WP  & 3.01 & 4.01 & 5.34 & 7.11 & 9.48 & 12.62  \\
\hline
\end{array}
$$
$WP$ will double in between 2 and 3 years.  ({\bf Note}:  The doubling time for an 
exponential function with $a = 1.322$ can also be found by the formula
in one of the review problems:
$$1.322 = 2^{1\over t_2}$$
where $t_2 = \frac{\log_{10}(2)}{\log_{10}(1.332)} \doteq 2.42$ years.)  
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\item[D.] (5) How many years will it take for wind power generation
to reach $20 \times 10^2$ gigawatts according to your model?
\item[] {\it Solution:}  The equation 
we want to solve is $20 = (3.01)(1.332)^t$.  This is true when 
$t = \frac{\log_{10}(20/3.01)}{\log(1.332)} \doteq 6.6$.  So the
projection is this value will be reached between 2022 and 2023.  
\end{enumerate}
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\noindent
III.  Suppose that a population of fast-reproducing insects in an area
has a natural growth rate of $5\%$ per month from births and deaths, and
that there is a net migration \emph{loss} of 20 individuals per month.
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\begin{enumerate}
\item[A.]  Which of the following difference equation models for $P(n) = $
population in month $n$ fits the description
above? (Place a check next to the correct one.)
\begin{enumerate}
\item[1)] $\underline{\ \ \ \ \ \ \ \ }$ $P(n) = 5 P(n - 1) + 20$
\item[2)] $\underline{\ \ \ \ \ \ \ \ }$ $P(n) = 1.05 P(n - 20)$
\item[3)] $\underline{\ \ \  X \ \ \ \ }$ $P(n) = 1.05 P(n - 1) - 20$
\item[4)] $\underline{\ \ \ \ \ \ \ \ }$ $P(n) = 1.05 P(n - 1) + 20$
\end{enumerate}
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\item[B.]   Using an initial value $P(0) = 500$, determine the populations
in months $1,2,3,4,5$ according to the model you picked in part A and
record the values in the following table (round any decimal values to the nearest
whole number)
\item[] {\it Solution:}
$$
\begin{array}{c|c|c|c|c|c|c|}
n & 0 & 1 & 2 & 3 & 4 & 5  \\
\hline
P(n)  & 500 & 505 &510 & 516 &522 & 528  \\
\hline
\end{array}
$$
\item[C.]  What happens to the population in the long run as $n$ increase?  Does
it tend to a definite value?
\item[] {\it Solution:}  It seems that the population will continue
increasing at an increasing rate as $n$ increases.  Note that if
$P(n-1) > 400$, then $P(n) = (1.05)P(n-1) - 20 > P(n-1)$ and the difference
$P(n) - P(n-1) = (.05)P(n-1) - 20$ is bigger the larger $P(n-1)$ is.
({\bf Note:} By the formula we discussed in class for the general solution
of affine first order difference equations, this difference equation has
an equilibrium solution at $P = \frac{20}{.05} = 400$.  However, it is an
{\it unstable equilibrium} since $1.05 > 1$.  This means that with an
initial value $P(0) = 500 > 400$, the solution will grow without bound
as $n$ increases.)
\end{enumerate}
\noindent
IV.  Answer any two of the following three questions (only the best two will be counted
if you answer more than two).
\begin{enumerate}
\item[A.]  If you are fitting an exponential model to a data set $(x_i,y_i)$ ``by hand,''
you start by transforming the data to $(X,Y) = (x_i,\log_{10}(y_i))$.  If the best fit
regression line for the transformed data is $Y = m X + b$, what is the corresponding
exponential model?  (Assume the logarithms have base 10 as we discussed in class.)
\item[] {\it Solution:}  The linear equation is equivalent to $\log_{10}(y) = m x + b$
in terms of the original variables.  So exponentiating both sides we get
$y = 10^b \cdot (10^m)^x$.  In other words, $10^m$ is the base $a$ of the 
exponential function, and $10^b$ is the constant multiplier.
\item[B.]  What does the $R^2$ statistic in measure in linear regression?  How
did we use it?  Explain what it would mean, for instance if $R^2 = 1$.
\item[] {\it Solution:}  The $R^2$ statistic measures the degree of 
linearity in a scatter plot (in other words, how close the data points come to 
lying on a single straight line).  If $R^2 = 1$, then all the points are on a 
single line.   We used this to measure the goodness
of fit even for exponential models.  When we
did this, we were looking at the correlation coefficient
for the \emph{transformed} data (the $(x_i,\log_{10}(y_i))$ in
the exponential case).  
\item[C.]  What type of chart (scatterplot, pie chart, bar chart, etc.) would be
most useful to describe the composition of a forest if there 5 different types of trees
present in different concentrations per acre?   Explain, and illustrate your answer
with a chart if a typical acre of forest contains 10 oaks, 12 maples, 5 pines, 2 hemlocks,
and 1 chestnut.
\item[] {\it Solution:}  For a chart indicating the composition of a whole made up 
of several parts, either a pie chart or a bar chart could be used.   But a \emph{pie chart} would
be a slightly superior choice to show the composition.  For a pie chart,
we would compute the percentages of the whole represented by each species:
$10 + 12 + 5 + 2 + 1 = 30$ trees.  So oaks account for $10/30 \times 100\% = 33.3\%$, 
maples account for $12/30\times 100\%$, or $40\%$, pines account for $5/30 \times 100\%$,
or $16.7\%$, hemlocks account for $6.7\%$, and chestnuts account for the remaining
$3.3\%$.  These would be shown as fractions of a whole circle in the pie chart.   
\end{enumerate}
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