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MONT 104N -- Modeling The Environment\\
Solutions for Final Examination Review Problems -- December 7, 2018\\
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\vskip 10pt
\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.
\vskip 10pt
\begin{enumerate}
\item[A.]  Determine the slope for the linear model of the forest area.
\item[] {\it Solution:}  The slope is
$$m = \frac{4.038 - 4.047}{2000 - 1990} = -.0009.$$
The slope is 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 $4.0 \times 10^9$ 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 $52$ years after 1990, so 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?
\vskip 10pt
$$
\begin{array}{c|c|c|c|c|c|c|c|}
Year & 2016 & 2017 & 2018 & 2019 & 2020 & 2021 \\
\hline
WP  & 3.01 & 4.01 & 5.34 & 7.11 & 9.48 & 12.62  \\
\hline
\end{array}
$$
$WP$ 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 problems sets:
$$1.322 = 2^{\frac{1}{t_2}}$$
where $t_2 = \frac{\log_{10}(2)}{\log_{10}(1.332)} \doteq 2.42$ years.)  
\vskip 10pt
\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.
\vskip 10pt
\begin{enumerate}
\item[A.]  Write a difference equation that models this situation.
\vskip 10pt
\item[] {\it Solution:}  The equation would be
$$P(n+1) = (1.05)\cdot P(n) - 20$$
\vskip 10pt
\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.  If
$P(n) > 400$, then $P(n+1) = (1.05)P(n) - 20 > P(n+1)$ and the difference
$P(n+1) - P(n) = (.05)P(n) - 20$ is bigger the larger $P(n)$ 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.  In a question like this, I could also ask you to 
find the equilibrium level.)
\end{enumerate}
\noindent
IV.  Answer the following questions with a few sentences each.
\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.]  If you are fitting a power law model to a data set $(x_i,y_i)$ ``by hand,''
you start by transforming the data to $(X,Y) = (\log_{10}(x_i),\log_{10}(y_i))$.  If the best fit
regression line for the transformed data is $Y = m X + b$, what is the corresponding
power law model?  (Assume the logarithms have base 10 as we discussed in class.)
\item[] {\it Solution:}  The linear equation is equivalent to $\log_{10}(y) = m \log_{10}(x) + b$
in terms of the original variables.  So exponentiating both sides we get
$y = 10^b \cdot x^m$.  In other words, $m$ is the exponent, and $10^b$ is the constant multipler
\item[C.]  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[D.]  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.  
\item[E.]  What difference equation would model a population undergoing logistic
growth if the population was growing at about $4\%$ per year when the population
is much smaller than the carrying capacity $M = 400$ of the habitat.
\item[] {\it Solution:}  We see $r = .04$, so equation is 
$$P(n+1) = (1+r)P(n) - \frac{r}{M} (P(n))^2 = 1.04 P(n) - \frac{.04}{400} (P(n))^2.$$
\item[F.]  Describe the SIR model for infectious disease outbreaks and give the 
corresponding difference equations.  What is an \emph{epidemic} and given 
the constants $\beta,\gamma$ and $S(0)$, what determines when an epidemic will
occur.
\item[] {\it Solution:}  In the SIR model, the population is divided
into three groups, $S = $ the susceptibles, $I = $ the infecteds, and $R = $
the removed (either recovered, or deaths because of the infection).  The basic
form of the model does not take population dynamics into account.  New infections
occur through contacts between susceptible and infected individuals and this
is modeled by terms containing the product $S(n)I(n)$.  The equations are
\begin{align*}
S(n+1) &= S(n) - \beta S(n)I(n)\\
I(n+1) &= I(n) + \beta S(n)I(n) - \gamma I(n)\\
R(n+1) &= R(n) + \gamma I(n).
\end{align*}
An epidemic is a situation where $I(n)$ is increasing for some range
of $n$ values starting at $0$.  This will be true when $\beta \cdot S(0) - \gamma > 0$,
or equivalently $\frac{\beta\cdot S(0)}{\gamma} > 1$.  
\item[G.]  What feature of the solutions of the Lotka-Volterra equations is considered
a confirmation that this model is capturing an important aspect of real-world
predator-prey interactions?
\item[] {\it Solution:}  The fact that the solutions tend to exhibit 
{\it cyclical, oscillatory} behavior is a confirmation of this.  The 
equations were originally developed to model the predator-prey behavior
of pairs of species (like the Canada lynx and snowshoe hares that we
saw in class).  Since those populations are observed to oscillate in
the wild, the model is (at the least) doing something similar.
\end{enumerate}
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