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MONT 104N -- Modeling The Environment\\
Solutions for Final Examination -- December 13, 2012\\
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\vskip 10pt 
\noindent
I.  Wind power has emerged as the fastest growing source of
energy for electrical power generation in recent years.
In 2004, the generating capacity of all wind turbines in use was about
$47,600$ megawatts and the generating capacity was increasing at about $26.8\%$ per
year.
\begin{enumerate}
\item[A.] (10) The typical English unit of power is the horsepower.
$1 {\rm\ horsepower} = .0007457$ megawatts.  Convert $47,600$ megawatts
to the equivalent number of horsepower.
\item[] {\it Solution:}  Since $1 {\rm\ horsepower} = .0007457$ megawatts, 
$1 {\rm\ megawatt} = 1/.000747 \doteq 1341$ horsepower.  Then 
$$47600 Mw = 47600 \times 1341 \doteq 6.383 \times 10^7 {\rm\ horsepower}.$$
\item[B.] (10) Using the information above, construct an 
exponential model for $WP = $ wind power generation as a function of
$t = $ years since 2004.  \emph{Use units of $10^4$ megawatts for WP}.
\item[] {\it Solution:}  From the given information, $r = .268$, so $M = 1 + r = 1.268$ and
the exponential model is
$$WP = 4.76 (1.268)^t.$$
\item[C.]  (15) Fill in the table of values for $WP$ below with
values predicted by
your model for the years $2004 - 2011$.   Round to 2 decimal places.
In what year did $WP$ reach approximately double the 2004 level?
\item[] {\it Solution:} 
$$
\begin{array}{c|c|c|c|c|c|c|c|c|}
Year & 2004 & 2005 & 2006 & 2007 & 2008 & 2009 & 2010 & 2011\\
\hline
WP  & 4.76  & 6.04 & 7.65 & 9.70 & 12.31 & 15.60 & 19.78 
& 25.09 \\ \hline
\end{array}
$$
$WP$ reached approximately double its 2004 value in 2007.  (Note that the
exact doubling time is the solution of $1.268^t = 2$, or $t = \log(2)/\log(1.268) \doteq 2.92$
years.)
\vskip 10pt
\item[D.] (10) How many years will it take for wind power generation
to reach $320,000$ megawatts according to your model?
\item[] {\it Solution:}  Expressing in units of $10^4$ megawatts, we want
to solve $32 = 4.76 (1.268)^t$,  or $t = \log(6.722)/\log(1.268) \doteq 8.025$.  About
8 years.  
\vskip 10pt 
\item[E.] (5)  The following graph (produced by the Global Wind Energy 
Council -- GWEC) shows the actual 
global wind electrical power generation capacity (estimated via surveys
of electrical power producers).  How do the 
actual figures compare with your model values?  Note:  The vertical 
scale of the graph is in \emph{gigawatts}.  $1$ gigawatt $ = 1000$ megawatts.  
%\begin{figure}
%\includegraphics[width=5in]{GlobalWindPowerCumulativeCapacity.png}
%\end{figure}
\item[] {\it Solution:}  To compare the values, we want to multiply the values in part C
above by 10 (since $10^4$ megawatts $= 10$ gigawatts).  The values are all quite
close.  For instance, the 2010 model prediction is $197.8$ gigawatts, while 
the actual value was $197.6$ gigawatts.  
\end{enumerate}
\noindent
II.  According to the United Nations Food and Agriculture
Organization, in 2000, forest area covered $4.038\times 10^9$ hectares of the Earth's surface.
The forest area in 2010 was $4.033 \times 10^9$ 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
$FA = $ (in units of $10^9$ hectares) remaining as a function of $t = $ years since 2000.
\vskip 10pt
\begin{enumerate}
\item[A.] (10)  Determine the slope for the linear model of the forest area.
\item[] {\it Solution:}  The slope is 
$$m = \frac{(4.033 - 4.038) \times 10^9}{2010 - 2000} \doteq -.0005 \times 10^9$$
hectares per year.  
\item[B.] (10)  What is the linear equation modeling the forest
area as a function of $t = $ years since 2000.
\item[] {\it Solution:}  The model is 
$$FA = -.0005 t + 4.038$$
\item[C.] (10) Use your equation to predict the amount of forest area
that will remain in 2020.
\item[] {\it Solution:}  The year 2020 is $t = 20$ years after 2000.  So the prediction is
$$FA = (-.0005)(20) + 4.038 = 4.028 (\times 10^9 {\rm\ hectares})$$
\item[D.]  (5)  According to your model, in what year will
the forest area reach $4.0 \times 10^9$ hectares?
\item[] {\it Solution:}  We solve for $t$ in the equation
$$4.0 = (-.0005)t + 4.038 \Rightarrow t = \frac{4.0 - 4.038}{-.0005} = 76 {\rm\ years}$$
\end{enumerate}
\vskip 10pt
\noindent
III.  Suppose that a population of fast-reproducing insects in an area
has a natural growth rate of $7\%$ per month from births and deaths, and
that there is a net migration \emph{loss} of 100 individuals per month.
\vskip 10pt
\begin{enumerate}
\item[A.] (5)  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)] $\underline{\ \ \ \ \ \ \ \ }$ $P(n) = 7 P(n - 1) - 100$
\item[2)] $\underline{\ \ \ \ \ \ \ \ }$ $P(n) = 1.07 P(n - 100)$
\item[3)] $\underline{\ \ \  X \ \ \ }$ $P(n) = 1.07 P(n - 1) - 100$
\item[4)] $\underline{\ \ \ \ \ \ \ \ }$ $P(n) = 1.07 P(n - 1) + 100$
\end{enumerate}
\vskip 10pt
\item[B.] (10)  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)
$$
\begin{array}{c|c|c|c|c|c|c|}
n & 0 & 1 & 2 & 3 & 4 & 5  \\
\hline
P(n)  & 500 & 435 & 365 & 291 & 211 & 126  \\
\hline
\end{array}
$$
(Note:  These values were computed from the difference equation in A 3) above.  I
kept the decimal places each time and rounded the result for the table value,
but I used the decimal places to compute the next value.  Your values would differ
slightly if you used the rounded values.  Either way is OK.)
\vskip 10pt
\item[C.] (10)  What happens to the population in the long run as $n$ increases?  Does
it tend to a definite value?  What is that value?
\item[] {\it Solution:}  This is an affine difference equation and the equilibrium
value is the solution of $E = 1.07 E - 100$, or $E = 1428.6$.  However that 
is an unstable equilibrium since $M = 1.07 > 1$.  Mathematically, the values of the 
solution equation would decrease
indefinitely with $n$.  When $p(n)$ reaches $0$, though, 
the population would have disappeared entirely (and negative values would not 
correspond to realistic population levels).   
\end{enumerate}
\vskip 10pt
\noindent
IV.  Answer any three of the following four questions (only the best three will be counted
if you answer more than three).
\begin{enumerate}
\item[A.]  (10) What does the correlation coefficient $r$ (or its square $r^2$) measure?  How
did we use it?  Explain what it would mean, for instance if $r^2 = 1$ or $r^2 = 0$.
\item[] {\it Solution:}  $r$ or $r^2$ measure the degree to which one quantity $y$ 
depends linearly on another quantity $x$ in an $(x,y)$ data
set.  The value $r^2 = 1$ would indicate perfect linearity (with positive slope if
$r = 1$ and negative slope if $r = -1$).  The value $r = 0$, would indicate
no linear trend at all.  We used this to measure ``goodness of fit'' for linear
models first, and then by extension to measure ``goodness of fit'' for 
exponential and power law models, since those become linear after a transformation
on the data.  
\item[B.]  (10) If you are fitting a power law model to a data set $(x_i,y_i)$ ``by hand,''
you would start by transforming the data to $(X,Y) = (\log(x_i),\log(y_i))$.  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:}  If $\log(y) = m\log(x) + b$, then exponentiating both sides with base 10,
we get $y = 10^b x^m$.  This is the power law model.  
\item[C.]  (10)   A population has unrestricted growth rate $r_{max} = .03$ and carrying
capacity $K = 1000$.  What is the corresponding logistic model?
\item[] {\it Solution:}  The model is given by 
$$p(n) = 1.03 p(n-1) - \frac{.03}{1000} (p(n-1))^2$$
\item[D.] (10)  The following graph shows $W = $ the word production of photovoltaic arrays
(used for solar power generation) in units of ``peak megawatts.''  
 Between 1998 and 2007, what type of model would be most appropriate for describing how
$W$ is growing.  Explain.  \emph{Look at the vertical axis scale carefully!}
\item[] {\it Solution:}  The vertical axis scale is logarithmic (because multiplying $y$ by a
factor of $10$ moves a fixed distance along the axis).  Since that portion of the 
graph is very close to linear, the relation is $\log(W) = m t + b$, so $W = (10^b) (10^m)^t$.
This is an \emph{exponential model}.  
\end{enumerate}
\noindent
V.  Essay.  (60)  In general terms, what is a mathematical model?  Describe what they are, 
how they are constructed, and how they are used.  Give examples of two 
different types of mathematical models we have studied in this course.  
Next, {\it why} do we try to build mathematical models of aspects 
of the real world?  Can {\it any} mathematical model be a 
completely accurate representation of some aspect of the natural world?  
As an example, why do scientists think it is important to 
understand how much $CO_2$ is present in the atmosphere?  What
tends to happen when $CO_2$ levels rise?  
Describe a key piece of evidence that suggests human activities might 
have changed atmospheric $CO_2$ levels over the past 50-200 years. 
Explain the case for saying the evidence points to that conclusion,
and relate your answer to the results of modeling 
exercises we did in this class.  
\vskip 10pt
\noindent
{\it Model Answer:}  A mathematical model is an equation, a system of equations, a diagram, 
or a relationship expressed in mathematical terms 
that aims to capture some aspect of the behavior of a real-world system.  
They are usually constructed by analyzing data from observations or experiments and
``fitting'' a mathematical form to the observed relationship between the 
measured quantities.    We have studied \emph{function models} including
linear, exponential, and power law forms, and difference equation models in this course.  
As mentioned above, models are usually used to develop some understanding of the real-world system
in cases or situations where data has not been collected.  They are often used to interpolate or extrapolate values for
quantities in cases where data has not been collected, for instance to make predictions
about future behavior of the system.  No mathematical model can reflect every aspect of a real world
system, though, since the model-building process always leaves out something.  

Modeling $CO_2$ levels in the atmosphere is of importance because $CO_2$ is a greenhouse
gas.  It tends to keep solar radiation reflected from the Earth's surface
from passing back out into space, so the higher the $CO_2$ level is, the higher 
the average surface temperature will get (other things being equal).  This has 
secondary effects such as rising sea levels, more
severe weather patterns, and long-term climate change.  

One key piece of evidence for why human activities may have changed $CO_2$ levels
over the past 50-200 years is the measurements made at the Mauna Loa observatory
in Hawaii since the 1950's.  In one of our projects we analyzed the monthly 
average measurements in this data set and we saw that, after factoring out 
a predictable annual seasonal variation (due to plant activity taking up 
$CO_2$ through the summer months in the northern hemisphere), the $CO_2$ levels
measured at Mauna Loa have been steadily increasing by about 2.0 ppm per year in recent
years.  (We also saw in our project that an exponential model with an annual
growth rate around $.5\%$ gave a marginally
better fit than a linear model, but in real terms the difference was not very
large over the short run.)  Since humans only began putting large quantities
of $CO_2$ into the atmosphere around the time of the Industrial Revolution
in the mid-1800's, and there have not been frequent volcanic eruptions or
other occurrences that would explain the $CO_2$ concentration increases over this
period, these measurements are often taken as evidence that humans have 
had an effect on the atmosphere.   
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