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\centerline{MONT 100N -- Modeling the Environment}
\centerline{Solutions for Practice Problems for the Final Exam}
\centerline{November 30, 2017}
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\noindent
Chapter 1
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\noindent
5.  
\item{(a)}  From the table of conversion factors, one kilometer $\doteq .6214$ mile, so
one square kilometer $\doteq (.6214)^2 \doteq .3861$ square miles.  Hence the Larsen 
C iceberg has an area of about $(.3861)\cdot 44200 \doteq 17067$ square miles.  
\item{(b)}  The thickness in kilometers is $350 \times .001 = .350$ kilometer.
So the volume in cubic kilometers would be approximately 
$$44200 \cdot .350 \doteq 15470\ {\rm cubic\ km}$$
This is approximately 
$$15470 \times (.6214)^3 = 3712$$
cubic miles.  (That is a lot of ice!)
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\noindent
11.  
\item{(a)}  $\log_{10}(5.34689) \doteq .728101$.  Then 
$\log_{10}(5.34689) \doteq 1.728101$ and $\log_{10}(53.4689) \doteq 2.728101$.  
What is happening here is that, for example 
$$53.4689 = 5.34689 \times 10$$
so 
$$\log_{10}(53.4689) = \log_{10}(5.34689) + \log_{10}(10) = 1 + \log_{10}(5.34689).$$
\item{(b)} No calculator I know of has a $\log_7$ key.  So this is one place
where you would need to use the conversion formula (1.2) on page 12 of the text:
$$\log_7(34.333) = {\log_{10}(34.333) \over \log_{10}(7)} \doteq 1.81720.$$
\item{(c)} $\ln(100.3) \doteq 4.60817$.  (For this one, you can use the $\ln$
key if your calculator has one, or use the conversion formula as in the previous
problem.)
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\noindent
14.  Since $M = {2\over 3}\log_{10}(S) - 10.7$, we can solve for $S$ like this:
$$\eqalign{{3\over 2}(M + 10.7) &= \log_{10}(S)\cr                                            
            10^{{3\over 2}(M + 10.7)} &= S.\cr}$$
\item{(a)}  Hence if the two quakes have moments $M_1$ and $M_2$, then 
$${S_1\over S_2} = {10^{{3\over 2}(M_1 + 10.7)} \over 10^{{3\over 2}(M_2 + 10.7)}}$$
Subtracting the exponents on the right, we get
$${S_1\over S_2} = 10^{{3\over 2}(M_1 - M_2)}.$$
\item{(b)}  The ratio is $10^{3/2} \doteq 31.6$.  Increasing the moment magnitude by 1
gives a quake that is $31.6$ times as strong in terms of the seismic moment.
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\noindent
Chapter 2
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\noindent
2.  
\item{(a)}  Percent change is 
$${324,700,000 - 308,745,538 \over 308,754,538} \times 100\% \doteq 5.2\%$$
(positive since it is an increase).  
\item{(b)}  Percent change per year would be ${5.2\over 7} \doteq .743\%$ per year.
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\noindent
7.   
\item{(a)}  The densities in people per square mile are (all rounded to whole numbers):
$$\eqalign{
&{\rm Bronx} \qquad  32900\cr
&{\rm Brooklyn} \qquad 35367\cr
&{\rm Manhattan} \qquad 69464\cr
&{\rm Queens} \qquad  20554\cr
&{\rm Staten Island} \qquad 8030\cr}$$
To get the people per square kilometer multiply each by $(1.609)^2 \doteq 2.59$.  
The $1.609 = 1/.6124$ as in Chapter 1, problem 5 above.
\item{(b)}  Not it is not.  The average of the densities in part (a) would be 
$${32900 + 35367 + 69464 + 20554 + 8030 \over 5} \doteq 33263.$$
But the actual overall NYC population density was 
$${1,385,108 + 2,504,700 + 1,585,873 + 2,230,722 + 468,730 \over 42.1 + 70.82 + 22.83 + 108.53 + 58.37} \doteq 27016.$$
Algebraically the reason these are not the same is that you don't add fractions
by adding the numerators and denominators and making a new fraction with the sums(!)  
In real world terms, Manhattan is over-represented in the sum of the densities
because, while it has a high population density, it also has a small area.
To get the city population density from the borough densities, you could 
multiply the borough density by the area for that borough, add, and divide
by the total land area.  That is the same as a weighted average where you 
multiply the borough density by the proportion of the total area in
that borough.   
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\noindent
Chapter 4
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\noindent
6.  
\item{(a)}
Parallel lines have equal slopes so $y - 5.2 = 7(x - 3.2)$.  It's fine to 
leave the equation in this form, or you could also rearrange it
to slope-intercept form:  $y = 7x - 17.2$.
\item{(b)}  Perpendicular lines have slopes that are negative reciprocals.
So the equation is $y - 1 = {-1\over 7}(x - 0)$ or $y = {-1\over 7}x + 1$.
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\noindent
8.  The only one where the slopes between successive pairs points are all the same
is $g(x)$.  The equation is 
$$g(x) = 1.525 + -2.25(x-1.1) = -2.25x + 4.$$
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\noindent
9.  Using the data points $(0,0)$ (that is, no kudzu in 1876) and 
$(141, 7400000)$ (that is, $7,400,000$ acres in 2017, which is 141 years after 1876), 
we get a linear model
$$k(t) \doteq 52482.3 t$$
where $k(t)$ is the land area covered by kudzu in year $t$.  The slope represents
the growth rate of the kudzu area; the units are acres per year.  
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\noindent
Chapter 5
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\noindent
2.  All these equations are solved by the same technique -- isolate a term with
the $x$ in the exponent on one side, take logarithms to get $x$ out of the exponent,
then solve the resulting equation for $x$.  
\item{(a)}  $(3.4)^{2x} = {28.3\over 4.5} \doteq 6.29$, so $2x \log_{10}(3.4) = \log_{10}(6.29)$,
so 
$$x \doteq {\log_{10}(6.29)\over 2\log_{10}(3.4)} \doteq .7513.$$
\item{(b)} Divide by $2^x$ first to get $2^x = 3.5$, then $x \log_{10}(2) = \log_{10}(3.5)$
and 
$$x = {\log_{10}(3.5) \over \log_{10}(2)} \doteq 1.807.$$
\item{(c)}  Subtract $5.6$ from both sides, divide by $2.8$, then take logs:
$$x = {\log_{10}(2.3/2.8)\over \log_{10}(7.4)} \doteq -0.0983.$$
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\noindent
6.  
\item{(a)}  $Q(0) = 5$, so we just need to solve the equation $10 = 5 (1.25)^t$
for $t$.  The answer is $t = {\log_{10}(2) \over \log_{10}(1.25)} \doteq 3.106$.
\item{(b)}  Same technique as (a).  The answer is $.5664$.  (Note that $3.4 > 1.25$
means that the function here is increasing faster than the one in part (a), so 
the doubling time is smaller.) 
7.  The equation to solve is $2Q(0) = Q(0) a^t$, so $2 = a^t$ and 
$$t = {\log_{10}(2) \over \log_{10}(a)}$$
follows.
8.   We get that when $t = t_2$, the factor $2^{t\over t_2}$
satisfies $2^{t_2/t_2} = 2^1 = 2$.  Hence $Q(t_2) = 2Q(0)$ and this exponential
function does have doubling time exactly $t_2$.  The value of $a$ is 
$$a = 2^{1\over t_2}, \quad {\rm since}\quad (2^{1\over t_2})^t = 2^{t\over t_2}$$
using rules for exponents.  
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\noindent
9. 
\item{(a)}  The most immediate way to write down the
 model is 
$$Q(t) = Q(0) (1/2)^{t/28.8}$$ 
(see equation (5.9) in the text).
\item{(b)}  We solve $.01\%$ is $0.0001$ so we solve
$$0.0001 Q(0) = Q(0) (1/2)^{t/28.8}$$
So $t = 28.8 \cdot {\log_{10}(.0001)\over \log_{10}(.5)} \doteq 382.7$ years.
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\noindent
Chapter 7
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\noindent
1.  We did this in class; consult your notes.  The idea is that the 
percent change is 
$${Q(n+1) - Q(n)\over Q(n)} \times 100\% = {r\over 100} \times 100\%.$$
Rearrange this equation to get the one you are asked to derive.  
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\noindent
4.  
\item{(a)}  $Q(n) = 3.4(1.8)^n$.
\item{(b)}  Use the general solution for affine first order linear equations (equation (7.4)
on page 114 of the text).  The solution is 
$$Q(n) = (Q(0) + {b\over a - 1})a^n - {b\over a - 1}$$
with $Q(0) = 4.3$, $a = .78$ and $b = -.03$, so
$$Q(n) = (4.3 + {-.03\over -.22})(.78)^n - {-.03\over -.22} \doteq 4.436 (.78)^n - 4.436$$
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\noindent
Chapter 8
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\noindent
6.  If you add the left, then the right sides of the SIR equations, you see 
$$S(n+1) + I(n+1) + R(n+1) = S(n) + I(n) + R(n)$$
because every other term that appears in one equation with a $+$ sign occurs in 
a second one with a $-$ sign, so they cancel.  This shows that the total population
is not changing from one time step to the next.
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\noindent
13.  
\item{(a)}  $r = .03$ and $r/M = .006$, so $M = 5$.  The $Q(0) < M$, so the solution
will increase and approach $M$ as a horizontal asymptote. 
\item{(b)}  $r = .34$ and $r/M = .0009$, so $M \doteq 378$.  Since $Q(0) > M$, the 
solution will decrease and tend toward $M$ as a horizontal asymptote.
\item{(c)}  $ = .86$ and $r/M = .0048$, so $M \doteq 179.2$.  Since $Q(0) < M$, the 
solution will look like the one in (a).   
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\noindent
8.  (a) and (b):  The modified SIR model would look like this.  Say $R_1$ 
represents the infected patients who {\it recover} and $R_2$ represents
the infected patients who die.  Then
$$\eqalign{
S(n+1) &= S(n) - \beta S(n) I(n)\cr
I(n+1) &= I(n) + \beta S(n) I(n) - \gamma_1 I(n) - \gamma_2 I(n)\cr
R_1(n+1) &= R_1(n) + \gamma_1 I(n)\cr
R_2(n+1) &= R_2(n) + \gamma_2 I(n).\cr}$$
As noted on the review sheet, parts (c) and (d) would not 
be suitable for the exam(!)
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\noindent
11.  To find equilibrium solutions we rewrite the system (8.8) as 
$$\eqalign{
P(n+1) - P(n) &= -a P(n) + b P(n)Q(n)\cr
Q(n+1) - Q(n) &= c Q(n) - cQ(n)^2/M - d P(n)Q(n)\cr}$$
For an equilibrium, the right sides of both equations should be {\it equal to zero}.
The first equation factors like this:
$$P(-a + bQ) = 0$$
so $P = 0$ or $Q = b/a$.  In the second equation, if $P = 0$, then we get $Q = 0$ or 
$Q = M$.  If $Q = b/a$, then we get $P = (1 - a/(bM))\cdot c/d$ as desired. 
\bye
