\magnification = \magstep1
\centerline{MONT 104N -- Modeling the Environment}
\centerline{Answers for additional practice problems for final exam}
\centerline{December 11, 2018}
\bigskip
\noindent
{\it Caution:} These answers were ``banged out'' at speed, so I cannot really guarantee that 
they are 100\% accurate.  Ask me if anything looks wrong.
\bigskip
\noindent
Chapter 1.
\bigskip
\item{5.}  
\itemitem{(a)}  $44,200\ {\rm km}^2 = 44,200\ {\rm km}^2 \cdot (.621)^2\ {\rm mi^2/km^2}
\doteq 17045\ {\rm mi}^2$.  
\itemitem{(b)}  The thickness in kilometers is $.350$ km, so the
volume in cubic kilometers $= 44,200 \cdot .350 \doteq 15470 {\rm km}^3$.  Then 
converting to cubic feet, 1 cubic kilometer is about $(5280/1.61)^3 \doteq (3280)^3$ cubic feet, so 
15470 cubic kilometers is $\doteq 3.53 \times 10^{10}$ cubic feet.
\bigskip
\item{11.}  
\itemitem{(a)} $\log_{10}(5.34689) \doteq .7281$, $\log_{10}(53.4689) \doteq 1.7281$ and 
$\log_{10}(534.689) \doteq 2.7281$.  What is happening is that when you multiply by a 
power $10^k$, then $\log_{10}(10^k) = k$ is {\it added} to the value of the logarithm, 
by part (1) of Proposition 1.8.
\itemitem{(b)} $\log_7(34.333) = \log_{10}(34.333)/\log_{10}(7) \doteq 1.8172$ (see formula
(1.2) on page 12.)
\itemitem{(c)} $\ln(100.3) \doteq 4.6082$.  (This could also be computed as
$\log_{10}(100.3)/\log_{10}(e)$, where $e$ is the base of the natural logarithms.)
\bigskip
\item{13.} 
\itemitem{(a)} $20\cdot \log_{10}(5000/20) \doteq 47.96\ {\rm dB}$.
\itemitem{(b)} If $L = 1\ {\rm dB}$, then $1 = 20\cdot \log_{10}(p_m/20)$, so 
$p_m = 20 \cdot 10^{1/20} \doteq 22.44$ micropascals.  Similarly if $L = 2\ {\rm dB}$, then
$p_m \doteq 25.18$ micropascals, and if $L = 10\ {\rm dB}$, then $p_m \doteq 63.25$ micropascals.
\itemitem{(c)}  About $20\cdot 10^{150/20} \doteq 6.32 \times 10^8$ micropascals.  This shows
a logarithmic scale in action!
\bigskip
\noindent
Chapter 2.
\bigskip
\item{2.} 
\itemitem{(a)} First student:  absolute error $= 3.44 - 3.40 = .04$ meter.  Percent
relative error $= {3.44 - 3.40\over 3.40} \times 100\% \doteq 1.2\%$.
Second student: absolute error = $.44 - .40 = .04$ meter.  Percent relative error
$= {.44 - .40\over .40}\times 100\% = 10\%$.
\itemitem{(b)}  The first student was more accurate because the percent relative
error was smaller.
\itemitem{(c)}  The absolute error tells you how far off the measurement is from
the true value.  Absolute errors are important one measurement at a time.
The percentage relative error expresses the error as a percent of
the exact value.  So, percentage relative errors are more useful for comparing 
accuracies of different measurements. 
\itemitem{(d)}  Precision of a set of measurements 
is not the same as accuracy.  Accuracy measures how
far the measurements are from an exact value.  Precision measures
how close the measurements are to each other.  
\bigskip
\item{7.}   
\itemitem{(a)}  Boston: $13321.3$ people per square mile $\doteq 34530.1$ people per square kilometer.  Chicago: $11868.1$ people per square mile $\doteq 30763.3$ people per square kilometer.
Miami: $11198.7$ people per square mile $\doteq 29028.1$ people per square kilometer.
NYC: $27016.3$ people per square mile $\doteq 70029$ people per square kilometer.
Philadelphia:  $11233.6$ people per square mile $\doteq 29118.7$ people per square kilometer.
San Francisco: $17246.4$ people per square mile $\doteq 44704.4$ people per square kilometer.
\itemitem{(b)} In 2016, from page 26, we see that the population of New York
was estimated at $8.54 \times 10^6$.  That gives a population density of 
$28222.1$ people per square mile.  The percent change in population density from 2010 (reference)
to 2016 (comparison) was
$${28222.1 - 27016.3\over 27016.3} \times 100\% \doteq 4.5\%$$
The numbers would come out the same if the densities per square kilometer were
used since the conversion factors from square miles to square kilometers in the 
numerator and denominator would cancel.  
\bigskip
\noindent
Chapter 4.
\bigskip
\item{6.} 
\itemitem{(a)} $y - 5.2 = 7\cdot (x - 3.2)$ or $y = 7x -17.2$.
\itemitem{(b)} $y - 1 = {-1/7}(x - 0)$ or $y = -x/7 + 1$.
\bigskip
\item{8.} $g(x)$ is the linear one because that is the only function for which 
the slopes between pairs of points are always the same $= -2.25$.  The equation
is $y - 1.525 = -2.25(x - 1.1)$ or $y = -2.25 x + 4$.  
\bigskip
\item{9.}  Kudzu area = ${7.4 - 0\over 2018 - 1876}(t - 1876)$ (in units of $10^6$ acres, $t$ in years).
\bigskip
\noindent
Chapter 5.
\bigskip
\item{2.} All of these are solved by taking logarithms:
\itemitem{(a)} $x = {1\over 2} {\log_{10}(28.3/4.5) \over \log_{10}(3.4)} \doteq .7513$.
\itemitem{(b)} $x = {\log_{10}(3.5) \over \log_{10}(4) - \log_{10}(2)} \doteq 1.8074$.
\itemitem{(c)} $x = {\log_{10}((7.9 - 5.6)/2.8)\over \log_{10}(7.4)} \doteq -.09828$.
\bigskip
\item{4.} 
\itemitem{(a)} $f(t) = 8.54\cdot (1.15)^t$.
\itemitem{(b)} $f(t) = 3.5711 \cdot (1.03)^t$.  
\bigskip
\item{6.}
\itemitem{(a)} doubling time $ = {\log_[10](10/5)\over \log_{10}(1.25)} \doteq 3.106$
\itemitem{(b)} doubling time $ = {\log_[10](36.6/18.3)\over \log_{10}(3.4)} \doteq .5664$.
\bigskip
\item{7.} As in the special cases in 6 above, we solve for $t$ in the equation
$$2\cdot Q(0) = Q(0)\cdot a^t$$
to find the doubling time, $t_2$.  Dividing both side by $Q(0)$, then taking logarithms, we get
$$t_2 = {\log_{10}(2)\over \log_{10}(a)}.$$
\bigskip
\item{8.} This follows from rules for exponents.  From the previous problem, 
$\log_{10}(a) = {\log_{10}(2)\over t_2}$, so 
$$a = 10^{\log_{10}(a)} = 10^{\log_{10}(2) \cdot {1\over t_2}} = 2^{1\over t_2}.$$
Then raising both sides to the $t$ power and multiplying by $Q(0)$ we get
$$Q(t) = Q(0)\cdot a^t = Q(0)\cdot \left(2^{1\over t_2}\right)^t = Q(0)\cdot 2^{t\over t_2}.$$
\bigskip
\item{9.}
\itemitem{(a)} Proceeding as in Example 5.8 in the text, from the half-life of this 
isotope: 
$$Q(28.8) = {1\over 2}\cdot Q(0) = Q(0)\cdot a^{28.8}$$
so $\log_{10}(a) = {\log_{10}(.5)\over 28.8} \doteq -.0105$ and 
$a = 10^{-.0105} \doteq .9762$.  Then the model is
$$Q(t) = Q(0)\cdot (.9762)^t.$$
\itemitem{(b)} We solve for $t$:
$$.01\cdot Q(0) = Q(0)\cdot (.9762)^t,$$
so $t = {\log_{10}(.01) \over \log_{10}(.9762)} \doteq 191.2$ years.
\bigskip
\noindent
Chapter 7.
\bigskip
\item{1.}  The percentage change gives ${Q(n+1) - Q(n)}{Q(n)} \times 100 = r$, 
so $Q(n+1) = \left(1 + {r\over 100}\right)\cdot Q(n)$.  Hence if we start from 
$t = 0$, we get
$$\eqalign{Q(1) &= \left(1 + {r\over 100}\right) Q(0)\cr
Q(2) &= \left(1 + {r\over 100}\right) Q(1) = \left(1 + {r\over 100}\right)^2 Q(0)\cr
Q(3) &= \left(1 + {r\over 100}\right) Q(2) = \left(1 + {r\over 100}\right)^3 Q(0)\cr
}$$
and so forth.  The general pattern is 
$$Q(n) = \left(1 + {r\over 100}\right)^n Q(0).$$
Note that this is an exponential function of $n$, with $a = \left(1 + {r\over 100}\right)$.
\bigskip
\item{4.} 
\itemitem{(a)} $Q(n) = 3.4\cdot (1.8)^n$ by problem 1 above.
\itemitem{(b)} Here we want to use the general solution for affine first order
equations from (7.4) on page 129:
$$Q(n) = \left(4.3 + {(-.03)\over (.78 - 1)}\right)(.78)^n - {(-.03)\over (.78 - 1)} 
= 4.4364\cdot (.78)^n - .1364.$$
\bigskip
\item{8.}
\itemitem{(a)} From the equation, $r = .03$ and $r/M = .006$, so $M = 5$.  Since $Q(0) = .8 < M$,
the solution will increase in a sort of $S$-shape and tend toward $M$ as a horizontal
asymptote as $n$ increases.  
\itemitem{(b)} $r = .34$ and $r/M = .0009$, so $M = 377.8$.  Since $Q(0) = 420 > M$, the
solution will decrease toward $M$.  
\itemitem{(c)} $r = .86$ and $r/M = .0048$, so $M = 179.2$.  Since $Q(0) < M$, this
is similar to part (a).
\bigskip
\noindent
Chapter 8.
\bigskip
\item{3.} The equations are
$$\eqalign{
A(n+1) &= .4 A(n) + .25 B(n) + .5 C(n)\cr
B(n+1) &= .4 B(n) + .4 A(n)\cr
C(n+1) &= .33 C(n) + .2 B(n)\cr}$$
\bigskip
\item{7.}
\itemitem{b.} Say the fraction of the $I$ group that recovers
is $\gamma_1 I(n)$ and the fraction that dies is $\gamma_2 I(n)$
for some constants $\gamma_1,\gamma_2$.  The $R$ compartment is correspondingly
split between $R_1$ and $R_2$. The equations would be
$$\eqalign{
S(n+1) &= S(n) - \beta S(n)I(n)\cr
I(n+1) &= I(n) + \beta S(n)I(n) - \gamma_1 I(n) - \gamma_2 I(n)\cr
R_1(n+1) &= R_1(n) + \gamma_1 I(n)\cr
R_2(n+1) &= R_2(n) + \gamma_2 I(n)\cr}$$
\end
