\magnification=\magstep1
\centerline{Mathematics 126 -- Calculus for Social Science 2}
\centerline{Solutions -- Exam 3}
\centerline{\it December 3, 2001}
\bigskip
\noindent
I.  This is a separable differential equation.  To
solve it, we can proceed as follows.  First we factor
$.016$ out of both terms on the right hand side:
$${dP\over dt} = .016(P - 210000/.016) = .016(P - 13125000)$$
Then divide by $P - 13125000$ and multiply by $dt$ 
to separate.  Integrating, then exponentiating both
sides, we have:
$$\eqalign{\int {dP\over P - 13125000} &= \int .016\ dt\cr
           \ln|P - 13125000| &= .016 t + c\cr
            P - 13125000 &= be^{.016t}\ ({\rm where}\ b = \pm e^c)\cr
            P &= 13125000 + be^{.016 t}\cr}$$
When $t = 0$ (1847), $P = 8000000$, so 
$b = 8000000-13125000 = -5125000$.  Then the population
in 1855 ($t = 8$) is 
$$P(8) = 13125000 - 5125000e^{(.016)(8)} \doteq 7300166.$$
(Because of the emigration, the total population of 
Ireland was decreasing over this time period.)
\bigskip
\noindent
II.  
\item{A)}  The probability that $x$ is in the range
$0 \le x \le 1$ is 
$$\eqalign{P(0 \le x \le 1) &= \int_0^1 ce^{-cx}\ dx\cr
&= -e^{-cx}|_0^1\cr
&= 1 - e^{-c}\cr}$$
Hence
$$\eqalign{.1 &= 1 - e^{-c}\cr
           e^{-c} &= .9\cr
           c &= -\ln(.9) \doteq .1053\cr}$$
\item{B)}  For the expected value of $x$, we must
take into account the {\it whole range of possible
nonzero values of $f(x)$, which is $x \ge 0$}.  Hence
integrating by parts (or using \#23 in the table of 
integrals):
$$\eqalign{E(x) &= \int_0^\infty x c e^{-cx}\ dx\cr
&= -xe^{-cx}|_0^\infty + \int_0^\infty e^{-cx}\ dx\cr
&= 0 + {-1\over c}e^{-cx}|_0^\infty\cr
&= {1\over c} \doteq 9.491\ {\rm with}\ c = .1053\ {\rm as\ above}\cr}$$
\item{C)} 
$$Var(x) = \int_0^\infty (x - 9.491)^2 (.1053)e^{-.1053 x}\ dx$$
\bigskip
\noindent
III.  The conversion from the car speed to the standard
normal is $z = {x - 68\over 5}$
\item{A)}  The probability that a randomly selected car
has speed $x \le 65$ is: 
$$\eqalign{P(x \le 65) &= P\left(z \le {65 - 68\over 5}\right)\cr
                       &= P(z \le -.6)\cr
                       &\doteq .2743\ {\rm (from\ the\ table)}\cr}$$
\item{B)}  The probability that a randomly selected 
car has speed $70 \le x \le 75$ is:
 $$\eqalign{P(70\le x \le 75) &= P\left(.4 \le z \le 1.4\right)\cr
                       &= P(z \le 1.4) - P(z \le .4)\cr
                       &\doteq .9192 - .6554\ {\rm (from\ the\ table)}\cr
                       &=.2638\cr}$$ 
\bigskip
\noindent
IV.  
\item{A)}
$$\eqalign{f_x = y - 4xy - 4y^2 \qquad &\qquad f_y = x - 2x^2 - 8xy\cr
f_{xx} = -4y \qquad f_{xy} &= 1 - 4x - 8y\qquad f_{yy} = -8x\cr}$$
\item{B)} The critical points occur where:
$$\eqalign{f_x &= y(1 - 4x - 4y) = 0\cr
           f_y &= x(1 - 2x - 8y) = 0\cr}$$
From these factored forms, we see either $y = 0$ or $4x + 4y = 1$,
and either $x = 0$ or $2x + 8y = 1$.  If
$y = 0$, then either $x = 0$, or $2x = 1$, so we have
two critical points with $y = 0$:  $(0,0)$ and $(1/2,0)$.
If $4x + 4y = 1$ and $x = 0$, then $y = 1/4$; if 
$4x + 4y = 1$ and $2x + 8y = 1$, then $(x,y) = (1/6,1/12)$
(subtract the first equation from 2 times the second).

For the second derivative test, we use the $f_{xx}$,
$f_{xy}$ and $f_{yy}$ from part A, together 
with 
$$H = f_{xx}f_{yy} - f_{xy}^2 = 32xy - (1-4x-8y)^2$$
We have 
$$\matrix{
{\rm point} & H(x,y) & f_{xx} & {\rm type}\cr
(0,0)       & -1     & 0 & {\rm saddle\ pt.}\cr
(1/2,0)     & -1     & 0 & {\rm saddle\ pt.}\cr
(0,1/4)     & -1     & -1 & {\rm saddle\ pt.}\cr
(1/6,1/12)  & 1/3    & -1/3 & {\rm local\ max.}\cr}$$
(For example,
$$H(1/6,1/12) = 32/72 - (1 - 2/3 - 2/3)^2 = 4/9 - 1/9 = 3/9 = 1/3)$$

 


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