\magnification=\magstep1
\centerline{Mathematics 126 -- Calculus for Social Science 2}
\centerline{Solutions -- Makeup Exam 3}
\centerline{\it December 4, 2001}
\bigskip
\noindent
I.  Let $Q$ be the inside temperature of the house 
as a function of time.  The statement of the problem gives
that 
$${dQ\over dt} = k(Q - 10)$$
for some constant $k$.  
This is a separable differential equation.  To
solve it, we can proceed as follows. 
$${dP\over dt} = k(Q - 10)$$
Then divide by $Q - 10$ and multiply by $dt$ 
to separate.  Integrating, then exponentiating both
sides, we have:
$$\eqalign{\int {dQ\over Q - 10} &= \int k\ dt\cr
           \ln|Q - 10| &= kt + c\cr
            Q - 10 &= be^{kt}\ ({\rm where}\ b = \pm e^c)\cr
            P &= 10 + be^{k t}\cr}$$
When $t = 0$ (10 am), $Q = 68$, so 
$b = 68-10 = 58$.  When $t = 2$ (noon), $Q = 58$, so 
$58 = 10+58e^{2k}$, or $k = {1\over 2}\ln(48/58) \doteq -.0946$.
Then the temperature at 3pm ($t = 5$) is 
$$Q(5) = 10 + 58 e^{(-.0946)(3)}\doteq 46.1$$
(Brr!! Where is that furnace repair guy?!!)
\bigskip
\noindent
II.  
\item{A)}  The probability that $t$ is in the range
$0 \le t \le 1$ is 
$$P(0 \le t \le 1) = \int_0^1 {9\over (9+t^2)^{3/2}}\ dt$$
Because we do not have $du = 2t$ in the numerator for
$u = 9 + t^2$, this is {\it not of the form} $\int u^{-3/2}\ du$.
But we can use entry \#12 from the table of integrals with $a = 3$:
$$\eqalign{
\int_0^1 {9\over (9+t^2)^{3/2}}\ dt &=
{9\over 9}\cdot {t\over (9+t^2)^{1/2}}|_0^1\cr
&= {1\over \sqrt{10}} - 0\cr
&\doteq .3162\cr}$$
\item{B)}  The expected value is 
$$\eqalign{E(t) &= \int_0^\infty t f(t)\ dt\cr
&= \int_0^\infty {9t\over (9 + t^2)^{3/2}}\ dt\cr}$$
With the factor of $t$ from 
the expected value formula, this {\it is of the form\/} $\int u^{-3/2}\ du$.
We have 
$$\int {9t\over (9 + t^2)^{3/2}}\ dt = {-9\over (9+t^2)^{1/2}}$$
So 
$$\int_0^\infty {9t\over (9 + t^2)^{3/2}}\ dt
= \lim_{b\to\infty} {-9\over (9 + t^2)^{1/2}}|_0^b = 3$$
(planned obsolescence!)
\item{C)}  The variance is 
$${\rm Var}(t) \int_0^\infty (t - 3)^2 {9\over (9+t^2)^{3/2}}\ dt$$
\bigskip
\noindent
III.  The conversion from the car speed to the standard
normal is $z = {x - 36\over 7}$
\item{A)}  The probability that a randomly selected car
has speed $x \le 35$ is: 
$$\eqalign{P(x \le 35) &= P\left(z \le {35 - 36\over 7}\right)\cr
                       &= P(z \le -.14)\cr
                       &\doteq .4443\ {\rm (from\ the\ table)}\cr}$$
\item{B)}  The probability that a randomly selected 
car has speed $40 \le x \le 45$ is:
 $$\eqalign{P(40\le x \le 45) &= P\left(.57 \le z \le 1.29\right)\cr
                       &= P(z \le 1.29) - P(z \le .57)\cr
                       &\doteq .9015 - .7157\ {\rm (from\ the\ table)}\cr
                       &=.1840\cr}$$ 
\bigskip
\noindent
IV.  
\item{A)}  Using the quotient rule:
$$\eqalign{f_x &= {1-x^2+y^2\over (1+x^2+y^2)^2}\cr
f_y &= {-2xy\over (1+x^2+y^2)^2}\cr
f_{xx} &={(1+x^2+y^2)^2(-2x) - (1-x^2+y^2)\cdot 2(1+x^2+y^2)(2x)\over
(1+x^2+y^2)^4}\cr
&= {(2x)(x^2-3y^2-3)\over (1+x^2+y^2)^3}\cr
f_{xy} &= {(2y)(3x^2-y^2-1)\over (1+x^2+y^2)^3}\cr 
f_{yy} &= {(2x)(3y^2-x^2-1)\over (1+x^2+y^2)^3}\cr}$$
\item{B)} The critical points occur where $f_x = f_y = 0$.
Look at $f_y$ first.  This is zero only when the numerator
is zero, so either $x = 0$ or $y = 0$.  If $x = 0$, then
$f_x(0,y) = {1\over 1+y^2} \ne 0$, so there are no critical
points with $x = 0$.  But if $y = 0$, then 
 $f_x(x,0) = {1-x^2\over (1+x^2)^2} = 0$ when $x =\pm 1$.
 The two critical points are $(1,0)$ and $(-1,0)$. 
$$\matrix{
{\rm point} & H(x,y) & f_{xx} & {\rm type}\cr
(1,0)       & 1/4     & -1/2 & {\rm local\ max.}\cr
(-1,0)     & 1/4     & 1/2 & {\rm local\ min.}\cr}$$
(For example,
$$H(1,0) = {(2)(1-3)\over 8}{(2)(-2)\over 8} - 0^2 = 1/4)$$

 


\bye   
