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\begin{document}
Math 196. Final Exam, May 7, 2001  \ \         \ \ \ \ \ \ Name \
\rule{80mm}{0.05mm}

1)  (12 pts.) You are given the function $f(x) = 2x^3 - 24x^2$.

\ i) Find the critical numbers for $f$.

\vspace{.8in}

\, ii) Determine the intervals over which $f$ is increasing and those over
which $f$ is
decreasing.

\vspace{1.2in}

2) (16 pts.) On October 12, 1999, the New York Times (in the Science Times
Section) reported
that a team of astronomers using a telescope on Mauna Kea, Hawaii, had
discovered a moon
orbiting the asteroid Eugenia (in the main asteroid belt between Mars and
Jupiter). The
moon is about $13$ kilometers across, and orbits Eugenia in a nearly
circular orbit at a distance
of $1120$ kilometers once every $4.7$ days.  (Use MKS in your computations
below.)

\ i) Compute (or, more appropriately, estimate) the speed with which the
moon orbits Eugenia.

\vspace{.8in}

ii) Compute (or, more appropriately, estimate) the mass of Eugenia (in
kilograms).

\newpage

3) (14 pts.) Give an explanation of the formula $F_P = \frac{8\kappa^2m}{L} \
\frac{1}{r_P^2}$ in the context of the orbit of Eugenia's moon. Include a
complete discussion of the
underlying physical situation, draw an appropriate diagram, and define the
constants $\kappa, L,$ and
$m$. Use the data from the previous problem to compute $\kappa$ for the
orbit and express the gravitational
force (in newtons) with which Eugenia pulls on its moon in terms of the
mass $m$ of this moon.

\vspace{3in}

4)	(16 pts.) The Tacoma Narrows Bridge was completely rebuilt in the
years 1948 to 1950.  The data for the
new bridge is as follows.  It has a total length of  5,000 feet and a
center span of  2,800 feet.  Its  2
main cables support a single deck that carries  4  lanes of automobile
traffic.  The dead load is 8,680
pounds per foot. Assume that it is designed for a live load capacity of
4,000 pounds per foot and that
the sag in the cable over the center span is 280 feet.

\ i)  Consider one of the main cables over the center span at a point where
it meets one of the
towers.  Compute the tension  $T$  of the cable at that point and compute
the angle  $\alpha$  that
the cable makes with the horizontal at that point.

\vspace{1in}

ii) Compute the contribution of this one cable to the compression forces on
the tower

\newpage

5) (24 pts.) The metal rim of a wagon wheel of radius 0.5 meters and mass
of 2 kilograms and a bowling ball of
10.8 centimeters in diameter and a mass of 5 kilograms are poised for a
race down a ramp
that is 10 meters long and makes an angle of $30^{\circ}$ with the
horizontal. The index of inertia of the
rim is given by the expression $mr^2$ and that of the ball by
$\frac{2}{5}mr^2$. Under the
assumption that the friction produced by the ramp is precisely that force
that rotates the two objects
do the following:

\ \ i) Draw pictures of the situations and use them as a guide towards your
solutions of the problems below.

\vspace{1.2in}

\ ii) In each case compute the frictional force on the object $f$ in terms
of the angular acceleration.

\ \ a) For the rim:

\vspace{.8in}

\ \ b) For the bowling ball:

\vspace{.8in}

iii) In each case compute the total force $F$ on the object and use $F =
ma$ to calculate the linear
acceleration $a$ of the motion of each of the two objects down the ramp.

\ \ a) For the rim:

\vspace{.8in}

\ \ b) For the bowling ball:

\vspace{.8in}

iv) Assuming that both the rim and the bowling ball start from rest
(without being given a push), which will win
the race?

\newpage

6)  (28 pts.) A travel agent offers the following group deal for a bus tour
to the Kentucky
Derby. It will only accept a booking of at least 25 people. For precisely
this number the cost is 120
dollars per person. For each additional person up to a maximum of 40 people
(this is the capacity of the buses
that the agency operates), there is a reduction of 2 dollars per person.
So the number of people going on the
trip is some number $x$ between 25 and 40. It costs the agency 2000 dollars
(fixed
cost) plus 18 dollars per person to operate a bus down to Louisville.

\ \ i) Express the price per person as a function of the number of
travellers $x$.

\vspace{.8in}

\ ii) For what number of travellers is the revenue a maximum?

\vspace{.8in}

iii) Express the cost to the agency as a function of the number of travellers.

\vspace{.8in}

iv) For what number of travellers is the profit at maximum level?

\vspace{.8in}

\ v) What changes in the pricing strategy will result in a situation where
the number of travellers
that maximizes the profit coincides with the capacity of the bus?  What
cost cutting strategy has the
same effect?

\vspace{.8in}

vi) Suppose that the trip takes place at the maximum capacity of 40
persons. Notice that in this case
the cost of the trip is $(120 - 15 \cdot 2) = 90$ dollars. What is the
total consumer surplus under the
assumption that all the travellers would have been willing to pay $120$ for
the trip.

\newpage

FORMULAS

$T_0  = T_x \cos \theta$  \ \ \ $wx = T_x \sin \theta$   \ \ \ $T_d  =  \sqrt{(
\frac{d}{2s})^2 \ + 1}$  \ \ \ $\tan \alpha = \frac{2s}{d}$


$\theta(t) = \frac{1}{r} s(t)$ \ \ \ $\omega(t) = \frac{1}{r} v(t)$ \ \ \
$\alpha(t) =
\frac{1}{r} a(t)$


Torque = (Index of Inertia)(Angular Acceleration)


$F_P = \frac{4\pi^2a^3m}{T^2} \ \frac{1}{r_P^2}$, $F_P = \frac{8\kappa^2m}{L} \
\frac{1}{r_P^2}$, $\frac{a^3}{T^2} = \frac{GM}{4\pi^2}$, $M =
\frac{4\pi^2a^3}{GT^2},$


$A_p = A_0(1 +  \frac{r}{n})^p   \ \ \ A(t)  =  A_{nt}  =  A_0(1 +
\frac{r}{n})^{nt}$


$A(t) = A_0e^{rt} \ \ \ S_p = \frac{12}{r}A_0(1 + \frac{r}{12}) ( (1 +
\frac{r}{12})^p - 1)$


$PV_p  = \frac{12B}{r}[1 -  (1 + \frac{r}{12})^{-p}] \ \ \ PV_p  =
\frac{2C}{r}[1 -  (1 +
\frac{r}{2})^{-p}]$


$p(t) = p(0) e^{kt} \ \ \  e_{D}(p) = p \frac{D'(p)}{D(p)}   \ \ \   e_S(p)
= p \frac{S'(p)}{S(p)}$

$\Pi(x) = R(x) - C(x)$

$G = 6.67 \times 10^{-11}
\frac{\hbox{\scriptsize{meters$^3$}}}{\hbox{\scriptsize{kilograms}}
\cdot \hbox{\scriptsize{seconds$^2$}}}$

\end{document}

