Honors
AP Calculus Quiz

(rev. 3/8/1999, 11/14/2002)

1. Woody is coasting backward down a hill. At time *t* = 0 sec., he slams the accelerator to
the floor, and the car's velocity is given by *v*(*t*) = –2 + Ö*t* for *t* ³ 0.

-- (a) Find the time at which *v* = 0.

-- (b) Find net displacement for 1 £ *t* £
9.

-- (c) Find total distance traveled for 1 £ *t* £
9.

2. A rectangle of length *L* and width *W* has constant area of 1200 sq. in. The length
changes at a rate of *dL*/*dt* in./min.

-- (a) Find *dW*/*dt* in terms of *W* and *dL*/*dt*.

-- (b) At a particular instant, the length is increasing at 6 in./min. and the width is decreasing at 2 in./min. Find the
dimensions of the rectangle at this instant.

-- (c) At the instant mentioned in part (b), is the length of the diagonal
increasing or decreasing? At what rate?

3. **(original version)** The U.S. Army Corps of
Engineers needs to construct a road over *y* = 3 + sin(*x*/2), where *x* is horizontal displacement (0 £ *x* £
2p)
and units are in km. Assume that the life-cycle cost of a surface road is 5*y*|*y*'|
million dollars per km, and assume that the life-cycle cost of a horizontal
tunnel is $30 million per km regardless of elevation. Draw a diagram and find
the elevation (to the nearest meter) of the tunnel entrance such that the
life-cycle cost of the route up and through

Note: The life-cycle cost of a surface road is assumed here to vary jointly
with elevation and absolute slope, using 5 as the constant of proportionality.

**SOLUTION TO #3 (original version)**

Almost everyone drew the same diagram, letting *x* denote the horizontal point at which the tunnel begins. This is
not really legitimate unless we use a different variable (say, *t*) to be the independent variable in the
elevation function. The other place where nearly everyone stumbled was in
failing to note that the cost function 5*y*|*y*'| uses elevation and slope information
that *change* continuously depending on horizontal position. In other
words, we will need to use an integral involving 5*y*|*y*'|, not simply a value
computed for a single *x*.

The use of arc length instead of horizontal distance turns out not to make a
huge difference. Let's work the problem first "naively," then more
intelligently using arc length.

If the tunnel enters at *x*, it exits
at 2p
– *x*, so the (inaccurate) cost
function is *B*(*x*)= [integral from 0 to x of 5y(t)y'(t)dt]
+ 30(2p
- x-x) + [integral from 2pi-x to 2pi of 5y(t)y'(t)dt]
= 2fnInt(5y(t)y'(t),t,0,x) + 60(p – *x*). To get
a sense of the behavior of this function, plot it over the only domain that
makes sense, namely 0 to p. What you'll see is that
the function is decreasing everywhere, so that the minimum occurs when x = p, y = 4, and tunnel length = 2p – 2x = 0; in other words,
the tunnel is not cost-effective at all and should not be built. The road
should be laid over the mountain. (Note: In the real world, the profile
function would show the road going over a mountain *pass,* not a
mountaintop. Also, the road is far too steep near the base to carry vehicle
traffic, since *y*'(0) = 0.5.)

TI-83 computational notes: By using the logically invalid notation fnInt(5Y_{1}Y_{2},X,0,X),
where Y_{2} denotes the function 0.5cos(X/2), you can roughly double
the speed of the graphing. Be sure to set *xres* = 5 in the WINDOW setting,
though, or you'll be waiting forever.

As mentioned above, the function *B*(*x*) is inaccurate because it assumes the
sloped, curved road is straight and flat. A better cost function, employing the
arc length differential *ds*
= Ö(1+(*y*')^{2}) *dx*, would be *C*(*x*) = 2*[integral from 0
to x of 5y(t)y'(t)sqrt(1+(y'(t))^2)] + 60(pi-x), or,
to use the logically sloppy but much faster TI-83 notation, 2fnInt(5Y_{1}Y_{2}*sqrt(1+Y_{2}^2),X,0,X)+60(p–X). Like the function *B*(*x*)
we plotted before, *C*(*x*) is also decreasing on [0, p]. Therefore, the conclusion is again that the tunnel is
not cost-effective.

But . . . should we even use the calculator at all? Using nothing more than
pencil and paper, we can get *dC*/*dx* = 10yy'sqrt(1+(y')^2) - 60, courtesy of the second form of the
FTC. We also know that y<=4 on the domain of interest, and 0<=y'<=0.5
on the domain of interest. Thus the first term in our expression for dC/dx can never possibly exceed 10*4*0.5*sqrt(1+0.25),
which is much less than 60. Thus *dC*/*dx* is always
negative. No calculator required! Or, if you insist on using your calculator,
please just graph dC/dx, which is very speedy--no
long wait for messy computations. (If *dC*/*dx* did have a
zero, graphing *dC*/*dx* would still be the way to go,
since you wouldn't be hunting for a flat spot on *C*(*x*) and wouldn't be
desperately trying to zoom in on it.)

3A. **(take-home version)** Change $30 million to $1 million; in other words,
a new tunnel-boring machine has been found that cuts the cost of tunnel
construction by a factor of 30. What is the optimal elevation of the tunnel
entrance, to the nearest meter? Also calculate the total cost of the road.

3B. **(Will Segal's "Problem With a Twist")** Suppose that the
cost per km of building the tunnel is unknown, but that for bureaucratic
reasons it has been predetermined that the elevation of the tunnel entrance is
to be 3500 m (i.e., 3.5 km, or halfway between the 3 km base of the mountain
and the 4 km summit). For what cost per km of tunnel construction would this
have been the optimal decision to make?

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*Last updated: 17 Oct 2004*