Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Topic: Course description from the college board web site
Replies: 0

 Joshua Zucker Posts: 710 Registered: 12/4/04
Course description from the college board web site
Posted: May 22, 1997 5:37 PM

board web site.

I still haven't finished organizing my thoughts on the differences in
the topics list, but I'll be sharing those with you soon. Here are my
thoughts on the changes in the sample questions.

The free-response questions are exactly the same as in the 1997
description, which is also the same as the contents of the 1995 AP
test.

The multiple choice questions are almost exactly the same as in the
1997 course description; in particular, they still include the WRONG
question, #6 from the AB sample, that they've been using for several

If dy/dx = 9y^4 and if y = 1 when x = 0, what is the value of y when
x = 1/3?

Ironically, the failure of this question is exactly the sort of thing
that the new AP syllabus is supposed to guard against -- brainlessly
plugging in to standard formulas and solutions rather than thinking

Of the new multiple choice questions, AB #1, 8, 9 are all interpreting
graphs, and #14 is a nice question that tests for understanding of
related rates quite well. Wups! #14 isn't really a new question, it
was just moved from the B part to the A part.

On the graphing calculator questions, AB #18 is "improved" so that you
actually need to put it in to the calculator now. That is, the old
and new questions both ask for the average value on [-1,1] of
e^(-x^2), but the old test gave answer choices of .37, .75, 1, 1.49,
and 1.81, while the new test gives .70, .75, .80, .85, .90. The old
question was easy to do by just noting that the function was bounded
between 1/e^2 and 1. And the old question caught the likely error,
which is to find the integral with your calculator but not divide by
two to find the average value.

The new question MUST be done by mechanically plugging in to a
calculator. The only error that might lead you to one of the other
answer choices is using too coarse a partition when you numerically
integrate. That is, they've replaced a question which allows thinking
and conceptual understanding OR mechanical plugging in to a calculator
with a question that tests ONLY ability to plug in to the calculator.
I think this is a very BAD change.

Another new question, #19, in my opinion asks for lack of
understanding. They give a table of values of f(1.7), f(1.8), f(1.9),
f(2.0). They tell you f is differentiable on [0,3] and ask for the
best approximation for f'(1.7). The correct answer SHOULD be that we
have no idea, because we don't know whether f is oscillating rapidly
course, the AP people just want us to plug in delta-f over delta-x

And not only that, but the wrong answers they give are foolish! They
do give the answer choice that would come from just taking delta-f
without dividing by delta-x. But they do NOT have the wrong answer
that would come from using f(2.0) - f(1.7) / .3 instead of
f(1.8) - f(1.7) / .1 ... that is, they don't test the one idea that
they should be, which is that the best approximation to the derivative
comes from using the point that's CLOSEST. They do have the answer
that comes from using 1.9, though.

The other new question looks pretty good: the rate of oil leakage
from a tanker is given, and you need to figure out that total amount
that's leaked out over a period of ten hours. A good way to check
that people understand that they need to integrate.

Now, on to the BC questions! Let's see, some minor changes like
replacing "The normal to" with "The line perpendicular to the tangent
of".

A L'Hopital's rule question from part A was switched with a Taylor
series question from part B. I guess the idea there is to allow
students to use the calculator to approximate the limit and figure out
a sensible answer (or at least check the answer) for the limit. Also,
they point out that some people have calculators that compute Taylor
series for them ... I guess they wanted to avoid that.

A really nice fundamental theorem question was moved from the B part
to the A part. (was #18, now #7)

There's also a slope field question which tests basic understanding of
the concept.

And of course, the epsilon-delta problem is gone, since that's no
longer in the AP curriculum at all.

There are some nice new questions on the B part: Find the limit as
x -> infinity of
integral from 1 to x of sqrt(4 - e^-t) dt, all over x.

I think this problem reflects the new ideas very well, because
understanding what this function looks like will get you far. It's an
easy question to answer just with intution and no calculation. And my
colleague Ted Alper points out that you could solve it very nicely
with L'Hopital's rule. What worries me is that people will do it by
approximating with a calculator (what's the value when x = 100?
1000?) and then deciding that it looks convergent to them. I'd rather
see this problem on the non-calculator part of the test.

#21 is also a nice, new problem, comparing the 7th degree Taylor
polynomial for sin x with the function itself, asking when it's
greater and when it's less.

And they've replaced what looks to me like a pretty nice graphing
question (which of the following could be the graph of the sum from
n = 0 to 25 of (sin x / 2)^n ?) with another fairly reasonable
question, #24: g(t) = 100 + 20 sin (pi t / 2) + 10 cos (pi t / 6).
For 0 <= t <= 8, g is decreasing most rapidly when t = ?
I don't have any particularly strong preference between these two
questions. The old one seems more original, the new one seems more
hackneyed. On the other hand, the new one seems more likely to come
up in real-world practice.

A couple of the new problems seem very traditional to me: find the
max area of a rectangle that fits under the curve of cos x. Find the
area of the region in the first quadrant enclosed by xe^x, x=0, and
x=k, in terms of k. Why are problems like these new for this year?

--Joshua

[just in case you didn't catch it: the problem with #6 on the AB test
is that the differential equation has a pole at x = 1/27 and so we
have no idea what the solution is beyond there. You can see this,
among other reasons, because y changes sign between the given x = 0
and the "answer" at x = 1/3; but the differential equation itself
gives slope zero when y is zero, so how can y change sign? Only by
going through positive infinity ...]