AP Calc BC - too easy?

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Discus: SAT/ACT Tests and Test Preparation: May 2004 Archive: AP Calc BC - too easy?
By Crappyjacks (Crappyjacks) on Saturday, May 08, 2004 - 09:09 pm: Edit

I read all of the posts on this site about the difficulty of the Ap Calc BC exam and I was shocked. The general consensus among the students who took the exam in my school was that it was jokingly easy. In fact I almost felt insulted by the exam. All of the questions were straightforward applications, there was little or no need to think on any of them.

Anyway, there must be some out there who feel the same way that I do about the exam.

By Smadans (Smadans) on Saturday, May 08, 2004 - 09:15 pm: Edit

I almost fully concur; however, my performance of numero 5 FR was way below average.

By Emmittsmith (Emmittsmith) on Saturday, May 08, 2004 - 09:39 pm: Edit

I agree

By Nabo (Nabo) on Saturday, May 08, 2004 - 09:46 pm: Edit

I think you need to assess the quality of your education relative to the vast majority of BC students out there. While I am sure there is a percentage of students in those classes that probably shouldn't have been there, I would wager there is a much larger percentage of students whose performance on this test does not reflect their actual abilities. Many people did not receive a strong educational background in the subject and as a result essentially had no way to perform well. For example, during my high school career I have taken Bio, Chem, Physics B + C AP's. However, I have not performed a single lab experiment in any of those AP courses. In this case, I don't think it's the students lack of ability or motivation that prevent them from reaching high marks on the AP exams. The only way I was able to get 5's on these exams [on bio, chem, and physics C] is through a tremendous amount of self study. I did not even bother attending my physics class last year because I found it better time allocation to study physics in the library. That, quite frankly, is pathetic.

Perhaps to add insult to injury, I would characterize my education as fairly decent, especially relative to many other's high school education. [Hey, my school actually has most of the AP courses available] This truly doesn't speak too well of the educational standards in the U.S.

All that being said, I would say the test indicative of the what should be learned in a Calc 1 and 2 course in college. The problems are designed to ensure that all major concepts are covered in the BC course to achieve a high mark. You certainly see the division between those who received a quality education and a subpar education on the calculator section of FR. Many kids don't even get to Taylor series or other later topics in the curriculum at high school. Consequently, they have no way of doing exceptionally well on this exam.

[Sorry, this isn't meant as a flame. Your post sorta instigated a whiny rant against the educational system]

To answer your question, I didn't find the exam particularly difficult. Much to my pleasure, I found I didn't need to use any fancy t-89 calculators to do most of the problems. [I had to waste a bit of time to integrate one function on the FR using a taylor approx, rather than quickly plugging it in a spiffy calculator I didn't have.] Most of the questions were extremely straight forward, plug and chug. As for "no need to think on any of them", I would say that was mostly true. [I do feel it's the AP's job to assess the general knowledge of calculus, not the ability to elegantly problem solve. So I won't complain on the CB's choice of problems on the exam]

If you don't mind me prodding, what kind of high school do you attend?

By Tri_N (Tri_N) on Saturday, May 08, 2004 - 09:55 pm: Edit

It depends on your idea of doing well.

If your idea of being easy is an average 70% while netting a five on the exam, then I guess the test is easy. In the end, I think I will receive a five on the exam but I didn't do as well as I had hope for.

In addition, I like to resonate Nabo's point. I don't think students are dumb when they complain that the exams are hard. Most of the complaints came from the fact that most AP teachers aren't qualified to teach the materials to HS students. I took several math classes in college and the level of instruction is much higher on the college level than at high school. When I returned to HS, I had a firm grasp of the materials unlike my peers who only know its general function.

Finally, Nabo is right when he said that the fancy calculators aren't needed. I brought my TI 83 with me and I rarely touch it during the three hours exam. In fact, you don't need calculators to do calculus. If you take calculus at a prestigious college, it's forbidden to use calculator during quizes, midterms, and final.

By Jason817 (Jason817) on Saturday, May 08, 2004 - 09:56 pm: Edit

Get over your arrogant self.

By Chrisk (Chrisk) on Saturday, May 08, 2004 - 10:00 pm: Edit

question: since we are allowed to discuss the FR now, what were some of the solutions that you got? I go to a rich private school, but our Calc BC class is even worse than the Calc AB class, so I was unsure about some of the harder FR questions.

By Tri_N (Tri_N) on Saturday, May 08, 2004 - 10:03 pm: Edit

Chrisk, what questions do you have for the exam?
I think I know the solution to all the free response questions with the exception of number 6. My teacher didn't cover Lagrange theory and I glimpse through it on the exam day. I know the general rule but not its application.

By Chrisk (Chrisk) on Saturday, May 08, 2004 - 10:08 pm: Edit

Well, I was pretty unsure about 5 and 6, along with finding the acceleration vector in problem 3...

For anyone who is curious, the FR section is posted on the CB's website at:

By Chrisk (Chrisk) on Saturday, May 08, 2004 - 10:15 pm: Edit

I i know i got 5c and d, btw, but i'm not sure if i got the right answers in 5ab or if i made a mistake in 6

By Tri_N (Tri_N) on Saturday, May 08, 2004 - 10:19 pm: Edit


BTW, I left the vector problem in number 3 blank but I just figure it out right now by glimpsing at it.

The tagent slope they give you is dy/dx.

The acceleration vector is:

(second deriv dy^2 + second deriv dx^2)^(1/2)

From the original problem, they give you dx/dt. To find dy/dt, you multiple dy/dx by dx/dt which should give you dy/dt. From here, I think it's self explanatory.

Damn, I left this subsection blank.

By Athlonmj (Athlonmj) on Saturday, May 08, 2004 - 10:19 pm: Edit

Mr Calculus has posted the answers to the 2004 FR.

By Nabo (Nabo) on Saturday, May 08, 2004 - 10:20 pm: Edit

I apologize for a long post, but this was to answer Tri_N's question.

Couldn't figure out how to link, so i did a quick copy past from the "Taylor's Remainder Theorem on Calc BC??" post.

This is quoting one of my earlier posts on my impression of the Lagrange theorem. [I taught myself this topic from my textbook, so i could be completely wrong. However it worked on the exam, so i figure i did something right.

The Lagrange theorem places a boundary on the possible values of an incomplete summation of a Taylor or Maclaurin series.

Let's say you are trying to approximate "e", using the first 3 "x" values [n=3] in the Taylor series.

e^x = 1 + x + (x^2)/2! + (x^3)/3! + ......

solving to find the value of e, you just set x=1

e = 1 + 1 + 1/2! + 1/3! + .....

Since you are only approximating to n=3, you only use up to 1/3! to approximate. However, you want to find out how close this approximation is to the actual value of e.

The Lagrange theorem places an upper bound [the highest possible value] on the value of the remaining terms in the sequence [the "+ ......"].

The theorem is quite simply:


Because n=3, we get (z)^4/4!.

Okey, what is "z"?

It's some number that can be used to find the exact value of the remainder. Unfortunately, it's impossible to calculate exactly, so we have to make an educated guess of its value. [If we could precisely calculate it, we could precisely calculate the value of the entire series. That would mean we could exactly calulate "e" or even pi.]

So what could "z" be?

Let's backtrack for a moment. Let's plug in the Lagrange Theorem as the first term [n=0].

e = z^(n+1)/(n+1)! = z^1/1 = z

Therefore, ideally "z" would "e". [If we knew what "e" was equal to we could then find the value of "e". Not very useful....]

However, we are trying to find an upper bound for the approximation using the "z" value, not the actual value of "e".

Therefore, we can select a number greater than "e" for "z", and we will know this will result in an upper boundary. [If we selected a number smaller than "e", we wouldn't know if this number was truly greater than the remainder, {the + ....."}].

So to be easy, let's set z=3
We know 3 is greater than "e", so this will work. However, bear in mind there are better values for z [numbers closer to "e"]. But the AP grading rubric accepts using a guestimated numnber, so using z=3 will earn you full credit.

So to return to the orginal problem:


z^(n+1)/(n+1)! = (z)^4/4! = 3^4/4! = "the + ....."

Consequently, we know that e must be less than
1 + 1 + 1/2! + 1/3! + 3^4/4!

but greater than e = 1 + 1 + 1/2! + 1/3!

And that's that.... [i fear i may have too much free time]

[a hint if you have to approximate sin(x):

z ideally equals sin(x). sin(x) is never greater than 1, therefore using z=1 will never lead you astray]

By Reality (Reality) on Saturday, May 08, 2004 - 10:34 pm: Edit

whats the url for Mr Calculus?

By Tri_N (Tri_N) on Saturday, May 08, 2004 - 10:36 pm: Edit

For number 5 part a and b, you must integrate the problem by partial fracion.

From the given equation, you must simplify it and receive the following equation:

The dt side is easy to integrate since you will receive 1t/60

As for the dP side, you must assign:

1/P(12-P) = A/P + B/(12-P)

If you multiply both sides by P(12=P), you should recceive the following equation:

1= A(12-P) + BP
Simplify it and receive:
1= 12A-AP+BP

12A must equal 1; A= 1/12
BP-AP must equal zero; B=1/12

Now, integrate the following:

1/12P + 1/12(12-P)

After integration, you should receive:

(1/12)(ln(P)) - (1/12)(ln(12-p)) = t/60

Simplify it:

(1/12)(ln(P/(12-P))) = t/60

ln(p/(12-P)) = 12t/60

e^(t/5) = P/12-P

Now, isolating the P.

Once you isolate the P:

You should receive

P = (12e^(t=5))/(1+e^(t/5)) + C

They give you different values for P(0). Those values are C. Once you plug them all in, find the limit. From here, i think it's self explanatory.

By Jason817 (Jason817) on Saturday, May 08, 2004 - 10:36 pm: Edit


By Chrisk (Chrisk) on Saturday, May 08, 2004 - 10:40 pm: Edit

aaaaaaah... i looked at Mr Calculus' website, and I got the taylor problem exactly right EXCEPT i forgot to multiply by 5 for each term, which screwed up every one of my answers... oh well, 0 points for me...

By Tri_N (Tri_N) on Saturday, May 08, 2004 - 10:42 pm: Edit

Thanks for the explanation, Nabo.

Theoretically, it sounds pretty easy to me. Damn, I miss some easy points because my teacher didn't think that this topic would be on the exam. Oh, well.

By Chrisk (Chrisk) on Saturday, May 08, 2004 - 11:05 pm: Edit

question: do you think that i will get 0 points for # 6 because i forgot to multiply by 5 for each term? I even did the lagrange remainder part right, dealing with the series...

for A i put 1/sq(2) + x/sq(2) - x^2/sq(2)*2! - x^3/sq(2)*3!
for B i put -1/22!sq(2), which is the correct 22nd coefficient for MY series
for C, solved it correctly, based on my incorrect series
for D, correctly integrated, based on my incorrect series

By Tri_N (Tri_N) on Saturday, May 08, 2004 - 11:16 pm: Edit

Hey, despite my complaint about the exam, I have 16 rights out of 21 sections. That's about a 76%.

It's not a hot score by any mean but not too shabby. I did well on the multiple choice. I should net at least 85% on the multiple choice.

Chrisk, I didn't know that you have to multiply by five each time. However, if you differentiate the equation to the third derivative, you will notice that the five x within the sine/cosine fuction will multiply the equation by five.

By Evil_Robot (Evil_Robot) on Saturday, May 08, 2004 - 11:26 pm: Edit

The free response was fairly easy. I managed to solve for P(t) to get the population equation, but I forgot to put in the constant of integration :D. Since I knew the Cs would cancel out, I put 12 for the limits of both. I missed part B for losing the constant though...and the Lagrange error term I was only vaguely familiar with so I made it up on the spot. I think I got most of the points.

However, the first multiple choice (without calculator) I didn't do so hot (it's been a year since I've done series :()

By Chrisk (Chrisk) on Saturday, May 08, 2004 - 11:29 pm: Edit

well, the 5 did have to be there, because the original was sin(5x+pi/4), so f'(x)=5cos(5x+pi/4), and so on... when you're doing a taylor series centered at 0, it's

f(0) + f'(0)x + f'(0)x^2/2!...

so that translates to


if you want to see a more formal solution, look at the link from Jason817 above...

So, back to my original question, do you think i'm going to get 0 points for this??

By Andrew123s (Andrew123s) on Sunday, May 09, 2004 - 12:42 am: Edit


By Stonedpanda (Stonedpanda) on Sunday, May 09, 2004 - 01:50 am: Edit

Yeah, my teacher never tought logistical diff EQs. I spent my time on that problem and never did the last one well. I did really well on the rest of the test. He mentioned passingly that Maximum Error Bounds were also called lagrunge Error bounds, but I knew that they were what they were because I'm one of those people who is a nerd on the inside. We still never really had to learn them.

This is all kinda funny because my calc teacher really rocks and my school is a pretty good one. But yeah, I think got a 5 and at the VERY VERY VERY least a 4.

Some of my friends who took the AB class said that the test totally sucked their dick i.e the test was really easy.

By Feuler (Feuler) on Sunday, May 09, 2004 - 02:15 am: Edit

I don't think Calc BC is too easy, if it were, fewer people would fail. But I do think it has an entirely incorrect and extremely detrimental focus.

I totally disagree with Nabo's last comment about how the CB should be testing knowledge of Calculus rather than problem solving ability- knowledge of Calculus, unless you go into a scientific field, is virtually worthless. Except for a few people, the value of studying mathematics at all is problem-solving skills, which have applications everywhere in life.

The reason so many people hate math and complain about learning it is that, the way math is taught, they are being taught something that may be entirely worthless. The average person will never need to take an integral in their life, and will certainly never use Taylor polynomials for everything. So why should we make people study them, unless they are gaining some applicable skill from them?

If problem-solving is forsaken for facts, then, save for specialists, there is no reason to learn math, nor history, nor much of science. The study of English or foreign language loses the majority of its purpose if it is pursues simply for the ability to speak.

By focused on the trivial, rote aspects of Calculus (and many other topics), Collegeboard is rendering what could be an extremely and beneficial topic of study worthless to students; it becomes only a hoop to jump through to go to college and study soemthing else. So many students hate calculus, and math itself, wanting only to finish it and move on, and that is a tragedy.

By Madd87 (Madd87) on Sunday, May 09, 2004 - 02:24 am: Edit

I definitely don't think you will get 0 points for missing the 5.

By Conker (Conker) on Sunday, May 09, 2004 - 03:56 am: Edit

Feuler, you are correct in saying that most jobs do not require Calculus. However, this does not suggest that teaching purely knowledge-based Calculus does not have its benefits. I think we are well aware that people, who do intend on going into a field that necessitates such knowledge, need these advanced courses. What about the people who aren't sure? What about the people who think that they are hotshots at English, but then realize that they are more adept at Physics and Calculus? And surely, there are people for which the converse is true as well. Exposing high school students to as much as possible is every bit as important as teaching them problem-solving skills. In fact, exposure may prove even more important than problem-solving because Physics and Math emphasize a different type of problem-solving that will prove virtually futile to lawyers and businessmen. Also, if our goal is to teach them problem-solving, why teach Calculus? I think that math contests have proven that mere Algebra and Geometry is sufficient to create intensely challenging problems.

Personally, I agree with you. I want classes based on problem-solving and not facts and rote memorization for the simple reason that classes would become more interesting. However, this does not mean that we should dismiss "rote" classes as pointless.

By Feuler (Feuler) on Sunday, May 09, 2004 - 04:38 am: Edit

Those are all excellent points. I did not meant to imply that facts and rote learning are unimportant, only that they cannot be all-important.

In fact, one reason I favor a more cerebral approach is simply because it makes rote learning easier. If one understands the reasoning behind Calculus, for example, they need not study it; it will stick naturally. I have found the same to be true with history. A problem-solving approach enables greater rote memorization with less effort and more interest and enjoyment.

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