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# OFFICIAL AP Calculus BC 2012 Test Thread

Registered User Posts: 43 Junior Member
edited May 2012
AP Calculus BC exam is in less than 4 days! If anyone has questions, post them in this thread. Also, let us use this thread to discuss the difficult of the test / questions after the actual exam.
Post edited by Chromedydx on

## Replies to: OFFICIAL AP Calculus BC 2012 Test Thread

• - Posts: 5,015 Senior Member
Dont forget to wait 48 hours to discuss FRQs.
• Registered User Posts: 43 Junior Member
Yes. What are some predictions for the topics on the Free Response?
• Registered User Posts: 46 Junior Member
Most likely:
1 volume or revolution
1 parametric or polar
1 with a table with data and you have to approximate/use riemann sums
1 differential equations
1 with a graph of f' and finding a bunch of points
1 taylor series at the end
• Registered User Posts: 624 Member
Keep in mind that this year College Board will administer several versions of FRQs, and only one version will be released. So some people taking the exam in May 9 will probably get FRQs that won't be released and therefore can never be discussed.

Important Changes: 2011-12

Anyway can someone help me on this? I never seem to understand questions involving volumes that are not solids of revolution. Like this question in 2011 Form B:

The region R is the base of a solid. For each y, where 0 <= y <= 2, the cross section of the solid taken perpendicular to the y-axis is a rectangle whose base lies in R and whose height is 2y. Write, but do not
evaluate, an integral expression that gives the volume of the solid.
• Registered User Posts: 68 Junior Member
1. taylor series
2. polar and parametrics
3. 2nd fundamental theorem of calculus
4. Euler's method
6. definite integral applications with area/volume
• Registered User Posts: 43 Junior Member
Techhexium:

You basically are building a solid (proper name - cross-sectional solid) whose base is the region R and whose height is 2y. Break this problem piece by piece:

a) "the cross section of the solid taken perpendicular to the y-axis" --> this means that the base of the solid is take to be g(y)-f(y), the horizontal length of the region R. Note that "perpendicular to y-axis" specifies that you are using horizontal slices. If it said "perpendicular to the x-axis", you would use vertical slices.

b) "rectangle whose base lies in R" --> this means that the solid you are building is a rectangle with a base that is perpendicular to the y-axis (we already established this in part A).

c) "whose height is 2y" --> this is given for the volume calculation. The area of the rectangle is base times the height. You know that the base is g(y)-f(y), and the height is 2y.

d) solve each equation for x (you need to do this since you are looking for the horizontal distances within R).

y=sqrt(x)
x=y^2

and y=6-x
x=6-y

integral ( base*height)
integral [ [ g(y)-f(y)] * 2y ]
integral [ [6-y-y^2]*2y*dy], from y=0 to y=2
• Registered User Posts: 43 Junior Member
Pseble,

What type of problem do you envision for the "advanced integration"? If you are basing the prediction on some past FRQs, can you mention the year? I think I have done all of the FRQs from the past 12 years, and I can't remember much advanced integration, except maybe integration by parts in some questions, but even that was rare.
• - Posts: 5,015 Senior Member
Advanced integration is typically integration by parts and partial fractions (for AP calculus). I say that because a considerable number of #5 FRQ requires integration by parts. That's it though.
• Registered User Posts: 64 Junior Member
The partial fractions and parts problems are easy - not really that hard. The AP exam doesn't really test difficult integrals that use these concepts. I am in AB and self studied BC and scored a 37/45 on the 2003 MC with less time.

I think the key things to know (Calc 2 wise) are parametric functions (vectors, velocity, acceleration, arc length, distance), testing whether a series diverges or converges (usually a roman numeral type question), interval of convergence for power series, maybe 1 polar question (I can kinda do area but the slope is a pain....), parts/partial fractions and like 1 improper integral.

Otherwise at this point its best to review everything especially the two parts of the Fundamental Theorem of Calculus and the Extreme value, Intermediate value and Mean value theorems.

Good luck everyone!
• Registered User Posts: 697 Member
Bad idea to take an AB class and self study BC.
I've got a good idea of everything, but I'm not very good with convergence. If I can use the ratio test, I've got it down, but alternate series and power series are still a pain in a butt. D:

And logistic growth... may the AP gods (College Board?) have mercy on me.
• Registered User Posts: 417 Member
Hey guys, so I learned in AP chem last year that my most efficient way of studying is answering other people's questions, so if any of you have any questions please ask them to me either here or via PM.
• Registered User Posts: 697 Member
Question:
How on earth do you write a general term... for anything? For Taylor/Mclaurin series. I get how to do the first x terms, but seriously, cannot do general anything to save my life. Like number 6a for this:
http://apcentral.collegeboard.com/apc/public/repository/ap10_calculus_bc_scoring_guidelines.pdf

HELP D:
• Registered User Posts: 43 Junior Member
Mascara,

According to the AP Calculus BC Course Description, you are required to know the expansion of four series: sine, cosine, e^x, and 1/(1-x). Part A tests this knowledge. Here are the expansions (MacLaurin, around c=0) :

sin(x)=x-x^3/3!+x^5/5!-x^7/7!.... (-1)^(n)*(x)^(2n+1) / (2n+1)!
cos(x)=1-x^2/2!+x^4/4!-x^6/6!.... (-1)^(n)*(x)^(2n) / (2n)!
e^(x)=1+x+x^2/2!+x^3/3!+x^4/4!.... x^(n)/n!
1/(1-x) = 1+x+x^2+x^3+x^4+x^5.... x^(n)

You need to know these by heart - including the general term.
• Registered User Posts: 43 Junior Member
In continuation of the part A, you subtract one from the cosine series, and get:

cos(x)-1=-x^2/2!+x^4/4!-x^6/6!....[(-1)^(n)*x^(2n)/(2n)! ==> this is the same series, started at n=1 instead of n=0. Then, we can rewrite the series so that it starts at n=0 by plugging in (n+1) anywhere you see "n". It then becomes:
....(-1)^(n+1)*x^(2n+2)/(2n+2)!]

Then dividing by x^2:
[cos(x)-1]/x^2= -1/2!+x^2/4!-x^4/6!...(-1)^(n+1)*(x)^(2n+2)/(2n+2)!*1/x^2
==> (-1)^(n+1)*x^(2n)/(2n+2)!
• Registered User Posts: 697 Member
-___-
You are a lifesaver. Why do math books have to make everything so complicated? The only thing I don't get is how when you divide by x^2, you get those specific first several terms. The general term I understand.

Thanks!
This discussion has been closed.