<p>I think they’re going to release the multiple choice this year since they released them in 2003 and in 2008.</p>
<p>@Vince-this has been going on for awhile, and CB doesn’t stop it</p>
<p>Where are the frqs being released at?</p>
<p>^
Futfsu1: Here is the web site for released Calculus AB exams: [AP</a> Central - The AP Calculus AB Exam](<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>AP Calculus AB Exam – AP Central | College Board)</p>
<p>The 2013 FRQs should be added around 4pm EST.</p>
<p>Ahh thank you. Will the scoring guideline be released also?</p>
<p>the frq are up.</p>
<p>2013 FRQ’s posted here: <a href=“http://media.collegeboard.com/digitalServices/pdf/ap/apcentral/ap13_frq_calculus_ab.pdf[/url]”>http://media.collegeboard.com/digitalServices/pdf/ap/apcentral/ap13_frq_calculus_ab.pdf</a></p>
<p>Scoring guidelines and other data won’t be posted for months</p>
<p>Anybody else get the Form B questions? I was reading through this thread and had absolutely no clue what you guys were talking about and now I see that the FRQ released are different than the ones I answered.</p>
<p>I had a question about a penguin population and another one was to just take a couple of limits. Am I the only one?</p>
<p>SOmeone make a post about all of the correct answers</p>
<p>I think I had form b… I had a slope field question for the frq</p>
<p>Whats 1B, 2B, 3D???</p>
<p>and what about number 2c</p>
<p>The questions from Form ‘O’ are posted on the Collegeboard website. Form O was the default test form. Unfortunately, those who had Form ‘G’ will probably never see their questions posted on the web.</p>
<p>Problem #1</p>
<p>A)G’(5)=-24.586 tons per hour^2. The amount of unprocessed gravel arriving per hour is decreasing at this rate.</p>
<p>B)Integration G(t) from t=0 to t=8 is equal to 825.551 tons</p>
<p>C)Amount of Gravel=G(5)-100=-1.8529. Thus the amount of unprocessed gravel is decreasing.</p>
<p>D)Critical points of G(t)-100 are at t=4.923, where t is a relative maximum. 500+Integration of G(t)-100 from t=0 to t=4.923 gets you the total amount of unprocessed gravel of 635.376 tons.</p>
<p>Problem #2</p>
<p>A) |2|=v(t). Solving for t=3.128 and t=3.473</p>
<p>B) s(t)=10+Integration of v(t) from 0 to t. When t=5, s(t) is equal to -9.207.</p>
<p>C) 0=v(t). Solving for t=0.536 and t=3.318. Point where velocity either goes from positive to negative or negative to positive.</p>
<p>D)Increasing because a(4)<0 and v(4)<0.</p>
<p>Problem #3</p>
<p>A) [C(4)-C(3)]/(4-3)=1.6 ounces per minute</p>
<p>B) Yes. According to the mean value theorem: Because the average rate of change from the interval (2,4) is equal to 2, there must be some time t in the interval (2,4) in which the instantaneous rate of change,C’(t), is also equal to 2.</p>
<p>C) (1/6)<em>2</em>(5.3+11.2+13.8)=10.1 ounces of coffee has dripped into the large cup since t=0.</p>
<p>Problem #4</p>
<p>A) x=6</p>
<p>B) 4+Integration of f’(x) from x=8 to 0 is equal to -8.</p>
<p>C) (0,1)U(3,4); f’(x)>0 signifies the graph is increasing, and slope of the line tangent to f’(x)<0 signifies the graph is concave down. These intervals satisfy both of these conditions.</p>
<p>D) g’(3)=3<em>f’(3)</em>f(3)^2=3<em>4</em>(-5/2)^2=75</p>
<p>Problem #5</p>
<p>A) Integration of g(x)-f(x) from x=0 to 2=16/PI-4/3</p>
<p>B) PI*Integration of (4-f(x))^2-(4-g(x))^2 from x=0 to 2</p>
<p>C) Integration of [g(x)-f(x)]^2 from x=0 to 2</p>
<p>Problem #6</p>
<p>a)f’(x)=e^0<em>(3</em>1^2-6*1)=-3. f(x)=-3(x-1); f(1.2)=0±3(0.2)=-.6</p>
<p>b)dy/(e^y)=(3x^2-6x)dx
(e^-y) dy=(3x^2-6x)dx
-e^(-y)=x^3-3x^2+c
-e^(0)=(1)^3-3(1)^2+c
c=1 (note that your c may be different depending on when you solved for it)
e^(-y)=-x^3+3x^2-1
-y=ln(-x^3+3x^2-1)
y=-ln(-x^3+3x^2-1)</p>
<p>Will form g have a different curve</p>
<p>Hell yeah, a lot of my answers matched yours. The value at 1.2 is -.6 not -.06. I can believe I got that last question of the solution of dy/dx. Feeling pretty good. I definitely got 635 on the gravel question.</p>
<p>Form G was a different test so yes it will have a different curve.</p>
<p>If i got 1a,1b,1c,2b,3a,3b,4a,4d,6a correct (plus a point here and there for sep of variables, integration, etc. About how many FRQ points is that? I probably got 20-25 right on multiple choice. Im hoping for a 3. Thanks</p>
<p>I took the BC test, but the gravel question was on there too. I got 635 as well, so I think we should be good. I also got about -24 for G’(5) and 825 tons for the integral from - to 8 and a critical point at t = 4.9.</p>
<p>isnt 4b (minimum value) -8. since the integral is -12 and adding the initial condition gives -8</p>