<p>@Niquii77 wouldn’t it only be [3,4] though cause it’s 0 < x < 8? I thought that meant 0 couldn’t be included in the interval. idk though. I made a couple stupid mistakes on free response cause the time limit messes with me man, it blows. but I think I’ll be fine for a 4 assuming I got around 60 points, we’ll see.</p>
<p>Fine then (0,1] I didn’t use bracket notation. I used these “<”. I didn’t underline the 0 to 1 arrow.</p>
<p>For some reason I got another form of the test.</p>
<p>My questions were:</p>
<p>1) Area between two graphs
2) Mars Rover modeled by parametric equations
3) Volume of water chart
4) Questions about the graph of xe^-x^2
5) Relating a graph of g’ to the graph of g
6) Taylor series for f’(x) = f(x)+x+1</p>
<p>Does anybody know why they did this? They put one form up and not this one. After looking at the one they put up (which seems to be the one that most people got), I’ve realized that the one I and some people in my class got was much easier.</p>
<p>Does anybody know anything about this? Is it a fluke? Will they put this form up? Will it be graded differently? Why did they do this? Any information would be nice.</p>
<p>Okay, so I’m a junior and I just took this test, and I thought it was hard! I had no idea about part a of the polar question, and I definitely didn’t do the last part of question 6 correctly. I signed up for Calculus 3 at the university next year because I thought I had a pretty good grip on this material, but now I’m not so sure! Anyone else in the same boat? I feel like a 4 is inevitable, and I’m not sure where that puts me as far as preparedness for Calc 3.</p>
<p>what was the ans to the polar question / how’d you guys do it?</p>
<p>I know this probably took more effort, but would it make sense to solve for the particular solution to the differential equation in part 5a (also in part c of course) in order to find the equation for f(x) and then use l’hopital’s rule to find the limit? Thanks</p>
<p>@JohnandTequila</p>
<p>College Board puts up the most frequent FRQs. That means the ones that you had will never be posted, which means technically we can never talk about it. If you remember some of the problems, you can always ask your teacher to work them out for you. Also, I think the teachers will have access to the less frequent FRQs eventually through their administrator account on College Board. Either way, your teacher might be your best source in getting answers to your FRQs.</p>
<p>It’s true that some classes had a mix of different forms. I got the easier one, while the majority of the others in my class received the one posted on College Board.</p>
<p>As far as I know, the FRQs will not be graded differently in that it will have different weights, but the scores received on the different forms of FRQs will influence the composite score cutoffs for this year. Perhaps the cutoff score will be lower, since many people with the FRQs posted on College Board thought they were hard and will likely earn fewer points than those who had the easier FRQs.</p>
<p>Wait concave down and increasing… Doesnt that mean f’ is positive and f’’ is neg? Isnt that only from six to 8?</p>
<p>No, because the given graph was f’. You’re thinking of it as f.</p>
<p>JohnAndTequila: I got that same set of FRQs too. They seemed a whole lot easier than the ones the College Board posted to the AP site today. Like mentioned above, starting this year (or maybe the year before) the College Board only puts up the most common FRQs. I find this highly annoying because I’ll never know what I did right or wrong. However, since it’s been two days since the exam, I think it’s fine to discuss what we got, since others who had the other form of the FRQs are discussing their questions right now. </p>
<p>For the first one, the graphs were y = 1 - x^3 and y = sin (x^2).
Part (a) asked for you to determine the area bounded by the graphs, which was very easy. It was just the integral from 0 to the intersection (x = 0.764972) of (1 - x^3) - sin (x^2) dx, which is about 0.534.
Part (b) asked if the two separate areas created by slicing the region found in part (a) by the line y = k (k is the y-value of the intersection, about 0.55235) were equal. So all you needed to do was find the area of the two regions by using integration and simple geometry. They turned out, for me, not to be equal.
Part (c) asked you to rotate the region found in part (a) about the line y = 3 I think. This was a simple washer method volume problem. I got around 11 for the volume. </p>
<p>Question two was, indeed, about the rover. The equations were x(t) = -12 sin (2t^2) and y(t) = 10 cos (1 + sqrt(t)).
Part (a) I think asked to find the acceleration and speed at t = 1. That was an easy calculation.
Part (b) asked you to find the distance traveled from 0 to 2. Just use the integral of the distance formula with the time differential.
Part (c) asked to find some coordinate position at some time. All you did was use the first FTOC and used your calculator and the initial position given, which was the point (5, 6) at t = 0.
Part (d) asked you to find when the slope of the path of the rover is a half on the interval [0, 2]. All you did was put y’(t) / x’(t) - (1/2) in your calculator and find where it crosses the x-axis. I only got one time that it had slope equal to a half, which was about t = 1.05 something.</p>
<p>Question 3 was a easy one. I forgot the specific parts of it, but I remember there was a Riemann sum and all that jazz. In part (c) I think you had to mention the intermediate value theorem. I do however remember part (d) where it said that the surface area of the water was proportional to the volume and gave you some equation relating the two. I think the equation was A(t) = .3W(t)^(2/3). it asked what was the rate of change of the area of the tank at t = 30. You then differentiating the equation with respect to time and got A’(t) = (1/5)<em>W(t)^(-1/3)</em>W’(t). Plug in t = 30, and you get A’(30) = (1/5)<em>W(30)^(-1/3)</em>W’(30). Then you referenced the data and saw W(30) was mentioned as being 125 and W’(30), from the tabular data, was 0.5 I think. Plugging in, you find your problem simplifies a lot. You end up getting A’(30) = (1/5)<em>(1/5)</em>(0.5) which is 1/50. I think that’s the answer. That one was fun.</p>
<p>Question 4 was a plug and chug one.
They gave you f(x) = x e^(-x^2).
Part (a) said to find the absolute maximum of f(x). Simple differentiation and finding where f’(x) changes sign gives you a maximum at x = 1/sqrt(2).
Part (b) asked you to find AN antiderivative of f(x). Integrate with a u-substitution (u = x^2) and you get an antiderivative of F(x) = -(1/2)e^(-x^2).
Part (c) was all about utilizing integration by parts. It said find the integral from 0 to infinity of x*f(x) dx given that the integral from 0 to infinity of e^(-x^2) (a famous onethe Gaussian integral) equals sqrt(pi)/2. Finding that the integral become unsolvable by u-substitution, you needed to use integration by parts. Your u was x and your dv was f(x) dx. After noting that a limit must be used, you get the limit as b goes to infinity of xF(x) evaluated at b and 0 - the integral from 0 to infinity of F(x) dx. The first limit goes to zero (L’Hospital). Your F(x) you found in part (b) becomes useful here. Take out the factor of (-1/2) and you end up with just (1/2) integral from 0 to infinity of e^(-x^2) dx, which, using what they give you, turns out to be sqrt(pi)/4. </p>
<p>Question 5 was very, very simple and wasn’t tricky at all. It was just functions defined by integrals and maximum and minimums and points of inflection. That should be a high-scoring FRQ. I can’t remember the specific answers or questions.</p>
<p>Question 6 was an easy Taylor polynomial FRQ. They gave you that f(0) = 2 and f’(x) = f(x) + x + 1.
Part (a) asked to find a simple tangent line at x = 0. Using the data they gave you, the tangent line is
y = 2 + 3x</p>
<p>Part (b) asked you to find the second degree Taylor Polynomial for f(x). Differentiate the equation they gave you and you get f’’(x) = f’(x) + 1. Using Taylor’s theorem, you get T<em>2 (x) = 3 + 2x + (4/2!)x^2, which simplifies to T</em>2(x) = 2 + 3x + 2x^2.</p>
<p>Part (c) asked you to find the fourth degree Taylor Polynomial for f(x). You just keep on differentiating and find that f’’’(x) = f’’(x), and f’’’’(x) = f’’’(x), so on and so on. You end up getting T_4 = 3 + 2x + 2x^2 + (2/3)x^3 + (1/6)x^4.</p>
<p>Part (d) asked you to find the n-th derivative of f(x) at x = 0 for n≥2. This, as you found out in previous parts of the question, turns out to be just 4. It also asked you to find a “polynomial expression” for the expression f(x) - 4e^x. Remembering the series for e^x, this simplifies really fast. Use your T_4 polynomial for f(x) to see the pattern. You get (3 + 2x + 2x^2 + (2/3)x^3 + (1/6)x^4 … ) - 4(1 + x + (1/2)x^2 + (1/6)x^3 + (1/24)x^4 …). All terms with a power greater than or equal to two cancel each other out, and you’re left with the expression -2 - x. </p>
<p>That’s all I can remember. I hope that helps.</p>
<p>I made solutions today for the tests that were on Collegeboard today. Enjoy, and sorry for any typos.</p>
<p>AB:
<a href=“Dropbox - Error - Simplify your life”>Dropbox - Error - Simplify your life;
<p>BC:
<a href=“Dropbox - Error - Simplify your life”>https://www.dropbox.com/s/nmo6mju19sqe05j/BC%20Solutions%202013.pdf</a></p>
<p>thanks for the solutions jamison ! its very refreshing to see someone get VERY similar answers to mine. However, can you rework the euler’s method problem? we seemed to get different answers. I re-did the problem and i got -11/32 or -0.34375</p>
<p>Do you think I would still receive credit if I did not use parentheses for my answers involving trig functions? Like instead of putting cos(t) I put cost.</p>
<p>Thanks PGD2013! Definitely scored high on the FRQs then. You have a really great memory by the way.</p>
<p>yes! i think i got the same answer…the denominator ended up being 32.</p>
<p>just checked the FRQ…did pretty well…then i went on to appass.com to calculate my ap score…i hypothetically set the MC as 25/45…still a 5.</p>
<p>It shouldn’t be a problem, although it’s a good habit to parenthasize trig so that something like cos x sec x can’t be interpreted as both cos (x) sec (x) and cos(x sec (x)).</p>
<p>@JamisonSloan
Thank you so much for your answers! Mine were very similar, except for a few careless mistakes I made here and there. You just made my day</p>
<p>There is an error in the solutions Jamison Sloan has posted. Number 5 is incorrect. He assumed that F(0) = 1, whereas on the 2013 free response, it said f(0) = -1</p>
<p>That might make a big difference</p>
<p>He assumed f(0)=-1.</p>