AP Calculus BC 2011 Thread

Test Preparation AP Mathematics & Computer Science
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<p>Ah I forgot that. It’s sqrt r^2 + (r’)^2 d#</p>

<p>f(x) inflection points: changes from curving down to curving up, or vice versa; it’s kinda hard to tell on a graph for f(x) so I don’t think they would ask that
f’(x) inflection points: changes from increasing to decreasing, or vice versa</p>

<p>f(x) absolute max/min: the highest point and the lowest point
f’(x) absolute max/min: first find relative max/min, then plug in the x’s you get into f(x) or integrals to find the absolute max/min</p>

<p>I don’t get 6b either. How did they get base = 1/n?</p>

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<p>edited, too slow.</p>

<p>@CORVIDS:
That’s a weird formula, I’ve never seen it before.</p>

<p>Thanks for the graph stuff.
When you say: “f’(x) inflection points: changes from increasing to decreasing, or vice versa” you mean when it goes from positive to negative? Like, crossing an axis?</p>

<p>For f’(x) absolute max/min, to find the relative max/min is that when the function crosses the axis?</p>

<p>Do we get part marks for consistent wrong answers?!</p>

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<p>I found it written as sqrt (r^2 + (dr/d#)^2) d#, but personally it’s easier for me to think of it the way I wrote it.</p>

<p>No, that doesn’t necessarily from positive to negative. Inflection points deal with the second derivative. So they’re basically the max/mins for f’(x). You know that a max/min is happening on a graph when it changes from increasing to decreasing or its derivative crosses the axis.</p>

<p>Yeah, max/mins happen when f’(x) crosses the axis.</p>

<p>And I’m pretty sure you get part marks.</p>

<p>You know I don’t think I’m good at explaining things haha. I hope I didn’t confuse you more.</p>

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<p>From the 2003 BC Exam MC Part A, #26:</p>

<p>Integral of 2x/[(x+2)(x+1)] dx</p>

<p>Answer is 4ln abs(x) - 2ln abs (x+1) + C</p>

<p>Should be integration by partial fractions, but I haven’t done that with anything but 1 in the numerator before. Can anyone explain how to do it with 2x in the numerator and/or how to solve the problem?</p>

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<p>^ “So they’re basically the max/mins for f’(x). You know that a max/min is happening on a graph when it changes from increasing to decreasing or its derivative crosses the axis.”
Increasing to decreasing SLOPE, or actual value?</p>

<p>Nope, very helpful!</p>

<p>If part marks aren’t awarded I"m failing. Hard.</p>

<p>@Quaker:
No idea, sorry! I’ve only done it with numbers, I think.</p>

<p>edit: Oh, I did it.
Just do it like normal, split it into two fractions with A and B on the top. (Just A and B, nothing else) and then do the usual thing.
You’ll end up with (A+B)x + (A+2B), so just set (A+B) to 2 and (A+2B) to 0 and you’ll be able to solve the equation.
You set it to zero because there are no constant terms, only a linear term (which is represented by (A+B)x.)</p>

<hr>

<p>I don’t understand 5.a) or b).
<a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;

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<p>For partial fractions, you know the step when normally you would be at “A + B = 0” because all of your “Ax + A + Bx - B . . . + Nx + N” are set equal to the coefficients? </p>

<p>Basically, if you have something like</p>

<p>(3x + 1)/[(x+2)(x-2)]</p>

<p>It decomposes into A / (x+2) + B / (x-2)</p>

<p>Ax + 2A + Bx - 2B</p>

<p>All of the terms with X’s are set equal to the terms in the fraction with x’s. So in our fraction, 3.</p>

<p>A + B = 3</p>

<p>And </p>

<p>2A - 2B = 1</p>

<p>Hope that helped!</p>

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<p>HEY, how do you know when to use the Cylindrical method for calculating volume vs Disk/Washer method?</p>

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<p>It’s possible to use both whenever. Use what’s easier for the specific question.</p>

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<p>MC question help please?</p>

<p>A population, P(t), satisfies the logistic differential equation dP/dt = (2/3)P(5-P/100). What is lim as t –> infinity P(t)?</p>

<p>I know you’re supposed to factor out the 5 to get dP/dt = (10/3)P(1-P/20)</p>

<p>I’m know k = 10/3 and M = 20
I’m trying to plug everything into M/(1+Ae^-kt) where A = (M-Po)/Po
I think Po is initial population</p>

<p>Any ideas?</p>

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<p>^ I saw that one too, and I didn’t understand it.</p>

<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;

<p>How did they do 5.b)?</p>

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<p>Anyone else feel majorly screwed for this…</p>

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<p>Totally.
Self-studied a 4 last year, didn’t learn any BC topics until like . . . a few days ago.
AWESOME. Not.
Going to fail majorly. I hope I get a 3.
You?</p>

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<p>I feel pretty screwed for the test too. I’m self-studying BC and I’m still having trouble with series and polar stuff. Hoping I get a 3 too, Abrayo.</p>

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<p>bout to learn sequences/series right now…■■■ im really screwed =/</p>

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<p>Hopin for a 4, but considering how my practice FRQs and MCs are going, I probably can’t get that >_< The MC especially… ughhh</p>

<p>I’m so awful with polar/parametric crap, and I make stupid mistakes so easily. Also what do we need to know about vectors? i didnt learn them</p>

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<p>position = s(t) = [vector form] s(t)i + s(t)j
velocity = s’(t) = v(t) = [vector form] v(t)i + v(t)j
acceleration = s’'(t) = a(t) = [vector form] a(t)i + a(t)j</p>

<p>speed = abs[sqrt((velocity.i)^2 + (velocity.j)^2)]</p>

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<p>awesome, thank you!</p>

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<p>Has anyone else taken the BC practice test? It’s pretty difficult, as in it asks you to do things in really creative ways…</p>

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<p>Anyone know the relative difficulty of the 2010 one? Was it easier than normal? There was only one series question on the FR, and I got a 4 on it. (On the FR only, when I just tested myself).
Seeing as there weren’t parametric/polars I’m assuming I won’t get a 4 on the real one.
I failed the series one, ha ha. . .</p>

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<p>BC exam is easy^ its like 41% 5s…well very easy after traditional classroom instruction,but i guess it be kind of hard to self study.</p>