Starting the AP calc thread.

<p>I did what gyros did at first but then I crossed it off!! This is partly because I shaded the region that included the graph being bounded by the x axis and so I assumed no holes and did the disk method from -3 to 3! I must have been on something...</p>

<p>Actually though, that still wouldn't have made sense because the the revolution would have still had a hole on the portion above the horizontal line...man I 'm so mad at myself</p>

<p>crapp im an idiot lol i put 37 first..which is the right answer...i realized that when i used my calculator it was giving me only the curve under y=2..crapp...okk my question is wat if i put the right formula and evaluation but my answer juss came out wrong (this case 12) will they deduct a lot of points? or a few? i sketched a graph too showing the region..i showed how i got my boundaries..and i showed the formula u needed to use to get to answer.</p>

<p>Why does everybody think #1 was so hard? I thought it was probably the easiest, except for 1c because it was really ambiguous. So how do you do 1c?</p>

<p>So I did pretty terribly on the FR section... I'm hoping I got most of the points for the area/volume question at least. Here's what I have stored in my little 89:</p>

<p>a) 2*integral(20/(1+x^2)-2),x,0,3) = 37.962</p>

<p>Don't ask why I bothered changing the limits. I'm lame.</p>

<p>b) pi*integral((20/(1+x^2))^2-4,x,-3,3) = 1871.190</p>

<p>c) For this... I can't remember what I did exactly... I believe I found the diameter, divided it by two, squared that, took the intregral of it from -3 to 3 and multiplied all of that by .5 pi </p>

<p>It appears that what I have in my calculator is the diameter squared.. so I guess I plugged it in incorrectly :(</p>

<p>Can anyone tell me if I did at least some of this problem correctly?</p>

<p>It's not that it was hard...it's that I didn't shade the correct region.</p>

<p>For 1c, the diameter has to lie inside the region R. If it were only the radius in R, then the other radius (to make the full diameter) would lie outside of R. Since R is clearly the base, the diameter has to lie in R.</p>

<p>Let f(x) = 20/(1+x^2) and g(x) = 2.</p>

<p>Then the diameter is 20/(1+x^2) - 2, and the radius is half of this, which is 10/(1+x^2) - 1.</p>

<p>The area of a typical semicircle is 1/2(pi)r^2.</p>

<p>So the formula for the volume of this region is found by 1/2(pi)*fnInt((10/(1+x^2)-1)^2,X,-3,3) = 55.471pi or 174.268.</p>

<hr>

<p>Also, my guess for points is this:</p>

<p>1 -- limits of -3 to 3 anywhere in either parts (a), (b), or (c).
(a) 1 -- integrand of f(x) - g(x)
(a) 1 -- solution of 37.961 or 37.962
(b) 2 -- integrand of [f(x)]^2 - [g(x)]^2
[0/2 if not of the form [f(x)]^2 - [g(x)]^2]
<a href="b">1/2 reversal</a> 1 -- pi as a constant, and solution of 595.618pi or 1871.190
(c) 2 -- proper integrand of the form [f(x) - g(x)]^2
[0/2 if not of this form]
<a href="c">1/2 if f(x) - g(x) is a factor</a> 1 -- constant of integration and solutions (as given above).</p>

<p>So if you have the correct formula for a part, and come up with the incorrect solution, my guess is that you get 2/3 for that part.</p>

<p>OMG 2397 don't feel bad you should see the thread I posted "Am I in trouble (AP Calculus AB)" I did what you did for part a except I said from -3 to 3 and for c I took the area of each cross section (plugged in the radius) and evaluated the integral. I messed up b so bad...I did what you did, but then I crossed it off!</p>

<p>Alright fine even though I think 1c was bs, whatever, I got the rest right. Anyways, how about #2? I'm pretty sure I got a right. But how do you do b and c? I'm so ****ed because out of all the FRQ I had to get leave b and c blank. I wasn't concentrating and I was freaking out, so I just couldn't do it. Yea it sucks, but oh well the test is over.</p>

<p>So on 1C how many people used D like me... because this is how I saw it.</p>

<p>In order to place a semi circle perpindicular to the x axis... let me try to draw it.</p>

<p>Actually no i won't. :D</p>

<p>So if our semicricle is Perpendicular to the X-axis that means its standing up like the letter C with the two graphs as end points. This is why I used D instead of R.</p>

<p>To do b.. the time when water was decreasing was when graph 2 was above graph 1.</p>

<p>to do c.. just think about it. Once graph 2 reaches 2000 there is never going to be a point where graph 1 will outdo graph 2 (at least on our interval 0< t < 7). But there are Points BEFORE it hits 2000 which is simply at 3 seconds (the last time its at 500). Try taking the integral from 0 to 3 and you will see you get a positive number. If you try from 0 to 7 you will get a neg. number. Remember to partition your integrals so that the right function is on top etc.</p>

<p>ARGH! I had a feeling that's how you were supposed to do b, but I was "spazzing" out during the exam. DAMN IT! Oh well, forget it the exam's over.</p>

<p>For 2b, since f(t) is the rate at which the water is entering the tank and g(t) is the rate at which the water leaves the tank, then the amount of water in the tank is decreasing when g(t) > f(t). This occurs on [0, 1.617] and [3, 5.076].</p>

<p>For 2c, the possible points when the amount of water in the tank could be greatest are the endpoints and critical values. Endpoints are 0 and 7, critical values are 1.617 (given), 3 (when f(x) - g(x) is undefined), and 5.076. Evaluate the function at all five points to determine that the amount of water in the tank is greatest when t = 3 hours, and the tank holds 5127 gallons (rounded to the nearest gallon).</p>

<p>For number 1, how many points do you guys think they will take off if I messed up the limits of integration for all three parts but got all the other parts of the integral right.. I wasn't thinking clearly during the test and only accounted for the area in the first quadrant... so I used [0, 3] as opposed to [-3, 3]. Of course, this gave me answers that were all 1/2 of the actual answer...</p>

<p>How lenient is college board regarding things like this? =/</p>

<p>Alright so for 2a, I just took the integral of f(t) from 0 to 7, since thats how I interpreted it...Is that right?</p>

<p>To mrniceguy:</p>

<p>I'm not 100% certain of this. As I'm guessing the points will break down, you'd lose 1 point. Even if each part has a limits of integration point, they usually won't triple-penalize you for making the same mistake. (Which is why I don't think they'll spend 3 points on that in the first place.)</p>

<p>So my guess is that if you got half of all the correct answers, you score 8/9 on the question. I don't know that for absolute sure, but it seems consistent with previous rubrics.</p>

<p>i think i got most of it right, except for maybe 3. Also.....WHY DID THEY HAVE TO PHRASE 1 SO ODDLY. I interpreted it as below y=2, so i integrated from 3 to infinity, then multiplied by 2. THat is so gay......w/e maybe i'll get some partial credit.</p>

<p>4a...I found the derived function, set it equal to zero, but only found t from 0 to pi instead of 2pi so I never found the answer...How many points do you think I will receive for getting that far?</p>

<p>That's what I did for 2a and it better be right.</p>

<p>To pearl16:</p>

<p>Looks good to me on 2a.</p>

<p>On 4a, I'm not sure of the points breakdown, but I'm leaning a loss of one point out of what I'm projecting to be either 4 or 5.</p>

<p>But in order to get those points, I think you'd have to also verify that you checked the endpoints and critical values and confirm that you're at your furthest left at t = pi on that interval, as well as considering and rejecting pi/4.</p>