<p>@powerbomb: The 387.5 is the total number of hours waited from time t = 1 hour to t = 2 hours from part (c). You want the total number of hours waited from time t = 0 hours to t = 2 hours for part (d), and divide that by the 980 people.</p>
<p>Dangit, how did I blank out on 2C… its just a regular fundamental theorem of calc problem! I was way to nervous during this test. I still can’t believe how simple it was… just a definite integral! and yet I blanked out during the test</p>
<p>for those of you that are rly experiecned, how is this 2009 calc bc FR and MC compared to that of that of 2003? like curve comparisons?</p>
<p>personally i thot 2003 MC relatively easier…FR, both are hard!</p>
<p>o my god. UGH i missed that one too!?!?!?!?!? UGGGGG! there goes my 5</p>
<p>oh haha laplast sorry - yeah that’s what i did on the test, but right now i dont why i was doing sqrt(.8^2 + (3.6 - 9.8 x 1.936)^2)</p>
<p>yeah i got your answer now</p>
<p>can anyone remember FRQ 6? would anybody like to go over it? </p>
<p>a) i think i just plugged in (x-1)^2 into the speical generic e^x series…</p>
<p>b) I took the first four terms from a) and subtracted 1 at the end which canceled out the 1’s and then put that all over x^2. my generic term came out to be something like (x-1)^2n/n!(x^2) </p>
<p>c) I don’t remember what this one asked…</p>
<p>d) POI. I said there were none. but didn’t provide justification. (half-credit?)</p>
<p>i know u did parts a and b right in 6</p>
<p>for FRQ 6, the series they listed was for e^x centered about x=0
for part
a) they asked for e^(x-1)^2 centered about x=1
so you have to use the formula
(nth derivative of f(a) * (x-a)^n) / (n)!
so you have to take the derivative of e^(x-1)^2 for 3 times because it asks for the 1st 3 terms
and then you just go on from there and plug it into the formula to get the terms im too lazy to list it out
b) you had the correct concept in b) , but since you did a) wrong, the answer in b) is not correct</p>
<p>^ Can’t you just plug in (X-1)^2 into anywhere where there is an X? Since we already have e^x and we’re looking for e^(x-1)^2</p>
<p>this was being debated at my high school as well. we did not remember exactly what we did or how to solve it, but half did the way above and the other half had the answers as shown on the other bc thread.</p>
<p>^ Above as in godbreath’s answer?</p>
<p>godbreath’s reasoning, not sure about answer, it was a discussion btw myself (first in calc class who your method) and the #2 in calc, we have been the top two for a while, so we were not sure which way to lean.</p>
<p>i hope it is my way, because that will go a long way to a 4/5</p>
<p>k. Can’t you just plug in (X-1)^2 into anywhere where there is an X? Since we already have e^x and we’re looking for e^(x-1)^2. We were taught to plug in to the 5 basic series</p>
<p>that is what i believe, the test seemed rather simple, with numerous “short-cuts” if you knew what you were doing, and i think this was one of the short cuts that they used. it only seems logical.</p>
<p>can anyone use the wording in the ap test book and clearly lay out an answer for the other method, just wondering…</p>
<p>number 6 wasn’t that hard :D</p>
<p>which answer did you get? the one by godbreath or by marcusf329</p>
<p>i did wat marcus did by just plugging in for every occurrence of x</p>
<p>ok, good that is what i did</p>
<p>"just plugging in for every occurrence of x "</p>
<p>you mean you plugged in (x-1)^2 into every x? like into the x^n/(n!) ??</p>
<p>i was wondering whether i had to recenter it around 1 or if it was centered already.</p>