<p>2010 AP Calculus AB Form A solutions.</p>
<p>Here are quick answers, let's check compare and check them.</p>
<p>1.
a) ∫(0→6) f(t) = 142.xxx cubic ft.</p>
<p>b) f(8) - g(8) = 48.417 - 108 = -59.58 cubic ft/hr</p>
<p>c) forgot already, but i did it</p>
<p>d) (∫(0→9) f(t)) - (125 + 108*2) = 26.335 cubic ft.</p>
<hr>
<p>man it's so uncomfortable to write here, will add others later, or u guys can add too...</p>
<p>i disagree with d) the intergral should be 0 - 9 as the snow is falling the whole time?</p>
<p>Did you have to write +c for the “set up, but do not evaluate” problems to earn the point?</p>
<p>ber1023, my bad. but the answer is still the same.
Action52, on calculator part, where the intervals are given, you just can write the answer.</p>
<p>I know I did (a) right. I think I did (b) wrong. I still don’t know about (c) or (d).</p>
<p>■■■. I better get a 5 on this, or I’ll be really mad. I was getting easy (and I mean easy) 5s on my practice tests, and in less time than allotted too.</p>
<p>2.
a) (E(7) - E(5)) / 7-5 = (21-13)/2 = 4 hund. of entries / hr.</p>
<p>b) ∫(0→8) E(t) = (2<em>4) + (3</em>13) + (2*21) + 23 = 112 hund. of entries
1/8(∫(0→8) E(t)) = 112/8 = 14 hund. of entries
1/8(∫(0→8) E(t)) represents the average number of entries, in hundreds, visitors deposited between noon - 8 PM.</p>
<p>c) 112 - (∫(8→12) P(t)) = 96 hund. of entries</p>
<p>d) at t=9.183; the max occurs</p>
<p>sh1t, i can’t edit the 1st post…</p>
<p>This is going very wrong, very fast. Did part of (b) wrong, and I did (c) wrong; I did (d) right but since (c) was wrong that one is wrong too. This is not good.</p>
<ol>
<li>d) should be 12 as p(t) is the rate so you dont do p’(t)</li>
</ol>
<p>and b) should be trapezoidal rule</p>
<p>I’m starting to get a little depressed now that I am seeing these. I know I got the concepts but I’m starting to think that the little things I messed up on are going to add up. I know I had to have aced the MCs though, so I don’t know why I am worrying.</p>
<p>FallenAngel9, I’m not 100% sure about (C), but 99% sure about others.</p>
<p>
Yes, find the max of p(t). You will get t=9.18xx</p>
<p>You know what, maybe you are right.</p>
<p>no at t = 12</p>
<p>maybe im ■■■■■■■■… but the question says “The number of entries in the box t hours after noon is modeled by a differentiable function E for 0£ t £8.” , so how do you intergreate e(t), i think it should say E(t) is the rate of entries going into the box,</p>
<p>am i ■■■■■■■■?</p>
<p>thats what i thought in the beginning, but then i found out in order to do part (a) we have to find E’(t), which there’s no way to do it.</p>
<p>i really hope E(t) should be the rate.</p>
<p>B.) is incorrect, that’s not the trapezoidal sum formula.</p>
<p>
Sure it is</p>
<p>I got 14 for the answer to the trapezoidal problem as well, but everyone was telling me I was wrong.</p>
<p>Also, the max should be at t = 12, not t = 9. It is an absolute maximum, not a local maximum. You need to check the endpoints.</p>