<p>points not percentage. like 65 pts out of 108</p>
<p>Just saw and took the 2007 Free Response. Here is my take, question-by-question:</p>
<ol>
<li><p>A pretty standard areas, volumes of revolution, and volumes by known cross-sections question. No surprise here.</p></li>
<li><p>a and b have been done before to some extent, although the piecewise function was a new wrinkle. While c had been done before, the piecewise wrinkle made this question a little more interesting than usual.</p></li>
<li><p>A pretty standard question if you understand the composition of functions. However, I find that most students understand the Chain Rule outside of the idea of understanding symbol notation, and accordingly, won't understand the idea that d/dx[f(g(x))] = f '(g(x))*g'(x).</p></li>
<li><p>I know that most of my students struggled with trig values, and so (a) was painfully hard for them. In order to do well on this question, I think you needed to be able to see that you could factor out an e^(-t) as you worked. If you saw this, you not only knew how to solve the equation at the end, but also had a significantly easier time taking the second derivative (you only needed the Product Rule once instead of twice).</p></li>
<li><p>I'm sure many students thought they needed a calculator on this one to calculate the actual volume at 5.4, but an understanding of concavity is sufficient to the relationship of a tangent line that you're fine on this one without it. However, this part and the other parts of this question have been asked on previous free response sections, and the only thing that's a little bit different is that usually they give you a table with R(t) values and ask you to do the Riemann sum for the integral of R(t) dt. Here, they give you a table with R'(t) values and ask you to do the Riemann sum for the integral of R'(t) dt. Not any harder when you see the connection, but I'm guessing the prime freaked people out.</p></li>
<li><p>The only thing that made this question more interesting than usual is the constant and the amount of algebra involved, which while higher than in most years, shouldn't be beyond the realm of most calculus students.</p></li>
</ol>
<p>Honestly, I have to say that while I think this Free Response exam was probably marginally harder than the exam set from '01 - '06, I don't think it was anywhere near on the level of "that was so impossible" that some folks wanted to give it.</p>
<p>I'm not terribly surprised that they went this direction, as I knew that conceptual questions of this nature was really more of the direction that the AP Exam was going in.</p>
<p>However, I have to admit that I even told my students that I thought that the slope fields and separable differential equations pieces was going to be on there, and I would have probably staked my house on that one. Good thing nobody was willing to take me up on it. =)</p>
<p>Not sure how accurate these are, but this is from the ap calc collegeboard forum that teachers use i guess?</p>
<p>omg, some of those answers actually look vaguely familiar. Maybe I didn't fail completely..</p>
<p>To admanrich's link, I disagreed on the answers to 4a, 5b, and 6c with those answers, but the rest look right (if truncated too much for credit on the explanations) to me.</p>
<p>My answers to those:
4a: t = 5pi/4 (pretty sure pi/4 is when it's furthest to the right
5b: dv/dt = 4(pi)r^2 dr/dt when r = 5 and dr/dt = 2 yields 7200pi cubic feet per minute
6c: Assuming f "(x) = 0, and f "(x) = [4 - k rt(x)]/4x^2, then 4 - k rt(x) = 0. Since the point is on the x-axis, k rt(x) = ln x, so 4 - ln x = 0, 4 = ln x, and x = e^4.</p>
<p>5b. dv/dt=4pi(r)^2(dr/dt)</p>
<p>t=5, so r=30, and dr/dt=2</p>
<p>4pi(30)^2(2)=7200pi</p>
<p>Ack! Good call on 5b. Yuck.</p>
<p>crap1! the answer to ab2-b is [0,1.617) and (3,5.067)...what if my answer was [0,1.617) and [3,5.067) ...................</p>
<p>Hey! Some of those match my answers....
Hooray!@!!!</p>
<p>i definitely agree with you on 4a, it was 5pi/4, not pi/4</p>
<p>couldn't you use f(x) + f'(X)(delta x) for the tangent line approximation one?</p>
<p>30 + 2(.4) = 30.8 so it is 0.8 greater than the original?</p>
<p>no, b/c that is eulers method and they specifically asked you to use a tangent</p>
<p>No it's not...
???</p>
<p>i think they wanted you to use the tangent equation</p>
<p>Hmmmmmmmmmmmmmmm</p>
<p>I dont remember my answers, but the one I bombed was the one asking for g-1(x) in order to find the tangent line. That night I realized g(g-1(x)) = x. Bummer. But I sorta messed up Part B of that question as well. In addition I'm not confident my answers to some sections of 1 and 2 were correct. Although I'm pretty sure my procedure was.</p>
<p>
[quote]
However, I have to admit that I even told my students that I thought that the slope fields and separable differential equations pieces was going to be on there, and I would have probably staked my house on that one. Good thing nobody was willing to take me up on it. =)
[/quote]
Yep... that's what my calc teacher told me. I would've taken that over #3 anyday.</p>
<p>I think pretty much everyone who took this exam is in agreement--MC wasn't impossible (I left 8 blank but that's just me ;__; ), but the FRQ were ridiculous. I got so confused on the first freaking question that I think I started to panic and lose focus. I was really hoping to do well--I even though I had a shot at a 5--but it doesn't look that way anymore.</p>
<p>The only hope I have is that the exam writers made it this way on purpose and that the curve will be more lenient/generous than usual to account for the unusual difficulty of the exam. <em>fingers crossed</em> Ah well, at least it's over.</p>
<p>Now I just to cram this weekend for Bio! w00t! I'm not worried about AP English Language & Composition at all, my teacher prepared us so well. Good luck to everyone with any exams you'll be taking next week!</p>
<p>I lost focus on the first question, lost my motivation and just couldn't answer the questions. If I don't do well in the beginning, I freak.</p>
<p>on 5 (b) it def was 5pi/4..</p>
<p>((4/3) pi r^3))</p>
<p>at time= 5, r=30</p>
<p>(dv/dt) = (4)(pi)(dr/dt) (r^2)</p>
<p>at time = 5, r' is 2.</p>
<p>(dv/dt) = (4)(pi)(2)(30)
= 240pi</p>
<p>??</p>