@mochilate1897 I don’t know. Maybe they were going by total area and wanted us to realize on our own that they must be negative. We know that the area under the x-axis is negative so if they suddenly decided to change that basic rule for the actual solutions to the problem, they’d have a big problem on their hands. Either way, the values on that interval of the graph were all negative, and the graph was of the derivative, so that alone tells us that the f(x) values to the left must be larger and values to the right must be smaller.
@Jamesesesess I’m pretty sure that’s what I did. I honestly don’t remember anymore.
I’m pretty sure I got a 5 at least. I definitely got a 1 on the AP Chem test.
@cocoabutter16 @mochilate1897 I got concave down in Quadrant 2. I used the same fact, that x-values are negative and y-positive but I thought that made the whole d2y/dx2 equation negative.
@Jamesesesess I see your point. Would that only affect part d?
For the area one I believe it was f(1)=3 so to find f(4) and f(-2) you do 3 + integral from 1 to 4 f’(x)dx and 3 + integral from 1 to -2 f’(x) dx which ended up being f(4)=-9 and f(-2) =12
@mochilate1897 Yup, that was the only part the area was used in. At first I wondered why the area values were even given until I realized what I had to do with part D lol.
@cocoabutter16 I ~think~ it was -6 and 15. Can’t really remember which side was which area.
@sunrisereader If i remember correctly, the equation made a positive product from the y part, which was subtracted by a negative product of the x part, and when you subtract by a negative, you get a positive. So that made the entire thing positive, if i remember correctly.
@Jamesesesess So i guess if we made the mistake of just taking what they put on the graph (darn you collegeboard), we’d probably get boat loads of partial credit but just 2 points off from the answers, and possibly 1-2 more from the procedure?
@Jamesesesess Hold up the answers were -6 and 15? Bruh I got that. But I think i got 15 for the first part and -6 for the second.
@mochilate1897 Hmm I remember the equation as -(x) - 2(y). If x is a negative value and y a positive value, then that would make the equation -(-x) - 2(+y) = x - 2y, so somehow that came out to be negative for me. I don’t remember that question too well though.
@mochilate1897 @Jamesesesess Yeah how does the scoring work? Are the graders usually pretty generous with credit on the FRQ?
@sunrisereader Your procedure is sound, but the problem is that I remember the equation as 2y-x.
How about the one part with the vertical distances this would be 2 c I think, not too sure, but it was right after the cross sections part dealing with regions R and S. Does anyone think they got that one completely correct
@cocoabutter16 I think my procedure was right.The vertical distance, h, was equal to f(x)-g(x). So I set it up like a related rates problem and used nDeriv on my calculator to find dh/dt. Does that help?
I have no idea how it would be scored… this was my only year doing AP exams since my school only has 2 real AP classes lol.
@mochilate1897 yeah I’m pretty sure I had the same answers. How did you get those if you used the areas as positive? You would’ve had 3-9=-6 and then 3+12=15.
I’m a B average student and I found it pretty easy. The test (in general) is all about remembering the little things!
When you people had to find all values that equal something, did you get y=blahblahblah? Or were there specific values?
@Jamesesesess im asking myself the same thing, but I distinctly remember getting 15 and -6. I guess it was alright after all!
@mochilate1897 you must’ve done it right without realizing, that’s happened to me a few times hahaha.