Looking back at them, I can remember how much I hated #2 and #3
@yeongil that’s odd, the 2016 BC frqs opened for me. but anyways, i’m going to start posting some of the answers i quickly reworked.
1 (a) R’(2) = R(3) - R(1)/2 = (950-1150)/2 = -100L/hr^2
(b) Approximating the integral with a left Riemann sum: 1(1340) + 2(1190) + 3(950) + 2(740) = 8050 Liters. This approximation would be an overestimate because the values are decreasing, indicating that the curve is decreasing and possibly concave up.
© 50,000 + ∫ W(t)dt from 0 to 8 = 57836
57836 - 8050 = 49786 Liters
(d) Did not do. All I specified on my exam was that W(t) - R(t) would have to = 0 for rates to be equal.
2 (a) 5 + ∫(dx/dt)dt from 0 to 3 = 14.377. Looking at the graph, we could see the y-position at t=3 was -1/2? So position at t=3 is <14.377, -1/2>
(b) dy/dx at t=3 is 19.913 because I got dx/dt at t=3 to be 9.956 and use rise over run from the graph, the slope at t=3 for dy/dt was 1/2.
© square root of (dx/dt)^2 + (dy/dt)^2 at t=3 would give speed
square root of (9.956)^2 + (-1/2)^2 = 9.969
(d) Idk, I did not know how to find total distance traveled in the y. All I wrote was the ∫ of square root (dx/dt)^2 + (dy/dt)2 from 0 to 2.
3 (a) I figured out that the key to this problem was g’(x) = f(x)
There is neither a max nor a min on the graph of g at t=10 because f(x) remains negative and did not change signs, indicating that g continued to decrease.
(b) Yes, there is a point of inflection at this point because f goes from increasing to decrease.
© Wrote a completely wrong answer because I misread the problem and did not have room to fix my errors
(d) This was kind of confusing? I said g(x) < 0 from [10,12] and [-4,2] and I they did not say to give an explanation in the problem, so I didn’t. Do you think they were looking for one? In addition, I found those intervals from plugging in values into the ∫ f(t)dt to see when the area would be negative but I doubt that’s right.
4 (a) I think I aced this one! d2y/dx^2 = 2x -(1/2)dy/dx and then plug in dy/dx because they gave it to us in the problem and they want this in terms of x and y. d2y/dx2 = 2x -(1/2)x^2 + (1/4)y
(b) dy/dx at (-2,8) was exactly zero, indicating that the f has a horizontal tangent there. d2y/dx^2 at (-2,8) was -4 < 0, indicating that at that point f was also concave down. Thus, this would have to be a maximum (am I right?)
© L’hopital twice gave me g’’(x) / 6. Plug in -1 into g’’(x), aka d2y/dx^2, and I got -6, so -6/2 gave me -1/3.
(d) After doing Euler’s Method I got 1.25 let’s hope I didn’t do special math here
5 (a) I hate my life for this. At max I got 3pts for this entire problem because the super complicated fractions made me think I was doing something wrong. For (a) I did (1/10) ∫ (3/20 + h^2/20)dh from 0 to 10 and after antideriving and applying the FTC I got 1090/600. Yes, I distributed the 1/20 out and yes I didn’t even bother simplifying my answer to 109/60. Might I still get credit for this problem?
(b) Once again, not a fan. A=pi x r^2. I did pi x ∫ (3/20 + h^20/20)^2 dh and left as is lmao. Hopefully I get setup points, if that was even right.
© Didn’t even try. Wrote V = (1/3)pi x r^2h, and then applied related rates techniques to get dV/dt = (1/3)pi x r^2dh/dt + (2/3)pi x r(dr/dt)h and moved on because I was scarce on time.
6 (a) Goodbye I hate series and this one was particularly hard. This was a mess. I couldn’t write a general term so the rest just didn’t happen for me. I did get the first four non-zero terms to be:
1 - (x-1)/2 + (x-1)^2/8 - (x-1)^3/24 so hopefully I at least get 1pt here! I did attempt to plug into 1.2 for © and set up the error bound for (d) but time ran out so.
Hopefully I get some similar answers to everyone else, especially for #4?
@writerzt71 I got most of the same answers, but not all.
I’m feeling confident about getting a 3 though 
Does anyone want to email this to their teacher and ask them to solve it?
@writerzt71 Some of mine different but based on how many you left off or know you got wrong I probably scored about the same as you. What do you think you got?
I would assume mrcalculus will post the solutions shortly
who can explain number 3 Part D in detail. I just write the value under X-axis that must be wrong
i thought question 2 part d we need separate the interval to 0-1 1-2 right?
@collagewhat Since the cutoff for calculus bc tends to be pretty low, as in 64-69 out of 108 for a 5, I think I may have just barely made it for a 5. MC was very easy and I predict I missed anywhere from 5-7 on it, so hopefully my MC will carry me to a 5. There are plenty of points tied to setup on these problems too, so I probably got 2-3pts just from setup on the funnel problem.
@Ethansports that’s exactly what I did
@writerzt71 Are you sure about 2b? looks like you did dx/dt over dy/dt and I thought it was the other way around
@yousefk Yes you’re right I mixed that up! Ah well, I think I still have some wiggle room. Then the actually answer would be .05022 or something in that ballpark
@writerzt71 how you think MC is easy. I actually think the non-calculation part is so hard!
@ethansports we did a ton of MC practice in class so I knew what to expect and a lot of the questions were familiar. inactive was slightly harder than active, but overall it was the frq and a lack of sleep that messed me up
@writerzt71 Minor correction in 1a:
R’(2) = [R(3) - R(1)]/2 = (950-1190)/2 = -120 L/hr^2
As for 1d, I would say: W(t) looks to be continuous and differentiable. Since R(t) is differentiable, it is also continuous. W(3) = 1275 and R(3) = 950, while W(6) = 330, and R(6) = 740. By the IVT there exists a t in [3, 6] such that W(t) = R(t).
(No, I’m not a Calculus teacher. I have tutored family and friends in AP Calc, however. Have to sign off now, so I can’t answer further questions ATM)
Did I have to mention the IVT for 1d? I just said that since both are continuous and W(t)'s initial value is higher than R(t)'s initial value, and W(t)'s final value is lower than R(t)'s final value, they have to cross at least once.
@yousefk generally when there is a question like this they are probing your knowledge of the big six theorems and thus there is a point tied to mentioning the theorem and tying it into your explanation. however i think you would still get the points because your explanation implies your knowledge of the IVT!
I don’t have to leave just yet. Have a couple of minutes. This is what I came up with for #5:
5a
(1/10) ∫ (from 0 to 10) (1/20)(3 + h^2) dh = 109/60
5b
pi * ∫ (from 0 to 10) ((1/20)(3 + h^2))^2 dh = 2209pi/40
5c
r = (1/20)(3 + h^2)
dr/dt = (1/20)(2h * dh/dt)
-1/5 = (1/20)(2*3 * dh/dt)
-4 = 6 dh/dt
dh/dt = -2/3
height is decreasing at a rate of 2/3 in/sec
(Now I’m leaving, for real this time 
)
^For 5b, i believe [(1/20)(3 + h^2)] has to be squared
@user34567 Fixed. This is what I get for typing too fast :">
For 3b is it a point of inflection even though f double prime is undefined?
3c. did you guys get min -8 and max 8