I see… Not very satisfying, but I see. And it’s staring us in the face in part a). I still don’t see how that rules out other possible solutions, though – what do you think?
@Bugiardi I used WolframAlpha to solve the differential equation and the solution is y = Ce^-x + 2x - 2. The solution to the differential equation will only be linear when C = 0, which indeed gives us y = 2x - 2.
If my final answer for the RRAM question was wrong because I did 40 *150 instead of 16 * 150 how many points would I lose? 
Because I remember writing down 11200 because of that mistake but everything was right I believe.
like 1 point?
Wow. Now that you put it that way, I believe you nailed it, and I believe this makes 4d) an extremely low-scoring problem. CB sure confounded me on this one. Kudos to you. Are the rest of my answers okay 
?
@Bugiardi Everything else looks fine except for 5b and 5d. The second interval for 5b should be (1, 3) and for 5d, f(4) = -9 and f(-2) = 12.
Thanks. Those were brain farts on my part.
@Bugiardi Ah okay thanks. That one really confused me. Kudos to you for getting that one.
dang it, i got 8 for 6c too haha @Hark123
So I had form G… My free response was different than what was posted. Anyone know when I can see those?
@goblazers hahah im not alone then! but as @zach1198 said we will get most of the credit just a careless number error.
How many forms of the test were there? I had form O and it seems like many others had O too. @BenX97 said he had G.
I got -1/64 for 6c too subtracting and multiplying is hard 
Just printed off the FRQ’s. I will work them out and post the answers when I am done.
Here is what I got:
-
a) int from 0 to 8 of R(t) = 76.570
b) R(3) = 5.086 and D(3) = 5.4; since D(3) > R(3) water is decreasing at t = 3
c) Amount of water in the pipe at any time t is 30 + int from 0 to x of R(t) - D(t); you must check endpoints and where R(t) = D(t) which is at t = 3.271; Therefore, A(0) = 30 (given), A(3.271) = 27.964, A(8) = 48.543 so absolute min occurs at t = 3.271
d) We know from part c that at t = 8 we have 48.543 cubic feet in the pipe. Therefore,
48.543 + int from 8 to w of R(t) - D(t) = 50 -
a) int from 0 to 1.032 of g(x) - f(x) + int from 1.032 to 2 of f(x) - g(x) = 2.004
b) V = int from 1.032 to 2 of [f(x) - g(x)]^2 = 1.283
c) h(x) = f(x) - g(x) therefore h’(x) = f’(x) - g’(x) and so h’(1.8) = f’(1.8) - g’(1.8) = -3.811 -
a) V’(16) = [V(20) - V(12)]/(20-12) = 5 m/min^2
b) It’s the total distance Johanna jogged; 12(200) + 8 (240) + 4(220) + 16(150) = 7600 m
c) B’(t) = 15 m/min^2
d) (1/10) * int from 0 to 10 of B(t) = 350 m/min -
a) draw slope lines
b) second deriv = 2 - dy/dx = 2 - (2x - y) = -2x + y + 2; since in Q II x < 0 and y > 0, the second deriv will be > 0 and thus all solutions in Q II will be concave up
c) dy/dx at (2,3) = 1 so since it’s not = 0 it is neither a rel min or rel max
d) dy/dx = y’ = m and also the second deriv = y’’ = 0 so using answer from b, -2x + y + 2 = 0 so y = 2x - 2 -
a) max at x = -2 since that is only point where f’(x) goes from positive to negative
b) (-2,1) and (1,3); these are only regions where f’(x) is decreasing which means f(x) is concave down and f’(x) < 0 which means f(x) is decreasing
c) x = -1, 1 and 3; points where f’(x) slopes change
d) f(x) = 3 + int from 1 to x of f’(t); f(4) = 3 + (-12) = -9 and f(-2) = 3 + (9) = 12; remember the value of these regions are negative. You must account for that. -
a) dy/dx at (-1,1) = 1/4; so y - 1 = (1/4)(x + 1)
b) set denominator = 0; 3y^2 - x = 0 so x = 3y^2; y^3 - (3y^2)y = 2 solve and you get y = -1; plug this in to equation and solve for x; (-1)^3 - x(-1) = 2 leads to x = 3 so point = (3, -1)
c) second deriv at (-1,1) = 1/32
Curve predictions anyone? Anyone else think it was pretty hard? Not impossible, but definitely different/harder than recent years
For 5b),I get (-2,-1), not (-2,1). Did I do something wrong?
@Markp16 no I think you’re right. I got that too
@MathToot, how would you allocate the points if you were grading for each part
For 1 d), would the barebones answer: 50 = 30 + int from 0 to w of R(t) – D(t) receive full credit?
How I felt I did: MC Section 1: Part A (no calc) - 50-60% at best; MC Section 1: Part B (calc allowed) - nailed it (only guessed on four of them)…also my personal graphing calc died halfway through but thank god my teacher gave the proctor spare graphing calculators. FR Part A (calc allowed) - Botched 1, possibly nailed part a and b and didn’t know how to do c at all. FR Part B (no calc) - Nailed #3, what the frick was (b) on #4?? and BS’d © on 4 as well, felt surprisingly decent on #5, definitely nailed #6. Probably got a 3 but maybe scathed a 4. Most of my classmates drew pictures on the free response and just died throughout the whole test so I feel good about myself because I actually tried!