only similar thing was polar curves
If CB is trying to lower the amount of students who got a 5 they will have to try harder because we AP Calculus students were not ready for this FRQ’s and if we all do worse on them praying to god that the curve is very generous
@putinator lol why you so salty. The polar one is pretty similar. On the course description, there’s also a question about improper integrals. And the series one also had the alternating series error bound
I had form c
Was I the only one with Form E? The FRQ’s were terrible. I probably got a 5/9 for the first and 2/9 for the second calc active.
Was that Form E? I had something similar that probably killed me.
Does anyone have any idea what a cutoff for score 5 be?
I think I have done form c
As I posted earlier, my D18 had Form E. She’s one of the top BC students in her class and she thought the FRQ’s were “impossible.”
If you believe the AP Pass site, to get a 5, then you’ll need 62%. In 2016, 48.4% of the students taking the BC test got a 5, so hopefully the odds are on all your sides. Good luck.
The FRQs have been released. My attempt at #1:
A) Volume ≈ 250.3 + 314.4 + 5*6.5 = 176.3 cu. ft.
B) overestimate because A(h) is decreasing
C) integral from 0 to 10 of f(h) dh ≈ 101.325
D) V(h) = integral from 0 to h of f(x) dx
dV/dt = f(h) (dh/dt)
at h = 5,
dV/dt = (50.3)/(e + 5) * 0.26 = 1.694 cu. ft. per min.
2:
A) A = 0.5 integral from 0 to pi/2 of (f(θ))^2 dθ = 0.648
B) integral from 0 to k of (g(θ))^2 - (f(θ))^2 dθ = integral from pi/2 to k of (g(θ))^2 - (f(θ))^2 dθ
C) w(θ) = g(θ) - f(θ)
w_a = (2/pi)*integral from 0 to pi/2 of g(θ) - f(θ) dθ = 0.485
D) g(θ) - f(θ) = 0.485
θ = 0.517 or 0.518
w’(0.518) = -0.581 or -0.582
w is decreasing since w’(0.518) < 0
3:
A) f(-6) = 7 - 0.524 = 3
f(5) = 7 - 0.5pi2^2 + 0.5 32 = 10 - 2pi
B) f is increasing on [-6, -2] and [2, 5] because f’ is positive
C) Check critical points (x = -2, 2) and endpoints (x = -6, 5)
f(-6) = 3
f(-2) = 7
f(2) = 7 - 0.5pi2^2 = 7 - 2pi
f(5) = 10 - 2pi
absolute min value is 7 - 2pi
D) f"(-5) = -1/2
f"(3) DNE because the one-sided limits of f"(x) do not agree
(as x -> 3-, lim f"(x) = 2, but as x -> 3+, lim f"(x) = -1)
Is this Form O?
@sushiritto I do not know. I am not a student. I just tutor family and friends on the side.
FYI, I sent your answers to my D18 and she said those weren’t her questions/answers. Bummer.
Yes, they are Form O @sushiritto
4:
A) @ t = 0 dH/dt = -0.25(91-27) - 16
Tangent line: y = -16t + 91
H(3) ≈ -48 + 91 = 43°C
B) d2H/dt2 = -0.25 dH/dt = -0.25(-0.25(H-27)) = (1/16)(H - 27)
Since d2H/dt2 > 0, H is concave upward. Therefore, answer in (A) is an underestimate
C) integral of (G - 27)^(-2/3) dG = integral of -dt => G = (-1/3 t + C)^3 + 27,
G(0) = 91 => C = 4
G(3) = (-1/3*3 + 4)^3 + 27 = 54°C
5:
A) f’(x) = -3(2x^2 - 7x + 5)^(-2) * (4x - 7)
f’(3) = -15/4
B) f’(x) = 0 in interval (1, 2.5) => x = 1.75
For (1, 1.75), f’(x) > 0, and for (1.75, 2.5), f’(x) < 0, so there is a relative max @ x = 1.75
C) lim as b -> inf of [integral from 5 to b of 2/(2x - 5) - 1/(x - 1) dx]
= lim as b -> inf of [ln(2x - 5) - ln(x - 1)] | from 5 to b
= lim as b -> inf of [ln (2b - 5) - ln 5] - [ln (b - 1) - ln 4]
= lim as b -> inf of [ln (2b - 5)/(b - 1) - ln 5 + ln 4]
= ln 2 - ln 5 + ln 4 = ln (8/5)
D) f is continuous, positive and decreasing on [5, inf);
integral in © converges
Therefore, by the Integral Test, the series also converges.
6:
A) f(x) ≈ 0 + 1x + (-11)/(2!)x^2 + (-2-1)/(3!)x^3 + (-3*2)/(4!)x^4 = x - (1/2)x^2 + (1/3)x^3 - (1/4)x^4
B) 1 - 1/2 + 1/3 - 1/4 + … is an alternating harmonic series, which converges
1 + 1/2 + 1/3 + 1/4 + … is a harmonic series, which diverges
Therefore, the series converges conditionally
C) g(x) = (1/2)x^2 - (1/6)x^3 + (1/12)x^4 - (1/20)x^5 + … + (-1)^n/[n(n-1)]x^n + …
D) alternating series error <= | (-1/20)(1/2)^5 | = 1/640 < 1/500
I had form c and the released frqs are entirely different than mine it makes me angry i really want to see the frqs and see if i answered them correctly i need to wait till july to see what i got on that test, and i will never be able to see these questions again in my lifeeee.