i am really worried because this year the curve is going to be high. i did most of the frq’s from every year and i feel like this year’s frq was the easiest. i made a lot of stupid mistakes and i lost a lot of point i am regretting it. one of the stupid mistake is on frq 1 for part where they wanted at time 8 how much water was in the tank. i integrated the rate going in- 8050. i forgot that 8050 was already done but i put that in my integral. btw i added initial but i got it wrong because 8050 was supposed to be outside the integral. smh another mistake was reading that g(X)<0 was positive but its actually negative looked at the graph where it was increasing instead i was suppose to look at the graph were it was decreasing. 
are the answers to them up
@desiboy1998 I got 49,786 gallons of water at the end of 8 hours. Is that correct?
For #1 I got
a.) -120 liters/hr^2 (edit, pretty sure i had it on the test though. oh god)
b.) 8050 liters. Overestimate bc LRAM overestimates decreasing functions
c.) Integral of W(t) from 0 to 8 - 8050 + 50000 = 49786.195 liters = 49,786 liters
d.) Yes
W(0) = 2000, W(8) = 81.524, R(0) = 1340), R(8) = 700
I felt like this is one of those questions where they want you to reference a theorem so I just used the Int. Value Theorem for continuous functions to explain that W(t) will take on every value between 81.524 and 2000.
But my real explanation (not sure if valid, though) was since W(0) > R(0) but W(8) < R(8), there has to be a point in the interval where R(t) becomes greater than W(t), and that point of intersection is where they equal.
@holyground I said Mean Value Theorem for that one. Ugh, i put IVT first, but then I erased it. It definitely wasn’t MVT right?
FRQ’s are released and on the CB site, so here is my first crack at solutions:
1a) -120liters/hr^2
b) 8050 liters. R(t) is decreasing => a left Riemann sum is an overestimate.
c) 50000 - 8050+ int(w(t) from 0 to 8 = 49786 liters
d) The question asks whether W(t)-R(t) must =0 on (0,8). W(0)-R(0)=660>0; W(8)-R(8)= -619<0; and W(t) and R(t) are both differentiable, hence continuous; therefore IVT guarantees that there must be a time t in (0,8) where W(t)-R(t)=0.
2a) v(4) =2.978 >0 and a(4)=v’(4) = -1.164 <0, therefore the particle’s speed is decreasing = particle is slowing down.
b) Particle changes direction => v(t) changes signs. This occurs at t=2.707
c) Position at t=0 = x(0)= x(4) - int(v(t)) from 0 to 4 = -3.815
d) Total distance traveled = int(absv(t)) = 5.301 or int(v(t)) from 0 tp 2.707 - int(v(t)) from 2.707 to 3
3a) Extrema of g => g’(x)=0 and changes sign => f(x)=0 and changes sign. g’(x)=f(x) does not change sign at x=10, therefore g has neither a relative min nor a relative max at x=10
b) POI of g => g’’ changes sign => f’ changes sign. At x=4, f’ changes from positive to negative, therefore g has a POI at x=4.
c) Consider endpoints and all values of x where g’(x)=f(x)=0. g(-4)=-4; g(-2)=-8; g(2)=0; g(6)=8; g(10)=0; g(12)=-4. Thus abs min of g is -8 at x=-2 and abs max of g is 8 at x=6.
d) g(x)<or=0 ==""> int(f(t) from 2 to x is <or=0; this occurs on [6,12] and [-2,2]?</or=0>
4a) slopes are 0, -1, -4 and 0, 1, 4 left to right, bottom to top
b) dy/dx=9 adn (2, 3) y-3=9(x-2) f(2.1) ~ 3.9
c) y=-1/(ln(abs(x-1)-1/3)
5a) Average value =1/10int(r(h)) from 0 to 10 = 109/60
b) volume = piint(r(h))^2 from 0-10 = 2209pi/40.
c) dr/dt=1/20(2hdh/dt). dh/dt = -2/3
6a) k’(3)=f’(g(3))*g’(3)=10 k(3)=f(g(3))=4 y-4=10(x-3)
b) too tedious to type. h’(1)=-3/2
c) f’(6)-f’(2) = 7
@Bugiardi
Same answers as you except for:
3.d) [-4,2] and [10,12]
5 has units, I think.
6.c) 7/2 bc isn’t the antiderivate 1/2f’(2x)?
But other than that, OMG YES I DID SOMETHING RIGHT
@holyground I got 7 on 6c
@holyground i got the same intervals as you for 3d
hey guys for the approximation just putting y=9(2.1-2)+3 get you the full points? cause apparently my teacher said never simply your non free response answers. so i left it at there. and yes 6 c) its 7/2 i messed it up too i forgot the chain rule. The stupid funnel question i started both he average value and the volume right the setup but then i didn’t move forward cause i thought i would get it wrong. oh well i guess i will get points for the setup.
@Bugiardi For the most part, I believe you got everything correct lol except for 6c. I got the same answer but it’s wrong since you have to apply chain rule. integral of f’’(2x) is not f’(2x) because if you take the derive of f’(2x) it would be (f’’(2x)) (2). Hence the answer is supposed to be 7/2.
frq 6 answer for b is h(X)=g(x)/f(x) they wanted h’(1). so use quotient rule you would get (g’(x)f(x))-(f’(x)g(x))/(f(x))^2 so then plug in the values using the table you would end up with ((8-6)-(32))/36 that would be -54/36 simply its -3/2
Can someone please explain to me the interval one for 3.d)
@Tajmahal @holyground Your intervals are wrong. -4 to 2 would be wrong since if you get the area of the graph from 4 to -2 it would be negative, but you have to multiply it by negative one since the integral is in the wrong form, meaning upper limit is smaller than lower limit in this case -2 and -4 and -2 and -3…
Can someone predict how many points are for each letter of each question lol?
@YoohooAddict do you think they will give full point for putting y=9(2.1-2)+3 and not simplifying. my teachers told us to not simplify
@YoohooAddict Why doesn’t 2 count as one of the intervals? Because isnt 2 to 2 on the g(x) = to 0 which fits the g(x) less than or equal to 0?
Ok well I really hope Bugiardi is right bc I had most of his answers lol.
I believe they might have a point for the right value for the answer. So I think yes
damn i am getting that 9 points for question 2 
feels good