Okay
Well once they come out i’m sure someone here will work them all out.
The FRQ is now available https://secure-media.collegeboard.org/digitalServices/pdf/ap/ap15_frq_calculus_ab.pdf
1c anyone?
just take R(t)-D(t)=0, so will be critical points on the volume function, then evaluate R(t)-D(t) from left and right an see if the volume minimized or maximized there.
@rockinman1 I got that the water in the pipe was at a minimum at time t=0. The amount of water going into the pipe is always more than is draining out, so the water level will never go below its original level, 30.
false, graph R(t)-D(t), it does go negative
@eyrar99 i got the same thing, when i graphed both r(t) and d(t) they never had any zeros therefore didn’t have the possibility of being a max or min. i then thought about the only 2 other options (0 and 8 since they are the endpoints and on a closed interval their has to be an abs max and min) and 0 was going to be the least amount of water because both functions were increasing and their was never a time where there would be less than 30. tbh though my explanation wasn’t so great for that question because i didn’t know how o prove that it would never be less than 30, i think i just wrote that since they would both be inc it would never go lower but i think i needed to justify it better
I put 8 because I said the difference between D(t) and R(t) would be the greatest there. (But I think its wrong)
@Kylemcgrogan but how do you know that it decreased long enough for the water to be less than 30?
@rockinman1 and @eyrar99 , please check out the wording of the question, the rate of change of volume is rate of change in input mine rate of change of draining.
@APScholar18 i thought that you were an ap scholar though how could you get it wrong… JK sure we all did amazing, were the tiny majority that cares enough after the test to make sure we did well because july is just too far out!
so what i wrote in one of my prior comments is find zeros of R(t)-D(t), these will be possible extrema of the volume of the container, then do a sign chart of R(t)-D(t) around the zeros to see if they were minima or maxima, then evaluate the integral of R(t)-D(t) (net change in volume) and see what point yields the lowest volume
@Kylemcgrogan i don’t exactly see what you mean but i do believe you.
edit: never you last post made it clear! thanks
@rockinman1 doesn’t nearly everyone get an AP Scholar award haha. Hopefully I get national ap scholar!
@APScholar18 hahaha i wish you the best of luck man im sure ull get it!
@kylemcgrogan your way of explaining sounds much better than mine! 
My biggest problem on the FRQs was always trying to explain my logic…
For 1c) did you guys get 3.2717?
if i remember correctly, yes, let me do the math again real quick to verify
just checked, to be exact is is 3.271658
Thats what i got