What did you guys get for the expression for f(x) assuming f(1) = 3?
@CollegeTester443 Really? YESSSS
no, it asked for all points on the equation with vertical tangents, it asked for the points, not just a set up
@Kylemcgrogan I think we’re talking about different ones. I was saying that the one with the water overflowing just asked for the equation
For the vertical tangent I set bottom to zero, then substituted x using original function to write equation in terms of y then solved for y.
I thought I was solid on the FRQ…and then I began part 2.
The no calculator section was horrible. Genuinely the hardest calc problems I’ve ever seen for college board. When we passed in the booklets at the end every single person in the room looked frazzled.
I tried that, i just couldn’t solve it, i think i see what you are saying though, you could eliminate x. did you get 1/32 for d^2y/dx^2
The separation of variables was a completely normal problem. If you complain, the collegeboard is just gonna be like “um… okay. well it was on the curriculum sooooooooo…ya”
and it was concave up for sure. put in points and youll see. I can’t talk about the question exactly obviously but just pay attention to the signs and there is no way that in the quadrant specified it could be concave down.
yeah 1/32 for last question cant remember if it was negative or positive
for the one with the integral and the pipe i think u had to do the integral from 8 to w of r(t) - 8 to w of d(t) = 50 because me and my friend were talking about it and it said for t>8
@rockinman1 if you do that you have to find what its value was at 8 (just doing what you did assumes it was at 0 at 8)
For the y=mx+b one it can’t be m=1 and b=1. If substitute in points on the line into dy/dx=2x-y, you’ll find that the slope is not the same for every point.
@RHSclassof16 OMFG! Thank goodness I’m not the only one who got m=1 b=1!
For the absolute min on the pipe on I remember the CV ended up being the absolute pin as opposed to the endpoints right?
@dogeater then what is it?
m=2 b=-2.
the right side needed the plus 30 for the initial condition, and it was 900/cos^2theta
for the limit and negative infinity did you guys get 3/4? (Or whatever the constant at the top divided by the constant by the bottom was)
@Kylemcgrogan Lol I think I got that but I completely guessed.