AP Calculus AB 2015 Thread

I’m talking about the H(t) from the multiple choice section.

@mochilate1897 Can you provide more detail? was it the question where they said h = f(x) - g(x) find h’(3) ? or something similar?

@mochilate1897 I got something like ln(H+1) = 2t+c but I don’t really remember.

@Eggyolk It was like dy/dt= H(t) +t + c and you had to integrate

@Eggyolk Yeah I multiplied each by their width. Got big numbers. I hate addition. T-T

@mochilate1897 Yeah i don’t remember that question.

@KaiserEric13 yeah same lol… but I think as long as we provide the details of how we got our answers and the process is right but we did some technical mistake we would only get -1 for an inaccurate answer and get credit for a correct process.

@KaiserEric13 7600 for sure

Does positive and decreasing ring a bell for anyone?

@mochilate1897 Yes, it had to do with velocity and acceleration of something. That’s the answer I put.

@sunrisereader Awesome. I’m counting on the MC to make up for the FRQs

@mochilate1897 Me too.

Also, does anyone remember the FRQ that gave the areas of two regions under the x-axis with f’(1)=3 or something like that? Why were the areas, 9 and 12, positive if they were below the x-axis?

@sunrisereader You’re totally right, but I think you just have to go off of what they give you and work the problem based on what they’ve established. So i ended up getting like -6 and 15

@sunrisereader I think they were positive to throw us off. Subtract the 9, add the 12 is what I did.

I think you added for both though because the negative area for one of them was going in the opposite direction meaning it was positive area.

I kept them positive.

@RHSclassof16 the slope was negative on both sides of the given point at x=1, which means that the values to the left of x=1 on f(x) were greater than 3 (decreasing towards 3) and the values to the right of x=1 were less than 3 (decreasing from 3). So you had to add the 12 because that point was to the left of x=1 and subtract the 9 because that point was to the right of x=1.

But it said that the area was positive 9 and positive 12. Why would they trip us up like that.

@mochilate1897 Okay thanks, it threw me off a bit. But yeah you’re right, you just have to use what’s given. I don’t know what answers I got but I think they were similar.

For the slope field on the FRQ what was the concavity for all solutions in quadrant 2. I said concave up but that was a hasty call on my part. I’ve been hearing discussion from side on it being down or vice versa.

@cocoabutter16 Yeah, that’s right, because in q2, all x values are negative and all y values are positive. when you put that into the d2y over dx2 equation, you can only get positive numbers.