<p>That was defiantly one of the easiest AP Calc exams I have taken. My teacher has been assigning entire AP exams from past years for us to do every night. And in comparison to those, this one was too easy… I hope I didn’t just jinx myself.</p>
<p>cuz its easy, i’m expecting a HARSH curve… get ready guys…</p>
<p>Thought it was pretty easy. Had to guess on few multiple choice and was unsure about few others. </p>
<p>I think I messed up on the area/volume. Pretty tricky.</p>
<p>I thought the exam was easy this year, which worries because I either messed up on some questions, or the curve will be really harsh this year. The usual curve for a 5 is 68, but it might be higher this year.</p>
<p>The curve will probably be somewhere in the low to mid 70s. I guessed on 2 mc with no idea of the right answer and made educated guesses on two more. There was only one part of one FRQ I didn’t fully know how to do but even then I knew how to set it up</p>
<p>I messed up solving the differential equation, 2 of the questions on FRQ #3, 1 of the questions on FRQ #2, and maybe another FRQ part or two.</p>
<p>If I got ~15 MC wrong, could I still get a 5?</p>
<p>I’ve done every test since 1985 or something like this and this definitely seemed tons easy especially compared to recent years…</p>
<p>But there’s still a high chance I could have messed it up and get a 2 or something.</p>
<p>I’m worried I screwed up on the easy problems like volume/area…</p>
<p>They can curve it?..I didn’t ttime myself and had to guess a lot of the MC but doesn’t it kinda suck how there only 2 grapher questions for free response?(there use to be 3)</p>
<p>Sent from my SPH-M910 using CC</p>
<p>runnerxc, You can get a five with -15 on MC if you get an average of at least 6 on your FR. It’d be cutting it real close though so it would depend on the curve. So assuming you averaged 6 on your FR you’d be within two or three points under/over the cutoff for a five.</p>
<p>the frq was definitely easier than the rest… i took like everyone from 2005 and i had to skip at least one question for all of them because i had no flipping clue *** was going on</p>
<p>and i don’t think the curve will be that bad. i mean, every year there are more test takers and the majority of them are derps who don’t know jack **** about calculus so the curve stays relatively constant. if top students like us have no idea what the hell is going on i doubt the rest of the country will.</p>
<p>…does that mean i got a 5?</p>
<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>Collegeboard released the FRQ! Would anybody mind making an answer key?</p>
<p>Correct me if I’m wrong here:
1a) W’(12) = 1.017 (approx.) This is the instantaneous rate of change of the temperature of the tub at 12 minutes (in degrees Fahrenheit per minute)
1b) 16 degrees Fahrenheit. It’s the change in temperature in degrees Fahrenheit.
1c)Approximate area/20 = 1215.8 / 20 = 60.79 degrees Fahrenheit. Underestimation because left riemann sums underestimate increasing functions.
1d) 73.043 degrees Fahrenheit</p>
<p>2a) You can either do two integrals with dx, or one with dy. I’ll show dx here since that’s what I did.
Intersection at x=3.693
Int(lnx, 1, 3.693) + Int(5-x, 3.693, 5) = Area = 2.986
b) See 2a, except each integrand is squared.
c) I have no clue.</p>
<p>3a) g(2) = int(f(t), 1, 2) = -1 * (1/2) * (1/2) = -1/4
g(-2) = int(f(t), 1, -2) or -int(f(t), -2, 1) = -(pi/2) *1^2 = -pi/2
b) g’(3) does exist, g’’(3) doesn’t exist (can’t have slope when there aren’t points on both sides of point)
c)g’(x) = f(x) (Second Fundamental Theorem of Calculus)
g’(x) = 0 when horizontal tangent line, so f(x) = 0 as well.
f(x) = 0 when x = -1, 1
When you test each point, x= -1 goes from positive to negative (increasing to decreasing for g(x)), while x=1 remains negative (decreasing for g(x)).
x = -1 is a maximum
x = 1 in neither a maximum or minimum</p>
<p>4a) f’(x) = -x/sqrt(25-x^2)
b)f’(-3) = 3/sqrt(25 - 9) = 3/4
at x=-3, f(3) = 4
y-4 = (3/4)(x+3) <---- This is an acceptable answer
c)lim g(x) = f(-3) = 4
x->-3-
lim g(x) = -3 + 7 = 4
x->-3+
Continuous, because the limit at x=-3 exists and is equal to g(-3).
d) antiderivative is [(25-x^2)^(3/2)]/3 (+ C, I know, but for definite integrals the constant of integration is useless), at x = 5, antiderivative = 0. At x = 0, antiderivative = 125/3. F(5) - F(0), so answer is -125/3.</p>
<p>5a) 40 grams, because dB/dt is larger at 40 grams than at 70 grams, which is the rate at which the bird gains weight.
b)(-1/5)dB/dt = (-1/25)(100-B)
For all values of B less than 100, B(t) is concave down. Since this graph stays below B = 100, it should always be concave down, but it is concave up towards the beginning.
c) ln(100 -B) = (t/5) + C
ln(80) = C
100 - B = 80e^(t/5)
B(t) = 100 - 80e^(t/5)</p>
<p>6a) Particle moves to left when velocity is negative.
cos(pi * t/6) = 0
t = 3, 9
v(t) is negative on the interval (3,9)
b)The definite integral from x=0 to x=6 of the ABSOLUTE VALUE of v(t) or cos(pi * t/6) (either should be acceptable)
c)a(t) = (-pi/6)sin(pi *t/6)
a(4) = -(pi/6)sin(2pi/3) = -(sqrt(3) * pi) /12
velocity is decreasing, since a(4) is negative, and a(4) is the instantaneous rate of change of the velocity.
d) x(t) = -2 + int(v(t), 0, 4)
antiderivative is 6sin(pi *t/6)/pi
integral = [6sin(2pi/3)/pi] - [6sin(0)/pi] = 3sqrt(3)/pi - 0 = 3sqrt(3)/pi
Answer is [3sqrt(3)/pi] - 2, because particle starts at x=-2</p>
<p>Again, say if anything is terribly wrong. I actually noticed a couple of mistakes I made in making this. >_></p>
<p>there’s a form b to the test right? Are they posting that because I didn’t have a bird question…and how easy are the graders on giving points on the frq?</p>
<p>On 1d)</p>
<p>I did the integral from 0 to 5 + 71. My thinking was that since the new function given represented the rate after t=20 of the original data, its x=0 value would be the rate of the new temperatures after t=20. The answer comes out 73.924 Anyone else do this?</p>
<p>If what FastNeutrino says is all right, then there is hope for me!
For 2a: Was it necessary for us to put the integral on the second part because wasn’t the second part simply a triangle that we could find without the integral?
LOL I probably only got 1 or 2 points on the dumb graph one. Goodness i hate those problems.</p>
<p>Various AP Calculus teachers are posting their solutions to the FRQs on this forum here:
[Math</a> Forum Discussions - ap-calculus](<a href=“Classroom Resources - National Council of Teachers of Mathematics”>Classroom Resources - National Council of Teachers of Mathematics)</p>
<p>@dood1234: The question says “20<t<25”, and t is the variable used in the derivative, so you have to use 20 to 25.
@Jordan134: I would accept that if I was the grader, but the integral is probably more of what they’re looking for.</p>
<p>Hey I’m taking the test this year and can someone answer this question. If you forget to put the estimation sign (the 2 squiggly equal signs ~) on the frq when estimating with a calculator, will you get any points off? thank you</p>
<p>what was the curve when you took it? what raw score did you get and what score did you get out of 5?</p>