<p>I’m still pretty confident I’ll get a 5 even after looking over those questions again. Except I realize that I forgot the domain on 6c and most of my justifications/reasons were bs.</p>
<p>
</p>
<p>How?..lol</p>
<p>You didn’t need the domain on 6c.</p>
<p>definitely, form B was much harder than form A.
the international one for AB.</p>
<p>Just how lenient are the graders? I feel like I seriously ****ed this up…my brain literally malfunctioned for the most part of the exam (damn mornings)</p>
<p>Can someone please be kind enough to solve those problems and show the work/explanations?</p>
<p>6c - separate the derivative
dy / y^3 = x dx
integrate
-1/(2y^2) = (x^2)/2 + C
i solved for C here and got -5/8
do some algebra and you get y= sqrt (-1/(x^2-5/4))
plug in 1 to check and you get 2, so it works</p>
<p>Edit: Also, for 5b, aren’t x=2 and x=3 inflection points as well as x=0? The derivative of the graph doesn’t exist at x=2 and x=3 (which means those points are critical points) and the slope changes signs at those points as well.</p>
<p>Here are my tentative solutions for BC. Ran through it in about 35 minutes today, so these might be off.</p>
<p>1.(a) 142.274 or 142.275
(b) -59.583
(c) h(t) = 0 for 0 <= t < 6 ; h(t) = 125(t-6) for 6 <= t < 7 ; h(t) = 125 + 108(t-7) for 7 <= t < 9
(d) 26.335</p>
<p>2.(a) E’(6) = [E(7) - E(5)]/(7-5) = 4 hundred entries per hour
(b) 10.6875; Between noon and 8 p.m., an average of 1068.75 entries per hour were received for the contest.
(c) 6950 entries not yet processed (approximately)
(d) t = 9.184, since P’(t) changes from positive to negative at this point</p>
<p>3.(a) 2sqrt(2) m/sec.
(b) 11.588
(c) t = 2.208, particle moving to the right since dx/dt|t = 2.208 > 0
(d) (i) At times t = 1 and t = 3
(ii) dy/dx| t= 1 = .432; dy/dx|t = 3 = 1
(iii) y(1) = y(3) = 4</p>
<ol>
<li><p>(a) 18
(b) pi * integral (0, 9) of [(7 - 2sqrt(x))^2 - 1^2] dx
(c) integral (0, 6) of (3y^4/16) dy</p></li>
<li><p>(a) y = -5/4
(b) 1/3
(c) y = 1 - e^(1-x)</p></li>
<li><p>(a) cos x = 1 - x^2/2! + x^4/4! + … + (-1)^n<em>x^(2n)/(2n)!
[cos x - 1]/x^2 = -1/2! + x^2/4! - x^4/6! + … + (-1)^n</em>x^(2n-2)/(2n)!</p></li>
</ol>
<p>(b) relative min since f’(x) = 0 and f"(x) = 2/4! > 0, by the Second Derivative Test</p>
<p>(c) g(x) = 1 - x/2! + x^3/(4!<em>3) - x^5/(6!</em>5)</p>
<p>(d) g(1) = 37/72. By the Alternating Series Remainder, the maximum error is less than the first omitted term, which is 1^5/(6!*5), which is less than 1/6!</p>
<p>I feel really bad after taking this test…but whatever.
BC answers:</p>
<p>1a) integral from 0 to 6, f(t)dt
b) f(8) - 108 haha probably wrong
c) 0 for 0<t<6
integral from 6 to t of 125 for 6<t<7
(integral from 6 to 7 of 125 for 6<t<7) +(integral from 7 to t of 108 for 7<t<9)
d) (integral from 0 to 9 f(t)dt) - (integral from 6 to 7 of 125 for 6<t<7) +(integral from 7 to 9 of 108 for 7<t<9)</p>
<p>2a) secant line of (5,13) and (7,21)
b) i don’t feel like writing this.
c) trapezoidal sum from b) minus integral from 8 to 12 of P(t)
d) set P(t) equal to 0 and found maximum.</p>
<p>3a) plugged 3 in for dy/dx (?)
b) arclength formula: integral from 0 to 4 of the sqrt of (dx/dt)^2 + (dy/dt)^2
c) i forget. haha it was probably ****ty though.
d) i. set x(t) equal to 5
ii. plug values of t into (dy/dx)
iii. I don’t think I had time to finish this, but I knew how to do it :(</p>
<p>4a) integral from 0 to 9 of (6 - 2rt(x))dx
b) pi*integral from 0 to 6 of {(7-2rt(x))^2 - 1}dx
c) I don’t feel like doing this…</p>
<p>5a) I remember I got -1.25
b) 1/3 (L’Hopital’s Rule)
c) I don’t remember. Probably screwed it up though…</p>
<p>6) ■■■.</p>
<p>Well there you go. Tell me if I failed?</p>
<p>On AP Calc BC for question 5, how did you do 5b, without doign 5c first? I was really confused</p>
<p>
I’ll fix these.</p>
<p>5.
A.) G(3) = 5 + pi + (3/2) = 13/10 + pi; G(-2) = 5 - pi
B.) Inflection points at x = 0, 2, 3. At these points g’‘(x) changes sign because the slope of the graph of g’(x) goes from either positive -> negative or negative -> positive.
C.) H(x) = G(x) - (1/2)x^2 so H’(x) = G’(x) - X. H’(x) = 0 at x = sqr(2) and 3. (Just envision the graph of y = x on the g’(x) graph and anytime the two graphs intersect is where H’(x) would equal 0). X = sqr(2) is a maximum because H’(x) goes from positive to negative and X = 3 is neither because H’(x) is negative on both sides.</p>
<p>@lemone, use the L’Hopital’s Rule. So you need to find lim( x goes to 1) of (dy/dx)/(3x^2). You know that f(1) = 0, so plug 0 into dy/dx and you get dy/dx = 1. The denominator is 3. So you get 1/3.</p>
<p>
</p>
<p>Lmao i did it the harder way lol, I actualyl did 5c first and then use that f(x) into the limit, but I got the answer nonetheless lol</p>
<p>Also @lemone, the two were totally independent. I did 5c before 5b because I wasn’t sure how to do 5b at first (then got it thankfully). But part 5c is just finding a particular solution to the differential equation. So just integrate and solve for C using the given initial values.</p>
<p>Yes!!! Thanks jersey, I really needed to know if I got 5 right. </p>
<p>What was 1c though? That was the only other place I could possibly have lost points.</p>
<p>what was 2d? I put 9.184, but apparently I forgot to check the endpoints and it’s actually t = 12? can anyone confirm this?</p>
<p>For 2d, I just graphed it and saw that the rate was greatest at t=12. Then I went back and looked over the calculator section when I was done with non-calculator and realized that I should probably have some justification. I found the critical points of the function which looks really sloppy without a calculator (used the quadratic equation). But basically, t=12 was the maximum and the rate was 8 hundreds of entries per hour. To find the amount of entries processed, take the integral of the equation from 8 to 12, and you will get 16. So 23-16= 7 hundreds of entries were still not processed.</p>
<p>1a)integral from 0 to 6 of f(t) dt= 142.275
1b) f(8) - 108 = -59.583
1c)I did this one wrong
1d) integral from 0 to 9 of f(t) -g(t) dt = 26.335
2a) (f(7) - f(5))/(7-5) = 400 entries per hour
2b) (4+25.5+34+22)/8 = 10.6875 did the meaning wrong
2c)Since P(t) is the the rate of entries being processed and E(t) is the amount of votes E(8) - integral from 8 to 12 of P(t) so 23-16 so 700 votes not processed
2d)Since P(t) is the rate you find where P(t) has a maximum which is at t= 12 reason: four candidates where P’(t) = 0 and the end points. I listed the four points and basically said that at P(12) the rate is the highest.
3a) integral from 0 to 3 of r(t) = 3200
3b) increasing because there are more people in line at t= 3 (1500) then at t=2 (1300)
3c) at t=3 there are 1500 people in line. reason: because after t=3 the rate of people arriving at the ride per hour is less than the amount of people riding the ride per hour
3d) integral from 0 to t of 800 dx - integral from 0 to t of r(x) dx = 0
4a) integral from 0 to 9 of 6 dx(54) - integral from 0 to 9 of 2sqrt(x) dx (36) so 54-36 = 18 (top function minus the bottom function)
4b)pi integral from 0 to 9 of (7-2sqrt(x))^2 - 1 dx (washer method bigger radius squared minus the smaller radius squared)
4c) x= (y/2)^2 H=3L so A=3L^2 so 3 multiplied by the integral from 0 to 6 of (y/2)^4 or y^4/16.
5a) g(3) = g(0) + integral from 0 to 3 of g’(x) dx = 6.5 + pi
g(-2) = g(0) - integral from -2 to 0 of g’(x) dx = 5-pi
5b)x= 0,2,3 reason: 0 inc to dec; 2 dec to inc; 3 inc to dec
5c)Critical points are at x = sqr(2) and 3. sqr(2) is a maximum and 3 is neither. There is no sign change at x = 3 for h’(x). (Jersey13)
6a) m = xy^3 f(1)=2 so m=8 y = 8x - 6
6b) (a) f(1.1) = 2.6 my reasoning is wrong
6c) 1/(-2y^2) = (x^2)/2 + C
y^-2 = -x^2 + c
y^2 = 1/(-x^2 +c)
y = sqrt(1/(-x^2 + c))
y = sqrt(1/(-x^2 + 5/4))
x is less than sqrt(5/4)</p>
<p>I just did all these out so sorry if there are any errors I’m eating dinner
I’ll edit errors out or if someone can add</p>
<p>
Critical points are at x = sqr(2) and 3. sqr(2) is a maximum and 3 is neither. There is no sign change at x = 3 for h’(x).</p>
<p>Ironically, I got the series question 100% correct, yet i missed a few easy parts of easier questions. boo</p>