2010 BC FRQ Solutions

Test Preparation AP Test Preparation
1

<p>First crack. Feel free to correct.</p>

<p>(1)
(a) int(f(t),0,6) = 142.274 ft^3
(b) f(8) = 48.417 ft^3/hr
(c) ...{ 0 0<t<5
h(t)={ 125(t-6) 6<t<7
........{ 125 + 108(t-7) 7<t<9
(d) int(f(t),0,9) - h(9) = 26.334 ft^3</p>

<p>(2)
(a) E'(6) = 4 hundred entries per hour
(b) 1/8<em>int(E(t),0,8) ~ 10.6875
</em><em>Over the interval 0<t<8, there was an average of 1068.75 entries in the box.
(c) int(P(t),8,12) = 16
*</em><em>8</em>1068.75 - 1600 = 6950 entries yet to process.
(d) P'(t) = 0 => t = 9.183 or t = 10.816
<em>*P(8) = 0
*</em><em>P(9.183) = 5.088
*</em><em>P(10.816) = 2.911
*</em><em>P(12) = 8
*</em>*Entries were processed most rapidly at t = 12.</p>

<p>(3)
(a) x'(t) = 2t - 4; x'(3) = 2; y'(3) = 2
<em>**|v(3)| = sqrt(x'(3)^2 + y'(3)^2) = sqrt(8)
(b) int(sqrt(x'(t)^2 + y'(t)^2),0,4) = 11.587
(c) y'(t) = 0 => t = 2.207
*</em><em>x'(2.207) = .415 > 0
*</em><em>Since x'(2.20) > 0 the particle is moving right.
(d)
(i) t^2 - 4t + 8 = 5 => t = 1 or t = 3
(ii) dy/dx|t=1 = y'(1)/x'(1) = .432
*</em>*dy/dx|t=3 = y'(3)/x'(3) = 1
(iii) int(y'(t),2,3) = y(2) - y(3) => y(3) = 4</p>

<p>(4)
(a) int(6 - 2x^.5,0,9) = 18
(b) pi<em>int(36-4x,0,9) = 162</em>pi
(c) 3/16*int(y^4,0,6) = 1458/5</p>

<p>(5)
(a) [Euler's method with step size -.5] f(0) ~ -5/4
(b) lim(f(x)/(x^3 + 1),x->1) = lim(f'(x)/3x^2,x->1) = 1/3
***because dy/dx|x=1 = dy/dx|y=0 = 1
(c) y = -e^-x + 1</p>

<p>(6)
(a) f(x) = -1/2! + 1/4!x^2 - 1/6!x^4+...+(-1)n<em>x^(2n)/(2n)!+...
(b) f'(0) = 0; f''(0) = 2!/4! = 1/12 > 0
</em><em>There is a relative minimum at x = 0 by the 2nd derivative test
(c) g(x) ~ 1 - 1/2!x + 1/(4!</em>3)x^3 - 1/(6!<em>5)x^5
(d) g(1) ~ 1 - 1/2 + 1/(24</em>3) = 37/72
*<em>|error| < 1/(6!</em>5) < 1/6!</p>

2

<p>there are several significant errors. See the math prof’s post in the other thread</p>

3

<p>

from Math Prof</p>

4

<p>He is correct on 2c) and 4b). I am fairly certain I am correct in the other discrepencies. For example, his result in 1b) is impossible since f(t)>) for t>0; his result in 2d) is a classic error in neglecting the endpoints; his interpretation in 2b) is incorrect (the units of the quantity are entries, not entries/hr), etc.</p>

<p>But thanks for pointing out 2c) and 4b). Will edit 1st post.</p>

5

<p>■■■ 10char</p>

6

<p>on 2c), I don’t see why you couldn’t just do 2300-1600= 7000 left. There were 2300 tickets in the box at 8pm</p>

7

<p>

This is what I did and what you were supposed to do.</p>

8

<p>^good,
****, i forgot the endpoints. Thanks for point that out OP</p>

9

<p>@pigs: You are absolutely correct on 2c). Wow. I blew that twice. Thanks for the the assist.</p>

<p>I can’t edit 1st post any more, so can’t correct 2c) 2300-1600 = 700 or 4b) pi*int([(7-2x^.5)^2 - 1],0,9)</p>

10

<p>What if for 4A I didn’t have time to simplify my answer down to 18? I got 54- 4/3 sqrt (719) which is essentially 18… will I be penalized?</p>

11

<p>You are fine if you meant sqrt(729). Numerical values need not be simplified on the ap test, so full credit if 719 was a typo.</p>

<p>I am seeing others report 1b) as -53ish, but I am certain that that is incorrect. I assume this is f’(8). However, f is clearly defined to be the rate at which snow is accumulating, so the correct answer is f(8). If this was not true, then all answers involving the integral of f would be incorrect. Can’t have it both ways.</p>

12

<p>I stand corrected. 1b) is f(8) - g(8) = -59.582</p>

13

<p>For 1b, it wanted the rate of change, I thought it was f(9) - f(7)/2</p>

14

<p>@tempacct</p>

<p>I’m pretty sure that’s right. Your initial answer for 2c) used the estimation from your trapezoidal sum, while the accurate value was just in the table, so anyone who did that would have gotten partial credit. 2c) will likely be worth 2 points.</p>

<p>btw, 5c) is 1-e^(-x+1)</p>

15

<p>@tempacct: In the other thread, I’d already caught my error with 2d, although that catch wasn’t transferred over here. Looking more closely at 2b, indeed, it appears that E(t) is the number of entries and not the number of entries/hour (which is nowhere near as much fun :slight_smile: ), which indeed supports your answer.</p>

<p>With regards to 1(b), I stand by my answer: the rate of change of the volume of the snow on the driveway at 8 a.m. is not f(8), but f(8) - g(8), since at this time, snow is both falling and being removed from the driveway.</p>

16

<p>I know it’s hard to say with the MC being added in too, but so far I think I might pull off about…</p>

<p>7, 9, 9, 9, 9, 2 maybe? Would that be good shape for a 5?</p>

<p>And have we confirmed 2c to be 700?</p>

17

<p>Funny that the only two parts I got wrong were the first two parts of 4. I didn’t notice the x-coordinate was 9 instead of 6 -___- I got the integrands right at least though</p>

<p>Also, if I left 6d unsimplified (as the sum of fractions) will that matter?</p>

18

<p>Thanks so much for putting this together. I put 700 for 2c and I think that is right.</p>

19

<p>Yeah, 2c) is 700. I misread all of #2 badly. :)</p>

<p>And if you left 6d as the sum of fractions, you’re fine.</p>

20

<p>Wait, can you post the updated answers to all of #2 then?</p>