<p>Can anyone do the FRQ’s (correctly) and show us the answers?</p>
<p>and maybe approximate the scoring distribution? :3</p>
<p>^^^^yes!!!
That would be greatly appreciated</p>
<p>WOW! This free response is like a million times easier than the international one -_- Lucky…</p>
<p>Anyways</p>
<p>Q1A) Pretty simple, A(30) - A(0)/30 lbs/day [y’all can do the arithmetic]
Q1B) A’(15) = -0.1636 lbs/day [TI-84 Evaluation]
Q1D) A’(30) = 0.004 and A(30) = -0.056
y + 0.056 = 0.004(x - 30)
0.556 = 0.004(x - 30)
x = 169</p>
<p>ILL <strong>SETUP</strong> the integrals, only SETUP</p>
<p>Q2A) V = pi * INT[36 - (f(x) + 2)^2 dx] from 0 to 2.3
Q2B) V = (1/2)INT[(4-f(x))^2 dx] from 0 to 2.3
Q2C) INT[4 - f(x) dx] from 0 to k = INT[4 - f(x) dx] from k to 2.3</p>
<p>NOTE* There are other ways to do these, but I just wrote the easier methods. You can use f(y) instead of f(x) for these.</p>
<p>Q3A) g(3) = (1/2)(4)(3) + (1/2)(4)(2) - (1/2)(1)(1) [Do the math]
Q3B) g’(x) = f(x) and g"(x) = f’(x)
g’(x) > 0 when f(x) > 0, thus the interval is -5 < x < 2
g"(x) < 0 when f’(x) < 0, thus the interval is -5 < x < -3 & 0 < x < 4
Since you need to consider both, g’(x) and g"(x) the final interval is -5 < x < -3 & 0 < x < 2 </p>
<p>Thats all for now, Im running out of time. </p>
<p>If anyone starts a document or something I will go over them all.</p>
<ol>
<li><p>a) (A(30) - A(0)) / (30-0) = -5.904/30 = -0.197</p>
<p>b) Using a calculator, graph A(x) and find the derivative at x = 15. The answer is -0.165 pounds/day. The context is: At time t = 15 days, the amount of grass remaining in the bin is decreasing at a rate of 0.165 pounds per day.</p>
<p>c) A(t) = 1/30 * integral from 0 to 30 of A(t)
6.687(0.931)^t = 2.753
t = 12.415 days</p>
<p>d) m = A’(30) = -0.0560
y = -0.0560x + b
A(30) = -0.0560*30 + b
b = 2.463
y = -0.0560x + 2.463
0.5 = -0.0560x + 2.463
x = 35.052 days</p></li>
</ol>
<p>^I think part c for number 1 is 12.428</p>
<p>I fixed it but I got 12.415. Gotta carry all your digits until the very end (which I didn’t do for some reason on the exam…)</p>
<p>oh. would they take off points for that difference?</p>
<p>2) a) Volume = pi * (integral from 0 to 2.3 of (36 - (x^4 - 2.3x^3 + 6)^2)dx = 98.868</p>
<pre><code>b) Volume = integral from 0 to 2.3 of A(x) where A(x) = 0.5 * (2.3x^3 - x^4)^2. The answer is 3.574
c) Integral from 0 to k of (2.3x^3 - x^4) = integral from k to 2.3 of (2.3x^3 - x^4) = .5 integral from 0 to 2.3 of (2.3x^3 - x^4)
</code></pre>
<p>@Rookie97 I REALLY want to say no because I lost like 3 points in total from being stupid, but I’m pretty sure it’s 1 pt for each incorrect answer. They are picky :(</p>
<p>One of my FRQ’s was about a zoo and logistic growth, and in fact none of the released questions match with the questions I did. Is it possible that I got a different form even though I live in the US and not in Hawaii or Alaska?</p>
<p>3) a) g(3) = integral from -3 to 3 of f(t) = 1/2 * 5 * 4 - 1/2 * 1 * 2 = 9</p>
<pre><code>b) If it’s increasing, g’(x) will be positive, meaning that f(t) will be positive. f(t) is positive on (-5,2)
g(x) is concave down when g’’(t) is negative, meaning that f’(t) will be negative. This is on (-5,-3) and (0,4)
Combining them both, g(x) is increasing and concave down on (-5,-3) and (0, 2).
c) h’(x) = (5xg’(x) - 5g(x)) / 25x^2
h’(3) = (15 * g’(3) - 5g(3)) / 225
= (15 * -2 - 5*9) / 225 = -75/225 = -1/3
d) I didn’t get that one sorry 
</code></pre>
<p>@nolasola The CollegeBoard administers different forms of each exam and typically only released Form O to the public. You got a different form. ;)</p>
<p>@kingofxbox99 thank you was freaking out for a moment</p>
<p><a href=“http://www.viewdocsonline.com/document/mryltt”>http://www.viewdocsonline.com/document/mryltt</a></p>
<p>I’ve typed up the solutions to BC. There are probably some mistakes – please, point them out!</p>
<p>4) a) Average acceleration = (v(8) - v(2)) / (8 - 2) = (-120 - 100) / 6 = -220/6 = -110/3 m/sec^2</p>
<pre><code>b) Yes, since at t = 5 the velocity is 40 and at t = 8 the velocity is -120. Since v(t) is differentiable (and therefore continuous), the intermediate value theorem dictates that there must be a time t inbetween 5 and 8 where the velocity is inbetween 40 and -120. -100 is in that interval.
c) Position = 300 + integral from 2 to 12 of v’(t)dt
Trapezoidal approx: 1/2 * 3 * (100 + 40) + 1/2 * 3 * (40 - 120) + 1/2 * 4 * (-120 + -150)
= 210 - 120 - 540 = - 450
Position = 300 - 450 = -150m from Origin Station.
d) d^2 = A^2 + B^2
2dd' = 2AA' + 2BB'
dd' = AA' + BB'
d = sqrt(300^2 + 400^2) = 500 (3-4-5 triangle)
A' = 100m/s
B' = -5(4) + 120 + 25 = 125m/s
500 * d' = 100 * 300 + 125 * 400
d' = 160m/s
</code></pre>
<p>BTW, my solutions are for AB. That link is for BC. :)</p>
<p>Yup – I took BC this year, hence made the solutions for it. </p>
<p>3) d) find p’(-1)
p’(x)= f’(x^2-x)(2x-1)
p’(-1)= f’(2)(-3)=(-2)(-3)=6
slope=6</p>