<p>I think since it says “in terms of k” there won’t be an exact number answer…so I don’t remember exactly if I had solved the integral or if I got it down to something with k.</p>
<p>1a. sqrt(y’(3)^2 + x’(3)^2) = 13.007
1b. slope = y’(3)/x’(3) = 0.0317
1c. position = ( 0 + ∫x’(t)dt, -4 + ∫y’(t)dt ) = (21, -3.226)
1d. distance = ∫sqrt(x’(t)^2 + y’(t)^2)dt = 21.091</p>
<p>2a. (F(b) - F(a))/(b-a) = (F(5) - F(2))/(5-2) = -8/3 °C/min
2b. Average temperature in °C of the teapot for the first ten minutes of its cooling phase; I’m too lazy to post the trapezoidal calculations but the answer is 52.95 °C
2c. Temperature difference in °C of the tea after 10 minutes of cooling; 43 - 66 = -23, so 23 °C temperature difference
2d. 100 + ∫B’(t)dt on the interval [0,10] = 34.183 °C; 43 - 34.183 = (to the nearest degree) 9 °C, so the biscuits are 9 °C cooler than the tea</p>
<p>3a. P(k) = 1 + k + e^2k + ∫sqrt(1 + 4e^(4x))dx on the interval [0,k]
3b. V(k) = pi∫(e^(2x))^2 dx on the interval [0,k]; V(x) = (pi/4)(e^(4k) - 1)
3c. dV/dt = pi<em>e^(4k)(dk/dt); dV/dt = pi</em>e^(2k)/3</p>
<p>4a. g(-3) = -6 - 9pi/4; g’(x) = 2 + f(x); g’(x) = 2
4b. x = 5/2, I don’t feel like showing the justifications for this one right now…
4c. g’‘(x) = f’(x), and the only place on f where slope changes from positive to negative is at x=0. Therefore, g has a point of inflection at x = 0.
4d. (F(b) - F(a))/(b-a); (F(3) - F(-4))/(-4-3); (-3 - -1)/(-4 -3) = -2/7; MVT is not violated since MVT requires that f be continuous on [a,b] and differentiable on (a,b); f is not differentiable at x=0</p>
<p>5a. W - 1400 = 44t; W = 1400 + 44(1/4); W = 1411 tons
5b. d^2W/dt^2 = (1/25)dW/dt; d^2W/dt^2 = (W-300)/625; d^2W/dt^2 is concave up on [0,20] since W(0) = 1400 and W is increasing on its domain, so the tangent line underapproximates the true amount of solid waste
5c. Separation of variables, blah blah blah, W(t) = 300 + 1100e^(t/25)</p>
<p>6a. sin(x): x - x^3/3! + x^5/5! - x^7/7!
sin(x^2): x^2 - x^6/3! + x^10/5! - x^14/7!
6b. cos(x): 1 - x^2/2 + x^4/4! - x^6/6!
sin(x^2) + cos(x): 1 + x^2/2 + x^4/4! - 121x^6/6!
6c. f(6)(0) = -121
6d. Blah blah blah Lagrange error = 1/4096 if using |f(5)(z)|max = 30 as your estimate, where z = 1/4</p>
<p>My (known) mistakes on the exam: forgetting about the arc length formula in 1d, although I still got 21.04 as an answer by using absolute value integrals and the Pythagorean theorem; dropping the negative sign on 6c</p>
<p>What if I put this for 6 part b:</p>
<p>cos(x)+sin(x^2) = 1+(1-1/(2!))x^2+x^4/(4!)-(1/(3!)+1/(6!))x^6</p>
<p>Also on part c I did -6!(1/6! + 1/3!) again I had a brain fart and didn’t think of a way to simplify it.</p>
<p>Same here i had less than 3 minutes left and I saw that I had to combine the fractions so I just went t the next part of the question haha</p>
<p>Guys, here are all the solutions:</p>
<p>[AP</a> 2011](<a href=“http://home.roadrunner.com/~askmrcalculus/11AP%202011.html]AP”>http://home.roadrunner.com/~askmrcalculus/11AP%202011.html)</p>
<p>Looking through the posted solutions makes me feel awful. I know I probably got around 7-8 points on FRQ #1, 7 points on #2, probably 6 on #3, 5 on #4, a grand total of 1 on #5 (lol ;)), and like 2 points for #6. Berkeley gives credit even for 3s, but I was hoping for as high a score as possible >.< Is a 5 totally out of the question for me? I think I’m being very harsh on myself in terms of the FRQs but I still want a chance at one…</p>
<p>I literally think I got every single one right…</p>
<p>^ Way to post that right after me and make me feel even more like an idiot… :p</p>
<p>Trust me…you aren’t the only one. FRQ 5 and 6 were bad for me as well. Praying for that 5 still…</p>
<p>That website is wrong for 3a, I’m almost positive. f’(x) = 2e^(2x), so squaring that would be (2e^2x)(2e^2x) = (4e^4x).</p>
<p>Yeah, my answers are all the same as Keasbey’s. I may have gotten a solid 54/54, barring any errors in my memory.</p>
<p>Actually, I just took a closer look at Keasbey’s answers and found out that I have basically the same things for #1! So assuming all of it is right, I have a 9 for that problem. 
Now I don’t feel so bad!</p>
<p>Dpes anyone rember thepolar curve mc? How do you do that one?</p>
<p>Look up the 2007 FRQ’s, there’s a question just like the polar curve one. Basically you had to add the area of the half circle to the smaller area of the polar curve.</p>
<p>but why they were just aksing for the shaded region shouldnt it just be the area of the cartioid</p>
<p>Think of it this way: the y-axis splits the shaded region into two parts, a half-circle and a section of a cartioid. You have to add the areas of each part together.</p>
<p>What were the bounds for the integral on the polar one?</p>
<p>-pi/2 to pi/2</p>
<p>so it was 2pi plus the integral of that thing?</p>
<p>Yeah it was</p>