<p>Sorry about the Euler mistake guys – I’ll fix that soon. I didn’t actually take the Calc BC test this year, I took it last year. Some of my friends wanted the answers so I worked through the whole test at home in about 40 minutes. Funny how arithmetic is still the most difficult area of math for most people including myself :P</p>
<p>Perhaps I can’t see the negative signs, but the answers should go as the following for 5</p>
<p>a) 2
b)-1/2 + 5/8
c) -1/ x^2 + 2x + 1</p>
<p>JamisonSloan, the dropbox pdf viewer doesn’t show the negative sign, but when i directly downloaded it, it showed it. #5 is correct but you have the download to pdf the see the correct answers</p>
<p>Jamison’s Euler solution is right guys…</p>
<p>for 3b I did it really weird. I found the rate of change at t = 2 to be (11.2 - 5.3) / (3-1) and got 2.95, then I found the rate of change at t = 4 to be 1.3.</p>
<p>Then I said that since derivative of C at 2 is 2.95 and derivative of C at 4 is 1.3, according to the Intermediate Value Theorem 1.3 < 2 < 2.95 so there has to be a value at which the derivative of C is 2.</p>
<p>Just wondering, would I receive any credit for this?</p>
<p>FRQs came out, but the scoring guide didn’t 
would’ve been nice</p>
<p>The answer to 5c. should be -11/32. Jamison’s solutions were perfect, he just forgot to multiply (1/4)(f’(1/4)) though he displayed his intention to do so. Thank you for creating them, jamison 
Your solutions confirmed my answers!</p>
<p>How could 5C be -11/32 when the problem asks for an equation??</p>
<p>Well they have to come up with a curve for the point distribution… apparently to set it up in the majority of students’ favor. right…</p>
<p>Scoring guidelines come out late July.</p>
<p>Do you guys think the cutoff for a 5 will be higher than 70 out of 108?</p>
<p>PGD2013:</p>
<p>First off, you’re my hero. Second, for #4, are you sure they asked a maximum? I remember finding the minimum, which was -sqrt(1/2)</p>
<p>His answer for Euler’s method was right</p>
<p>Edit:
The answer to 5b. should be -11/32 ((-1/2)+((1/4)(5/8)). Jamison’s solutions were nearly perfect, he just forgot to multiply (1/4)(f’(1/4)) though he displayed his intention to do so. Thank you for creating them, jamison*Your solutions confirmed my answers!</p>
<p>Euler answer revised, and some formatting cleaned up a bit.</p>
<p><a href=“Dropbox - Error - Simplify your life”>https://www.dropbox.com/s/nmo6mju19sqe05j/BC%20Solutions%202013.pdf?m</a></p>
<p>@Xterminate:
That’s how I did it too! I don’t know if they’ll count it, though. I didn’t even think of using MVT.</p>
<p>Also, how much credit would I get if I screwed up the Euler’s problem by getting -1 + 1/2 = -3/2? I can’t do basic addition…</p>
<p>----- 5b -----</p>
<p>dy/dx = y^2(2x + 2)</p>
<p>delta x = 1/4
delta y = m * delta x</p>
<p>first point - (0, -1)
m = 2
delta x = 1/4
delta y = 2 * 1/4 = 1/2</p>
<p>second point - (1/4, -1/2)
m = (-1/2)^2 * (1/2 + 2) = 1/4 * 5/2 = 5/8
delta x = 1/4
delta y = (5/8) * (1/4) = 5/32</p>
<p>f(1/2) = (-1/2) + (5/32) = -11/32</p>
<p>I can’t remember if I did this right on the actual test and I’m dying right now, I neeed a 4 
oh well YOLO man I give 0</p>
<p>JohnAndTequila:
I’m pretty sure they asked for the absolute maximum. However, you would be correct in showing that -1/sqrt(2) was the absolute minimum. Here are a few corrections on my answers that I posted: in question two (the rover one) the equations given were velocity equations, not position equations. So it should be x’(t) = -12 sin (2t^2) and y’(t) = 10 cos (1 + sqrt(t)). But all in all, I hope I did good. I have a quick question though. Do you remember getting 2pi - (1/2) or 2pi - 1 on question 5? It was part (a) I think. I asked my friend what he got and he said he got 2pi - 1 and I was scared he was right. I’m curious as to what you remember putting. Thanks.</p>
<p>@some11no nice, Intermediate Value Theorem’s where it’s at!! I mean they have to give us full credit since it works, right? They better. And if the Euler problem is worth two points, you would probably earn 1 for identifying Euler’s method and showing the right way to do it and stuff, you wouldn’t get the other point though because of a wrong answer.</p>
<p>I’m probably gonna have nightmares tonight about having to sit through a whole two semesters of Calc 1 and 2 at college all because I screwed up a few problems on a stupid test</p>