<p>I made a stupid mistake on 1c and i wanna kick myself for not checking over it or else i woulda caught it..should only be 1 pt tho...</p>
<p>2a, what do you mean by straightforward?</p>
<p>I confirm everything else.</p>
<p>it was just the integral of the line from 0 to 18...pretty basic...use calc</p>
<h1>3</h1>
<p>A)
g = 16
g' = 0
g'' = -2</p>
<p>then
B) i got neither cause my slope was -2</p>
<p>C) then i gpt G(10) = 15</p>
<p>and G(108) = y-2 = 224 (x-108)</p>
<p>i think i got b and c wrong but any one for A?</p>
<p>got owned by 3, missed 1 on number 1, missed 3 on number 5</p>
<p>hugh, the slope at x=108 was 3, not 224. Remember that the graph is a period, so it repeats?</p>
<p>I thought the slope was 2... I believe I got y = 2x-296 or something, but I am pretty sure I messed it up.</p>
<p>yah i bit the bullet on #3 b and c</p>
<p>how bout a though</p>
<p>yeah two that's what I meant.</p>
<p>108 is 21 times of 0 and three remaining, so the slope is the same as on x=3.</p>
<p>so no one for question 3 part a?</p>
<p>3a.</p>
<p>g(4)=3
g'(4)=0
g''(4)=-2</p>
<p>Also 3b is a relative minimum, not neither.</p>
<p>if I got 38/45 MC and say 34/54 FR, will that be a 5?</p>
<p>
[quote]
3a.</p>
<p>g(4)=3
g'(4)=0
g''(4)=-2</p>
<p>
[/quote]
</p>
<p>Yep. :)</p>
<p>I worked hard in Calc this year but did not do too much real practice beyond doing/looking through no more than two MC practice tests in my AP book. I've read through this entire board and although feeling down immediately following the test, today I feel pretty confident. Did not get all the MC or FR, and did not think that it was easy.. though overall it could have been much worse.</p>
<p>I'm convinced I would've been close to perfect if I had even studied one hour more the day before. But oh well.</p>
<p>Don't remember problems anymore, and have no motivation to look through FR questions again at the moment. Will contribute tomorrow.</p>
<p>Yeah I got what you got, giants.</p>
<p>crosscurrent yes thats a 5 with room to spare if you did that well on the MC.</p>
<p>how did u get that for a) i got a+2</p>
<p>How did you guys do the THE DIFFERENTIAL EQUATION QUESTION!!!</p>
<p>Omg, everyone I asked said they ended up with a LINEAR equation, I being ONE OF THEM...
and the stupid thing was that I realized LATER that the SLOPE FIELD was nowhere NEAR that of a linear function! I was kinduv mad at myself, but really, what WAS IT? and how did you solve it</p>
<p>Can someone go thru the process...I remember the problems:</p>
<p>(integral) (dy / ( 1 + y)) = (integral) (dx / (1 + x))</p>
<p>PLEASE? thanks</p>
<p>wat else did u guys put for #6?</p>
<p>Manu, the equation was actually dy/dx = (y+1)/X so.. and we were given f(-1) = 1</p>
<p>Int(dy/(y+1)) = Int(dx/x))
ln|y+1| = ln|x| + C</p>
<p>Plug in our given values:</p>
<p>ln2 = ln1 + C
ln2 = C</p>
<p>ln|y+1| = ln|x| + ln2</p>
<p>y-1 = e^(ln|x| + ln2)
y -1 = 2e^(ln|x|)
y = 2e^(ln|x|) -1</p>
<p>looking at the question again i revise my tangent line to equal y - 44 = 2(x-108)</p>
<p>g(10) = 4</p>
<p>Each period of 5 has an area of 2 so.. 108/5 = 21 remainder 3.</p>
<p>21(2) + int from 0 to 3 of f(t) = 42+2 = 44.</p>
<p>Slope = F(108) well because its periodic of 5 the remainder 3 is what we care about. F(3) = 2 so... slope = 2</p>
<p>therefore, y - 44 = 2(x - 108)</p>