<p>It's Time People!!!!!!!!!!!!!</p>
<p>I'll start. On question 5b, you end up with ln(abs(y-1)) = -X^(-1) +C after taking the antiderives seperately. After the exam, my teacher said that there was never a FR problem that used that absolute value for the rest of the problem, so all the practice tests I took did not use it. So I just dropped it habitually to get ln(y-1)=-X^(-1) + C, y=e^(-1/X+C) + 1, and 0=e^(-.5 + C) + 1. So basically I spent 5-10 minutes trying to figure out how to get e^(-.5 + C) to equal -1! That was probably my biggest mistake during FR, maybe besides the whole oil problem.</p>
<p>Other tests have dealt with the ln |y-1| before, particularly the (2004</a> FR #6). Technically, the absolute value shouldn't be automatically dropped, but should only be dropped when the initial condition supports that y-1 > 0 (as it does in '08).</p>
<p>I've found that it's often easier to solve for the initial condition when you do so immediately after integrating, rather than waiting until C is solved for.</p>
<p>On that problem, you have to substitute in the initial conditions right after using implicit differentiation. So, I remember this from the test, I had:</p>
<p>-1/x + c= ln|y-1| Plugging in (2,0):
-.5 + c = ln|-1|
c=.5 (ln(1) = 0)</p>
<p>And then, you can solve for y. </p>
<p>That gave you... umm.. (1/2)e^(-1/x) + 1 i THINK, I am doing this from memory so I don't remember exactly).
And, so the next limit question, the answer was 3/2. Limit as x->infinity e^0 is 1, so 1/2 + 1 = 3/2.</p>
<p>The answer I have for that is e^(-1/x)*e^(1/2) + 1, and therefore that the limit is sqrt(e) + 1.</p>
<p>I know how you got that answer. You simplified and THEN put initial conditions--i know because i got that at first too. However, you cannot do that. You have to plug in the initial conditions right after differentiating, and after doing so, you get c=1/2 -- Giving you the answer that i got</p>
<p>You're right about the C = 1/2, but you didn't take it all the way.</p>
<p>dy/dx = (y-1)/x^2
dy/(y-1) = 1/x^2 dx
ln |y-1| = -1/x + C
C = 1/2 (as indicated above)
ln |y-1| = -1/x + 1/2
|y-1| = e^(-1/x + 1/2)
y - 1 = e^(-1/x + 1/2) (because of initial condition (0,2))
y = e^(-1/x + 1/2) + 1
y = e^(-1/x)*e^(1/2) + 1</p>
<p>And technically, you can simplify and put in the initial conditions, but you have to be a lot more careful with the algebra.</p>
<p>Are the official answers up as well? I'm looking thru AP central, and answers dont seem to be out yet...</p>
<p>I see. In the very last problem, 6d, I put does not exist. Is that the same as negative infinity, the actual answer? I.E., would I get credit for it?</p>
<p>The official answers won't be up until July.</p>
<p>As far as 6d goes, I believe they will accept either answer (negative infinity or does not exist). They have in the past anyway.</p>
<p>If for 6d i just wrote DNE (not Does Not Exist/Negative INfinity), is that OK?</p>
<p>DNE is universally recognized in math circles as Does Not Exist, especially with regard to limits.</p>
<p>MathProf,
I believe the initial condition for the differential equation problem is (2,0) instead of (0,2)</p>
<p>I got what WantIvy got (3/2 for 5c). It took me a while to remember that absolute value...</p>
<p>Oooh, good catch, Hannah.ng. :) So used to seeing an initial condition with x = 0. :)</p>
<p>So that would change the solution above to 1 - sqrt(e)*e^(-1/x), and the limit to 1 - sqrt(e).</p>
<p>What were some answers you guys got for the oil spill problem?</p>
<p>Do you guys think the FRQ are not super hard ? I think it's not bad, kind of straightforward.</p>
<p>However, my friends and a lot of people think they are hard. Am I trapped the test ?</p>
<p>I'm pretty sure 6d is negative infinity, not DNE because it is a limit where x apporaches 0 only from the right.</p>
<p>These are the answers I got for #3.</p>
<p>a) I got dh/dt = (1700 - 250pi)/10000pi = .287
b) V has a maximum volume at t = 25 minutes
c) V = 60000 + integral(0, 25) (2000 - R(t)) dt</p>
<p>vestige, some calculus texts refer to all limits that are "equal" to positive or negative infinity as limits that do not exist, and the AP exam will accept that answer.</p>
<p>The foundation for this lies in the definition of a limit, where it says that a limit is a "value" a function approaches as x approaches another value. Accordingly, since infinity is not a "value" (in the sense that it's not a countable number), a limit cannot be equal to infinity. At least according to those texts.</p>
<p>The texts that claim that a limit can be equal to infinity recognize that what is really going on is that the limit does not exist, but by saying the limit approaches infinity, we are giving more information about why that limit does not exist.</p>