Sheed30

It's Time People!!!!!!!!!!!!!

http://apcentral.collegeboard.com/ap...lus_ab_frq.pdf

jlinn12

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.

TheMathProf

Other tests have dealt with the ln |y-1| before, particularly the (2004 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).

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.

WantIvy

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:


-1/x + c= ln|y-1| Plugging in (2,0):
-.5 + c = ln|-1|
c=.5 (ln(1) = 0)

And then, you can solve for y.

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.

TheMathProf

The answer I have for that is e^(-1/x)*e^(1/2) + 1, and therefore that the limit is sqrt(e) + 1.

WantIvy

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

TheMathProf

You're right about the C = 1/2, but you didn't take it all the way.

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

And technically, you can simplify and put in the initial conditions, but you have to be a lot more careful with the algebra.

Kowloon

Are the official answers up as well? I'm looking thru AP central, and answers dont seem to be out yet...

WantIvy

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?

TheMathProf

The official answers won't be up until July.

As far as 6d goes, I believe they will accept either answer (negative infinity or does not exist). They have in the past anyway.

AzRocks

If for 6d i just wrote DNE (not Does Not Exist/Negative INfinity), is that OK?

TheMathProf

DNE is universally recognized in math circles as Does Not Exist, especially with regard to limits.

Hannah.ng

MathProf,
I believe the initial condition for the differential equation problem is (2,0) instead of (0,2)

figment42

I got what WantIvy got (3/2 for 5c). It took me a while to remember that absolute value...

TheMathProf

Oooh, good catch, Hannah.ng. So used to seeing an initial condition with x = 0.

So that would change the solution above to 1 - sqrt(e)*e^(-1/x), and the limit to 1 - sqrt(e).