chronicfuture12

What were some answers you guys got for the oil spill problem?

nns91

Do you guys think the FRQ are not super hard ? I think it's not bad, kind of straightforward.

However, my friends and a lot of people think they are hard. Am I trapped the test ?

vestige

I'm pretty sure 6d is negative infinity, not DNE because it is a limit where x apporaches 0 only from the right.

TheMathProf

These are the answers I got for #3.

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

TheMathProf

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.

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.

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.

wendyling

for 5b this is what i did:
ln ly-1l = -1/x + k
e^(lnly-1l) = e^(-1/x) times e^(k)
let C = e^(k)
ly-1l = C(e^(-1/x))
then l0-1l=C(e^(-1/2)
so C = e^(1/2)
so y = e^(-1/x) times e^(1/2) + 1= e^(.5 - 1/x) +1

then for 5c: you'd get lim x-> inf. of the above
so e^(-1/infinity) equals 0 so the limit would be e^(.5 - 0) equal e^(.5)
and therefore the final answer is e^(.5) +1 or = sqrt(e) + 1

TheMathProf

Since the initial condition is indeed (2, 0), then the absolute value sign cannot simply be removed, and therefore, when the absolute value sign is removed, the sign of the right-hand side of the equation will have to become negative.

wendyling

lost a point there. they'll probably give the points for everything else though, like they did on the past exams. hope so! i wonder if they'll give credit for doing part c with the wrong f(x)

Sixthsense

what did u guys get for numbers 1 and 2?

10isairman

how do you do # 6

TheMathProf

wendyling, I would imagine they would give the point for a part (c) limit, so long as the function you came up with in part (b) is exponential.

Sixthsense, I got the following:

#1
Let f(x) = sin(pi*x) - (x^3 - 4x)
a) Area = integral(0,2) f(x) dx = 4
b) Area = integral(.539,1.675) (-2 - (x^3 - 4x)) dx
c) Volume = integral (0, 2) [f(x)]^2 dx = 9.978
d) Volume = integral (0, 2) [f(x)]*(x-3) dx = 8.369 or 8.370

#2
a) L'(5.5) = [L(7) - L(4)]/(7-4) = 8 people/hour
b) (156 + 120)/2 + 2*(176 + 156)/2 + (176 + 126)/2 = 621 = integral (0, 4) L(t) dt, therefore 1/4 L(t) dt = 155.25
c) Three, one where L changes from increasing to decreasing on [1, 4], one where L changes from decreasing to increasing on [3, 7], and one where L changes from increasing to decreasing on [4, 8].
d) integral (0, 3) r(t) dt = 973

wendyling

for #6 i got:
a) y-(2/e^2)=(-1/e^4)(x-e^2)
b) x=e, relative maximum, because f'x changes from positive to negative
c) x=e^(3/2)
d) got wrong lol dont want to say what i wrote

december

For number 2 this I what I think I got

a) 8 people/hr
b) 151.75
c) 3
d) 973

TheMathProf

I agree with wendyling's answers for parts (a)-(c), and part (d) has been previously answered.

AKittka

wow i didn't see average and forgot to divide by 4.. i feel like an idiot