linduhcow

Haha well I feel dumb asking this, but I'd be too paranoid if I didn't...

On the free-response, were we only supposed to write in the boxes labeled "work"? We didn't have to transfer our final answers anywhere, right?

Oh and how much weight do colleges put on the AB score vs. the AB subscore from the BC test?

TheMathProf

I have .039, but I believe .038 will be accepted for problem 3(a).

justaspirant

could someone explain 1d to me? i took the integral of h(x) from 0 to 2 and then multiplied it with the previously calculated area... and got an answer of 16.

dchow08

Justaspirant:

Why did you take the integral of h(x) from 0 to 2 and then multiply it with the previously calculated area?

Visualize the region R as the base of a swimming pool. At x=0, it's deepest, and it gets shallower and shallower as x gets to 2, where the depth is 1.
At x=1, at every point y, the depth is 2. Imagine a vertical line going down x=1, down into the paper 2 feet deep. What you have is a rectangle with height 2 and side length equalling the distance of the vertical line. So what the pool really is is a bunch of rectangular slices from x=0 to x=2. This question is basically a hidden slice method question, with the width being the vertical distances along R and the height being 3-x.

The area of a rectangle is base times height. So the volume of the whole pool is the integral from 0 to 2 of [f(x)][3-x]dx, where f(x) is the sin(pi)x - (x^3-4x)

dchow08

And yes, I got 0.039 for my answer for 3a.

Peytoncline

i'm so mad about #6... for letter c, i knew that the point of inflection is where the 2nd deriv = 0 or undefined and changes sign (or when the first deriv has a relative extrema)... i wrote that down.. and then i commenced to write down x = e for my answer... and then on 6(a) i wrote 0 as one of my critical points, which isn't really one b/c it's not included in the domain... those two things i believe are the only 2 things i missed on the FR section... just makes me mad because i hope i didn't fall into any traps on the MC (because the MC seemed, for the most part, pretty easy)

Sheed30

I'm pretty sure you all got 2d wrong.....you have to add 120 tickets i think because there were 120 pre-sold tickets....so the answer would be 1093.....right?


i dont remember what i put..but i was looking it over and i think im right!

Peytoncline

and on the oil problem (last part), i got..

Volume = 60,000 + [integral from 0 to 25 of (2000 - R(t)) dt]


didn't the question say that when the machine was hooked up and the oil began to removed, the volume was already 60,000, and then it asked how much was the total volume when the oil reached its max (which would make my integral right)?

justaspirant

wow, ok. i was stumped on that one for some reason so i went back to it at the end, when i realized that i still had no idea what i was supposed to do. so i just saw that they wanted volume.. volume is depth*area.. and it's calculus so i decided to stick in integral in there. ha..
but i understand it now. thanks for the explanation!

Sheed30

wait..it says t could be equal to zero...maybe im wrong....dammit

Peytoncline

and (one more question lol).. on the one with the lake, it gave the area of the surface of the lake represented by area bounded by sin(pi*x) and x^3 - 4x (thus making A = sin(pi*x) - (x^3 - 4x))... and it gave the depth of the water to be represented by h(x) = 3 - x

that just looked like a cross-sectional problem to me.. it said find the volume, so i just set it up like a cross-sectional problem made up of rectangles perpendicular to the x-axis

so, my equation was [integral from 0 to 2 of (3 - x)(sin(pi*x) - (x^3 - 4x)) dx], where 3-x = h.. and sin(pi*x) - (x^3 - 4x) = b

that was right, wasn't it?

Sheed30

anyone want to say for sure...for 2 d????? explain really quick?

Peytoncline

sorry, i lied.. one more question.. on #2, it said to use a Reimann sum (i think it was midpoint.. but not sure) to find the average number of people in the line at one time within the first 4 hours

i did this and got 621, which with units is 621 people * hours

and since it said average amount of people, i assumed it meant average value, so i did 1/(b-a) times my answer (to cancel out the hours) and got 155.25 people at one time as my answer

how does this compare to others' answers?

seanbow

Wait... Where does it mention pre-sold tickets? I only see people waiting in line and a rate of tickets being sold.

dchow08

I took the AP test last year, but I still know how to do these problems. I'll explain how I did questions 1-3.

1. (a) The distance from the top of R to the bottom of R is sin(pi x) - (x^3-4x). The area then, would be the integral of this equation from 0 to 2.

(b) The distance from the x-axis to y=x^3-4x is x^3 - 4x. The distance from the x-axis to y=-2 is 2. The distance from the top of the small region to the bottom of the region, then, is x^3-4x+2, since you're subtracting -2. Integrate that function from 0 to 2.

(c) The length of one side of the rectangle is sin(pi*x)-x^3+4x, so the area of each square is that function squared. To find the volume of the solid, integrate the equation for the area from 0 to 2.

(d) I explained this above. It's a slice method question except with rectangles of varying heights and widths. You get V=integral from 0 to 2 of [sin(pi x)-x^3+4x][3-x]dx

2.) (a) You could draw a diagram of the x and y values from x=4 to x=7. Connect the two points. Finding the slope of that line gives a general approximation of the rate of change at x=5.5. So the slope is the change in y over the change in x, which is (150-126)/(7-4) = 3.

(b) Draw trapezoids to connect each of the points, then find the area of each trapezoid and add them up. That's the total number of tickets, which is like the area of a rectangle. The width of the rectangle is how much x spanned, so divide that from the total area to get the height of the rectangle, which is the average.

(c) The answer is 3. There must be at least 3 points where the derivative is zero. From x = 1 to x = 3, the slope is positive (there is some point where the slope is positive, or else how would you increase y from x=1 to x=3?). Now from x=3 to x=4, for similar reasons, there must be a point where the slope is negative, and you can't go from a positive slope to a negative slope without the slope being zero at some point in between. Apply the same reasoning for x=3 to x=7 and from x=5 to x=8, and you'll see that there must be at least 3 points where the derivative is 0.

(d) It's just the integral of the function from 0 to 3. Remember how the antiderivative of the derivative is the function itself? So if you're given the equation for rate, to find the original function, take the integral.

3.) Here's what you're given: dv/dt = 2000, V=pi r^2 h.
(a) r=100, h=0.5, dr/dt = 2.5, what is dh/dt?

V=pi r^2 h
dV/dt = pi (r^2 dh/dt + h2rdr/dt)
dV/dt = Pi r^2 dh/dt + 2pi hrdr/dt
dV/dt - 2pi hr dr/dt = pi r^2 dh/dt
dh/dt = [dv/dt - 2pi hr dr/dt] / [pi r^2]
Plug in values given to find dh/dt = 0.039.

(b) I won't give a formal explanation, but a common-sense and logical one.

The rate of change of the volume of the oil spill is 2000-400sqrt t.
Graph that function. You will see that all the function is positive until a certain point where the function crosses the x-axis. In plain terms, the volume keeps increasing (the function is positive), until a certain point when the volume starts decreasing (the rate of change in volume is negative). Find that point. Well, set 2000-400sqrt t equal to 0 and solve for t. You get that t = 25 minutes.

(c). The integral of 2000-400sqrt t from 0 to 25 is how much the volume increased from t=0 to t=25. So the total volume at t= 25 is 6000 + that integral.