Official 2008 Calculus AB FRQ Discussion

dchow08
Fri, 09 May 2008 20:32:40 +0000

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.


(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.