Solutions to the 2008 AP Calculus AB Exam Free Response Questions

Louis A. Talman
Department of Mathematical & Computer Sciences
Metropolitan State College of Denver

Part A

Problem 1.

a)

The area of the given region R is ∫_0^2[sin (π x) - (x^3 - 4 x)] x.

∫_0^2 (Sin[π x] - (x^3 - 4 x)) x

4

b)

We must first solve the equation x^3 - 4 x = -2 to find the limits of integration:

Solve[x^3 - 4 x  -2, x]

{{x (-9 +  111^(1/2))^(1/3)/3^(2/3) + 4/(3 (-9 +  111^(1/2)))^(1/3)}, ... ; 111^(1/2))^(1/3))/(2 3^(2/3)) - (2 (1 +  3^(1/2)))/(3 (-9 +  111^(1/2)))^(1/3)}}

N[%]

{{x1.67513 - 1.11022*10^-16 }, {x0.539189 + 2.22045*10^-16 }, {x -2.21432 - 5.55112*10^-17 }}

Although the solutions appear to be complex, the imaginary parts are very tiny and result from numerical errors.  We need the roots that lie in the interval [0, 2], so we will ignore the third solution.

The area of that part of the region R which lies below the horizontal line y = -2 is given by ∫_0.539^1.675[-2 - (x^3 - 4 x)] x.

c)

The area of a cross-section of the solid perpendicular to the x-axis at x = t is [sin(π t) - (t^3 - 4 t)]^2.  Thus, the volue of the solid is ∫_0^2[sin(π t) - (x^3 - 4 t)]^2t:

∫_0^2 (Sin[π t] - (t^3 - 4 t))^2t

1129/105 - 24/π^3

Numerically, this is

N[%]

9.97834

d)

Under the conditions given, the pool is a region in 3-space whose base is R and whose cross-section perpendicular to the x-axis at x = t has area [sin(π t) - (t^3 - 4 t)] (3 - t).  The volume is thus ∫_0^2[sin(π t) - (t^3 - 4 t)] (3 - t) t:

∫_0^2 (Sin[π t] - (t^3 - 4 t)) (3 - t) t

116/15 + 2/π

Numerically,

N[%]

8.36995

Problem 2.

a)

At 5:30pm, the rate at which the number of people standing in line was changing was approximately [L(7) - L(4)]/(7 - 4) = (150 - 126)/(7 - 4) = 8 people per hour.

b)

The average number of people standing in line during the first four hours that tickets were on sale was 1/(4 - 0) ∫_0^4L(t) t.  From the data in the table and the trapezoid rule, that is approximately

1/(4 - 0) ((120 + 156)/2 (1 - 0) + (156 + 176)/2 (3 - 1) + (176 + 126)/2 (4 - 3))

621/4

c)

By the Mean Value Theorem, there must be a point ξ_1∈ (1, 3) such that L ' (ξ_1) = (L(3) - L(1))/(3 - 1) = (176 - 156)/2> 0 and there must be a point ξ_2∈ (3, 4) such that L ' (ξ_2) = (L(4) - L(3))/(4 - 3) = (126 - 176)/1< 0.  Now L'' (t) is defined for all t∈[0, 9], and this means that L ' must be a continuous function on [0, 9].  By the Intermediate Value Theorem for Continuous Functions, there must be a point η_1∈ (ξ_1, ξ_2) where L ' (η_1) = 0.  By similar reasoning, there must be ξ_3∈ (4, 7) where L ' (ξ_3) >0, and so η_2∈ (ξ_2, ξ_3) where L ' (η_2) = 0.  Further, there must be FormBox[RowBox[{ξ_4, ∈, RowBox[{(7, 8), Cell[]}]}], TraditionalForm] for which L ' (ξ_4) <0, and therefore η_3∈ (ξ_3, ξ_4) where L ' (η_3) = 0.

We conclude that L ' (t) takes on the value 0 at least 3 times in the interval (0, 9).

Note:  It actually suffices to know that L is differentiable throughout (0, 9) to make this argument:  Derivatives have the Intermediate Value Property even when they are not continuous.

d)

If T(t) denotes the number of tickets sold at time t,  we are given T(0) = 0 and T ' (t) = 550 t ^(-t/2).  Consequently,

T[t_] = ∫_0^t550 t ^(-t/2) t

1100 (2 - ^(-t/2) (2 + t))

Thus,

T[3.]

972.784

To the nearest whole number, 973 tickets have been sold by 3:00pm.

Problem 3.

a)

We write V(t) = π [r(t)]^2h(t), and find that V ' (t) = 2 π r(t) h(t) r ' (t) + π [r(t)]^2h ' (t).  We are given that V ' (t) = 2000 cc/min for all t, and that when t = t_0 we have r(t_0) = 100 cm, h(t_0) = 0.5 cm, and r ' (t_0) = 2.5 cm/min.  Thus 2000 = 2 π · 100 · 0.5 · 2.5 + π 100^2 · h ' (t_0), so that

Solve[2000  2 π 100 0.5 2.5 + π 100^2h '[t_0], h '[t_0]]

{{h^′[t_0] 0.038662}}

At the given instant, h ' (t_0) = 0.039 cm/min.

b)

Taking t = 0 to be the moment when the recovery device goes into action, we have V ' (t) = 2000 - 400t^(1/2).  Thus, V(t) has a critical point at t = 25, when V ' (t) = 0.  Because V ' (t) > 0 for t < 25, while V ' (t) < 0 when t > 25, it follows from the First Derivative Test that V(t) is maximal when t = 25.

c)

If there were 60,000 cc of oil in the slick at the moment t = 0, when the recovery device began to operate, then by the Fundamental Theorem of Calculus we must have V(t) - 60000 = V(t) - V(0) = ∫_0^tV ' (τ) τ.  From what we saw in part b) above, we must therefore have FormBox[RowBox[{V(t), , =, , RowBox[{60000, , +, , RowBox[{∫_0^t, RowBox[{(2000 - 400τ^(1/2) ), τ, Cell[.]}]}]}]}], TraditionalForm]

Part B

Problem 4.

a)

By the Fundamental Theorem of Calculus and what we are given, we must have x(t) = -2 + ∫_0^tv(τ) τ.  This means that x(3) = -10, x(5) = -7, and x(6) = -9.  From the figure and the information given, we have x ' (t) = v(t) < 0 for 0<t<3 and for 5<t<6, while x ' (t) > 0 for 3 < t < 5.  Thus x is decreasing when 0≤t≤3 and when 5≤t≤6, while x is increasing when 3≤t≤5.  Thus, the particle is farthest to the left when t = 3 and its position at that instant is x = -10.

b)

Because x(0) = -2 and x(3) = -10, (see part a)), the particle moves through x = -8 once (leftward bound) when 0 < t < 3.  Because x(3) = -10, and x(5) = -7 (see part a)), it moves through x = -8 again (rightward bound) at some time in the interval (3, 5).  Because x(5) = -7 and x(6) = -9 (see part a) again) it moves through x = -8 still again (now leftward bound) at some time in the interval FormBox[Cell[TextData[Cell[BoxData[(5, 6)]]]], TraditionalForm].  The existence of these times is guaranteed, in each case, because the differentiable function x must be continuous on [0, 6], and continuous functions have the intermediate value property.  That these three instances are the only instances is guaranteed by the fact that x must be monotonic on each of the intervals [0, 3], [3, 5], and [5, 6].  We conclude that the particle passes through x = -8 just three times.

c)

Speed is the magnitude of velocity.  On the interval (2, 3), velocity is—from the graph—increasing, but negative.  On this interval, we obtain the magnitude of velocity by reflecting the relevant portion of the curve about the t-axis.  So speed is decreasing on (2, 3).

d)

Acceleration is v ' (t).  Thus, acceleration is negative on intervals where v(t) is decreasing.  From the graph and what we have been given about it, acceleration is negative on [0, 1) and on (4, 6].

Problem 5.

a)

[Graphics:HTMLFiles/index_136.gif]

b)

We have y ' (x)/[y(x) - 1] = 1/x^2, with y(2) = 0. Hence ∫_2^x[y ' (τ)/[y(τ) - 1]] τ = ∫_2^xτ^(-2) τ.  Integrating, we find that ln[1 - y(x)] - ln[1 - y(2)] = 1/2 - 1/x.  But y(2) = 0 is given, so ln[1 - y(x)] = 1/2 - 1/x.  Thus, 1 - y(x) = exp(1/2 - 1/x), or y = 1 - ^(1/2)/^(1/x).

c)

lim_ (x  ∞) y(x) = lim_ (x  ∞) (1 - ^(1/2)/^(1/x)) = 1 - ^(1/2)/lim_ (x  ∞) ^(1/x) = 1 - e^(1/2).

Problem 6.

a)

f ' (^2) = (1 - ln ^2)/(^2)^2 = (1 - 2)/^4 = -^(-4), so an equation of the tangent line at the point which corresponds to x = ^2 is y = 2 ^(-2) - ^(-4)(x - ^2).

b)

f ' (x) = 0 when (1 - ln x)/x^2 = 0, or when x = , so the x-coordinate of the critical point of f is x = .  We note that f ' (x) > 0 when x<, but that f ' (x) <0 when x > , and it follow that f has a local maximum at x = .

c)

From f ' (x) = (1 - ln x)/x^2, we have f'' (x) = (2 ln x - 3)/x^3.  We note that f'' (x) < 0 when x <^(3/2) but that f'' (x) >0 when x > ^(3/2).  The function f therefore has an inflection point at x = ^(3/2).  

d)

lim_ (x  0^+) f(x) = lim_ (x  0^+) (ln x)/x = -∞ .  

Created by Mathematica  (March 24, 2010)