Solutions to the
2006 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 curves intersect where ln x = x - 2.

a = x /.FindRoot[Log[x]  x - 2, {x, 0.1}][[1]]

0.158594

b = x/.FindRoot[Log[x]  x - 2, {x, 1.6}][[1]]

3.14619

∫_a^b (Log[x] - (x - 2)) x

1.94909

Answer:  1.949

b)

π∫_a^b ((Log[x] - (- 3))^2 - ((x - 2) - (-3))^2) x

34.1986

Answer:  34.199

c)

Answer:  Either of the integrals π∫_ (a - 2)^(b - 2)[(y + 2)^2 - e^(2 y)] y or 2 π ∫_a^bx  [ln x - (x - 2)] x will do.  Evaluation was not required; however

2 π ∫_a^bx (Log[x] - (x - 2)) x

17.0993

and, of course,

π ∫_ (a - 2)^(b - 2) ((y + 2)^2 - ^(2y)) y

17.0993

Problem 2:

a)

The rate at which left turns happen is:

L[t_] = 60 t^(1/2) (Sin[t/3])^2

60 t^(1/2) Sin[t/3]^2

The number of turns when 0≤t≤18 is therefore

NIntegrate[L[t], {t, 0, 18}]

1657.82

Answer:  To the nearest whole number, this is 1658.

b)

From the graph, we see that L[t] ≥150  on [a, b] where a ~ 12 and b ~ 16.  Solving numerically for a and b, we find

a = t/.FindRoot[L[t]  150, {t, 12}][[1]]

12.4283

b = t/.FindRoot[L[t]  150, {t, 16}][[1]]

16.1217

1/(b - a) ∫_a^bL[t] t

199.426

Answer:  There are 150 or more left turns per hour approximately when 12.428 ≤ t ≤16.122, where t is measured in hours.  The average during this interval is 199.426 left turns per hour.

c)

During the two-hour interval 13 ≤ t ≤ 15,

NIntegrate[L[t], {t, 13, 15}]

431.931

cars make left turns.  500 oncoming cars pass straight through the intersection in this two-hour period.  The product of these two numbers is FormBox[Cell[215,967.7], TraditionalForm], and this exceeds the threshhold of 200, 000.

Answer:  The intersection requires a traffic signal. The reasoning is given in the preceding paragraph.

Problem 3:

a)

Answer:  g(4) = ∫_0^4f(t) t = 3;
g ' (4) = f(4) = 0;
g'' (4) = f ' (4) = -2.

b)

Answer:  By the Fundamental Theorem of Calculus, g ' (x) = f(x).  Thus, g ' (1) = f(1) = 0, while g ' (x) = f(x) < 0 for 0<x<1, and g ' (x) = f(x) >0 for 0<x<1.  By the First Derivative Test, g has a relative minumum at x = 1.

c)

By periodicity, g(10) - g(5) = g(5) - g(0) = g(5) = 2.Hence, g(10) = g(10) - g(0) = [g(10) - g(5)] + [g(5) - g(0)] = 2 + 2 = 4.  Reasoning as in the previous sentence, we find that for any positive integer k, we must have g(5 k) - g(5 k - 5) = 2, so that g(105) = ∑_ (k = 1)^212 = 42.  Also, ∫_105^108f(t) t = ∫_0^3f(t) t = 2, by inspection of the graph.  Hence g(108) = ∫_0^108f(t) t = ∫_0^105f(t) t + ∫_105^108f(t) t = g(105) + ∫_105^108f(t) t = 44.  Also, the first of the following equalities being a consequence of the Fundamental Theorem of Calculus, and the second being a consequence of periodicity, g ' (108) = f(108) = f(3) = 2.  So the equation of the line tangent to y = g(x) at x = 108 is y = 44 + 2 (x - 108).

Answer:   g(10) =   4, and the equation of the tangent line at x = 108 is y = 2 x - 172.

Part B

Problem 4:

a)

Answer:  Average acceleration is (v(80) - v(0))/(80 - 0) = (49 - 5)/80 = 11/20feet per second per second.

b)

Answer:  ∫_10^70v(t) t measures the distance in feet between the rocket's position at time t = 10 to its position at time t = 70.  The midpoint Riemann sum with three subdivisions of equal length is v(20) · 20 + v(40) · 20 + v(60) · 20 = 440 + 700 + 880 = 2020 feet.

c)

Answer:  For rocket B, we have, by the Fundamental Theorem of Calculus, v(80) = v(0) + ∫_0^80a(τ) τ = 2 + 3∫_0^80 ( τ)/(τ + 1)^(1/2) = 50 ft/sec.  Thus Rocket A is traveling at 49 ft/sec when t = 80, so rocket B is traveling faster.

Problem 5:

a)

[Graphics:HTMLFiles/index_71.gif]

b)

If y ' (x) = [1 + y(x)]/x, then y ' (x)/[1 + y(x)] = 1/x.  Thus, ∫_ (-1)^x {y ' (s)/[1 + y(s)]} s = ∫_ (-1)^x (1/s) s, as long as x < 0 and y(x) ≠ -1.  Consequently, ln | 1 + y(x) | - ln | 1 + y(-1) | = ln | x | - ln | -1 |, or ln | 1 + y(x) | - ln 2 = ln | x |.  Now we have assumed that x < 0, so the latter equation is equivalent to | 1 + y(x) | = -2x.  When x is near -1, y(x) must be near 1, so FormBox[RowBox[{1 + y(x), , >, , RowBox[{0, Cell[]}]}], TraditionalForm] for such x.  Hence, 1 + y(x) = -2 x, or y(x) = - 2 x - 1.   We may not have either x = 0 or y(x) = -1, so the domain of this solution is -∞<x<0.

Answer:  y(x) = - 2 x - 1, when -∞<x<0.

Problem 6:

a)

If g(x) = e^(a x) + f(x), then g ' (x) = a e^(a x) + f ' (x), and g'' (x) = a^2e^(a x) + f'' (x).

Answer:  g ' (0) = a - 4 ; g'' (0) = a^2 + 3.

b)

If h(x) = f(x) cos k x, then h ' (x) = f ' (x) cos k x   - k f(x) sin k x.  Thus, h(0) = 2, while h ' (0) = f ' (0) cos (k · 0) - k f(0) sin (k · 0) = -4.  

Answer:  h ' (x) = f ' (x) cos k x - k f(x) sin k x.  The equation of the line tangent to the graph of h at x = 0 is y = h(0) + h ' (0) (x - 0), or y = -4 x + 2.

Created by Mathematica  (May 8, 2006)