Solutions to the
1998 AP Calculus BC Exam
Free Response Questions

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

Problem 1.

a.

The area of the region R is

∫_0^4 (8 - x^(3/2)) x

96/5

b.

The volume of the solid generated by revolving R about the x-axis is

π ∫_0^4 (8 - x^(3/2))^2x

(576 π)/5

c.

The value of k is given by

FindRoot[π∫_0^k (8 - x^(3/2))^2x == π /2∫_0^4 (8 - x^(3/2))^2x, {k, 1}]

{k0.994904}

  k = 0.995, to three digits beyond the decimal.

Problem 2

a.

lim_ (x -∞) 2 x ^(2x) = lim_ (x -∞) (2x)/^(-2x).  L'Hôpital's Rule is applicable to the latter expression.  Thus, lim_ (x -∞) (2x)/^(-2x) = lim_ (x -∞) 2/(-2^(-2x)) = 0.

b.

If f[x] = 2 x ^(2x), then f '[x] = (2 + 4 x) ^(2x).  Consequently, f '[x] = 0 only when x = -1/2.  Now (by part a, above) lim_ (x -∞) f[x] = 0, while lim_ (x∞) f[x] = ∞.  Consequently there are numbers x_1 and x_2, x_1< -1/2<x_2, such that x≤x_1 implies that f[x] ≥ -1/100>f[-1/2] = -^(-1) and x_2≤x implies that f[x] ≥ -1/100>f[-1/2].  But f must have an absolute minimum in the interva l [x_1, x_2], and it cannot be located at either x_1 or x_2.  Because x = -1/2 is the only critical point in this interval, it must give the absolute minimum for f[x] when x_1≤x≤x_2, and therefore for -∞<x<∞.

c.

By the observations we have made in part b. above, the range of f is [-^(-1), ∞).

d.

Let us assume, for the moment, that b>0.  Then, arguing as we have in parts a. and b. above, we find that f[x] = b x ^(b x)  has an absolute minimum at x = -1/b.  This minimum value is f[-1/b] = -e^(-1), which is independent of b.  If b<0, we obtain the same result after the change of variables u = -x, which amounts to a reflection about the y-axis.

Problem 3.

a.

The third-degree Taylor polynomial for f about x = 0 is f[0] + f '[0] x + f''[0]/2x^2 + f'''[0]/6x^3 or 5 - 3 x + 1/2x^2 + 2/3x^3.  Thus, f[0.2] is approximately

5 - 3 x + 1/2x^2 + 2/3x^3/.x0.2

4.42533

b.

We can obtain the third-degree Taylor polynomial for g[x] = f[x^2]  about x = 0 by substituting x^2 for x in the Taylor polynomial for f and then truncating.  This gives 5 - 3 x^2 + 1/2x^4.

c.

We can obtain the third-degree Taylor polynomial for h[x] = ∫_0^xf[t] t by integrating that of f term by term and truncating.  We obtain 5 x - 3/2x^2 + 1/6x^3.

d.

We cannot determine h[1] = ∫_0^1f[t] t from what is given.  It is possible that f[x] = 5 - 3 x + 1/2x^2 + 2/3x^3, in which case we would have h[t] given by

∫_0^1 (5 - 3 t + 1/2t^2 + 2/3t^3) t

23/6

However, it is also consistent with what has been given that f[x] = 5 - 3 x + 1/2x^2 + 2/3x^3 + x^4, and if this is the case, then h[1] would be given by

∫_0^1 (5 - 3 t + 1/2t^2 + 2/3t^3 + t^4) x

5 - 3 t + t^2/2 + (2 t^3)/3 + t^4

Problem 4

a.

[Graphics:HTMLFiles/index_70.gif]

b.

Euler's Method is given by

F[x_, y_] = (x y)/2

(x y)/2

EulerStep[{x_, y_}, h_] = {x + h, y + F[x, y] h}

{h + x, y + (h x y)/2}

The first Euler step with h = 0.1 gives

EulerStep[{0, 3}, 0.1]

{0.1, 3}

The second gives

EulerStep[%, 0.1]

{0.2, 3.015}

Thus, f[0.2] is approximately 3.015.

c.

If f is the solution to y/x = (x y)/2 for which f[0] = 3, then ∫_0^xf '[t]/f[t] t = ∫_0^xt/2t, or ∫_f[0]^f[x] u/u = ∫_0^xt/2t. Hence, ln(f[x]/3) = x^2/4, and f[x] = 3 ^(x^2/4).  This gives, for f[0.2],

3 ^(x^2/4)/.x0.2

3.03015

Problem 5

a.

[Graphics:HTMLFiles/index_92.gif]

b.

Average temperature is 1/(14 - 6) ∫_6^14F[t] t:

1/(14 - 6) ∫_6^14 (80 - 10 Cos[(π t)/12]) t

1/8 (640 + 180/π)

N[%]

87.162

To the nearest degree, this is 87 degrees.

c.

FindRoot[80 - 10 Cos[(π t)/12] 78, {t, 6}]

{t5.23087}

a = t/.%

5.23087

FindRoot[80 - 10 Cos[(π t)/12] 78, {t, 18}]

{t18.7691}

b = t/.%

18.7691

The air conditioner ran when FormBox[RowBox[{5.231, ≤, t, ≤, RowBox[{18.769, Cell[.]}]}], TraditionalForm]

d.

The approximate total cost is 0.05 ∫_a^b (2 - 10 Cos[(π t)/12]) t, or

0.05 ∫_a^b (2 - 10 Cos[(π t)/12]) t

5.09637

To the nearest cent, this is $5.10.

Problem 6

a.

If x '[t] = 1/(2 t + 1)^(1/2), with x[0] = -4, then ∫_0^tx '[τ] t = ∫_0^tτ/(2 τ + 1)^(1/2), or ∫_x[0]^x[t] u = ∫_0^tτ/(2τ + 1)^(1/2).  Thus, x[t] + 4 = (2t + 1)^(1/2) - 1, so that x[t] = (2t + 1)^(1/2) - 5.

b.

y[t] = x[t]^3 - 3 x[t], so y '[t] = 3 x[t]^2x '[t] - 3 x '[t] = (3 ((2t + 1)^(1/2) - 5)^2 - 3) · 1/(2t + 1)^(1/2).

c.

When t = 4, x[t] = x[4] = (2 · 4 + 1)^(1/2) - 5 = -2 and y = x^3 - 3 x = (-2)^3 - 3 (-2) = -8 + 6 = -2.  Also, x '[4] = 1/(2 · 4 + 1)^(1/2) = 1/3, while y '[t] = (3 (3 - 5)^2 - 3) · 1/3 = 3.  Hence, when t = 4, speed is (x '[t]^2 + y '[t]^2)^(1/2) = (1/9 + 9)^(1/2) = 82^(1/2)/3.  At time t = 4, the particle is at (-2, -2) with speed 82^(1/2)/3.

Created by Mathematica  (May 5, 2009)