Solutions to the
2004 AP Calculus BC Exam (Form B)
Free Response Questions

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

Problem 1.  (Thanks to Bill Kehlenbeck for pointing out a careless error that rendered the first three parts of this one wrong.)

a.

vx[t_] = (t^4 + 9)^(1/2)

(9 + t^4)^(1/2)

vy[t_] = 2 ^t + 5 ^(-t)

5 ^(-t) + 2 ^t

Speed:

s[t_] = (vx[t]^2 + vy[t]^2)^(1/2)

(9 + (5 ^(-t) + 2 ^t)^2 + t^4)^(1/2)

s[0]

58^(1/2)

Numerically,

N[s[0]]

7.61577

Acceleration:

a[t_] = {vx '[t], vy '[t]}

{(2 t^3)/(9 + t^4)^(1/2), -5 ^(-t) + 2 ^t}

a[0]

{0, -3}

b.

The tangent vector is

T[t_] = {vx[t], vy[t]}

{(9 + t^4)^(1/2), 5 ^(-t) + 2 ^t}

At time t = 0,

T[0]

{3, 7}

The slope of a line parallel to this vector is 7/3, so an equation of the line tangent to the curve at (4, 1), the point corresponding to t = 0, is y = 1 + 7/3 (x - 4).

c.

Total distance travelled over the interval 0≤t≤3 is ∫_0^3 (v_x[t]^2 + v_y[t]^2)^(1/2) t.  Integrating numerically, we obtain

NIntegrate[(vx[t]^2 + vy[t]^2)^(1/2), {t, 0, 3}]

45.2268

d.

The x-coordinate of the particle at time t = 3 is x[3] = x[0] + ∫_0^3x '[t] t = 4 + ∫_0^3 (t^4 + 9)^(1/2) t.  Numeric integration gives

4 + NIntegrate[(t^4 + 9)^(1/2), {t, 0, 3}]

17.9308

Problem 2

We are given T[x] = 7 - 9 (x - 2)^2 - 3 (x - 2)^3 as the third-degree Taylor polynomial for a certain function f about x = 2.

a.

In any Taylor polynomial, the coefficient a_k of (x - a)^k is f^(k)[a]/k !.  Hence f[2] = 7 and f''[2] = 2 · (-9) = -18.

b.

Reasoning as in part a.), we find that f '[2] = 0, so that f has a critical point at x = 2.  Because f''[2] = -18, the Second Derivative Test allows us to conclude that f has a local maximum at x = 2.

c.

f[0] ≃7 - 9 · (-2)^2 - 3 · (-2)^3 = -5.  There is not enough information to determine whether or not f has a critical point at x = 0.  This is because the third-degree Taylor polynomial carries no information about derivatives at any point other than the point about which the expansion has been done; it is determined solely by the values of the function and its first three derivatives at that point.

d.

The Lagrange Remainder for the third-degree Taylor polynomial at x = 2 has the form f^(4)[ξ]/4 ! (x - 2)^4, where ξ  is some unknown number in the interval between x and 2.  Thus, f[0] = T[0] + 1/24 f^(4)[ξ] (0 - 2)^4 for a certain ξ∈ (0, 2).  Because | f^(4)[x] | ≤6 for all x∈[0, 2], this means that | f[0] - T[0] | ≤6/24 (2)^4 = 4.  But, from part c.) above, we know that T[0] = -5.  Hence, -4≤f[0] - (-5) ≤4, whence -9≤f[0] ≤ -1.

Problem 3.

We are given

Inner[Set, Map[v, {0, 5, 10, 15, 20, 25, 30, 35, 40}], {7., 9.2, 9.5, 7., 4.5, 2.4, 2.4, 4.3, 7.3}, List]

{7., 9.2, 9.5, 7., 4.5, 2.4, 2.4, 4.3, 7.3}

(This arcane syntax assigns the correct values to v[0], v[5], etc.)

a.

The Mid-Point Rule with four subintervals of equal length gives for ∫_0^40v[t] t the approximate value

Underoverscript[∑, k = 1, arg3] 10v[5 + 10 (k - 1)]

229.

v[t] is given in miles per minute, so the integral gives miles travelled during the time interval over which the integral is taken.

b.

By Rolle's Theorem, acceleration--which is v '[t]--must be zero at least once in the interval 0≤t≤15, because v[0] = v[15].  Similarly, v '[t] must be zero at least once in the interval 25≤t≤30, because v[25] = v[30].  Thus, acceleration must vanish at least twice in the interval 0≤t≤40.

c.

f[t_] = 6 + Cos[t/10] + 3Sin[7 t/40]

6 + Cos[t/10] + 3 Sin[(7 t)/40]

If the function f models velocity, then acceleration is f '.  Thus

f '[t]

21/40 Cos[(7 t)/40] - 1/10 Sin[t/10]

%/.t23.

-0.407694

At time t = 23, acceleration is -0.408 miles/minute^2.

d.

Average velocity over 0≤t≤40 is 1/40∫_0^40f[t] t:

1/40∫_0^40f[t] t

1/40 (240 + 240/7 Sin[7/2]^2 + 10 Sin[4])

N[%]

5.91627

Problem 4:

a.

Inflection points are to be found where f ' has relative extrema.  Consequently, the function f whose derivative is pictured has inflection points at x = 1 and at x = 3.

b.  Thanks to Tammy Brown, Becky Myers, and Sue Wall, who all pointed out that I'd blown this one completely.

The function f is decreasing on the interval [-1, 4] and increasing on the interval [4, 5] because f ' is non-positive on the first of these intervals, and non-negative on the second.  Consequently, f takes on its absolute minimum value for the interval [-1, 5] at x = 4.

The function f has its absolute maximum at one of the points where x = -1 or x = 5.  (There can be no absolute maximum for f at any point interior to (-1, 5) because f ' has in that interval no zeroes at which it undergoes a sign change from positive to negative as x increases.)    The area bounded by f and the x-axis on the interval [-1, 4] is clearly larger than the area bounded by f and the x-axis on the interval [4, 5], so ∫_4^(-1) f '[t] t = f[-1] - f[4] >f[5] - f[4] = ∫_4^5f '[t] t.  This means that f[-1] > f[5], so that the absolute maximum value taken on by f in the interval [-1, 5] is f[-1].

c.

We are given g[x] = x f[x], so g '[2] = f[2] + 2 · f '[2] = 6 + 2 · (-1) = 4.  Because g[2] = 2 f[2] = 12, this means that an equation for the line tangent to the graph of g at x = 2 is FormBox[RowBox[{y, , =, , RowBox[{12, , +, , RowBox[{4, (x - 2), Cell[]}]}]}], TraditionalForm].

Problem 5

a.

The average value of g[x] = 1/x^(1/2) on the interval [1, 4] is 1/(4 - 1) ∫_1^4x/x^(1/2) = 2/3 x^(1/2) Underoverscript[|, 1, arg3] = 2/3.

b.

The volume of the solid generated when the region bounded by the graph of y = g[x], the vertical lines x = 1 and x = 4, and the x-axis is revolved about the x-axis is π∫_1^41/xx = π ln 4.

c.

The average value of the areas of the cross sections perpendicular to the x-axis is π/(4 - 1) ∫_1^41/xx = π/3ln 4.

d.

The improper integral ∫_4^∞g[x] x is Underscript[lim, b∞] ∫_4^bx/x^(1/2) = lim_ (b∞) 2x^(1/2) Underoverscript[|, 4, arg3] = lim_ (b∞) (2 b^(1/2) - 4) = ∞, so the improper integral diverges.  However, lim_ (b∞) 1/(b - a) ∫_4^bx/x^(1/2) = lim_ (b∞) (2b^( ... #8734;) (b^(1/2) - 2)/((b^(1/2) - 2) (b^(1/2) + 2)) = 2lim_ (b∞) 1/(b^(1/2) + 2) = 0.

Problem 6

a.

We have ∫_0^1x^nx = x^(n + 1)/(n + 1) Underoverscript[|, 0, arg3] = 1/(n + 1).

b.

If y = x^n, then y ' = n x^(n - 1), so that the equation of the line ℓ is y = 1 + n(x - 1).  This line meets the x-axis  when x = 1 - 1/n, so that the base of the triangle T has length 1/n.  Because the altitude of the triangle T is 1, the area of T is 1/(2n).

c.

From what we have seen in parts a.) and b.), the area A[n] of the region S is A[n] = 1/(n + 1) - 1/(2n) = (n - 1)/(2n(n + 1)).  Then A '[n] = -(n^2 - 2 n - 1)/(2n^2(n + 1)^2), which is zero for n>1 only when n = 1 + 2^(1/2) (by the Quadratic Formula).  Noting that A '[n] >0 for 1≤n<1 + 2^(1/2), while A '[n] <0 for 1 + 2^(1/2) <n, we conclude that the maximal area occurs when n = 1 + 2^(1/2).

Created by Mathematica  (May 13, 2004)