when t≥0. Thus, speed is 
. When t = 3, this is 
. The acceleration vector is 
. At t = 3, this is <4, 6 cos 9>.
Solutions to the
2011 AP Calculus BC
Free Response Questions
Louis A. Talman, Ph.D.
Department of Mathematical & Computer Sciences
Metropolitan State College of Denver
Problem 1
(a)
Speed is the magnitude of the velocity vector, which we are given as when t≥0. Thus, speed is . When t = 3, this is . The acceleration vector is . At t = 3, this is <4, 6 cos 9>.
(b)
The slope of the line tangent to the path at t = 3 is 
.
(c)
The position of the particle at t = 3 is 
. We must integrate numerically to obtain the second coordinate:
Position at time t = 3 is therefore approximately (21.000, -3.226).
(d)
Total distance traveled is the definite integral from 0 to 3 of speed. Once again, numerical integration is necessary.
Total distance traveled over 0≤ t ≤ 3, is therefore about 21.091.
Problem 2
(a)
The rate at which the temperature of the tea is changing at time t = 3.5 is given approximately by the difference quotient (H[3.5+1.5]-H[3.5-1.5])/[(3.5+1.5)-(3.5-1.5)] = (52-60)/3=-8/3 degrees per minute.
(b)
The integral 
is the average value of the temperature of the tea, in degrees Celsius.
The trapezoidal approximation for this integral is 
, or
(c)
By the Fundamental Theorem of Calculus, 
degrees, which is the amount by which the temperature of the tea has changed over the time-interval 0≤t≤10.
(d)
B(t) is given by 
. Therefore B(10) is
We seek H(10) - B(10) = 43 - B(10) =
The biscuits are about 8.817 degrees Celsius cooler than the tea at time t = 10.
Problem 3
(a)
The perimeter of the region shown consists of three line segments and the piece of the curve 
corresponding to 0 ≤ x ≤ k. The perimeter of the region shown is 
, where we have used the arc-length integral to find the length of the portion of the perimeter tht is not a straight line.
(b)
The area of a cross section of the volume perpendicular to the x-axis at x = t is 
The volume of the solid described is therefore 
.
(c)
From (b), above, we have 
. Thus, 
. when 
, this is 
.
Problem 4
(a)
.
.
g'(-3) = 2 + f(-3) = 2.
(b)
The absolute maximum of g must occur at an endpoint of the interval [-4,3] or at a critical point. But g'(x) = 2 + f(x), and this is simply the curve y = f(x) shifted 2 units upward. Note that all of the shifted curve that lies to the left of the y-axis lies above the x-axis, so that g'(x) is positive to the left of the y-axis, and for a substantial interval just to the right of that axis. For 0<x < 3, we then have g'(x) = 5 - 2 x, so that g'(x) = 0 when 
. Thus, g'(x) is positive for 
, negative for 
, and zero when 
. The latter value is the only critical value for g. It is clear, on geometric ground, that the area under the g' curve on the interval 
is positive and exceeds, in magnitude, the area between the g' curve and the x-axis on the interval 
. Consequently 
, and 
. The absolute maximum for g therefore occurs at 
.
(c)
g' [see part (b) above for an explicit description of g'] is increasing on [-4,0], decreasing on [0,3]. Inflection points are to be found where the monotonicity of the derivative changes, so x= 0 is the location of the only inflection point for this curve.
(d)
We have f(-4) = -1 and f(3) = -3. The average rate of change of f on the interval [-4, 3] is therefore 
.
This doesn't contradict the Mean Value Theorem because f'(0) does not exist. The hypotheses of the Mean Value Theorem require, among other things, that f be differentiable on (-4,3) before we may apply that theorem to this function. Because this is not so for our f, there is no contradiction.
Problem 5
(a)
Because 
is given, we have 
, and the equation for the line tangent to the solution of the initial value problem at t = 0 is W = W(0) + W'(0)(t - 0) = 1400 + 44t. When 
, this gives w = 1400 + 11 = 1411 tons for the approximate amount of solid waste at the end of the first three months of 2010.
(b)
, so 
. When t =0, therefore, W''(0) = 
. The solution curve is therefore concave upward near t = 0, and so the tangent line to the curve at t = 0 lies below the curve. this means that the estimate given in part (a) is an underestimate.
(c)
We have 
, whence either W=300 or 
for a certain constant c. We may discard the constant solution because W(0) ≠ 300. Consequently, 
for a certain positive constant C, or 
, where the constant C may now be any non-zero real number. (In fact C may be zero, but this would give the constant solution we've already rejected.).
Inserting the initial condition W(0) = 1400, which we have been given, into our general solution, we find that 1400 = 300 + C, so that C = 1100. The particular solution we seek is therefore 
.
Problem 6
(a)
The first four nonzero terms of the Taylor series for sin x about x = 0 are 
. The first four terms of the Taylor series for 
about x = 0 are therefore 
.
(b)
The first four nonzero terms of the Taylor series for cos x about x =0 are 
. Consequently, the first four nonzero terms of the Taylor series for 
are 
.
(c)
The coefficient of 
in the Taylor series about x = 0 for a function g is 
. From (b), above, we see that 
, so that 
.
(d)
The Taylor polynomial of degree 4 in powers of x for f(x) approximates f(x) to within 
, where 
on the interval from 0 to x. (This is Taylor's Theorem with Lagrange Remainder.) From the graph, we see that 
on any interval of the form [0, x], where, say, 
. Consequently, 
.
