1. No category

Solutions to the
2015 AP Calculus BC
Free Response Questions
Louis A. Talman, Ph.D.
Department of Mathematical & Computer Sciences
Metropolitan State University of Denver
Problem 1
Ÿ a.
The rate of change, in cubic feet per hour, of the volume of water in the pipe at time t is rHtL - dHtL, where
t2
r@t_D = 20 SinB
F
35
t2
20 SinB
F
35
d@t_D = - 0.04 t3 + 0.4 t2 + 0.96 t
0.96 t + 0.4 t2 - 0.04 t3
v@t_D = r@tD - d@tD
- 0.96 t - 0.4 t2 + 0.04 t3 + 20 SinB
t2
F
35
The amount of water, in cubic feet, that flows into the tank in the eight-hour time interval 0 £ t £ 8
8
is Ù0 rHtL â t :
NIntegrate@r@tD, 8t, 0, 8<D
76.5703529473
Ÿ b.
The rate of change of water in the pipe at time t = 3 is vH3L:
v@3D
- 0.313632157643
2
BC2015Solutions.nb
This is negative, so volume of water in the tank is decreasing when t = 3.
Ÿ c.
Plot@v@tD, 8t, 0, 8<D
6
4
2
2
4
6
8
FindRoot@v@tD Š 0, 8 t, 3<D
8t ® 3.27165844976<
We see from a plot of v HtL that volume is decreasing until about 3.27, after which it increases. Thus, volume is
minimal when t = 3.272 hours, approximately.
Ÿ d.
t
Solution of the equation 30 + Ù0 vHΤL â Τ = 50 for t gives the time at which the pipe will begin to overflow.
Problem 2
Ÿa
If vHtL = Xx ' HtL, y ' HtL\ = Ycos t2 , e0.5 t ], and the particle is at H3, 5L when t = 1, then the particle’s position
t
2
vector at time t is given by rHtL = XxHtL, yHtL\ = X3, 5\ + Ù1 vHΤL â Τ. Thus, xH2L = 3 + Ù1 cos Τ2 â Τ. By
numerical integration, this is
3 + NIntegrateACosAΤ2 E, 8Τ, 1, 2<E
2.55693722453
Ÿ b.
The slope of the particle’s curve at time t is y ' HtL  x ' HtL = e0.5 t ‘ Icos t2 M. This is to be 2, so we set the quotient
equal to 2 and solve numerically:
ã0.5 t
FindRootB
Š 2, 8t, 0.5<F
CosAt2 E
8t ® 0.840164472691<
BC2015Solutions.nb
Ÿ c.
@x ' HtLD2 + @y ' HtLD2 . This is 3 when cos2 t2 + et = 32 . We solve numerically:
The particle’s speed is
2
FindRootAICosAt2 EM + ãt Š 9, 8t, 0, 2.2<E
8t ® 2.1958951534<
Ÿ d.
The total distance traveled by the particle from time t = 0 to time t = 1 is the integral of speed over that time
interval. Numerical integration gives
NIntegrateB
1.59460884988
2
ICosAt2 EM + ãt , 8t, 0, 1<F
3
4
BC2015Solutions.nb
Problem 3
Ÿ a.
The approximate value of the derivative v ' H16L, using data from the table, is
@vH20L - vH12LD  H20 - 12L = @240 - 200D  @20 - 12D = 5 meters/minute.
Ÿ b.
40
The definite integral Ù0
vHtL â t gives the actual distance, in meters, that Johanna traveled in the time
interval 0 £ t £ 40. The right Riemann sum approximation of this distance is 12 × 200 + 8 × 240 + 4 × H- 220L
+ 16 ·150 = 5840 m
Ÿ c.
If Bob's velocity at time t is BHtL = t3 - 6 t2 + 300 meters per minute, then his acceleration at time t is
B ' HtL = 3 t2 - 12 t. Thus, his acceleration at time t = 5 is 15 meters per minute per minute.
Ÿ d.
Bob's average velocity, in meters per minute, over the interval @0, 10< is
1
10
10
Ù0 BHΤL â Τ = 350 .
Problem 4
Ÿ a.
The required slope field looks like this:
2
1
0.5
-0.5
1.0
1.5
-1
Ÿ b.
If y ' = 2 x - y, then y '' = 2 - y ' = 2 - H2 x - y) = 2 - 2 x + y. In the second quadrant, x < 0 and y > 0,
so y '' > 0 throughout that quadrant. It follows that any solution must be concave upward in the second
quadrant.
BC2015Solutions.nb
5
If y ' = 2 x - y, then y '' = 2 - y ' = 2 - H2 x - y) = 2 - 2 x + y. In the second quadrant, x < 0 and y > 0,
so y '' > 0 throughout that quadrant. It follows that any solution must be concave upward in the second
quadrant.
Ÿ c.
If f is a solution of this differential equation for which f H2L = 3, then f ' H2L = 2 × 2 - 3 = 1, so f doesn't have
a critical point at x = 2. Thus, f has neither a maximum nor a minimum at x = 2.
Ÿ d.
If y = m x + b is a solution of the differential equation y ' = 2 x - y, then, on the one hand, direct
differentiation of the solution shows that we must have y ' = m, while, on the other hand, the differential
equation implies that we must also have y ' = 2 x - Hm x + bL = H2 - mL x - b. Consequently,
m = H2 - mL x - b, and this latter must be so for all values of x. It follows, then, by equating coefficients of
poweres of x, that m = 2 and from this that b = - 2.
Ÿ Remark
The differential equation y ' = 2 x - y can be rewritten as y ' + y = 2 x. This equation has the form
y ' + pHxL y = qHxL, which is a first order linear differential equation. Such an equation can be solved by
choosing a function PHxL such that P ' HxL = pHxL and multiplying the differential equation through by ePHxL .
d
After doing so, the new equation can always be put into the form d x Ay ePHxL E = qHxL ePHxL . All that remains then
is to carry out the integration on the right side.
When this technique is applied to the current equation, we find that it becomes y ' ex + y ex = 2 x ex , or,
d
equivalently, d x Hy ex L = 2 x ex . Integration now yields y ex = 2 x ex - 2 ex + C, or y = 2 x - 2 + C e-x ,
which is the general solution of the differential equation.
The initial value problem of part c can now be solved by substituting 3 for y and 2 for x, to learn that we must
then take C = e2 . This gives, as solution for the initial value problem, y = 2 x - 2 + e2 - x . We can use this
explicit representation of the solution of the initial value problem to confirm the conclusions obtained above for
part c of the problem.
Note also that we can obtain the linear solution required in part d by taking C = 0 in the general solution.
Problem 5
Ÿ a.
2
If f ' HxL = Hk - 2 xL ‘ Ix2 - k xM N and k = 3, then f ' H4L = H3 - 8L ‘ H16 - 12L2 = - 5  16. Because
f H4L = 1  H16 - 12L = 1  4, an equation for the desired tangent line is y = H1  4L - H5  16L Hx - 4L.
Ÿ b.
2
With k = 4, f ' HxL = H4 - 2 xL ‘ Ix2 - 4 xM so f ' H2L = 0. The denominator of the derivative is never
negative, so the sign of the derivative when x is near 2 is the same as the sign of its numerator. Thus, f ' HxL is
positive when x is in the region immediately to the left of x = 2, and f ' HxL is negative when x is in the region
immediately to the right of x = 2. Consequently, f has a relative maximum at x = 2. (Note: It is also possible
to draw this conclusion by way of the Second Derivative Test, but doing so is inadvisable on account of the
heavy computational burden.)
Ÿ c.
If f ' H- 5L = 0, then @k - 2 H- 5LD ‘ AH- 5L2 - k H- 5LE = 0 , so that k = - 10.
Ÿ d.
If A  x + B  Hx - 6L = @AHx - 6L + B xD ‘ Ix2 - 6 xM = 1 ‘ Ix2 - 6 xM, then A + B = 0 and - 6 A = 1. Thus,
A = - 1  6 and B = 1  6. It follows that
2
Ù Aâ x ‘ Ix - 6 xME = H1  6L Ù @1  Hx - 6L - 1  xD â x = H1  6L@ln x - 6 - ln x D + C.
6
BC2015Solutions.nb
If A  x + B  Hx - 6L = @AHx - 6L + B xD ‘ Ix2 - 6 xM = 1 ‘ Ix2 - 6 xM, then A + B = 0 and - 6 A = 1. Thus,
A = - 1  6 and B = 1  6. It follows that
2
Ù Aâ x ‘ Ix - 6 xME = H1  6L Ù @1  Hx - 6L - 1  xD â x = H1  6L@ln x - 6 - ln x D + C.
Problem 6
Ÿ a.
To take the limit of [ÈH- 3Ln xn+1  Hn + 1LE ‘ A H- 3Ln-1 xn  nE as n ® ¥, we reduce the fraction to
3 x n  Hn + 1L, and the is easily seen to be 3 x , which is less than one when x
< 1  3. By the Ratio
Test, the desired radius of convergence is thus 1  3.
Ÿ b.
The first four nonzero terms of the Maclaurinn series for f ' can be obtained by differentiating the original
series term by term. The radius of convergence of a derived power series is the same as the radius of
convergence of the original power series— that is, 1  3 in this case. The derivative of the nth term of the given
series here is H- 3Ln-1 xn-1 , so the first four terms of the derived series are 1 - 3 x + 9 x2 - 27 x3 . This is the
geometric series with common ratio - 3 x, so f ' HxL = 1  H1 + 3 xL when x < 1  3.
Ÿ c.
The first four nonzero terms of the Maclaurin series for ãx are 1 + x + x2 ‘ 2 + x3 ‘ 6. We can obtain the
third degree Maclaurin series for gHxL = ãx f HxL by multiplying the series for ãx by the series for f given
above, and recording just the terms through degree three. We obtain x - x2 ‘ 2 + 2 x3 .