1. No category

Solutions to the
2015 AP Calculus AB
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
AB2015Solutions.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
Ÿ a.
We solve first for the root of f HxL = gHxL that lies near x = 1.
f@x_D = 1 + x + ExpAx2 - 2 xE
2
1 + ã-2 x+x + x
g@x_D = x4 - 6.5 x2 + 6 x + 2
2 + 6 x - 6.5 x2 + x4
FindRoot@f@xD Š g@xD, 8x, 1<D
8x ® 1.03283188836<
x0 = x . %@@1DD
1.03283188836
The desired area is
x0
à
2
Hg@xD - f@xDL â x + à Hf@xD - g@xDL â x
0
2.00434566313
x0
AB2015Solutions.nb
Ÿ b.
The volume of the solid is :
2
2
à Hf@xD - g@xDL â x
x0
1.28316333919
Ÿ c.
We have
h@x_D = f@xD - g@xD
2
- 1 + ã-2 x+x - 5 x + 6.5 x2 - x4
So the rate at which h changes when x = 1.8 is
h '@1.8D
- 3.81171787829
3
4
AB2015Solutions.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 × H220L +
16 ·150 = 7600 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.
AB2015Solutions.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.
The function f has a relative maximum at x = - 2, where f ' HxL changes sign from positive to negative. There
are no other relative maxima.
Ÿ b.
The graph of f is both concave downward and decreasing on any open interval where f ' is both decreasing and
negative. The open intervals where f ' displays such behavior are H- 2, - 1L and H1, 3L. (And any open subinterval of either of these two intervals.)
Ÿ c,
Points of inflection are to be found whereever the derivative changes its behavior from increasing to decreasing
or from decreasing to increasing. These things happen at x = - 1, and x = 1, and at x = 3, so these are the xcoordinates of points of inflection for f .
Ÿ d.
x
If f H1L = 3, then, by the Fundamental Theorem of Calculus, f HxL = 3 + Ù1 f HΤL â Τ . Thus,
4
-2
f H4L = 3 + Ù1 f HΤL â Τ = 3 - 12 = - 9 and f H- 2L = 3 + Ù1 f HΤL â Τ = 3 + 9 = 12.
Problem 6
6
AB2015Solutions.nb
Problem 6
Ÿ a.
At H- 1, 1L, we have y ' = 1 ‘ A3 H1L2 - H- 1LE = 1  4, so an equation for the line tangent to this curve at the
point H- 1, 1L is y = 1 + H1  4L Hx + 1L.
Ÿ b.
If the tangent to this curve is to be vertical, y ' must not exist. This can be so only if 3 y2 - x = 0, or x = 3 y2
(But see the Remarks below). Substituting this for x in the equation of the curve, we obtain y3 - I3 y2 M y = 2,
or y3 = - 1. Thus, y = - 1. But if y = - 1 and y3 - x y = 2, then x = 3. Thus, the only point on this curve
where the tangent line can be vertical is the point with coordinates H3, - 1L,
Ÿ c.
Differentiating the equation y ' = y ‘ I3 y2 - x) with respect to x while treating y as a function of x yields the
2
equation y '' = Ay ' I3 y2 - xM - yH6 y y ' - 1LE ‘ I3 y2 - xM . Setting x = - 1, y = 1, and y ' = 1  4, as found in
part a., above, we obtain y '' = 1  32 at H- 1, 1L.
Ÿ Remarks:
I’ve given here the solution for Part (b) of this problem as I expect that most successful examinees will answer
it, and as I think the Exam Committee intended it to be answered. But the solution as given above is devoid of
logical support, because the statement that the derivative is not defined at a point need not imply that the
tangent line to the curve at that point is vertical.
The solutions works, but it works because of things, including the Implicit Function Theorem, that beginning
students of calculus don’t know— and shouldn’t be expected to know. Implicit differentiation with respect to x
applied to an equation FHx, yL = 0, which is the unnamed source of the expression the problem gives for the
derivative here, is itself meaningless at a point where the partial derivative F2 Hx, yL vanishes. That’s because
the equation FHx, yL = 0 may fail to define, anywhere near such a point, a unique differentiable function Φ of x
that satisfies F@x, ΦHxLD = 0.
That’s exactly what happens in this problem, where FHx, yL = y3 - x y - 2, at the point H3, - 1L: There is no
open interval centered at x = 3 on which there is a differentiable function Φ such that ΦH3L = - 1 and
F@x, ΦHxLD = 0. In fact, if x is a number just smaller than 3, there aren’t even any values of y near - 1 that
satisfy FHx, yL = 0. So the fact that the denominator in this derivative vanishes at H3, - 1L really signifies the
meaninglessness of the question “What’s y ' at H3, - 1L?”and not what the examiners probably wanted it to
signify.
The solution given above works, but not because of anything I’ve said yet, and not because of anything that
examinees are likely to say. It works because the partial derivative F1 H3, - 1L ¹ 0, and, on that account, the
Implicit Function Theorem guarantees the existence, on some open interval centered at y = - 1, of a unique
differentiable function Ψ of y such that ΨH- 1L = 3 and F@ΨHyL, yD = 0. Moreover, the derivative Ψ ' HyL is
given there by Ψ ' HyL = - F2 @ΨHyL, yD  F1 @ΨHyL, yD. It is easily checked that Ψ ' H- 1L = 0, and, this being a
derivative with respect to y, it signifies a vertical tangent at H3, - 1L.
Thus, the “solution”that I gave above works because the partial derivative F1 H3, - 1L doesn’t vanish, while the
partial derivative F2 H3, - 1L does. This is something we shouldn’t expect beginners to know— let alone
understand.
The reader of this remark can get a better feeling for the difficulties that underlie the “solution”above by
considering the same question for the curve FHx, yL = 0 at the point H0, 0L, where FHx, yL = y2 - x2 . Is there,
or is there not, a vertical tangent to this curve at that point? What does reasoning as in the “solution”given
above say about this?