AP Calc AB BC 2014 FRQ Answers .pdf

AP Calculus AB & BC Solutions, 2014 Form A

This guide contains all the solutions to the free-response questions from both the AB and BC exams. Note
that some of the problems are common to both tests, but have only been printed once.

Calculus AB-only problems
2(a) We need to find the points of intersection so we set y = f (x) =⇒ x4 − 2.3x2 + 4 = 4 =⇒ x = 0 and
x = 2.3. So the volume is
Z 2.3
Z 2.3
Vy=−2 = π
[4 − (−2)]2 − [f (x) − (−2)]2 dx = π
36 − [f (x) + 2]2 dx = 98.868
0

0

2(b) The length of a leg is y − f (x) =⇒ 4 − (x4 − 2.3x3 + 4) = 2.3x3 − x4 . Since we are dealing with right
triangles, both legs are congruent and the area of each cross section is just 21 × leg × leg. So the volume
is
Z 2.3
1
(2.3x3 − x4 )2 dx = 3.574
2
0
Z
2(c) The area of R is

2.3

[4 − (x4 − 2.3x3 + 4)] dx. We simplify and split the integral:

0

Z

k

2.3x3 − x4 dx =

0

Z

2.3

2.3x3 − x4 dx

k

Solving the equation above will find the value of k the problem seeks.

Page 2

5(a) f has a relative minimum whenever the sign of f 0 changes from a negative number to a positive number.
This occurs only at x = 1.
5(b) Since

f 0 (1) − f 0 (−1)
= 0, the mean value theorem guarantees some c such that f 00 (c) = 0.
1 − (−1)

5(c) We just use the chain-rule:
h0 (x) =
From the table, we find that h0 (3) =
Z
5(d) The key is to recognize that
easy:
Z

1
2

7

=

f 0 (x)
1
· f 0 (x) =
f (x)
f (x)

1
.
14

f 0 (g(x)) g 0 (x) dx = f (g(x)) + C by the chain-rule. Now the problem is

3

f 0 (g(x)) g 0 (x) dx = f (g(3)) − f (g(−2)) = f (1) − f (−1) = 2 − 8 = −6

−2

Page 3

6(a) I don’t know how to make a graph using this software, but just use a graphing calculator to plot the
solution to part (c) and see how your sketch compares.
6(b) We find the slope using

dy
:
dx

dy 
= (3 − 1) cos 0 = 2

dx (0,1)

The equation of the tangent line is y − 1 = 2x =⇒ y = 2x + 1 . At x = 0.2 we find that y(0.2) =
f (0.2) = 2(0.2) + 1 = 1.4 .
6(c) Separation of variables and then algebra:
dy
dx
dy
3−y
Z
dy
3−y
− ln |3 − y|

=

(3 − y) cos x

=

cos x dx
Z
=
cos x dx
=

sin x + C

Plug in (0, 1) to find that − ln |3 − 1| = sin 0 + C =⇒ C = − ln 2. We now deal with the algebra:
ln |3 − y| =
e

ln 2 − sin x

ln |3−y|

=

eln 2−sin x

3−y

=

2e− sin x

Rearranging we get the final solution to be y = 3 − 2e− sin x

Page 4

Calculus AB & BC Solutions
1(a) The average rate of change of A(t) on 0 ≤ t ≤ 30 is

A(30) − A(0)
= −0.197 pound/day.
30 − 0

1(b) A0 (15) = −0.164. This means that the amount of grass clippings in the bin at the beginning of day
15, in pounds, is decreasing at the rate of 0.164 pound/day.
Z 30
Z 30
1
1
A(t) dt. Using our calculator we find that
A(t) dt = 2.7526.
1(c) We want t such that A(t) =
30 0
30 0
Thus, we have A(t) = 2.7526 =⇒ 6.687(0.931)t = 2.7526 =⇒ t = 12.415 .
1(d) We have A0 (30) = −0.05598 and A(30) = 0.78293. Thus, we have L(t) − 0.78293 = −0.05598(t − 30)
and want to find t such that L(t) = 0.5; so we have 0.5 − 0.78293 = −.05598(t − 30) =⇒ t = 35.054 .

Page 5

2(a) The area of the shaded region is
9π
+
4
2(b) x = r cos θ = cos θ(3 − 2 sin 2θ);

π/2

Z
0

1 2
r dθ = 9.708
2

dx 
= −2.366 .

dθ θ= π6

p
2(c) Solution 1. - Rectangular The distance between the two curves is (x1 − x2 )2 + (y1 − y2 )2 . Let
r1 = 3 and r2 = 3 − 2 sin(2θ). We have x1 = 3 cos θ and y1 = 3 sin θ. We have x2 = cos θ(3 − 2 sin(2θ))
and y2 = sin θ(3 − 2 sin(2θ)). So x1 − x2 = 2 sin 2θ cos
√ θ = a and y1 − y2 = 2 sin θ sin 2θ = b. Enter
a2 + b2 into Y1 and take the numerical derivative of Y1 at θ = π3 to get that the rate of change is
−2.000 .

Solution 2. - Polar Note that r1 − r2 = 2 sin 2θ = distance between the two curves at θ =


derivative of 2 sin 2θ at θ = π3 is 4 cos 2 · π3 = −2 .

π
3.

The

 π
dr dθ
dr
dr
2(d) Note that
·
=
. Thus,
= −4 cos(2θ) · 3 = −12 cos 2θ =⇒ −12 cos 2 ·
= −6 =⇒
dθ dt
dt
dt
6

dr 
= −6 .

dθ θ= π6

Page 6

Z

3

f (t) dt; this is the area bounded by the graph of f and the x-axis on [−3, 3]. Thus

3(a) g(3) =
−3

g(3) = 12 (5)(4) + 12 (1)(−2) =⇒ g(3) = 9
3(b) g is increasing when f (t) is positive and concave down when the slope of the line tangent to f (t) is
negative. This occurs on (−5, −3) and on (0, 2). (Note: x = −5 meets these conditions, but is excluded
because the problem specifies “open intervals”)
3(c) h0 (x) =

−4 − 4
5xg 0 (x) − 5g(x)
5(3)(−2) − 5(9)
1
; g(3) = 9 and g 0 (3) =
= −2; so h0 (3) =
= − .
25x2
4−0
25(32 )
3

3(d) p0 (x) = f 0 (x2 − x) · (2x − 1) =⇒ p0 (−1) = f 0 (2) · −3 = −2 · −3 = 6

Page 7

4(a) The average acceleration of train A on 2 ≤ t ≤ 8 is

vA (8) − vA (2)
2
= − 110
3 meters/minute .
8−2

4(b) Yes – since vA is continuous, vA (5) = 40 and vA (8) = −120 it follows from the intermediate value
theorem that vA (t) = −100 at some time 5 ≤ t ≤ 8.
Z

12

vA (t) dt + 300 , where the +300 comes from the fact

4(c) The position of train A at t = 12 is given by
2

that Va (2) = 300.

A trapezoidal approximation gives 12 (3)(100 + 40) + 12 (3)(40 − 120) + 12 (4)(−120 − 150) = −450. So,
train A is approximately 450 meters west of its starting point at t = 12.
4(d) Let d be the distance between train A and train B. We want d0 at t = 2. Let dA be the east-west
distance of train A from the origin, and let dB be the north-south distance of train B from the origin.
By the distance formula
d2 = d2A + d2B
Taking the derivative of each term with respect to time t and cancelling factors of 2:
dd0 = dA d0A + dB d0B
We know that d = 500 because d is the hypotenuse of a right triangle with legs dA = 300 and dB = 400.
We know from the table that d0A = vA = 100 at t = 2, and we know from the function vB = d0B that
vB (2) = 125. We just substitute and solve for d0 :
500d0 = 300(100) + 400(125) =⇒ 5d0 = 300 + 4(125) =⇒ d0 = 60 + 100 = 160
The rate of change of the distance between the trains is 160 meters/minute .

Page 8

Z

1

xe

5(a) R =

x2

Z
− (−2x) dx =

2

1

xe

x2

+ 2x dx =

0

0

Z
5(b) Vy=−2 =

1

ex
+ x2
2

!

1
e+1

 =
2
0

h
i
2
(xex + 2)2 − (−2x + 2)2 dx

0

5(c) We can consider each of the 3 parts of the perimeter separately.

For y = −2x: The graph forms
a right triange with the x-axis with legs 1 and 2; thus the hypotenuse
√
(part of the perimiter) is 5.
2

For x = 1: At x = 1 the value of xex is e and the value of y = −2x is −2. Thus the perimeter
is e − (−2) = e + 2.
Z

x2

For y = xe : The formula for the length of this curve is
Z 1q

2
1 + 2x2 ex2 + ex2 dx.

1

p
1 + f 0 (x)2 dx; so the perimeter is

0

0

Z
So the total perimeter P is P =

1

q

1 + 2x2 ex2 + ex2

0

Page 9


2

dx + e + 2 +

√

5.