# AP Calc AB BC 2014 FRQ Answers .pdf

### File information

Original filename:

**AP Calc AB_BC 2014 FRQ Answers.pdf**

This PDF 1.5 document has been generated by TeX / pdfTeX-1.40.12, and has been sent on pdf-archive.com on 11/05/2014 at 18:20, from IP address 72.197.x.x.
The current document download page has been viewed 1324 times.

File size: 125 KB (10 pages).

Privacy: public file

### Share on social networks

### Link to this file download page

### Document preview

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.

### Link to this page

#### Permanent link

Use the permanent link to the download page to share your document on Facebook, Twitter, LinkedIn, or directly with a contact by e-Mail, Messenger, Whatsapp, Line..

#### Short link

Use the short link to share your document on Twitter or by text message (SMS)

#### HTML Code

Copy the following HTML code to share your document on a Website or Blog