5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 16:20

AP Calculus AB Practice Exam 2 397

Solutions to AB Practice Exam 2 Section II

Section II Part A

1.

2

0 π 2

R

y = 2 cos x

x

y

(A) Area ofR=

∫π/ 2

0

2 cosxdx

=2 sinx]π/ 02

=2 sin

(

π

2

)

−2 sin(0)= 2.

(B)

∫a

0

2 cosxdx= 1

2 sinx

]a

0 =2 sina−2 sin(0)=2 sina

2 sina= 1 ⇒sina=

1

2

⇒a=sin−^1

(

1

2

)

=

π

6

(C) Volume=π

∫π/ 2

0

(2 cosx)^2 dx

=π

∫π/ 2

0

4 cos^2 xdx

= 4 π

∫π/ 2

0

cos^2 dx

= 4 π

∫π/ 2

0

1 +cos(2x)

2

dx

= 2 π

∫π/ 2

0

[

1 +cos(2x)

]

dx

= 2 π

[

x +

sin(2x)

2

]π/ 2

0

= 2 π

[(

π

2

+

sinπ

2

)

− 0

]

=π^2.

(D) Area of cross section=

1

2

π

(

2 cosx

2

) 2

=

1

2

πcos^2 x.

V=

∫π/ 2

0

1

2

πcos^2 xdx

=

1

2

π

∫π/ 2

0

cos^2 xdx

=

1

2

π

∫π/ 2

0

1 +cos(2x)

2

dx

=

π

4

∫π/ 2

0

(1+cos(2x)dx)

=

π

4

[

x +

sin(2x)

2

]π/ 2

0

=

π

4

[(

π

2

+

sinπ

2

)

− 0

]

=

π^2

8

2. (A)

0

80

82

84

86

88

90

5101520

t (minutes)

degrees (Fahrenheit)

g(t) = 90 – 4 tan( 20 t (

g(10)= 90 −4 tan

(

10

20

)

= 90 −4 tan

(

1

2

)

≈ 87. 81 ◦For87. 82 ◦F