5 Steps to a 5 AP Calculus BC 2019

428 STEP 5. Build Your Test-Taking Confidence

Solutions to BC Practice Exam 2---Section II

Section II Part A

1. The correct answer is (A).

0

80

82

84

86

88

90

5101520

t (minutes)

degrees (Farenheit)

g(t) = 90 – 4 tan( 20 t (

g(10)= 90 −4 tan

(

10

20

)

= 90 −4 tan

(

1

2

)

≈ 90 −4(0.5463)≈ 90 − 2. 1852

≈ 87. 81 ◦or 87. 82 ◦F

(B) g′(t)=−4 sec^2

(

t

20

)(

1

20

)

g′(10)=−

1

5

sec^2

(

10

20

)

≈− 0. 26 ◦/min.

(C) Set the temperature of the liquid equal to

86 ◦F. Using your calculator, let

y 1 = 90 −4 tan

(

x

20

)

; andy 2 = 86.

To find the intersection point ofy 1 and

y 2 , lety 3 =y 1 −y2 and find the zeros of

y 3.

Using the [Zero] function of your

calculator, you obtainx= 15 .708. Since

y 1 <y 2 on the interval 15. 708 <x≤20,

the temperature of the liquid is below

86 ◦F when 15. 708 <t≤20.

Alternatively, you could use the

Intersection Function of your calculator

and find the intersection of the graphs of

y 1 , andy 2.

(D) Average temperature below 86◦

=

1

20 − 15. 708

∫ 20

15. 708

(

90 −4 tan

(

x

20

))

dx.

Using your calculator, you obtain:

Average temperature =

1

4. 292

(364.756)

≈ 84. 9851

≈ 84. 99 ◦F.

2. (A) Velocity vector〈
   x′(t),y′(t)

〉

=

〈

6 t,−2 sin(2t−4)

〉

and

the acceleration vector〈

x′′(t),y′′(t)

〉

=

〈

6,−4 cos(2t−4)

〉

, and

att=2, the acceleration vector is

〈

6,− 4

〉

.

(B) The speed of the particle at√ t=2 equals

(x′(2))^2 +(y′(2))^2 =

√

(12)^2 +(0)^2 =12.

(C) Att=2 the position vector is

〈

16, 1

〉

.

The slope of the line tangent to the path

of the particle att=2is

y′(2)

x′(2)

=

0

12

=0,

which means you have a horizontal

tangent. Thus, the equation isy=1.

(D) The total distance traveled by the particle

for 0≤t≤2 is the length of the path,

which is

∫ 2

0

√(

dx

dt

) 2

+

(

dy

dt

) 2

dt=

∫ 2

0

√

(6t)^2 +(−2 sin(2t−4))^2 dt= 12 .407.