428 STEP 5. Build Your Test-Taking Confidence
Solutions to BC Practice Exam 2---Section II
Section II Part A
- 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.
- (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.