AP Calculus BC Practice Exam 1 395
- The correct answer is (C).
Temperature of metal= 100 −
∫ 6
0
10 e−^0.^1 tdt.
Using your calculator, you obtain:
Temperature of metal= 100 − 45. 1188
= 54. 8812 ≈ 55 ◦F.
- The correct answer is (C).
The position of the particle is
y(t)= 5 t^3 − 9 t^2 + 2 t−1, and the velocity is
v(t)=y′(t)= 15 t^2 − 18 t+ 2 .At the moment
the particle changed direction, its velocity was
zero, so 15t^2 − 18 t+ 2 =0. Solving tells us that
the particle changes direction twice, first at
t≈ 0 .124 and later att= 1 .076. Taking the
first of these, and evaluating the position
function,y≈− 0 .881. At the moment when
the particle first changes direction, its position
is (0,− 0 .881). Remember that the particle is
moving on they-axisand thus thex-coordinate
is always 0.
- The correct answer is (D).
Use the ratio test for absolute convergence.
nlim→∞
∣∣
∣∣(x−3)
n+ 1
(n+1)^2
·
n^2
(x−3)n
∣∣
∣∣
=nlim→∞
∣
∣∣
∣
(x−3)n^2
(n+1)^2
∣
∣∣
∣
=
∣∣
x− 3
∣∣
nlim→∞
(
n
n+ 1
) 2
=
∣∣
x− 3
∣∣
Set
∣
∣x− 3
∣
∣< 1 ⇒− 1 <(x−3)< 1
⇒ 2 <x<4. Atx=4, the series becomes
∑∞
n= 1
1
n^2
, which is ap-series withp=2. The
series converges. Atx=2, the series becomes
∑∞
n= 1
(−1)n
n^2
, which converges absolutely. Thus,
the interval of convergence is [2,4].
- The correct answer is (C).
Area of a cross section=
√
3
4
(2x)^2 =
√
3 x^2.
Using your calculator, you have:
Volume of solid=
∫ 4
0
√
3(x^2 )dx=
64
√
3
3
.
- The correct answer is (B).
∫ 6
0
f(x)dx≈
6 − 0
2(3)
·[0+2(1)+2(2.25)+ 6 .25]
≈ 12. 75
- The correct answer is (C).
dy
dx
=ky⇒y=y 0 ekt
Triple in 10 hours⇒y= 3 y 0 att=10.
3 y 0 =y 0 e^10 k⇒ 3 =e^10 k⇒ln 3=ln(e^10 k)
⇒ln 3= 10 kork=
ln 3
10
≈ 0. 109861 ≈ 0. 110
- The correct answer is (A).
Use the intersection function to find that the
points of intersection ofy=cosx+1 and
y= 2 + 2 x−x^2 are (0, 2) and (2.705, 0.094).
The area enclosed by the curves is
∫ 2. 705
0
[
2 + 2 x−x^2 −(cosx+1)
]
dx
=x+x^2 −
x^3
3
−sinx
∣∣
∣
∣
2. 705
0
≈ 3. 002.
