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Math Problems
Calculus
Find derivatives of sine and cosine functions
Taima was given this problem:
\newline
The radius
r
(
t
)
r(t)
r
(
t
)
of the base of a cone is decreasing at a rate of
2
2
2
centimeters per minute. The height of the cone is fixed at
12
12
12
centimeters. At a certain instant
t
0
t_{0}
t
0
, the radius is
13
13
13
centimeters. What is the rate of change of the surface area
S
(
t
)
S(t)
S
(
t
)
of the cone at that instant?
\newline
Which equation should Taima use to solve the problem?
\newline
Choose
1
1
1
answer:
\newline
(A)
S
(
t
)
=
π
[
r
(
t
)
]
2
+
2
π
⋅
r
(
t
)
⋅
12
S(t)=\pi[r(t)]^{2}+2 \pi \cdot r(t) \cdot 12
S
(
t
)
=
π
[
r
(
t
)
]
2
+
2
π
⋅
r
(
t
)
⋅
12
\newline
(B)
S
(
t
)
=
π
[
r
(
t
)
]
2
⋅
12
S(t)=\pi[r(t)]^{2} \cdot 12
S
(
t
)
=
π
[
r
(
t
)
]
2
⋅
12
\newline
(C)
S
(
t
)
=
π
[
r
(
t
)
]
2
+
π
⋅
r
(
t
)
[
r
(
t
)
]
2
+
1
2
2
S(t)=\pi[r(t)]^{2}+\pi \cdot r(t) \sqrt{[r(t)]^{2}+12^{2}}
S
(
t
)
=
π
[
r
(
t
)
]
2
+
π
⋅
r
(
t
)
[
r
(
t
)
]
2
+
1
2
2
\newline
(D)
S
(
t
)
=
π
[
r
(
t
)
]
2
⋅
12
3
S(t)=\frac{\pi[r(t)]^{2} \cdot 12}{3}
S
(
t
)
=
3
π
[
r
(
t
)
]
2
⋅
12
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The function
b
(
t
)
b(t)
b
(
t
)
gives the number of books sold by a store by time
t
t
t
(in days) of a given year.
\newline
What does
∫
45
50
b
′
(
t
)
d
t
\int_{45}^{50} b^{\prime}(t) d t
∫
45
50
b
′
(
t
)
d
t
represent?
\newline
Choose
1
1
1
answer:
\newline
(A) The number of books sold between day
45
45
45
and day
50
50
50
\newline
(B) The number of days it takes to sell
50
50
50
books
\newline
(C) The change in the rate of selling books between
t
=
45
t=45
t
=
45
and
t
=
50
t=50
t
=
50
\newline
(D) The total number of books sold by day
50
50
50
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The function
c
(
t
)
c(t)
c
(
t
)
gives the number of cars produced in a factory by time
t
t
t
(in hours) on a given day.
\newline
What does
∫
0
6
c
′
(
t
)
d
t
\int_{0}^{6} c^{\prime}(t) d t
∫
0
6
c
′
(
t
)
d
t
represent?
\newline
Choose
1
1
1
answer:
\newline
(A) The average rate of change of the car production over the first
6
6
6
hours.
\newline
(B) The time it takes to produce
6
6
6
cars.
\newline
(C) The instantaneous rate of production at
t
=
6
t=6
t
=
6
.
\newline
(D) The number of cars produced over the first
6
6
6
hours.
Get tutor help
The function
c
(
t
)
c(t)
c
(
t
)
gives the number of cars produced in a factory by time
t
t
t
(in hours) on a given day.
\newline
What does
∫
0
6
c
′
(
t
)
d
t
\int_{0}^{6} c^{\prime}(t) d t
∫
0
6
c
′
(
t
)
d
t
represent?
\newline
Choose
1
1
1
answer:
\newline
(A) The average rate of change of the car production over the first
6
6
6
hours.
\newline
(B) The time it takes to produce
6
6
6
cars.
\newline
(C) The number of cars produced over the first
6
6
6
hours.
\newline
(D) The instantaneous rate of production at
t
=
6
t=6
t
=
6
.
Get tutor help
The number of people who have adopted a new fashion trend is increasing at a rate of
r
(
t
)
r(t)
r
(
t
)
people per month (where
t
t
t
is the time in months).
\newline
What does
∫
5
6
r
(
t
)
d
t
\int_{5}^{6} r(t) d t
∫
5
6
r
(
t
)
d
t
represent?
\newline
Choose
1
1
1
answer:
\newline
(A) The instantaneous rate of change of the number of people to adopt the fashion trend when
t
=
6
t=6
t
=
6
months
\newline
(B) The total number of people who have adopted the fashion trend by
t
=
6
t=6
t
=
6
months
\newline
(C) The growth in the number of people to adopt the fashion trend during the sixth month
\newline
(D) The time it took to change from
5
5
5
to
6
6
6
people who had adopted the fashion trend
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Consider the following problem:
\newline
The air gap (the distance from the bottom of the bridge to the surface of the water) under a bridge is changing at a rate of
r
(
t
)
=
0.6
sin
(
π
t
6
)
r(t)=0.6 \sin \left(\frac{\pi t}{6}\right)
r
(
t
)
=
0.6
sin
(
6
π
t
)
meters per hour (where
t
t
t
is the time in hours). At time
t
=
13
t=13
t
=
13
, the air gap is
62
62
62
meters. What is the air gap when
t
=
15
t=15
t
=
15
hours?
\newline
Which expression can we use to solve the problem?
\newline
Choose
1
1
1
answer:
\newline
(A)
62
+
∫
13
15
r
′
(
t
)
d
t
62+\int_{13}^{15} r^{\prime}(t) d t
62
+
∫
13
15
r
′
(
t
)
d
t
\newline
(B)
r
′
(
15
)
r^{\prime}(15)
r
′
(
15
)
\newline
(C)
r
′
(
13
)
+
62
r^{\prime}(13)+62
r
′
(
13
)
+
62
\newline
(D)
62
+
∫
13
15
r
(
t
)
d
t
62+\int_{13}^{15} r(t) d t
62
+
∫
13
15
r
(
t
)
d
t
Get tutor help
The function
b
(
t
)
b(t)
b
(
t
)
gives the number of books sold by a store by time
t
t
t
(in days) of a given year.
\newline
What does
∫
45
50
b
′
(
t
)
d
t
\int_{45}^{50} b^{\prime}(t) d t
∫
45
50
b
′
(
t
)
d
t
represent?
\newline
Choose
1
1
1
answer:
\newline
(A) The number of books sold between day
45
45
45
and day
50
50
50
\newline
(B) The total number of books sold by day
50
50
50
\newline
(C) The change in the rate of selling books between
t
=
45
t=45
t
=
45
and
t
=
50
t=50
t
=
50
\newline
(D) The number of days it takes to sell
50
50
50
books
Get tutor help
The function
s
(
t
)
s(t)
s
(
t
)
gives the number of students enrolled in a school by time
t
t
t
(in years).
\newline
What does
∫
15
18
s
′
(
t
)
d
t
=
20
\int_{15}^{18} s^{\prime}(t) d t=20
∫
15
18
s
′
(
t
)
d
t
=
20
mean?
\newline
Choose
1
1
1
answer:
\newline
(A) There were
20
20
20
more students enrolled in year
18
18
18
than in year
15
15
15
.
\newline
(B) Between years
15
15
15
and
18
18
18
, the cumulative number of years of schooling of all of the enrolled students is
20
20
20
years.
\newline
(C) There are
20
20
20
students enrolled in year
18
18
18
.
\newline
(D) The rate of change of enrollment is
20
20
20
students per year more in year
18
18
18
than it was in year
15
15
15
.
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h
(
x
)
=
(
5
−
6
x
)
5
h
′
(
x
)
=
?
\begin{aligned} h(x) & =(5-6 x)^{5} \\ h^{\prime}(x) & =? \end{aligned}
h
(
x
)
h
′
(
x
)
=
(
5
−
6
x
)
5
=
?
\newline
Choose
1
1
1
answer:
\newline
(A)
−
30
(
5
−
6
x
)
4
-30(5-6 x)^{4}
−
30
(
5
−
6
x
)
4
\newline
(B)
(
−
6
)
5
(-6)^{5}
(
−
6
)
5
\newline
(C)
−
6
x
5
+
5
x
4
(
5
−
6
x
)
-6 x^{5}+5 x^{4}(5-6 x)
−
6
x
5
+
5
x
4
(
5
−
6
x
)
\newline
(D)
5
(
5
−
6
x
)
4
5(5-6 x)^{4}
5
(
5
−
6
x
)
4
Get tutor help
Find
lim
θ
→
π
4
cos
(
2
θ
)
2
cos
(
θ
)
−
1
\lim _{\theta \rightarrow \frac{\pi}{4}} \frac{\cos (2 \theta)}{\sqrt{2} \cos (\theta)-1}
lim
θ
→
4
π
2
c
o
s
(
θ
)
−
1
c
o
s
(
2
θ
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
2
2
2
\newline
(B)
1
2
\frac{1}{2}
2
1
\newline
(C)
2
\sqrt{2}
2
\newline
(D) The limit doesn't exist
Get tutor help
Find
lim
x
→
π
4
cos
(
2
x
)
cos
(
x
)
−
sin
(
x
)
\lim _{x \rightarrow \frac{\pi}{4}} \frac{\cos (2 x)}{\cos (x)-\sin (x)}
lim
x
→
4
π
c
o
s
(
x
)
−
s
i
n
(
x
)
c
o
s
(
2
x
)
.
\newline
Choose
1
1
1
answer:
\newline
(A)
2
\sqrt{2}
2
\newline
(B)
2
2
2
\newline
(C)
4
4
4
\newline
(D) The limit doesn't exist
Get tutor help
Let
f
(
x
)
=
−
x
ln
2
(
x
−
1
)
f(x)=\frac{-x}{\ln ^{2}(x-1)}
f
(
x
)
=
l
n
2
(
x
−
1
)
−
x
.
\newline
Select the correct description of the one-sided limits of
f
f
f
at
x
=
2
x=2
x
=
2
.
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
lim
x
→
2
+
f
(
x
)
=
+
∞
and
lim
x
→
2
−
f
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow 2^{+}} f(x)=+\infty \text { and } \\ \lim _{x \rightarrow 2^{-}} f(x)=+\infty \end{array}
lim
x
→
2
+
f
(
x
)
=
+
∞
and
lim
x
→
2
−
f
(
x
)
=
+
∞
\newline
(B)
\newline
lim
x
→
2
+
f
(
x
)
=
+
∞
and
lim
x
→
2
−
f
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow 2^{+}} f(x)=+\infty \text { and } \\ \lim _{x \rightarrow 2^{-}} f(x)=-\infty \end{array}
lim
x
→
2
+
f
(
x
)
=
+
∞
and
lim
x
→
2
−
f
(
x
)
=
−
∞
\newline
(C)
\newline
lim
x
→
2
+
f
(
x
)
=
−
∞
and
lim
x
→
2
−
f
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow 2^{+}} f(x)=-\infty \text { and } \\ \lim _{x \rightarrow 2^{-}} f(x)=+\infty \end{array}
lim
x
→
2
+
f
(
x
)
=
−
∞
and
lim
x
→
2
−
f
(
x
)
=
+
∞
\newline
(D)
\newline
lim
x
→
2
+
f
(
x
)
=
−
∞
and
lim
x
→
2
−
f
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow 2^{+}} f(x)=-\infty \text { and } \\ \lim _{x \rightarrow 2^{-}} f(x)=-\infty \end{array}
lim
x
→
2
+
f
(
x
)
=
−
∞
and
lim
x
→
2
−
f
(
x
)
=
−
∞
Get tutor help
Let
f
(
x
)
=
−
1
(
x
−
1
)
2
f(x)=-\frac{1}{(x-1)^{2}}
f
(
x
)
=
−
(
x
−
1
)
2
1
.
\newline
Select the correct description of the one-sided limits of
f
f
f
at
x
=
1
x=1
x
=
1
.
\newline
Choose
1
1
1
answer:
\newline
(A)
\newline
lim
x
→
1
+
f
(
x
)
=
+
∞
and
lim
x
→
1
−
f
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow 1^{+}} f(x)=+\infty \text { and } \\ \lim _{x \rightarrow 1^{-}} f(x)=+\infty \end{array}
lim
x
→
1
+
f
(
x
)
=
+
∞
and
lim
x
→
1
−
f
(
x
)
=
+
∞
\newline
(B)
\newline
lim
x
→
1
+
f
(
x
)
=
+
∞
and
lim
x
→
1
−
f
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow 1^{+}} f(x)=+\infty \text { and } \\ \lim _{x \rightarrow 1^{-}} f(x)=-\infty \end{array}
lim
x
→
1
+
f
(
x
)
=
+
∞
and
lim
x
→
1
−
f
(
x
)
=
−
∞
\newline
(C)
\newline
lim
x
→
1
+
f
(
x
)
=
−
∞
and
lim
x
→
1
−
f
(
x
)
=
+
∞
\begin{array}{l} \lim _{x \rightarrow 1^{+}} f(x)=-\infty \text { and } \\ \lim _{x \rightarrow 1^{-}} f(x)=+\infty \end{array}
lim
x
→
1
+
f
(
x
)
=
−
∞
and
lim
x
→
1
−
f
(
x
)
=
+
∞
\newline
(D)
\newline
lim
x
→
1
+
f
(
x
)
=
−
∞
and
lim
x
→
1
−
f
(
x
)
=
−
∞
\begin{array}{l} \lim _{x \rightarrow 1^{+}} f(x)=-\infty \text { and } \\ \lim _{x \rightarrow 1^{-}} f(x)=-\infty \end{array}
lim
x
→
1
+
f
(
x
)
=
−
∞
and
lim
x
→
1
−
f
(
x
)
=
−
∞
Get tutor help
Find the derivative of
f
(
x
)
f(x)
f
(
x
)
.
\newline
f
(
x
)
=
cos
(
x
)
f(x) = \cos(x)
f
(
x
)
=
cos
(
x
)
\newline
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
______
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