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Math Problems
Grade 8
Interpreting Linear Expressions
The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of people on an island is modeled by the following differential equation:
\newline
d
P
d
t
=
17753
53132
P
(
1
−
P
866
)
\frac{d P}{d t}=\frac{17753}{53132} P\left(1-\frac{P}{866}\right)
d
t
d
P
=
53132
17753
P
(
1
−
866
P
)
\newline
At
t
=
0
t=0
t
=
0
, the number of people on the island is
148
148
148
and is increasing at a rate of
41
41
41
people per hour. At what value of
P
P
P
is
P
(
t
)
P(t)
P
(
t
)
growing the fastest?
\newline
Answer:
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The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of people on a beach is modeled by the following differential equation:
\newline
d
P
d
t
=
595
2664
P
(
1
−
P
510
)
\frac{d P}{d t}=\frac{595}{2664} P\left(1-\frac{P}{510}\right)
d
t
d
P
=
2664
595
P
(
1
−
510
P
)
\newline
At
t
=
0
t=0
t
=
0
, the number of people on the beach is
222
222
222
and is increasing at a rate of
28
28
28
people per hour. At what value of
P
P
P
is
P
(
t
)
P(t)
P
(
t
)
growing the fastest?
\newline
Answer:
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The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of students who heard a rumor is modeled by the following differential equation:
\newline
d
P
d
t
=
6500
19153
P
(
1
−
P
500
)
\frac{d P}{d t}=\frac{6500}{19153} P\left(1-\frac{P}{500}\right)
d
t
d
P
=
19153
6500
P
(
1
−
500
P
)
\newline
At
t
=
0
t=0
t
=
0
, the number of students who heard the rumor is
179
179
179
and is increasing at a rate of
39
39
39
students per hour. Find
lim
t
→
∞
P
(
t
)
\lim _{t \rightarrow \infty} P(t)
lim
t
→
∞
P
(
t
)
.
\newline
Answer:
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Target heart rate is determined by multiplying the reserve heart rate by training intensity level,
p
p
p
, where
0
≤
p
≤
1
0 \leq p \leq 1
0
≤
p
≤
1
, and adding the resting heart rate. An adult woman with a resting heart rate of
65
65
65
beats per minute (
bpm
\text{bpm}
bpm
) and a reserve heart rate of
125
bpm
125\text{bpm}
125
bpm
is told by a trainer that her target heart rate should not exceed
140
bpm
140\text{bpm}
140
bpm
. What is the maximum training intensity level,
p
p
p
, that is consistent with her trainer's advice?
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The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of fox on an island is modeled by the following differential equation:
\newline
d
P
d
t
=
1386
3791
P
(
1
−
P
616
)
\frac{d P}{d t}=\frac{1386}{3791} P\left(1-\frac{P}{616}\right)
d
t
d
P
=
3791
1386
P
(
1
−
616
P
)
\newline
At
t
=
0
t=0
t
=
0
, the number of fox on the island is
170
170
170
and is increasing at a rate of
45
45
45
fox per day. At what value of
P
P
P
is
P
(
t
)
P(t)
P
(
t
)
growing the fastest?
\newline
Answer:
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The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of students who heard a rumor is modeled by the following differential equation:
\newline
d
P
d
t
=
30804
90965
P
(
1
−
P
906
)
\frac{d P}{d t}=\frac{30804}{90965} P\left(1-\frac{P}{906}\right)
d
t
d
P
=
90965
30804
P
(
1
−
906
P
)
\newline
At
t
=
0
t=0
t
=
0
, the number of students who heard the rumor is
115
115
115
and is increasing at a rate of
34
34
34
students per hour. Find
lim
t
→
∞
P
′
(
t
)
\lim _{t \rightarrow \infty} P^{\prime}(t)
lim
t
→
∞
P
′
(
t
)
.
\newline
Answer:
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Today, there were
2
2
2
members absent from the band. The present members folded
25
25
25
programs each, for a total of
525
525
525
programs.
\newline
What question does the equation
\newline
525
=
25
(
x
−
2
)
525=25(x-2)
525
=
25
(
x
−
2
)
help answer?
\newline
Choose
1
1
1
answer:
\newline
(A) How many programs did each member fold?
\newline
(B) How many programs would the members fold if no one were absent?
\newline
(C) How many members are in the band when no one is absent?
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Assume that
x
x
x
and
y
y
y
are both differentiable functions of
t
t
t
and find the required values of
d
y
d
t
\frac{dy}{dt}
d
t
d
y
and
d
x
d
t
\frac{dx}{dt}
d
t
d
x
.
x
y
=
6
xy=6
x
y
=
6
\newline
(a) Find
d
y
d
t
\frac{dy}{dt}
d
t
d
y
, given
x
=
6
x=6
x
=
6
and
d
x
d
t
=
13
\frac{dx}{dt}=13
d
t
d
x
=
13
.
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The fraction
p
p
p
of the population who has heard a breaking news story increases at a rate proportional to the fraction of the population who has not yet heard the news story.
\newline
Which equation describes this relationship?
\newline
Choose
1
1
1
answer:
\newline
(A)
d
p
d
t
=
−
k
p
\frac{d p}{d t}=-k p
d
t
d
p
=
−
k
p
\newline
(B)
d
p
d
t
=
1
−
k
p
\frac{d p}{d t}=1-k p
d
t
d
p
=
1
−
k
p
\newline
(C)
d
p
d
t
=
k
p
\frac{d p}{d t}=k p
d
t
d
p
=
k
p
\newline
(D)
d
p
d
t
=
k
(
1
−
p
)
\frac{d p}{d t}=k(1-p)
d
t
d
p
=
k
(
1
−
p
)
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The fraction
p
p
p
of the population who has heard a breaking news story increases at a rate proportional to the fraction of the population who has not yet heard the news story.
\newline
Which equation describes this relationship?
\newline
Choose
1
1
1
answer:
\newline
(A)
d
p
d
t
=
k
p
\frac{d p}{d t}=k p
d
t
d
p
=
k
p
\newline
(B)
d
p
d
t
=
1
−
k
p
\frac{d p}{d t}=1-k p
d
t
d
p
=
1
−
k
p
\newline
(C)
d
p
d
t
=
k
(
1
−
p
)
\frac{d p}{d t}=k(1-p)
d
t
d
p
=
k
(
1
−
p
)
\newline
(D)
d
p
d
t
=
−
k
p
\frac{d p}{d t}=-k p
d
t
d
p
=
−
k
p
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A puppy gains weight,
w
w
w
, at a rate approximately inversely proportional to its age,
t
t
t
, in months.
\newline
Which equation describes this relationship?
\newline
Choose
1
1
1
answer:
\newline
(A)
d
w
d
t
=
k
w
\frac{d w}{d t}=\frac{k}{w}
d
t
d
w
=
w
k
\newline
(B)
d
w
d
t
=
k
t
\frac{d w}{d t}=\frac{k}{t}
d
t
d
w
=
t
k
\newline
(C)
d
w
d
t
=
k
t
\frac{d w}{d t}=k t
d
t
d
w
=
k
t
\newline
(D)
d
w
d
t
=
k
w
\frac{d w}{d t}=k w
d
t
d
w
=
k
w
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The water level of a strait is changing at a rate of
3
2
sin
(
2
−
t
2
)
\frac{3}{2} \sin \left(2-\frac{t}{2}\right)
2
3
sin
(
2
−
2
t
)
centimeters per hour (where
t
t
t
is the hours since midnight).
\newline
By approximately how many centimeters does the water level change between
t
=
1
t=1
t
=
1
and
t
=
4
t=4
t
=
4
?
\newline
Choose
1
1
1
answer:
\newline
(A)
1
1
1
.
4
4
4
\newline
(B)
1
1
1
.
5
5
5
\newline
(C)
2
2
2
.
8
8
8
\newline
(D)
3
3
3
.
0
0
0
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The population of a town grows at a rate of
2
e
0.2
t
−
t
2 e^{0.2 t}-t
2
e
0.2
t
−
t
people per year (where
t
t
t
is the number of years).
\newline
Approximately by how many people did the population grow between
t
=
0
t=0
t
=
0
and
t
=
5
t=5
t
=
5
?
\newline
Choose
1
1
1
answer:
\newline
(A)
5
5
5
\newline
(B)
10
10
10
\newline
(C)
15
15
15
\newline
(D)
20
20
20
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A water tank is filled at a rate of
r
(
t
)
r(t)
r
(
t
)
liters per minute (where
t
t
t
is the time in minutes).
\newline
What does
∫
1
7
r
′
(
t
)
d
t
\int_{1}^{7} r^{\prime}(t) d t
∫
1
7
r
′
(
t
)
d
t
represent?
\newline
Choose
1
1
1
answer:
\newline
(A) The rate at which the tank was filled at
t
=
7
t=7
t
=
7
.
\newline
(B) The amount of water filled between
t
=
1
t=1
t
=
1
and
t
=
7
t=7
t
=
7
.
\newline
(C) The change in the rate of filling between
t
=
1
t=1
t
=
1
and
t
=
7
t=7
t
=
7
.
\newline
(D) The average rate of filling between
t
=
1
t=1
t
=
1
and
t
=
7
t=7
t
=
7
.
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Water fills a bathtub at a rate of
r
(
t
)
r(t)
r
(
t
)
liters per minute (where
t
t
t
is the time in minutes).
\newline
What does
∫
2
3
r
(
t
)
d
t
=
6
\int_{2}^{3} r(t) d t=6
∫
2
3
r
(
t
)
d
t
=
6
mean?
\newline
Choose
1
1
1
answer:
\newline
(A) The rate of the water filling the bathtub increased by
6
6
6
liters per minute between minutes
2
2
2
and
3
3
3
.
\newline
(B) The bathtub fills with an additional
6
6
6
liters of water every minute.
\newline
(C) There were
6
6
6
liters of water in the bathtub by the end of
3
3
3
minutes.
\newline
(D) The water in the bathtub increased by
6
6
6
liters during the third minute.
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The processing speeds of Peach brand computers is increasing at a rate of
r
(
t
)
r(t)
r
(
t
)
megahertz per year (where
t
t
t
is the time in years). When
t
=
4
t=4
t
=
4
, the computers had a processing speed of
2500
2500
2500
megahertz.
\newline
What does
2500
+
∫
4
7
r
(
t
)
d
t
2500+\int_{4}^{7} r(t) d t
2500
+
∫
4
7
r
(
t
)
d
t
represent?
\newline
Choose
1
1
1
answer:
\newline
(A) The processing speed of the computers when
t
=
7
t=7
t
=
7
\newline
(B) The average processing speed of the computers between
t
=
4
t=4
t
=
4
and
t
=
7
t=7
t
=
7
\newline
(C) The net change in processing speeds of the computers between
t
=
4
t=4
t
=
4
and
t
=
7
t=7
t
=
7
\newline
(D) The average rate of change in the processing speed of the computers between
t
=
4
t=4
t
=
4
and
t
=
7
t=7
t
=
7
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The greater the daily dose of vitamin
C
C
C
patients take, the greater their plasma concentration of vitamin
C
\mathrm{C}
C
becomes.
\newline
This relationship increases at a rate of
0.1
e
−
0.002
x
0.1 e^{-0.002 x}
0.1
e
−
0.002
x
micromoles per milligram of daily dose.
\newline
By approximately how many micromoles does the plasma concentration increase between
x
=
100
x=100
x
=
100
and
x
=
200
x=200
x
=
200
?
\newline
Choose
1
1
1
answer:
\newline
(A)
0
0
0
.
067
067
067
\newline
(B)
0
0
0
.
15
15
15
\newline
(C)
7
7
7
.
42
42
42
\newline
(D)
33
33
33
.
5
5
5
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Water fills a bathtub at a rate of
r
(
t
)
r(t)
r
(
t
)
liters per minute (where
t
t
t
is the time in minutes).
\newline
What does
∫
2
3
r
(
t
)
d
t
=
6
\int_{2}^{3} r(t) d t=6
∫
2
3
r
(
t
)
d
t
=
6
mean?
\newline
Choose
1
1
1
answer:
\newline
(A) The bathtub fills with an additional
6
6
6
liters of water every minute.
\newline
(B) The water in the bathtub increased by
6
6
6
liters during the third minute.
\newline
(C) The rate of the water filling the bathtub increased by
6
6
6
liters per minute between minutes
2
2
2
and
3
3
3
.
\newline
(D) There were
6
6
6
liters of water in the bathtub by the end of
3
3
3
minutes.
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A school charges
$
6
\$6
$6
per ticket to a dance. They spend
$
300
\$300
$300
on music and decorations. The expression
6
n
−
300
6n-300
6
n
−
300
describes the amount of profit the school makes from the dance if
n
n
n
students buy tickets. What does
6
n
6n
6
n
represent in this context?
\newline
Choices:
\newline
(A)The amount of money the school charges for all `\n` tickets
\newline
(B)The amount of money the school spends on the dance
\newline
(C)The number of students who buy tickets
\newline
(D)The profit the school earns per ticket
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