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Math Problems
Precalculus
Find instantaneous rates of change
One leg of a right triangle is decreasing at a rate of
5
5
5
kilometers per hour and the other leg of the triangle is increasing at a rate of
14
14
14
kilometers per hour.
\newline
At a certain instant, the decreasing leg is
3
3
3
kilometers and the increasing leg is
9
9
9
kilometers.
\newline
What is the rate of change of the area of the right triangle at that instant (in square kilometers per hour)?
\newline
Choose
1
1
1
answer:
\newline
(A)
1
1
1
.
5
5
5
\newline
(B)
111
111
111
\newline
(C)
−
1
-1
−
1
.
5
5
5
\newline
(D)
−
111
-111
−
111
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How does
f
(
t
)
=
1
0
t
f(t) = 10^t
f
(
t
)
=
1
0
t
change over the interval from
t
=
9
t = 9
t
=
9
to
t
=
10
t = 10
t
=
10
?
\newline
Choices:
\newline
(A)
f
(
t
)
f(t)
f
(
t
)
increases by
1
,
000
%
1,000\%
1
,
000%
\newline
(B)
f
(
t
)
f(t)
f
(
t
)
decreases by a factor of
10
10
10
\newline
(C)
f
(
t
)
f(t)
f
(
t
)
increases by
10
%
10\%
10%
\newline
(D)
f
(
t
)
f(t)
f
(
t
)
increases by
t
=
9
t = 9
t
=
9
0
0
0
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Recall
π
\pi
π
is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is
π
=
c
d
\pi=\frac{c}{d}
π
=
d
c
. This seems to contradict the fact that
π
\pi
π
is irrational. How will you resolve this contradiction?
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Show for all
N
∈
N
∖
{
0
}
N \in \mathbb{N}\setminus\{0\}
N
∈
N
∖
{
0
}
and all
a
≠
1
a \neq 1
a
=
1
:
\newline
1
1
−
a
=
a
N
1
−
a
+
∑
n
=
0
N
−
1
a
n
\frac{1}{1-a} = \frac{a^N}{1-a} + \sum_{n=0}^{N-1}a^n
1
−
a
1
=
1
−
a
a
N
+
∑
n
=
0
N
−
1
a
n
.
\newline
Use this result to find a formula for
\newline
∑
n
=
0
∞
a
n
\sum_{n=0}^{\infty}a^n
∑
n
=
0
∞
a
n
.
\newline
What assumption did you have to make to obtain a converging result?
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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 bacteria in a tank is modeled by the following differential equation:
\newline
d
P
d
t
=
2
9849
P
(
598
−
P
)
\frac{d P}{d t}=\frac{2}{9849} P(598-P)
d
t
d
P
=
9849
2
P
(
598
−
P
)
\newline
At
t
=
0
t=0
t
=
0
, the number of bacteria in the tank is
196
196
196
and is increasing at a rate of
16
16
16
bacteria per minute. At what value of
P
P
P
does the graph of
P
(
t
)
P(t)
P
(
t
)
have an inflection point?
\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 algae in a tank is modeled by the following differential equation:
\newline
d
P
d
t
=
2317
10614
P
(
1
−
P
662
)
\frac{d P}{d t}=\frac{2317}{10614} P\left(1-\frac{P}{662}\right)
d
t
d
P
=
10614
2317
P
(
1
−
662
P
)
\newline
At
t
=
0
t=0
t
=
0
, the number of algae in the tank is
174
174
174
and is increasing at a rate of
28
28
28
algae per minute. At what value of
P
P
P
is
P
(
t
)
P(t)
P
(
t
)
growing the fastest?
\newline
Answer:
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