1
MATH 141B
QUIZ#8 Solutions
Date: Tues March 25, 2008
1.
Separating and integrating gives
−
20
integraltext
c
−
2
dc
=
integraltext
dt
, or
20
c
=
t
+
K,
where
K
is a constant.
Since
c
(0) = 0
.
1 = 1
/
10
,
we get
K
= 200. Therefore,
20
c
=
t
+ 200 and hence
c
20
=
1
t
+ 200
.
This gives
c
(
t
) =
20
t
+ 200
.
Setting this equal to 0
.
05 = 1
/
20
,
gives
20
t
+ 200
=
1
20
.
Solving for
t
, we get
t
= 200 seconds, which is equal to 3 minutes and 20 seconds.
Therefore, A) is the answer.
2.
We need to solve
N
′
(
t
) = 0
.
The fast way of solving this is to realize we are given the
derivative
dN
dt
=
kN
·
sin(2
πt
)
.
Since
N
negationslash
= 0
,
this happens when sin(2
πt
) = 0
.
This means
2
πt
= 0 or
t
= 2
πt
=
π.
In other words,
t
= 0 or
t
= 1
/
2
.
Recall from last semester, that any
t
for which
N
′
(
t
) = 0 can be a max, a min, or neither. From the graph of sin, we conclude
that
t
= 0 gives a minimum, and
t
= 1
/
2 gives rise to a maximum.
The long way of solving this is to separate and integrate. Going through these details,
you get the function
N
(
t
) =
N
0
e

k
π
[cos(2
πt
)+1]
.
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 Spring '08
 MARCFABBRI
 Math, Logic, Derivative, Trigraph, Constant of integration, Solutions c2 dc

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