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Math 3301
Homework Set 3 – Solutions
10 Points
1. (3p ts)
Here’s the IVP’s we need for this problem.
( ) ( )
() () ()
0
5
0
3
150
7
4
9
4
8
0
50
P
r
P
PP
P
r
P
r
¢
=
==
¢
=
+
=
=
Solving the first gives,
( )
( )
3
1
3
5
0
15
0
5
0
l
n3
r
tr
P
=
ee
The second IVP is now,
( ) ( )
1
3
l
n
3
8
0
50
P
¢
=
The solution is,
()
ln3
3
24
ln3
28.1542
t
Pt
=+
e
From this we can see that they will survive to take over the world!
2. (3 pts)
Here’s the IVP’s for this case.
( )
( )
10
2
11
7
5
15
30
2
2 2
22
7
55
9.
8
9.
8
02
9.
8
9.
8
30
c
vv
v v
v
vt
¢
= =
¢
The solution to the first one is,
2
15
1
73.
5
71.5
t

=
e
and we’ll need to set this equal to 30 and solve to determine the time when the second parachute
opens.
1
5
15
3
0
73.
5
71.
5
0.608
4
3.7269
tt
t

=

ﬁ=
&=
Eventually we’ll need the “height” function for this so let’s get that and the “height” of the sky diver at
this time.
() ()
1
5
15
1
73.
5
71.
5
0
0
73.
5
536.2
5
536.25
s
t
d
t
s
s
=

=
&
=
±
The sky diver is now
( )
3.726
9
63.9325
sm
=
below the plane.
The second IVP is now,
( )
2
222
5
9.
8
3.726
9
30
vvv
¢
=
The solution to this one is,
2
5
2
24.
5
24.4226
t

e
Next we’ll need the “height” function.
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 Spring '08
 Deng
 Math, Differential Equations, Equations

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