Section 3.8, problem 6.
+1 =
xt
3
xt
xt
20
3x2
t
Start with initial
guess 0 = 3.
3 20
3
1=3
333 2 = 2 7407
1 20
2= 1
3 21 = 2 7147
3 20
2
3= 1
3 22 = 2 7144
x
:
x
x
x
x
x
x
:
x
x
:
x
Section 3.8, problem 28.
= ( ) +1
1. To nd an equilibrium we need to solve:
Mt
M
p Mt Mt
= p( M ) M + 1;
which is equivalent to nding a zero of
( )=
f x
( )
x
p x x
1
:
(0) = 1 0, and (10) = 5 69 0, so by the intermediate value theorem
there is an equilibrium of the above system between = 0 and = 10.
2. Start with 0 = 10. Then 1 = 4 3109, 2 = 3 5211, 3 = 3 2284,
4 = 3 1039, 5 = 3 0481, 6 = 3 0225, 7 = 3 0107, 8 = 3 0052,
7 = 3 0026, 9 = 3 0014. According to this is close to 3.
3. NOw do the same thing with the Newton's method.
( )
1 =
0 9 01
1
+1 =
0
0
1
0 9 01
1 ( )
( )
1 + 0 09
Start with 0 = 10.
Then 1 = 4 3109, 2 = 3 1273, 3 = 3 0020, 4 = 3 0020, 5 = 3 0004.
This sytem coverges to = 3 much faster.
f
<
f
:
>
M
M
M
:
M
M
:
xt
M
:
M
:
M
M
:
xt
xt
M
:
M
:
:
M
p xa xt
p
:
M
:
M
xt xt
xt
xt
p xt
: xt
: e
:
e
: xt
xt
xt
: e
: xt
:
x
x
:
x
:
x
:
x
:
x
:
M
p.128, 2
a. = (3 16) + 2, where is time in hours from 5PM on Tuesday and
is the number of bacteria in hundreds of millions.
b. 11 = (3 16) + 2. Solving for : = 48, i.e at 5pm on Thursday.
y
=
t
=
t
t
t
t
1
y
c. Their culrure is described with 2 = (1 4 16) + 2. Need to solve:
y
: =
t
(3 16) + 2 = 2((1 4 16) + 2)
=
t
t
: =
t
:
= 32 2 = 160 hours, i.e. at 9am on tuesday a week later.
=:
p.128, 4
a. 3 , where time step is 2 hours
b. 162 107
c. 0 3
d. 2 107 3 = 109, i.e 3 = 50, = 3 57 time units or = 3 57 2 = 7 14
hours
xt
x
t
t
t
t
:
t
:
:
p.233, 28
b.
c.
d.
(2) = 2, (2 5) = 3 25. Secant line = 2 + 2 5( 2).
2 + 2 . 0 (2) = 2 thousands of cubic microns per minute.
It is growing ( 0(2) 0)
V
V
0
V (t) =
:
t
V
:
y
:
t
V
>
p. 234,34
a. (1) = 1 3; (2) = 4. Average rate of change (.4-1/3)/1=1/15
b. = 4 + 0 07( 2)
P
y
=
:
P
:
:
t
2