Exercises 3.4
⇒
x
(
t
) = 98
.
1
t
+ 981
e
−
t/
10
−
981
.
To find out when the object will hit the ground, we solve
x
(
t
) = 1000 for
t
. Therefore, we
have
1000 = 98
.
1
t
+ 981
e
−
t/
10
−
981
⇒
98
.
1
t
+ 981
e
−
t/
10
= 1981
.
In this equation, if we ignore the term 981
e
−
t/
10
we will find that
t
≈
20
.
2. But this means
that we have ignored the term similar to 981
e
−
2
≈
132
.
8 which we see is to large to ignore.
Therefore, we must try to approximate
t
. We will use Newton’s method on the equation
f
(
t
) = 98
.
1
t
+ 981
e
−
t/
10
−
1981 = 0
.
(If we can find a root to this equation, we will have found the
t
we want.) Newton’s method
generates a sequence of approximations given by the formula
t
n
+1
=
t
n
−
f
(
t
n
)
f
(
t
n
)
.
Since
f
(
t
) = 98
.
1
−
98
.
1
e
−
t/
10
= 98
.
1
(
1
−
e
−
t/
10
)
, the recursive equation above becomes
t
n
+1
=
t
n
−
t
n
+ 10
e
−
t
n
/
10
−
(1981
/
98
.
1)
1
−
e
−
t
n
/
10
.
(3.10)
To start the process, let
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Numerical Analysis, Equations, Approximation, 10g, 5g, 18.6437 sec

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