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Fig. 1.12 The figure
illustrates a circle with unit
radius. The marked segment
has length =4
1
0.8
0.6
0.4
0.2
0
−0.2
−0.4
−0.6
−0.8
−1
−1 −0.8−0.6−0.4−0.2
0
0.2 0.4 0.6 0.8
1
Fig. 1.13
One patch of the
200
cable
180
160
140
120
100
80
60
40
20
0
0
50
100
150
200
250
y.x/ D 0:003x 2 ;
where x goes from x D 0 to x D 250 meters, see Fig. 1.13 . Use formula (1.51)
to show that the total length of the entire cable, counting both sides of the bridge,
is given by
L D 4 Z 250
0
p 1 C .0:006x/ 2 dx:
(i) Use the composite trapezoidal scheme to estimate 14 L:
14 Using n
D
100,wegetL
4
324:91
D
1299:6 m.
 
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