
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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this is a practice exam question given to me by my lecturer, he is on holiday so im reaching out over here. he has provided the answer and some working out, however i am struggling to follow his method. particularly part a. im struggling to understand why he has put -ls^2 and 4a^2 as the

Transcribed Image Text:(ii)
(a)
-
y² 12 = a² => (a + h)²−(LAB/2)² = a² for Is
=
LAB/2=10m
-2a+ √4a²-4(-12)
That is h² + 2ah−(LAB/2)² = 0 => h = ·
at point B, we have y = H = a + h = 10 +4.14 = 14.14 m, and a=10 m.
= 4.14m
2
(b) as, y = acosh(), and a=10 m and y = 14.14 m
D
x
2
=
acosh¹ (#) = 10cosh 1 (14.14) = 8.88 m,
D = 17.6 m

Transcribed Image Text:A
LAB
H
a
D
y
B
s(x, y)
h
X
Figure 5.1
The graph of a catenary in the Cartesian coordinates is shown in Figure 5.1.
The two poles, A and B, have the same height to the x-axis, and the same
distance to the y-axis. For a given point on the catenary, s(x, y), we have,
y = acosh().
(1)
It is also known that the length of the catenary segment between C (the
lowest point of the catenary) and point s, i.e., Is, is given by,
y² - 12 = a².
(2)
If the total length of the catenary between A and B, i.e., LAB, is 20 m and a is
10 m, determine
(i) the height of the pole, i.e. H, as shown in the figure.
(ii) the distance between A and B, i.e., D, as shown in the figure.
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