(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 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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10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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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 quadratic equation is 2a for b and Lab/2 for c ? i can understand to an extent up until he puts it in the quadratic equation. many thanks for your help ! 

(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:(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
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.
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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