A flexible cable suspended between two vertical supports is hanging under its own weight. The weight of a horizontal roadbed is distributed evenly along the x-axis with the density p = 0.1 lb/ft. The coordinate system is chosen so that the y-axis passes through the lowest point P₁ and the x-axis is a = 2 ft below P₁. If the absolute value of tension force in the cable at point P₁ is T = 3.5, then the initial value problem that governs the shape of the cable is (Use the notation dy/dx for the first derivative) d'y dx² y(0) = (0) = The solution of this problem is y = y(x), where y(x) =

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
6th Edition
ISBN:9781337111348
Author:Bruce Crauder, Benny Evans, Alan Noell
Publisher:Bruce Crauder, Benny Evans, Alan Noell
ChapterA: Appendix
SectionA.2: Geometric Constructions
Problem 9P: A soda can is made from 40 square inches of aluminum. Let x denote the radius of the top of the can,...
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A flexible cable suspended between
two vertical supports is hanging under its own
weight. The weight of a horizontal roadbed is
distributed evenly along the x-axis with the
density p = 0.1 lb/ft. The coordinate system is
chosen so that the y-axis passes through the
lowest point P₁ and the x-axis is a = 2 ft
below P₁. If the absolute value of tension force
in the cable at point P₁ is T = 3.5, then the
initial value problem that governs the shape of
the cable is (Use the notation dy/dx for the first
derivative)
d'y =
dx²
y(0) =
(0) =
The solution of this problem is y = y(x), where
y(x) =
Transcribed Image Text:A flexible cable suspended between two vertical supports is hanging under its own weight. The weight of a horizontal roadbed is distributed evenly along the x-axis with the density p = 0.1 lb/ft. The coordinate system is chosen so that the y-axis passes through the lowest point P₁ and the x-axis is a = 2 ft below P₁. If the absolute value of tension force in the cable at point P₁ is T = 3.5, then the initial value problem that governs the shape of the cable is (Use the notation dy/dx for the first derivative) d'y = dx² y(0) = (0) = The solution of this problem is y = y(x), where y(x) =
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