Chapter 2: Differentiation
Section 2.9: The Hyperbolic Functions and Their Derivatives
Example 2.9.4
It can be shown that for a sufficiently idealized wire cable of length c, hanging between two supports at x1,y1 and x2,y2 in a vertical xy-plane, the equation describing the shape of the cable is of the form yx=k coshd+x/k−λ, with c constrained by the equation c=k sinhd+x2/k−sinhd+x1/k. In this context, the curve is called a catenary, from the Latin catina (chain). If such a cable of length c=2 hangs between the points 0,1 and 1,3/2, find the equation of the resulting catenary, and draw its graph.
(In the typical North American pronunciation of catenary, the accent is on the first syllable; in the British, it is on the second: thus, cat'-ěn-ary and că-tēn'-ery, respectively.)
Solution
Control-drag the equation yx=… Context Panel: Assign Function
yx=k coshd+x/k−λ→assign as functiony
Implement the following steps
Write the three equations y0=1,y1=3/2, and c=… Press the Enter key.
Context Panel: Solve≻Numerically Solve
Expression palette: Evaluation template≻yx evaluated at numeric solution (reference by equation label). Press the Enter key.
Context Panel: Plot Builder≻2-D plot Domain: 0≤x≤1 view: axis[2] ≻ 0,3/2
y0=1,y1=3/2,2=k sinhd+1/k−sinhd
k⁢cosh⁡d−λ=1,k⁢cosh⁡d+1k−λ=32,2=k⁢sinh⁡d+1k−sinh⁡d
→solve
d=−1.864503925,k=0.2358583199,λ=−0.2207586498
yxx=a|f(x)
0.2358583199⁢cosh⁡−1.864503925+4.239833475⁢x+0.2207586498
→
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