Chapter 6 | Applications of Integration
753
Figure 6.84 A hyperbolic cosine function forms the shape of a catenary.
Example 6.51 Using a Catenary to Find the Length of a Cable
Assume a hanging cable has the shape 10cosh( x /10) for −15≤ x ≤15, where x is measured in feet. Determine the length of the cable (in feet).
Solution Recall from Section 2.4 that the formula for arc length is Arc Length = ∫ a b 1+ ⎡ ⎣ f ′( x ) ⎤ ⎦ 2 dx . We have f ( x ) =10cosh( x /10), so f ′( x ) = sinh( x /10). Then Arc Length = ∫ a b 1+ ⎡ ⎣ f ′( x ) ⎤ ⎦ 2 dx = ∫ −15 15 1+sinh 2 ⎛ ⎝ x 10 ⎞ ⎠ dx . Now recall that 1+sinh 2 x =cosh 2 x , so we have Arc Length = ∫ −15 15 1+sinh 2 ⎛ ⎝ x 10 ⎞ ⎠ dx = ∫ −15 15 cosh ⎛ ⎝ x 10 ⎞ ⎠ dx =10sinh ⎛ ⎝ x 10 ⎞ ⎠ | −15 15 =10 ⎡ ⎣ sinh ⎛ ⎝ 3 2 ⎞ ⎠ −sinh ⎛ ⎝ − 3 2
⎞ ⎠ ⎤ ⎦ =20sinh ⎛
⎞ ⎠
⎝ 3 2
≈ 42.586 ft.
6.51 Assume a hanging cable has the shape 15cosh( x /15) for −20≤ x ≤20. Determine the length of the cable (in feet).
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