The correct answer for the ratio of the area under the catenary to its arc length is 12 : 1 .
y = 12cosh(x/12) Since the анс length from FGN) on the interval [c, d] is Arc length =
Let y' =f'(x) prime = derivative of 12cosh(x/12) for x = Sinh(x/12) .
. A(x) length = ∫√1+ Sinh² (1/2). dx
= int c ^ d sqrt(cosh^2 (alpha/12)) * dx
= integrate d from c to d cosh( x 12 )* dx
=[12. sinh(x/12) ] e ^ d
= 12[sinh(d/12) - sinh(c/12)]
and area under f(x) on the interval [c, d]
Area = ∫ f(x).dx from [c,d ]
= 144[sinh(d/12) - sinh(c/12)]
the area ratio of the area under the catenary to it's length is
= 144[sinh((d)/12) - sinh(c/12))/ 12[sinh( d/ 12 ) -sinh( c /12 )] .
=12
So, ratio is 12:1
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