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Question Number 75463 by indalecioneves last updated on 11/Dec/19

	Answered by mr W last updated on 12/Dec/19

	$s h a p e o f c a b l e i s c a t e n a r y :$ $\frac{L}{d} = \frac{2 \sinh \frac{d}{x}}{\frac{d}{x}}$ $\frac{f}{d} = \frac{\cosh \frac{d}{x} - 1}{\frac{d}{x}}$ $w i t h t = \frac{d}{x} a s p a r a m e t e r :$ ${\begin{cases} \frac{L}{d} = \frac{2 \sinh t}{t} \\ \frac{f}{d} = \frac{\cosh t - 1}{t} \end{cases}$ $w i t h \frac{f}{d} = \frac{100}{200} = 0.5$ $w e g e t \frac{L}{d} \approx 2.3 f r o m d i a g r a m$ $\Rightarrow L \approx 2.3 \times 200 = 460 m$
	Commented by mr W last updated on 11/Dec/19

	Commented by indalecioneves last updated on 11/Dec/19

	$T h a n k y o u, S i r!$ $G o d b l e s s y o u!$
	Answered by mr W last updated on 11/Dec/19
	$Approximation as parabola: y=f((x/d))^2 y′=((2fx)/d^2 ) L=2∫_0 ^d (√(1+(((2fx)/d^2 ))^2 ))dx (L/d)=2∫_0 ^d (√(1+(((2fx)/d^2 ))^2 ))d((x/d)) (L/d)=2∫_0 ^1 (√(1+(((2fu)/d))^2 ))du (L/d)=(d/f)∫_0 ^1 (√(1+(((2fu)/d))^2 ))d(((2fu)/d)) ⇒(L/d)=(1/((2f)/d)){ln [((2f)/d)+(√(1+(((2f)/d))^2 ))]+((2f)/d)(√(1+(((2f)/d))^2 ))} with ((2f)/d)=((2×100)/(200))=1 ⇒(L/d)=ln (1+(√2))+(√2)=2.296 ⇒L=2.296×200≈460 m$
	$A p p r o x i m a t i o n a s p a r a b o l a :$ $y = f {(\frac{x}{d})}^{2}$ $y^{'} = \frac{2 f x}{d^{2}}$ $L = 2 \int_{0}^{d} \sqrt{1 + {(\frac{2 f x}{d^{2}})}^{2}} d x$ $\frac{L}{d} = 2 \int_{0}^{d} \sqrt{1 + {(\frac{2 f x}{d^{2}})}^{2}} d (\frac{x}{d})$ $\frac{L}{d} = 2 \int_{0}^{1} \sqrt{1 + {(\frac{2 f u}{d})}^{2}} d u$ $\frac{L}{d} = \frac{d}{f} \int_{0}^{1} \sqrt{1 + {(\frac{2 f u}{d})}^{2}} d (\frac{2 f u}{d})$ $\Rightarrow \frac{L}{d} = \frac{1}{\frac{2 f}{d}} {\ln [\frac{2 f}{d} + \sqrt{1 + {(\frac{2 f}{d})}^{2}}] + \frac{2 f}{d} \sqrt{1 + {(\frac{2 f}{d})}^{2}}}$ $w i t h \frac{2 f}{d} = \frac{2 \times 100}{200} = 1$ $\Rightarrow \frac{L}{d} = \ln (1 + \sqrt{2}) + \sqrt{2} = 2.296$ $\Rightarrow L = 2.296 \times 200 \approx 460 m$
	Commented by mr W last updated on 11/Dec/19

	Commented by mr W last updated on 11/Dec/19
	$we see the approximation as parabola is exact enough if the cable is flat. the advantage is that we can get L/d directly from f/d using the formula: (L/d)=(1/((2f)/d)){ln [((2f)/d)+(√(1+(((2f)/d))^2 ))]+((2f)/d)(√(1+(((2f)/d))^2 ))}$
	$w e s e e t h e a p p r o x i m a t i o n a s$ $p a r a b o l a i s e x a c t e n o u g h i f t h e$ $c a b l e i s f l a t .$ $t h e a d v a n t a g e i s t h a t w e c a n g e t$ $L / d d i r e c t l y f r o m f / d u s i n g t h e$ $f o r m u l a :$ $\frac{L}{d} = \frac{1}{\frac{2 f}{d}} {\ln [\frac{2 f}{d} + \sqrt{1 + {(\frac{2 f}{d})}^{2}}] + \frac{2 f}{d} \sqrt{1 + {(\frac{2 f}{d})}^{2}}}$
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