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Question Number 40906 by Penguin last updated on 29/Jul/18

f(x) = 5.687cosh((x/(5.687)))−5.687  L=∫_(-11) ^(11) (√(1+[f ′(x)]^2 ))dx

f(x)=5.687cosh(x5.687)5.687L=11111+[f(x)]2dx

Commented by maxmathsup by imad last updated on 29/Jul/18

f(x)=λch((x/λ))−λ with λ=5,687 ⇒f^′ (x)=sh((x/λ)) ⇒  L=∫_(−11) ^(11) (√(1+sh^2 ((x/λ))))dx= ∫_(−11) ^(11)  ch((x/λ))dx =_((x/λ)=t)   ∫_(−((11)/λ)) ^((11)/λ)  ch(t)λdt  =2λ ∫_0 ^((11)/λ)  ch(t)dt =2λ [sh(t)]_0 ^((11)/λ)  =2λ sh(((11)/λ))=2λ  ((e^((11)/λ)  −e^(−((11)/λ)) )/2)  =5,687( e^((11)/(5,687))  −e^(−((11)/(5,687))) ) .

f(x)=λch(xλ)λwithλ=5,687f(x)=sh(xλ)L=11111+sh2(xλ)dx=1111ch(xλ)dx=xλ=t11λ11λch(t)λdt=2λ011λch(t)dt=2λ[sh(t)]011λ=2λsh(11λ)=2λe11λe11λ2=5,687(e115,687e115,687).

Answered by tanmay.chaudhury50@gmail.com last updated on 29/Jul/18

f(x)=acosh((x/a))−a  f(x)=a((e^(x/a) +e^((−x)/a) )/2)−a  f′(x)=(a/2)(e^(x/a) .(1/a)+e^(((−x)/a) ) .−(1/a))  f′(x)=((e^(x/a) −e^((−x)/a) )/2)  [f′(x)]^2 =((e^(((2x)/a) ) +e^((−2x)/a) −2)/4)  1+[f′(x)]^2 =(((e^(x/a) +e^((−x)/a) )/2))^2   L=∫_(−11) ^(11) ((e^((x/a) ) +e^((−x)/a) )/2)dx  =(1/2)∣e^(x/a) .a−e^(((−x)/a) ) .a ∣_(−11) ^(11)   =(a/2){(e^((11)/a) −e^(((−11)/a)  ) )−(e^(((−11)/a) ) −e^((11)/a) )}  =a.(e^(((11)/a) ) −e^(((−11)/a) ) )  =5.687(e^(((11)/(5.687)) ) −e^(((−11)/(5.687)) ) )  plscheck...

f(x)=acosh(xa)af(x)=aexa+exa2af(x)=a2(exa.1a+exa.1a)f(x)=exaexa2[f(x)]2=e2xa+e2xa241+[f(x)]2=(exa+exa2)2L=1111exa+exa2dx=12exa.aexa.a1111=a2{(e11ae11a)(e11ae11a)}=a.(e11ae11a)=5.687(e115.687e115.687)plscheck...

Answered by MJS last updated on 29/Jul/18

(d/dx)[acosh (x/a) −a]=sinh (x/a)  (√(1+sinh (x/a)))=cosh (x/a)  ∫cosh (x/a) dx=asinh (x/a)  ⇒ L≈38.525

ddx[acoshxaa]=sinhxa1+sinhxa=coshxacoshxadx=asinhxaL38.525

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