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](https://www.tinkutara.com/question/Q40906.png)
$${f}\left({x}\right)\:=\:\mathrm{5}.\mathrm{687cosh}\left(\frac{{x}}{\mathrm{5}.\mathrm{687}}\right)−\mathrm{5}.\mathrm{687} \\ $$$${L}=\int_{-\mathrm{11}} ^{\mathrm{11}} \sqrt{\mathrm{1}+\left[{f}\:'\left({x}\right)\right]^{\mathrm{2}} }{dx} \\ $$
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))) ) .](https://www.tinkutara.com/question/Q40921.png)
$${f}\left({x}\right)=\lambda{ch}\left(\frac{{x}}{\lambda}\right)−\lambda\:{with}\:\lambda=\mathrm{5},\mathrm{687}\:\Rightarrow{f}^{'} \left({x}\right)={sh}\left(\frac{{x}}{\lambda}\right)\:\Rightarrow \\ $$$${L}=\int_{−\mathrm{11}} ^{\mathrm{11}} \sqrt{\mathrm{1}+{sh}^{\mathrm{2}} \left(\frac{{x}}{\lambda}\right)}{dx}=\:\int_{−\mathrm{11}} ^{\mathrm{11}} \:{ch}\left(\frac{{x}}{\lambda}\right){dx}\:=_{\frac{{x}}{\lambda}={t}} \:\:\int_{−\frac{\mathrm{11}}{\lambda}} ^{\frac{\mathrm{11}}{\lambda}} \:{ch}\left({t}\right)\lambda{dt} \\ $$$$=\mathrm{2}\lambda\:\int_{\mathrm{0}} ^{\frac{\mathrm{11}}{\lambda}} \:{ch}\left({t}\right){dt}\:=\mathrm{2}\lambda\:\left[{sh}\left({t}\right)\right]_{\mathrm{0}} ^{\frac{\mathrm{11}}{\lambda}} \:=\mathrm{2}\lambda\:{sh}\left(\frac{\mathrm{11}}{\lambda}\right)=\mathrm{2}\lambda\:\:\frac{{e}^{\frac{\mathrm{11}}{\lambda}} \:−{e}^{−\frac{\mathrm{11}}{\lambda}} }{\mathrm{2}} \\ $$$$=\mathrm{5},\mathrm{687}\left(\:{e}^{\frac{\mathrm{11}}{\mathrm{5},\mathrm{687}}} \:−{e}^{−\frac{\mathrm{11}}{\mathrm{5},\mathrm{687}}} \right)\:. \\ $$
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...](https://www.tinkutara.com/question/Q40914.png)
$${f}\left({x}\right)={acosh}\left(\frac{{x}}{{a}}\right)−{a} \\ $$$${f}\left({x}\right)={a}\frac{{e}^{\frac{{x}}{{a}}} +{e}^{\frac{−{x}}{{a}}} }{\mathrm{2}}−{a} \\ $$$${f}'\left({x}\right)=\frac{{a}}{\mathrm{2}}\left({e}^{\frac{{x}}{{a}}} .\frac{\mathrm{1}}{{a}}+{e}^{\frac{−{x}}{{a}}\:} .−\frac{\mathrm{1}}{{a}}\right) \\ $$$${f}'\left({x}\right)=\frac{{e}^{\frac{{x}}{{a}}} −{e}^{\frac{−{x}}{{a}}} }{\mathrm{2}} \\ $$$$\left[{f}'\left({x}\right)\right]^{\mathrm{2}} =\frac{{e}^{\frac{\mathrm{2}{x}}{{a}}\:} +{e}^{\frac{−\mathrm{2}{x}}{{a}}} −\mathrm{2}}{\mathrm{4}} \\ $$$$\mathrm{1}+\left[{f}'\left({x}\right)\right]^{\mathrm{2}} =\left(\frac{{e}^{\frac{{x}}{{a}}} +{e}^{\frac{−{x}}{{a}}} }{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${L}=\int_{−\mathrm{11}} ^{\mathrm{11}} \frac{{e}^{\frac{{x}}{{a}}\:} +{e}^{\frac{−{x}}{{a}}} }{\mathrm{2}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mid{e}^{\frac{{x}}{{a}}} .{a}−{e}^{\frac{−{x}}{{a}}\:} .{a}\:\mid_{−\mathrm{11}} ^{\mathrm{11}} \\ $$$$=\frac{{a}}{\mathrm{2}}\left\{\left({e}^{\frac{\mathrm{11}}{{a}}} −{e}^{\frac{−\mathrm{11}}{{a}}\:\:} \right)−\left({e}^{\frac{−\mathrm{11}}{{a}}\:} −{e}^{\frac{\mathrm{11}}{{a}}} \right)\right\} \\ $$$$={a}.\left({e}^{\frac{\mathrm{11}}{{a}}\:} −{e}^{\frac{−\mathrm{11}}{{a}}\:} \right) \\ $$$$=\mathrm{5}.\mathrm{687}\left({e}^{\frac{\mathrm{11}}{\mathrm{5}.\mathrm{687}}\:} −{e}^{\frac{−\mathrm{11}}{\mathrm{5}.\mathrm{687}}\:} \right) \\ $$$${plscheck}… \\ $$
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](https://www.tinkutara.com/question/Q40922.png)
$$\frac{{d}}{{dx}}\left[{a}\mathrm{cosh}\:\frac{{x}}{{a}}\:−{a}\right]=\mathrm{sinh}\:\frac{{x}}{{a}} \\ $$$$\sqrt{\mathrm{1}+\mathrm{sinh}\:\frac{{x}}{{a}}}=\mathrm{cosh}\:\frac{{x}}{{a}} \\ $$$$\int\mathrm{cosh}\:\frac{{x}}{{a}}\:{dx}={a}\mathrm{sinh}\:\frac{{x}}{{a}} \\ $$$$\Rightarrow\:{L}\approx\mathrm{38}.\mathrm{525} \\ $$