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Question Number 68145 by Joel122 last updated on 06/Sep/19

Find the arc length, given the curve x(t) = sin (πt), y(t) = t , 0 ≤ t ≤ 1

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{arc}\:\mathrm{length},\:\mathrm{given}\:\mathrm{the}\:\mathrm{curve} \\ $$$${x}\left({t}\right)\:=\:\mathrm{sin}\:\left(\pi{t}\right),\:\:{y}\left({t}\right)\:=\:{t}\:,\:\:\mathrm{0}\:\leqslant\:{t}\:\leqslant\:\mathrm{1} \\ $$

Commented by Joel122 last updated on 06/Sep/19

x′(t) = π cos (πt), y′(t) = 1 L = ∫_0 ^1 (√((x′(t))^2 + (y′(t))^2 )) dt = ∫_0 ^1 (√(π^2 cos^2 (πt) + 1)) dt I′m stuck with the integral. Please help

$${x}'\left({t}\right)\:=\:\pi\:\mathrm{cos}\:\left(\pi{t}\right),\:{y}'\left({t}\right)\:=\:\mathrm{1} \\ $$$$ \\ $$$${L}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{\left({x}'\left({t}\right)\right)^{\mathrm{2}} \:+\:\left({y}'\left({t}\right)\right)^{\mathrm{2}} }\:{dt} \\ $$$$\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{\pi^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\left(\pi{t}\right)\:+\:\mathrm{1}}\:\:{dt} \\ $$$$ \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{stuck}\:\mathrm{with}\:\mathrm{the}\:\mathrm{integral}.\:\mathrm{Please}\:\mathrm{help} \\ $$

Commented by MJS last updated on 06/Sep/19

this can′t be solved using elementary calculus it′s an elliptic integral

$$\mathrm{this}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{using}\:\mathrm{elementary}\:\mathrm{calculus} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{an}\:{elliptic}\:{integral} \\ $$

Commented by MJS last updated on 06/Sep/19

https://en.m.wikipedia.org/wiki/Elliptic_integral

Commented by Joel122 last updated on 06/Sep/19

thank you Sir

$${thank}\:{you}\:{Sir} \\ $$

Commented by MJS last updated on 06/Sep/19

the path is u=πt → dt=(du/π) this leads to ((√(1+π^2 ))/π)∫(√(1−(π^2 /(1+π^2 ))sin^2 u))=((√(1+π^2 ))/π)E (u∣(π^2 /(1+π^2 ))) = =((√(1+π^2 ))/π)E (πt∣(π^2 /(1+π^2 ))) +C

$$\mathrm{the}\:\mathrm{path}\:\mathrm{is} \\ $$$${u}=\pi{t}\:\rightarrow\:{dt}=\frac{{du}}{\pi} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\frac{\sqrt{\mathrm{1}+\pi^{\mathrm{2}} }}{\pi}\int\sqrt{\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\pi^{\mathrm{2}} }\mathrm{sin}^{\mathrm{2}} \:{u}}=\frac{\sqrt{\mathrm{1}+\pi^{\mathrm{2}} }}{\pi}\mathrm{E}\:\left({u}\mid\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\pi^{\mathrm{2}} }\right)\:= \\ $$$$=\frac{\sqrt{\mathrm{1}+\pi^{\mathrm{2}} }}{\pi}\mathrm{E}\:\left(\pi{t}\mid\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\pi^{\mathrm{2}} }\right)\:+{C} \\ $$

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