Question Number 68145 by Joel122 last updated on 06/Sep/19
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{arc}\:\mathrm{length},\:\mathrm{given}\:\mathrm{the}\:\mathrm{curve} \\ $$$${x}\left({t}\right)\:=\:\mathrm{sin}\:\left(\pi{t}\right),\:\:{y}\left({t}\right)\:=\:{t}\:,\:\:\mathrm{0}\:\leqslant\:{t}\:\leqslant\:\mathrm{1} \\ $$
Commented by Joel122 last updated on 06/Sep/19
$${x}'\left({t}\right)\:=\:\pi\:\mathrm{cos}\:\left(\pi{t}\right),\:{y}'\left({t}\right)\:=\:\mathrm{1} \\ $$$$ \\ $$$${L}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{\left({x}'\left({t}\right)\right)^{\mathrm{2}} \:+\:\left({y}'\left({t}\right)\right)^{\mathrm{2}} }\:{dt} \\ $$$$\:\:\:\:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{\pi^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\left(\pi{t}\right)\:+\:\mathrm{1}}\:\:{dt} \\ $$$$ \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{stuck}\:\mathrm{with}\:\mathrm{the}\:\mathrm{integral}.\:\mathrm{Please}\:\mathrm{help} \\ $$
Commented by MJS last updated on 06/Sep/19
$$\mathrm{this}\:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{using}\:\mathrm{elementary}\:\mathrm{calculus} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{an}\:{elliptic}\:{integral} \\ $$
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
$${thank}\:{you}\:{Sir} \\ $$
Commented by MJS last updated on 06/Sep/19
$$\mathrm{the}\:\mathrm{path}\:\mathrm{is} \\ $$$${u}=\pi{t}\:\rightarrow\:{dt}=\frac{{du}}{\pi} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\frac{\sqrt{\mathrm{1}+\pi^{\mathrm{2}} }}{\pi}\int\sqrt{\mathrm{1}−\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\pi^{\mathrm{2}} }\mathrm{sin}^{\mathrm{2}} \:{u}}=\frac{\sqrt{\mathrm{1}+\pi^{\mathrm{2}} }}{\pi}\mathrm{E}\:\left({u}\mid\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\pi^{\mathrm{2}} }\right)\:= \\ $$$$=\frac{\sqrt{\mathrm{1}+\pi^{\mathrm{2}} }}{\pi}\mathrm{E}\:\left(\pi{t}\mid\frac{\pi^{\mathrm{2}} }{\mathrm{1}+\pi^{\mathrm{2}} }\right)\:+{C} \\ $$