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

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

Commented by Joel122 last updated on 06/Sep/19

$$x^{\prime }\left(t\right)=\pi \cos \left(\pi t\right),y^{\prime }\left(t\right)=1$$$$$$$$L={\int }_0^1\sqrt{{\left(x^{\prime }\left(t\right)\right)}^2+{\left(y^{\prime }\left(t\right)\right)}^2}dt$$$$={\int }_0^1\sqrt{{\pi }^2{\cos }^2\left(\pi t\right)+1}dt$$$$$$$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{stuck}\mathrm{with}\mathrm{the}\mathrm{integral}.\mathrm{Please}\mathrm{help}$$

Commented by MJS last updated on 06/Sep/19

$$\mathrm{this}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{be}\mathrm{solved}\mathrm{using}\mathrm{elementary}\mathrm{calculus}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{an}ellipticintegral$$

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

$$thankyouSir$$

Commented by MJS last updated on 06/Sep/19

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

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