Find-the-arc-length-given-the-curve-x-t-sin-pit-y-t-t-0-t-1-

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

Find the arc length, given the curve

x (t) = \sin (π t), y (t) = t, 0 ⩽ t ⩽ 1

Commented by Joel122 last updated on 06/Sep/19

x^{'} (t) = π \cos (π t), y^{'} (t) = 1

L = \int_{0}^{1} \sqrt{{(x^{'} (t))}^{2} + {(y^{'} (t))}^{2}} d t

= \int_{0}^{1} \sqrt{π^{2} \cos^{2} (π t) + 1} d t

I^{'} m stuck with the integral . Please help

Commented by MJS last updated on 06/Sep/19

this {can}^{'} t be solved using elementary calculus

{it}^{'} s an e l l i p t i c i n t e g r a l

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

t h a n k y o u S i r

Commented by MJS last updated on 06/Sep/19

the path is

u = π t \to d t = \frac{d u}{π}

this leads to

\frac{\sqrt{1 + π^{2}}}{π} \int \sqrt{1 - \frac{π^{2}}{1 + π^{2}} \sin^{2} u} = \frac{\sqrt{1 + π^{2}}}{π} E (u ∣ \frac{π^{2}}{1 + π^{2}}) =

= \frac{\sqrt{1 + π^{2}}}{π} E (π t ∣ \frac{π^{2}}{1 + π^{2}}) + C

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