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Find the length of the curve $x=a(t-\sin t),y=a(1-\cos t)$ between $t=0$and $\pi$.

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  • Area bounded by the curve $t=f(x),$ the x-axis and the ordinates $x=a,x=b$ is $\int \limits_a^b f(x) dx $ or $ \int \limits _a^b y dx $
  • If the curve lies below the x-axis for $a \leq x \leq b,$ then the area is $\int \limits_a^b (-y) dx=\int \limits_a^b (-f(x))dx$
$x=a(t -\sin t) \qquad y= a(1-\cos t)$
Step 1:
$\large\frac{dx}{dt}$$=a(1-\cos t)= 2 a \sin ^2 t/2$
$\large\frac{dy}{dt}$$=a \sin t= 2 a \sin \large\frac{t}{2} $$ \cos \large\frac{t}{2}$
$\bigg(\large\frac{dx}{dt}\bigg)^2+\bigg(\large\frac{dy}{dt}\bigg)^2$$=4a^2 \sin ^4 \large\frac{t}{2} $$+4a ^2 \sin ^2 \large\frac{t}{2} $$ \cos ^2 \large\frac{t}{2}$
$\qquad=4a^2 \sin ^2 \large\frac{t}{2} $$\bigg[\sin ^2 \large\frac{t}{2}$$+ \cos ^2 \large\frac{t}{2}\bigg]$
$\qquad=4a^2 \sin ^2 \large\frac{t}{2} $
Step 2:
$\therefore \sqrt{\bigg(\large\frac{dx}{dt}\bigg)^2+\bigg(\large\frac{dy}{dt}\bigg)^2}$$=2a \sin \large\frac{t}{2} $
Length of arc of the curve between $t=0$ and $t=\pi$
$L= \int \limits_0^{\pi} \sqrt{\bigg(\large\frac{dx}{dt}\bigg)^2+\bigg(\large\frac{dy}{dt}\bigg)^2}$$dt=\int \limits_0^{\pi}\;2a \sin \large\frac{t}{2} $$dt$
$\qquad=2a-2 \cos \large\frac{t}{2} \bigg]_0 ^{\pi}$
$\qquad=4a [-0+1]$
$\qquad= 4a \;units$
answered Aug 17, 2013 by meena.p

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