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

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Solution

We have,
$\frac{d x}{d t} = a (1 - cos t) = y$

$\Rightarrow d x = y d t = a (1 - cos t) d t$

$\int_{0}^{x} d x = \int_{0}^{π} a (1 - cos t) d t x = [a t - sin t]_{0}^{π} = π a$

Length of the curve $= π a$ units

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Similar questions

x = a (t - sin t), y = a (1 - cos t),

d y / d x

t = π

x = sin (t + \frac{7 π}{12}) + sin (t - \frac{π}{12}) + sin (t + \frac{3 π}{12}),

y = cos (t + \frac{7 π}{12}) + cos (t - \frac{π}{12}) + cos (t + \frac{3 π}{12})

\frac{d}{d t} (\frac{x}{y} - \frac{y}{x})

t = \frac{π}{8}

x = sin (t + \frac{7 π}{12}) + sin (t - \frac{π}{12}) + sin (t + \frac{3 π}{12}),

y = cos (t + \frac{7 π}{12}) + cos (t - \frac{π}{12}) + cos (t + \frac{3 π}{12})

\frac{d}{d t} (\frac{x}{y} - \frac{y}{x})

t = \frac{π}{8}

x = sin (t + \frac{7 π}{12}) +

sin (t - \frac{π}{12}) + sin (t + \frac{3 π}{12}),

y = cos (t + \frac{7 π}{12}) + cos (t - \frac{π}{12}) + cos (t + \frac{3 π}{12})

\frac{d}{d t} (\frac{x}{y} - \frac{y}{x})

t = \frac{π}{8}

x = a (t - sin t)

y = a (1 + cos t)

y_{2}

t = \frac{π}{2}

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