Question

Найдите производные у'=dy/dx и y"=d^2y/dx^2
$x = t - \sin(t) \\ y = 1 - \cos(t)$
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Answer 1

Ответ:

$y'x = \frac{y't}{x't}$

$y''xx = \frac{(y'x)'t}{x't}$

$y't = 0 - ( - \sin(t)) = \sin(t)$

$x't = 1 - \cos(t)$

$y'x = \frac{ \sin(t) }{ 1 - \cos(t) }$

$(y'x)'t = \frac{ \cos(t)(1 - \cos(t)) - \sin(t) \times \sin(t) }{ {(1 \cos(t)) }^{2} } = \\ = \frac{ \cos(t) - { \cos }^{2} (t) - { \sin }^{2} (t)}{ {(1 - \cos(t)) }^{2} } = \\ = \frac{ \cos(t) - 1}{ {(1 - \cos(t)) }^{2} } = \frac{1}{ \cos(t) - 1 }$

$y''xx = \frac{1}{ \cos(t) - 1} \times \frac{1}{1 - \cos(t) } = \\ = - \frac{1}{ {(1 - \cos(t)) }^{2} }$