Explanation
$$x=f''(t)\cos { t } +f'(t)\sin { t } \\ \dfrac { dx }{ dt } =f'''(t)\cos { t } -f''(t)\sin { t } +f''(t)\sin { t } +f'(t)\cos { t } \\ \dfrac { dx }{ dt } =f'''(t)\cos { t } +f'(t)\cos { t } \\ \dfrac { dx }{ dt } =(f'''(t)+f'(t))\cos { t } ..........(1)\\ y=-f''(t)\sin { t } +f'(t)\cos { t } \\ \dfrac { dy }{ dt } =-f'''(t)\sin { t } -f''(t)\cos { t } +f''(t)\cos { t } -f'(t)\sin { t } \\ \dfrac { dy }{ dt } =-f'''(t)\sin { t } -f'(t)\sin { t } \\ \dfrac { dy }{ dt } =-\sin { t } (f'''(t)+f'(t))........(2)\\ add\quad squares\quad of\quad eq.1\quad and\quad 2\quad \\ { \left( \dfrac { dx }{ dt } \right) }^{ 2 }+{ \left( \dfrac { dy }{ dt } \right) }^{ 2 }={ (f'''(t)+f'(t)) }^{ 2 }\\ { \left( { \left( \dfrac { dx }{ dt } \right) }^{ 2 }+{ \left( \dfrac { dy }{ dt } \right) }^{ 2 } \right) }^{ \dfrac { 1 }{ 2 } }=f'''(t)+f'(t)\\ integrate\quad both\quad sides\quad \\ \int { { \left( { \left( \dfrac { dx }{ dt } \right) }^{ 2 }+{ \left( \dfrac { dy }{ dt } \right) }^{ 2 } \right) }^{ \dfrac { 1 }{ 2 } }dt } =f''(t)+f(t)+c$$
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