If x=f′′(t)cos⁡t+f′(t)sin⁡t,y=−f′′(t)sin⁡t+f′(t)cos⁡t then ∫dxdt2+dydt212dt

x=f′′(t)cos⁡t+f′(t)sin⁡tdxdt=f′′′(t)cos⁡t−f′′(t)sin⁡t+f′′(t)sin⁡t+f′(t)cos⁡tdxdt=f′′′(t)cos⁡t+f′(t)cos⁡tdxdt=f′′′(t)+f′(t)cos⁡t………..(1)y=−f′′(t)sin⁡t+f′(t)cos⁡tdydt=−f′′′(t)sin⁡t−f′′(t)cos⁡t+f′′(t)cos⁡t−f′(t)sin⁡tdydt=−f′′′(t)sin⁡t−f′(t)sin⁡tdydt=−sin⁡tf′′′(t)+f′(t)……..(2)dxdt2+dydt2=f′′′(t)+f′(t)2dxdt2+dydt212=f′′′(t)+f′(t) ⇒∫dxdt2+dydt212dt=f′′(t)+f(t)+c

If x=f′′(t)cos⁡t+f′(t)sin⁡t,y=−f′′(t)sin⁡t+f′(t)cos⁡t then ∫dxdt2+dydt212dt

x=f′′(t)cos⁡t+f′(t)sin⁡tdxdt=f′′′(t)cos⁡t−f′′(t)sin⁡t+f′′(t)sin⁡t+f′(t)cos⁡tdxdt=f′′′(t)cos⁡t+f′(t)cos⁡tdxdt=f′′′(t)+f′(t)cos⁡t………..(1)y=−f′′(t)sin⁡t+f′(t)cos⁡tdydt=−f′′′(t)sin⁡t−f′′(t)cos⁡t+f′′(t)cos⁡t−f′(t)sin⁡tdydt=−f′′′(t)sin⁡t−f′(t)sin⁡tdydt=−sin⁡tf′′′(t)+f′(t)……..(2)dxdt2+dydt2=f′′′(t)+f′(t)2dxdt2+dydt212=f′′′(t)+f′(t) ⇒∫dxdt2+dydt212dt=f′′(t)+f(t)+c

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If $x = f^{''} (t) \cos t + f^{'} (t) \sin t, y = - f^{''} (t) \sin t + f^{'} (t) \cos t then \int {({(\frac{dx}{dt})}^{2} + {(\frac{dy}{dt})}^{2})}^{\frac{1}{2}} dt$

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$f^{'} (t) + f^{''} (t) + c$

$f^{''} (t) + f^{'''} (t) + c$

$f (t) + f^{''} (t) + c$

$f^{'} (t) - f^{''} (t) + c$

answer is C.

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Detailed Solution

$\begin{array}{l} x = f^{''} (t) \cos t + f^{'} (t) \sin t \\ \frac{dx}{dt} = f^{'''} (t) \cos t - f^{''} (t) \sin t + f^{''} (t) \sin t + f^{'} (t) \cos t \\ \frac{dx}{dt} = f^{'''} (t) \cos t + f^{'} (t) \cos t \\ \frac{dx}{dt} = (f^{'''} (t) + f^{'} (t)) \cos t \dots \dots \dots . . (1) \end{array}$
$\begin{array}{l} y = - f^{''} (t) \sin t + f^{'} (t) \cos t \\ \frac{d y}{d t} = - f^{'''} (t) \sin t - f^{''} (t) \cos t + f^{''} (t) \cos t - f^{'} (t) \sin t \\ \frac{dy}{dt} = - f^{'''} (t) \sin t - f^{'} (t) \sin t \\ \frac{dy}{dt} = - \sin t (f^{'''} (t) + f^{'} (t)) \dots \dots . . (2) \end{array}$
$\begin{aligned} {(\frac{dx}{dt})}^{2} + {(\frac{dy}{dt})}^{2} = {(f^{'''} (t) + f^{'} (t))}^{2} \\ {({(\frac{dx}{dt})}^{2} + {(\frac{dy}{dt})}^{2})}^{\frac{1}{2}} = f^{'''} (t) + f^{'} (t) \end{aligned}$
$\Rightarrow \int {({(\frac{d x}{d t})}^{2} + {(\frac{d y}{d t})}^{2})}^{\frac{1}{2}} d t = f^{''} (t) + f (t) + c$