If x=a(cos⁡t+tsin⁡t), y=a(sin⁡t−tcos⁡t) then dxdt2+dydt2=

x=a(cos⁡t+tsin⁡t),y=a(sin⁡t−tcos⁡t)⇒dxdt=a(−sin⁡t+1⋅sin⁡t+tcos⁡t)=atcos⁡t,dydt=a[cos⁡t−(1⋅cos⁡t−tsin⁡t)]=atsin⁡t∴dxdt2+dydt2=a2t2cos2⁡t+a2t2sin2⁡t=a2t2cos2⁡t+sin2⁡t=at

If x=a(cos⁡t+tsin⁡t), y=a(sin⁡t−tcos⁡t) then dxdt2+dydt2=

x=a(cos⁡t+tsin⁡t),y=a(sin⁡t−tcos⁡t)⇒dxdt=a(−sin⁡t+1⋅sin⁡t+tcos⁡t)=atcos⁡t,dydt=a[cos⁡t−(1⋅cos⁡t−tsin⁡t)]=atsin⁡t∴dxdt2+dydt2=a2t2cos2⁡t+a2t2sin2⁡t=a2t2cos2⁡t+sin2⁡t=at

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If $x = a (\cos t + tsin t), y = a (\sin t - tcos t) then \sqrt{{(\frac{dx}{dt})}^{2} + {(\frac{dy}{dt})}^{2}} =$

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$a^{2} t^{2}$

$a^{2} t$

answer is B.

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Detailed Solution

$\begin{array}{l} x = a (\cos t + tsin t), y = a (\sin t - tcos t) \\ \Rightarrow \frac{dx}{dt} = a (- \sin t + 1 \cdot \sin t + tcos t) = atcos t, \frac{dy}{dt} = a [\cos t - (1 \cdot \cos t - tsin t)] = atsin t \\ ∴ \sqrt{{(\frac{dx}{dt})}^{2} + {(\frac{dy}{dt})}^{2}} = \sqrt{a^{2} t^{2} \cos^{2} t + a^{2} t^{2} \sin^{2} t} = \sqrt{a^{2} t^{2} (\cos^{2} t + \sin^{2} t)} = at \end{array}$