If x=a(cos t+t sin t) and y=a(sin t-t cos t), | vedclass

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(B) Given $x = a(\cos t + t \sin t)$.Differentiating with respect to $t$:$\frac{dx}{dt} = a(-\sin t + \sin t + t \cos t) = at \cos t$.Given $y = a(\si

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$at$

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(B) Given $x = a(\cos t + t \sin t)$.Differentiating with respect to $t$:$\frac{dx}{dt} = a(-\sin t + \sin t + t \cos t) = at \cos t$.Given $y = a(\sin t - t \cos t)$.Differentiating with respect to $t$:$\frac{dy}{dt} = a(\cos t - (\cos t - t \sin t)) = a(\cos t - \cos t + t \sin t) = at \sin t$.Now,calculate $\sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}$:$= \sqrt{(at \cos t)^2 + (at \sin t)^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)}$$= \sqrt{a^2 t^2 (1)}$$= at$.

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

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