Let the curve be represented by x=2(cos t+t s | vedclass

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(C) Given the parametric equations of the curve are $x=2(\cos t+t \sin t)$ and $y=2(\sin t-t \cos t)$.First,we find the derivatives with respect to $t

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(C) Given the parametric equations of the curve are $x=2(\cos t+t \sin t)$ and $y=2(\sin t-t \cos t)$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt} = 2(-\sin t + \sin t + t \cos t) = 2t \cos t$$\frac{dy}{dt} = 2(\cos t - \cos t + t \sin t) = 2t \sin t$Now,the slope of the tangent is $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t \sin t}{2t \cos t} = 	an t$.The slope of the normal is $-\frac{1}{dy/dx} = -\frac{1}{	an t} = -\frac{\cos t}{\sin t}$.The equation of the normal at point '$t$' is given by $y - y_1 = m_n(x - x_1)$:$y - 2(\sin t - t \cos t) = -\frac{\cos t}{\sin t} [x - 2(\cos t + t \sin t)]$Multiplying by $\sin t$:$y \sin t - 2 \sin^2 t + 2t \sin t \cos t = -x \cos t + 2 \cos^2 t + 2t \sin t \cos t$$x \cos t + y \sin t = 2(\sin^2 t + \cos^2 t)$$x \cos t + y \sin t = 2$The distance of the line $Ax + By + C = 0$ from the origin $(0,0)$ is given by $\frac{|C|}{\sqrt{A^2 + B^2}}$.Here,$A = \cos t$,$B = \sin t$,and $C = -2$.Distance $= \frac{|-2|}{\sqrt{\cos^2 t + \sin^2 t}} = \frac{2}{1} = 2$ units.

Let the curve be represented by $x=2(\cos t+t \sin t)$ and $y=2(\sin t-t \cos t)$. Then the normal at any point '$t$' of the curve is at a distance of . . . . . . units from the origin.

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