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(A) Given the parametric equations of the curve:$x = 2\cos t + 2t\sin t$$y = 2\sin t - 2t\cos t$First,find the derivatives with respect to $t$:$\frac{

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$2$

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(A) Given the parametric equations of the curve:$x = 2\cos t + 2t\sin t$$y = 2\sin t - 2t\cos t$First,find the derivatives with respect to $t$:$\frac{dx}{dt} = -2\sin t + 2(\sin t + t\cos t) = 2t\cos t$$\frac{dy}{dt} = 2\cos t - 2(\cos t - t\sin t) = 2t\sin t$Now,find the slope of the tangent $\frac{dy}{dx}$:$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2t\sin t}{2t\cos t} = 	an t$At $t = \frac{\pi}{4}$,the slope of the tangent is $	an(\frac{\pi}{4}) = 1$.Therefore,the slope of the normal is $m = -\frac{1}{1} = -1$.Find the point $(x, y)$ at $t = \frac{\pi}{4}$:$x = 2\cos(\frac{\pi}{4}) + 2(\frac{\pi}{4})\sin(\frac{\pi}{4}) = 2(\frac{1}{\sqrt{2}}) + \frac{\pi}{2}(\frac{1}{\sqrt{2}}) = \sqrt{2} + \frac{\pi}{2\sqrt{2}}$$y = 2\sin(\frac{\pi}{4}) - 2(\frac{\pi}{4})\cos(\frac{\pi}{4}) = 2(\frac{1}{\sqrt{2}}) - \frac{\pi}{2}(\frac{1}{\sqrt{2}}) = \sqrt{2} - \frac{\pi}{2\sqrt{2}}$The equation of the normal is $y - y_1 = m(x - x_1)$:$y - (\sqrt{2} - \frac{\pi}{2\sqrt{2}}) = -1(x - (\sqrt{2} + \frac{\pi}{2\sqrt{2}}))$$y - \sqrt{2} + \frac{\pi}{2\sqrt{2}} = -x + \sqrt{2} + \frac{\pi}{2\sqrt{2}}$$x + y = 2\sqrt{2}$The distance of the line $Ax + By + C = 0$ from the origin $(0, 0)$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.Here,$x + y - 2\sqrt{2} = 0$,so $A=1, B=1, C=-2\sqrt{2}$.$d = \frac{|-2\sqrt{2}|}{\sqrt{1^2 + 1^2}} = \frac{2\sqrt{2}}{\sqrt{2}} = 2$.

The distance,from the origin,of the normal to the curve,$x = 2\cos t + 2t\sin t, y = 2\sin t - 2t\cos t$ at $t = \frac{\pi}{4}$,is

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