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:

We are given the parametric curve $$x = 2\cos t + 2t\sin t$$ and $$y = 2\sin t - 2t\cos t$$. First we compute the derivatives: $$\frac{dx}{dt} = -2\sin t + 2\sin t + 2t\cos t = 2t\cos t$$ and $$\frac{dy}{dt} = 2\cos t - 2\cos t + 2t\sin t = 2t\sin t$$.

So $$\dfrac{dy}{dx} = \dfrac{2t\sin t}{2t\cos t} = \tan t$$. At $$t = \dfrac{\pi}{4}$$, the slope of the tangent is $$\tan\dfrac{\pi}{4} = 1$$, and the slope of the normal is $$-1$$.

The point on the curve at $$t = \dfrac{\pi}{4}$$ is: $$x_0 = 2\cos\frac{\pi}{4} + 2 \cdot \frac{\pi}{4} \cdot \sin\frac{\pi}{4} = \sqrt{2} + \frac{\pi\sqrt{2}}{4} = \sqrt{2}\!\left(1 + \frac{\pi}{4}\right)$$ and $$y_0 = 2\sin\frac{\pi}{4} - 2 \cdot \frac{\pi}{4} \cdot \cos\frac{\pi}{4} = \sqrt{2} - \frac{\pi\sqrt{2}}{4} = \sqrt{2}\!\left(1 - \frac{\pi}{4}\right)$$.

The equation of the normal at this point is $$y - y_0 = -1(x - x_0)$$, which simplifies to $$x + y = x_0 + y_0 = \sqrt{2}\!\left(1 + \frac{\pi}{4}\right) + \sqrt{2}\!\left(1 - \frac{\pi}{4}\right) = 2\sqrt{2}$$.

The normal line is $$x + y - 2\sqrt{2} = 0$$. The distance from the origin $$(0, 0)$$ to this line is $$\dfrac{|0 + 0 - 2\sqrt{2}|}{\sqrt{1^2 + 1^2}} = \dfrac{2\sqrt{2}}{\sqrt{2}} = 2$$.