Explanation
$$x = 2 \cos t + 2t \sin t$$ $$\cfrac { dx }{ dt } =2t\cos t+2\sin t-2\sin t=2t\cos t$$ $$y = 2 \sin t - 2t \cos t $$ $$\cfrac { dy }{ dt } =2\cos t+2t\sin t-2\cos t=2t\sin t$$ $$\cfrac { dy }{ dx } =\cfrac { 2t\sin t }{ 2t\cos t } (t=\dfrac{\pi }4)$$ $$\cfrac { dy }{ dx } =1$$ $$\cfrac { dx }{ dy } =-1$$ for slope of normal. $$x\left( \cfrac { \pi }{ 4 } \right) =2\cos \left( \cfrac { \pi }{ 4 } \right) +2.\cfrac { \pi }{ 4 } \sin \left( \cfrac { \pi }{ 4 } \right) =\sqrt { 2 } \left( 1+\cfrac { \pi }{ 4 } \right) $$ $$y\left( \cfrac { \pi }{ 4 } \right) =2\sin \left( \cfrac { \pi }{ 4 } \right) -2.\cfrac { \pi }{ 4 } \cos \left( \cfrac { \pi }{ 4 } \right) =\sqrt { 2 } \left( 1-\cfrac { \pi }{ 4 } \right)$$ Equation of normal with given point $$x+y-2\sqrt { 2 } =0$$ $$d=\cfrac { (2\sqrt { 2) } }{ \sqrt { 2 } } =2$$
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