Question

$r\left(t\right)=\{t\sin t,t\cos t\}$

Short answer

The normal component of acceleration,

$$\begin{gathered} a_N=\frac{\Vert v\times a\Vert }{\Vert v\Vert } \\ =\frac{t^2+2}{\sqrt{t^2+1}} \\ \text{Thus}{a}_T=\frac{t}{\sqrt{t^2+1}}\text{and}{a}_N=\frac{t^2+2}{\sqrt{t^2+1}} \end{gathered}$$

Step-by-step solution

Introduction 

Consider $r\left(t\right)=⟨t\sin t,t\cos t⟩$

The objective is to find the tangential and normal components of acceleration for $r\left(t\right)$.

The tangential and normal components of Acceleration

$r\left(t\right)$ is a twice - differentiable vector function with $r^{\text{'}}\left(t\right)=v\left(t\right)$ and $r^{\text{'}\text{'}}\left(t\right)=a\left(t\right)$.

The tangential component of acceleration, $a_T=\frac{v·a}{\Vert v\Vert }$. The normal component of acceleration,

$a_N=\frac{\Vert v\times a\Vert }{\Vert v\Vert }$

$r\left(t\right)=⟨t\sin t,t\cos t⟩$

$v\left(t\right)=r^{\text{'}}\left(t\right)$

$=⟨t\cos t+\sin t+\cos t,-t\sin t\cos t⟩$

$a\left(t\right)=r^{\text{'}\text{'}}\left(t\right)$

$=⟨-t\sin t+\cos t+\cos t,-t\cos t-\sin t-\sin t⟩$

$=⟨-t\sin t+2csot,-t\cos t-2\sin t⟩$

$\Vert v\Vert =\Vert ⟨t\cos t+\sin t,-t\sin t+\cos t⟩\Vert$

$=\sqrt{(t\cos t+\sin t{)}^2+(-t\sin t+\cos t{)}^2}$

$=\sqrt{t^2\left({\cos }^{2}t+{\sin }^{2}t\right)+\left({\sin }^{2}t+{\cos }^{2}t\right)}$

$=\sqrt{t^2+1}$

Given information 

$$\begin{gathered} v·a=v\left(t\right)·a\left(t\right) \\ =⟨t\cos t+\sin t,-t\sin t+\cos t⟩·⟨-t\sin t+2\cos t,-t\cos t-2\sin t⟩ \\ =⟨t\cos t+\sin t⟩(-t\sin t+2\cos t)+(-t\sin t+\cos t)(-t\cos t-2\sin t) \\ =-t^2\sin t\cos t+2t{\cos }^{2}t-t{\sin }^{2}t+2\sin t\cos t+t^2\sin t\cos t \\ +2t{\sin }^{2}t-t{\cos }^{2}t-2\sin t\cos t \\ =2t\left({\sin }^{2}t+{\cos }^{2}t\right)-t\left({\sin }^{2}t+{\cos }^{2}t\right) \\ =2t-t \\ =t \end{gathered}$$

$v\times a=v\left(t\right)\times a\left(t\right)$

$=\left|\begin{array}{c}ijk\\ t\cos t+\sin t-t\sin t+\cos t0\\ -t\sin t+2\cos t-t\cos t-2\sin t0\end{array}\right|$

$=i\left(0\right)-j\left(0\right)+k\left(\begin{array}{l}(t\cos t+\sin t)(-t\cos t-2\sin t)-(-t\sin t+\cos t)\\ (-t\sin t+2\cos t)\end{array}\right)$

$=k\left(\begin{array}{l}\left.-t^2{\cos }^{2}t-2t\sin t\cos t-t\sin t\cos t-2{\sin }^{2}t-t^2{\sin }^2+2t\sin t\cos t\right)\\ +t\sin t{\cos }^{2}t-2{\cos }^{2}t\end{array}\right)$

$=k\left(-t^2-2\right)$

$\Vert v\times a\Vert =\sqrt{{\left(-t^2-2\right)}^2}$

$=\sqrt{-{\left(t^2+2\right)}^2}$

$=t^2+2$

Explanation 

The tangential component of acceleration,

$$\begin{gathered} a_r=\frac{a·v}{\Vert v\Vert } \\ =\frac{t}{\sqrt{t^2+1}} \end{gathered}$$

The normal component of acceleration,

$$\begin{gathered} a_N=\frac{\Vert v\times a\Vert }{\Vert v\Vert } \\ =\frac{t^2+2}{\sqrt{t^2+1}} \end{gathered}$$

Thus $a_T=\frac{t}{\sqrt{t^2+1}}$and $a_N=\frac{t^2+2}{\sqrt{t^2+1}}$