Answer
$$
\begin{aligned}
a_{T}&=\frac{4 t}{\sqrt{4 t^{2}+5}}
\end{aligned}
$$
$$
\begin{aligned}
a_{N}&=\frac{2 \sqrt{5}}{\sqrt{4 t^{2}+5}}
\end{aligned}
$$
Work Step by Step
We have the position function:
$$
\begin{aligned}
\mathbf{r}(t)&=t \mathbf{i}+2t \mathbf{j}+ t^{2} \mathbf{k} \\
&=\left\langle t, 2t, t^{2} \right\rangle
\end{aligned}
$$
$\Rightarrow$
$$
\left|\mathbf{r}^{\prime}(t)\right|=\sqrt{1+4+4 t^{2}}=\sqrt{4 t^{2}+5}\\
$$
The velocity is
$$
\begin{aligned}
\mathbf{v}(t) &=\mathbf{r}^{\prime}(t)\\
&=\mathbf{i}+2 \mathbf{j}+2 t \mathbf{k} \\
&=\left\langle 1, 2, 2 t \right\rangle
\end{aligned}
$$
The acceleration is
$$
\begin{aligned}
\mathbf{a}(t)&=\mathbf{r}^{\prime\prime}(t)=\mathbf{v}^{\prime}(t)\\
&=2 \mathbf{k},\\
&=\left\langle 0, 0, 2 \right\rangle
\end{aligned}
$$
Therefore, Equation 9 gives the tangential component as,
$$
\begin{aligned}
a_{T}&=\frac{\mathbf{r}^{\prime}(t) \cdot \mathbf{r}^{\prime \prime}(t)}{\left|\mathbf{r}^{\prime}(t)\right|}\\
&=\frac{4 t}{\sqrt{4 t^{2}+5}} ,
\end{aligned}
$$
where,
$$
\begin{aligned}
\mathbf{r}^{\prime}(t) \cdot \mathbf{r}^{\prime \prime}(t)=
\left\langle 1, 2, 2 t \right\rangle \cdot \left\langle 0, 0, 2 \right\rangle= 4t
\end{aligned}
$$
and Equation 10 gives the normal component as
$$
\begin{aligned}
a_{N}&=\frac{\left|\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)\right|}{\left|\mathbf{r}^{\prime}(t)\right|} \\
&=\frac{|4 \mathbf{i}-2 \mathbf{j}|}{\sqrt{4 t^{2}+5}} \\
&=\frac{2 \sqrt{5}}{\sqrt{4 t^{2}+5}}
\end{aligned}
$$
where,
$$
\begin{aligned}
\mathbf{r}^{\prime}(t) \times \mathbf{r}^{\prime \prime}(t)&=
\left\langle 1, 2, 2 t \right\rangle \times \left\langle 0, 0, 2 \right\rangle \\
&=\left[\begin{array}{ccc}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 & 1 & 2t \\
0, & 0 & 2 \\
\end{array}\right]\\
&=\left\langle 4, -2, 0 \right\rangle \\
&=4 \mathbf{i}-2 \mathbf{j}
\end{aligned}
$$
