Find the centripetal acceleration of $\vec{r}(t)=(2\cos{t}-1)\hat{i}+(4\sin{t}+2)\hat{j}$
Using first derivative to find the velocity and we'll found :
$\vec{v}(t)=-2\sin(t)\hat{i}+4cos(t)\hat{j}$
We know the equation of centripetal acceleration is $a_c=\frac{v^2}{r}$. R is the value of $\frac{1}{\kappa}$ where $\kappa(t)=\frac{||\vec{T'}(t)||}{||\vec{v}(t)||}$
Let $\vec{T}(t)=\frac{\vec{v}(t)}{||\vec{v}(t)||}$. Then :
$\vec{T}(t)=\frac{1}{\sqrt{12\cos^2(t)+4}}(-2\sin(t)\hat{i}+4cos(t)\hat{j})\\ \vec{T'}(t)=\frac{1}{\sqrt{12\cos^2(t)+4}}(-2\cos(t)\hat{i}+4sin(t)\hat{j})\\$
Length of $\vec{T}(t)$ is 1 and $||\vec{v}(t)||=\sqrt{12cos^2{t}+1}$. And I found that $\kappa=\frac{1}{\sqrt{12cos^2{t}+1}}$. Then the radius is $\sqrt{12cos^2{t}+4}$.
$\begin{align*}a_c(t)&=\frac{\sqrt{(12cos^2{t}+4})^2}{\sqrt{12cos^2{t}+4}}\\ &=\sqrt{12cos^2{t}+4} \end{align*}$.
The result isn't a vector.