# Can't see why parallel curve is regular when $0 < r < \frac{1}{2}$

Let $\gamma: \mathbb{R} \rightarrow \mathbb{R}^2$ be given by $\gamma(t) = (t, t^2)$. Find an explicit paramterization of the parallel curves $\delta_r$. Showw that $\delta_r$ is regular for $0 < r < \frac{1}{2}$ and has exactly two points of non-regularity for $r > \frac{1}{2}$.

I know that a parallel curve is given by $\delta_r(t) = \gamma(t) + r \cdot U(t)$ where $U(t)$ is the unit normal of the curve $\gamma$. So I worked out

$$\gamma'(t) = (1, 2t) \implies | \gamma ' (t)| = \sqrt{1 + 4t^2}.$$

So from here, I got

$$U(t) = \left( \frac{-2t}{\sqrt{1 + 4t^2}}, \frac{1}{\sqrt{1 + 4t^2}} \right)$$

and so we get

$$\delta_r(t) = \left( t - \frac{2tr}{\sqrt{1 + 4t^2}}, t^2 + \frac{r}{\sqrt{1 + 4rt^2}}\right).$$

Now to show its regular or not, we differentiate and show that $\delta_r ' (t) \neq 0$. So differentiating gives $\delta_r'(t) = \gamma'(t) + r U'(t)$. We have already worked out $\gamma'(t)$. I got

$$U'(t) = \left(\frac{8t^2}{\left(\sqrt{1 + 4t^2} \right)^3} + \frac{2}{\sqrt{1 + 4t^2}}, -\frac{4t}{\left(\sqrt{1 + 4t^2}\right)^3} \right)$$

and so we get

$$\delta_r'(t) =\left( 1 + r\left(\frac{8t^2}{\left(\sqrt{1 + 4t^2} \right)^3} + \frac{2}{\sqrt{1 + 4t^2}} \right), 2t - r \left(\frac{4t}{\left(\sqrt{1 + 4t^2}\right)^3} \right) \right).$$

From here, how do I now see that the curve is regular for $0 < r < \frac{1}{2}$? Surely it depends on what $t$ is aswell, doesn't it?

-