Hi

evinda,

The parametrization $(at^2, t)$, $t\in \Bbb R$, is not a parametrization of the parabola $y = ax^2$, but rather the parabola $x = ay^2$. An appropriate parametrization would be $(t,at^2)$, $t\in \Bbb R$. Then

$$\kappa(t) = \frac{\lvert \dot{x}\ddot{y}-\dot{y}\ddot{x}\rvert}{(\dot{x}^2+\dot{y}^2)^{3/2}} = \frac{|(1)(2a) - (2at)(0)\rvert}{(1 + 4a^2t^2)^{3/2}} = \frac{2\lvert a\rvert}{(1 + 4a^2t^2)^{3/2}}$$

Therefore

$$\dot{\kappa}(t) = 2|a|\cdot \left(-\frac{3}{2}\right)(1 + 4a^2t^2)^{-5/2}\cdot (8a^2 t) = -24|a|a^2t(1 + 4a^2t^2)^{-5/2}$$

There is only one zero of $\frac{d\kappa}{dt}$, namely, $\kappa = 0$. It follows that $\kappa$ has a unique critical point at $t = 0$. Since $\dot{\kappa}(t)$ is positive for negative $t$ and negative for positive $t$, it follows from the first derivative test ensures that $\kappa$ has a maximum at $t = 0$ and no minimum (after all, $\kappa$ has no other extrema besides the one at $t = 0$). The maximum value of $\kappa$ is $\kappa_{\operatorname{max}} = \kappa(0) = 2\lvert a\rvert$. Keep in mind that the vertex of the parabola $y = ax^2$ is $(0,0)$, so the maximum indeed occurs at the vertex where $t = 0$.

Edit: Here's an important side note. If $f(x)$ is a twice-differentiable function, the curvature of the graph of $y = f(x)$ is given by

$$\kappa(x) = \frac{\lvert f''(x)\rvert}{(1 + [f'(x)]^2)^{3/2}}$$