Determine the angle between the left and right tangent

Determine the angle between the left and right tangent of the curve $y=\sqrt{1-e^{-(ax)^{2}}}$ at the point $x=0$.

Step 1: A corner at the origin

At $x=0$, $y(0)=\sqrt{1-e^{0}}=0$. For $x\neq0$ the radicand is strictly positive, and it vanishes only at $x=0$; this hints at a corner (angular point), so the one-sided derivatives must be computed separately.

Step 2: Right-hand derivative

Using $1-e^{-u}\sim u$ as $u\to0$ with $u=(ax)^{2}$, so $\sqrt{1-e^{-(ax)^2}}\sim\sqrt{a^2x^2}=|a|\,x$ for $x\gt 0$: \[y_{+}'(0) = \lim_{x\to0^{+}}\frac{\sqrt{1-e^{-(ax)^{2}}}}{x} = \lim_{x\to0^{+}}\frac{|a|x}{x} = |a|. \tag{1}\]

Step 3: Left-hand derivative

For $x\lt 0$, $|x|=-x$, so $\sqrt{1-e^{-(ax)^{2}}}\sim -|a|x$, and

\[ y_{-}'(0) = \lim_{x\to0^{-}}\frac{-|a|x}{x} = -|a|. \tag{2} \]

The two one-sided slopes, $k_{1}=-|a|$ and $k_{2}=|a|$, confirm the corner.

Step 4: Angle between the two tangent lines

With $\tan\alpha=\left|\dfrac{k_{2}-k_{1}}{1+k_{1}k_{2}}\right|$, and $k_{2}-k_{1}=2|a|$, $1+k_{1}k_{2}=1-a^{2}$: \[\tan\alpha = \frac{2|a|}{|1-a^{2}|}. \tag{3}\]

If $|a|\lt 1$, $\tan\alpha=\dfrac{2|a|}{1-a^{2}}$; if $|a|=1$, the denominator vanishes and $\alpha=\tfrac\pi2$ (perpendicular tangents); if $|a|\gt 1$, $\tan\alpha=\dfrac{2|a|}{a^{2}-1}$. Since $\tan\alpha$ has the form $\tan(2\theta)$ with $\tan\theta=|a|$, equivalently $\alpha=2\arctan|a|$ — the full opening angle of the corner.

Step 5: Verification

\[a=0.5\text{: } y_+'(0)\approx0.5000,\ y_-'(0)\approx-0.5000 \text{ (numerically)}\ \checkmark\]

\[a=2.0\text{: } y_+'(0)\approx2.0000,\ y_-'(0)\approx-2.0000\ \checkmark\]