Circle Inscribed in Bell Curves

Since the two curves are symmetrical about the origin $O(0,0)$ , let us consider only the first quadrant. Let the distance from $O$ to a point $P(x,y)$ on the bell curve be $r$ . Then by Pythagorean theorem, $r^2 = x^2+y^2$ . The smallest $r$ is the radius of the circle inscribed within the bell curves. Since $r$ is smallest when $r^2$ is smallest, we have:

$\begin{aligned} r^2 & = x^2 + y^2 = x^2 + e^{-2x^2} & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \\ \frac {d(r^2)}{dx} & = 2x - 4x e^{-2x^2} & \small \color{#3D99F6} \text{Equate }\frac {d(r^2)}{dx} = 0 \\ \implies 4x e^{-2x^2} & = 2x \\ e^{-2x^2} & = \frac 12 \\ \implies x^2 & = \frac {\ln 2}2 \end{aligned}$

Note that $\dfrac {d^2(r^2)}{dx^2} = 2 - 4 e^{-2x^2} + 16x^2 e^{-2x^2}$ $\implies \dfrac {d^2(r^2)}{dx^2}\bigg|_{x^2=\frac {\ln 2}2} > 0$ and $r^2$ is minimum when $x^2 = \dfrac {\ln 2}2$ .

The area of the inscribed circle is $\pi r^2 = \pi \left(x^2 + e^{-2x^2} \right) = \boxed{\dfrac \pi 2\left(1+\ln 2\right)}$ .

Solution in video