War vs Maths

The closest distance between the helicopter and the soldier is when the normal to the curve passes $(3,7)$ .

Equation of normal to any point $(x, y)$ on the curve passing through $(3,7)$ is:

$\frac{dy}{dx}(y) + (x) = \frac{dy}{dx}(7) + (3)$

But for a point on the curve, $y = x^2 + 7$ , substituting this and solving for $x$ :

$\displaystyle \Rightarrow 2x( x^2 + 7) + (x) = 2x(7) + (3)$

$\displaystyle \Rightarrow 2x^3 + x - 3 = 0$

$\displaystyle \Rightarrow (x-1)(2x^2 + 2x + 3)=0$

The quadratic equation has no real roots, so $x=1$ is the only solution.

Hence, $(1,8)$ is the point where the helicopter is closest to the soldier and the square of the distance between them is:

$\displaystyle \Rightarrow (8-7)^2 + (1-3)^2 = \boxed{5}$

By translating the problem by $(0, - 7)$ , it is equivalent to minimize the distance between the points $(x, x^2)$ and $(3,0)$ . Then, the minimum euclidian distance squared between these points is : $\min_{x \in R} (x-3)^2 + x^4$ .

By denoting $g(x) = (x-3)^2 + x^4$ , its derivative is :

$\dfrac{dg(x)}{dx} = (4x^3 + 2x - 6) = (x-1)(4x^2 + 4x +6)$ , which has only one root at $x = 1$ . Consequently, since $g$ is strictly convex (the derivative of $g$ is positive), the minimum distance squared is given by :

$(3 - 1)^2 + 1^4 = 2^2 + 1 = \boxed{5}$ .