Application of derivative - helicopter problem

Solution 1:

The square distance $d$ helicopter - soldier is: $$d^2=(x^2+7-7)^2+(x-3)^2$$ The minimum of this distance is given when the derivative of $d^2$ is zero. So: $$(d^2)'=4x^3+2x-6$$ which is zero for $x=1$ Putting $x=1$ in the equation of the helicopter trajectory, you have: $$P_m=(1,8)$$ which is the helicopter coordinates in wich the distance is minimum.

Solution 2:

A related problem. Assume the point that gives the nearest distance that lies on the curve is $(x,y)$, so the distance between the soldier and the plane is

$$ d=\sqrt{(x-3)^2+(y-7)^2}= \sqrt{(x-3)^2+x^4}$$

$$ \implies d' = \frac{ 4x^3 + 2(x-3) }{2\sqrt{(x-3)^2+x^4}}=0 \implies x=1.$$

Checking the second derivative at $x=1$ gives $d''>0$ which means it is minima. So, $y=x^2+7=8$ and the minimum distance will be at the point $(x,y)=(1,8)$ and it is equal to $d=\sqrt{5}$.