Application of derivative - helicopter problem A helicopter of enemy is flying along the curve given by $y =x^2+7$. A soldier, placed at (3,7) wants to shoot down the helicopter when it is nearest to him. Find the nearest distance. 
Please guide how to proceed for this problem.. This is an example of application of derivative
 A: 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.
A: Let $A=(3, 7)$. The tangent at the nearest point, say $P:=(x, x^2+7)$, is perpendicular to $PA$. Since the slope of tangent at $P$ is $(x^2+7)'=2x$, we have
$$
2x\times \frac{x^2+7-7}{x-3}=-1
$$
or 
$$
2x^3+x-3=0
$$
the rest is yours.
A: 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 $d=\sqrt{5}$. 
