Question

An apache 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. Then the nearest distance is___ (in units)

• A. $\sqrt{3}$
• B. 5
• C. 0
• D. $\sqrt{5}$

Answer 1

Answer: (D)

For each value of $x$, the helicopter's position is at a point $\left(x,x^2+7\right)$.
Therefore, the distance between the helicopter and soldier placed at $(3,7)$ is $\sqrt{(x-3{)}^2+{\left(x^2+7-7\right)}^2}$, i.e., $\sqrt{(x-3{)}^2+x^4}$.
Let $f(x)=(x-3{)}^2+x^4$ or $f^{\prime }(x)=2(x-3)+4x^3=2(x-1)\left(2x^2+2x+3\right)$
Thus $f^{\prime }(x)=0$ gives $x=1$ or $2x^2+2x+3=0$ for which there are no real roots. Also there are no end points of the interval to be added to the set for which $f^{\prime }(x)=0$.
i.e., there is only one point, namely, $x=1$ at which $f^{\prime }(x)=0$.
The value of $f$ at this point is given by $f(1)=(1-3{)}^2+(1{)}^2=5$.
Thus, the distance between the solider and the helicopter is $\sqrt{f(1)}=\sqrt{5}$
Note that $\sqrt{5}$ is either a maximum value or a minimum value. Since $\sqrt{f(0)}=\sqrt{(0-3{)}^2+(0{)}^2}=3>\sqrt{5}$
It follows that $\sqrt{5}$ is the minimum value of $\sqrt{f(x)}$. Hence, $\sqrt{5}$ is the minimum distance between the soldier and helicopter.