A helicopter flies along $y=x^2+7$. A soldier at $(3,7)$ wants the nearest distance. The nearest distance is

1. Distance squared: $f(x)=(x-3)^2+(x^2+7-7)^2=(x-3)^2+x^4$. 2. $f'(x)=2(x-3)+4x^3=2(x-1)(2x^2+2x+3)$. 3. The quadratic factor has no real roots, so $x=1$ is the only critical point. 4. $f(1)=(1-3)^2+1=4+1=5$, so distance $=\sqrt{5}$. 5. Check $f(0)=9>5$, confirming a minimum. _Source: NCERT Class 12 Mathematics Ch 6 "Application of Derivatives", p.27_