JEE Main 2019 (Online) 10th January Evening ShiftQuestion 1
A helicopter is flying along the curve given by $y-x^{3/2}=7$, $(x\ge0)$. A soldier positioned at the point $\left(\frac12,7\right)$, who wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is :
Differentiation and its Application
Solution
Concept. Minimise $d^2$ instead of $d$; on the curve $y-7=x^{3/2}$. Formulas used. (1) $d^2=(x-x_0)^2+(y-y_0)^2$Step 1 - $d^2$ in $x$. With $y-7=x^{3/2}$, by (1): $$d^2(x)=\left(x-\tfrac12\right)^2+x^3,\qquad x\ge0.$$ Step 2 - Critical point. $\frac{d}{dx}d^2=2x-1+3x^2=(3x-1)(x+1)=0$, so $x=\frac13$ (in range). It is a minimum: the derivative changes from $-$ to $+$, and $d^2(0)=\frac14$ is larger. Step 3 - Distance. $$d^2=\frac1{36}+\frac1{27}=\frac{3+4}{108}=\frac{7}{108},\qquad d=\frac{\sqrt7}{6\sqrt3}=\frac16\sqrt{\frac73}.$$ Answer: (A)