Question

A helicopter is flying along the curve given by $y - x^{3/2} = 7, (x \ge 0)$. A soldier positioned at the point $\left(\frac{1}{2}, 7\right)$ wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is :

• A. $\frac{1}{2}$
• B. $\frac{1}{3} \sqrt{\frac{7}{3} }$
• C. $\frac{1}{6} \sqrt{\frac{7}{3} }$
• D. $\frac{\sqrt{5}}{6}$

Answer 1

Answer: (C)

$\frac{1}{6} \sqrt{\frac{7}{3} }$

Answer 2

$y-x^{3/2} =7\left(x\ge0\right)$
$\frac{dy}{dx} =\frac{3}{2}x^{1/2}$
$\left(\frac{3}{2} \sqrt{x}\right)\left(\frac{7-y}{\frac{1}{2}-x}\right)=-1$
$\left(\frac{3}{2} \sqrt{x}\right) \left(\frac{-x^{3/2}}{\frac{1}{2}-x}\right) =-1$
$\frac{3}{2}.x^{2} =\frac{1}{2}-x$
$3x^{2} =1-2x$
$3x^{2}+2x-1=0$
$3x^{2}+3x-x-1=0$
$\left(x+1\right)\left(3x-1\right)=0$
$\therefore x=-1$ (rejected)
$x=\frac{1}{3}$
$y=7+x^{3/2} =7+\left(\frac{1}{3}\right)^{3/2}$
$\ell_{AB} = \sqrt{\left(\frac{1}{2} -\frac{1}{3}\right)^{2} +\left(\frac{1}{3}\right)^{3}} = \sqrt{\frac{1}{36}+\frac{1}{27}}$
$= \sqrt{\frac{3+4}{9\times12}}$
$=\sqrt{\frac{7}{108}} = \frac{1}{6} \sqrt{\frac{7}{3}}$