Q. 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 :
Solution:
$y-x^{3/2} =7\left(x\ge0\right)$
$ \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}} $
