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)

$y-x^{3/2}=7\left(x\ge 0\right)$
$\frac{dy}{dx}=\frac{3}{2}{x}^{1/2}$
$\left(\frac{3}{2}\sqrt{x}\right)\left(\frac{7-y}{\frac{1}{2}-x}\right)=-1$
$\left(\frac{3}{2}\sqrt{x}\right)\left(\frac{-x^{3/2}}{\frac{1}{2}-x}\right)=-1$
$\frac{3}{2}.x^2=\frac{1}{2}-x$
$3x^2=1-2x$
$3x^2+2x-1=0$
$3x^2+3x-x-1=0$
$\left(x+1\right)\left(3x-1\right)=0$
$\therefore x=-1$ (rejected)
$x=\frac{1}{3}$
$y=7+x^{3/2}=7+{\left(\frac{1}{3}\right)}^{3/2}$
${\ell }_{AB}=\sqrt{{\left(\frac{1}{2}-\frac{1}{3}\right)}^2+{\left(\frac{1}{3}\right)}^3}=\sqrt{\frac{1}{36}+\frac{1}{27}}$
$=\sqrt{\frac{3+4}{9\times 12}}$
$=\sqrt{\frac{7}{108}}=\frac{1}{6}\sqrt{\frac{7}{3}}$