A helicopter is flying along the curve given

Question:

A helicopter is flying along the curve given by $y-x^{3 / 2}=7,(x \geq 0)$. A soldier positioned at the

$\operatorname{point}\left(\frac{1}{2}, 7\right)$ wants to shoot down the helicopter

when it is nearest to him. Then this nearest distance is :

 

  1. $\frac{1}{2}$

  2. $\frac{1}{3} \sqrt{\frac{7}{3}}$

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

  4. $\frac{\sqrt{5}}{6}$


Correct Option: , 3

Solution:

$y-x^{3 / 2}=7(x \geq 0)$

$\frac{d y}{d x}=\frac{3}{2} x^{1 / 2}$

$\left(\frac{3}{2} \sqrt{x}\right)\left(\frac{7-y}{\frac{1}{2}-x}\right)=-1$

$\left(\frac{3}{2} \sqrt{x}\right)\left(\frac{-x^{3 / 2}}{\frac{1}{2}-x}\right)=-1$

$\frac{3}{2} \cdot x^{2}=\frac{1}{2}-x$

$3 x^{2}=1-2 x$

$3 x^{2}+2 x-1=0$

$3 x^{2}+3 x-x-1=0$

$(x+1)(3 x-1)=0$

$\therefore \quad x=-1($ rejected $)$

$x=\frac{1}{3}$

$y=7+x^{3 / 2}=7+\left(\frac{1}{3}\right)^{3 / 2}$

$\ell_{\mathrm{AB}}=\sqrt{\left(\frac{1}{2}-\frac{1}{3}\right)^{2}+\left(\frac{1}{3}\right)^{3}}=\sqrt{\frac{1}{36}+\frac{1}{27}}$

$=\sqrt{\frac{3+4}{9 \times 12}}$

$=\sqrt{\frac{7}{108}}=\frac{1}{6} \sqrt{\frac{7}{3}}$

Option (3)

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now