Application Of Derivatives Ques 21
21. A helicopter is flying along the curve given by $y-x^{3 / 2}=7,(x \geq 0)$. A soldier positioned at the point $\frac{1}{2}, 7$ wants to shoot down the helicopter when it is nearest to him. Then, this nearest distance is
(2019 Main, 10 Jan II)
(a) $\frac{1}{3} \sqrt{\frac{7}{3}}$
(b) $\frac{\sqrt{5}}{6}$
(c) $\frac{1}{6} \sqrt{\frac{7}{3}}$
(d) $\frac{1}{2}$
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Answer:
Correct Answer: 21.(c)
Solution:
Formula:
Shortest distance between two curves :
- The helicopter is nearest to the soldier, if the tangent to the path, $y=x^{3 / 2}+7,(x \geq 0)$ of helicopter at point $(x, y)$ is perpendicular to the line joining $(x, y)$ and the position of soldier $(\frac{1}{2}, 7)$.
$\because$ Slope of tangent at point $(x, y)$ is
$ \frac{d y}{d x}=\frac{3}{2} x^{1 / 2}=m_{1}(\mathrm{let}) ……(i) $
and slope of line joining $(x, y)$ and $(\frac{1}{2}, 7)$ is
$ m_{2}=\frac{y-7}{x-\frac{1}{2}}……(ii) $
Now, $\quad m_{1} \cdot m_{2}=-1$
$\Rightarrow \quad \frac{3}{2} x^{1 / 2} (\frac{y-7}{x-(1 / 2)})=-1$ [from Eqs. (i) and (ii)]
$\Rightarrow \quad \frac{3}{2} x^{1 / 2} \frac{x^{3 / 2}}{x-\frac{1}{2}}=-1$ $\left[\because y=x^{3 / 2}+7\right]$
$\Rightarrow \quad \frac{3}{2} x^{2}=-x+\frac{1}{2}$
$\Rightarrow \quad 3 x^{2}+2 x-1=0$
$\Rightarrow \quad 3 x^{2}+3 x-x-1=0$
$\Rightarrow 3 x(x+1)-1(x+1)=0$
$\Rightarrow \quad x=\frac{1}{3},-1$
Thus, the nearest point is $\frac{1}{3}, \frac{1}{3}^{3 / 2}+7$
Now, the nearest distance
$ \begin{aligned} & =\sqrt{(\frac{1}{2}-\frac{1}{3})^{2}+(7-(\frac{1}{3})^{3 / 2}-7)}=\sqrt{(\frac{1}{6})^{2}+(\frac{1}{3})^{3}} \\ & =\sqrt{\frac{1}{36}+\frac{1}{27}}=\sqrt{\frac{3+4}{108}}=\sqrt{\frac{7}{108}}=\frac{1}{6} \sqrt{\frac{7}{3}} \end{aligned} $
BETA
