Application of Derivatives 1 Question 5
5. 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: 5. (d)
Solution:
- 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
$$ \frac{d y}{d x}=\frac{3}{2} x^{1 / 2}=m _1(let) $$
and slope of line joining $(x, y)$ and $\frac{1}{2}, 7$ is
$$ m _2=\frac{y-7}{x-\frac{1}{2}} $$
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
$$ \begin{aligned} & =\sqrt{\frac{1}{2}-\frac{1}{3}^{2}+7-\frac{1}{3}^{3 / 2}-7}=\sqrt{\frac{1}{6}^{2}+\frac{1}{3}^{3}} \\ & =\sqrt{\frac{1}{36}+\frac{1}{27}}=\sqrt{\frac{3+4}{108}}=\sqrt{\frac{7}{108}}=\frac{1}{6} \sqrt{\frac{7}{3}} \end{aligned} $$
