Question

A particle move from right to left along the parabola $y=\sqrt{-x}$ in such a way that $x$-coordinate (measured in meters) decreases at the rate of $8\mathrm{m}/\sec$. At the moment when $\mathrm{x}=-4$ the rate at which the angle of inclination $\theta$ of the line joining the particle to the origin is changing, is

• A. -0.4
• B. -0.2
• C. -0.5
• D. -1

Answer 1

Answer: (A)

$y=\sqrt{-x};\tan \theta =\frac{y}{x}$
$\tan \theta =\frac{\sqrt{-x}}{x}=\frac{1}{\sqrt{-x}}$ ${\sec }^2\theta \frac{\mathrm{d}\theta }{\mathrm{dt}}=+\frac{1}{2}\frac{1}{(-\mathrm{x}{)}^{3/2}}\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1}{16}\cdot (-8)($ where $\mathrm{x}=-4)$
${\sec }^2\theta \frac{\mathrm{d}\theta }{\mathrm{dt}}=-\frac{1}{2}$
Also where $\mathrm{x}=-4,\tan \theta =\frac{2}{-4}=-\frac{1}{2}$
$\therefore \left(1+{\tan }^2\theta \right)\frac{\mathrm{d}\theta }{\mathrm{dt}}=-\frac{1}{2}$
$\frac{5}{4}\frac{\mathrm{d}\theta }{\mathrm{dt}}=-\frac{1}{2}\Rightarrow \frac{\mathrm{d}\theta }{\mathrm{dt}}=\frac{-2}{5}=-0.4$.