Chapter 4.6, Problem 26E

To determine

To find: How quickly does the line connecting the particle to the origin change its $\theta$ angle of inclination when $x=-4$ .

Answer to Problem 26E

The answer: Decreasing at $0.4\text{ radians/sec }$ .

Explanation of Solution

Given information:

Peripheral Motion A particle moves along the parabolic curve $y=\sqrt{-x}$ from right to left in such a way that the rate at which its $x$ -coordinate $($ in metres $)$ drops is $x=-4$ .

Calculation:

  

Get an equation relating $\theta$ and $x$ :

  $\begin{array}{c}\frac{dx}{dt}=8\\ y=\sqrt{-x}\\ \theta =\arctan \frac{y}{x}\\ =\arctan \frac{\sqrt{-x}}{x}\\ \frac{d\theta }{dt}=\frac{1}{{\left(\frac{\sqrt{-x}}{x}\right)}^2+1}\left(\frac{x\left((1/2){(-x)}^{-1/2}(-1)\right)-\sqrt{-x}}{x^2}\right)\frac{dx}{dt}\\ =\frac{1}{\frac{-x}{x^2}+1}\cdot \frac{\frac{-x}{2\sqrt{-x}}-\sqrt{-x}}{x^2}\frac{dx}{dt}\\ =\frac{\frac{-x}{2\sqrt{-x}}-\sqrt{-x}}{x^2-x}\cdot \frac{2\sqrt{-x}}{2\sqrt{-x}}\frac{dx}{dt}\\ =\frac{-x-2(-x)}{\left(x^2-x\right)2\sqrt{-x}}\frac{dx}{dt}\\ =\frac{1}{2(x-1)\sqrt{-x}}\frac{dx}{dt}\\ \frac{d\theta }{dt}=\frac{1}{2((-4)-1)\sqrt{-(-4)}}(8)\\ =-0.4\end{array}$

Therefore, the required decreasing at $0.4\text{ radians/sec }$ the line connecting the particle to the origin change its $\theta$ angle of inclination when $x=-4$ .