Chapter 5.6, Problem 26E

To determine

To find: the angle of inclination of the line joining the particle to the origin changing when $x=-4$ .

Answer to Problem 26E

The angle of inclination is $-0.4\text{ radian/sec}$ .

Explanation of Solution

Given information:

A particle moves from right to left along the parabolic curve $y=\sqrt{-x}$ .

  $\frac{dx}{dt}=8\text{ m}\text{./sec }\left[\begin{array}{l}\text{We use positive }\frac{dx}{dt}\text{because}\\ \text{length along }x\text{ axis is increasing}\text{.}\end{array}\right]$

All variables are differentiable functions of t .

Calculation :

We have to calculate the angle of inclination of the line joining the particle to the origin changing when $x=-4$ .

Therefore,

  $\begin{array}{l}y=\sqrt{-x}\\ \theta ={\text{tan}}^{-1}\left(\frac{y}{x}\right)\left[\begin{array}{l}\text{Since in parabolic}\\ \text{equation,}\theta ={\text{tan}}^{-1}\left(\frac{y}{x}\right)\end{array}\right]\end{array}$ .

Putting the value of $y=x^2$ in above equation,

  $\begin{array}{l}\theta ={\text{tan}}^{-1}\left(\frac{\sqrt{-x}}{x}\right)\\ \theta ={\text{tan}}^{-1}\left[\frac{{\left(-x\right)}^{\frac{1}{2}}}{x}\right]\end{array}$

Using change rule and quotient rule.

Differentiate with respect to t .

  $\begin{array}{l}\frac{d\theta }{dt}=\frac{1}{{\left(\frac{\sqrt{-x}}{x}\right)}^2+1}\left[\frac{x\cdot \frac{1}{2}{\left(-x\right)}^{\frac{1}{2}-1}\cdot \left(-1\right)\frac{dx}{dt}-\sqrt{-x}\frac{dx}{dt}}{x^2}\right]\\ \frac{d\theta }{dt}=\frac{1}{\frac{-x}{x^2}+1}\left[\frac{x\cdot \frac{1}{2}{\left(-x\right)}^{\frac{1}{2}}-\sqrt{-x}}{x^2}\right]\frac{dx}{dt}\\ \frac{d\theta }{dt}=\frac{1}{\frac{-x}{x^2}+1}\left[\frac{\frac{-x}{2\sqrt{-x}}-\sqrt{-x}}{x^2}\right]\frac{dx}{dt}\end{array}$

Putting the value of $x=-4$ and $\frac{dx}{dt}=8$ in above equation,

  $\begin{array}{l}\frac{d\theta }{dt}=\frac{1}{\frac{-\left(-4\right)}{{\left(-4\right)}^2}+1}\left[\frac{\frac{-\left(-4\right)}{2\sqrt{-\left(-4\right)}}-\sqrt{-\left(-4\right)}}{{\left(-4\right)}^2}\right]\left(8\right)\\ \frac{d\theta }{dt}=\frac{1}{\frac{4}{16}+1}\left[\frac{\frac{4}{2\sqrt{4}}-\sqrt{4}}{16}\right]\left(8\right)\\ \frac{d\theta }{dt}=\frac{1}{\frac{4+16}{16}}\left[\frac{\frac{4}{2\cdot \left(2\right)}-2}{16}\right]\left(8\right)\\ \frac{d\theta }{dt}=\frac{16}{20}\left[\frac{\frac{4}{4}-2}{16}\right]\left(8\right)\\ \frac{d\theta }{dt}=\frac{1}{20}\left[\frac{1-2}{1}\right]\left(8\right)\\ \frac{d\theta }{dt}=\frac{1}{20}\left[-1\right]\left(8\right)\\ \frac{d\theta }{dt}=\frac{-8}{20}\\ \frac{d\theta }{dt}=-0.4\text{ radains/sec}\text{.}\end{array}$

Hence, the angle of inclination is $-0.4\text{ radian/sec}$ .