Chapter 4.6, Problem 34E

To determine

To calculate: Find the changing angle.

Answer to Problem 34E

Thus that the angle $\theta$ is changing at the rate of ${5.73}^{°}/\text{sec}.$

Explanation of Solution

Given information:

Speed of $A$ approaches the intersection $2m/sec$ .

Speed of $B$ away the intersection $1m/sec$ .

At what rate is the angle $\theta$ changing when $A$ is $10meter$ from the intersection and $B$ is $20meter$ from the intersection.

Calculation:

The goal of the exercise is to compute the rate at which the angle $\theta$ changes in the given figure at the instant when the person $A$ is at $10 \text{m}$ distance from the point $O$ and the person $B$ is at $20 \text{m}$ from $O$ .

It is given that $A$ approaches the point $O$ at the rate of $2 \text{m}/\text{sec}$ and $B$ is moving away from the point $O$ at the rate of $1 \text{m}/\text{sec}$

  

Let's denote the distance of person $B$ from point $O$ by $b$ and the distance of person $A$ from point $O$ by $a$ .

It is given that

  $\begin{array}{c}\frac{da}{dt}=2 \text{m}/\text{sec},\\ \frac{db}{dt}=-1 \text{m}/\text{sec}\end{array}$

To obtain the rate of change of $\theta$ need to find a relation between a, b and $\theta$ using the trigonometric ratio in the triangle $△ABC$ .

The most appropriate trigonometric ratio to relate a, b and $\theta$ will be the tangent of $\theta$

  $\begin{array}{c}\tan \theta =\frac{OA}{OB}\\ =\frac{a}{b}.\end{array}$

Now will differentiate the obtained equation with respect to $t$

  $\begin{array}{c}\frac{d(\tan \theta )}{dt}=\frac{d}{dt}\frac{a}{b}\\ {\sec }^2\theta \frac{d\theta }{dt}=\frac{\frac{da}{dt}\cdot b-a\cdot \frac{db}{dt}}{b^2}\end{array}$ or equally

  $\frac{d\theta }{dt}=\frac{1}{b^2{\sec }^2\theta }b\frac{da}{dt}-a\frac{db}{dt}$

Now from Eq. (1) let's observe that to find the required rate of change of angle $\theta$ at the given instant need $a,b,\theta ,\frac{da}{dt}$ and $\frac{db}{dt}$ .

Note that have $a,b,\frac{da}{dt}$ and $\frac{db}{dt}$ so need to compute $\theta$ at the given instant.

Let's substitute $a=10 \text{m}$ and $b=20 \text{m}$ in the tangent of $\theta$

  $\begin{array}{c}\tan \theta =\frac{10}{20}\\ =\frac{1}{2}\end{array}$

which gives that $\theta$ is equal to

  $\theta ={\tan }^{-1}\frac{1}{2}$

Now let's substitute $a=10,b=20,\theta ={\tan }^{-1}\frac{1}{2},\frac{da}{dt}=2 \text{m}/\text{sec}$ and $\frac{db}{dt}=-1 \text{m}/\text{sec}$ in Eq. (1) to obtain $\frac{d\theta }{dt}$ at the given instant

  $\begin{array}{c}\frac{d\theta }{dt}=\frac{1}{{(20)}^2{\sec }^2\left({\tan }^{-1}1/2\right)}(20\cdot 2-10\cdot (-1))\\ =\frac{{\cos }^2\left({\tan }^{-1}1/2\right)}{400}\cdot (50)\\ =0.1\text{rad}/\text{sec}\end{array}$

The next task of the problem is to convert the $\frac{d\theta }{dt}$ into degrees per second.

Know that $1\text{rad}=\frac{{180}^{°}}{\pi }$ so to convert the given rate $\frac{d\theta }{dt}$ into degrees per second let' multiply $\frac{{180}^{°}}{\pi }$ in the rate $\frac{d\theta }{dt}$

  $\begin{array}{c}\frac{d\theta }{dt}=0.1\cdot \frac{{180}^{°}}{\pi }/\text{sec}\\ \approx {5.73}^{°}\text{sec}.\end{array}$

Thus that the angle $\theta$ is changing at the rate of ${5.73}^{°}/\text{sec}.$