Chapter 5.6, Problem 34E

To determine

To find: the rate at which the angle $\theta$ will change when A is 10m. and B is 20m. from the intersection.

Answer to Problem 34E

The rate at which the angle $\theta$ will change is $-\frac{18}{\pi }\text{ degree/second}$

Explanation of Solution

Given information :

From the given diagram in the question, we can see that

  $\theta$ changes in a right triangle at a given moment.

  $\begin{array}{l}\text{A}\frac{d\text{A}}{dt}=-2\text{m}\text{./sec}\text{. }\left[\begin{array}{l}\text{It is negative because it}\\ \text{approaches the intersection}\text{.}\end{array}\right]\\ \frac{d\text{B}}{dt}=1\text{m}\text{./sec}\text{. }\left[\begin{array}{l}\text{It is positive because it moves}\\ \text{away from the intersection}\text{.}\end{array}\right]\end{array}$

A = 10m.

B = 20m.

All variables are differentiable functions of t .

Calculation:

We have to calculate the rate at which the angle $\theta$ will change when A is 10m. and B is 20m. from the intersection.

First find the value of C.

From Pythagoras theorem,

  $\begin{array}{l}{\text{A}}^2+{\text{B}}^2={\text{C}}^2\\ \text{C}=\sqrt{{\text{A}}^2+{\text{B}}^2}\\ \text{C}=\sqrt{{10}^2+{20}^2}\\ \text{C}=\sqrt{100+400}\left[\begin{array}{l}\text{Since, A}=\text{10m}\text{.}\\ \text{B}=\text{20m}\text{.}\end{array}\right]\\ \text{C}=\sqrt{500}\end{array}$

Again from trigonometry equation,

  $\begin{array}{l}\text{tan}\theta =\frac{\text{perpendicular}}{\text{base}}=\frac{\text{A}}{\text{B}}\\ \text{tan}\theta ={\text{AB}}^{-1}\end{array}$

Differentiate with respect to t using product rule.

  $\begin{array}{l}\frac{1}{\cos \theta }\cdot \frac{d\theta }{dt}=\text{A}\cdot \left(-1\right)\cdot {\text{B}}^{-1-1}\frac{d\text{B}}{dt}+{\text{B}}^{-1}\frac{d\text{A}}{dt}\\ \frac{1}{\cos \theta }\cdot \frac{d\theta }{dt}=-{\text{AB}}^{-2}\frac{d\text{B}}{dt}+{\text{B}}^{-1}\frac{d\text{A}}{dt}\\ \frac{1}{\cos \theta }\cdot \frac{d\theta }{dt}=-\frac{\text{A}}{{\text{B}}^2}\frac{d\text{B}}{dt}+\frac{1}{\text{B}}\frac{d\text{A}}{dt}\\ \frac{d\theta }{dt}={\cos }^2\theta \left[-\frac{\text{A}}{{\text{B}}^2}\frac{d\text{B}}{dt}+\frac{1}{\text{B}}\frac{d\text{A}}{dt}\right]\mathrm{.....}\left(1\right)\end{array}$

Again from trigonometry equation,

  $\begin{array}{l}\text{cos}\theta =\frac{\text{base}}{\text{hypotenuse}}\\ {\text{cos}}^2\theta ={\left(\frac{\text{20}}{\sqrt{500}}\right)}^2\left[\begin{array}{l}\text{Since, base is B that is 20}\text{.}\\ \text{Hypotenuse is c that is }\sqrt{500}.\end{array}\right]\end{array}$

Putting the value of $\text{A}=10$ , $\text{B}=20$ , $\frac{d\text{A}}{dt}=-2$ , $\frac{d\text{B}}{dt}=1$ and ${\text{cos}}^2\theta ={\left(\frac{\text{20}}{\sqrt{500}}\right)}^2$ in equation (1),

  $\begin{array}{l}\frac{d\theta }{dt}={\left(\frac{\text{20}}{\sqrt{500}}\right)}^2\left[-\frac{10}{{\left(20\right)}^2}\left(1\right)+\frac{1}{20}\left(-2\right)\right]\\ \frac{d\theta }{dt}=\frac{\text{400}}{500}\left[-\frac{10}{400}-\frac{1}{10}\right]\\ \frac{d\theta }{dt}=\frac{\text{400}}{500}\times \left(-\frac{10}{400}\right)-\frac{1}{10}\times \frac{\text{400}}{500}\\ \frac{d\theta }{dt}=-\frac{\text{1}}{50}-\frac{4}{50}\\ \frac{d\theta }{dt}=-\frac{5}{50}\\ \frac{d\theta }{dt}=-\frac{\text{1}}{10}\text{ radians/second}\\ \frac{d\theta }{dt}=-\frac{\text{1}}{10}\cdot \frac{180}{\pi }\text{ degree/second }\left[\text{Since, 1 radian}=\frac{180}{\pi }\text{ degree/second}\right]\\ \frac{d\theta }{dt}=-\frac{18}{\pi }\text{ degree/second}\end{array}$

Hence, the rate at which the angle $\theta$ will change is $-\frac{18}{\pi }\text{ degree/second}$