Chapter 5.6, Problem 31E

To determine

To find: the speed in which the camera angle $\theta$ will change when the car is right in front of you and a half seconds later.

Answer to Problem 31E

The speed in which the camera angle $\theta$ will change right in front of you is $-2\text{ radians/sec}\text{.}$ and a half seconds later is $1\text{ radians/sec}\text{.}$

Explanation of Solution

Given information :

Let x be the distance of car from point right in front of you. We use negative $\frac{dr}{dt}$ , because r is decreasing.

  $\frac{dr}{dt}=-264\text{ft}\text{./sec}\text{.}$

From the given diagram in the question,

We can see that

  $\begin{array}{l}\text{tan}\theta =\frac{\text{perpendicular}}{\text{base}}=\frac{r}{132}\\ r=132\text{tan}\theta \end{array}$

All variables are differentiable functions of t .

Calculation:

We have to calculate the speed in which the camera angle $\theta$ will change when the car is right in front of you and a half seconds later.

Therefore,

  $r=132\text{tan}\theta$

Differentiate with respect to t .

  $\begin{array}{l}\frac{dr}{dt}=132{\text{sec}}^2\theta \frac{d\theta }{dt}\left[\text{Since,}\frac{d\left(\text{tan}\theta \right)}{dt}={\text{sec}}^2\theta \right]\\ \frac{d\theta }{dt}=\frac{1}{132{\text{sec}}^2\theta }\cdot \frac{dr}{dt}\\ \frac{d\theta }{dt}=\frac{{\text{cos}}^2\theta }{132{\text{sec}}^2\theta }\cdot \frac{dr}{dt}\mathrm{....}\left(1\right)\left[\text{Since,}\frac{1}{{\text{sec}}^2\theta }={\cos }^2\theta \right]\end{array}$

When the car is right in front of you,

  $\theta =0$

Putting the value of $\theta =0$ and $\frac{dr}{dt}=-264$ in equation (1),

  $\begin{array}{l}\frac{d\theta }{dt}=\frac{{\text{cos}}^2\theta }{132}\cdot \left(-264\right)\\ \frac{d\theta }{dt}=\frac{1}{132}\cdot \left(-264\right)\left[{\text{Since, cos}}^2\theta =1\right]\\ \frac{d\theta }{dt}=-2\text{ radians/sec}\text{.}\end{array}$

Again, when the car is right a half seconds later,

  $r=132$ and $\frac{dr}{dt}$ is positive.

So,

  $\begin{array}{l}\theta ={\text{tan}}^{-1}\left(\frac{r}{132}\right)\\ \theta ={\text{tan}}^{-1}\left(\frac{132}{132}\right)\\ \theta ={\text{tan}}^{-1}\left(1\right)\\ \theta ={45}^{\circ }\end{array}$

Putting the value of $\theta ={45}^{\circ }$ and $\frac{dr}{dt}=264$ in equation (1),

  $\begin{array}{l}\frac{d\theta }{dt}=\frac{{\text{cos}}^{2}45}{132}\cdot \left(264\right)\\ \frac{d\theta }{dt}=\frac{{\left(\frac{\sqrt{2}}{2}\right)}^2}{132}\cdot \left(264\right)\left[\text{Since, cos45}=\frac{\sqrt{2}}{2}\right]\\ \frac{d\theta }{dt}=\frac{2}{4\times 132}\cdot \left(264\right)\\ \frac{d\theta }{dt}=\frac{4}{4}\\ \frac{d\theta }{dt}=1\text{ radians/sec}\text{.}\end{array}$

Hence, the value of $\frac{d\theta }{dt}$ when the car is right in front of you is $-2\text{ radians/sec}\text{.}$ and a half seconds later is $1\text{ radians/sec}\text{.}$