Question

The traffic police has installed Over Speed Violation Detection (OSVD) system at various locations in a city. These cameras can capture a speeding vehicle from a distance of 300 m and even function in the dark. A camera is installed on a pole at the height of 5 m. It detects a car travelling away from the pole at the speed of 20 m/s. At any point, \(x\) m away from the base of the pole, the angle of elevation of the speed camera from the car C is \(\theta\). 
On the basis of the above information, answer the following questions: 
(i)Express \(\theta\) in terms of the height of the camera installed on the pole and x.
(ii) Find \(\frac{d\theta}{dx}\).
(iii) (a) Find the rate of change of angle of elevation with respect to time at an instant when the car is 50 m away from the pole.
(iii) (b) If the rate of change of angle of elevation with respect to time of another car at a distance of 50 m from the base of the pole is \(\frac{3}{101} \, \text{rad/s}\), then find the speed of the car.

Hint:

For related rates problems involving trigonometric functions, differentiate using the chain rule, substitute given values, and simplify systematically.

Answer 1

A camera is installed on a pole at the height of \(5 \, \text{m}\). It detects a car traveling away from the pole at the speed of \(20 \, \text{m/s}\). At any point, \(x \, \text{m}\) away from the base of the pole, the angle of elevation of the camera to the car \(C\) is \(\theta\). 

(i) Express \(\theta\) in terms of the height of the camera and \(x\). From the given setup, we can use the right triangle formed by the height of the pole and the distance of the car from the base. 

The tangent of the angle \(\theta\) is given by: \[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{x}. \] 
Thus: \[ \theta = \tan^{-1}\left(\frac{5}{x}\right). \] 

(ii) Find \(\frac{d\theta}{dx}\). 
Differentiating \(\theta = \tan^{-1}\left(\frac{5}{x}\right)\) with respect to \(x\), we use the chain rule: \[ \frac{d\theta}{dx} = \frac{1}{1 + \left(\frac{5}{x}\right)^2} \cdot \frac{d}{dx}\left(\frac{5}{x}\right). \] 

Simplify \(\frac{5}{x}\): \[ \frac{d}{dx}\left(\frac{5}{x}\right) = -\frac{5}{x^2}. \] 

Substitute: \[ \frac{d\theta}{dx} = \frac{1}{1 + \frac{25}{x^2}} \cdot \left(-\frac{5}{x^2}\right). \] 

Simplify further: \[ \frac{d\theta}{dx} = \frac{-\frac{5}{x^2}}{1 + \frac{25}{x^2}} = \frac{-5}{x^2 + 25}. \]
 
(iii) (a) Find the rate of change of angle of elevation with respect to time when the car is \(50 \, \text{m}\) away from the pole. 
Let \(x = 50 \, \text{m}\) and the car's speed be \(\frac{dx}{dt} = 20 \, \text{m/s}\). 

The rate of change of the angle of elevation with respect to time is: \[ \frac{d\theta}{dt} = \frac{d\theta}{dx} \cdot \frac{dx}{dt}. \] 
From part (ii), \(\frac{d\theta}{dx} = \frac{-5}{x^2 + 25}\). 

Substitute \(x = 50\): \[ \frac{d\theta}{dx} = \frac{-5}{50^2 + 25} = \frac{-5}{2500 + 25} = \frac{-5}{2525} = \frac{-1}{505}. \] 

Now: \[ \frac{d\theta}{dt} = \frac{-1}{505} \cdot 20 = \frac{-20}{505} = \frac{-4}{101} \, \text{rad/s}. \] 

(iii) (b) If the rate of change of angle of elevation with respect to time for another car at \(50 \, \text{m}\) from the base is \(\frac{3}{101} \, \text{rad/s}\), find the speed of the car. 

Let the speed of the car be \(\frac{dx}{dt} = v\). 

From part (ii): \[ \frac{d\theta}{dt} = \frac{d\theta}{dx} \cdot \frac{dx}{dt}. \] 

Substitute \(\frac{d\theta}{dt} = \frac{3}{101}\) and \(\frac{d\theta}{dx} = \frac{-1}{505}\): \[ \frac{3}{101} = \frac{-1}{505} \cdot v. \] 

Solve for \(v\): \[ v = \frac{3}{101} \cdot 505 = \frac{1515}{101} = 15 \, \text{m/s}. \] 

Final Answers: 
1. \(\theta = \tan^{-1}\left(\frac{5}{x}\right)\), 
2. \(\frac{d\theta}{dx} = \frac{-5}{x^2 + 25}\), 
3. (a) \(\frac{d\theta}{dt} = \frac{-4}{101} \, \text{rad/s}\), 
(b) Speed of the car is \(15 \, \text{m/s}\).