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 mounted on a pole at a height of \(5 \, \text{m}\). It observes a car moving away from the pole at a speed of \(20 \, \text{m/s}\). Let \(x \, \text{m}\) be the distance of the car from the base of the pole at any given moment. The angle of elevation from the camera to the car is denoted by \(\theta\).

(i) Determine an expression for \(\theta\) in terms of the camera's height and \(x\). Utilizing the right triangle formed by the pole's height and the car's distance from the base: 

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

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

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

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

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

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

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

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

(iii) (b) For a different car, the rate of change of the angle of elevation with respect to time is \(\frac{3}{101} \, \text{rad/s}\) when it is \(50 \, \text{m}\) from the base. Determine the speed of this car. 

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

Using the chain rule from part (ii): \[ \frac{d\theta}{dt} = \frac{d\theta}{dx} \cdot \frac{dx}{dt}. \] 

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

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

Summary of Results: 
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) The car's speed is \(15 \, \text{m/s}\).