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. \vspace{0.3cm} \includegraphics[width=\linewidth]{latex2.png} \vspace{0.3cm} 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 \(\theta\). \vspace{0.3cm} On the basis of the above information, answer the following questions: \begin{enumerate} Express \( \theta \) in terms of the height of the camera installed on the pole and \( x \). Find \( \frac{d\theta}{dx} \). \begin{enumerate} 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. OR 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 \( \frac{3}{101} \) rad/s, then find the speed of the car. \end{enumerate} \end{enumerate}

Hint:

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

Answer 1

Step 1: Express \( \theta \) in terms of the height of the camera and \( x \). Using the right triangle formed by the height of the pole and the distance of the car from the base, we have: \[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{5}{x} \] Thus, \[ \theta = \tan^{-1}\left(\frac{5}{x}\right) \] Step 2: Find \( \frac{d\theta}{dx} \). Differentiating \( \theta = \tan^{-1}\left(\frac{5}{x}\right) \) with respect to \( x \): \[ \frac{d\theta}{dx} = \frac{1}{1 + \left(\frac{5}{x}\right)^2} \cdot \left(-\frac{5}{x^2}\right) \] \[ \frac{d\theta}{dx} = \frac{-5}{x^2 + 25} \] Step 3: Find \( \frac{d\theta}{dt} \) at \( x = 50 \). Given \( \frac{dx}{dt} = 20 \, \text{m/s} \), using the chain rule: \[ \frac{d\theta}{dt} = \frac{d\theta}{dx} \cdot \frac{dx}{dt} \] \[ \frac{d\theta}{dt} = \frac{-5}{50^2 + 25} \cdot 20 \] \[ \frac{d\theta}{dt} = \frac{-5}{2525} \cdot 20 = \frac{-20}{505} = \frac{-4}{101} \, \text{rad/s} \] Step 4: Find the speed of another car if \( \frac{d\theta}{dt} = \frac{3}{101} \). Using the given value of \( \frac{d\theta}{dt} \): \[ \frac{3}{101} = \frac{-5}{2525} \cdot v \] \[ v = \frac{3}{101} \cdot 505 = 15 \, \text{m/s} \] Final Answers: \begin{enumerate} \( \theta = \tan^{-1}\left(\frac{5}{x}\right) \) \( \frac{d\theta}{dx} = \frac{-5}{x^2 + 25} \) \begin{enumerate} \( \frac{d\theta}{dt} = \frac{-4}{101} \) rad/s The speed of the car is \( 15 \) m/s \end{enumerate} \end{enumerate}