CBSE Board Examination 2024 · Section E · 4 marks

Q36: Case study: a speed camera and the rate of change of the angle of elevation

Board 2024Ch 6 · Application of Derivatives Ch 5 · Continuity and Differentiability

The 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 height of the camera installed on the pole and $x$ ⁠. (1 mark)

(ii) Find $\dfrac{d\theta}{dx}$ ⁠. (1 mark)

(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. (2 marks)

(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 $\dfrac{3}{101}$ rad/s, then find the speed of the car. (2 marks)

Exam answer

(i) The camera is 5 m high and the car is $x$ m from the base of the pole, so

Why?

The pole, the ground and the line of sight form a right-angled triangle. Seen from the car, the angle of elevation $\theta$ has the 5 m pole opposite it and the $x$ m of ground next to it:

\tan\theta = \frac{\text{opposite}}{\text{adjacent}}

$\tan\theta = \frac{5}{x} \Rightarrow \theta = \tan^{-1}\left(\frac{5}{x}\right)$

Why?

$\theta$ is the angle whose tangent is $\frac{5}{x}$ ⁠. As $x > 0$ ⁠, $\frac{5}{x} > 0$ ⁠, and $\tan^{-1}$ gives an angle between 0 and $\frac{\pi}{2}$ ⁠, as an angle of elevation should be.

(ii)

\begin{aligned} \frac{d\theta}{dx} &= \frac{1}{1 + \frac{25}{x^2}}\cdot\left(-\frac{5}{x^2}\right) \\ &= \frac{x^2}{x^2 + 25}\cdot\left(-\frac{5}{x^2}\right) \end{aligned}

Why?

Chain rule with $u = \frac{5}{x}$ ⁠:

\frac{d}{dx}\tan^{-1}u = \frac{1}{1 + u^2}\cdot\frac{du}{dx}

Here $u^2 = \frac{25}{x^2}$ ⁠, and $\frac{du}{dx} = -\frac{5}{x^2}$ since $\frac{5}{x} = 5x^{-1}$ ⁠.

\begin{aligned} u &= \frac{5}{x} = 5x^{-1} \\ \frac{du}{dx} &= 5(-1)x^{-2} = -\frac{5}{x^2} \\[4pt] u^2 &= \Big(\frac{5}{x}\Big)^2 = \frac{25}{x^2} \end{aligned}

For the second line, multiply the top and the bottom of the first fraction by $x^2$ ⁠. The $x^2$ then cancels.

\begin{aligned} &\frac{1}{1 + \frac{25}{x^2}} \\ &= \frac{1 \times x^2}{\big(1 + \frac{25}{x^2}\big) \times x^2} \\ &= \frac{x^2}{x^2 + 25} \\[4pt] &\frac{x^2}{x^2 + 25}\cdot\Big(-\frac{5}{x^2}\Big) \\ &= \frac{-5x^2}{x^2(x^2 + 25)} \\ &= \frac{-5}{x^2 + 25} \quad (\div x^2) \end{aligned}

$\therefore\ \frac{d\theta}{dx} = \frac{-5}{25 + x^2}$

Why?

Watch out: in $1 + u^2$ the whole of $u = \frac{5}{x}$ is squared, giving $\frac{25}{x^2}$ ⁠.

The result is negative for every $x$ ⁠: the further the car, the smaller the angle.

(iii) (a) The car moves away from the pole at 20 m/s, so $\frac{dx}{dt} = 20$ ⁠.

Why?

Moving away means the distance $x$ is increasing, so its rate $\frac{dx}{dt}$ is positive.

\begin{aligned} \frac{d\theta}{dt} &= \frac{d\theta}{dx}\times\frac{dx}{dt} \\ &= \frac{-5}{25 + x^2}\times 20 \end{aligned}

Why?

$\theta$ depends on $x$ ⁠, and $x$ depends on time, so the chain rule links the two rates. We use $\frac{d\theta}{dx}$ from part (ii).

At $x = 50$ ⁠,

\frac{d\theta}{dt} = \frac{-100}{25 + 2500} = \frac{-100}{2525}

Why?

$50^2 = 2500$ and $-5\times 20 = -100$ ⁠. Dividing the top and the bottom by 25 gives $\frac{-4}{101}$ ⁠.

\begin{aligned} &\frac{-5}{25 + (50)^2}\times 20 \\ &= \frac{-5 \times 20}{25 + 2500} \\ &= \frac{-100}{2525} \\ &= \frac{-100 \div 25}{2525 \div 25} = \frac{-4}{101} \end{aligned}

The sign is negative: as the car moves away, the camera is seen lower, so the angle decreases.

$\therefore\ \frac{d\theta}{dt} = -\frac{4}{101}$ rad/s (the angle decreases at $\frac{4}{101}$ rad/s).

(iii) (b)

\begin{aligned} \frac{d\theta}{dt} &= \frac{d\theta}{dx}\times\frac{dx}{dt} \\ \Rightarrow \frac{3}{101} &= \frac{-5}{25 + 50^2}\times\frac{dx}{dt} \\ &= \frac{-5}{2525}\times\frac{dx}{dt} \end{aligned}

Why?

The same chain rule, but now the rate of the angle is known and $\frac{dx}{dt}$ is the unknown. Put $x = 50$ in $\frac{d\theta}{dx}$ ⁠.

\Rightarrow \frac{dx}{dt} = -\frac{3 \times 2525}{5 \times 101} = -15

Why?

Multiply both sides by $-\frac{2525}{5}$ ⁠. Since $2525 = 25\times 101$ ⁠, the 101 cancels and $\frac{25}{5} = 5$ ⁠, so the value is $-3\times 5 = -15$ ⁠.

\begin{aligned} \frac{dx}{dt} &= \frac{3}{101}\times\Big(-\frac{2525}{5}\Big) \\ &= -\frac{3 \times 2525}{101 \times 5} \\ &= -\frac{3 \times 25 \times 101}{101 \times 5} \\ &= -\frac{3 \times 25}{5} \quad (\div 101) \\ &= -3 \times 5 = -15 \end{aligned}

Speed is the size of the rate, without its sign, so the speed is 15 m/s.

The negative sign means this car is moving towards the pole.

$\therefore$ The speed of the car is 15 m/s.

Where this comes from

Chapter 6, Application of Derivatives: Rates of change: angle of elevation.

Question type: Rates in time

How to spot it: click to see

Type 2. Rates in time: how fast is it changing now?

Spot it: words like "per second", "how fast", "at the rate of", and one moment, such as "when the radius is 10 cm".

Here: The car moves away at 20 m/s, so $\frac{dx}{dt} = 20$ ⁠. In (iii)(a) multiply $\frac{d\theta}{dx} = -\frac{5}{25 + x^2}$ by $\frac{dx}{dt}$ at $x = 50$ ⁠; in (iii)(b) $\frac{d\theta}{dt} = \frac{3}{101}$ is given and you solve for $\frac{dx}{dt}$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 5, Continuity and Differentiability.

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