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 $\dfrac{d\theta}{dx}$. (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.

> [!TLDR] > This problem involves related rates, where we first establish a trigonometric relationship between the angle of elevation and the car's distance from the pole. We then differentiate this relationship to find the rate of change of the angle with respect to distance, and finally use the chain rule to determine the rate of change of the angle with respect to time. The rate of change of the angle of elevation with respect to time when the car is 50 m away is $\boxed{-\frac{4}{101} \text{ rad/s}}$. The problem describes a scenario where a speed camera on a pole observes a car moving away. We need to analyze how the angle of elevation from the car to the camera changes with the car's position and, ultimately, with time. This is a classic application of related rates, where quantities that depend on each other also change over time. The core idea is to: 1. Establish a geometric relationship between the angle of elevation ($\theta$), the height of the camera ($h$), and the car's distance from the pole ($x$). 2. Differentiate this relationship with respect to the relevant variable (either $x$ or $t$) to find the rates of change. 3. Use the chain rule to connect these rates when necessary. Let's break down the problem into the requested parts. ### 1. Visualizing the Setup and Establishing the Relationship Imagine a right-angled triangle formed by: * The pole (vertical side) with the camera at its top. * The ground (horizontal side) from the base of the pole to the car. * The line of sight from the car to the camera (hypotenuse). Let: * $h$ be the height of the camera on the pole. Given $h = 5$ m. * $x$ be the horizontal distance of the car from the base of the pole. * $\theta$ be the angle of elevation of the camera from the car. From the diagram, we can see that the height $h$ is opposite to the angle $\theta$, and the distance $x$ is adjacent to $\theta$. > [!TIP] > Always draw a clear diagram for related rates problems. It helps in correc

The traffic police has installed Over Speed Violation Detection (OSVD) system at various locations in a city.

Question

x

(i) Express

\theta

in terms of the height of the camera installed on the pole and

x

(ii) Find

\dfrac{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.

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Concept understanding — Related Rates

Related Rates

The idea: quantities that change together

Many real situations involve two or more quantities that all vary with time, linked by a fixed relationship. Inflate a balloon and its radius and volume both grow; slide a ladder down a wall and the top's height and the foot's distance both change. A related-rates problem gives you the rate at which one quantity is changing and asks for the rate of another, at some instant.

The key insight: if the quantities are tied together by an equation, then their rates are tied together too. We uncover that link by differentiating the equation with respect to time $t$ .

The core mechanism: differentiate with respect to time

Every variable is a function of $t$ , so differentiating brings in the chain rule — each variable's derivative picks up a factor of its own rate. For example, if the volume of a sphere is $V=\tfrac{4}{3}\pi r^{3}$ , then differentiating both sides with respect to $t$ gives

\frac{dV}{dt} = 4\pi r^{2}\,\frac{dr}{dt}.

This single equation connects the rate the volume grows, $\dfrac{dV}{dt}$ , to the rate the radius grows, $\dfrac{dr}{dt}$ . Knowing one (and the current $r$ ) gives the other.

The standard procedure

t

Worked example

Air is pumped into a spherical balloon at $\dfrac{dV}{dt}=100\ \text{cm}^3/\text{s}$ . How fast is the radius increasing when $r=5\ \text{cm}$ ?

From $\dfrac{dV}{dt}=4\pi r^{2}\dfrac{dr}{dt}$ , substitute $\dfrac{dV}{dt}=100$ and $r=5$ :

$100 = 4\pi(5)^2\,\frac{dr}{dt} = 100\pi\,\frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{\pi}\ \text{cm/s}.$ …

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This problem involves related rates, where we first establish a trigonometric relationship between the angle of elevation and the car's distance from the pole. We then differentiate this relationship to find the rate of change of the angle with respect to distance, and finally use the chain rule to determine the rate of change of the angle with respect to time. The rate of change of the angle of elevation with respect to time when the car is 50 m away is $\boxed{-\frac{4}{101} \text{ rad/s}}$ .

The problem describes a scenario where a speed camera on a pole observes a car moving away. We need to analyze how the angle of elevation from the car to the camera changes with the car's position and, ultimately, with time. This is a classic application of related rates, where quantities that depend on each other also change over time.

The core idea is to:

Establish a geometric relationship between the angle of elevation ( $\theta$ ), the height of the camera ( $h$ ), and the car's distance from the pole ( $x$ ).
Differentiate this relationship with respect to the relevant variable (either $x$ or $t$ ) to find the rates of change.
Use the chain rule to connect these rates when necessary.

Let's break down the problem into the requested parts.

1. Visualizing the Setup and Establishing the Relationship

Imagine a right-angled triangle formed by:

The pole (vertical side) with the camera at its top.
The ground (horizontal side) from the base of the pole to the car.
The line of sight from the car to the camera (hypotenuse).

Let:

$h$ be the height of the camera on the pole. Given $h = 5$ m.
$x$ be the horizontal distance of the car from the base of the pole.
$\theta$ be the angle of elevation of the camera from the car.

From the diagram, we can see that the height $h$ is opposite to the angle $\theta$ , and the distance $x$ is adjacent to $\theta$ .

Tip

Always draw a clear diagram for related rates problems. It helps in correctly identifying the variables and the trigonometric or geometric relationships.

(i) Express $\theta$ in terms of the height of the camera installed on the pole and $x$ .

Using the tangent function in the right-angled triangle:

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

Given that the height of the camera $h = 5$ m, we substitute this value:

\tan \theta = \frac{5}{x}

To express $\theta$ explicitly, we take the inverse tangent (arctangent) of both sides:

\theta = \arctan\left(\frac{5}{x}\right)

This is the required expression for $\theta$ in terms of $x$ .

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

To find $\frac{d\theta}{dx}$ , we differentiate the expression for $\theta$ with respect to $x$ .

We have $\theta = \arctan\left(\frac{5}{x}\right)$ .

\arctan(u)

Here, $u = \frac{5}{x}$ .

First, find $\frac{du}{dx}$ :

u = 5x^{-1}

\frac{du}{dx} = 5(-1)x^{-2} = -\frac{5}{x^2}

Now, apply the chain rule for $\arctan(u)$ :

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

Simplify the expression:

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

Combine the terms in the denominator:

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

Invert and multiply the first term:

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

The $x^2$ terms cancel out:

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

This is the rate of change of the angle of elevation with respect to the distance $x$ . The negative sign indicates that as $x$ increases (car moves away), the angle $\theta$ decreases.

(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. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 15 on this concept.

CBSE 2026Set ANNUAL1 mark
$Q. The edge of a variable cube is increasing at the rate of 3 cm/s. The volume of the cube is increasing at the rate of __________ while the edge is 10 cm long.$
›Reveal solutionSolution

Use $V=e^3$ and the chain rule $dV/dt = 3e^2\,de/dt$ .

Let $e$ be the edge; $V=e^3$ , so $\dfrac{dV}{dt}=3e^2\dfrac{de}{dt}$ .

Given $\dfrac{de}{dt}=3$ cm/s and $e=10$ cm:

…

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CBSE 2026Set ANNUAL1 mark
$1/2$
›Reveal solutionSolution

Use $V=\dfrac43\pi r^3$ and $\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$ .

Given $\dfrac{dr}{dt}=\dfrac12$ cm/s, at $r=1$ cm:

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Radius of a circle is increasing at the rate of 1/π m/s. Rate of change of its circumference is:$
(a) 4π m/s
(b) 2 m/s
(c) 2π m/s
(d) 4 m/s
›Reveal solutionSolution

Since $C = 2\pi r$ , differentiating both sides w.r.t. time gives $\frac{dC}{dt} = 2\pi \frac{dr}{dt}$ directly.

The circumference of a circle of radius $r$ is $C = 2\pi r$ .

Differentiating with respect to time $t$ :

$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$

Given $\frac{dr}{dt} = \frac{1}{\pi}$ m/s, substitute: …

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CBSE 2025Set 65/2/11 markMCQ
$100\pi$
›Reveal solutionSolution
The volume of a cylinder is $V = \pi r^2 h$ . Since the radius is constant, the rate of change of volume with respect to time is $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$ . Given $\frac{dV}{dt} = 100\pi$ cm³/s and $r = 10$ cm, solving gives $\frac{dh}{dt} = 1$ cm/s. The correct option is (C).

This is a classic Related Rates problem. The core idea is that when two quantities are linked by a geometric formula (here, volume and height of a cylinder), their rates of change with respect to time are also linked. You differentiate the relationship with respect to time, plug in what you know, and solve for the unknown rate.

The key insight: the tank’s radius is fixed at 10 cm. So as sugar pours in, the height increases, but the cross-sectional area stays the same. That means the volume increases at a constant rate per unit height — specifically, each 1 cm rise in height adds $\pi (10)^2 = 100\pi$ cm³ of volume. Since sugar is being added at exactly $100\pi$ cm³/s, the height must be rising at 1 cm/s.

Let’s work it out formally.

Write the relationship between volume and height. For a cylinder, $V = \pi r^2 h$ . Here $r = 10$ cm, so

$V = \pi (10)^2 h = 100\pi h.$

Differentiate both sides with respect to time $t$ .

Since $r$ is constant, $\frac{dV}{dt} = 100\pi \frac{dh}{dt}$ .

This is the related rates equation — it tells us how fast the volume changes in terms of how fast the height changes.

Substitute the given rate.

We know $\frac{dV}{dt} = 100\pi$ cm³/s. So:

$100\pi = 100\pi \frac{dh}{dt}.$

Solve for $\frac{dh}{dt}$ . Divide both sides by $100\pi$ : …
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CBSE 2025Set ANNUAL1 markMCQ
$Q. Radius of a circle is increasing at the rate of 2 m/s. Rate of change of its circumference is:$
(a) 4π m/s
(b) 2 m/s
(c) 2π m/s
(d) 4 m/s
›Reveal solutionSolution

Differentiate the circumference formula $C=2\pi r$ with respect to time and plug in $\dfrac{dr}{dt}$ .

Given $\dfrac{dr}{dt} = 2$ m/s. Circumference $C = 2\pi r$ .

Differentiating both sides with respect to time $t$ : …

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CBSE 2024Set ANNUAL1 mark
$Q. The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.$
›Reveal solutionSolution

Use related rates: differentiate $A=\pi r^2$ w.r.t. time and substitute the given $\dfrac{dr}{dt}$ and $r$ .

Given $\dfrac{dr}{dt} = 3$ cm/s, find $\dfrac{dA}{dt}$ at $r = 10$ cm.

$A = \pi r^2 \Rightarrow \dfrac{dA}{dt} = 2\pi r \dfrac{dr}{dt}$

…

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CBSE 2024Set ANNUAL1 mark
$0.7$
›Reveal solutionSolution

Differentiate the circumference formula $C=2\pi r$ with respect to time.

The circumference of a circle of radius $r$ is

$C = 2\pi r.$

Given $\dfrac{dr}{dt} = 0.7$ cm/s. Differentiating with respect to $t$ : …

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CBSE 2023Set ANNUAL1 markMCQ
$0.3$
(a) $0.4\pi$ cm/sec
(b) $0.6\pi$ cm/sec
(c) $0.8\pi$ cm/sec
(d) none of these
›Reveal solutionSolution

This is a related-rates problem: differentiate the perimeter formula w.r.t. time.

Perimeter (circumference) $P = 2\pi r$ . Differentiating w.r.t. time $t$ : $\dfrac{dP}{dt} = 2\pi\dfrac{dr}{dt}$ .

…

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CBSE 2023Set ANNUAL1 mark
$3$
›Reveal solutionSolution

Differentiate $A=\pi r^2$ with respect to time and substitute $r=10$ , $dr/dt=3$ .

Area of circle: $A=\pi r^2$ . Differentiating both sides with respect to $t$ :

$\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}$

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. The radius of a circle is increasing at the rate of 0.7 cm/s. The rate of increase of its circumference is -$
(a) 7.1 cm/s
(b) 4.0 cm/s
(c) 3.9 cm/s
(d) 4.4 cm/s
›Reveal solutionSolution

Differentiate $C = 2\pi r$ with respect to time and substitute the given rate of change of the radius.

Circumference $C = 2\pi r$ . Differentiating with respect to time $t$ :

$\frac{dC}{dt} = 2\pi\frac{dr}{dt}.$

Given $\dfrac{dr}{dt} = 0.7$ cm/s: …

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CBSE 2018Set ANNUAL1 markMCQ
$cm^3$
(a) 5 $cm^2$ /sec.
(b) 10 $cm^2$ /sec.
(c) 15 $cm^2$ /sec.
(d) 20 $cm^2$ /sec.
›Reveal solutionSolution

related rates: relate dV/dt to dr/dt, then dS/dt to dr/dt

Volume $V=\dfrac43\pi r^3\Rightarrow \dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}$ .

At $r=5$ : $25=4\pi(25)\dfrac{dr}{dt}\implies \dfrac{dr}{dt}=\dfrac{25}{100\pi}=\dfrac{1}{4\pi}$ .

Surface area $S=4\pi r^2\Rightarrow \dfrac{dS}{dt}=8\pi r\dfrac{dr}{dt}$ .

…

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CBSE 2018Set ANNUAL1 markMCQ
$x$
(a) $\dfrac{\pi}{3}$
(b) $\dfrac{\pi}{2}$
(c) $\pi$
(d) $\dfrac{3\pi}{2}$
›Reveal solutionSolution

Translate "x increases twice as fast as sin x" into $dx/dt = 2\,d(\sin x)/dt$ and solve for x.

"x increases twice as fast as its sine" means $\dfrac{dx}{dt} = 2\dfrac{d(\sin x)}{dt} = 2\cos x\dfrac{dx}{dt}$ …

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1. Visualizing the Setup and Establishing the Relationship

(i) Express θ\thetaθ in terms of the height of the camera installed on the pole and xxx.

(ii) Find dθdx\dfrac{d\theta}{dx}dxdθ​.

(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. …

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(i) Express $\theta$ in terms of the height of the camera installed on the pole and $x$ .

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

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