Sample 2026-27 Class 12 Maths Q27 Solution: Rate at which a surveillance drone rises

The question

The circular area of vision on the ground by a surveillance drone fitted with camera increases at the rate of 44 m² per metre as it gains vertical height. The radius of the circle on the ground increases at the rate of 6 m/s. Find the rate at which the height of the drone increases with time, at the instant when the radius of the circular area on the ground is 14 m. $\left(\text{Use } \pi = \dfrac{22}{7}\right)$

Exam answer

Let A be the area of the circular region covered by the drone, r its radius and h the height of the drone at any time t.

Given $\frac{dA}{dh} = 44$ m² per metre and $\frac{dr}{dt} = 6$ m/s. To find $\frac{dh}{dt}$ when $r = 14$ m.

Why?

Read each rate carefully. "44 m² per metre as it gains height" is a rate with respect to height, $\frac{dA}{dh}$ ⁠. "6 m/s" is a rate with respect to time, $\frac{dr}{dt}$ ⁠. The question asks for a rate with respect to time, $\frac{dh}{dt}$ ⁠.

A = \pi r^2 \Rightarrow \frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}

Why?

The area of a circle is $\pi r^2$ ⁠. Differentiate with respect to $t$ ⁠. Since $r$ changes with time, the chain rule gives

\frac{d}{dt}(r^2) = 2r\,\frac{dr}{dt}

\Rightarrow \frac{dA}{dh}\cdot\frac{dh}{dt} = 2\pi r\,\frac{dr}{dt}

Why?

The area depends on the height, and the height depends on time. So, by the chain rule,

\frac{dA}{dt} = \frac{dA}{dh}\cdot\frac{dh}{dt}

This brings in the given $\frac{dA}{dh}$ and the unknown $\frac{dh}{dt}$ ⁠.

At $r = 14$ ⁠,

\begin{aligned} 44\,\frac{dh}{dt} &= 2 \times \frac{22}{7} \times 14 \times 6 \\ &= 528 \\ \Rightarrow \frac{dh}{dt} &= \frac{528}{44} = 12 \end{aligned}

Why?

Use $\pi = \frac{22}{7}$ as the question says. Then $\frac{22}{7} \times 14 = 44$ ⁠, so the right side is $2 \times 44 \times 6 = 528$ ⁠.

\begin{aligned} \frac{22}{7} \times 14 &= 22 \times \frac{14}{7} \\ &= 22 \times 2 = 44 \\[4pt] 2 \times 44 &= 88 \\ 88 \times 6 &= 528 \end{aligned}

Divide by 44: $528 \div 44 = 12$ ⁠. (Check: $44 \times 12 = 528$ ⁠.) The units are metres per second, the units of $\frac{dh}{dt}$ ⁠.

$\therefore$ The height of the drone increases at the rate of 12 m/s at that instant.

Where this comes from

Chapter 6, Application of Derivatives: Rate of change: related rates.

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 radius grows at 6 m/s, so $\frac{dr}{dt} = 6$ ⁠, and the moment is $r = 14$ ⁠. The 44 m² per metre is $\frac{dA}{dh}$ ⁠: differentiate $A = \pi r^2$ with respect to $t$ and write $\frac{dA}{dt} = \frac{dA}{dh}\cdot\frac{dh}{dt}$ ⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Q27: Rate at which a surveillance drone rises