MathsCBSE Class 12 · Application of Derivatives

Rate of Change of Quantities and Related Rates (Class 12)

Q: What are related rates?

Rates of change of two or more quantities that are connected by an equation. Differentiating the equation with respect to time links their rates, so one known rate gives the other.

Q: Why can't I substitute values before differentiating?

Because the given values hold only at one instant. If you substitute first, the variables become constants and their derivatives become zero, which is wrong.

Q: What is the difference between marginal cost and average cost?

Marginal cost is dC/dx, the extra cost of producing one more unit at the current level. Average cost is C(x)/x, the cost per unit over all units. Only marginal cost is a rate of change.

Q: Which NCERT section covers this topic?

Section 6.2 (Rate of Change of Quantities) of Chapter 6, Application of Derivatives, in NCERT Mathematics Part I for Class 12.

Q: Are approximations and tangents part of this chapter for 2026-27?

No. Tangents and normals and approximations were dropped. The chapter covers rate of change, increasing and decreasing functions, and maxima and minima.

Updated 15 October 2026 · Fact-checked

The rate of change of y with respect to x is dy/dx; with respect to time it is dy/dt. When two changing quantities are linked by an equation, differentiate the equation with respect to time and use the chain rule, dy/dt = (dy/dx)(dx/dt). Substitute the given instant's values only after differentiating.

As the foot slides out (dx/dt > 0) the top slides down (dy/dt < 0); differentiate x² + y² = L² first, then substitute.

Illustration drawn for QuestionBuddy.

Understand Rate of Change of Quantities

The derivative $\frac{d y}{d x}$ measures how fast $y$ changes as $x$ changes. If $A = π r^{2}$ is the area of a circle, $\frac{d A}{d r} = 2 π r$ is the rate of change of area with respect to radius; at $r = 5$ cm it is $10 π$ cm² per cm. When the change happens over time, we use $\frac{d}{d t}$ , and a negative rate means the quantity is decreasing.

Related rates. Often two or more quantities change with time and are tied together by a geometric or physical relation: the sides of a sliding ladder ( $x^{2} + y^{2} = L^{2}$ ), the radius and volume of a balloon ( $V = \frac{4}{3} π r^{3}$ ), the water depth and volume in a cone. Knowing one rate, we find another by the chain rule:
$\frac{d V}{d t} = \frac{d V}{d r} \cdot \frac{d r}{d t}$ .

The method that never fails.

1. Draw a diagram and name every changing quantity with a letter; write constants as numbers.
2. Write what is given and what is asked as derivatives with respect to $t$ (with signs: decreasing means negative).
3. Find an equation linking the quantities, valid at every instant. If there are extra variables, remove them using fixed ratios (similar triangles in a cone or a shadow).
4. Differentiate the whole equation with respect to $t$ .
5. Only now substitute the values at the given instant, and solve for the unknown rate. Give units.

Substituting before differentiating is the classic mistake: if you put $r = 10$ into $V = \frac{4}{3} π r^{3}$ first, $V$ becomes a constant and its derivative is zero.

Marginal cost and revenue. In economics, if $C (x)$ is the total cost of producing $x$ units, the marginal cost $M C = \frac{d C}{d x}$ is the approximate cost of producing one more unit. Similarly, marginal revenue $M R = \frac{d R}{d x}$ for revenue $R (x)$ (often $R =$ price $\times$ quantity). These are rates of change with respect to quantity, not time.

The 2026-27 syllabus keeps rate of change, increasing and decreasing functions, and maxima and minima in this chapter; tangents, normals and approximations were dropped. Related-rates questions are frequent as 2- and 3-mark questions and as 4-mark case studies (NCERT Section 6.2).

Rate of Change of Quantities: formulas to remember

Chain rule for rates: $\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$; Links one rate to another.
Circle: $A = π r^{2}, \frac{d A}{d t} = 2 π r \frac{d r}{d t}; C = 2 π r, \frac{d C}{d t} = 2 π \frac{d r}{d t}$; Ripples, oil spills, wheels.
Sphere: $V = \frac{4}{3} π r^{3}, \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}; S = 4 π r^{2}, \frac{d S}{d t} = 8 π r \frac{d r}{d t}$; Balloons, bubbles, melting ice balls.
Cone: $V = \frac{1}{3} π r^{2} h$; Use the fixed ratio $\frac{r}{h}$ to remove one variable first.
Cube: $V = a^{3}, \frac{d V}{d t} = 3 a^{2} \frac{d a}{d t}; S = 6 a^{2}, \frac{d S}{d t} = 12 a \frac{d a}{d t}$; Expanding or melting cubes.
Marginal cost and revenue: $M C = \frac{d C}{d x}, M R = \frac{d R}{d x}$; Rate of change per extra unit produced or sold.

How to solve Rate of Change of Quantities questions

Use this five-step routine for every related-rates question.

1Draw and label: name each changing length, area or volume with a letter.
2Write the given rate and the required rate as $\frac{d ( \cdot )}{d t}$ , with a minus sign for anything decreasing.
3Write a relation between the variables that holds at every moment; use similar triangles or a fixed ratio to remove extra variables.
4Differentiate both sides with respect to $t$ using the chain rule.
5Substitute the values at the given instant (find any missing one from the relation), solve for the unknown rate, and state units and whether the quantity is increasing or decreasing.

Rate of Change of Quantities in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Rate of change of the radius of a melting sphere

Suppose a spherical ball of ice in a drink melts so that its volume decreases at 2 cm³ per minute, keeping its spherical shape.

How fast is the radius decreasing when the radius is 5 cm?

Try it first, then show the solutionSolution

$V = \frac{4}{3} π r^{3}$ and $\frac{d V}{d t} = - 2$ cm³/min (decreasing).
Differentiate: $\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ .
At $r = 5$ : $- 2 = 4 π (25) \frac{d r}{d t}$ , so $\frac{d r}{d t} = - \frac{2}{100 π} = - \frac{1}{50 π}$ cm/min.
$\frac{1}{50 π} \approx 0.0064$ cm/min.

Answer: The radius decreases at $\frac{1}{50 π} \approx 0.0064$ cm per minute.

Memory hook: Volume falls fast, radius falls slowly: divide by $4 π r^{2}$ .

Marginal revenue

Imagine a phone-cover seller who finds that she can sell $x$ covers a week if she charges $(100 - \frac{x}{2})$ rupees each.

Write the weekly revenue $R (x)$ and find the marginal revenue when she sells 60 covers. Interpret it.

Try it first, then show the solutionSolution

$R (x) = x (100 - \frac{x}{2}) = 100 x - \frac{x ^{2}}{2}$ .
$M R = \frac{d R}{d x} = 100 - x$ .
At $x = 60$ : $M R = 40$ .
Selling the 61st cover (at the lower price needed) adds about Rs 40 to revenue.

Answer: $R (x) = 100 x - \frac{x ^{2}}{2}$ ; $M R = 40$ rupees per cover at $x = 60$ .

Memory hook: Marginal means 'one more': the derivative per extra unit.

Tricks to remember Rate of Change of Quantities

Short memory aids. Say each one aloud once, then use it on a question the same day.

Differentiate first, substitute last
Values at an instant are photographs; the relation is the film. Differentiate the film, then look at the photograph.
Use it when: Avoiding the zero-derivative trap.
Minus for shrinking
Melting, leaking, sliding down: give the rate a minus sign. A negative answer then means 'decreasing'.
Use it when: Getting signs right.
Similar triangles kill extra letters
Cones and shadows have fixed ratios; use them to write everything in one variable before differentiating.
Use it when: Cone and lamp-post questions.
Marginal = derivative
Marginal cost, revenue or profit is just $\frac{d}{d x}$ of the total.
Use it when: Economics-flavoured MCQs.

Rate of Change of Quantities: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Area of an expanding circle · 2 marks · Very short answerBoard level

The radius of a circle increases at 0.5 cm/s. Find the rate of increase of its area when the radius is 8 cm.

Try it first, then show the solutionSolution

$A = π r^{2} \Rightarrow \frac{d A}{d t} = 2 π r \frac{d r}{d t}$ . (1 mark)
At $r = 8$ : $\frac{d A}{d t} = 2 π (8) (0.5) = 8 π$ cm²/s. (1 mark)

Answer: $8 π$ cm²/s

Example 2: Marginal revenue and marginal cost · 2 marks · Very short answerBoard level

A firm's revenue from selling $x$ units is $R (x) = 40 x - 0.1 x^{2}$ and its cost of producing $x$ units is $C (x) = 500 + 12 x + 0.03 x^{2}$ (in rupees). Find the marginal revenue at $x = 100$ and the marginal cost at $x = 50$ .

Try it first, then show the solutionSolution

$M R = R^{'} (x) = 40 - 0.2 x$ ; at $x = 100$ : $40 - 20 = 20$ . (1 mark)
$M C = C^{'} (x) = 12 + 0.06 x$ ; at $x = 50$ : $12 + 3 = 15$ . (1 mark)

Answer: MR = Rs 20 per unit; MC = Rs 15 per unit.

Example 3: A sliding ladder · 3 marks · Short answerHarder board level

A 5 m ladder leans against a wall. Its foot is pulled away from the wall at 0.3 m/s. How fast is the top sliding down when the foot is 3 m from the wall?

Try it first, then show the solutionSolution

Let the foot be $x$ m from the wall and the top $y$ m high: $x^{2} + y^{2} = 25$ , $\frac{d x}{d t} = 0.3$ . (1 mark)
Differentiate: $2 x \frac{d x}{d t} + 2 y \frac{d y}{d t} = 0$ . (1 mark)
At $x = 3$ : $y = 4$ , so $3 (0.3) + 4 \frac{d y}{d t} = 0 \Rightarrow \frac{d y}{d t} = - 0.225$ m/s. (1 mark)

Answer: The top slides down at 0.225 m/s.

Example 4: Filling an inverted cone · 3 marks · Short answerHarder board level

Water is poured at 4 cm³/s into an inverted conical funnel of base radius 5 cm and height 10 cm. How fast is the water level rising when the water is 8 cm deep?

Try it first, then show the solutionSolution

By similar triangles, at depth $h$ the water surface radius is $r = \frac{5}{10} h = \frac{h}{2}$ . (1 mark)
$V = \frac{1}{3} π r^{2} h = \frac{π h ^{3}}{12}$ , so $\frac{d V}{d t} = \frac{π h ^{2}}{4} \frac{d h}{d t}$ . (1 mark)
At $h = 8$ : $4 = \frac{64 π}{4} \frac{d h}{d t} = 16 π \frac{d h}{d t}$ , so $\frac{d h}{d t} = \frac{1}{4 π}$ cm/s $\approx 0.08$ cm/s. (1 mark)

Answer: $\frac{1}{4 π}$ cm/s

Example 5: A lengthening shadow · 3 marks · Short answerHOTS / competency

A boy 1.6 m tall walks away from a 4 m high lamp-post at 5 km/h. Find the rate at which his shadow lengthens, and the speed of the tip of his shadow.

Try it first, then show the solutionSolution

Let the boy be $x$ from the post and his shadow length $s$ . Similar triangles: $\frac{4}{x + s} = \frac{1.6}{s}$ , so $4 s = 1.6 x + 1.6 s$ and $s = \frac{2 x}{3}$ . (1 mark)
$\frac{d s}{d t} = \frac{2}{3} \frac{d x}{d t} = \frac{2}{3} (5) = \frac{10}{3}$ km/h. (1 mark)
The tip is at distance $x + s$ , so it moves at $5 + \frac{10}{3} = \frac{25}{3}$ km/h. (1 mark)

Answer: Shadow lengthens at $\frac{10}{3}$ km/h; the tip moves at $\frac{25}{3}$ km/h.

Example 6: Case study: inflating a balloon · 4 marks · Case-basedHOTS / competency

Suppose a decorator pumps air into a spherical balloon at 400 cm³/s.
(i) Write $\frac{d V}{d t}$ in terms of $r$ and $\frac{d r}{d t}$ . (1 mark)
(ii) Find the rate of increase of the radius when $r = 10$ cm. (1 mark)
(iii) Find the rate of increase of the surface area at that moment. (1 mark)
(iv) Will the radius grow faster or slower as the balloon gets bigger? Why? (1 mark)

Try it first, then show the solutionSolution

(i) $V = \frac{4}{3} π r^{3} \Rightarrow \frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ . (1 mark)
(ii) $400 = 4 π (100) \frac{d r}{d t} \Rightarrow \frac{d r}{d t} = \frac{1}{π}$ cm/s. (1 mark)
(iii) $\frac{d S}{d t} = 8 π r \frac{d r}{d t} = 8 π (10) (\frac{1}{π}) = 80$ cm²/s. (1 mark)
(iv) $\frac{d r}{d t} = \frac{400}{4 π r ^{2}}$ decreases as $r$ increases, so the radius grows more slowly. (1 mark)

Answer: (i) $4 π r^{2} \frac{d r}{d t}$ (ii) $\frac{1}{π}$ cm/s (iii) 80 cm²/s (iv) slower.

Where students lose marks in Rate of Change of Quantities

The slips examiners see most often, and what to write instead.

Mistake: Substituting before differentiating
Putting the instant's value in first turns a variable into a constant.
Wrong working
$r = 10$ : $V = \frac{4000 π}{3}$ , so $\frac{d V}{d t} = 0$ .
Right working
$\frac{d V}{d t} = 4 π r^{2} \frac{d r}{d t}$ , then put $r = 10$ .
Do this: Differentiate the general relation first; substitute last.
Usually costs: The whole question.
Mistake: Ignoring the sign of a decreasing rate
A leak or a melt is a negative rate.
Wrong working
Volume decreases at 2 cm³/min: $\frac{d V}{d t} = 2$ .
Right working
$\frac{d V}{d t} = - 2$ , giving $\frac{d r}{d t} < 0$ (radius decreasing).
Do this: Write the sign with every given rate and interpret the sign of the answer.
Usually costs: Half to one mark.
Mistake: Keeping two variables in a cone
Without the fixed ratio, $V = \frac{1}{3} π r^{2} h$ has two unknown rates.
Wrong working
$\frac{d V}{d t} = \frac{1}{3} π r^{2} \frac{d h}{d t}$ (treating r as constant).
Right working
Use $r = \frac{h}{2}$ : $V = \frac{π h ^{3}}{12}$ , then differentiate.
Do this: Use similar triangles to express r in terms of h first.
Usually costs: 2 marks.
Mistake: Missing units in the answer
Rates need units such as cm/s or cm²/s.
Wrong working
$\frac{d A}{d t} = 8 π$ .
Right working
$\frac{d A}{d t} = 8 π$ cm²/s.
Do this: Write units with every final rate.
Usually costs: Sometimes half a mark.

Exam tips

Related-rates questions appeared in the 2023, 2024, 2025 and 2026 papers (including 4-mark case studies); draw a labelled figure even if not asked.
List 'given' and 'to find' as derivatives in the first line; this earns credit and organises the work.
For marginal cost or revenue, differentiate the total function and substitute; no time variable is involved.
State whether the quantity is increasing or decreasing in the final sentence.
Keep answers exact (like $\frac{1}{4 π}$ ) unless a decimal is requested.

Practice questions on Rate of Change of Quantities

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The side of a square increases at 3 cm/s. The rate of increase of its area when the side is 10 cm is:
(a) 30 cm²/s
(b) 60 cm²/s
(c) 100 cm²/s
(d) 6 cm²/s
Hint
$\frac{d A}{d t} = 2 a \frac{d a}{d t}$ .
Answer
(b) $2 (10) (3) = 60$ cm²/s.
Q2 · 2 marks · Very short answerBoard level
The volume of a cube increases at 12 cm³/s. How fast is its surface area increasing when an edge is 4 cm long?
Hint
First find $\frac{d a}{d t}$ from $\frac{d V}{d t} = 3 a^{2} \frac{d a}{d t}$ .
Answer
$12 = 3 (16) \frac{d a}{d t} \Rightarrow \frac{d a}{d t} = \frac{1}{4}$ cm/s. $\frac{d S}{d t} = 12 a \frac{d a}{d t} = 12 (4) (\frac{1}{4}) = 12$ cm²/s.
Q3 · 2 marks · Very short answerBoard level
A particle moves along the curve $y = x^{2} + 2 x$ . Find the point at which its y-coordinate changes twice as fast as its x-coordinate.
Hint
$\frac{d y}{d t} = (2 x + 2) \frac{d x}{d t}$ ; set this equal to $2 \frac{d x}{d t}$ .
Answer
$(2 x + 2) \frac{d x}{d t} = 2 \frac{d x}{d t} \Rightarrow x = 0$ , so the point is $(0, 0)$ .
Q4 · 3 marks · Short answerHarder board level
A kite flies at a constant height of 60 m and moves horizontally away from the girl flying it at 4 m/s. How fast is the string being let out when 100 m of string is out (string straight)?
Hint
$s^{2} = x^{2} + 6 0^{2}$ ; find $x$ when $s = 100$ .
Answer
$s^{2} = x^{2} + 3600 \Rightarrow s \frac{d s}{d t} = x \frac{d x}{d t}$ . When $s = 100$ , $x = 80$ , so $\frac{d s}{d t} = \frac{80 \times 4}{100} = 3.2$ m/s.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If the radius of a circle increases at a constant rate, its area also increases at a constant rate.
Reason (R): $\frac{d A}{d t} = 2 π r \frac{d r}{d t}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Does $2 π r \frac{d r}{d t}$ depend on r?
Answer
(d). R is true, and it shows $\frac{d A}{d t}$ grows with $r$ , so the area increases faster and faster; A is false.
Q6 · 2 marks · Very short answerHarder board level
A company's profit from selling $x$ units is $P (x) = - 0.5 x^{2} + 60 x - 400$ rupees. Find the marginal profit at $x = 40$ , and the number of units at which the marginal profit is zero.
Hint
Marginal profit $= P^{'} (x)$ .
Answer
$P^{'} (x) = - x + 60$ ; at $x = 40$ it is Rs 20 per unit. $P^{'} (x) = 0$ when $x = 60$ .
Q7 · 3 marks · Short answerHOTS / competency
Two cars leave a crossing at the same time, one going north at 40 km/h and the other east at 30 km/h. How fast is the distance between them increasing after 2 hours? Show that this rate does not change with time.
Hint
$D^{2} = x^{2} + y^{2}$ with $y = 40 t$ (north) and $x = 30 t$ (east).
Answer
After 2 h, $y = 80$ , $x = 60$ , $D = 100$ . $D \frac{d D}{d t} = x \frac{d x}{d t} + y \frac{d y}{d t}$ (with $\frac{d y}{d t} = 40$ north, $\frac{d x}{d t} = 30$ east) gives $\frac{d D}{d t} = \frac{60 ( 30 ) + 80 ( 40 )}{100} = 50$ km/h. In general $D = (40 t)^{2} + (30 t)^{2} = 50 t$ , so $\frac{d D}{d t} = 50$ always.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write $\frac{d V}{d t}$ for a sphere.Show answer
$4 π r^{2} \frac{d r}{d t}$ .
2.What does a negative rate of change mean?Show answer
The quantity is decreasing.
3.Define marginal cost.Show answer
$\frac{d C}{d x}$ , the rate of change of total cost with respect to output.
4.For a ladder of length L, what relation links x and y?Show answer
$x^{2} + y^{2} = L^{2}$ , so $x \frac{d x}{d t} + y \frac{d y}{d t} = 0$ .
5.In a cone with $r = \frac{h}{2}$ , what is V in terms of h?Show answer
$\frac{π h ^{3}}{12}$ .
6.When should you substitute the given values?Show answer
After differentiating.

Rate of Change of Quantities: frequently asked questions

What are related rates?

Rates of change of two or more quantities that are connected by an equation. Differentiating the equation with respect to time links their rates, so one known rate gives the other.

Why can't I substitute values before differentiating?

Because the given values hold only at one instant. If you substitute first, the variables become constants and their derivatives become zero, which is wrong.

What is the difference between marginal cost and average cost?

Marginal cost is dC/dx, the extra cost of producing one more unit at the current level. Average cost is C(x)/x, the cost per unit over all units. Only marginal cost is a rate of change.

Which NCERT section covers this topic?

Section 6.2 (Rate of Change of Quantities) of Chapter 6, Application of Derivatives, in NCERT Mathematics Part I for Class 12.

Are approximations and tangents part of this chapter for 2026-27?

No. Tangents and normals and approximations were dropped. The chapter covers rate of change, increasing and decreasing functions, and maxima and minima.