medium difficulty10 min

Rate of Change of Quantities

If y is a function of x, then dy/dx gives the instantaneous rate of change of y with respect to x at a given value of x.

In one line: In related rates, keep the variable alive: formula first, d/dt next, given instant last.

Practice This Concept

Understand it

In rate-of-change questions, the derivative must be read in the context of the variables.
If y = f(x), then dy/dx measures how many units y changes for one unit change in x at that instant.
In related-rate problems, two or more quantities usually depend on time, so each derivative must be taken with respect to t even if the formula first contains radius, side, height, or distance.
The main Class 12 skill is to keep the relation variable-based until differentiation is complete, then substitute the instant mentioned in the question.
A positive rate means the quantity is increasing, a negative rate means it is decreasing, and the unit of the derivative must combine the units of the two quantities, such as cm²/s or m/s.

Example

If the radius r of a circle increases at 2 cm/s, then the area A = π r² changes at dA/dt = 2 π r dr/dt. At r = 5 cm, dA/dt = 2 π(5)(2) = 20 π cm²/s.

Formula and rule

For y = f(x), instantaneous rate = dy/dx.

If y = f(t), rate with respect to time = dy/dt.

Chain rule condition: if y depends on x and x depends on t, then dy/dt = (dy/dx)(dx/dt).

See it worked out

Follow the example line by line, then use the same method on new questions.

Worked example

A spherical balloon has volume V = (4/3)π r³.

1Its radius increases at 0.5 cm/s.
2Find dV/dt when r = 6 cm.
3Differentiate with respect to t: dV/dt = 4 π r² dr/dt.
4Substitute r = 6 and dr/dt = 0.5: dV/dt = 4 π(36)(0.5) = 72 π cm³/s.
5The volume is increasing at 72 π cm³/s.

Method you can reuse

Identify the changing quantities; write the formula connecting them; differentiate with respect to the required variable; substitute the given value at the end; attach correct units and direction such as increasing or decreasing.

Visual tip

Usually no diagram is necessary, but a labelled sketch of the shape can help identify the variables in area, volume, or distance questions.

Check yourself

Answer the question in your head before opening the answer.

Try it yourself

The side x of a square increases at 3 cm/s. What is the rate of change of its area when x = 4 cm?

Think first, then show answer

Area A = x², so dA/dt = 2x dx/dt = 2(4)(3) = 24 cm²/s.

Common wrong answer

Using dV/dr as the final answer when the question asks for dV/dt.

Don't mix this up

Substituting the given numerical value before differentiating, which removes the variable and makes the derivative zero.

Exam tip

Write the relation between quantities first, differentiate both sides with respect to time or the given independent variable, then substitute values with units.

Write it in the exam

Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.

How CBSE asks it

Usually appears as a short-answer or case-based problem involving area, volume, distance, or height changing with time.

1
Enough for 1 mark
If y is a function of x, then dy/dx gives the instantaneous rate of change of y with respect to x at a given value of x.
2
Add for 2 marks
Example: If the radius r of a circle increases at 2 cm/s, then the area A = π r² changes at dA/dt = 2 π r dr/dt. At r = 5 cm, dA/dt = 2 π(5)(2) = 20 π cm²/s.
3
Add for 3 marks
If y = f(x), then dy/dx measures how many units y changes for one unit change in x at that instant.

MCQ Quiz

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