Rates of change: Class 12 practice

What you should know first

• Chain rule

The idea behind rates of change

If y depends on x and x changes with time, dy/dt = (dy/dx)(dx/dt). Typical problems: a growing circle's area, a balloon's volume, or a ladder sliding down a wall.

Step-by-step method

1. Write the formula linking the quantities.
2. Differentiate with respect to t.
3. Substitute the given values.

Worked example 1 (Basic)

The radius of a circle grows at 1 cm/s. How fast is the area increasing when r = 11 cm?

1. A = πr² → dA/dt = 2πr dr/dt
2. = 2π × 11 × 1 = 22π cm²/s
3. Answer: 22π cm²/s

Worked example 2 (Standard)

A spherical balloon's radius increases at 5 cm/s. How fast is its volume increasing when r = 3 cm?

1. V = (4/3)πr³ → dV/dt = 4πr² dr/dt
2. = 180π cm³/s
3. Answer: 180π cm³/s

Worked example 3 (Challenge)

A 15 m ladder leans on a wall. Its foot slides away at 3 m/s. How fast is the top sliding down when the foot is 9 m from the wall?

1. x² + y² = 225 → x dx/dt + y dy/dt = 0
2. y = 12; dy/dt = −9 × 3/12 = -2.25 m/s
3. Answer: 2.25 m/s downwards

Common mistakes and misconceptions

• Substituting values before differentiating
• Forgetting the units per second

Practice questions

1. (Basic) The radius of a circle grows at 2 cm/s. How fast is the area increasing when r = 15 cm?
2. (Basic) The radius of a circle grows at 1 cm/s. How fast is the area increasing when r = 8 cm?
3. (Standard) A spherical balloon's radius increases at 4 cm/s. How fast is its volume increasing when r = 13 cm?
4. (Standard) A spherical balloon's radius increases at 2 cm/s. How fast is its volume increasing when r = 7 cm?
5. (Challenge) A 13 m ladder leans on a wall. Its foot slides away at 2 m/s. How fast is the top sliding down when the foot is 5 m from the wall?
6. (Challenge) A 15 m ladder leans on a wall. Its foot slides away at 1 m/s. How fast is the top sliding down when the foot is 9 m from the wall?

Answer key with full working

• Q1: A = πr² → dA/dt = 2πr dr/dt → = 2π × 15 × 2 = 60π cm²/s → Answer: 60π cm²/s
• Q2: A = πr² → dA/dt = 2πr dr/dt → = 2π × 8 × 1 = 16π cm²/s → Answer: 16π cm²/s
• Q3: V = (4/3)πr³ → dV/dt = 4πr² dr/dt → = 2704π cm³/s → Answer: 2704π cm³/s
• Q4: V = (4/3)πr³ → dV/dt = 4πr² dr/dt → = 392π cm³/s → Answer: 392π cm³/s
• Q5: x² + y² = 169 → x dx/dt + y dy/dt = 0 → y = 12; dy/dt = −5 × 2/12 = -0.83 m/s → Answer: 0.83 m/s downwards
• Q6: x² + y² = 225 → x dx/dt + y dy/dt = 0 → y = 12; dy/dt = −9 × 1/12 = -0.75 m/s → Answer: 0.75 m/s downwards

Syllabus note

This skill is listed under "Application of Derivatives" for Class 12 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

• Application of Derivatives chapter notes (Class 12)
• Increasing and decreasing intervals
• Local maxima and minima
• Tangents and normals
• Derivatives of polynomials and slope of a tangent
• Definite integrals of polynomials
• Evaluating 2×2 and 3×3 determinants
• Matrix multiplication
• All Class 12 practice

Frequently asked questions

Which chapter is rates of change in?

"Application of Derivatives", Class 12 Maths (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.