Question 81 markMCQSection A

The rate of change of the area of a circle with respect to its radius r (in cm2/s\text{cm}^2/\text{s}), when r=6r = 6 cm, is :

Calculus (Applications of Derivatives)
Rate of Change — application of derivatives

Options

(A)10π10\pi
(B)12π12\pi
(C)8π8\pi
(D)11π11\pi

Official answer

The correct option is B) 12π12\pi.


Working:


  • Area of a circle: A=πr2A = \pi r^2.
  • Rate of change with respect to r: dAdr=2πr\frac{dA}{dr} = 2\pi r.
  • At r=6r = 6 cm: dAdr=2π(6)=12π cm2/cm\frac{dA}{dr} = 2\pi(6) = 12\pi \text{ cm}^2/\text{cm}.
Answer keywords:rate of changearea of circleA = pi r^2dA/dr = 2 pi r12 piapplication of derivativesr = 6

Marking scheme

  1. 1.1 mark: dAdr=2πr\frac{dA}{dr} = 2\pi r and value 12π12\pi at r=6r = 6 (option B).

Hint

Differentiate A=πr2A = \pi r^2 to get dAdr=2πr\frac{dA}{dr} = 2\pi r, then substitute r=6r = 6.

Quick oral answer

Since A=πr2A = \pi r^2, dAdr=2πr\frac{dA}{dr} = 2\pi r, and at r=6r = 6 this equals 12π cm212\pi \text{ cm}^2 per cm.

Common mistakes

  1. 1.Using circumference 2πr2\pi r as the area formula or vice versa.
  2. 2.Substituting a wrong radius value (5 or 4) yielding 10π10\pi or 8π8\pi.
  3. 3.Forgetting the factor 2 from differentiating r2r^2, giving πr=6π\pi r = 6\pi.

Analysis and explanation

This tests the basic rate-of-change application of derivatives.


Concept:


  • The instantaneous rate of change of area with respect to radius is dAdr\frac{dA}{dr}.
  • For A=πr2,dAdr=2πrA = \pi r^2, \frac{dA}{dr} = 2\pi r, which is numerically the circumference.

Why B is correct:


  • dAdr=2πr=2π(6)=12π\frac{dA}{dr} = 2\pi r = 2\pi(6) = 12\pi.

Why the distractors are wrong:


  • A (10π10\pi): from r=5r = 5 instead of 6.
  • C (8π8\pi): from r=4r = 4, or from mis-differentiating.
  • D (11π11\pi): a non-standard value with no valid derivation.

Interesting facts

The derivative of a circle's area (πr2\pi r^2) with respect to radius equals its circumference (2πr2\pi r) — a beautiful geometric fact.

This same idea generalises: the derivative of a sphere's volume (43πr3\frac{4}{3}\pi r^3) with respect to r is its surface area (4πr24\pi r^2).

Question details

Q8 of 44 questions

Board
CBSE
Class
Class 12
Subject
Applied Mathematics
Year
2026
Section
Section A
Marks
1
Type
MCQ
Chapter
Calculus (Applications of Derivatives)
Topic
Rate of Change — application of derivatives
Bloom's level
apply
Syllabus topic
Unit 3: Calculus — Applications of Derivatives: rate of change of quantities
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why does dAdr\frac{dA}{dr} equal the circumference?

Because A=πr2A = \pi r^2 differentiates to 2πr2\pi r, which is exactly the circumference formula. Geometrically, growing the radius adds a thin ring of length 2πr2\pi r.

What are the units of dAdr\frac{dA}{dr} here?

Area is in cm2\text{cm}^2 and radius in cm, so dAdr\frac{dA}{dr} is in cm2\text{cm}^2/cm, i.e. cm — numerically 12π12\pi when r=6r = 6 cm.