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Q3:
The radius of a circle is increasing uniformly at the rate of $3$ cm/s. Find the rate at which the area of the circle is increasing when the radius is $10$ cm.

Solution :

Given Variables & Initial Setup

Let $r$ represent the radius of the circle at any time $t$, and $A$ represent the area of the circle. We are provided with the following information:

  • Rate of change of the radius with respect to time: $\frac{dr}{dt} = 3 \text{ cm/s}$
  • Specific radius at which the rate of change of area is required: $r = 10 \text{ cm}$

Step 1: Establishing the Geometric Relationship

The area $A$ of a circle in terms of its radius $r$ is given by the formula [Per the standard formula for the area of a circle]:

$A = \pi r^2$

Step 2: Differentiating with respect to Time ($t$)

To find the rate of change of the area with respect to time, we apply the Chain Rule of calculus [By the Chain Rule: $\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}$]:

$\frac{d}{dt}(A) = \frac{d}{dt}(\pi r^2)$

$\frac{dA}{dt} = \pi \cdot \frac{d}{dr}(r^2) \cdot \frac{dr}{dt}$

$\frac{dA}{dt} = \pi \cdot (2r) \cdot \frac{dr}{dt}$

$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$

Step 3: Calculating the Result at $r = 10 \text{ cm}$

Substitute the given values $r = 10 \text{ cm}$ and $\frac{dr}{dt} = 3 \text{ cm/s}$ into the derived equation:

$\frac{dA}{dt} = 2 \cdot \pi \cdot (10 \text{ cm}) \cdot (3 \text{ cm/s})$

$\frac{dA}{dt} = 60\pi \text{ cm}^2/\text{s}$

Visual Representation of the Expansion

r = 10 cm dr/dt = 3 cm/s

Summary of Calculation

Variable Value
Radius ($r$) $10 \text{ cm}$
Rate of Radius Change ($\frac{dr}{dt}$) $3 \text{ cm/s}$
Rate of Area Change ($\frac{dA}{dt}$) $60\pi \text{ cm}^2/\text{s}$

Final Solution: The area of the circle is increasing at the rate of $60\pi \text{ cm}^2/\text{s}$.


More Questions from Class 12 Mathematics Application of Derivatives EXERCISE 6.1


CBSE Solutions for Class 12 Mathematics Application of Derivatives


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