Q.The radius of a circle is increasing uniformly at the rate of $3 \text{ cm/s}$. Find the rate at which the area of the circle is increasing when the radius is $10 \text{ cm}$.

Concept: Related Rates — we differentiate the area formula with respect to time, using the chain rule.

Step 1: Area of a circle: $A = \pi r^2$.

Step 2: Differentiate both sides with respect to $t$:

Step 3: Given $\frac{dr}{dt} = 3 \text{ cm/s}$ and $r = 10 \text{ cm}$:

This is a classic related rates problem. The core idea: when two quantities are linked by a formula (here, area and radius of a circle), their rates of change with respect to time are also linked. If you know how fast one is changing, you can find how fast the other is changing — provided you know the relationship at the instant in question.

The key tool is the chain rule from calculus. If $A = \pi r^2$, then differentiating both sides with respect to time $t$ gives $\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = 2\pi r \cdot \frac{dr}{dt}$.

Let’s walk through it step by step.

1. Identify the given and required rates.

   We are told: $\frac{dr}{dt} = 3$ cm/s (the radius increases at this constant rate).

   We need: $\frac{dA}{dt}$ when $r = 10$ cm.

2. Write the relationship between area and radius.

   For a circle, $A = \pi r^2$.

3. Differentiate with respect to time.

   Since $A$ depends on $r$, and $r$ depends on $t$, use the chain rule:

4. Substitute the known values. At the instant when $r = 10$ cm and $\frac{dr}{dt} = 3$ cm/s:

5. Interpret the result. The area is increasing at $60\pi$ cm²/s at that moment. Since $\pi \approx 3.14$, this is roughly $188.4$ cm²/s. The rate itself will keep increasing as the radius grows, because $\frac{dA}{dt}$ depends on $r$.

Method: Related Rates for a Single Geometric Relation

This is the base pattern for related-rates problems where one quantity (here, area) depends on a single other quantity (radius) that is itself changing with time.

Steps

Step 1: Write the formula linking the two quantities

For a circle, $A=\pi r^2$, where both $A$ and $r$ are functions of time $t$ even though $t$ doesn't appear explicitly.

Step 2: Differentiate both sides with respect to time

Because $r$ is a function of $t$, differentiating $r^2$ requires the chain rule:

Step 3: Substitute the given rate and the given instantaneous value

Substitute the known $\frac{dr}{dt}$ and the value of $r$ at the instant asked about — both go in only after differentiating, never before.

Step 4: Attach the correct units

Since $r$ is in cm and $\frac{dr}{dt}$ is in cm/s, the result for $\frac{dA}{dt}$ carries units of cm²/s.

Common Mistakes

Mistake 1: Substituting the radius before differentiating

Plugging $r=10$ into $A=\pi r^2$ first gives a fixed number, $A=100\pi$, whose derivative is zero — this destroys the relationship between the rates entirely. The formula must be differentiated while $r$ is still treated as a variable.

Mistake 2: Dropping the factor of $2$ from differentiating $r^2$

$\frac{d}{dt}(r^2) = 2r\frac{dr}{dt}$, not $r\frac{dr}{dt}$ — forgetting the exponent's factor of 2 halves the final answer.