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To solve this problem, we need to find the rate of change of the area of the circle, given that the radius is increasing at a rate of 5 cm/min when the radius \( r = 29 \) cm.

The area \( A \) of a circle is related to the radius \( r \) by the formula:

\[
A = \pi r^2
\]

To find how fast the area is changing (i.e., \(\frac{dA}{dt}\)), we differentiate both sides of this equation with respect to time \( t \):

\[
\frac{dA}{dt} = \frac{d}{dt} (\pi r^2) = 2\pi r \frac{dr}{dt}
\]

We are given:
- \( \frac{dr}{dt} = 5 \) cm/min (the rate at which the radius is increasing),
- \( r = 29 \) cm (the current radius).

Substitute these values into the formula:

\[
\frac{dA}{dt} = 2\pi (29) (5)
\]

\[
\frac{dA}{dt} = 290\pi \ \text{cm}^2/\text{min}
\]

Approximating \( \pi \approx 3.1416 \):

\[
\frac{dA}{dt} \approx 290 \times 3.1416 = 911.6 \ \text{cm}^2/\text{min}
\]

Thus, the rate of change of the area is approximately \( 911.6 \ \text{cm}^2/\text{min} \).

Would you like more details or have any questions?

Here are five related questions:
1. How does the formula for the area of a circle change if the radius is doubled?
2. What happens to the rate of change of the area if the radius increases at a slower rate?
3. How would this approach change for finding the rate of change of the circumference?
4. Can this method be extended to find the rate of change for other shapes?
5. How does the value of \(\pi\) influence calculations like this in real-world applications?

**Tip:** When solving related rates problems, always start by identifying the known rates and the relationship between the variables before differentiating.

The radius r of a circle is increasing at a rate of 5 centimeters per minute. Find the rate of change of the area (in cm²/min) when r = 29 centimeters.

Learn how to find the rate of change of a circle's area when its radius increases. The problem involves related rates, differentiation, and geometry, using the formula A = πr². Suitable for students in Grades 10-12.

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