A water pipe mounted to the ceiling has a leak and is dripping onto the floor below, creating a circular puddle of water. The area of the circular puddle is increasing at a constant rate of 11 cm2/hour.
(a) Find the area and radius of the puddle after 1 minute, 95 minutes, 4 hours, and 1 day. (Round your answers to four decimal places as needed.)

To solve this, we can use the formula for the area of a circle, \( A = \pi r^2 \), where \( A \) is the area of the circle and \( r \) is its radius. Given that the area of the puddle is increasing at a constant rate of \( \frac{dA}{dt} = 11 \, \text{cm}^2/\text{hour} \), we need to calculate the area after different time periods and use this to find the corresponding radius.

### Step-by-Step Solution:

#### 1. Rate of area increase:
- \( \frac{dA}{dt} = 11 \, \text{cm}^2/\text{hour} \).

Since the rate of area increase is constant, the area at any time \( t \) (in hours) is:
\[
A(t) = 11 \times t \, \text{cm}^2.
\]

Now, let’s calculate the area at different times:

#### (a) After 1 minute:
- Convert 1 minute to hours: \( t = \frac{1}{60} \, \text{hours} \).
\[
A\left(\frac{1}{60}\right) = 11 \times \frac{1}{60} = 0.1833 \, \text{cm}^2.
\]

- To find the radius, use the formula for the area of a circle: \( A = \pi r^2 \).
\[
r = \sqrt{\frac{A}{\pi}} = \sqrt{\frac{0.1833}{\pi}} \approx 0.2416 \, \text{cm}.
\]

#### (b) After 95 minutes:
- Convert 95 minutes to hours: \( t = \frac{95}{60} \, \text{hours} \).
\[
A\left(\frac{95}{60}\right) = 11 \times \frac{95}{60} = 17.4167 \, \text{cm}^2.
\]

- Radius:
\[
r = \sqrt{\frac{17.4167}{\pi}} \approx 2.3554 \, \text{cm}.
\]

#### (c) After 4 hours:
\[
A(4) = 11 \times 4 = 44 \, \text{cm}^2.
\]

- Radius:
\[
r = \sqrt{\frac{44}{\pi}} \approx 3.7437 \, \text{cm}.
\]

#### (d) After 1 day (24 hours):
\[
A(24) = 11 \times 24 = 264 \, \text{cm}^2.
\]

- Radius:
\[
r = \sqrt{\frac{264}{\pi}} \approx 9.1644 \, \text{cm}.
\]

### Summary of results:
- **After 1 minute**: Area = \( 0.1833 \, \text{cm}^2 \), Radius = \( 0.2416 \, \text{cm} \).
- **After 95 minutes**: Area = \( 17.4167 \, \text{cm}^2 \), Radius = \( 2.3554 \, \text{cm} \).
- **After 4 hours**: Area = \( 44 \, \text{cm}^2 \), Radius = \( 3.7437 \, \text{cm} \).
- **After 1 day**: Area = \( 264 \, \text{cm}^2 \), Radius = \( 9.1644 \, \text{cm} \).

Would you like more details or have any questions?

### Related questions:
1. How would the radius change if the rate of area increase was doubled?
2. What is the derivative of the radius with respect to time?
3. How long would it take for the puddle to reach a radius of 5 cm?
4. Can you find the time at which the area reaches 100 cm²?
5. What is the relationship between the rate of area increase and the rate of radius increase?

### Tip:
The relationship between the radius and the area of a circle is quadratic, so small changes in area lead to smaller changes in radius for large circles.

A water pipe mounted to the ceiling has a leak and is dripping onto the floor below, creating a circular puddle of water. The area of the circular puddle is increasing at a constant rate of 11 cm²/hour. (a) Find the area and radius of the puddle after 1 minute, 95 minutes, 4 hours, and 1 day. (Round your answers to four decimal places as needed.)

Learn how to calculate the area and radius of a circular puddle that increases at a constant rate of 11 cm²/hour. This problem involves geometric formulas for the area of a circle and related rates concepts. Suitable for high school students in Grades 10-12.

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