A circle is inside a square. The radius of the circle is decreasing at a rate of 3 meters per hour and the sides of the square are increasing at a rate of 2 meters per hour. When the radius is 2 meters, and the sides are 19 meters, then how fast is the AREA outside the circle but inside the square changing? The rate of change of the area enclosed between the circle and the square is square meters per hour.
A circle is inside a square. The radius of the circle is decreasing at a rate of 3 meters per hour and the sides of the square are increasing at a rate of 2 meters per hour. When the radius is 2 meters, and the sides are 19 meters, then how fast is the AREA outside the circle but inside the square changing? The rate of change of the area enclosed between the circle and the square is square meters per hour.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![**Problem:**
A circle is inside a square.
The radius of the circle is decreasing at a rate of 3 meters per hour and the sides of the square are increasing at a rate of 2 meters per hour.
When the radius is 2 meters, and the sides are 19 meters, then how fast is the area outside the circle but inside the square changing?
The rate of change of the area enclosed between the circle and the square is [ ] square meters per hour.
**Options:**
- Add Work
- Check Answer
**Explanation:**
To solve this problem, consider the following steps:
1. **Calculate the area of the square and the circle:**
- The area of the square is given by the square of its side length, \( A_{\text{square}} = s^2 \).
- The area of the circle is given by \( A_{\text{circle}} = \pi r^2 \).
2. **Determine the rate of change of the areas:**
- Differentiate the area of the square with respect to time to find the rate of change of the square’s area.
- Differentiate the area of the circle with respect to time to find the rate of change of the circle’s area.
3. **Find the rate of change of the area between the circle and the square:**
- The rate of change of the area between the circle and the square is the difference between the rate of change of the square and the circle areas.
Use these principles along with the given rates and dimensions to solve the problem.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3ef879d6-d05a-4de3-b7a5-e661a4d5bec1%2F24e38013-1483-4dc5-b121-4c5b777fffae%2F7taivem_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem:**
A circle is inside a square.
The radius of the circle is decreasing at a rate of 3 meters per hour and the sides of the square are increasing at a rate of 2 meters per hour.
When the radius is 2 meters, and the sides are 19 meters, then how fast is the area outside the circle but inside the square changing?
The rate of change of the area enclosed between the circle and the square is [ ] square meters per hour.
**Options:**
- Add Work
- Check Answer
**Explanation:**
To solve this problem, consider the following steps:
1. **Calculate the area of the square and the circle:**
- The area of the square is given by the square of its side length, \( A_{\text{square}} = s^2 \).
- The area of the circle is given by \( A_{\text{circle}} = \pi r^2 \).
2. **Determine the rate of change of the areas:**
- Differentiate the area of the square with respect to time to find the rate of change of the square’s area.
- Differentiate the area of the circle with respect to time to find the rate of change of the circle’s area.
3. **Find the rate of change of the area between the circle and the square:**
- The rate of change of the area between the circle and the square is the difference between the rate of change of the square and the circle areas.
Use these principles along with the given rates and dimensions to solve the problem.
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