A circle is inside a square. The radius of the circle is increasing at a rate of 1 meter per minute and the sides of the square are increasing at a rate of 5 meters per minute. When the radius is 5 meters, and the sides are 16 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 minute.

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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**Problem Statement:**

A circle is inside a square.

The radius of the circle is increasing at a rate of 1 meter per minute and the sides of the square are increasing at a rate of 5 meters per minute.

When the radius is 5 meters, and the sides are 16 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 minute.

**Explanation:**

1. **Variables Involved:**
   - Let \( r \) be the radius of the circle.
   - Let \( s \) be the side length of the square.
   - Let \( A_s \) be the area of the square.
   - Let \( A_c \) be the area of the circle.

2. **Given Rates:**
   - \(\frac{dr}{dt} = 1\) meter per minute (rate of change of the radius of the circle).
   - \(\frac{ds}{dt} = 5\) meters per minute (rate of change of the side length of the square).

3. **Areas:**
   - Area of the square: \( A_s = s^2 \)
   - Area of the circle: \( A_c = \pi r^2 \)

4. **Rate of Change of Areas:**
   - Differentiating \( A_s \) with respect to time \( t \):
     \[
     \frac{dA_s}{dt} = \frac{d}{dt} (s^2) = 2s \frac{ds}{dt}
     \]
   - Differentiating \( A_c \) with respect to time \( t \):
     \[
     \frac{dA_c}{dt} = \frac{d}{dt} (\pi r^2) = 2\pi r \frac{dr}{dt}
     \]

5. **Current Values:**
   - When \( r = 5 \) meters, and \( s = 16 \) meters.

6. **Calculation:**
   - Substituting the given values into the equations for the rate of change of areas:
     \[
     \frac{dA_s}{dt} = 2 \cdot 16 \cdot 5 = 160 \text{ square meters per minute}
     \]
     \
Transcribed Image Text:**Problem Statement:** A circle is inside a square. The radius of the circle is increasing at a rate of 1 meter per minute and the sides of the square are increasing at a rate of 5 meters per minute. When the radius is 5 meters, and the sides are 16 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 minute. **Explanation:** 1. **Variables Involved:** - Let \( r \) be the radius of the circle. - Let \( s \) be the side length of the square. - Let \( A_s \) be the area of the square. - Let \( A_c \) be the area of the circle. 2. **Given Rates:** - \(\frac{dr}{dt} = 1\) meter per minute (rate of change of the radius of the circle). - \(\frac{ds}{dt} = 5\) meters per minute (rate of change of the side length of the square). 3. **Areas:** - Area of the square: \( A_s = s^2 \) - Area of the circle: \( A_c = \pi r^2 \) 4. **Rate of Change of Areas:** - Differentiating \( A_s \) with respect to time \( t \): \[ \frac{dA_s}{dt} = \frac{d}{dt} (s^2) = 2s \frac{ds}{dt} \] - Differentiating \( A_c \) with respect to time \( t \): \[ \frac{dA_c}{dt} = \frac{d}{dt} (\pi r^2) = 2\pi r \frac{dr}{dt} \] 5. **Current Values:** - When \( r = 5 \) meters, and \( s = 16 \) meters. 6. **Calculation:** - Substituting the given values into the equations for the rate of change of areas: \[ \frac{dA_s}{dt} = 2 \cdot 16 \cdot 5 = 160 \text{ square meters per minute} \] \
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