3. The area of a circle increases at a rate of 47 cm²/s. How fast is the radius changing when the radius is 2 cm? A=TTY² 21Tr 4π = 2T (2) 4IT=4 TT

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:**

The area of a circle increases at a rate of \(4\pi \, \text{cm}^2/\text{min}\). How fast is the radius changing when the radius is 2 cm?

**Solution:**

1. **Given equation for area of a circle:**
   \[
   A = \pi r^2
   \]

2. **Differentiate with respect to time \(t\):**
   \[
   \frac{dA}{dt} = 2\pi r \frac{dr}{dt}
   \]

3. **Given \(\frac{dA}{dt} = 4\pi\), and we need to find \(\frac{dr}{dt}\) when \(r = 2\):**

4. **Substitute the values into the differentiated equation:**
   \[
   4\pi = 2\pi (2) \frac{dr}{dt}
   \]

5. **Simplify:**
   \[
   4\pi = 4\pi \frac{dr}{dt}
   \]

6. **Solve for \(\frac{dr}{dt}\):**
   \[
   \frac{dr}{dt} = 1
   \]

Thus, the radius is changing at a rate of 1 cm/min when the radius is 2 cm.
Transcribed Image Text:**Problem:** The area of a circle increases at a rate of \(4\pi \, \text{cm}^2/\text{min}\). How fast is the radius changing when the radius is 2 cm? **Solution:** 1. **Given equation for area of a circle:** \[ A = \pi r^2 \] 2. **Differentiate with respect to time \(t\):** \[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \] 3. **Given \(\frac{dA}{dt} = 4\pi\), and we need to find \(\frac{dr}{dt}\) when \(r = 2\):** 4. **Substitute the values into the differentiated equation:** \[ 4\pi = 2\pi (2) \frac{dr}{dt} \] 5. **Simplify:** \[ 4\pi = 4\pi \frac{dr}{dt} \] 6. **Solve for \(\frac{dr}{dt}\):** \[ \frac{dr}{dt} = 1 \] Thus, the radius is changing at a rate of 1 cm/min when the radius is 2 cm.
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