The radius r of a circle is increasing at a rate of 7 centimeters per minute. Find the rate of change of the area when r = 29 centimeters. cm²/min Need Help? Read It
The radius r of a circle is increasing at a rate of 7 centimeters per minute. Find the rate of change of the area when r = 29 centimeters. cm²/min Need Help? Read It
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter3: Linear And Nonlinear Functions
Section3.3: Rate Of Change And Slope
Problem 55PPS
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![**Problem:**
The radius \( r \) of a circle is increasing at a rate of \( 7 \) centimeters per minute. Find the rate of change of the area when \( r = 29 \) centimeters.
**Solution:**
To find the rate of change of the area, we can use the relationship between the area \( A \) of a circle and its radius \( r \). The area of a circle is given by:
\[ A = \pi r^2 \]
Given the rate of change of the radius \( \frac{dr}{dt} = 7 \ \text{cm/min} \), we want to find the rate of change of the area \( \frac{dA}{dt} \) when \( r = 29 \) cm.
By differentiating both sides of the area formula with respect to time \( t \), we get:
\[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \]
Substitute \( r = 29 \) cm and \( \frac{dr}{dt} = 7 \ \text{cm/min} \) into the formula:
\[ \frac{dA}{dt} = 2\pi (29) (7) \]
Now calculate the value:
\[ \frac{dA}{dt} = 2\pi (29) (7) \]
\[ \frac{dA}{dt} = 2\pi (203) \]
\[ \frac{dA}{dt} = 406\pi \]
Therefore, the rate of change of the area when \( r = 29 \) centimeters is:
\[ 406\pi \ \text{cm}^2/\text{min} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0711daf6-3b19-448b-99c0-3e7fe08f4fe3%2Feb838629-096c-438a-8823-09415059080b%2F3voei1_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem:**
The radius \( r \) of a circle is increasing at a rate of \( 7 \) centimeters per minute. Find the rate of change of the area when \( r = 29 \) centimeters.
**Solution:**
To find the rate of change of the area, we can use the relationship between the area \( A \) of a circle and its radius \( r \). The area of a circle is given by:
\[ A = \pi r^2 \]
Given the rate of change of the radius \( \frac{dr}{dt} = 7 \ \text{cm/min} \), we want to find the rate of change of the area \( \frac{dA}{dt} \) when \( r = 29 \) cm.
By differentiating both sides of the area formula with respect to time \( t \), we get:
\[ \frac{dA}{dt} = 2\pi r \frac{dr}{dt} \]
Substitute \( r = 29 \) cm and \( \frac{dr}{dt} = 7 \ \text{cm/min} \) into the formula:
\[ \frac{dA}{dt} = 2\pi (29) (7) \]
Now calculate the value:
\[ \frac{dA}{dt} = 2\pi (29) (7) \]
\[ \frac{dA}{dt} = 2\pi (203) \]
\[ \frac{dA}{dt} = 406\pi \]
Therefore, the rate of change of the area when \( r = 29 \) centimeters is:
\[ 406\pi \ \text{cm}^2/\text{min} \]
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