Question

the area of a square is increasing at a rate of 15 square inches per minute. at the time when the area of the square is 1, what is the rate of change of the perimeter of the square? round your answer to three decimal places (if necessary).

Explanation:

Step1: Define variables and relationships

Let \( A \) be the area of the square, \( s \) be the side length, and \( P \) be the perimeter. We know \( A = s^2 \) and \( P = 4s \). We are given \( \frac{dA}{dt}=15 \) square inches per minute, and we need to find \( \frac{dP}{dt} \) when \( A = 1 \).

First, when \( A = 1 \), since \( A = s^2 \), we solve for \( s \): \( s=\sqrt{A}=\sqrt{1} = 1 \) inch.

Step2: Differentiate area with respect to time

Differentiate \( A = s^2 \) with respect to \( t \) (using the chain rule):
\( \frac{dA}{dt}=2s\frac{ds}{dt} \)

We know \( \frac{dA}{dt}=15 \) and \( s = 1 \), so substitute these values in:
\( 15 = 2(1)\frac{ds}{dt} \)
Solve for \( \frac{ds}{dt} \): \( \frac{ds}{dt}=\frac{15}{2}=7.5 \) inches per minute.

Step3: Differentiate perimeter with respect to time

Differentiate \( P = 4s \) with respect to \( t \):
\( \frac{dP}{dt}=4\frac{ds}{dt} \)

Now substitute \( \frac{ds}{dt}=7.5 \) into this equation:
\( \frac{dP}{dt}=4(7.5)=30 \)

Answer:

\( 30.000 \) (or simply \( 30 \))