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The area of a square is increasing at a rate of 2 square inches per second. 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).

The area of a square is increasing at a rate of $2$ square inches per second. 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).

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Find the rate of change of one variable when rate of change of other variable is given

Full solution

Q. The area of a square is increasing at a rate of

2

square inches per second. 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).

Area of Square: Relate the area of the square to its side length. $\newline$ Area of square, $A = s^2$ , where $s$ is the side length. $\newline$ Given $A = 1$ , find $s$ . $\newline$ $s^2 = 1$ $\newline$ $s = 1$
Differentiate Area Formula: Differentiate the area formula with respect to time. $\newline$ $\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}$ $\newline$ Given $\frac{dA}{dt} = 2$ square inches per second, substitute and solve for $\frac{ds}{dt}$ . $\newline$ $2 = 2\cdot 1 \cdot \frac{ds}{dt}$ $\newline$ $\frac{ds}{dt} = 1$ inch per second
Perimeter of Square: Relate the perimeter of the square to its side length. $\newline$ Perimeter, $P = 4s$ $\newline$ Differentiate the perimeter formula with respect to time. $\newline$ rac{dP}{dt} = 4 \times rac{ds}{dt} $\newline$ Substitute $ds/dt$ from Step $2$ . $\newline$ rac{dP}{dt} = 4 \times 1 $\newline$ rac{dP}{dt} = 4 inches per second

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