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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 22 square inches per second. At the time when the area of the square is 11 , 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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Q. The area of a square is increasing at a rate of 22 square inches per second. At the time when the area of the square is 11 , what is the rate of change of the perimeter of the square? Round your answer to three decimal places (if necessary).
  1. Area of Square: Relate the area of the square to its side length.\newlineArea of square, A=s2A = s^2, where ss is the side length.\newlineGiven A=1A = 1, find ss.\newlines2=1s^2 = 1\newlines=1s = 1
  2. Differentiate Area Formula: Differentiate the area formula with respect to time.\newlinedAdt=2sdsdt\frac{dA}{dt} = 2s \cdot \frac{ds}{dt}\newlineGiven dAdt=2\frac{dA}{dt} = 2 square inches per second, substitute and solve for dsdt\frac{ds}{dt}.\newline2=21dsdt2 = 2\cdot 1 \cdot \frac{ds}{dt}\newlinedsdt=1\frac{ds}{dt} = 1 inch per second
  3. Perimeter of Square: Relate the perimeter of the square to its side length.\newlinePerimeter, P=4sP = 4s\newlineDifferentiate the perimeter formula with respect to time.\newline rac{dP}{dt} = 4 \times rac{ds}{dt}\newlineSubstitute ds/dtds/dt from Step 22.\newline rac{dP}{dt} = 4 \times 1\newline rac{dP}{dt} = 4 inches per second

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