The area of a square is increasing at a rate of 20 square meters per hour. At a certain instant, the area is 49 square meters. What is the rate of change of the perimeter of the square at that instant (in meters per hour)? Choose 1 answer: (A) 7 (B) (40)/(7) (C) 28 (D) 2sqrt5

Question

The area of a square is increasing at a rate of 20 square meters per hour. At a certain instant, the area is 49 square meters. What is the rate of change of the perimeter of the square at that instant (in meters per hour)? Choose 1 answer: (A) 7 (B)  (40)/(7) (C)  28 (D)  2sqrt5

Bytelearn · Accepted Answer

Find Side Length: First, let's find the length of one side of the square when the area is 49 square meters. Since the area of a square is side squared (s^2), we can find the side length by taking the square root of the area. So, s = sqrt(49) = 7 meters.Calculate Perimeter: Now, the perimeter of a square is 4 times the side length, so the perimeter P when the area is 49 square meters is P = 4 * s = 4 * 7 = 28 meters.Differentiate Perimeter: To find the rate of change of the perimeter, we need to differentiate the perimeter with respect to time. The perimeter P = 4s, and we know that the area A = s^2 is changing at a rate of 20 square meters per hour. So, we need to find ds/dt when A = 49 square meters.Use Chain Rule: Differentiating A = s^2 with respect to time t gives us 2s * (ds/dt). We know that dA/dt = 20 square meters per hour, so we can set up the equation 2s * (ds/dt) = 20.Calculate Rate of Change: Plugging in the side length s = 7 meters into the equation gives us 2 * 7 * (ds/dt) = 20. Solving for ds/dt gives us ds/dt = 20 / (2 * 7) = 20 / 14 = 10 / 7 meters per hour.Calculate dP/dt: Finally, since the perimeter P = 4s, the rate of change of the perimeter dP/dt is 4 times the rate of change of the side length ds/dt. So, dP/dt = 4 * (10 / 7) = 40 / 7 meters per hour.

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