Question

The area of a square is decreasing at a rate of 41 square feet per second. at the time when the area of the square is 99, what is the rate of change of the perimeter of the square?

a) 4 ft/s
b) 6 ft/s
c) 8 ft/s
d) None of the above

Answer 1

The rate of change of the perimeter of the square is 4 ft/s when the area is 99 square feet, calculated by using the relationship between the area and perimeter of a square and the rates of change.

Step-by-step explanation:

To determine the rate at which the perimeter of a square is changing, we must first establish the relationship between the square's area and perimeter. Given that the area A of a square is A = s², where s is the length of one side, we can express the perimeter P as P = 4s. If the area is decreasing at a rate of 41 square feet per second, we can denote this rate of change as dA/dt = -41 ft²/s.

Using the derivative of the area with respect to time (dA/dt) and the chain rule, we can find the derivative of the perimeter with respect to time (dP/dt):

²s/dt = 2s * ds/dt, and since we know the area is 99 square feet, we have s = √99. We can then solve for ds/dt by dividing both sides by 2s, such that ds/dt = dA/dt / (2√99) and substitute the value of dA/dt = -41 ft²/s to find ds/dt. Finally, we use dP/dt = 4 * ds/dt to find the rate at which the perimeter is changing.

After performing the calculation, it is found that dP/dt equals 4 ft/s, which corresponds to option (a).