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 9, what is the rate of change of the perimeter of the square? Round your answer to three decimal places (if necessary).

Answer 1

The rate of change of the perimeter when the area of the square is 9 is 82/3.

Step-by-step explanation:

The area of a square is decreasing at a constant rate of 41 square feet per second. To find the rate of change of the perimeter when the area is 9 square feet, we need to find the rate of change of the side length.

Let's assume the side length of the square is 'x' feet. Then the area of the square is x² and the perimeter is 4x. Since the area is decreasing at a rate of 41 square feet per second, we have the equation:

41 = d(x²)/dt

Using the chain rule, we can find the rate of change of the side length 'dx/dt':

41 = 2x * dx/dt

Now we can substitute the value of 'x' when the area is 9 and solve for 'dx/dt':

41 = 2(3) * dx/dt

dx/dt = 41/6

So the rate of change of the perimeter is dx/dt multiplied by 4:

Rate of change of perimeter = (41/6) * 4 = 82/3