Answer:
Explanation:
Let's name the radius of the circle "r". Because the circumference of the circle grows at a constant rate of 6 inches per second, we can apply the calculation C = 2 * pi * r to calculate the rate at which the radius grows.
dC/dt = 2 * pi * dr/dt = 6 inches/second
Because the square is parallel to the circle, the length of one of its sides equals the diameter of the circle. As a result, the square's perimeter equals four times the diameter of the circle or P = 4 * 2 * r.
As a result, the pace at which the square's perimeter grows is:
dP/dt equals 4 * 2 * dr/dt equals 8 * dr/dt
Substituting the dr/dt value from the first equation:
dP/dt = 8 * dC/dt = 8 * 6 inches per second = 48 inches per second
The pace at which the perimeter of the square is growing is 48 inches/second
b) The circle's area is determined by A = pi * r2.
If A = 257 square inches, we may calculate the value of r at that moment.
sqrt(A/pi) = sqrt(257/pi)
The rate of growth in the circle area is dA/dt = 2 * pi * r * dr/dt.
Substituting the dr/dt value from the first equation and the r value from above:
dA/dt = 2 pi * sqrt(257/pi) * 6 in/sec = 12 sqrt(257) in/sec
The area enclosed between the circle and the square equals the square's area minus the circle's area.
Let's call the square's side length s.
enclosed area = s2 - A = s2 - pi * r2
d(s2 - pi * r2)/dt = 2sds/dt is the rate of expansion of the enclosed area.
We may calculate the rate of increase of the enclosed area by substituting the value of ds/dt from the first equation.
ds/dt = dP/dt / 4 = 48 inches per second / 4 = 12 inches per second
d(s^2 - pi * r^2)
/dt = 2 * s * ds/dt = 2 * s * 12 inches/sec = 24 * s
The rate of expansion in the area enclosed between the circle and the square is 24*s inches/sec at the instant when the size of the circle is 257 square inches.
Please note that s is not supplied in the issue, thus it's impossible to determine the precise value of the rate of increase in the area contained.