circle is inscribed in a square. the circumference of the circle is increasing at a constant rate of 6 inches per second. as the circle expands, the square expands to maintain the condition of tangency.
1. find the rate at which the perimeter of the square is increasing. indicate units of measure.
2. at the instant when the area of the circle is 25pi square inches, find the rate of increase of the area enclosed between the circle and the square. indicate units of measure.
can someone please explain how to do this?!
1. find the rate at which the perimeter of the square is increasing. indicate units of measure.
2. at the instant when the area of the circle is 25pi square inches, find the rate of increase of the area enclosed between the circle and the square. indicate units of measure.
can someone please explain how to do this?!
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circle, c = pi/d or d = c/pi
so the diameter of the circle is increasing at a rate of (6/pi) in/sec
the perimeter of the square = 4 * side or 4 * diameter
so the perimeter of the square is increasing at a rate of (6*4)/pi = 24/pi inches/sec
If circle area = 25(pi) then the radius must be 5. The diameter is 10 = side of the square.
Circle Area = (pi*d^2)/4
Square area = d^2
Area difference = d^2(1 - pi/4)
d/dt area = 2d(1-pi/4)*dd/dt, dd/dt = 6/pi
= 12/pi(1-pi/4) sq in/sec
so the diameter of the circle is increasing at a rate of (6/pi) in/sec
the perimeter of the square = 4 * side or 4 * diameter
so the perimeter of the square is increasing at a rate of (6*4)/pi = 24/pi inches/sec
If circle area = 25(pi) then the radius must be 5. The diameter is 10 = side of the square.
Circle Area = (pi*d^2)/4
Square area = d^2
Area difference = d^2(1 - pi/4)
d/dt area = 2d(1-pi/4)*dd/dt, dd/dt = 6/pi
= 12/pi(1-pi/4) sq in/sec