The edge of a square is increasing at the rate of 3 cm/sec. At what rate is the square
a.) perimeter changing
b.) area changing
when the edge of the square is 10 cm.
O a) 12 cm/sec
b) 60 cm^2/sec
O a) 10 cm/sec
b) 50 cm^2/sec
a) 14 cm/sec
b) 65 cm^2/sec
a) 15 cm/sec
b) 70 cm^2/sec
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To solve this problem, we can use the formulas for the perimeter and area of a square.
a) The perimeter of a square is given by the formula P = 4s, where s is the length of a side of the square. Since the edge of the square is increasing at a rate of 3 cm/sec, the rate of change of the side length is also 3 cm/sec. When the side length is 10 cm, the rate of change of the perimeter can be calculated as follows:
dP/dt = 4(ds/dt)
dP/dt = 4(3 cm/sec)
dP/dt = 12 cm/sec
Therefore, the rate at which the perimeter is changing is 12 cm/sec.
b) The area of a square is given by the formula A = s^2. Using the same logic as in part a, the rate of change of the side length is 3 cm/sec. When the side length is 10 cm, the rate of change of the area can be calculated as follows:
dA/dt = 2s(ds/dt)
dA/dt = 2(10 cm)(3 cm/sec)
dA/dt = 60 cm^2/sec
Therefore, the rate at which the area is changing is 60 cm^2/sec.
Therefore, the correct answer is:
a) 12 cm/sec
b) 60 cm^2/sec
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