A constant volume of pizza dough is formed into a cylinder with a relatively small height and large radius. The dough is spun and tossed into the air in such a way that the height of the dough decreases the radius increases, it retains its cylindrical shapeAt time of the dough 1/3 inch, the radius of the dough is 12 inches, and the radius of the dough is increasing at a rate of 2 inches per minute at the area of the circular surface of the dough increasing with respect to time? Show answer and numerical answer and units of measure.

The rate of change of each circular surface of the dough is approximately 150.796 in.²/minute

Step-by-step explanation:

The given parameters are;

When the radius of the dough, r = 12 inches, the radius is increasing at 2 inches per minute

Therefore, we have;

dr/dt = 2 in./min

The area of the circular surface of the dough, A = π·r²

The rate of change of each (top or bottom) circular surface area of the dough dA/dt is given as follows;

dA/dt = d(π·r²)/dt = π·2·r·dr/dt

Where;

r = 12 in.

dr/dt = 2 in./min

Substituting, we have;

dA/dt = π·2·r·dr/dt = π × 2 × 12 in. × 2 in./min ≈ 150.796 in.²/minute.

The rate of change of each circular surface of the dough = dA/dt ≈ 150.796 in.²/minute.

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