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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.

Sagot :

Answer:

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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