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A pizza chef begins to spin a constant volume of dough to make a pie by spinning and tossing the dough into the air, such that the dough takes on a cylindrical shape. The dough's radius increases while the height decreases, but the dough remains a cylinder. At time t = tı, the height of the dough is 1/2 inch, the radius of the dough is 16 inches, and the radius of the dough is increasing at a rate of 2 inches per minute.

Required:
a. At time tı, at what rate is the area of the "top" of the pizza (the part with the toppings) increasing with respect to time? Show the computations that lead to your answer, and indicate units of measure.
b. At time tı, at what rate is the height of the dough decreasing with respect to time?

Respuesta :

Answer:

a ) dAt/dt Ā = Ā 50,24 in/min

dh/dt Ā = Ā - Ā 0,125 in/min

Step-by-step explanation:

The area of the top is At :

At = Ļ€*rĀ²

a) Ā Tacking derivatives with respect to time:

dAt/dt Ā = Ā 2* Ļ€*r * dr/dt

At Ā  t Ā = Ā tā‚ Ā  Ā  Ā r Ā = 16 in Ā  Ā  and Ā dr/dt = Ā 0,5

Then

dAt/dt Ā = Ā 2*3,14*16*0,5 Ā  in/min

a ) dAt/dt Ā = Ā 50,24 in/min

b) The volume of the cylinder is:

Vc = Ā Ļ€*rĀ²*h Ā  Ā  ( where h is the heigh of the cylinder )

Tacking derivatives with respect to time

dVc/dt Ā = Ā 2* Ļ€*r*h*dr/dt Ā + Ā Ļ€*rĀ²*dh/dt

But Ā dVc/dt Ā = 0 Ā since the volume remains constant, then:

Ļ€*rĀ²*dh/dt Ā = - Ā 2* Ļ€*r*h*dr/dt

r*dh/dt Ā = Ā - Ā 2*h*dr/dt

dh/dt Ā = - 2*0,5*2/16 Ā in/min

dh/dt Ā = Ā - Ā 0,125 in/min

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