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Use differentials to estimate the amount of material in a closed cylindrical can that is 60 cm high and 24 cm in diameter if the metal in the top and bottom is 0.1 cm thick, and the metal in the sides is 0.05 cm thick. Note, you are approximating the volume of metal which makes up the can (i.e. melt the can into a blob and measure its volume), not the volume it encloses. The differential for the volume is

Sagot :

Answer:

dV ≈ 927.4 cm³

Step-by-step explanation:

Formula for volume of a cylinder is;

V = πr²h

Using differentials, we have;

dV = (dV/dr)dr + (dV/dh)dh

Now, dV/dr = 2πrh

dV/dh = πr²

Thus, at r = 24/2 = 12 cm

And at h = 60 cm, we have;

dV/dr = 2π(12 × 60)

dV/dr = 4523.89 cm²

Also;

dV/dh = π(12)²

dV/dh = 452.39 cm²

the metal in the top and bottom is 0.1 cm thick. Thus, dr = 2 × 0.1 = 0.2

the metal in the sides is 0.05 cm thick. Thus; dh = 0.05

Thus;

dV = (4523.89 × 0.2) + (452.39 × 0.05)

dV ≈ 927.4 cm³

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