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A conical cup is 4 cm across and 8 cm deep. Water leaks out of the bottom at the rate of 3cm/s. What is the rate of change of the water's level when the height of the water is 4 cm? Please enter your answer in decimal format with three significant digits after the decimal point.

1 Answer

2 votes

Answer:

-0.955 cm/s

Explanation:

You want to know the rate of change of the water level in a conical cup 4 cm in diameter and 8 cm high when the water level is 4 cm and the volume is decreasing at 3 cm³/s.

Area

When the water level in the cup is half the height of the cup, the diameter of its surface will be half the diameter of the cup. That diameter will be ...

(4 cm)/2 = 2 cm

The area of the surface of the water is ...

A = (π/4)d² = (π/4)(2 cm)² = π cm²

Rate of change

The rate of change of height will be the rate of change of volume divided by the area of the surface:

dh/dt = (dV/dt)/A = (-3 cm³/s)/(π cm²) ≈ -0.955 cm/s

The height of the water is decreasing at 0.955 cm/s.

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Additional comment

You can differentiate the volume function and arrive at the same conclusion. For that, you need to express the water volume as a function of height only: V = (π/48)h³. Then V' = (π/16)h²·h'. At h=4, h' = V'/ π, as above.

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A conical cup is 4 cm across and 8 cm deep. Water leaks out of the bottom at the rate-example-1
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