A can of soda measures 2.5 inches in diameter and 5 inches in height. If a fullcan of soda gets spilled at a rate of 4 in^3/sec, how is the level of soda changing at the moment when the can of soda is halffull?
The volume of liquid in the full can is given by:
The rate of loss of liquid is given by:
So
This is constant.
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To find how the level of soda is changing at the moment when the can of soda is halffull:
 Determine the volume of the soda in the can when it is halffull.
 Differentiate the volume equation with respect to time to find the rate of change of volume.
 Use the given rate of change of volume to find the rate of change of the level of soda.
The volume V of the soda in the can is given by the formula for the volume of a cylinder: V = π * r^2 * h
Given that the diameter of the can is 2.5 inches, the radius r is 1.25 inches, and the height h is 5 inches.

When the can is halffull, the volume of the soda is half the volume of the can: V_half = (1/2) * π * (1.25)^2 * 5

Differentiate the volume equation with respect to time t: dV/dt = π * (2 * 1.25 * dr/dt * h + r^2 * dh/dt)

At the moment when the can is halffull, dV/dt = 4 in^3/sec, and we need to find dh/dt when V = V_half.

Plug in the known values and solve for dh/dt: 4 = π * (2 * 1.25 * dr/dt * 5 + (1.25)^2 * dh/dt)
Solving for dh/dt gives: dh/dt = (4  5π * 1.25^2 * dr/dt) / (2 * π * 1.25 * 5)
Substituting dr/dt = 4/(π * 1.25^2): dh/dt = (4  5π * 1.25^2 * (4/(π * 1.25^2))) / (2 * π * 1.25 * 5)
Simplify to find dh/dt.
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