Question

a conical cup is 4 \text{ cm} across and 8 \text{ cm} deep. water leaks out of the bottom at the rate of 3 \textrm{ cm}^3/\textrm{s}. what is the rate of change of the water's level when the height of the water is 4 \text{ cm}? please enter your answer in decimal format with three significant digits after the decimal point.

Answer 1

The rate of change of the water's level is -4 cm/s when the height of the water is 4 cm.

Step-by-step explanation:

To find the rate of change of the water's level, we can use the concept of similar triangles. Since the cup is conical in shape, the cross-section of the cup at any height is similar to the cross-section at the bottom. This means that the ratio of the height to the diameter is constant at all heights. Let's denote the rate of change of the water's level as dh/dt and the rate of change of the diameter as dd/dt. We have:

dh/dt = (hd - hr) / (dd/dt)

where hd is the height of the water, hr is the height of the cup, and dd/dt is the rate of change of the diameter. Given that hd = 4 cm and hr = 8 cm, we can substitute these values into the equation. Since we're asked for the rate of change in cm/s, we need to convert the rate of change of the diameter from cm/s to cm:

dd/dt = (4/8) * 3 cm/s = 1.5 cm/s

Substituting these values back into the equation, we get:

dh/dt = (4 - 8) / 1.5 = -4 cm/s

The rate of change of the water's level when the height of the water is 4 cm is -4 cm/s.