Final answer:
The depth of the water changes at a rate of -1/9 ft/min when the water is 10 ft high.
Step-by-step explanation:
To find the rate at which the depth of the water changes, we can use the concept of volume and rate of change. The volume of a cone can be given by the formula V = (1/3)πr^2h, where r is the radius of the base and h is the height of the cone. In this case, the height of the cone represents the depth of the water, and the volume rate of change can be given by dV/dt, where dt represents the change in time.
Since the cone leaks water at a rate of 10 ft³/min, we can equate dV/dt to -10 ft³/min, because the volume is decreasing due to the leak. We are given that the cone has a radius of 10 ft, and we need to find the height when the depth is 10 ft. We can substitute the given values into the volume formula and solve for h, since dV/dt and r are known.
By plugging in the known values in the volume formula and solving for h, we find that when the water depth is 10 ft, the height of the cone (or the depth of the water) changes at a rate of -1/9 ft/min.