Question

a conical salt spreader is spreading salt at a rate of 3 cubic feet per minute. the diameter of the base of the cone is 4 feet and the height of the cone is 5 feet. how fast is the height of the salt in the spreader decreasing when the height of the salt in the spreader (measured from the vertex of the cone upward) is 3 feet? give your answer in feet per minute.

Answer 1

-25/(12π) ft/min ≈ -0.663 ft/min

Explanation:

You want the rate of change of height of the salt in a cone with height 5 ft and diameter 4 ft, filled 3 ft up from the vertex, and decreasing in volume at the rate of 3 cubic feet per minute.

Rate of change of height

The rate of change of height is the rate of change of volume, divided by the surface area.

Surface area

The given cone dimensions tell you the radius is 2/5 of the height of the cone, so the radius of the surface when it is filled to a depth of 3 ft is ...

r = (2/5)(3 ft) = 1.2 ft

The area of the surface is the area of a circle with this radius:

A = πr² = π(1.2 ft)² = 1.44π ft²

Application

Then the rate of change of height is ...

h' = V'/A

h' = (-3 ft³/min)/(1.44π ft²) = -25/(12π) ft/min ≈ -0.663 ft/min