Question

Related rates. A water bucket is shaped like the frustum of a cone with height 2 feet, base radius 6 inches, and top radius 12 inches. The bucket is slowly being filled with water at 8 in3/minute. At what rate is the water level rising when the depth of the water in the bucket is 4 inches?

The volume of a conical frustum with height h and base radii a and b
V= ((pi * h)/3) * (a^2 +ab +b^2)
Use geometry to establish a relationship between the remaining two variables and use that relationship to substitute, thereby turning the volume into an equation with just one variable.

Answer 1

The volume of a conical frustum with height h and base radii a and b is

V= ((? * h) / 3) * (a^2 +ab +b^2)Hint #2: Establish the fact that one of the dimensions (h, a, or b) is a constant.
Hint #3: Use geometry to establish a relationship between the remaining two variables and use that relationship to substitute, thereby turning the volume into an equation with just one variable. To establish the relationship extend the bottom of the bucket to form a right circular cone and use similar triangles

Answer 2

The the radius at the bottom of the bucket is 6 inches. in the span of 2ft (the complete height of the bucket) the radius increases to 12 inches. Therefore the the upper radius of the Volume equation (as the bucket is filled) will increase wrt to height:
\[Radius_{upper} = 6 + 6(\frac{h}{24})\]
Where the radius will be 6" at height of zero, and 12" at height of 2feet (24inches)
\[Radius_{upper} = 6 + \frac{h}{4}\]
\[Radius_{upper} = \frac{24+h}{4}\]