Cylinder rate problem?
Gravel is being dumped from a conveyor belt at a rate of 10 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 8 ft high? (Round your answer to two decimal places
- wy125Lv 48 years agoFavorite Answer
The volume of a cone is given by: V = 1/3 PI * R^2 * h.
This particular cone is such that its base diameter and height are equal so the volume of this cone is given by:
V = 1/3 PI * h^3
Now let's take the derivative with respect to time:
d/dt (V = 1/3 PI * h^3)
dV/dt = 1/3 * PI * 3 * h^2 * dh/dt
dV/dt = PI * h^2 * dh/dt
solve for dh/dt (how fast the pile is increasing in height)
dh/dt = dV/dt / (PI * h^2)
plug in the values: dV/dt = 10 ft^3/min, h = 8 ft
dh/dt = 10 ft^3/min / (PI * (8 ft)^2)
dh/dt = 0.05 ft / min
- 4 years ago
leaking at value of 7000 cubic cm a min top 6m diameter 4 m radius 2 increasing at 0.2 cubic m a minute section = n x R x H squared = section = 3.14 x 2 x 6 squared = a million,419.7824 wait, perhaps 7020 cubic centimeters a minute?