Gravel is being dumped from a conveyor belt at a rate of 10 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 14 feet high?
Recall that the volume of a right circular cone with height h and radius of the base r is given by V=1/3πr2h
____ft/min
v = 1/3 πr^2h = 1/12 πh^3
dv/dt = 1/4 πh^2 dh/dt
To find how fast the height of the pile is increasing, we can differentiate the volume of the cone with respect to time.
Given:
Rate of change of volume, dV/dt = 10 cubic feet per minute
Height of the pile, h = 14 feet
We need to find:
Rate of change of height, dh/dt (in ft/min)
The volume of a cone is given by V = (1/3)πr^2h, where r is the radius of the base and h is the height.
To relate the height to the radius, we need to consider the geometry of the cone. The diameter of the base is equal to the height of the cone, so the radius is half of the height. Therefore, r = h/2.
Substituting this expression for r into the volume formula, we have:
V = (1/3)π(h/2)^2h
= (1/12)πh^3
Differentiating both sides of the equation with respect to t:
dV/dt = (1/4)πh^2 * dh/dt
We are given that dV/dt = 10 cubic ft/min, and we need to find dh/dt when h = 14 ft.
Substituting the known values into the equation, we have:
10 = (1/4)π(14^2) * dh/dt
Simplifying and solving for dh/dt:
10 = 1.54π * dh/dt
dh/dt = 10 / (1.54π)
dh/dt ≈ 2.05 ft/min
Therefore, the height of the pile is increasing at a rate of approximately 2.05 ft/min when the pile is 14 feet high.
To find how fast the height of the pile is increasing, we need to use the concept of related rates. We are given that gravel is being dumped at a rate of 10 cubic feet per minute. We need to determine how fast the height of the pile is changing when the pile is 14 feet high.
Let's denote the height of the pile as h and the radius of the base as r. Since the pile is in the shape of a right circular cone, the height and the radius of the base are always equal.
We are given that the rate of change of volume, dV/dt, is 10 cubic feet per minute. We need to find dh/dt, the rate at which the height is changing when the height is 14 feet.
The volume of a cone is given by the formula V = (1/3) * π * r^2 * h.
We can differentiate both sides of the equation with respect to time t to find the relationship between the rates of change:
dV/dt = (1/3) * π * (2r * dr/dt) * h + (1/3) * π * r^2 * dh/dt
Since the radius of the base and the height of the cone are always equal, we can substitute r = h into the equation:
dV/dt = (1/3) * π * (2h * dr/dt) * h + (1/3) * π * h^2 * dh/dt
Simplifying the equation:
10 = (2/3)πh^2 * dr/dt + (1/3)πh^2 * dh/dt
We are asked to find dh/dt when h = 14. Substitute h = 14 into the equation:
10 = (2/3)π * 14^2 * dr/dt + (1/3)π * 14^2 * dh/dt
We can solve this equation to find the value of dh/dt, which represents the rate at which the height of the pile is changing when the height is 14 feet. Since we know the value of dr/dt (which is not given in the problem statement), we can solve for dh/dt.
Is there any other information or value given in the question from which we can determine the value of dr/dt?