Gravel is being dumped from a conveyor belt at a rate of 20 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always equal to each other. How fast is the height of the pile increasing when the pile is 23 feet high?
volume of cone = (1/3)pi(r^2)h
but h=2r
or r = h/2
then volume = (1/12)pi(h^3)
dV/dt = (1/4)pi (h^2) dh/dt
20 = (1/4)pi (23^2)dh/dt
etc
To find the rate at which the height of the pile is changing, we need to use related rates. The problem provides information about the rate at which gravel is being dumped, and we need to find the rate at which the height of the pile is changing.
Let's denote the height of the pile as h and the base diameter as d. Since the base diameter and the height are always equal, we can write d = h.
We are given that gravel is being dumped at a rate of 20 cubic feet per minute. This means the volume of the cone is increasing at a rate of 20 cubic feet per minute.
We can use the formula for the volume of a cone to relate the rate of change of volume to the rate of change of height.
The volume of a cone is given by V = (1/3)πr²h, where r is the radius of the base. Since the base diameter is equal to the height, we have r = d/2 = h/2.
Substituting the values, the volume of the cone becomes V = (1/3)π(h/2)²h = (1/12)πh³.
Now, we differentiate both sides of the equation with respect to time t to find the rate of change of volume:
dV/dt = (1/4)πh² dh/dt.
We know dV/dt = 20 ft³/min, and we need to find dh/dt when h = 23 ft. Substituting these values into the equation, we have:
20 = (1/4)π(23)² dh/dt.
Now, solve for dh/dt:
dh/dt = (4/π(23)²) * 20.
Using a calculator, we can evaluate this expression to find the rate at which the height of the pile is changing.