Gravel is being dumped from a conveyor belt at a rate of 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 feet high?
Hint: The volume of a right circular cone with height and radius of the base is given by .
Diameter is equal to height, or
2r = h ---> r = h/2
V= (1/3)πr^2 h
= (1/3)π(h^2/4)(h)
= (1/12πh^3
dV/dt = (1/4)π h^2 dh/dt
You did not list the given data in your post.
Once you have them, just sub in to finish the question.
To find how fast the height of the pile is increasing, we need to use the concept of related rates.
First, let's assign some variables:
- Let V be the volume of the cone (pile of gravel).
- Let r be the radius of the base of the cone.
- Let h be the height of the cone (pile of gravel).
From the given information, we know that the base diameter and height of the cone are always equal. Since the base diameter is twice the radius, the base diameter is 2r. Therefore, the base radius is r.
The volume of a cone is given by the formula:
V = (1/3)πr²h
We need to find dh/dt, the rate at which the height of the cone is changing with respect to time.
To solve for dh/dt, we need to relate the variables in terms of the given information. We are given that the gravel is being dumped from a conveyor belt at a rate of ______ cubic feet per minute. Let's call this rate dV/dt.
Now, we can differentiate the formula for the volume of the cone with respect to time:
dV/dt = (1/3)π(2rh(dr/dt) + r²(dh/dt))
Since we are looking for the rate at which the height of the pile is changing, we can rearrange the equation and solve for dh/dt:
dh/dt = (dV/dt - (1/3)π(2rh(dr/dt)) / πr²
Now, substitute the given values into the equation and solve for dh/dt when the height of the pile is given as feet.