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. How fast is the height of the pile increasing when the pile is 11 feet high?
Recall that the volume of a right circular cone with height h and radius of the base r is given by
V=(pi*(r^2)h)/3
If you are in calculus...
V=PI *r^2 * r/3 because r and h are the same
or
V=PI *h^3 /3
dV/dt=PI*h^2 dh/dt
You know dv/dt as 20ft^3/min
solve for dh/dt.
To find how fast the height of the pile is increasing, we need to take the derivative of the volume function with respect to time and then substitute the given values.
Let V be the volume of the cone, r be the radius of the base, and h be the height of the cone.
Given:
The rate of change of volume with respect to time (dv/dt) is 20 cubic feet per minute.
The height of the pile h is 11 feet.
The volume formula for a right circular cone is given as:
V = (π * r^2 * h) / 3
Now, we can differentiate both sides of the equation with respect to t (time):
dV/dt = (d/dt)((π * r^2 * h) / 3)
Using the product rule, we can differentiate each term separately:
dV/dt = (π/3) * [2r * dr/dt * h + r^2 * dh/dt]
Since we are given that dh/dt is the rate at which the height is changing (which we need to find), we can rearrange the equation as follows:
dh/dt = (3/dt) * (dV/dt) / (π * r^2) - 2r * dr/dt
Now, let's substitute the given values into the equation:
dh/dt = (3/π * r^2) * (20) / (π * r^2) - 2r * dr/dt
Simplifying the equation:
dh/dt = 60/π - 2r * dr/dt
We have the value of r as half of the base diameter. Since the base diameter equals the height of the cone, r = h/2.
Substituting for r:
dh/dt = 60/π - 2(h/2) * dr/dt
dh/dt = 60/π - h * dr/dt
Now, substitute the value of h = 11 and solve for dr/dt.
dh/dt = 60/π - 11 * dr/dt
To calculate dr/dt, we need more information or an additional equation relating the rate of change of radius with respect to time.