Gravel is being dumped from a conveyor belt at a rate of 30 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 14 feet high?
radius of cone --- r
height of cone ---h
vol = (1/3)πr^2 h
but h = 2r or r = h/2
so
Vol = (1/3)π (h^2/4)(h)
= (1/12)π h^3
d(Vol)/dt = (1/4)π h^2 dh/dt
so when h = 14
30 = (1/12)π(196) dh/dt
solve for dh/dt
To find how fast the height of the pile is increasing, we need to use the concept of related rates. We are given information about the rate at which gravel is being dumped and we need to find the rate at which the height is changing.
Let's denote the height of the pile as h (in feet) and the radius of the base as r (in feet). Since the base diameter and height are always equal, we have r = h.
The volume of a cone is given by the formula V = (1/3)πr^2h. The rate at which the gravel is being dumped is 30 cubic feet per minute, so we can write dV/dt = 30, where dV/dt represents the rate of change of volume with respect to time.
Now, let's differentiate the formula for the volume of the cone with respect to time t.
dV/dt = (1/3)π(2rh)(dh/dt) + (1/3)πr^2(dh/dt)
Since r = h, we can simplify the equation to:
30 = (1/3)π(2h^2)(dh/dt) + (1/3)πh^2(dh/dt)
Next, we can factor out (1/3)π(dh/dt) from both terms:
30 = (1/3)π((2h^2) + h^2)(dh/dt)
Simplifying further:
30 = (1/3)π(3h^2)(dh/dt)
Now, we can solve for dh/dt by isolating it:
dh/dt = (30 * 3) / (π * 3h^2)
Simplifying further:
dh/dt = 10 / h^2
So, the rate at which the height of the pile is increasing is given by the formula dh/dt = 10 / h^2.
To find how fast the height of the pile is increasing when the pile is 14 feet high, we substitute h = 14 into the formula:
dh/dt = 10 / (14^2)
= 10 / 196
= 0.051 feet per minute
Therefore, when the pile is 14 feet high, the height of the pile is increasing at a rate of approximately 0.051 feet per minute.