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 byV=13πr2h
Answer: ____Ft/ Min
V = (1/3) pi r^2 h I think you mean
h = 2 r given
V = (2/3) pi r^3
dV/dt = (2 pi/3) 3 r^2 dr/dt = 2 pi r^2 dr/dt
10 = 2 pi (14)^2 dr/dt
solve for dr/dt
dh /dt = 2 dr/dt
I'm confused
To find the rate at which the height of the pile is increasing, we need to use the formula for the volume of a right circular cone and apply the chain rule to differentiate with respect to time.
Given:
Rate of gravel being dumped, dV/dt = 10 cubic feet per minute
We are asked to find:
Rate at which the height of the pile is increasing, dh/dt (in ft/min)
Formula for the volume of a right circular cone:
V = (1/3)πr^2h
To find dh/dt, let's differentiate the equation with respect to time (t), using the chain rule:
dV/dt = (1/3)πr^2(dh/dt) + (2/3)πrh(dr/dt)
Since the question states that the base diameter and the height are equal, we know that the radius is half of the diameter. So, let's replace r with h/2 in the equation:
dV/dt = (1/3)π(h/2)^2(dh/dt) + (2/3)π(h/2)(dh/dt)
Simplifying further:
10 = (1/3)π(h^2/4)(dh/dt) + (2/3)π(h/2)(dh/dt)
10 = (1/3)π(h^2/4 + h^2/2)(dh/dt)
Multiply both sides by 3/π and simplify:
30/π = (h^2/4 + h^2/2)(dh/dt)
Simplify further:
30/π = (3h^2/4)(dh/dt)
Now, substitute the given height value of h = 14 feet into the equation:
30/π = (3(14)^2/4)(dh/dt)
Simplifying:
30/π = 3(196/4)(dh/dt)
30/π = 147(3/4)(dh/dt)
30/π = 110.25(dh/dt)
Finally, solve for dh/dt:
dh/dt = (30/π) / 110.25
Calculating this expression will give you the answer in feet per minute.