ive asked this already but the person did not do it right and i've already attempted to do it my way a few times by taking the derivative of the V formula and pluging it in for r. Here is the question
Gravel is being dumped from a conveyor belt at a rate of 50 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 25 feet high?
Recall that the volume of a right circular cone with height h and radius of the base r is given by
V=(1/3)pi(r^2)h
r=radius
h=height.
Gravel is being dumped from a conveyor belt at a rate of 50 cubic feet per minute. It forms a pile in the shape of a right circular cone whose base diameter and height are always the same. How fast is the height of the pile increasing when the pile is 25 feet high?
Recall that the volume of a right circular cone with height h and radius of the base r is given by
V=(1/3)pi(r^2)h
r=radius
h=height.
-
h = d = 2r
r = h/2
V = (1/3) * pi * r^2 * h
V = (1/3) * pi * (h/2)^2 * h
V = (1/3) * pi * (1/4) * h^3
V = (pi / 12) * h^3
dV/dt = 3 * (pi / 12) * h^2 * dh/dt
dV/dt = (pi / 4) * h^2 * dh/dt
dV/dt = 50
h = 25
dh/dt = ?
50 = (pi/4) * 25^2 * dh/dt
50 * 4 / (pi * 25 * 25) = dh/dt
4 / (25 * pi) = dh/dt
r = h/2
V = (1/3) * pi * r^2 * h
V = (1/3) * pi * (h/2)^2 * h
V = (1/3) * pi * (1/4) * h^3
V = (pi / 12) * h^3
dV/dt = 3 * (pi / 12) * h^2 * dh/dt
dV/dt = (pi / 4) * h^2 * dh/dt
dV/dt = 50
h = 25
dh/dt = ?
50 = (pi/4) * 25^2 * dh/dt
50 * 4 / (pi * 25 * 25) = dh/dt
4 / (25 * pi) = dh/dt