Water is leaking out of an inverted conical tank at a rate of 9400.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 14.0 meters and the diameter at the top is 4.0 meters. If the water level is rising at a rate of 28.0 centimeters per minute when the height of the water is 3.0 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute. [answer]
Note: Let "R" be the unknown rate at which water is being pumped in. Then you know that if V is volume of water, dVdt=R−9400.0. Use geometry (similar triangles?) to find the relationship between the height of the water and the volume of the water at any given time. Recall that the volume of a cone with base radius r and height h is given by 13πr2h.
let the input rate be x cm^3/min
effective rate of change of volume = x-9400 cm^3/min
or
dV/dt = x-9400
At any time of t min, let the height of water be h cm and the radius of the water level be r cm
As suggested, by similar triangles,
h/r = 14/2 = 7
h = 7r or r = h/7
V = (1/3)π r^2 h = (1/3)π(h^2/49)(h)
V = (2/147) π h^3
dV/dt = (2/49)π h^2 dh/dt
given: when h = 300, dh/dt = 28
x - 9400 = (2/49)π(300^2)(28)
solve for x, it will be in cm^3/min
check my arithmetic, should have written it out on paper first
To solve this problem, we will set up an equation that relates the rate at which the water level is rising to the rate at which water is being pumped into the tank.
Let's start by finding the relationship between the height of the water and the volume of the water at any given time.
The volume of a cone with base radius r and height h is given by 1/3 * π * r^2 * h. In this case, the radius of the top of the inverted cone is half of the diameter, which is 4.0 meters. So the radius, r, is 2.0 meters.
We want to relate the height of the water, h, to the volume of the water, V. Let's call the proportionality constant "k".
V = k * h
We can solve for k by considering the situation when the height of the water is 14.0 meters, which is the height of the tank. The volume of the water at this point is the maximum volume of the tank, which is a full cone.
V = 1/3 * π * (2.0)^2 * 14.0
V = 1/3 * π * 4.0 * 14.0
V = 1/3 * π * 56.0
V = 18.66π
So, when h = 14.0 meters, V = 18.66π cubic meters.
Now, let's differentiate both sides of the equation V = k * h with respect to time t to find the rate at which the volume is changing.
dV/dt = k * dh/dt
We are given that dh/dt = 28.0 centimeters per minute when h = 3.0 meters. We want to find dV/dt, the rate at which the volume is changing.
dV/dt = k * dh/dt
To find k, we can use the values of V and h when h = 14.0 meters.
V = k * h
18.66π = k * 14.0
Solving for k:
k = (18.66π) / 14.0
k = 1.333π
Now we can substitute the given values of k, dh/dt, and h into the equation dV/dt = k * dh/dt to find the rate at which the volume is changing.
dV/dt = (1.333π) * 28.0
dV/dt = 37.3π cubic centimeters per minute
So, the rate at which water is being pumped into the tank is 37.3π cubic centimeters per minute.