Water is leaking out of an inverted conical tank at a rate of 0.0142 m3/min. At the same time water is being pumped into the tank at a constant rate. The tank has height 13 meters and the diameter at the top is 3.5 meters. If the water level is rising at a rate of 0.17 m/min when the height of the water is 3.5 meters, find the rate at which water is being pumped into the tank.
let the height of the water level be h m
let the radius of the water level be r m
we are told r/h = 1.75/13 = 7/52
52r = 7h
r = 7h/52
let the rate at which water is pumped be x m^3/min
V = (1/3)π r^2 h
= (1/3)π(49h^2/2704)h
= (49π/8112) h^3
dV/dt = (49π/2704) h^2 dh/dt
for our given data
x - .0142 = (49π/2704)(3.5)^2 (.17)
I get x = .1327562.. metres/min
check my arithmetic
To solve this problem, we can use the concept of related rates. Let's assume that the radius of the conical tank at any given time is "r" meters, and the height of the water inside the tank at the same time is "h" meters.
Given information:
- The height of the tank (h) is 13 meters.
- The diameter at the top of the tank is 3.5 meters, which means the radius (r) of the tank is 3.5/2 = 1.75 meters.
- The rate at which water is leaking out of the tank is 0.0142 m^3/min.
- The rate at which water is being pumped into the tank is what we need to find.
- The rate at which the water level is rising (dh/dt) is 0.17 m/min.
- At the time when the water level is 3.5 meters, we can assume the radius of the water level is the same as the radius of the tank, which is 1.75 meters.
To find the rate at which water is being pumped into the tank, we can use the equation of the volume of a cone:
V = (1/3) * π * r^2 * h
Taking the derivative of both sides with respect to time (t), we get:
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Now, let's substitute the given values and find the rate at which water is being pumped into the tank:
Given:
dr/dt = 0 (since the radius of the tank is constant)
dh/dt = 0.17 m/min
h = 3.5 m
r = 1.75 m
dV/dt = (1/3) * π * (2 * 1.75 * 0 + (1.75)^2 * 0.17)
= (1/3) * π * (0 + 1.75^2 * 0.17)
= (1/3) * π * (0 + 0.8046875)
= (1/3) * π * 0.8046875
≈ 0.845 m^3/min
Therefore, the rate at which water is being pumped into the tank is approximately 0.845 m^3/min.
To find the rate at which water is being pumped into the tank, we need to understand the relationship between the volume of water in the tank and the rate of change of the height.
First, let's establish some notation. Let:
- V be the volume of water in the tank
- r be the radius of the water surface
- h be the height of the water in the tank
We are given that the height of the water is 3.5 meters, and it is rising at a rate of 0.17 m/min. This implies that dh/dt = 0.17 m/min.
Now, let's derive a relationship between V and h using the geometry of a cone. We know that the volume of a cone can be calculated using the formula:
V = (1/3) * π * r^2 * h
Since the diameter at the top of the tank is 3.5 meters, the radius is half of that, so r = 3.5/2 = 1.75 meters.
Now let's differentiate both sides of the equation with respect to time t:
dV/dt = (1/3) * π * [(2r * dr/dt * h) + (r^2 * dh/dt)]
We know that dh/dt = 0.17 m/min. We need to find the rate at which water is being pumped into the tank, dV/dt.
The volume of water leaking out of the tank is given as 0.0142 m3/min. Therefore, dV/dt = -0.0142 m3/min (since the water is being lost).
We can substitute the known values into the equation:
-0.0142 = (1/3) * π * [(2 * 1.75 * dr/dt * 3.5) + (1.75^2 * 0.17)]
Simplifying the equation:
-0.0142 = (1/3) * π * [12.25 * dr/dt + 0.5 * 0.17]
Next, we solve for dr/dt:
dr/dt = [-0.0142 / ((1/3) * π * 12.25)] - (0.5 * 0.17)
dr/dt ≈ -0.001095 m/min
Therefore, the rate at which water is being pumped into the tank is approximately 0.001095 m/min.