Water is leaking out of an inverted conical tank at a rate of 0.0109 {\rm m}^3{\rm /min}. At the same time water is being pumped into the tank at a constant rate. The tank has height 9 meters and the diameter at the top is 4.5 meters. If the water level is rising at a rate of 0.24 m/min when the height of the water is 4 meters, find the rate at which water is being pumped into the tank. Must be accurate to the fifth decimal place.
I keep getting 3.0268 but WebWorks will not accept that answer.
To find the rate at which water is being pumped into the tank, we can use the principle of conservation of volume. The rate of change of volume in the tank is equal to the sum of the rate at which water is being pumped into the tank and the rate at which water is leaking out of the tank. We can set up an equation as follows:
Rate of Change of Volume = Rate of Inflow - Rate of Outflow
Let's denote the height of the water in the tank as "h" and the radius of the water level as "r". We are given that the height of the tank is 9 meters and the diameter at the top is 4.5 meters. Therefore, the radius of the tank is 4.5/2 = 2.25 meters.
We are also given the rates at which water is leaking out of the tank and the rate at which the water level is rising. Let's denote the rate of water leakage as "L" and the rate of water level rise as "R".
Given values:
L = 0.0109 m^3/min (rate of water leakage)
R = 0.24 m/min (rate of water level rise)
Now, let's find expressions for the rate of inflow and the rate of outflow:
Rate of Inflow: The volume of water being pumped into the tank per unit time is equal to the cross-sectional area of the tank multiplied by the rate of water level rise. Since the tank has a conical shape, the cross-sectional area of the water level is a function of height. We can use similar triangles to find this relationship.
At height "h", the corresponding radius "r" can be found using similar triangles:
r/h = 2.25/9
r = (2.25/9) * h = 0.25h
Therefore, the cross-sectional area of the water level is:
A = πr^2 = π(0.25h)^2 = 0.0625πh^2
So, the rate of inflow is:
Rate of Inflow = A * R = 0.0625πh^2 * R
Rate of Outflow: The rate of water leakage is given as L = 0.0109 m^3/min.
Now, let's substitute the given values and the expressions for the rate of inflow and outflow into our conservation equation and solve for the rate of water being pumped into the tank:
Rate of Change of Volume = Rate of Inflow - Rate of Outflow
0 = 0.0625πh^2 * R - 0.0109
Solving for R gives:
R = 0.0109 / (0.0625πh^2)
Substituting the given height of the water level as 4 meters:
R = 0.0109 / (0.0625π(4)^2)
R = 0.0109 / (0.0625π(16))
R = 0.0109 / (1π)
R = 0.0109 / 3.14159
R ≈ 0.0034667
Therefore, the rate at which water is being pumped into the tank is approximately 0.00347 m^3/min, accurate to the fifth decimal place.
To solve this problem, we can apply similar triangles and the volume formula of a cone. Let's break down the problem into steps:
Step 1: Find the rate at which the water level is changing (dh/dt).
Given that the height of the water (h) is changing at a rate of 0.24 m/min, we have dh/dt = 0.24 m/min.
Step 2: Find the rate at which the radius of the water surface is changing (dr/dt).
Since the tank is conical, the radius (r) is related to the height (h) by similar triangles. The ratio of the radius to the height remains constant.
Given that the diameter at the top is 4.5 meters, the radius at the top is 4.5/2 = 2.25 meters.
So, at any height h, the radius r can be found using the proportion:
(2.25 meters)/(9 meters) = (r meters)/(h meters)
Simplifying, we get: r = (2.25/9) * h
To find the rate at which the radius is changing (dr/dt), we differentiate both sides of the equation with respect to time:
dr/dt = (2.25/9) * (dh/dt)
Substituting dh/dt = 0.24 m/min, we get: dr/dt = (2.25/9) * 0.24
Step 3: Find the volume of the cone V and its rate of change dV/dt.
The volume of a cone can be calculated using the formula: V = (1/3) * π * r^2 * h
Substituting the expression for r in terms of h, we have: V = (1/3) * π * [(2.25/9) * h]^2 * h
Simplifying, we get: V = (1/27) * π * h^3
To find the rate of change of the volume (dV/dt), we differentiate both sides of the equation with respect to time:
dV/dt = (1/27) * π * 3 * h^2 * (dh/dt)
Substituting π = 3.14159 and dh/dt = 0.24 m/min, we get: dV/dt = (1/27) * 3.14159 * 3 * h^2 * 0.24
Step 4: Find the rate at which water is being pumped into the tank.
The rate at which water is being pumped into the tank is given by dV/dt.
Now, let's substitute the given values into the formulas to find the rate at which water is being pumped into the tank:
dr/dt = (2.25/9) * 0.24
dr/dt = 0.06
dV/dt = (1/27) * 3.14159 * 3 * (4)^2 * 0.24
dV/dt = 0.2688π
Therefore, the rate at which water is being pumped into the tank is 0.2688π m^3/min.