Water is leaking out of an inverted conical tank at a rate of 9,500 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank. (Round your answer to the nearest integer.)
You know that when the water has depth y, the surface of the water has radius (4/6)y
So, at depth y, the volume of water is
v = π/3 r^2 y = π/3 (4/9) y^3
dv/dt = 4π/9 y^2 dy/dt
That's if no water is draining out. SO, adjust dv/dt when you get your answer above.
To find the rate at which water is being pumped into the tank, we can use the principle of related rates.
Let's denote the height of the water in the tank as h and the radius of the water at level h as r. Since the tank is inverted, the relationship between the height and the radius can be described by similar triangles.
Using similar triangles, we have the following equation:
h / r = (6 - h) / (4 / 2)
Simplifying this equation, we find:
2h = 12 - 2h
Combining like terms, we get:
4h = 12
Dividing both sides by 4, we obtain:
h = 3
Now, let's define V(t) as the volume of water in the tank at time t. The rate at which water is being pumped into the tank is given by dV(t) / dt.
We are given that the water level is rising at a rate of 20 cm/min when the height of the water is 2 m. Since the height is given in meters, we need to convert the given rate to the same units:
20 cm/min * (1 m / 100 cm) = 0.2 m/min
We can now set up an equation using related rates:
dV(t) / dt - 9500 cm^3/min = 0.2 m/min * (pi * r^2)
To solve for the rate at which water is being pumped into the tank, we need to find the value of r when h = 2 m.
Using the equation h / r = (6 - h) / (4 / 2) and substituting h = 2, we get:
2 / r = (6 - 2) / (4 / 2)
Simplifying, we find:
2 / r = 8 / 4
Cross-multiplying, we obtain:
8r = 8
Dividing by 8, we get:
r = 1
Now, substituting h = 2 m and r = 1 into the related rates equation, we have:
dV(t) / dt - 9500 cm^3/min = 0.2 m/min * (pi * (1)^2)
Simplifying further, we find:
dV(t) / dt - 9500 cm^3/min = 0.2 * pi m^3/min
To convert the units, we know that:
1 m^3 = 1000000 cm^3
So, we have:
0.2 * pi m^3/min = (0.2 * pi * 1000000) cm^3/min
Simplifying, we get:
0.2 * pi m^3/min = 200000 * pi cm^3/min
Finally, we can solve for dV(t) / dt by adding 9500 cm^3/min to both sides:
dV(t) / dt = 9500 cm^3/min + 200000 * pi cm^3/min
Calculating this value, we find:
dV(t) / dt ≈ 631853 cm^3/min
Therefore, the rate at which water is being pumped into the tank is approximately 631853 cm^3/min (rounded to the nearest integer).