Water is leaking out of an inverted conical tank at a rate of 1100.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 15.0 meters and the diameter at the top is 5.0 meters. If the water level is rising at a rate of 300 centimeters per minute when the height of the water is 2.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Question

Water is leaking out of an inverted conical tank at a rate of 1100.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 15.0 meters and the diameter at the top is 5.0 meters. If the water level is rising at a rate of 300 centimeters per minute when the height of the water is 2.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Answer 1

Can you double check if the water going out is 1100 litres per minute (1.1 m³/min.) or 1100 cc (0.0011 m³/min)?

Answer 2

It's cubic centimeter per minute

Answer 3

Assume "inverted" conical tank is wider at the top, with a diameter of 5m.

At the base, r=0, h=0.
At the top, r=2.5, h=15
radius at height h:
r(h)=h/6
Area of surface of water, Aw
=πr(h)^2
=πh²/36

Equate input (Q m³/min.) and output (x):

dh/dt = (Q-x)/Aw

At h=2.5m,
dh/dt=300cm/min = 3m/min.
x=0.0011 m³/min.
Therefore
Q=(dh/dt)*Aw+x
=3 m/min. * πh²/36
=0.521π+0.0011 m³/min
=1.637 m³/min.

Please check my calculations.

Answer 4

To solve this problem, we need to use the concept of related rates. We can start by finding the relationship between the changing variables.

Let's assume that the radius of the water level in the conical tank is r and the height of the water level is h. We are given that the height of the tank is 15.0 meters and the diameter at the top is 5.0 meters.

From the given information, we know that the water level in the tank is rising at a rate of 300 centimeters per minute when the height is 2.5 meters. This gives us the related rate:

dh/dt = 300 cm/min

We also know that the volume of a cone is given by the formula V = (1/3)πr²h. Since the tank is inverted, the water fills the tank in the opposite direction. Therefore, the rate at which water is being pumped into the tank is the negative of the rate at which the water level is rising:

dV/dt = -dh/dt

Now, we can calculate the rate at which water is being pumped into the tank by relating the variables:

V = (1/3)πr²h

To express r in terms of h, we can observe that as the water level rises, the ratio of r to h remains constant. This ratio can be found using similar triangles.

Since the diameter of the top of the tank is 5.0 meters, the radius at the top is 2.5 meters. Therefore, the ratio of r to h is 2.5/15.0 = 1/6.

We can write r in terms of h:

r = (1/6)h

Now we can substitute this expression for r into the volume equation:

V = (1/3)π((1/6)h)²h
V = (1/108)πh³

Differentiating both sides of the equation with respect to t, we get:

dV/dt = (1/36)πh²(dh/dt)

Now, we can substitute the given values into the equation:

1100 = (1/36)π(2.5)²(dh/dt)

Simplifying the equation, we can solve for dh/dt:

dh/dt = (1100 * 36) / (π * 2.5²)
≈ 794.77 cm/min

So, the rate at which water is being pumped into the tank is approximately 794.77 cubic centimeters per minute.