a) The slope-intercept form of a linear equation, we get the following function rule: y = -45x + 225 .
b) the volume of water in Tank 2 with respect to time as the tank drains:
V = A(900 - 2.5kt + 1.25k^2)^2.
c) The intersection point is approximately (2.5, 450).
(a) Since the tank drains at a constant rate, the volume of water in the tank over time is a linear function.
We can use the slope-intercept form of a linear equation to model this function:
y = mx + b
where:
y is the volume of water in the tank in liters
m is the rate of pumping in liters per hour
x is the time in hours
b is the initial volume of water in the tank in liters
We know that the tank initially contains 225 liters of water and drains completely in 5 hours. Therefore, the initial volume of water in the tank is 225 liters and the rate of pumping is 225/5 = 45 liters per hour.
Substituting these values into the slope-intercept form of a linear equation, we get the following function rule:
y = -45x + 225
This function rule models the volume of water in Tank 1 with respect to time as the tank drains.
(b) Torricelli's Law states that the rate of flow of a liquid through an opening in the bottom of a tank is proportional to the square root of the height of the liquid in the tank.
Therefore, the volume of water in the tank over time is governed by the following differential equation:
dV/dt = -kV√h
where:
V is the volume of water in the tank in liters
t is the time in hours
k is a constant of proportionality
h is the height of the water in the tank in meters
To solve this differential equation, we can use separation of variables:
dV/√h = -k dt
Integrating both sides of the equation, we get:
2√h = -kt + C
where C is a constant of integration.
We know that the tank initially contains 900 liters of water and drains completely in 5 hours.
Therefore, the initial volume of water in the tank is 900 liters and the time it takes to drain the tank completely is 5 hours.
Substituting these values into the equation above, we can solve for C:
2√900 = -5k + C
C = 1800 + 5k
We can also substitute the initial volume of water in the tank and the time it takes to drain the tank completely into the equation above to get:
2√h = -kt + 1800 + 5k
h = (900 - 2.5kt + 1.25k^2)^2
This equation models the height of the water in Tank 2 as a function of time.
To get the volume of water in the tank as a function of time, we can substitute this equation into the following equation:
V = Ah
where A is the area of the base of the tank in square meters.
Therefore, the following function rule models the volume of water in Tank 2 with respect to time as the tank drains:
V = A(900 - 2.5kt + 1.25k^2)^2
where A is the area of the base of the tank in square meters and k is a constant of proportionality.
(c) The two tanks have the same volume at the intersection of the two graphs.
The intersection point is approximately (2.5, 450). This means that the two tanks have the same volume after 2.5 hours and the volume in each tank is 450 liters.
(d) Find the volume in each tank at each solution.
To find the times when the two tanks hold the same amount of water, we need to set the two function rules equal to each other and solve the resulting equation