Develop a research plan in order to find out the time it takes the water tank in supplying the school garden.
A rectangular water tank was built with square base 3m long and height 5m. Water tank is initially full of water: water will leak out through the base tap for irrigation at a rate proportional to the total area in contact with water. When the depth of water is 4m, water is filling at the rate of 0.4m/h.
H denotes the depth of water in tank after t, hours.
A. Using Calculus, model this by a differential equation, stating clearly the variables considered. (15marks)
B. Solve the differential equation in [A] giving H in terms of t. Find how long it takes the tank to be half full 2. Find how long the tank irrigates the school garden [10]
C. Give the recommendations on the size of the school garden or side of the water tank or rate of irrigating
surely the area of water in contact with the sides cannot affect tha flow rate.
And since the base is rectangular, the area in contact is constant.
Am I missing something here? What is the shape of the base tap, and where is it located?
A. To model the situation using calculus, let's consider the following variables:
- H: represents the depth of water in the tank after t hours.
- A: represents the total area in contact with water in the tank.
According to the problem, the rate at which water is leaking is proportional to the total area in contact with water. Hence, we can express it using the derivative:
dH/dt = -k * A
Now, let's find an expression for A in terms of H. The tank has a square base with a side length of 3m, and the height of the tank is 5m. So, the area of the base is A_base = (3m)^2 = 9m^2, and the area of each of the four vertical sides is A_sides = 3m * 5m = 15m^2.
Since the tank is rectangular, the area in contact with the water changes depending on the water depth. By considering the water depth H, we can express the area A in terms of H:
A = A_base + 2 * A_sides = 9m^2 + 2 * (3m * H) = 9m^2 + 6mH
Thus, the differential equation becomes:
dH/dt = -k * (9m^2 + 6mH)
B. To solve the differential equation, we need to integrate both sides:
∫ (1/(9m^2 + 6mH)) dH = -k * ∫ dt
After integration, we get:
(1/6m) * ln|9m^2 + 6mH| = -kt + C
Where C is the constant of integration. Let's solve for H:
ln|9m^2 + 6mH| = -6kmt + C'
Where C' is another constant obtained by merging constants.
Now, let's solve for H:
|9m^2 + 6mH| = e^(-6kmt + C')
We can drop the absolute value since the tank is initially full (H = 5m) and the depth cannot be negative. Solving for H:
9m^2 + 6mH = e^(-6kmt + C')
H = (e^(-6kmt + C') - 9m^2) / 6m
Now, let's address the questions:
1. To find the time it takes for the tank to be half full (H = 2.5m), we substitute H = 2.5 and solve for t:
2.5 = (e^(-6kmt + C') - 9m^2) / 6m
Solving the equation for t will give you the required time.
2. To find how long the tank irrigates the school garden, we need to find the time it takes for the tank to become empty (H = 0m). Set H = 0 and solve for t.
C. Based on the rate of irrigating and the desired amount of watering, we can adjust the size of the school garden or side of the water tank. If the tank isn't providing enough water, increasing the tank's size could help. Similarly, if the tank is providing an excessive amount of water, reducing the tank's size would be appropriate. Additionally, adjusting the rate of irrigation can also control the amount of water supplied to the garden. These adjustments depend on the specific requirements and needs of the school garden.
Research Plan:
1. Define the variables:
- H: Depth of water in the tank after t hours (in meters)
- t: Time elapsed (in hours)
2. Formulate Differential Equation:
Based on the information given, the rate of water filling up the tank is 0.4m/h when the depth of water is 4m. We can assume that the leak rate is directly proportional to the total area in contact with water. Since the tank has a square base, the total area in contact with water is H^2 (square meters).
Thus, the rate of water leaking out of the tank is k * H^2, where k is a constant.
The rate of change of the depth of water in the tank is given as the difference between the rate of water filling up and the rate of water leaking out:
dH/dt = 0.4 - k * H^2
3. Solve the Differential Equation:
Rearranging the equation, we get:
dH/(0.4 - k * H^2) = dt
Integrate both sides with respect to their respective variables:
∫dH/(0.4 - k * H^2) = ∫dt
This integration will give us the solution to the differential equation, H in terms of t.
4. Find the time it takes for the tank to be half full:
Substitute H = 5/2 (half of the height of the tank) into the equation obtained in step 3. Solve for t to find how long it takes for the tank to be half full.
5. Find how long the tank can irrigate the school garden:
Substitute H = 0 (empty tank) into the equation obtained in step 3. Solve for t to find how long the tank can irrigate the school garden.
6. Recommendations on the size of the school garden or side of the water tank or rate of irrigating:
- To increase the time the tank can irrigate the school garden, the size of the water tank can be increased. This can be achieved by increasing the side length of the square base or the height of the tank.
- To decrease the time it takes for the tank to be half full, the rate of water filling up the tank can be increased. This can be achieved by increasing the leak rate constant (k) or the rate of water input.
- The size of the school garden should be based on the desired irrigation duration and the rate at which the water tank fills up. Adjustments may be made based on the available resources and the amount of water required for irrigation.
Research Plan:
1. Determine the necessary variables: In order to find out the time it takes for the water tank to supply the school garden, we need to consider the following variables:
- H: Depth of water in the tank after time t (hours)
- t: Time in hours
- A: Total area in contact with water
- R: Rate of leakage through the base tap for irrigation
- V: Volume of water in the tank
2. Use calculus to model the situation by a differential equation: We can use the concept of related rates to model the rate of change of the depth of water in the tank with respect to time. Since the rate of leakage is proportional to the total area in contact with water, we can write the following differential equation:
- dH/dt = -R * A
3. Solve the differential equation: To find the solution to the differential equation, we need to determine the constant of proportionality R and the total area in contact with water A. This information is not provided in the given problem statement. It is possible that additional data or assumptions are required to solve the differential equation and determine these values.
4. Calculate time for the tank to be half full: Once the differential equation has been solved and the values of R and A are determined, we can substitute these values into the equation and solve for the time it takes for the water tank to be half full (H = 2.5m).
5. Calculate the time for the tank to irrigate the school garden: We need to determine the time it takes for the water level to reach zero (H = 0m) in order to calculate the time for the tank to irrigate the school garden. This can be done by solving the differential equation for H = 0.
6. Make recommendations: Based on the calculations and the desired time for the tank to irrigate the school garden, recommendations can be made regarding the size of the school garden, the side length of the water tank, or the rate of irrigating. These recommendations will depend on the specific requirements and constraints of the project.