Problem 21
Question
Water leaks from a vertical cylindrical tank through a small hole in its base at a rate proportional to the square root of the volume of water remaining. If the tank initially contains 200 liters and 20 liters leak out during the first day, when will the tank be half empty? How much water will there be after 4 days?
Step-by-Step Solution
Verified Answer
The tank will be half empty in a little over 6 days. After 4 days, approximately 120 liters will remain.
1Step 1: Understand the problem
We have a cylindrical tank with water leaking out at a rate proportional to the square root of the remaining volume. Initially, the tank has 200 liters. We need to determine when half the initial amount (100 liters) remains and find the volume after 4 days.
2Step 2: Set up the differential equation
The rate of water leakage is given by:\[ \frac{dV}{dt} = -k\sqrt{V} \]where \( V \) is the volume of water at time \( t \) and \( k \) is a constant of proportionality.
3Step 3: Solve the differential equation
Separate variables and integrate:\[ \int \frac{1}{\sqrt{V}} \, dV = -k \int dt \]Solving these integrals results in:\[ 2\sqrt{V} = -kt + C \]where \( C \) is an integration constant to be determined.
4Step 4: Use initial conditions to find constants
Initially, \( t = 0 \), \( V = 200 \) liters, which gives:\[ 2\sqrt{200} = C \]Thus, \( C = 2\sqrt{200} = 20\sqrt{2} \).
5Step 5: Use the fact that 20 liters leak in the first day
At \( t = 1 \), \( V = 180 \) liters:\[ 2\sqrt{180} = -k(1) + 20\sqrt{2} \]Solve for \( k \) from:\[ 2\sqrt{180} - 20\sqrt{2} = -k \]This calculation shows the proportional constant \( k \).
6Step 6: Find when the tank will be half empty
Half empty means \( V = 100 \):\[ 2\sqrt{100} = -kt + 20\sqrt{2} \]Solve for \( t \) using the \( k \) value found in the previous step.
7Step 7: Determine the volume after 4 days
Use \( t = 4 \) in the derived equation:\[ 2\sqrt{V(4)} = -4k + 20\sqrt{2} \]Solve for \( V(4) \) to find how much water remains after 4 days.
Key Concepts
Cylindrical TankRate of ChangeIntegrationInitial Conditions
Cylindrical Tank
A cylindrical tank is a common shape used in various storage applications, particularly for storing liquids like water. These tanks are characterized by their circular cross-section and uniform depth, making them geometrically simple but functionally effective.
The exercise focuses on a vertical cylindrical tank, which implies that the tank’s circular base is horizontal and the sides extend vertically. This orientation ensures that as liquid leaves from the base, the water level drops evenly.
Understanding the cylindrical shape helps us visualize how the water level changes with time since any change in volume directly correlates with a change in height. This relationship is crucial when solving differential equations related to fluid dynamics in cylindrical tanks.
The exercise focuses on a vertical cylindrical tank, which implies that the tank’s circular base is horizontal and the sides extend vertically. This orientation ensures that as liquid leaves from the base, the water level drops evenly.
Understanding the cylindrical shape helps us visualize how the water level changes with time since any change in volume directly correlates with a change in height. This relationship is crucial when solving differential equations related to fluid dynamics in cylindrical tanks.
Rate of Change
In this exercise, the rate of change refers to how quickly the water level in the cylindrical tank decreases over time. The problem states that the rate of water leakage is proportional to the square root of the remaining water volume.
Mathematically, this is described through the differential equation: \[ \frac{dV}{dt} = -k\sqrt{V} \]where:
Mathematically, this is described through the differential equation: \[ \frac{dV}{dt} = -k\sqrt{V} \]where:
- \( \frac{dV}{dt} \) represents the rate of change of volume with respect to time.
- \( k \) is a proportionality constant.
- \( \sqrt{V} \) indicates that the rate depends on the square root of the current volume \( V \).
Integration
Integration is a fundamental concept in calculus often used to solve differential equations. In this problem, we need to integrate to find the relationship between time and the volume of water left in the tank.
Upon setting up the differential equation for the leakage rate, we separate variables by dividing both sides by \( \sqrt{V} \) and then integrate both sides:\[ \int \frac{1}{\sqrt{V}} \, dV = -k \int dt \]Integrating solves for the function \( V(t) \). Solving the left integral deals with the form \( \int x^{-1/2} \), which results in \( 2\sqrt{V} \). The expression becomes:\[ 2\sqrt{V} = -kt + C \] where \( C \) is the integration constant. This integration process helps us determine how variables are related and gives us the necessary equation to analyze the tank's behavior over time.
Upon setting up the differential equation for the leakage rate, we separate variables by dividing both sides by \( \sqrt{V} \) and then integrate both sides:\[ \int \frac{1}{\sqrt{V}} \, dV = -k \int dt \]Integrating solves for the function \( V(t) \). Solving the left integral deals with the form \( \int x^{-1/2} \), which results in \( 2\sqrt{V} \). The expression becomes:\[ 2\sqrt{V} = -kt + C \] where \( C \) is the integration constant. This integration process helps us determine how variables are related and gives us the necessary equation to analyze the tank's behavior over time.
Initial Conditions
Initial conditions are crucial for solving differential equations as they allow us to find unknown constants and make the solution specific to a given scenario. In this context, since the problem starts with 200 liters of water, we use this as our initial condition at \( t = 0 \).
This initial setting gives:\[ 2\sqrt{200} = C \] which simplifies to \( 20\sqrt{2} \). Such values define \( C \), ensuring that our general solution fits the specific scenario described.
The initial condition also helps in finding the constant \( k \) when additional conditions, such as the volume at a specific time (e.g., one day), are given. Knowing these aspects permits us to refine our equation to calculate the water's future state, like finding when the tank will be half empty or the remaining volume after 4 days.
This initial setting gives:\[ 2\sqrt{200} = C \] which simplifies to \( 20\sqrt{2} \). Such values define \( C \), ensuring that our general solution fits the specific scenario described.
The initial condition also helps in finding the constant \( k \) when additional conditions, such as the volume at a specific time (e.g., one day), are given. Knowing these aspects permits us to refine our equation to calculate the water's future state, like finding when the tank will be half empty or the remaining volume after 4 days.
Other exercises in this chapter
Problem 21
Warfarin is a drug used as an anticoagulant. After administration of the drug is stopped, the quantity remaining in a patient's body decreases at a rate proport
View solution Problem 21
Give the solution to the logistic differential equation with initial condition. $$\frac{d P}{d t}=0.8 P\left(1-\frac{P}{8500}\right) \text { with } P_{0}=500$$
View solution Problem 21
In Exercises \(2-28,\) use separation of variables to find the solutions to the differential equations subject to the given initial conditions. $$\frac{d y}{d t
View solution Problem 21
(a) For what values of \(C\) and \(n\) (if any) is \(y=C x^{n}\) a solution to the differential equation $$x \frac{d y}{d x}-3 y=0 ?$$ (b) If the solution satis
View solution