Problem 65
Question
. Water leaks out of a 200-gallon storage tank (initially full) at the rate \(V^{\prime}(t)=20-t\), where \(t\) is measured in hours and \(V\) in gallons. How much water leaked out between 10 and 20 hours? How long will it take the tank to drain completely?
Step-by-Step Solution
Verified Answer
50 gallons leaked; the tank drains completely in 20 hours.
1Step 1: Understand the given rate of leakage
The rate of leakage is given by the function \( V^{\prime}(t) = 20 - t \), where \( t \) is in hours. This function tells us the amount of water leaking from the tank per hour.
2Step 2: Integrate the rate to find the volume function
To determine the total volume of water leaked over time, integrate the rate of leakage. The integral of \( V^{\prime}(t) = 20 - t \) is \( V(t) = \int (20 - t) \ dt = 20t - \frac{t^2}{2} + C \). The constant \( C \) will be determined based on initial conditions.
3Step 3: Determine the constant of integration
Use the initial condition that the tank is full at time \( t = 0 \), so \( V(0) = 200 \). Therefore, \( 200 = 20(0) - \frac{0^2}{2} + C \), which implies \( C = 200 \). Thus, the volume function is \( V(t) = 20t - \frac{t^2}{2} + 200 \).
4Step 4: Calculate the water leaked between 10 and 20 hours
The water leaked between 10 and 20 hours is given by \( V(20) - V(10) \). Calculate \( V(20) = 20(20) - \frac{20^2}{2} + 200 \) and \( V(10) = 20(10) - \frac{10^2}{2} + 200 \).
5Step 5: Simplify calculations
Calculate \( V(20) = 400 - 200 + 200 = 400 \) and \( V(10) = 200 - 50 + 200 = 350 \). Thus, the amount of water leaked is \( 400 - 350 = 50 \) gallons.
6Step 6: Determine time taken to completely drain the tank
Set \( V(t) = 0 \) for the complete drain. Solve \( 20t - \frac{t^2}{2} + 200 = 0 \) to find \( t \). This quadratic equation simplifies to \( t^2 - 40t + 400 = 0 \). Use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
7Step 7: Solve the quadratic equation
Using \( a = 1, b = -40, c = 400 \), calculate \( t = \frac{40 \pm \sqrt{(-40)^2 - 4 \times 1 \times 400}}{2} = \frac{40 \pm \sqrt{0}}{2} = 20 \). Thus, it takes 20 hours to drain the tank completely.
Key Concepts
Rate of ChangeIntegrationQuadratic EquationInitial Conditions
Rate of Change
The **Rate of Change** in a scenario describes how a quantity changes over time. In the water leak problem, the rate of change is provided by the derivative \( V'(t) = 20 - t \). Here, \( V'(t) \) is the rate at which water leaks out from the tank per hour.
This rate indicates how fast the tank is diminishing in volume with respect to time \( t \). As time progresses from hour to hour, the expression "20 - t" shows that initially, the water leaks out faster, at 20 gallons an hour.
However, as \( t \) increases, the rate decreases linearly. By 10 hours, the rate is down to 10 gallons per hour. Understanding such a rate of change is vital in predicting when the tank will be empty if no action is taken to stop the leak.
This rate indicates how fast the tank is diminishing in volume with respect to time \( t \). As time progresses from hour to hour, the expression "20 - t" shows that initially, the water leaks out faster, at 20 gallons an hour.
However, as \( t \) increases, the rate decreases linearly. By 10 hours, the rate is down to 10 gallons per hour. Understanding such a rate of change is vital in predicting when the tank will be empty if no action is taken to stop the leak.
Integration
**Integration** is a fundamental concept in calculus that allows us to determine the accumulated quantity from a rate of change. Applied to our problem, we use integration to find the volume of water that has leaked out over time.
The integral of the rate function \( V'(t) = 20 - t \) leads to the volume function \( V(t) = \int (20 - t) \, dt = 20t - \frac{t^2}{2} + C \).
Integration essentially "sums up" the infinitesimally small amounts of water leaking every instant over a given time frame, providing the total volume of water that has leaked.
The constant \( C \) in the integral represents an arbitrary constant that accounts for any initial quantity or value present before we count the accumulated changes. In our case, it is the initial amount of water in the tank.
The integral of the rate function \( V'(t) = 20 - t \) leads to the volume function \( V(t) = \int (20 - t) \, dt = 20t - \frac{t^2}{2} + C \).
Integration essentially "sums up" the infinitesimally small amounts of water leaking every instant over a given time frame, providing the total volume of water that has leaked.
The constant \( C \) in the integral represents an arbitrary constant that accounts for any initial quantity or value present before we count the accumulated changes. In our case, it is the initial amount of water in the tank.
Quadratic Equation
A **Quadratic Equation** is an equation of the form \( at^2 + bt + c = 0 \). In our problem, the function for water volume \( V(t) = 20t - \frac{t^2}{2} + 200 \) is represented by a quadratic expression.
To find out how long it takes for the tank to fully drain, we set \( V(t) = 0 \), yielding the quadratic equation \( \frac{t^2}{2} - 20t + 200 = 0 \).
This equation can be simplified to \( t^2 - 40t + 400 = 0 \), from which we solve for \( t \) using the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
Solving this formula requires calculating the discriminant \( b^2 - 4ac \). For our particular problem, the discriminant is zero, resulting in only one solution, \( t = 20 \). This means the tank drains out completely in 20 hours.
To find out how long it takes for the tank to fully drain, we set \( V(t) = 0 \), yielding the quadratic equation \( \frac{t^2}{2} - 20t + 200 = 0 \).
This equation can be simplified to \( t^2 - 40t + 400 = 0 \), from which we solve for \( t \) using the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
Solving this formula requires calculating the discriminant \( b^2 - 4ac \). For our particular problem, the discriminant is zero, resulting in only one solution, \( t = 20 \). This means the tank drains out completely in 20 hours.
Initial Conditions
When solving calculus problems involving integration, **Initial Conditions** are used to determine the constant of integration \( C \). These conditions tell us the status of the system at the beginning of the examined period.
In the water tank problem, the initial condition is that the tank is full at \( t = 0 \). Thus, when \( t = 0, V(0) = 200 \).
By substituting these into the equation \( V(t) = 20t - \frac{t^2}{2} + C \), we determine that \( C = 200 \).
Initial conditions set the stage for calculations, enforcing specific scenarios in physical, economic, or other processes that begin from a known state. They are crucial for solving integrals that model real-world systems, ensuring our solutions align with reality.
In the water tank problem, the initial condition is that the tank is full at \( t = 0 \). Thus, when \( t = 0, V(0) = 200 \).
By substituting these into the equation \( V(t) = 20t - \frac{t^2}{2} + C \), we determine that \( C = 200 \).
Initial conditions set the stage for calculations, enforcing specific scenarios in physical, economic, or other processes that begin from a known state. They are crucial for solving integrals that model real-world systems, ensuring our solutions align with reality.
Other exercises in this chapter
Problem 63
Let \(f\) be continuous on \([a, b]\) and thus integrable there. Show that $$ \left|\int_{a}^{b} f(x) d x\right| \leq \int_{a}^{b}|f(x)| d x $$
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