Problem 23 - Calculus Early Transcendentals

Question

A tank with a 1000 gal capacity initially contains 500 gal of water that is polluted with 50 lb of particulate matter. At time \(t=0,\) pure water is added at a rate of \(20 \mathrm{gal} / \mathrm{min}\) and the mixed solution is drained off at a rate of \(10 \mathrm{gal} / \mathrm{min}\). How much particulate matter is in the tank when it reaches the point of overflowing?

Step-by-Step Solution

Answer

There are 25 lb of particulate matter in the tank when it overflows.

Step 1: Determine the volume of the water in the tank

Initially, the tank contains 500 gallons of water. At time \( t = 0 \), water is added at a rate of 20 gallons per minute, and it is being drained at 10 gallons per minute. Therefore, the net rate at which water is added is \( 20 - 10 = 10 \) gallons per minute. We will let \( V(t) \) be the volume of the water at time \( t \). Thus, \( V(t) = 500 + 10t \). The tank overflows when \( V(t) = 1000 \) gallons.

Step 2: Calculate overflow time

To find the time \( t \) when the tank is full, set \( V(t) = 1000 \). Therefore, we solve: \[ 500 + 10t = 1000 \]Solving for \( t \), we get:\[ 10t = 500 \]\[ t = 50 \] minutes.

Step 3: Determine differential equation for particulate matter

Let \( P(t) \) be the amount of particulate matter in the tank at time \( t \). The rate of change of particulate matter in the tank is given by the rate of outgoing particulate matter. The concentration of particulate matter is \( \frac{P(t)}{500 + 10t} \) lb/gal. Since the mixture is drained at 10 gallons per minute, the rate of change is:\[ \frac{dP}{dt} = -10 \times \frac{P(t)}{500 + 10t} \].

Step 4: Solve the differential equation

The differential equation is separable:\[ \frac{dP}{P} = -\frac{10}{500 + 10t} \, dt \]Integrating both sides:\[ \ln|P| = -\ln|500 + 10t| + C \]Exponentiating both sides yields:\[ P(t) = C(500 + 10t)^{-1} \].

Step 5: Apply initial condition to find C

Given that initially, \( P(0) = 50 \) lb, substitute into the equation:\[ 50 = C(500 + 10 \cdot 0)^{-1} \]\[ 50 = \frac{C}{500} \]\[ C = 25000 \].

Step 6: Find amount of particulate matter when tank overflows

Substitute \( t = 50 \) minutes into the equation for \( P(t) \):\[ P(50) = 25000(500 + 10 \cdot 50)^{-1} \]\[ P(50) = 25000(1000)^{-1} \]\[ P(50) = 25 \] lb.

Key Concepts

Particulate Matter

In environmental science, particulate matter (PM) refers to tiny particles and droplets in the air that can harm human health and the environment. These particles may include dust, dirt, soot, and smoke. In the context of the problem, we are dealing with particulate matter mixed in water within a tank. Initially, the tank contains 50 lb of particulate matter distributed in 500 gallons of water.
Understanding how particulate matter behaves in systems like these is crucial in various fields, including environmental engineering and chemistry. Here, as water is continuously added and drained, the concentration of particulate matter changes, affecting overall pollution levels. This dynamic process is modeled using calculus, particularly differential equations.
By accurately determining how the particulate concentration changes over time, we can predict and manage pollution levels effectively. This is crucial for maintaining safe and healthy environmental standards.

Rate of Change

The rate of change is a fundamental concept in calculus, describing how a quantity varies concerning another. In this scenario, we're looking at how the amount of particulate matter changes over time. To understand it better, consider:

• The inflow rate: Pure water is added at 20 gallons per minute, increasing the volume of the solution.
• The outflow rate: The mixture is drained at 10 gallons per minute, which affects the concentration of particulate matter.

We find that the net inflow rate is the difference between these two, leading to a change in the tank's volume at a rate of 10 gallons per minute.
With the volumetric rate of change established, the concentration of particulate matter alters accordingly. As time progresses, the rate at which particulate matter exits the tank depends on the concentration at that instant, providing a real-time insight into the shifting balance between incoming clean water and the outgoing polluted mixture.

Separable Differential Equations

Understanding separable differential equations is vital for solving real-world problems where variables are interdependent. In simple terms, such equations can be rewritten so that all terms involving the same variable are on one side of the equation.
In our example, we apply this method to find the concentration of particulate matter over time. The differential equation describing the process is:\[ \frac{dP}{dt} = -10 \times \frac{P(t)}{500 + 10t} \]This equation is separable, meaning we can rearrange it as:\[ \frac{dP}{P} = -\frac{10}{500 + 10t} \, dt \]By integrating both sides, we find a relationship between the amount of particulate matter and time. This allows us to solve for the constant using given initial conditions, thus providing a complete model of the system.Ultimately, successfully solving separable differential equations like this one enables us to tailor our model to predict real-world behaviors accurately. These solutions are crucial, especially in scenarios where environmental and health regulations rely on controlled substance levels.