Problem 58 - Applied Calculus

Question

Pollution Contamination is leaking from an underground waste-disposal tank at the rate of \(t\) ln \(t\) thousand gallons per month, where \(t\) is the number of months since the leak began. Find the total leakage from the end of month 1 to the end of month 4 .

Step-by-Step Solution

Answer

The total leakage is approximately 7.338 thousand gallons.

Step 1: Identify the Problem

We need to calculate the total leakage between the end of month 1 and the end of month 4, given the rate of leakage is \( t \ln t \) thousand gallons per month.

Step 2: Determine the Formula

To find the total leakage, we should integrate the rate of leakage function, \( f(t) = t \ln t \), from \( t = 1 \) to \( t = 4 \).

Step 3: Set Up the Integral

The definite integral that represents the total leakage is given by \[ \int_{1}^{4} t \ln t \, dt \].

Step 4: Integration by Parts

For the integral \( \int t \ln t \, dt \), we can use integration by parts where \( u = \ln t \) and \( dv = t \, dt \). Then \( du = \frac{1}{t} \, dt \) and \( v = \frac{t^2}{2} \). Apply the integration by parts formula: \[ \int u \, dv = uv - \int v \, du \].

Step 5: Apply Integration by Parts

Plug in the values into the formula: \[ \int t \ln t \, dt = \frac{t^2}{2} \ln t - \int \frac{t^2}{2} \cdot \frac{1}{t} \, dt \].Simplify to:\[ = \frac{t^2}{2} \ln t - \frac{1}{2} \int t \, dt \].

Step 6: Evaluate the Integral

Further evaluating, the integral \( \frac{1}{2} \int t \, dt \) becomes \( \frac{1}{2} \cdot \frac{t^2}{2} = \frac{t^2}{4} \). Combine to get: \[ \int t \ln t \, dt = \frac{t^2}{2} \ln t - \frac{t^2}{4} + C \].

Step 7: Compute the Definite Integral

Evaluate the definite integral \[ \left[ \frac{t^2}{2} \ln t - \frac{t^2}{4} \right]_{1}^{4} \].First, substitute \( t = 4 \):\[ \frac{4^2}{2} \ln 4 - \frac{4^2}{4} = 8 \ln 4 - 4 \].Next, substitute \( t = 1 \):\[ \frac{1^2}{2} \ln 1 - \frac{1^2}{4} = 0 - \frac{1}{4} = -\frac{1}{4} \].Now find the difference:\[ \left(8 \ln 4 - 4\right) - \left(-\frac{1}{4}\right) = 8 \ln 4 - 4 + \frac{1}{4} \].

Step 8: Simplify the Result

Finally, combine terms and calculate the numerical value:\[ 8 \ln 4 - \frac{16}{4} + \frac{1}{4} = 8 \ln 4 - \frac{15}{4} \].Using \( \ln 4 \approx 1.386 \),\[ 8 \times 1.386 - 3.75 = 11.088 - 3.75 = 7.338 \].

Key Concepts

Integration by Parts

Integration by parts is a technique used in calculus to solve integrals that are products of two functions. It is particularly useful when one of the functions is easy to differentiate and the other is easy to integrate.To apply integration by parts, we use the formula:\[ \int u \, dv = uv - \int v \, du \]

• First, you choose which parts of the integral to assign to \( u \) and \( dv \). A common strategy is to set \( u \) to the function that becomes simpler when differentiated, such as \( \ln t \) in our case.
• Then, differentiate \( u \) to find \( du \), and integrate \( dv \) to find \( v \).
• Substitute these values into the integration by parts formula.

In this exercise, we applied this method to \( \int t \ln t \, dt \), where \( u = \ln t \) and \( dv = t \, dt \). The resulting integral became simpler and was easier to manage through further integration.

Pollution Leakage Calculation

Calculating pollution leakage involves determining the total amount of a substance that has leaked over a specific period. This process requires integrating the leakage rate over that duration.The exercise presents us with the leakage rate function \( t \ln t \), where \( t \) is in months since the leakage began. To find the total leakage from month 1 to month 4, we integrate this function:\[ \int_{1}^{4} t \ln t \, dt \]

• This integral gives us the total volume of leaked pollutant in thousands of gallons.
• The integration by parts technique simplifies the math and allows us to compute the definite integral over the given limits.
• Evaluating the result provides a numerical value representing the volume of pollution leaked during the specified period.

Such calculations are crucial in environmental management, helping identify and quantify potential issues efficiently.

Calculus Application in Real World

Calculus is not just a theoretical branch of mathematics but an essential tool in many real-world applications. It helps us solve problems related to:

• Environmental management such as calculating the total pollution leaking over months, helping authorities to take timely action.
• Physics for determining change over time, such as velocity and acceleration.
• Economics for optimizing costs and efficiency by modeling changes in economic variables.
• Medicine for analyzing the growth of bacteria or the decay of substances in pharmacology.

In this exercise about pollution leakage, calculus allowed us to use integration to sum up the effects of leakage over time, showing its practical application in addressing environmental concerns. Understanding and applying these calculus concepts can provide powerful insights and solutions in various fields.