Problem 98 - Applied Calculus

Question

After \(t\) hours of work, a medical technician can carry out T-cell counts at the rate of \(2 t^{2} e^{-t / 4}\) tests per hour. How many tests will the technician process during the first eight hours (time 0 to time 8 )?

Step-by-Step Solution

Answer

The technician processes approximately 204 tests in the first eight hours.

Step 1: Identify the function

The rate of T-cell counts per hour is given by the function \( f(t) = 2t^2 e^{-t/4} \). We need to find the total number of tests performed in the first 8 hours.

Step 2: Set up the integral

To find the total number of tests performed, we calculate the definite integral of the rate function from \( t = 0 \) to \( t = 8 \). The integral is given by:\[ \int_{0}^{8} 2t^2 e^{-t/4} \, dt \].

Step 3: Integrate by parts setup

Apply integration by parts: Choose \( u = t^2 \) and \( dv = 2e^{-t/4} dt \). Then compute \( du = 2t \, dt \) and \( v = -8e^{-t/4} \).

Step 4: Apply integration by parts formula

Using the integration by parts formula, \( \int u \, dv = uv - \int v \, du \), compute:\[ uv = -8t^2 e^{-t/4} \] and \[ \int v \, du = \int -16t e^{-t/4} \, dt \].

Step 5: Second integration by parts

For \( \int -16t e^{-t/4} \, dt \), use integration by parts again. Set \( u = t \) and \( dv = -16e^{-t/4} dt \), then \( du = dt \) and \( v = 64e^{-t/4} \).

Step 6: Solve second integration by parts

Apply the formula again: \[ uv = 64t e^{-t/4} \]\[ \int v \, du = \int 64 e^{-t/4} \, dt = -256 e^{-t/4} \].

Step 7: Combine results and evaluate

Combine results from steps 4 and 6: the integral becomes \[ -8t^2 e^{-t/4} - (64t e^{-t/4} - 256e^{-t/4}) \].Simplify and combine these terms, evaluate from \( t = 0 \) to \( t = 8 \).

Step 8: Calculate definite integral

Plug in the limits and evaluate the expression:At \( t=8 \), compute: \[ -8(8)^2 e^{-8/4} - (64(8) e^{-8/4} + 256e^{-8/4}) \].At \( t=0 \), the terms simplify as the exponential term goes to one, ultimately resulting in \( 0 \) after simplification. Use these to find the final result.

Step 9: Simplify and calculate result

Perform the calculations from the previous step to find the exact number of tests processed, totaling all terms evaluated.

Key Concepts

Understanding Integration by Parts

Integration by parts is a useful technique in calculus utilized to integrate products of two functions. It is particularly effective when dealing with integrals where the product of functions complicates the process of direct integration. This technique is derived from the product rule of differentiation and is expressed with the formula:

• \( \int u \, dv = uv - \int v \, du \)

In our exercise, we apply integration by parts twice to handle the complex product of the polynomial function \( t^2 \) and the exponential function \( e^{-t/4} \). To employ this method effectively:

• Choose one part of the product to differentiate (let's call it \( u \)) and another to integrate (denoted as \( dv \)).
• Differentiate \( u \) to find \( du \) and integrate \( dv \) to get \( v \).
• Plug these into the integration by parts formula to simplify and solve the original integral.

This systematic approach breaks down the problem into smaller, manageable integrals, as seen in our solution.

Calculating Definite Integrals

A definite integral represents the accumulation of quantities, which in this context is the total number of T-cell counts performed over a specified time interval. The process involves:

• Setting the boundaries of integration, which are from \( t = 0 \) to \( t = 8 \) in this exercise.
• Evaluating the antiderivative of the function at these bounds, essentially giving the area under the curve of the function within the specified interval.

The definite integral is symbolized as \( \int_{a}^{b} f(t) \, dt \), where \( f(t) \) is the function being integrated. Here, it efficiently calculates the total tests performed by evaluating \( 2t^2 e^{-t/4} \) from \( t = 0 \) to \( t = 8 \). Evaluating definite integrals entails finding the antiderivative and calculating the difference in its values at the upper and lower limits, providing the total quantity in question.

Exploring Rate of Change

The rate of change is a fundamental concept in calculus, describing the speed at which a variable changes over a specific time. In this problem, the rate of T-cell counts is given by the function \( 2t^2 e^{-t/4} \). This model uses the variable \( t \) representing time in hours to quantify how many tests the technician completes per hour.

• The rate of change indicates the immediate rate at which T-cell counts increase or decrease as hours progress.
• A positive rate indicates an increase in T-cell counts per hour over time.

Understanding the function's behavior over the interval \( 0 \leq t \leq 8 \) requires calculating its definite integral. This helps in understanding the cumulative effect of this rate. Essentially, the rate of change describes how dynamic the situation is, while integration focusses on these changes' additive effect over time, thereby providing the total output of T-cell counts over the specified duration.