Problem 85 - Calculus: Early Transcendentals

Question

Dialysis treatment removes urea and other waste products from a patient's blood by diverting some of the bloodflow externally through a machine called a dialyzer. The rate at which the urea is removed from the blood (in mg/min) is often well described by the equation $$ u(t) = \frac{r}{V} C_0 e^{-rt/V} $$ where \( r \) is the rate of flow of blood through the dialyzer (in mL/min), \( V \) is the volume of the patient's blood (in mL), and \( C_0 \) is the amount of urea in the blood (in mg) at time \( t = 0 \). Evaluate the integral \( \displaystyle \int^{30}_0 u(t) \, dt \) and interpret it.

Step-by-Step Solution

Answer

The total urea removed is \( \frac{C_0 V}{r}(1 - e^{-\frac{30r}{V}}) \).

Step 1: Understand the Integral

The given function is \( u(t) = \frac{r}{V} C_0 e^{-rt/V} \), which represents the rate of urea removal over time. We are asked to evaluate \( \int_0^{30} u(t) \, dt \). This integral calculates the total amount of urea removed from the patient's blood over 30 minutes.

Step 2: Set Up the Integral

Set up the integral to evaluate: \[ \int_0^{30} \frac{r}{V} C_0 e^{-rt/V} \, dt \].

Step 3: Apply Integration Techniques

The function inside the integral is an exponential function, so use substitution to solve it. Let \( u = -\frac{rt}{V} \), giving \( \frac{du}{dt} = -\frac{r}{V} \), hence \( dt = -\frac{V}{r} du \).

Step 4: Substitute and Change Limits

Change the limits of integration based on the substitution. When \( t = 0 \), \( u = 0 \). When \( t = 30 \), \( u = -\frac{30r}{V} \). Substitute into the integral: \[ C_0 \int_0^{30} \frac{r}{V} e^{-rt/V} \, dt = C_0 \int_0^{-rac{30r}{V}} e^u \left( -\frac{V}{r} \right) \, du \].

Step 5: Solve the Integral

Solve the integral: \[ = -C_0 \frac{V}{r} \int e^u \, du = -C_0 \frac{V}{r} e^u \Bigg|_0^{-\frac{30r}{V}} \].

Step 6: Evaluate the Definite Integral

Evaluate the expression: \[ -\frac{C_0 V}{r} \left( e^{-\frac{30r}{V}} - e^0 \right) = \frac{C_0 V}{r}(1 - e^{-\frac{30r}{V}}) \].

Step 7: Interpretation of the Result

The resulting expression, \( \frac{C_0 V}{r}(1 - e^{-\frac{30r}{V}}) \), represents the total amount of urea removed from the blood over 30 minutes during the dialysis process. This takes into account the rate of flow, initial urea concentration, and blood volume.

Key Concepts

Integration Techniques

Integration techniques are essential for solving calculus problems involving functions that change over time. In the context of calculus applications, integration is used to find the area under a curve, which can represent many real-world situations like total accumulation.

When dealing with complex functions, substitution is a common method to simplify integration. This technique involves identifying a part of the function to substitute with a new variable, transforming the integral into a simpler form. In our exercise, we used substitution to change the variables of integration to better handle the exponential decay function for urea removal.

• Choose a substitution: Identify an expression within the integral to replace.
• Find the differential: Calculate the differential of your new variable.
• Adjust limits: Recalculate the limits if dealing with definite integrals.
• Solve the transformed integral: Integrate using the new variable, often finding it more manageable.

These techniques simplify the problem and make calculus more applicable to real-world situations, such as the urea removal rate.

Exponential Functions

Exponential functions are a type of mathematical expression where a constant base is raised to a variable exponent. They are vital in representing natural processes where changes occur proportionally over time, like radioactive decay or population growth.

In our context, the function for urea removal involves an exponential component, written as \( e^{-rt/V} \). This expression indicates that the urea removal rate decreases exponentially over time.

• The base \( e \) is a constant approximately equal to 2.718 and is the natural logarithm base.
• The exponent dictates the rate of decay or growth, shaped by parameters like blood flow rate \( r \) and blood volume \( V \).
• Exponential decay occurs when the exponent is negative, depicting a decrease over time.

Understanding these concepts allows us to interpret the significant change dynamics in natural processes, vital for analyzing the efficacy of medical treatments like dialysis.

Definite Integrals

Definite integrals are used to calculate the accumulated value over a specific interval, represented by the area under a curve between the given limits. In calculus applications, they are crucial for determining quantities where the rate of change is not constant.

In our exercise, the definite integral \( \int_0^{30} u(t) \ dt \) helps us find the total amount of urea removed from the blood during dialysis over 30 minutes.

• Lower and upper limits denote the start and end of the interval; here, it's 0 to 30 minutes.
• The outcome expresses a total value, different from the indefinite integral that results in a function.
• Calculating definite integrals often involves evaluating the antiderivative at both limits and finding the difference.

This method provides a vital tool for analyzing real-world problems, allowing for precision in calculating total quantities like urea removal in a medical process.

Urea Removal Rate

The urea removal rate in dialysis is a measure of how effectively the dialysis process cleans the patient's blood by removing waste products like urea. It is crucial for assessing the treatment's success and patient health.

The function \( u(t) = \frac{r}{V} C_0 e^{-rt/V} \) describes how this rate changes over time, illustrating the decrease in urea as blood passes through a dialyzer.

• \( r \) is the blood flow rate through the dialyzer, impacting how quickly the blood is cleaned.
• \( V \) is the total blood volume; larger volumes mean more urea to remove and can affect treatment time.
• \( C_0 \) is the initial concentration of urea at the start of treatment, influencing the total urea to be removed.

The knowledge of these parameters allows healthcare providers to customize dialysis treatment and improve patient outcomes.

Dialysis Process

The dialysis process is a medical treatment that filters waste products from the blood when the kidneys are not functioning properly. It replicates the kidney's function by removing toxins and maintaining safe levels of chemicals in the body.

The process described in the exercise is typical of hemodialysis, where blood is diverted through a machine containing a dialyzer. This machine utilizes semipermeable membranes to separate waste from blood cells.

• Blood flows through the dialyzer, where urea and other waste pass through the membrane.
• Cleaned blood returns to the body, ensuring the removal of excess water and toxins.
• Dialysis requires careful monitoring of fluid levels and chemical balances.

This treatment is crucial for patients with kidney failure, ensuring they remain healthy by simulating the natural filtration process.