Problem 74 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

A liquid carries a drug into an organ of volume \(V \mathrm{~cm}^{3}\) at the rate of \(a \mathrm{~cm}^{3} / \mathrm{sec}\) and leaves at the same rate. The concentration of the drug in the entering liquid is \(c \mathrm{~g} / \mathrm{cm}^{3}\). Letting \(x(t)\) denote the concentration of the drug in the organ at any time \(t\), we have \(x(t)=c\left(1-e^{-a t / V}\right)\). a. Show that \(x\) is an increasing function on \((0, \infty)\). b. Sketch the graph of \(x\).

Step-by-Step Solution

Answer

a) Based on the first derivative of x(t), \(\frac{dx}{dt}\) = -c\(\frac{a}{V}e^{-at/V}\), we observe that the first derivative is negative for all \(t>0\). Hence, the function x(t) is an increasing function on the interval (0, ∞). b) The graph of x(t) = \(c(1-e^{-a t / V})\) is a saturating exponential curve, starting at (0,0) and approaching c as t -> ∞. This implies that it is an increasing concave function within the given interval.

Step 1: Find the first derivative of x(t)

To determine if the function is increasing or decreasing, we need to find its first derivative and identify the critical points. Given x(t) = \(c(1-e^{-a t / V})\), let's find the derivative: \(\frac{dx}{dt}\) = \(\frac{d}{dt}\left(c(1-e^{-a t / V})\right)\) Using the chain rule, \(\frac{d}{dt}\left(e^{-at/V}\right) = -\frac{a}{V}e^{-at/V}\), we have: \(\frac{dx}{dt}\) = \(c(-\frac{a}{V}e^{-at/V})\).

Step 2: Determine critical points for the function

To determine if the function is increasing, we need to analyze its first derivative in the given interval (0, ∞). Since the concentration of the drug, c, is positive and \(\frac{-a}{V}e^{-at/V}\) has always a negative value, (\(-\frac{a}{V} <0\) and \(e^{-at/V} >0\)), the first derivative of x(t) can be expressed as follows: \(\frac{dx}{dt}\) = -c\(\frac{a}{V}e^{-at/V}\), where \(\frac{dx}{dt}\) is negative for all \(t>0\). Since the first derivative is negative for all positive values of t, the function x(t) is an increasing function in the interval (0, ∞).

Step 3: Sketch the graph of x(t)

Now, let's sketch the graph of the function x(t) = \(c(1-e^{-a t / V})\): 1) As t approaches infinity, we have \(e^{-\frac{a t}{V}}\) -> 0, which means x(t) -> c. 2) At t=0, \(x(0)=c(1-e^{0})=c(1-1)=0\) Considering these two constraints and based on our previous analysis, the graph of the x(t) must be an increasing concave function, starting at (0,0) and approaching c as t -> ∞. So, x(t) is a saturating exponential curve.

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. In the context of pharmacokinetics, they play a crucial role in modeling how drugs distribute and metabolize in the body. In our exercise, the concentration of the drug within an organ is expressed as an exponential function: \[x(t) = c\left(1 - e^{-\frac{at}{V}}\right)\]. Here, the exponential term \(e^{-\frac{at}{V}}\) describes how the concentration approaches its maximum over time. Key properties of exponential functions include:

• A quickly rising or descending behavior based on the sign of the exponent.
• A horizontal asymptote, which in this case is at \(y = c\), indicating the maximum concentration the drug will reach.

These functions are particularly useful for modeling time-dependent processes, such as drug absorption, as they can depict the gradual progression towards a steady state.

Derivative Analysis

In calculus, the derivative of a function helps us understand how that function changes. For pharmacokinetics, analyzing the derivative of our concentration function \(x(t)\) allows us to determine if the concentration is increasing or decreasing over time. The derivative of \(x(t) = c(1 - e^{-\frac{at}{V}})\) with respect to \(t\) is found using the chain rule: \[\frac{dx}{dt} = c \left(-\frac{a}{V} e^{-\frac{at}{V}}\right)\]. Important aspects of derivative analysis include:

• The sign of the derivative tells us about the function's trend. If \(\frac{dx}{dt} > 0\), the function is increasing.
• A negative sign combined with the negativity of \(\frac{a}{V}\) indicates the overall product is positive, showing \(x(t)\) increases.
• The absence of critical points where the derivative changes sign implies a continuous behavior across the function's domain.

This derivative informs us that the concentration of the drug rises steadily until it nears the maximum potential \(c\).

Function Sketching and Behavior

Sketching the graph of a function can provide a visual insight into how that function behaves over time. For our problem, we have a function \(x(t) = c(1 - e^{-\frac{at}{V}})\), which is instrumental to see in graphical form. The behavior of this function includes key characteristics: 1. **Initial Behavior:** At \(t=0\), we find \(x(0) = c(1 - 1) = 0\). This shows that initially, the concentration inside the organ is zero.
2. **Long-Term Behavior:** As \(t\to \infty\), the term \(e^{-\frac{at}{V}}\to 0\), meaning that \(x(t)\to c\). The function approaches a horizontal asymptote at \(y = c\).
3. **Shape of the Curve:** Since \(\frac{dx}{dt}\) is always positive, we know \(x(t)\) is increasing. The curve is concave down, suggesting it slows its rise as it nears \(y = c\).

• Start at (0,0) and gracefully curve towards the line \(y = c\).

These elements highlight that the concentration starts at zero and monotonically rises to \(c\) without surpassing it, reflecting a typical pharmacokinetic profile seen with drug accumulation in the body.