Problem 43 - Advanced Engineering Mathematics

Question

A mathematical model for the rate at which a drug disseminates into the bloodstream is given by \(d x / d t=r-k x\), where \(r\) and \(k\) are positive constants. The function \(x(t)\) describes the concentration of the drug in the bloodstream at time \(t\). (a) Since the \(\mathrm{DE}\) is autonomous, use the phase portrait concept of Section \(2.1\) to find the limiting value of \(x(t)\) as \(t \rightarrow \infty\) (b) Solve the \(\mathrm{DE}\) subject to \(x(0)=0\). Sketch the graph of \(x(t)\) and verify your prediction in part (a). At what time is the concentration one- half this limiting value?

Step-by-Step Solution

Answer

(a) Limiting value: \( \frac{r}{k} \). (b) \( x(t) = \frac{r}{k}(1 - e^{-kt}) \); concentration is half at \( t = \frac{\ln 2}{k} \).

Step 1: Understanding the Differential Equation

The differential equation given is \( \frac{dx}{dt} = r - kx \). This equation describes the rate of change of the drug concentration \( x(t) \) in the bloodstream. The equation is linear and autonomous, meaning \( r \) and \( k \) are constants and it does not explicitly depend on \( t \).

Step 2: Finding the Limiting Value of \( x(t) \)

To find the limiting value of \( x(t) \) as \( t \rightarrow \infty \), we set \( \frac{dx}{dt} = 0 \) since we want to find equilibrium points. Solving \( r - kx = 0 \) gives \( x = \frac{r}{k} \). This shows that as \( t \rightarrow \infty \), \( x(t) \) approaches \( \frac{r}{k} \).

Step 3: Solving the Differential Equation

To solve the differential equation \( \frac{dx}{dt} = r - kx \), we rearrange it as \( \frac{dx}{r - kx} = dt \) and integrate both sides. The left side becomes \( -\frac{1}{k} \ln|r-kx| \) and the right side is \( t + C \). Solving the equation yields \( x(t) = \frac{r}{k}(1 - e^{-kt}) \) by using the condition \( x(0) = 0 \) to find \( C \).

Step 4: Graphing \( x(t) \)

The function \( x(t) = \frac{r}{k}(1 - e^{-kt}) \) is an exponential growth curve starting from 0 at \( t = 0 \) and asymptotically approaching \( \frac{r}{k} \) as \( t \rightarrow \infty \). The graph is an increasing function that approaches its limiting value \( \frac{r}{k} \).

Step 5: Determining Half of Limiting Concentration

To find when \( x(t) \) is half the limiting value, solve \( \frac{r}{2k} = \frac{r}{k}(1 - e^{-kt}) \). Simplify to find \( e^{-kt} = \frac{1}{2} \), leading to \( -kt = \ln \frac{1}{2} \). Therefore, \( t = \frac{\ln 2}{k} \).

Key Concepts

Phase Portrait

A phase portrait is a visual tool used to understand differential equations. It helps us see where equilibrium points are and how solutions to the equation evolve over time.
The phase portrait for an autonomous differential equation like \( \frac{dx}{dt} = r - kx \) involves plotting the direction of \( dx/dt \) against the function value \( x \).
You can visualize how the system behaves over time by looking at the arrows or directions in the portrait. This shows whether \( x(t) \) increases or decreases and moves towards an equilibrium.

• A positive value for \( dx/dt \) implies the function \( x(t) \) is increasing.
• A negative value means \( x(t) \) is decreasing.
• At equilibrium points, \( dx/dt = 0 \), hence the flow is neither increasing nor decreasing.

Using a phase portrait, we can see how \( x(t) \) tends to stabilize at certain values, providing insights into the system's behavior without solving the differential equation explicitly.

Limiting Behavior

Limiting behavior helps us predict the long-term state of a differential equation. When studying an autonomous differential equation like \( \frac{dx}{dt} = r - kx \), we are often interested in the behavior of the solution as time \( t \) goes to infinity.
In the given equation, we set \( \frac{dx}{dt} = 0 \) to identify equilibrium points, as the rate of change becomes zero when the system stabilizes. Solving \( r - kx = 0 \) gives \( x = \frac{r}{k} \).
This equilibrium point \( \frac{r}{k} \) reflects the limiting value \( x(t) \) approaches as \( t \to \infty \).

• This tells us that the concentration of the drug in the blood will stabilize at \( \frac{r}{k} \), assuming the system continues indefinitely without changes to \( r \) or \( k \).
• The limiting behavior offers a clear target value even if initial conditions (like the starting concentration) differ.

The effect of the parameters \( r \) and \( k \) is crucial in dictating how the system evolves and the nature of its equilibrium.

Exponential Growth

Exponential growth describes how quantities increase rapidly over time. In our differential equation \( \frac{dx}{dt} = r - kx \), solving gives us the solution \( x(t) = \frac{r}{k}(1 - e^{-kt}) \).
This represents a form of exponential growth behavior with respect to time.

• The expression \( 1 - e^{-kt} \) accounts for the exponential part, modulating \( 1 \) by the quickness of the process \( e^{-kt} \) diminishing.
• As \( t \) increases, \( e^{-kt} \) approaches zero, making \( x(t) \) approach the equilibrium \( \frac{r}{k} \).

The characteristic of an exponential growth function here is that the growth rate proportionally decreases as \( x \) gets closer to its maximum potential (\( \frac{r}{k} \)).
This is distinguished from pure exponential growth where the rate remains unchanged as the quantity increases. Overall, it's important to recognize the graphs' signature curve, which rises sharply initially, then flattens as it reaches its limiting value.

Equilibrium Points

Equilibrium points refer to the stable states a differential equation can tend toward. These points occur when the rate of change is zero, implying no change in the system as time progresses.

• For the differential equation \( \frac{dx}{dt} = r - kx \), solving for \( r - kx = 0 \) identifies \( x = \frac{r}{k} \) as an equilibrium point.
• Equilibrium points often signify systems settling into unchanging conditions as \( t \to \infty \).

In our example, this suggests that over time the concentration of a drug will stabilize to the level \( \frac{r}{k} \).
Equilibrium analysis is central when examining autonomous differential equations since understanding these points helps predict long-term behavior without going through explicit integration for every possible initial value.
Systems can have stable or unstable equilibriums. In this case, any deviation from the equilibrium \( \frac{r}{k} \) tends to return to it, indicating stability in the long run.