Problem 4 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Consider the differential equation \(\frac{d P}{d t}=k P(L-P)\), where \(k\) and \(L\) are positive constants. (a) For what values of \(P\) is \(\frac{d P}{d t}\) zero? (b) Show that \(P(t)=\frac{L}{1+C e^{-k L t}}\), where \(C\) is a constant, is a solution of the logistic equation above.

Step-by-Step Solution

Answer

The values of \(P\) for which \(\frac{d P}{d t}\) is zero are \(P=0\) and \(P=L\). Furthermore, \(P(t)=\frac{L}{1+C e^{-k L t}}\) is a solution of the logistic differential equation.

Step 1: Solve for \(\frac{d P}{d t}=0\)

Set \(\frac{d P}{d t}=k P(L-P)=0\) and solve for \(P\). This gives two solutions: \(P=0\) and \(P=L\).

Step 2: Differentiate \(P(t)\)

Differentiate the given function \(P(t)=\frac{L}{1+C e^{-k L t}}\) with respect to \(t\) using the chain rule.

Step 3: Substitute \(P(t)\) and its derivative into the logistic equation

Check whether substituting \(P(t)\) and its derivative into the logistic equation yields an identity.

Step 4: Simplify

If necessary, simplify the result such that it becomes clear whether or not the substitution is successful. If the left-hand and right-hand sides of the equation match, then \(P(t)=\frac{L}{1+C e^{-k L t}}\) is indeed a solution to the differential equation.

Key Concepts

Logistic Equation

The logistic equation is a widely used model in biological and ecological studies. It describes how a population grows over time under natural resource constraints. The standard form of the logistic equation is given as \( \frac{dP}{dt} = kP(L - P) \), where:

• \( P \) is the population size at time \( t \).
• \( k \) is the growth rate constant.
• \( L \) is the carrying capacity, or the maximum population size that the environment can sustain.

The logistic equation models the population growth by assuming that the growth rate is proportionate to both the current population and the remaining capacity. This means that as the population gets closer to the carrying capacity, the growth slows down.
For example, when \( P \) approaches \( L \), the term \( L - P \) becomes small, reducing the growth rate. Conversely, if \( P \) is much less than \( L \), the growth is nearly exponential. These characteristics make the logistic equation quite effective in representing real-world phenomena, such as species population growth or saturation levels.

Chain Rule

The chain rule is a fundamental concept in calculus. It is particularly useful when dealing with the differentiation of composite functions. In the context of the logistic equation, we apply the chain rule to differentiate the proposed solution \( P(t) = \frac{L}{1 + Ce^{-kLt}} \).
To use the chain rule here, we identify the outer and inner functions. The outer function is \( \, \frac{L}{1 + u} \), and the inner function is \( u = Ce^{-kLt} \).

To differentiate \( P(t) \), we first find the derivative of the outer function with respect to the inner function \( u \), and then multiply it by the derivative of the inner function with respect to \( t \). This process involves:

• Finding \( \frac{d}{du} \left( \frac{L}{1 + u} \right) = -\frac{L}{(1+u)^2} \).
• Finding \( \frac{du}{dt} \) for \( u = Ce^{-kLt} \). Differentiating gives \( \frac{du}{dt} = -kLCe^{-kLt} \).

Finally, we multiply these derivatives according to the chain rule: \[ \frac{dP}{dt} = -\frac{L}{(1 + Ce^{-kLt})^2} \times (-kLCe^{-kLt}) \]. This result is used to show that \( P(t) \) satisfies the logistic equation.

Solution Verification

Verifying a solution to a differential equation is critical to ensure the proposed function satisfies the original equation. In this exercise, the function \( P(t) = \frac{L}{1 + Ce^{-kLt}} \) was proposed as a solution to the logistic equation \( \frac{dP}{dt} = kP(L - P) \).
To verify this, perform the following steps:

• Calculate \( \frac{dP}{dt} \) using the chain rule, as described earlier.
• Substitute \( P(t) \) and its derivative \( \frac{dP}{dt} \) into the logistic equation.
• Check if both sides of the equation simplify to the same expression.

Upon substitution and simplification, if the equation holds true (meaning the left and right sides are equal), then \( P(t) \) is indeed a valid solution. This confirmation step underlines the importance of correctly applying calculus techniques in the verification process and shows that \( P(t) \) satisfies the logistic differential equation fully, demonstrating its applicability in modeling growth scenarios.