Problem 1 - Calculus

Question

(a) Show that \(P=1 /\left(1+e^{-t}\right)\) satisfies the logistic equation $$\frac{d P}{d t}=P(1-P)$$ (b) What is the limiting value of \(P\) as \(t \rightarrow \infty ?\)

Step-by-Step Solution

Answer

(a) The equation satisfies \(\frac{dP}{dt} = P(1-P)\). (b) As \(t \to \infty\), \(P\) approaches 1.

Step 1: Differentiate P with respect to t

Given that \(P = \frac{1}{1 + e^{-t}}\), first, differentiate \(P\) with respect to \(t\). Rewrite \(P\) as \((1 + e^{-t})^{-1}\) and apply the chain rule. The derivative of \((1 + e^{-t})^{-1}\) is \(-1 \times (1 + e^{-t})^{-2} \times (-e^{-t})\). This simplifies to \(\frac{e^{-t}}{(1 + e^{-t})^2}\).

Step 2: Express P(1-P) and Simplify

Compute \(P(1-P)\). Substitute \(P = \frac{1}{1 + e^{-t}}\), so \(1-P = 1 - \frac{1}{1 + e^{-t}} = \frac{e^{-t}}{1 + e^{-t}}\). Multiply \(P\) and \(1-P\): \(\frac{1}{1 + e^{-t}} \times \frac{e^{-t}}{1 + e^{-t}} = \frac{e^{-t}}{(1 + e^{-t})^2}\).

Step 3: Confirm Equality to Prove Logistic Equation

Compare \(\frac{dP}{dt} = \frac{e^{-t}}{(1 + e^{-t})^2}\) with \(P(1-P) = \frac{e^{-t}}{(1 + e^{-t})^2}\). Both expressions are identical, confirming that the differential equation \(\frac{dP}{dt} = P(1-P)\) is satisfied.

Step 4: Find the Limiting Value of P as t Approaches Infinity

To find \(\lim_{{t \to \infty}} P\), use the expression \(P = \frac{1}{1 + e^{-t}}\). As \(t \to \infty\), \(e^{-t} \to 0\), so \(P = \frac{1}{1 + 0} = 1\).

Key Concepts

Differential Equation

A differential equation is a mathematical equation that relates some function with its derivatives. In our case, we are dealing with the **logistic equation** which is written as \(\frac{dP}{dt} = P(1-P)\). This particular equation models population growth where \(P\) represents the population in relative terms.

Solving a differential equation can involve finding a function, \(P(t)\), that satisfies the equation for every value of \(t\). For the logistic equation, the function \(P = \frac{1}{1 + e^{-t}}\) happens to be one that works. Differentiating \(P\), as done in the original solution steps, shows that \(\frac{dP}{dt}\) equals the logistic equation's right side \(P(1-P)\).

Understanding and solving differential equations is key in fields like physics, engineering, and biology, where changes are continuously related to rates of change.

Chain Rule

The chain rule is a fundamental formula in calculus used for finding the derivative of the composition of two or more functions. When you have a function inside another function, you apply the chain rule to differentiate it.

In the problem at hand, we consider \(P = (1 + e^{-t})^{-1}\). To differentiate \(P\) with respect to \(t\), the chain rule is applied as follows:

• Start with the outer function, which is \((1 + e^{-t})^{-1}\). The derivative here is \(-1\times(1+e^{-t})^{-2}\).
• Then differentiate the inner function \(1 + e^{-t}\), which gives \(-e^{-t}\).

Finally, multiply these results, which simplifies neatly in our solution: \(\frac{e^{-t}}{(1 + e^{-t})^2}\).
This shows how versatile and necessary the chain rule is in calculus for solving real-world scenarios that involve nested functions.

Exponential Function

The exponential function, denoted here as \(e^{-t}\), plays a central role in the logistic equation. It's important to understand how exponential functions behave and why they're important.

Exponential functions are defined by the constant \(e\), approximately 2.718, and they exhibit continuous growth or decay. In the context of the logistic equation, \(e^{-t}\) represents an exponential decay because of the negative exponent.

• As \(t\) increases, \(e^{-t}\) approaches zero, which significantly simplifies expressions like \(P = \frac{1}{1 + e^{-t}}\).
• This decay explains why, as time goes to infinity, the factor \(e^{-t}\) becomes negligible, and thus the population \(P\) stabilizes at 1.

An understanding of exponential functions is crucial across disciplines, as they model a wide range of phenomena including population growth, radioactive decay, and financial investments.