Problem 58 - Calculus - AP Edition

Question

One model for the spread of information over time \(t\) is given by $$\ln y-\ln (1-y)=\alpha t$$ where \(y\) is the fraction of the population with the information and \(\alpha>0\) is a constant. Find an expression for \(d y / d t\) and explain why \(y\) is growing most rapidly at the moment when half the population has the information.

Step-by-Step Solution

Answer

The expression for \(\frac{dy}{dt}\) is \(\alpha y(1-y)\). \(y\) grows most rapidly when \(y = \frac{1}{2}\).

Step 1: Differentiate Both Sides with Respect to Time

We start by differentiating the given equation \(\ln y - \ln(1-y) = \alpha t\) with respect to time \(t\). Using the chain rule, we get:\[\frac{1}{y} \cdot \frac{dy}{dt} - \frac{1}{1-y} \cdot \left(-\frac{dy}{dt}\right) = \alpha\] Simplify this to:\[\frac{dy}{dt} \left( \frac{1}{y} + \frac{1}{1-y} \right) = \alpha\]

Step 2: Simplify the Expression

The expression inside the parentheses can be simplified as follows:\[\frac{1}{y} + \frac{1}{1-y} = \frac{1-y + y}{y(1-y)} = \frac{1}{y(1-y)}\]So, the equation becomes:\[\frac{dy}{dt} \cdot \frac{1}{y(1-y)} = \alpha\]

Step 3: Solve for dy/dt

We solve for \(\frac{dy}{dt}\) by multiplying both sides by \(y(1-y)\):\[\frac{dy}{dt} = \alpha y(1-y)\]This gives us the rate of change of \(y\) with respect to \(t\).

Step 4: Analyze When y is Growing Most Rapidly

Since \(\frac{dy}{dt} = \alpha y(1-y)\), the growth rate of \(y\) is directly related to the product \(y(1-y)\). To maximize this, examine the expression \(y(1-y)\). It reaches its maximum when the derivative with respect to \(y\) is zero.\[\frac{d}{dy}[y(1-y)] = 1 - 2y = 0\]Solving for \(y\) gives \(y = \frac{1}{2}\).Thus, \(y\) grows most rapidly when half the population has the information.

Key Concepts

Chain Rule

The chain rule is a vital tool in differential calculus used to differentiate composite functions. It's especially helpful when dealing with functions nested inside each other. In simple terms, the chain rule states that if you have a function inside another function, the derivative of that composite function is the product of the derivative of the outer function evaluated at the inner function times the derivative of the inner function. In the exercise, we had to differentiate the equation \( \ln y - \ln(1-y) = \alpha t \) with respect to time \( t \). Here, \( y \) is a function of \( t \), so using the chain rule is necessary to handle this differentiation. For \( \ln y \), the derivative is \( \frac{1}{y} \cdot \frac{dy}{dt} \) since \( y \) is a function of \( t \). Similarly, \( \ln (1-y) \) becomes \( -\frac{1}{1-y} \cdot \frac{dy}{dt} \). These steps illustrate the chain rule by showing how the rate of change of \( y \) over time is the focus, reflecting how intertwined functions are handled neatly using this principle.
Use the chain rule next time when you find a similar nested relation.

Rate of Change

The rate of change is a fundamental idea in calculus that measures how one quantity changes in relation to another.quantities do not stay the same but vary as other quantities change. In our context, we're looking at how \( y \), the fraction of the population with information, changes over time. The derivative \( \frac{dy}{dt} \) provides this rate of change, indicating how quickly or slowly \( y \) is increasing or decreasing as time goes on.In the solved exercise, we found the rate of change of \( y \) by rearranging the equation to \( \frac{dy}{dt} = \alpha y(1-y) \). Here, \( \alpha \) is a constant, and the term \( y(1-y) \) is vital because it affects the speed of growth.

• When \( y \) is close to 0 or 1, \( y(1-y) \) becomes very small, leading to a slower rate of change.
• When \( y \) is around 0.5, \( y(1-y) \) reaches its maximum value, resulting in the fastest growth rate.

Understanding the rate of change helps comprehend how dynamic processes evolve, exemplified here by how a population quickly or slowly becomes informed.

Exponential Growth Model

Exponential growth models describe how quantities change over time by becoming larger at a consistent relative growth rate. These are common in natural and social sciences, characterizing processes such as populations, investments, or, as in this case, information spread within a community.In the exercise, the model \( \ln y - \ln (1-y) = \alpha t \) captures how information permeates among people over time. This equation is a form of the logistic model, which eventually behaves like exponential growth at certain phases of its progression. The expression \( \frac{dy}{dt} = \alpha y (1-y) \) signifies not just growth but also regulation as the spread increases. As more people become informed, the value of \( y \) approaches 1, slowing the rate down due to the \( 1-y \) factor, typical in logistic-like growth models. This model is different from simple exponential growth where the derivative \( \frac{dy}{dt} = \alpha y \) would suggest constant relative growth. Instead, the term \( y(1-y) \) ensures the growth rate adapts as \( y \) changes. Exponential growth models offer insight into real-world scenarios where rapid advancement is checked by limitations, making them more reflective of actual dynamics.