Problem 15 - Applied Calculus

Question

One theory on the speed an employee learns a new task claims that the more the employee already knows, the more slowly he or she learns. Suppose that the rate at which a person learns is equal to the percentage of the task not yet learned. If \(y\) is the percentage learned by time \(t\), the percentage not yet learned by that time is \(100-y\), so we can model this situation with the differential equation $$ \frac{d y}{d t}=100-y $$ (a) Find the general solution to this differential equation. (b) Sketch several solutions. (c) Find the particular solution if the employee starts learning at time \(t=0\) (so \(y=0\) when \(t=0\) ).

Step-by-Step Solution

Answer

The general solution is \(y = 100 - Ae^{-t}\). Particular solution: \(y = 100 - 100e^{-t}\).

Step 1: Setup the Differential Equation

The differential equation given is \( \frac{d y}{d t} = 100 - y \). This represents the rate of change of the learned percentage \( y \) with respect to time \( t \).

Step 2: Solve the Differential Equation

We start by separating variables. Rewrite the equation as \( \frac{d y}{100 - y} = dt \). Integrate both sides: \( \int \frac{1}{100 - y} \, dy = \int 1 \, dt \). This gives us: \(-\ln|100-y| = t + C\), where \(C\) is the integration constant.

Step 3: Exponentiate to Solve for y

Exponentiate both sides to solve for \(y\): \( |100-y| = e^{-t-C} = Ae^{-t} \) where \(A = e^{-C}\). Thus, \( y = 100 - Ae^{-t} \). This is the general solution to the differential equation.

Step 4: Sketch Several Solutions

Each solution curve \( y = 100 - Ae^{-t} \) represents a different value of \(A\). As \(t\to \infty\), \( y \to 100 \). Sketch how the graph starts at different initial positions for different \(A\) and asymptotically approaches \(y=100\).

Step 5: Find the Particular Solution

We look for the particular solution when the employee starts learning at time \(t=0\), i.e., \(y=0\) when \(t=0\). Substitute into the general solution: \(0 = 100 - A\cdot e^{0} \). This gives \(A = 100\). Thus, the particular solution is \( y = 100 - 100e^{-t} \).

Key Concepts

Rate of Change

The rate of change in a problem describes how a quantity alters as time progresses. In the given exercise, the rate at which an employee learns a task is modeled using the differential equation \( \frac{d y}{d t} = 100 - y \). This equation indicates that the change in knowledge over time is proportional to the fraction of the task that remains unlearned.

In simple terms:

• The more there is to learn, the faster the person learns initially.
• As the individual learns and the percentage unlearned decreases, the rate of learning also decreases.

This inverse relationship characterizes the learning process: quick at first and slower as the person gains more knowledge. Understanding the rate of change is crucial, as it reveals the dynamics between time and learning.

Separation of Variables

This is a fundamental technique used to solve differential equations, especially those representing real-world phenomena like learning. "Separation of Variables" involves rearranging the equation to isolate variables on different sides so they can be integrated separately.

For our equation \( \frac{d y}{d t} = 100 - y \), we separate the variables by rewriting it as \( \frac{d y}{100 - y} = dt \). This step is key because:

• It splits the problem into two manageable parts, integrating dy and dt separately.
• It simplifies the differential equation into a form that can be solved through integration.

By tackling each side independently, we turn a complex differential equation into simpler integrals, bringing us closer to finding the general solution.

Integration Constant

During the integration process, an integration constant \(C\) emerges, representing an indefinite integral's arbitrary constant addition. When integrating equations derived from real-world conditions, this constant must be identified to find a specific solution.

In our example, after integrating both sides, we find \(-\ln|100-y| = t + C\). The constant \(C\) encapsulates initial conditions of the problem, which help tailor the solution to specific circumstances, such as our initial condition \(y=0\) when \(t=0\).

The importance of the integration constant:

• It ensures that solutions fit the specific context of the problem.
• It allows multiple solution curves, each corresponding to different starting conditions in real scenarios.

Using initial values helps determine \(C\), ensuring the mathematical model accurately reflects the situation.

Exponential Growth

Exponential growth describes how quantities increase at a rate proportional to their current value. Our exercise provides an exponential "decay" in the amount remaining to be learned, effectively modeling learning progress over time.

The general solution, \( y = 100 - Ae^{-t} \), embodies exponential growth in reverse, showing learning slows as time goes on. Important points about this type of growth in our context:

• The term \(Ae^{-t}\) decreases exponentially, hence more is learned initially than later.
• The solution approaches an asymptote, indicating that 100% learning is approached but never instantaneously achieved.

This understanding clarifies why learning is initially rapid but slows over time, a clear depiction of diminishing returns as mastery approaches. By recognizing exponential trends, learners can better anticipate realistic progress in varied learning tasks.