Problem 15 - Essential Calculus

Question

Find the function \(y=f(t)\) passing through the point (0,10) with the given first derivative. Use a graphing utility to graph the solution. $$ \frac{d y}{d t}=-\frac{1}{2} y $$

Step-by-Step Solution

Answer

The particular solution to the differential equation is \(y = 10 e^{-\frac{1}{2} t}\).

Step 1: Understand the given information

The given equation is a simple differential equation. The derivative of \(y\) with respect to \(t\) is \(-\frac{1}{2} y\). The goal is to find \(y\) as a function of \(t\) (also denoted as \(y=f(t)\)). The function \(y=f(t)\) passes through the point (0,10).

Step 2: Solve the differential equation

Firstly, rearrange \(-\frac{1}{2} y = \frac{d y}{d t}\) to \(-\frac{2}{y} dy = dt\. integrating both sides get \(-2 \ln |y| = t + C_1\), where \(C_1\) is the constant of integration. Apply the exponential function to both sides to remove the natural logarithm. The result is \(y = e^{-\frac{1}{2} t + C_1}\) or \(y = e^{C_1} e^{-\frac{1}{2} t}\). By observing that \(e^{C_1}\) is just a constant itself, which can be denoted by C, the general solution to the differential equation is \(y = C e^{-\frac{1}{2} t}\).

Step 3: Find the particular solution

To find the constant \(C\), substitute the point (0,10) into the general solution. Thus \(10 = C e^{-\frac{1}{2} \cdot 0}\), which simplifies to \(10 = C e^0\), then \(10 = C\). So the particular solution to the differential equation is \(y = 10 e^{-\frac{1}{2} t}\).

Step 4: Graph the function

The final step is graphically representing the function. Use a graphing utility to produce a graph for \(y = 10 e^{-\frac{1}{2} t}\). The graph should start at (0,10) and decay exponentially as \(t\) increases, reaching asymptotically towards the \(t\)-axis.

Key Concepts

First Derivative

The first derivative of a function is a fundamental concept in calculus. It represents the rate at which the function's value changes with respect to a change in the independent variable, often time or space. In this exercise, the first derivative is given as \( \frac{d y}{d t} = -\frac{1}{2} y \). This indicates that the rate of change of the function \( y \) concerning \( t \) is negative, scaling with \( y \) itself. Such a relationship is typical in processes described by exponential decay, where the amount decreases over time. In our problem, the negative coefficient \( -\frac{1}{2} \) signifies a proportional reduction in \( y \) with each small increase in \( t \). Understanding this is crucial in interpreting how the dependent variable behaves over time.

Exponential Decay

Exponential decay is a process where a quantity decreases at a rate proportional to its current value. In our differential equation \( \frac{d y}{d t} = -\frac{1}{2} y \), the decay factor is \(-\frac{1}{2}\), meaning that for each unit increase in \( t \), the function \( y \) is multiplied by the exponential term \( e^{-\frac{1}{2} t} \). This results in a graph that starts at an initial value and drops rapidly at first, then more slowly as it approaches closer to zero but never quite touches it. The notion of exponential decay is pivotal for understanding phenomena like radioactive decay, depreciation of assets, or cooling of objects over time. Each has a decay constant, in this case, \(-\frac{1}{2}\), dictating how quickly the process occurs.

Integration

Integration is the mathematical process of finding the original function given its derivative. In this exercise, integration helps us solve the differential equation \( \frac{d y}{d t} = -\frac{1}{2} y \). By rearranging and integrating both sides, we establish a relationship between the function \( y \) and \( t \). More specifically,

• Firstly, rearrange the equation to isolate \( dy \) and \( dt \) on opposite sides.
• Integrate both sides to find \(-2 \ln |y| = t + C_1\), where \( C_1 \) is the constant of integration.
• Exponentiate results to solve for \( y \), which becomes \( y = e^{C_1} e^{-\frac{1}{2} t} \).

By applying integration, we effectively reverse the differentiation process, uncovering the function that models the situation. This method is vital in solving real-world problems where we need to find the relationship governing observed changes, such as in physics, economics, and biology. Through integration, the abstract concept of a changing rate becomes a tangible function describing a system's behavior over time.