Problem 14 - Calculus

Question

(a) Use ten steps of Euler's method to approximate \(y-\) values for \(d y / d t=1 / t,\) starting at (1,0) and using \(\Delta t=0.1\) (b) Using integration, solve the differential equation to find the exact value of \(y\) at the end of these ten steps. (c) Is your approximate value of \(y\) at the end of ten steps bigger or smaller than the exact value? Use a slope field to explain your answer.

Step-by-Step Solution

Answer

The approximate value 0.7187 is larger than the exact value 0.6931.

Step 1: Understanding Euler's Method

Euler's method provides an approximation of the function values by iterating over an initial point using the differential equation. Given \( \frac{dy}{dt} = \frac{1}{t} \), we'll start at (1,0) and take ten steps with \( \Delta t = 0.1 \). We'll substitute values to find successive \( y \)-values.

Step 1: Initial Values

Start with the initial values given: \( y_0 = 0 \) at \( t_0 = 1 \).

Step 2: First Iteration

Use Euler's formula: \( y_{n+1} = y_n + \Delta t \cdot f(t_n, y_n) \). For the first iteration: \( y_1 = y_0 + 0.1 \cdot \frac{1}{1} = 0 + 0.1 \cdot 1 = 0.1 \).

Step 3: Second Iteration

Calculate the next value. \( y_2 = y_1 + 0.1 \cdot \frac{1}{1.1} \approx 0.1 + 0.1 \cdot 0.909 = 0.1909 \).

Step 4: Third Iteration

Continue to the next step. \( y_3 = y_2 + 0.1 \cdot \frac{1}{1.2} \approx 0.1909 + 0.1 \cdot 0.833 = 0.2742 \).

Step 5: Fourth Iteration

Calculate the next value. \( y_4 = y_3 + 0.1 \cdot \frac{1}{1.3} \approx 0.2742 + 0.1 \cdot 0.769 = 0.3511 \).

Step 6: Fifth Iteration

Continue with the next iteration. \( y_5 = y_4 + 0.1 \cdot \frac{1}{1.4} \approx 0.3511 + 0.1 \cdot 0.714 = 0.4225 \).

Step 7: Sixth Iteration

Calculate the next value. \( y_6 = y_5 + 0.1 \cdot \frac{1}{1.5} \approx 0.4225 + 0.1 \cdot 0.667 = 0.4892 \).

Step 8: Seventh Iteration

Continue with the next iteration. \( y_7 = y_6 + 0.1 \cdot \frac{1}{1.6} \approx 0.4892 + 0.1 \cdot 0.625 = 0.5517 \).

Step 9: Eighth Iteration

Calculate the next value. \( y_8 = y_7 + 0.1 \cdot \frac{1}{1.7} \approx 0.5517 + 0.1 \cdot 0.588 = 0.6105 \).

Step 10: Ninth Iteration

Calculate the next value. \( y_9 = y_8 + 0.1 \cdot \frac{1}{1.8} \approx 0.6105 + 0.1 \cdot 0.556 = 0.6661 \).

Step 11: Tenth Iteration

Calculate the next value. \( y_{10} = y_9 + 0.1 \cdot \frac{1}{1.9} \approx 0.6661 + 0.1 \cdot 0.526 = 0.7187 \).

Step 13: Exact Solution Using Integration

Integrate \( \frac{dy}{dt} = \frac{1}{t} \) to find the exact function. The integral \( \int 1/t \, dt = \ln |t| + C \). By solving \( y(1) = 0 \), we find \( C = 0 \), so \( y(t) = \ln t \). At the end of ten steps, \( y(2) = \ln 2 \approx 0.6931 \).

Step 14: Comparing Approximate and Exact Values

The approximate \( y_{10} = 0.7187 \) is greater than the exact \( y(2) = 0.6931 \). Euler's method overestimates the value as the slope field indicates that the slope is gradually decreasing, meaning the method accumulates the gradient.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. These equations are fundamental in describing various physical, biological, and engineering systems where rates of change are present. In the context of Euler's method, we specifically look at the first-order differential equation \( \frac{dy}{dt} = \frac{1}{t} \). This equation suggests that the rate of change of \( y \) with respect to \( t \) is inversely related to \( t \). This kind of equation can model scenarios where growth rates change over time, like population growth or radioactive decay. By solving differential equations, we can understand the behavior of dynamic systems.

• They are crucial for modeling dynamic systems.
• Represent rates of change.
• Solving them helps predict future outcomes of the system.

Numerical Analysis

Numerical analysis involves developing and analyzing techniques that yield approximate solutions to mathematical problems. Euler's method is a classic numerical approach used to solve ordinary differential equations (ODEs). It involves iterative steps to approximate function values at discrete points. This is particularly helpful when a solution cannot be easily obtained analytically. In our exercise, with the differential equation \( \frac{dy}{dt} = \frac{1}{t} \), we use ten steps of Euler's method starting from (1,0) with a step size \( \Delta t = 0.1 \).

• Provides approximate solutions to complex equations.
• Used when analytic solutions are hard or impossible to find.
• Includes techniques like Euler's method for solving ODEs.

Integration

Integration is the process of finding functions whose derivative is known. It is the inverse operation to differentiation. In solving differential equations analytically, integration helps find the exact solution by integrating the given derivative. In our case, the differential equation \( \frac{dy}{dt} = \frac{1}{t} \) is integrated to yield \( y(t) = \ln |t| + C \). By applying initial conditions, such as \( y(1) = 0 \), we can determine the constant of integration, \( C \), and find the specific solution to the equation.

• Inverse of differentiation.
• Helps to find the exact solution of differential equations.
• Determines unknown constants using initial conditions.

Slope Field

A slope field is a visual representation of a differential equation. It shows short line segments, or slopes, that illustrate the direction of the solution at any given point in a plane. By examining a slope field for the equation \( \frac{dy}{dt} = \frac{1}{t} \), one can observe how solutions behave and reason about the accuracy of numerical approximations like Euler's method. In our exercise, by observing the slope field, we notice that Euler's method ultimately overestimates the exact solution due to the slopes becoming shallower as \( t \) increases. This causes the accumulated estimate to be slightly higher than the actual exact solution, as seen by the difference in values at \( t = 2 \).

• Visual tool for understanding differential equations.
• Shows direction of solutions across a grid of points.
• Helps compare and validate numerical approximations.