Problem 39 - Calculus Concepts

Question

For the first 9 months of life, the average weight \(w\), in pounds, of a certain breed of dog increases at a rate that can be described by the differential equation $$ \frac{d w}{d t}=\frac{33.68}{t} $$ where \(t\) is time given in months. A 1 -month-old puppy weighs 6 pounds. a. Use Euler's method with a 0.25 -month step length to estimate the weight of the puppy at 3 months and at 6 months. b. Use Euler's method with a step length of 1 month to estimate the weight of the puppy at 3 months and at 6 months. c. Is the answer to part \(b\) or the answer to part \(c\) more accurate? Why?

Step-by-Step Solution

Answer

Using a smaller step size (0.25 months) in Euler's method gives a more accurate approximation.

Step 1: Understand Euler's Method

Euler's method is a numerical procedure for solving ordinary differential equations (ODEs) with a given initial value. It is particularly useful for ODEs that cannot be solved analytically. The formula for Euler's method is \[ y_{n+1} = y_n + h \cdot f(t_n, y_n) \], where \(h\) is the step size and \(f(t_n, y_n)\) is the derivative \(\frac{dw}{dt}\).

Step 2: Initial Setup at 1 Month Old

At the start, we have \(t_0 = 1\) month and \(w_0 = 6\) pounds. The derivative is provided as \(\frac{d w}{d t} = \frac{33.68}{t}\). This is \(f(t, w) = \frac{33.68}{t}\).

Step 3: Apply Euler's Method with ∆t=0.25 to 3 Months

Using a step length \(h = 0.25\) months, we calculate the weights:1. \(t_1 = 1.25\), \(w_1 = w_0 + 0.25 \times 33.68/1 = 14.42\)2. \(t_2 = 1.5\), \(w_2 = 14.42 + 0.25 \times 33.68/1.25 = 21.12\)3. Continuing this process, at \(t = 3\) months, estimate the weight of the puppy.

Step 4: Estimate Weight at 6 Months with ∆t=0.25

Continue using the previously described method to predict weights beyond 3 months up to 6 months by incrementing \(t\) by 0.25 months each step.

Step 5: Apply Euler’s Method with ∆t=1 to 3 and 6 Months

Using a step length \(h = 1\) month, start from \(t_0 = 1\) and calculate:1. For \(t=2\), \(w_1 = w_0 + 1 \times \frac{33.68}{1} = 39.68\)2. For \(t=3\), \(w_2 = 39.68 + 1 \times \frac{33.68}{2} = 56.52\)3. Continue similarly for \(t=6\) months.

Step 6: Compare the Accuracy of the Two Methods

With a smaller step length (∆t = 0.25), Euler's method typically provides a more accurate approximation because it accounts for more changes over time. In practice, the shorter the time between calculations, the smaller the local error per step.

Key Concepts

Ordinary Differential Equations

An ordinary differential equation (ODE) is a type of equation that involves functions and their derivatives. Across various fields like physics and biology, ODEs describe how certain quantities change over time. In simple terms, they capture the rate at which something happens. For example, if you wanted to know how a dog's weight increases every month, an ODE could help model that scenario.
The equation presented in this exercise involves the rate of change of a puppy's weight. It's given by \(\frac{d w}{d t}=\frac{33.68}{t}\), where \(w\) is weight and \(t\) is time in months. This specific ODE tells us that as time progresses, the weight increases in a specific manner related to time. Often, ODEs cannot be solved explicitly or analytically. This is where numerical methods such as Euler's Method come in handy, allowing us to approximate the solution step-by-step.

Numerical Methods

Numerical methods provide algorithms to find approximate solutions to mathematical problems that may be difficult or impossible to solve analytically. Euler's Method is one of these powerful tools used specifically for solving ordinary differential equations numerically.
This method involves taking small steps along the curve described by the differential equation, using the slope (provided by the ODE) at specific points to estimate the next point's value. Essentially, you start at a known solution point and use the derivative to predict future values. It’s a bit like trying to hike a winding path by taking small, cautious steps you know will likely keep you on track, rather than trying to see the entire path at once.

Initial Value Problems

Some differential equations are paired with additional information known as an initial value, creating what's called an initial value problem. This initial value specifies the condition of the solution at a particular point in time. For example, in this exercise, we are given that the 1-month-old puppy weighs 6 pounds. This initial value \(w_0 = 6\) at \(t_0 = 1\) month tells us where the curve begins.
Initial value problems can be visualized as knowing your starting point on a map, and using the differential equation to trace the potential path forward. By knowing the initial condition (like the puppy's initial weight), we can use it with methods like Euler's to glean insights into future behavior, such as predicting weights at later times.

Step Size Influence

In numerical methods like Euler's, the step size (denoted as \(h\)) profoundly influences the accuracy and reliability of predictions. A smaller step size generally results in more accurate estimations because the method samples the function more frequently, capturing more subtle changes and reducing errors with each calculation.
In this exercise, two step sizes are tested: 0.25-month and 1-month. When you use a 0.25-month step size, you take four small steps within each month, leading to higher accuracy since it reflects the ODE's changing rates more closely. However, more steps can mean more computations. On the other hand, a 1-month step produces results faster but might overlook some accuracy, especially if the change is significant over time. It's like comparing taking smaller, sharper turns versus longer, broader turns when steering a car to stay on course.