Problem 31 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(27-32\) for \(r\) as a vector function of \(t .\) $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d^{2} \mathbf{r}}{d t^{2}}=-32 \mathbf{k}} \\ {\text { Initial conditions: }} & {\mathbf{r}(0)=100 \mathbf{k} \text { and }} \\ {} & {\left.\frac{d \mathbf{r}}{d t}\right|_{t=0}=8 \mathbf{i}+8 \mathbf{j}}\end{array} $$

Step-by-Step Solution

Answer

The solution is \( \mathbf{r}(t) = 8t \mathbf{i} + 8t \mathbf{j} + (100 - 16t^2) \mathbf{k} \).

Step 1: Integrate the Differential Equation

We begin with the differential equation \( \frac{d^2 \mathbf{r}}{dt^2} = -32 \mathbf{k} \). Since this is a second-order differential equation, integrate to find the velocity vector function \( \frac{d\mathbf{r}}{dt} \). Upon integrating, we get \( \frac{d\mathbf{r}}{dt} = -32t \mathbf{k} + \mathbf{C_1} \), where \( \mathbf{C_1} \) is a constant vector of integration.

Step 2: Apply Initial Velocity Condition

Use the initial condition \( \left.\frac{d \mathbf{r}}{dt}\right|_{t=0} = 8 \mathbf{i} + 8 \mathbf{j} \) to determine \( \mathbf{C_1} \). Substitute \( t = 0 \) into the velocity function: \( 8 \mathbf{i} + 8 \mathbf{j} = \mathbf{C_1} \). Thus, \( \mathbf{C_1} = 8 \mathbf{i} + 8 \mathbf{j} \). The velocity function becomes \( \frac{d\mathbf{r}}{dt} = 8 \mathbf{i} + 8 \mathbf{j} - 32t \mathbf{k} \).

Step 3: Integrate the Velocity Function

Integrate the velocity function \( \frac{d\mathbf{r}}{dt} = 8 \mathbf{i} + 8 \mathbf{j} - 32t \mathbf{k} \) to find the position vector function \( \mathbf{r}(t) \). We have \( \mathbf{r}(t) = 8t \mathbf{i} + 8t \mathbf{j} - 16t^2 \mathbf{k} + \mathbf{C_2} \), where \( \mathbf{C_2} \) is another constant vector of integration.

Step 4: Apply Initial Position Condition

Using the initial condition \( \mathbf{r}(0) = 100 \mathbf{k} \), substitute \( t = 0 \) into the position function: \( \mathbf{C_2} = 100 \mathbf{k} \). Thus, the position vector function is \( \mathbf{r}(t) = 8t \mathbf{i} + 8t \mathbf{j} - 16t^2 \mathbf{k} + 100 \mathbf{k} \).

Step 5: Simplify and Write the Final Solution

Combine terms in the position vector function \( \mathbf{r}(t) = 8t \mathbf{i} + 8t \mathbf{j} + (100 - 16t^2) \mathbf{k} \). This is the solution to the initial value problem.

Key Concepts

Initial Value Problem

An initial value problem is a type of problem in mathematics, primarily in calculus, where the solution to a differential equation must satisfy specific conditions at the initial point. These conditions are called initial conditions and set the state of the system at time zero. For example, in this exercise, we have a differential equation that describes the motion of a vector function and initial conditions given for both position and velocity. This helps determine the unique solution to our problem.

Understanding initial value problems is crucial because:

• They are commonly used in physics and engineering to describe systems over time.
• Initial conditions help us find specific solutions tailored to the scenario, instead of a general one.

By solving an initial value problem, we can predict the behavior of a system based on its initial state and the governing law represented by a differential equation. This predictability is what makes it powerful for modeling real-world phenomena.

Differential Equations

Differential equations are equations that relate a quantity with its rates of change. In the exercise, we use a second-order differential equation, involving the second derivative of the vector function \( \mathbf{r}(t) \). Understanding differential equations is essential in contexts where rates of change are involved, like physics, biology, and engineering.

There are two main types of differential equations:

• Ordinary Differential Equations (ODEs): These involve one independent variable, as seen here with time \( t \).
• Partial Differential Equations (PDEs): These involve multiple independent variables.

Second-order differential equations, like the one in this exercise, typically describe systems with acceleration, since they involve the second derivative of position or another quantity. Solving differential equations often involves integrating to find functions, satisfying given conditions, such as initial or boundary conditions.

Integration

Integration is the process of finding a function from its derivative, essentially "undoing" differentiation. In this exercise, we took two integrals: first, to find the velocity vector function from the acceleration, and second, to find the position vector function from the velocity.

Integration is vital for solving differential equations because:

• It allows us to reconstruct original functions from known rates of change.
• Definite integration can give exact changes over an interval, while indefinite integration is concerned with finding general antiderivatives.

Constant vectors like \( \mathbf{C_1} \) and \( \mathbf{C_2} \) appear during this process as indefinite integration yields families of functions, not unique solutions. These constants are determined using the initial conditions to tailor the solution to the specific scenario, such as the initial position or velocity of the system.

Vector Functions

A vector function is a function that takes a variable, often time \( t \), and returns a vector, which means its components are functions of \( t \). In this exercise, \( \mathbf{r}(t) \) is a vector function indicating the position of an object in space.

Vector functions are integral in modeling motions and forces since they're naturally suited to describe quantities with direction and magnitude. They offer a rich representation of:

• Position: Where an object is in space (e.g., \( \mathbf{r}(t) \)).
• Velocity: How fast and in which direction the position changes (its rate of change, \( \frac{d\mathbf{r}}{dt} \)).
• Acceleration: How the velocity changes (its rate of change, \( \frac{d^2\mathbf{r}}{dt^2} \)).

In applying initial conditions, vector functions allow us to find specific solutions, giving a complete picture of the system's evolution over time.