Problem 11 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(11-20\) for \(\mathbf{r}\) as a vector function of \(t .\) $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d \mathbf{r}}{d t}=-t \mathbf{i}-t \mathbf{j}-t \mathbf{k}} \\ {\text { Initial condition: }} & {\mathbf{r}(0)=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}}\end{array} $$

Step-by-Step Solution

Answer

\(\mathbf{r}(t) = (1 - \frac{t^2}{2})\mathbf{i} + (2 - \frac{t^2}{2})\mathbf{j} + (3 - \frac{t^2}{2})\mathbf{k}\).

Step 1: Integrate the Differential Equation

We start with the given differential equation: \(\frac{d \mathbf{r}}{dt} = -t \mathbf{i} - t \mathbf{j} - t \mathbf{k}\). To find \(\mathbf{r}(t)\), we integrate each component with respect to \(t\). The integral of \(-t\) with respect to \(t\) is \(-\frac{t^2}{2}\). Thus, \(\mathbf{r}(t) = -\frac{t^2}{2} \mathbf{i} - \frac{t^2}{2} \mathbf{j} - \frac{t^2}{2} \mathbf{k} + \mathbf{C}\), where \(\mathbf{C}\) is a vector of integration constants.

Step 2: Apply the Initial Condition

Using the initial condition \(\mathbf{r}(0) = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}\), substitute \(t = 0\) into \(\mathbf{r}(t)\): \(0\cdot \mathbf{i} + 0\cdot \mathbf{j} + 0\cdot \mathbf{k} + \mathbf{C} = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}\). This implies that \(\mathbf{C} = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}\).

Step 3: Combine Results for the Solution

Substitute \(\mathbf{C}\) back into the general solution from Step 1: \(\mathbf{r}(t) = -\frac{t^2}{2} \mathbf{i} - \frac{t^2}{2} \mathbf{j} - \frac{t^2}{2} \mathbf{k} + (\mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k})\). Simplify to obtain: \(\mathbf{r}(t) = (1 - \frac{t^2}{2})\mathbf{i} + (2 - \frac{t^2}{2})\mathbf{j} + (3 - \frac{t^2}{2})\mathbf{k}\). This is the vector function \(\mathbf{r}(t)\).

Key Concepts

Differential Equations

Differential equations are equations involving a function and its derivatives. They are widely used to describe various physical phenomena, such as motion and change over time. In our exercise, the differential equation describes how a vector function \( \mathbf{r}(t) \) changes with respect to time \( t \).

To solve it, we need to integrate with respect to \( t \). Integration reverses the effect of differentiation, allowing us to recover the original function \( \mathbf{r}(t) \). Consider the equation
\[ \frac{d \mathbf{r}}{dt} = -t \mathbf{i} - t \mathbf{j} - t \mathbf{k} \]
This indicates how each component of the vector \( \mathbf{r}(t) \) is changing at any time \( t \). Once we integrate each component, we find the vector function that describes the behavior of the system over time.

Initial Value Problem

An initial value problem in the context of differential equations is a problem where you are given a differential equation along with an initial condition. The initial condition specifies the value of the function (or vector function here) at a particular point in time, usually \( t = 0 \).

In our exercise, the initial condition is \( \mathbf{r}(0) = \mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k} \). This tells us where the vector starts at time \( t = 0 \).

• The initial condition helps us determine the specific constants of integration.
• It allows us to refine the general solution to meet specific scenarios or constraints described by the problem.

To solve the initial value problem, we first integrate to obtain the general solution, and then use the initial condition to find any necessary constants.

Vector Functions

Vector functions describe quantities that have both magnitude and direction as they depend on a variable, usually time. In physics and engineering, vector functions are used to model various phenomena such as velocity and force.

In our specific exercise, \( \mathbf{r}(t) \) is a vector function of time \( t \) where each component varies as a function of time. The expression we derived
\[ \mathbf{r}(t) = (1 - \frac{t^2}{2})\mathbf{i} + (2 - \frac{t^2}{2})\mathbf{j} + (3 - \frac{t^2}{2})\mathbf{k} \]
shows how \( \mathbf{r}(t) \) is expressed in terms of its components, offering insights into how each dimension of the vector changes over time.

• The \( \mathbf{i}, \mathbf{j}, \text{ and } \mathbf{k} \) components can be seen as standard unit vectors along the x, y, and z axes respectively.
• This vector function provides a complete picture of the motion and position of an object in three-dimensional space over time.

Integration Constants

Whenever you integrate a function to find its antiderivative, an integration constant is introduced. This constant accounts for the fact that there are potentially infinite functions that can have the same derivative.

In our problem, after integrating the differential equation, we initially obtain a general solution for \( \mathbf{r}(t) \) that includes a vector of constants \( \mathbf{C} \):
\[ \mathbf{r}(t) = - \frac{t^2}{2} \mathbf{i} - \frac{t^2}{2} \mathbf{j} - \frac{t^2}{2} \mathbf{k} + \mathbf{C} \]

• \( \mathbf{C} \) is determined by the initial condition, thus specifying the precise solution that meets the specified conditions of the problem.
• By evaluating the initial condition at \( t = 0 \), we solve for \( \mathbf{C} \), ensuring the general solution aligns with the initial data provided.

This makes the solution more specific, guiding us to real world applications where exact starting values are known or measured.