Problem 30 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(27-32\) for \(r\) as a vector function of \(t .\) $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d \mathbf{r}}{d t}=\left(t^{3}+4 t\right) \mathbf{i}+t \mathbf{j}+2 t^{2} \mathbf{k}} \\\ {\text { Initial condition: }} & {\mathbf{r}(0)=\mathbf{i}+\mathbf{j}}\end{array} $$

Step-by-Step Solution

Answer

\( \mathbf{r}(t) = \left(\frac{t^4}{4} + 2t^2 + 1\right) \mathbf{i} + \left(\frac{t^2}{2} + 1\right) \mathbf{j} + \frac{2t^3}{3} \mathbf{k} \) is the solution.

Step 1: Understand the Problem

We need to find the vector function \( \mathbf{r}(t) \) for given vector differential equation \( \frac{d \mathbf{r}}{d t}=(t^{3}+4t) \mathbf{i}+t \mathbf{j}+2t^{2} \mathbf{k} \) and initial condition \( \mathbf{r}(0)=\mathbf{i}+\mathbf{j} \).

Step 2: Integrate the Differential Equation

To find \( \mathbf{r}(t) \), integrate the differential equation component-wise. The integral of the \( \mathbf{i} \) component is \( \int (t^3 + 4t) \, dt \), the \( \mathbf{j} \) component is \( \int t \, dt \), and the \( \mathbf{k} \) component is \( \int 2t^2 \, dt \).

Step 3: Calculate the Integrals

Calculate each integral:- For \( \mathbf{i} \): \( \int (t^3 + 4t) \, dt = \frac{t^4}{4} + 2t^2 + C_1 \).- For \( \mathbf{j} \): \( \int t \, dt = \frac{t^2}{2} + C_2 \).- For \( \mathbf{k} \): \( \int 2t^2 \, dt = \frac{2t^3}{3} + C_3 \).

Step 4: Form the General Solution

Combine the results to form \( \mathbf{r}(t) = \left( \frac{t^4}{4} + 2t^2 + C_1 \right) \mathbf{i} + \left( \frac{t^2}{2} + C_2 \right) \mathbf{j} + \left( \frac{2t^3}{3} + C_3 \right) \mathbf{k} \).

Step 5: Apply the Initial Condition

Substitute \( t = 0 \) into \( \mathbf{r}(t) \) for \( \mathbf{r}(0) \) and equate it to \( \mathbf{i} + \mathbf{j} \) to find the constants: - \( C_1 \) corresponds to \( \mathbf{i} \): \( C_1 = 1 \).- \( C_2 \) corresponds to \( \mathbf{j} \): \( C_2 = 1 \).- \( C_3 \) corresponds to \( \mathbf{k} \): \( C_3 = 0 \).

Step 6: Write the Final Solution

Substitute \( C_1 = 1 \), \( C_2 = 1 \), and \( C_3 = 0 \) back into the general solution to obtain the final vector function: \[ \mathbf{r}(t) = \left( \frac{t^4}{4} + 2t^2 + 1 \right) \mathbf{i} + \left( \frac{t^2}{2} + 1 \right) \mathbf{j} + \frac{2t^3}{3} \mathbf{k} \]

Key Concepts

Vector Differential Equations

A vector differential equation is an extension of ordinary differential equations (ODEs) in which the unknown function is a vector function. Instead of solving for a single variable, we are solving for multiple components or dimensions at once. In the exercise, the vector differential equation is given by \( \frac{d \mathbf{r}}{dt}=(t^{3}+4t) \mathbf{i}+t \mathbf{j}+2t^{2} \mathbf{k} \). Here, \( \mathbf{r}(t) \) represents a vector function of the variable \( t \), and it consists of three components along the \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) axes. Each of these components is a function that can be solved separately, which leads us to the next important concept: component-wise integration.

Integration of Vector Functions

To solve vector differential equations, one common method used is the integration of vector functions. This involves integrating each component of the vector function separately. Here, we aim to find a vector \( \mathbf{r}(t) \), which means integrating its components

• \( \int (t^3 + 4t) \, dt \) for the \( \mathbf{i} \)-component,
• \( \int t \, dt \) for the \( \mathbf{j} \)-component,
• \( \int 2t^2 \, dt \) for the \( \mathbf{k} \)-component.

After finding each indefinite integral, we introduce constants of integration \( C_1, C_2, \) and \( C_3 \) corresponding to each component. This way, we obtain a general form of the vector \( \mathbf{r}(t) \) that includes these constants.

Initial Conditions

Initial conditions in differential equations help determine the specific solution out of the infinite possibilities generated by indefinite integration. The initial condition provided in the problem is \( \mathbf{r}(0)=\mathbf{i}+\mathbf{j} \). This means when \( t = 0 \), the vector function \( \mathbf{r}(t) \) must satisfy this condition. By substituting \( t = 0 \) into the general form of \( \mathbf{r}(t) \), we get equations for solving the constants:

• \( C_1 \) is adjusted for the \( \mathbf{i} \)-component resulting in \( C_1 = 1 \).
• \( C_2 \) is adjusted for the \( \mathbf{j} \)-component resulting in \( C_2 = 1 \).
• \( C_3 \) is adjusted for the \( \mathbf{k} \)-component, where any value is possible since it doesn't impact the initial condition \( \mathbf{r}(0) = \mathbf{i} + \mathbf{j} \), thus \( C_3 = 0 \).

The integration constants are crucial, as they allow the solution to stay true to the initial conditions.

Component-wise Integration

Component-wise integration refers to the process of integrating each component of a vector differential equation separately. Each component of the vector \( \mathbf{r}(t) \) is treated as its own integral problem:

• The \( \mathbf{i} \)-component is integrated as \( \int (t^3 + 4t) \, dt \), yielding \( \frac{t^4}{4} + 2t^2 + C_1 \).
• The \( \mathbf{j} \)-component is integrated as \( \int t \, dt \), yielding \( \frac{t^2}{2} + C_2 \).
• The \( \mathbf{k} \)-component is integrated as \( \int 2t^2 \, dt \), yielding \( \frac{2t^3}{3} + C_3 \).

These integrations allow us to construct a complete vector function by combining the components, which results in \( \mathbf{r}(t) \). The general solution \( \mathbf{r}(t) = \left( \frac{t^4}{4} + 2t^2 + 1 \right) \mathbf{i} + \left( \frac{t^2}{2} + 1 \right) \mathbf{j} + \frac{2t^3}{3} \mathbf{k} \) fulfills both the differential equation and the initial conditions through component-wise analysis.