Problem 14 - Thomas Calculus

Question

Solve the initial value problemsfor \(\mathbf{r}\) 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

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

Step 1: Understand the Problem

We need to find the vector function \( \mathbf{r}(t) \) that satisfies both the differential equation and the given initial condition.

Step 2: Identify Components of the Vector Function

The differential equation is given in terms of \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) components: \( \frac{d \mathbf{r}}{dt} = (t^3 + 4t) \mathbf{i} + t \mathbf{j} + 2t^2 \mathbf{k} \). This implies:\( \frac{dx}{dt} = t^3 + 4t \) \( \frac{dy}{dt} = t \) \( \frac{dz}{dt} = 2t^2 \).

Step 3: Integrate Each Component

Calculate the integral of each component separately to find \( x(t), y(t), \) and \( z(t) \). 1. \( x(t) = \int (t^3 + 4t) \, dt \) = \( \frac{t^4}{4} + 2t^2 + C_1 \)2. \( y(t) = \int t \, dt \) = \( \frac{t^2}{2} + C_2 \)3. \( z(t) = \int 2t^2 \, dt \) = \( \frac{2t^3}{3} + C_3 \)

Step 4: Apply Initial Conditions

Given \( \mathbf{r}(0) = \mathbf{i} + \mathbf{j} \), we can substitute \( t = 0 \) into our integrated functions:1. \( x(0) = \frac{0^4}{4} + 2 \cdot 0^2 + C_1 = 1 \rightarrow C_1 = 1 \)2. \( y(0) = \frac{0^2}{2} + C_2 = 1 \rightarrow C_2 = 1 \)3. \( z(0) = \frac{2 \cdot 0^3}{3} + C_3 = 0 \rightarrow C_3 = 0 \)

Step 5: Write the Final Vector Function

Substitute the values of \( C_1, C_2, \) and \( C_3 \) back into the integral results:\[ \mathbf{r}(t) = \left( \frac{t^4}{4} + 2t^2 + 1 \right) \mathbf{i} + \left( \frac{t^2}{2} + 1 \right) \mathbf{j} + \left( \frac{2t^3}{3} \right) \mathbf{k} \]

Key Concepts

Initial Value Problem

When dealing with an initial value problem in vector calculus, it's crucial to understand two main components: a differential equation and an initial condition. The differential equation describes how the vector function changes over time, while the initial condition provides a specific starting point for the function. This approach ensures that the solution is not just any function but one that passes through a defined point, giving us a unique and customized answer to our problem.

In this exercise, you are asked to find the vector function \( \mathbf{r}(t) \) that satisfies:

• the differential equation \( \frac{d \mathbf{r}}{d t}=\left(t^{3}+4 t\right) \mathbf{i}+t \mathbf{j}+2 t^{2} \mathbf{k} \)
• the initial condition \( \mathbf{r}(0)=\mathbf{i}+\mathbf{j} \)

The initial condition specifies that at time \( t = 0 \), the vector function \( \mathbf{r}(t) \) should equal \( \mathbf{i}+\mathbf{j} \). This means any solution must start from this initial vector, ensuring continuity and correctness of the solution from the very beginning.

Vector Function

A vector function is a function that assigns a vector to each value of its input, often representing quantities with multiple dimensions. In this context, the input is time \( t \), and the output is the vector \( \mathbf{r}(t) \), which has components in the directions of \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \). These components usually represent dimensions in three-dimensional space, like \( x, y, \) and \( z \).

The task is to find a specific vector function \( \mathbf{r}(t) \) that aligns with the given differential equation and satisfies the initial condition. Here, the vector function is expressed as:

• The \( \mathbf{i} \) component (\( x(t) \)) is influenced by \( t^{3} + 4t \)
• The \( \mathbf{j} \) component (\( y(t) \)) is changed by \( t \)
• The \( \mathbf{k} \) component (\( z(t) \)) is affected by \( 2t^{2} \)

Each directional component can be independently derived and then combined to represent the whole vector function \( \mathbf{r}(t) \), which captures how this vector evolves over time.

Differential Equation

A differential equation involves derivatives of a function and provides a way to connect rates of change with the values of the function. In vector calculus, it often helps describe how each part or component of a vector function changes with respect to time.

For this exercise, the differential equation is:\[ \frac{d \mathbf{r}}{dt} = (t^3 + 4t) \mathbf{i} + t \mathbf{j} + 2t^2 \mathbf{k} \]This tells us that:

• The rate of change of the \( x \)-component is \( \frac{dx}{dt} = t^3 + 4t \)
• The rate of change of the \( y \)-component is \( \frac{dy}{dt} = t \)
• The rate of change of the \( z \)-component is \( \frac{dz}{dt} = 2t^2 \)

Integration of these equations provides each component's function over time, crucial for constructing the vector function \( \mathbf{r}(t) \) that adheres to both the equation and the initial condition.

Integration of Components

Integration is the process used to determine the original function from its derivative, essentially reversing differentiation. In this context, each component of the vector function is integrated separately to find the vector function \( \mathbf{r}(t) \).

By integrating each derivative from the differential equation, we establish the individual component functions:

• \( x(t) = \int (t^3 + 4t) \, dt = \frac{t^4}{4} + 2t^2 + C_1 \)
• \( y(t) = \int t \, dt = \frac{t^2}{2} + C_2 \)
• \( z(t) = \int 2t^2 \, dt = \frac{2t^3}{3} + C_3 \)

The constants \( C_1, C_2, \text{ and } C_3 \) are constants of integration. These are determined by using the initial condition, in this case, \( \mathbf{r}(0) = \mathbf{i} + \mathbf{j} \), which means you calculate the values of the constants so the function aligns perfectly with the initial condition at \( t = 0 \). This step ensures that your solution accurately reflects both the change described by the differential equation and the specific starting point given in the problem.