Problem 14 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(11-20\) for \(\mathbf{r}\) as a vector function of \(t .\) $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d \mathbf{r}}{d t}=\left(t^{3}+4 t\right) \mathbf{i}+t \mathbf{j}+2 t^{2} \mathbf{k}} \\\ {\text { Initial condition: }} & {\mathbf{r}(0)=\mathbf{i}+\mathbf{j}}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to solve the vector differential equation \( \frac{d \mathbf{r}}{dt} = (t^3 + 4t) \mathbf{i} + t \mathbf{j} + 2t^2 \mathbf{k} \) with the initial condition \( \mathbf{r}(0) = \mathbf{i} + \mathbf{j} \). This involves finding the vector function \( \mathbf{r}(t) \).

Step 2: Integrate the Component Functions

Integrate each component function separately with respect to \( t \). For the \( \mathbf{i} \)-component, \( \int (t^3 + 4t) \, dt = \frac{t^4}{4} + 2t^2 + C_1 \). For the \( \mathbf{j} \)-component, \( \int t \, dt = \frac{t^2}{2} + C_2 \). For the \( \mathbf{k} \)-component, \( \int 2t^2 \, dt = \frac{2t^3}{3} + C_3 \).

Step 3: Apply the Initial Condition

Use the initial condition \( \mathbf{r}(0) = \mathbf{i} + \mathbf{j} \) to calculate the constants. At \( t = 0 \), the expressions become: \( \frac{0^4}{4} + 2(0)^2 + C_1 = 1 \), giving \( C_1 = 1 \); \( \frac{0^2}{2} + C_2 = 1 \), giving \( C_2 = 1 \); \( \frac{2(0)^3}{3} + C_3 = 0 \), giving \( C_3 = 0 \).

Step 4: Write the Solution

Combine the results from each component to write the solution in vector form. \( \mathbf{r}(t) = \left( \frac{t^4}{4} + 2t^2 + 1 \right) \mathbf{i} + \left( \frac{t^2}{2} + 1 \right) \mathbf{j} + \left( \frac{2t^3}{3} \right) \mathbf{k} \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function to its derivatives. They are crucial in modeling situations where quantities change at rates that are dependent on other variables. In this exercise, we are dealing with a vector differential equation, which means we have multiple equations that relate the components of the vector function to their rates of change with respect to time, \( t \).

In our problem, the vector differential equation given is \( \frac{d \mathbf{r}}{dt} = (t^3 + 4t) \mathbf{i} + t \mathbf{j} + 2t^2 \mathbf{k} \). This can be interpreted as a vector whose direction and magnitude depend on the functions \( t^3 + 4t \), \( t \), and \( 2t^2 \) along the \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) axes, respectively.

To solve such equations, our goal is to determine a function \( \mathbf{r}(t) \) that satisfies the equation. This involves determining all possible paths or trajectories of the vector as \( t \) varies. This fundamental understanding is essential since many real-world phenomena can be described using differential equations, from physical forces to economic models.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation that comes with additional constraints known as initial conditions. These conditions specify the value of the unknown function at a given point, and they are critical in finding a unique solution to the differential equation.

In the exercise, we have the initial condition \( \mathbf{r}(0) = \mathbf{i} + \mathbf{j} \). This tells us the position of the vector at time \( t = 0 \). Initial conditions are vital as they help in pinpointing the specific trajectory out of the infinite possible solutions generated by the differential equation.

When solving an IVP, after finding the general solution to the differential equation, the initial condition is applied to find the constants that make up the specific solution to the problem. This customization of the solution to satisfy the initial condition is what makes it an initial value problem rather than a general one.

Integration

Integration is a powerful tool in calculus that is used to solve differential equations. It is essentially the reverse process of differentiation, focusing on finding a function given its derivative. In the context of vector-valued functions, each component of the vector is integrated separately to solve the system.

For our differential equation \( \frac{d \mathbf{r}}{dt} = (t^3 + 4t) \mathbf{i} + t \mathbf{j} + 2t^2 \mathbf{k} \), we integrate each component:

• \( \mathbf{i} \)-component: \( \int (t^3 + 4t) \, dt = \frac{t^4}{4} + 2t^2 + C_1 \)
• \( \mathbf{j} \)-component: \( \int t \, dt = \frac{t^2}{2} + C_2 \)
• \( \mathbf{k} \)-component: \( \int 2t^2 \, dt = \frac{2t^3}{3} + C_3 \)

Applying the initial condition \( \mathbf{r}(0) = \mathbf{i} + \mathbf{j} \), we solve for the constants \( C_1, C_2, \text{ and } C_3 \). Integration allows us to go from the derivative form of a differential equation to a general solution, which then becomes specific when initial conditions are applied. This is how integration seamlessly complements the solving of differential equations.