Problem 16 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(11-20\) for \(\mathbf{r}\) as a vector function of \(t .\) $$ \begin{array}{c}{\text { Differential equation: }} \\ {\frac{d \mathbf{r}}{d t}=\left(\frac{t}{t^{2}+2}\right) \mathbf{i}-\left(\frac{t^{2}+1}{t-2}\right) \mathbf{j}+\left(\frac{t^{2}+4}{t^{2}+3}\right) \mathbf{k}} \\ {\text { Initial condition: } \mathbf{r}(0)=\mathbf{i}-\mathbf{j}+\mathbf{k}}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Integrate Each Component

To solve the problem, integrate each component of the vector function separately. For the i-component=\(\int \left(\frac{t}{t^2 + 2}\right) dt\), for the j-component = \(\int -\left(\frac{t^2 + 1}{t - 2}\right) dt\), and for the k-component = \(\int \left(\frac{t^2 + 4}{t^2 + 3}\right) dt\). Integrate each term independently.

Step 2: Integrate the i-component

For the i-component \( \int \left( \frac{t}{t^2 + 2} \right) dt \). Using substitution, let \( u = t^2 + 2 \), then \( du = 2t dt \), and the integral becomes \( \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \ln |u| + C_1 \). Substitute back \( u = t^2 + 2 \) to get \( \frac{1}{2} \ln |t^2 + 2| + C_1 \).

Step 3: Integrate the j-component

For the j-component \( \int -\left( \frac{t^2 + 1}{t - 2} \right) dt \). Break it into partial fractions or polynomial division may be required. After simplification and integration, it gives \(-t - 4 \ln |t - 2| + C_2\).

Step 4: Integrate the k-component

For the k-component \( \int \left( \frac{t^2 + 4}{t^2 + 3} \right) dt \). This can be simplified by polynomial division because the numerator has the same degree as the denominator. The integral simplifies to \( t + \frac{1}{3} \ln |t^2 + 3| + C_3 \).

Step 5: Combine to Form Vector Function

Combine the results from the integrals to form the general solution: \( \mathbf{r}(t) = \left( \frac{1}{2} \ln |t^2 + 2| + C_1 \right) \mathbf{i} + \left( -t - 4 \ln |t - 2| + C_2 \right) \mathbf{j} + \left( t + \frac{1}{3} \ln |t^2 + 3| + C_3 \right) \mathbf{k} \).

Step 6: Substitute Initial Condition

Use the initial condition \( \mathbf{r}(0) = \mathbf{i} - \mathbf{j} + \mathbf{k} \) to determine the constants \( C_1, C_2, \text{ and } C_3 \). Substitute \( t = 0 \) and equate to the initial condition components to find: \( C_1 = 0 \), \( C_2 = 1 \), \( C_3 = 0 \).

Step 7: Final Vector Function Solution

With the constants determined, the solution for \( \mathbf{r}(t) \) is: \( \mathbf{r}(t) = \left( \frac{1}{2} \ln |t^2 + 2| \right) \mathbf{i} + \left( -t - 4 \ln |t - 2| + 1 \right) \mathbf{j} + \left( t + \frac{1}{3} \ln |t^2 + 3| \right) \mathbf{k} \).

Key Concepts

Initial Value Problem

An initial value problem is a type of differential equation that comes with an additional set of conditions known as initial conditions. These conditions specify the values of the function or its derivatives at a particular point. This helps in finding a unique solution to the differential equation problem. In our case, the initial condition specifies that the vector function \( \mathbf{r}(0) \) equals \( \mathbf{i} - \mathbf{j} + \mathbf{k} \). This condition allows us to solve for the constants in the vector function after integration.

Understanding the initial value ensures the solution of a differential equation doesn't just float without a solid foundation. Without it, we could have infinitely many solutions, as the integration process introduces constant values (usually noted as \( C \)). By substituting the initial conditions into the integrated function, these constants are pinned down, giving us the particular solution.

Differential Equations

Differential equations are equations that relate a function to its derivatives. They play a crucial role in modeling real-world phenomena where change is persistent, like physics, engineering, and economics. In the context of vector calculus, we often see vector differential equations, where the functions involve vectors.

In this exercise, we deal with a vector differential equation \( \frac{d \mathbf{r}}{dt} = \left(\frac{t}{t^2+2}\right) \mathbf{i} - \left(\frac{t^2+1}{t-2}\right) \mathbf{j} + \left(\frac{t^2+4}{t^2+3}\right) \mathbf{k} \). Here, the three components of the vector \( \mathbf{r}(t) \) are functions in themselves and are derived independently during the integration process.

Differential equations can often look intimidating, but breaking them down into their component parts, as shown in the step-by-step solution, makes them much more approachable and simpler to solve.

Vector Functions

Vector functions are functions where each input, often a real number \( t \), corresponds to a vector output. In three dimensions, this output is expressed in terms of unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \). Such functions are useful for parametrizing curves and surfaces in space, describing trajectories, and much more.

In this exercise, the vector function \( \mathbf{r}(t) \) is defined in a three-dimensional space with components \( (x(t), y(t), z(t)) \). Each component is solved independently using integration techniques and then combined back into a vector form.

Think of vector functions as instructions that specify where to go in space over time, while differentiation and integration tell you how to move and adjust your path.

Integration Techniques

Integration techniques are crucial in solving differential equations, especially when dealing with complex mathematical expressions. When integrating, we often apply different strategies such as substitution, partial fraction decomposition, or polynomial division.

In the walkthrough of this exercise:

• The \( i \)-component is integrated using substitution, which simplifies the fraction \( \int \left( \frac{t}{t^2+2} \right) dt \).
• The \( j \)-component necessitates partial fraction decomposition or polynomial division, useful for expressions like \( \int -\left( \frac{t^2+1}{t-2} \right) dt \).
• For the \( k \)-component, polynomial division simplifies the integral \( \int \left( \frac{t^2+4}{t^2+3} \right) dt \).

Each technique has its rules of thumb, but the goal remains: simplify the expression to a form where integration is straightforward. Mastery of these techniques allows for the resolution of more complex integrals across various calculus problems.