Problem 28 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(27-32\) for \(r\) as a vector function of \(t .\) $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d \mathbf{r}}{d t}=(180 t) \mathbf{i}+\left(180 t-16 t^{2}\right) \mathbf{j}} \\ {\text { Initial condition: }} & {\mathbf{r}(0)=100 \mathbf{j}}\end{array} $$

Step-by-Step Solution

Answer

\( \mathbf{r}(t) = 90t^2 \mathbf{i} + (90t^2 - \frac{16}{3}t^3 + 100) \mathbf{j} \)

Step 1: Integrate the i-component

The differential equation provides the vector function derivative: \( \frac{d \mathbf{r}}{dt} = (180t) \mathbf{i} + (180t - 16t^2) \mathbf{j} \). To find \( \mathbf{r}(t) \), start by integrating the \( \mathbf{i} \)-component: \( \int 180t \, dt \). This yields \( 90t^2 + C_1 \), where \( C_1 \) is the constant of integration.

Step 2: Integrate the j-component

Next, integrate the \( \mathbf{j} \)-component given by \( 180t - 16t^2 \). Perform the integration: \( \int (180t - 16t^2) \, dt \), which results in \( 90t^2 - \frac{16}{3}t^3 + C_2 \), with \( C_2 \) as a constant of integration.

Step 3: Include the initial condition

Given the initial condition \( \mathbf{r}(0) = 100 \mathbf{j} \), substitute \( t = 0 \) into the expressions for the \( \mathbf{i} \) and \( \mathbf{j} \)-components. For the \( \mathbf{i} \)-component, \( 90(0)^2 + C_1 = 0 \) gives \( C_1 = 0 \). For the \( \mathbf{j} \)-component, \( 90(0)^2 - \frac{16}{3}(0)^3 + C_2 = 100 \) results in \( C_2 = 100 \).

Step 4: Write the complete solution

Combine the results from the integrations and the initial condition to write the complete solution for \( \mathbf{r}(t) \): \( \mathbf{r}(t) = (90t^2)\mathbf{i} + (90t^2 - \frac{16}{3}t^3 + 100)\mathbf{j} \).

Key Concepts

Initial Value Problem

An initial value problem in vector calculus involves finding a function that satisfies a given differential equation and fulfills specified initial conditions. In the problem stated here, we were given a differential equation

• \( \frac{d \mathbf{r}}{dt} = (180t) \mathbf{i} + (180t - 16t^2) \mathbf{j} \)
• and an initial condition \( \mathbf{r}(0) = 100 \mathbf{j} \).

The main goal is to find \( \mathbf{r}(t) \) such that when \( t = 0 \), the vector \( \mathbf{r}(t) \) equals the initial condition given, which ensures the solution is uniquely determined. By knowing the value of the function at this starting point, you have a pathway to trace the entire path of the function as it follows the given rate of change.

Integration

Integration is a technique in calculus used to find the antiderivative of a function. In solving our vector problem, integration helped us find the original vector function \( \mathbf{r}(t) \).

In this scenario, we perform separate integrations for each component of the vector function.

• For the \( \mathbf{i} \)-component, the integral \( \int 180t \, dt \) is evaluated to \( 90t^2 + C_1 \).
• The \( \mathbf{j} \)-component involves the integral \( \int (180t - 16t^2) \, dt \), resulting in \( 90t^2 - \frac{16}{3}t^3 + C_2 \).

Understanding integration is crucial because these integrals reverse the effect of differentiation, bringing us back to the original functions we started from before differentiation.

Constants of Integration

When finding antiderivatives through integration, a constant of integration appears. Since differentiation erases constant terms, every antiderivative comes with an arbitrary constant added to it. In vector functions, each component has its own constant of integration:

• \( C_1 \) for the \( \mathbf{i} \)-component, and
• \( C_2 \) for the \( \mathbf{j} \)-component.

These constants are critical because they allow us to tailor the general solution to satisfy specific initial conditions. Once the initial condition is given, these constants can be determined, which effectively adds a specific shift or translates the function to meet the initial conditions.

Differential Equations

Differential equations involve derivatives and their solutions that are functions representing how quantities change. In our example, the differential equation \( \frac{d \mathbf{r}}{dt} = (180t) \mathbf{i} + (180t - 16t^2) \mathbf{j} \) specifies how the change in \( \mathbf{r}(t) \) occurs over time.

Differential equations can be challenging because they encompass a broad range of possible functions. However, the initial conditions narrow down the possibilities to a specific function. Learning to handle differential equations is vital as they appear in numerous fields like physics, engineering, and economics.
By solving the differential equations, we determined how the position vector \( \mathbf{r}(t) \) evolves over time, complying with the predefined rate of change for each of its components.