Problem 12 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

\(\mathbf{r}(t) = (180 \mathrm{h}t) \mathbf{i} + \left(90 t^2 - \frac{16}{3}t^3 + 100\right) \mathbf{j}\)

Step 1: Integrate the Differential Equation

Given that \(\frac{d \mathbf{r}}{dt}=(180 \mathrm{h}) \mathbf{i}+(180 t-16 t^{2}) \mathbf{j}\), integrate with respect to \(t\) to find \(\mathbf{r}(t).\)The integral of \(180 \mathrm{h}\) with respect to \(t\) is \(180 \mathrm{ht} \mathbf{i}.\)The integral of \(180t - 16t^2\) with respect to \(t\) is \(90t^2 - \frac{16}{3}t^3\) for the \(\mathbf{j}\) component. Thus,\[\mathbf{r}(t) = (180 \mathrm{h}t) \mathbf{i} + (90 t^2 - \frac{16}{3}t^3 + C) \mathbf{j}\] for a constant \(C.\)

Step 2: Apply the Initial Condition

Use the initial condition \(\mathbf{r}(0) = 100\mathbf{j}\) to find the constant \(C.\)Substitute \(t = 0\) into \(\mathbf{r}(t)\):\[\mathbf{r}(0) = (180 \cdot 0 \cdot \mathrm{h}) \mathbf{i} + (90 \cdot 0^2 - \frac{16}{3}\cdot 0^3 + C) \mathbf{j} = C \mathbf{j}\]Since \(\mathbf{r}(0) = 100 \mathbf{j}\), we have \(C = 100.\)

Step 3: Write the Final Solution

Substitute \(C = 100\) back into the expression for \(\mathbf{r}(t)\):\[\mathbf{r}(t) = (180 \mathrm{h}t) \mathbf{i} + \left(90 t^2 - \frac{16}{3}t^3 + 100 \right) \mathbf{j}\]Thus, \(\mathbf{r}(t)\) is the vector function representing the solution to the initial value problem.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are essential in describing various phenomena in engineering, physics, economics, and other fields where rates of change are significant. In a basic sense, a differential equation connects a function with its rate of change. The main aim when working with differential equations is to find the function that satisfies the equation. This is often achieved through integration, subject to any given initial conditions.
Understanding the types of differential equations can help in solving them effectively. Differential equations can be classified into:

• Ordinary Differential Equations (ODEs): These involve functions with a single independent variable.
• Partial Differential Equations (PDEs): These consist of functions with multiple independent variables.

Our original problem involves an ordinary differential equation with respect to time \(t\), meaning it tells us how the vector function \(\mathbf{r}(t)\) changes over time.

Initial Value Problem

An initial value problem is a type of differential equation that specifies the value of the solution at a particular point, often at the starting point \(t = 0\). Solving an initial value problem involves finding a function that satisfies both the differential equation and the initial condition.
In our exercise, the vector function \(\mathbf{r}(t)\) has been given an initial condition: \(\mathbf{r}(0) = 100\mathbf{j}\). This initial condition is instrumental because it allows us to determine any constants that appear after integrating. Without it, we can only find a general solution, but the initial condition provides the unique solution to the problem.
Honing in on the initial condition ensures that the solution fits the specific physical situation or real-world scenario described by the problem.

Integration

Integration is a fundamental concept in calculus, particularly useful in solving differential equations. It is the process of finding a function when its derivative is given. In many physical problems, the act of integration helps determine the accumulated or total quantity from a rate of change. In our vector calculus problem, integration is used to find the vector function \(\mathbf{r}(t)\) from its rate of change given by \(\frac{d \mathbf{r}}{dt}\).

• The \(\mathbf{i}\) component is integrated to \(180ht\), as it's a constant term relative to \(t\).
• The \(\mathbf{j}\) component is more complex and integrates to \(90t^2 - \frac{16}{3}t^3\) with an additional constant \(C\).

Remember, after integrating, applying any initial conditions is crucial to determining the constant (in this case, \(C = 100\)). This creates the unique solution fitting the initial value problem.