Problem 19 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(11-20\) for \(\mathbf{r}\) as a vector function of \(t .\) Differential equation: $$\quad \frac{d^{2} \mathbf{r}}{d t^{2}}=e^{t} \mathbf{i}-e^{-t} \mathbf{j}+4 e^{2} \mathbf{k}$$ Initial conditions: $$\mathbf{r}(0)=3 \mathbf{i}+\mathbf{j}+2 \mathbf{k} and \left.\frac{d \mathbf{r}}{d t}\right|_{t=0}=-\mathbf{i}+4 \mathbf{j}$$

Step-by-Step Solution

Answer

The solution steps are incomplete, so I cannot provide the final answer.

Step 1: Integrate the Acceleration

Given the differential equation \( \frac{d^{2} \mathbf{r}}{d t^{2}}=e^{t} \mathbf{i}-e^{-t} \mathbf{j}+4 e^{2} \mathbf{k} \), we integrate each component with respect to \( t \) to find the velocity vector \( \frac{d\mathbf{r}}{dt} \). - \( \int e^{t} \, dt = e^{t} + C_1 \) in the \( \mathbf{i} \) direction.- \( \int -e^{-t} \, dt = e^{-t} + C_2 \) in the \( \mathbf{j} \) direction.- \( \int 4 e^{2t} \, dt = 2 e^{2t} + C_3 \) in the \( \mathbf{k} \) direction.

Step 2: Determine the Velocity Function

Combine the integrated terms to form the velocity vector:\[ \mathbf{v}(t) = (e^{t} + C_1) \mathbf{i} + (e^{-t} + C_2) \mathbf{j} + (2 e^{2t} + C_3) \mathbf{k} \] Use the initial condition \( \left.\frac{d\mathbf{r}}{dt}\right|_{t=0} = -\mathbf{i} + 4 \mathbf{j} \) to find the constants. When \( t = 0 \), - \( e^{0} + C_1 = -1 \Rightarrow C_1 = -2 \)- \( e^{0} + C_2 = 4 \Rightarrow C_2 = 3 \)- \( 2 \times e^{0} + C_3 = 0 \Rightarrow C_3 = -2 \).

Key Concepts

Differential Equations

A differential equation is a mathematical equation that relates a function with its derivatives. In the context of vector calculus, differential equations often describe how a vector function changes with respect to one or more variables. Here, the differential equation involves the second derivative of a vector function \( \mathbf{r} \) with respect to time \( t \). The equation is given as:

• \(\frac{d^{2} \mathbf{r}}{d t^{2}} = e^{t} \mathbf{i} - e^{-t} \mathbf{j} + 4 e^{2} \mathbf{k}\)

This means that the acceleration vector of an object, given by the second derivative of its position vector, can have different behaviors in each of the \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) directions. The solution of a differential equation is a function, or a set of functions, that satisfies the equation. In this case, we're trying to determine the vector function \( \mathbf{r}(t) \). We start by integrating the given function to find the velocity and further integrate to find the position.

Initial Value Problems

Initial value problems involve finding a function that satisfies a differential equation and also meets specified conditions at a given point, known as initial conditions. In this example, our task is to find \( \mathbf{r}(t) \) such that it satisfies both the given differential equation and the initial conditions:

• \( \mathbf{r}(0) = 3 \mathbf{i} + \mathbf{j} + 2 \mathbf{k} \)
• \( \left.\frac{d \mathbf{r}}{dt}\right|_{t=0} = -\mathbf{i} + 4 \mathbf{j} \)

These initial conditions help determine the arbitrary constants that arise when we integrate the differential equation. Essentially, they "ground" the solution in real-world terms, specifying the state of the system at a starting point. This is crucial in applications where the physical position or velocity at an initial time is known.

Integration

Integration is a mathematical technique used to find functions when their derivative is known. In this scenario, integration is used to solve the differential equation to find the velocity vector from the acceleration vector. The process often involves constant terms represented here by \( C_1, C_2, \) and \( C_3 \). These are found by applying the initial conditions.

• For the \( \mathbf{i} \) direction, integrate \( e^{t} \) to get \( e^{t} + C_1 \).
• For the \( \mathbf{j} \) direction, integrate \( -e^{-t} \) to get \( e^{-t} + C_2 \).
• For the \( \mathbf{k} \) direction, integrate \( 4e^{2t} \) to get \( 2e^{2t} + C_3 \).

When we integrate each component of the acceleration vector, we stitch these results together into a single vector function for velocity. After applying the initial velocity conditions, we solve for the constants \( C_1, C_2, \) and \( C_3 \). This method allows us to accurately model the motion described by the differential equation.

Vector Functions

A vector function describes a vector in terms of one or more variables, typically over a parameter, such as time \( t \). In this exercise, \( \mathbf{r}(t) \) is a vector function that represents a particle's position at any given time. Vector functions are powerful as they allow us to express complex motions and trajectories in terms of simpler, scalar functions, which define each component of the vector separately.
The integration of the differential equation and respecting initial conditions helps transform these individual components back into the vector \( \mathbf{r} \). The velocity vector function is derived as:

• \( \mathbf{v}(t) = (e^{t} - 2) \mathbf{i} + (e^{-t} + 3) \mathbf{j} + (2 e^{2t} - 2) \mathbf{k} \)

In physics and engineering, such vector functions are essential to describe and predict the motion of objects under various forces. By using vector functions, we can perform operations like differentiation and integration globally on vector components, thus solving complex real-world problems with elegance.