Problem 29 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(27-32\) for \(r\) as a vector function of \(t .\) $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d \mathbf{r}}{d t}=\frac{3}{2}(t+1)^{1 / 2} \mathbf{i}+e^{-t} \mathbf{j}+\frac{1}{t+1} \mathbf{k}} \\ {\text { Initial condition: }} & {\mathbf{r}(0)=\mathbf{k}}\end{array} $$

Step-by-Step Solution

Answer

\( \mathbf{r}(t) = ((t+1)^{3/2} - 1) \mathbf{i} + (-e^{-t} + 1) \mathbf{j} + (\ln|t+1| + 1) \mathbf{k} \).

Step 1: Integrate the i-Component

The differential equation for the i-component is \( \frac{dr_i}{dt} = \frac{3}{2}(t+1)^{1/2} \). Integrate both sides with respect to \( t \) to find \( r_i(t) \):\[ r_i(t) = \int \frac{3}{2}(t+1)^{1/2} \, dt = (t+1)^{3/2} + C_i\]where \( C_i \) is the constant of integration.

Step 2: Integrate the j-Component

For the j-component, the differential equation is \( \frac{dr_j}{dt} = e^{-t} \). Integrating both sides with respect to \( t \):\[ r_j(t) = \int e^{-t} \, dt = -e^{-t} + C_j\]where \( C_j \) is the constant of integration.

Step 3: Integrate the k-Component

The equation for the k-component is \( \frac{dr_k}{dt} = \frac{1}{t+1} \). Integrate both sides with respect to \( t \):\[ r_k(t) = \int \frac{1}{t+1} \, dt = \ln|t+1| + C_k\]where \( C_k \) is the constant of integration.

Step 4: Apply Initial Condition to Determine Constants

The initial condition \( \mathbf{r}(0) = \mathbf{k} \) means \( r_i(0) = 0 \), \( r_j(0) = 0 \), and \( r_k(0) = 1 \).- For \( r_i(0) = 0 \): \[ 1 + C_i = 0 \Rightarrow C_i = -1 \]- For \( r_j(0) = 0 \): \[ -e^{0} + C_j = 0 \Rightarrow C_j = 1 \]- For \( r_k(0) = 1 \): \[ \ln|1| + C_k = 1 \Rightarrow C_k = 1 \]

Step 5: Write Final Solution for Vector Function

With constants \( C_i = -1 \), \( C_j = 1 \), and \( C_k = 1 \) found from the initial condition, substitute back into the expressions for each component:\[\mathbf{r}(t) = ((t+1)^{3/2} - 1) \mathbf{i} + (-e^{-t} + 1) \mathbf{j} + (\ln|t+1| + 1) \mathbf{k}\]

Key Concepts

Initial Value Problem

An initial value problem in mathematics is a type of problem that seeks to find a function that satisfies a differential equation and meets certain specified values at a given point. In the exercise, we have a differential equation for a vector function \( \mathbf{r}(t) \) and a specified initial condition \( \mathbf{r}(0) = \mathbf{k} \). This condition tells us the specific values the vector components must take when \( t = 0 \).

The importance of the initial value lies in its role to uniquely determine a solution to the differential equation. Without it, multiple solutions could exist. Solving the initial value problem means integrating the differential equation and adjusting the constants of integration to satisfy the given initial condition.

In practice, you would:

• First, solve the differential equations by integrating them.
• Then, use the initial conditions to determine any constants of integration that arise from integrating.

Differential Equation Integration

Differential equation integration is the process of finding a function from its derivative. In vector calculus, this often involves integrating each component of a vector separately. Given a differential equation like \( \frac{d\mathbf{r}}{dt} = \mathbf{F}(t) \), you integrate each component function of \( \mathbf{F}(t) \) with respect to \( t \).

In this exercise:

• For the \( i \)-component, integrate \( \frac{dr_i}{dt} = \frac{3}{2}(t+1)^{1/2} \).
• For the \( j \)-component, integrate \( \frac{dr_j}{dt} = e^{-t} \).
• For the \( k \)-component, integrate \( \frac{dr_k}{dt} = \frac{1}{t+1} \).

Each integration results in a function plus a constant of integration. These constants are crucial, as they allow us to adjust the solution to meet the initial conditions.

Vector Function

A vector function assigns a vector to each point in its domain. Unlike scalar functions, vector functions such as \( \mathbf{r}(t) \) consist of multiple components, each of which is a scalar function of the variable \( t \). These functions are crucial in physics and engineering, representing quantities with both magnitude and direction that change over time.

In this problem, the vector function \( \mathbf{r}(t) \) is composed of three separate component functions of \( t \). By integrating each component separately and combining them, we construct the solution to the initial value problem. The final form of \( \mathbf{r}(t) \) captures not only the path traced by the vector but also how it changes at each moment \( t \).

Vector functions are helpful in modeling phenomena such as:

• The path of a moving body.
• Changing forces over time.
• Flow of fluids with varying velocities.

Components of Vectors

Vectors have components that define their behavior in different directions, typically in the \( \mathbb{i} \), \( \mathbb{j} \), and \( \mathbb{k} \) directions in three-dimensional space. Each component is like a piece of the vector that points in one of the coordinate axes. Solving the differential equation for each component separately is a key step in handling vector calculus problems.

For our vector function \( \mathbf{r}(t) \), the components are:

• \( \mathbf{i} \)-component: Represents the component of the vector in the x-direction.
• \( \mathbf{j} \)-component: Represents the component in the y-direction.
• \( \mathbf{k} \)-component: Represents the component in the z-direction (or along the depth).

By working with each component independently, we can apply the rules of integration to simpler scalar functions rather than dealing with the entire vector equation as one complex entity. This step-by-step component integration allows for a clearer understanding of how each part of the vector changes over time.