Problem 13 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(11-20\) for \(\mathbf{r}\) 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: Set up the Integration

To find the vector function \( \mathbf{r}(t) \), integrate each component of the given differential equation separately. The differential equation is \( \frac{d \mathbf{r}}{d t}=\frac{3}{2}(t+1)^{1/2} \mathbf{i}+e^{-t} \mathbf{j}+\frac{1}{t+1} \mathbf{k} \).We will integrate each of these component functions: 1. \( \frac{3}{2}(t+1)^{1/2} \) with respect to \( t \) for the \( \mathbf{i} \) component.2. \( e^{-t} \) with respect to \( t \) for the \( \mathbf{j} \) component.3. \( \frac{1}{t+1} \) with respect to \( t \) for the \( \mathbf{k} \) component.

Step 2: Integrate \(\mathbf{i}\) Component

Integrate the first component: \( \int \frac{3}{2}(t+1)^{1/2} \, dt \).Using substitution let \( u = t+1 \), then \( du = dt \).The integral becomes: \( \int \frac{3}{2} u^{1/2} \, du = \frac{3}{2} \cdot \frac{2}{3} u^{3/2} + C_1 = (t+1)^{3/2} + C_1 \).

Step 3: Integrate \(\mathbf{j}\) Component

Integrate the second component: \( \int e^{-t} \, dt \).The integral of \( e^{-t} \) is \( -e^{-t} + C_2 \).

Step 4: Integrate \(\mathbf{k}\) Component

Integrate the third component: \( \int \frac{1}{t+1} \, dt \).This is a simple logarithmic integration:\( \int \frac{1}{t+1} \, dt = \ln|t+1| + C_3 \).

Step 5: Construct the General Solution

Combine the integrated components to form the vector function: \[ \mathbf{r}(t) = \left((t+1)^{3/2} + C_1\right) \mathbf{i} + \left(-e^{-t} + C_2\right) \mathbf{j} + \left(\ln|t+1| + C_3\right) \mathbf{k} \].

Step 6: Apply Initial Condition

Use the initial condition \( \mathbf{r}(0) = \mathbf{k} \) to find the constants.At \( t = 0 \), every component of \( \mathbf{r} \) equals that of \( \mathbf{k} = 0 \mathbf{i} + 0 \mathbf{j} + 1 \mathbf{k} \), leading to:\((1)^{3/2} + C_1 = 0 \Rightarrow C_1 = -1\)\( -e^{0} + C_2 = 0 \Rightarrow C_2 = 1\)\( \ln|1| + C_3 = 1 \Rightarrow C_3 = 1\).

Step 7: Write the Particular Solution

Substitute \( C_1 = -1 \), \( C_2 = 1 \), and \( C_3 = 1 \) into the general solution to obtain the particular solution:\[ \mathbf{r}(t) = ((t+1)^{3/2} - 1) \mathbf{i} + (-e^{-t} + 1) \mathbf{j} + (\ln|t+1| + 1) \mathbf{k} \]

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. In this exercise, we are given a differential equation involving a vector function \( \mathbf{r}(t) \). Solving such equations helps us understand how quantities change over time or any other variable. Here, the goal is to find a function \( \mathbf{r}(t) \) whose rate of change is governed by the equation provided. This particular differential equation is given as \( \frac{d \mathbf{r}}{dt} = \frac{3}{2}(t+1)^{1/2} \mathbf{i} + e^{-t} \mathbf{j} + \frac{1}{t+1} \mathbf{k} \). Each part of this equation corresponds to a specific direction vector \( \mathbf{i}, \mathbf{j}, \mathbf{k} \).

• The \( \mathbf{i} \) component’s change is governed by \( \frac{3}{2}(t+1)^{1/2} \)
• The \( \mathbf{j} \) component’s change by \( e^{-t} \)
• The \( \mathbf{k} \) component’s change by \( \frac{1}{t+1} \)

Our task is to integrate each part to find the complete vector function \( \mathbf{r}(t) \). This converts a differentiation problem into an integration challenge, which we'll explore through different integration techniques.

Vector Calculus

Vector calculus is a field of mathematics that deals with vector fields and differentiating and integrating vector functions. It's essential for solving problems in physics and engineering, among others. In our task, \( \mathbf{r}(t) \) is a vector function, with \( \mathbf{i}, \mathbf{j}, \) and \( \mathbf{k} \) components, each function of \( t \). This is consistent with vector calculus as these methods often break down multi-dimensional problems into solvable parts.
By examining each component separately, we simplify the process of solving the differential equation. The challenge lies in understanding how vector quantities like \( \mathbf{r}(t) \) behave as functions of time. Calculating the vector derivatives and integrals often requires applying rules and methods specific to vector calculus.

• We converted the task into integrating scalar functions related to each vector component.
• Each integration step provides a piece of the vector function, which, when combined, gives us the complete solution.

This structured approach helps manage complexity and derive the complete vector solution accurately.

Integration Techniques

Integration is the process of finding a function given its derivative, which is what we are doing when solving differential equations. In this context, we use specific integration techniques for each component of the vector function.

• For the \( \mathbf{i} \) component, \( \int \frac{3}{2}(t+1)^{1/2} \, dt \), we employed substitution, setting \( u = t+1 \), leading to \( \int \frac{3}{2} u^{1/2} \, du \). The result is \( (t+1)^{3/2} + C_1 \).
• The \( \mathbf{j} \) component involves an exponential function, \( \int e^{-t} \, dt \), which integrates to \( -e^{-t} + C_2 \).
• For the \( \mathbf{k} \) component, \( \int \frac{1}{t+1} \, dt \), recognizing it as a natural logarithm function simplifies it to \( \ln|t+1| + C_3 \).

Integration techniques vary based on the function's form. By employing substitution, exponential rules, and logarithmic properties, we can systematically tackle each component of the vector differential equation.

Vector Functions

Vector functions are functions that take one or more variables and return a vector as a result. They are central to describing paths or trajectories in space. In our exercise, \( \mathbf{r}(t) \) is a vector function of \( t \) that we determined by integrating the differential equation.
Our vector function for this exercise becomes \( \mathbf{r}(t) = ((t+1)^{3/2} - 1) \mathbf{i} + (-e^{-t} + 1) \mathbf{j} + (\ln|t+1| + 1) \mathbf{k} \). Each component of this function represents the position in the \( x \), \( y \), and \( z \) directions at any time \( t \).

• \( \mathbf{i} \) component: represents the trajectory in the \( x \)-direction.
• \( \mathbf{j} \) component: changes in the \( y \)-direction over time.
• \( \mathbf{k} \) component: describes movement in the \( z \)-direction.

These components together form a path in three-dimensional space, starting from our initial condition \( \mathbf{r}(0) = \mathbf{k} \). By applying initial conditions, we align our functions with real-world scenarios, ensuring they describe the exact path or behavior needed.