91Ó°ÊÓ

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 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 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 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.