Problem 11 - Thomas Calculus

Question

Solve the initial value problemsfor \(\mathbf{r}\) as a vector function of \(t .\) $$\begin{array}{ll}{\text { Differential equation: }} & {\frac{d \mathbf{r}}{d t}=-t \mathbf{i}-t \mathbf{j}-t \mathbf{k}} \\ {\text { Initial condition: }} & {\mathbf{r}(0)=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}}\end{array}$$

Step-by-Step Solution

Answer

\( \mathbf{r}(t) = \left(-\frac{t^2}{2} + 1\right) \mathbf{i} + \left(-\frac{t^2}{2} + 2\right) \mathbf{j} + \left(-\frac{t^2}{2} + 3\right) \mathbf{k} \).

Step 1: Integrate the Differential Equation

To solve the vector differential equation \( \frac{d \mathbf{r}}{dt} = -t \mathbf{i} - t \mathbf{j} - t \mathbf{k} \), we integrate each component with respect to \( t \). This gives us: \[ \mathbf{r}(t) = \int (-t)\, dt \cdot \mathbf{i} + \int (-t)\, dt \cdot \mathbf{j} + \int (-t)\, dt \cdot \mathbf{k}. \] Performing the integration, we obtain: \[ \mathbf{r}(t) = \left(-\frac{t^2}{2} + C_{i}\right) \mathbf{i} + \left(-\frac{t^2}{2} + C_{j}\right) \mathbf{j} + \left(-\frac{t^2}{2} + C_{k}\right) \mathbf{k}, \] where \( C_{i}, C_{j}, \) and \( C_{k} \) are integration constants.

Step 2: Apply Initial Conditions

Use the initial condition \( \mathbf{r}(0) = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \) to solve for the constants \( C_{i}, C_{j}, \) and \( C_{k} \). Substitute \( t = 0 \) into the general solution: \[ \mathbf{r}(0) = \left(-\frac{0^2}{2} + C_{i}\right) \mathbf{i} + \left(-\frac{0^2}{2} + C_{j}\right) \mathbf{j} + \left(-\frac{0^2}{2} + C_{k}\right) \mathbf{k} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}. \] This implies \( C_{i} = 1 \), \( C_{j} = 2 \), and \( C_{k} = 3 \).

Step 3: Write the Final Solution

Substitute the constants back into the expression derived from the integration to find the specific solution. This gives us: \[ \mathbf{r}(t) = \left(-\frac{t^2}{2} + 1\right) \mathbf{i} + \left(-\frac{t^2}{2} + 2\right) \mathbf{j} + \left(-\frac{t^2}{2} + 3\right) \mathbf{k}. \] This is the vector function \( \mathbf{r}(t) \) that satisfies both the differential equation and the initial condition.

Key Concepts

Vector Calculus

Vector calculus extends the traditional operations of calculus into the realm of vectors. A vector has both magnitude and direction, making them useful for describing physical phenomena. In vector calculus, we often face problems involving vector fields, which assign a vector to every point in a region. This can be particularly useful in physics to represent fields such as velocity or electromagnetic fields.

One common problem in vector calculus is solving differential equations that involve vectors. A differential equation for vectors will involve derivatives with respect to a variable, such as time \( t \). The goal is to find a vector function \( \mathbf{r}(t) \) that satisfies both the differential equation and any provided initial conditions. The problem at hand asks us to find a vector function \( \mathbf{r} \) as a function of time \( t \) that adheres to a given differential equation and satisfies a specified condition at \( t = 0 \).

• Vectors are directional quantities.
• Vector calculus helps in modeling real-world scenarios involving multiple dimensions.
• Understanding the geometry of vector functions is crucial for solving differential equations.

Differential Equations

Differential equations play a vital role in numerous engineering and scientific problems. These equations involve unknown functions and their derivatives, and they describe a variety of phenomena such as motion, growth, and decay processes.

In the context of vector functions, solving a differential equation typically involves finding the function that describes how vectors change over time or space. The process involves integrating the given derivative to recover the original function. In our case, the differential equation \( \frac{d \mathbf{r}}{dt} = -t \mathbf{i} - t \mathbf{j} - t \mathbf{k} \) provides the rate of change of the vector with respect to time. By integrating, we find the vector function \( \mathbf{r}(t) \) which encompasses the integral constant that gets determined by the initial conditions.

• Differential equations can be ordinary or partial, based on the type of derivatives involved.
• Solutions to these equations often involve integration.
• Initial conditions are necessary to pinpoint exact solutions from general ones.

Integration in Calculus

Integration is a core operation in calculus used to determine the cumulative values like areas under curves, total quantities, or in our specific case, the overall vector function from its rate of change.

When dealing with vectors, integration operates component-wise, meaning each component of the vector is integrated separately. In our example, we individually integrate \(-t\) with respect to \(t\) to find each component of the vector \( \mathbf{r}(t) \). This results in an expression involving constants \(C_i, C_j,\) and \(C_k\), which are determined using initial conditions.

• Integration is often seen as the inverse operation of differentiation.
• For vectors, component-wise integration provides a means to recover the full vector function.
• Constants of integration are solved using initial conditions to provide unique solutions.