Problem 70 - Calculus 3

Question

Find the antiderivative \(\mathbf{r}^{\prime}(t)=\cos (2 t) \mathbf{i}-2 \sin t \mathbf{j}+\frac{1}{1+t^{2}} \mathbf{k}\) that satisfies the initial condition \(\mathbf{r}(0)=3 \mathbf{i}-2 \mathbf{j}+\mathbf{k}\)

Step-by-Step Solution

Answer

\( \mathbf{r}(t) = \left(\frac{1}{2}\sin(2t) + 3\right) \mathbf{i} + (2\cos(t) - 4) \mathbf{j} + (\tan^{-1}(t) + 1) \mathbf{k} \)

Step 1: Find the Antiderivative of Each Component

The derivative function is \( \mathbf{r}^{\prime}(t)=\cos (2 t) \mathbf{i}-2 \sin t \mathbf{j}+\frac{1}{1+t^{2}} \mathbf{k} \). To find the antiderivative, integrate each component separately.For \( \mathbf{i} \), integrate \( \cos(2t) \) with respect to \( t \). The antiderivative is \( \frac{1}{2}\sin(2t) + C_1 \).For \( \mathbf{j} \), integrate \( -2\sin(t) \) with respect to \( t \). The antiderivative is \( 2\cos(t) + C_2 \).For \( \mathbf{k} \), integrate \( \frac{1}{1+t^2} \) with respect to \( t \). The antiderivative is \( \tan^{-1}(t) + C_3 \).

Step 2: Form the Antiderivative Function

Combine the components with their corresponding integration constants to form the vector function:\[ \mathbf{r}(t) = \left(\frac{1}{2}\sin(2t) + C_1\right) \mathbf{i} + (2\cos(t) + C_2) \mathbf{j} + (\tan^{-1}(t) + C_3) \mathbf{k} \]

Step 3: Apply the Initial Condition

Use the initial condition \( \mathbf{r}(0) = 3 \mathbf{i} - 2 \mathbf{j} + \mathbf{k} \) to solve for constants.1. Substituting \( t = 0 \) into the \( \mathbf{i} \)-component: \( \frac{1}{2}\sin(0) + C_1 = 3 \rightarrow C_1 = 3 \).2. Substituting \( t = 0 \) into the \( \mathbf{j} \)-component: \( 2\cos(0) + C_2 = -2 \rightarrow 2 + C_2 = -2 \rightarrow C_2 = -4 \).3. Substituting \( t = 0 \) into the \( \mathbf{k} \)-component: \( \tan^{-1}(0) + C_3 = 1 \rightarrow C_3 = 1 \).

Step 4: Write the Final Antiderivative Function

Substitute the constants \( C_1 = 3 \), \( C_2 = -4 \), and \( C_3 = 1 \) back into the general antiderivative function:\[ \mathbf{r}(t) = \left(\frac{1}{2}\sin(2t) + 3\right) \mathbf{i} + (2\cos(t) - 4) \mathbf{j} + (\tan^{-1}(t) + 1) \mathbf{k} \]

Key Concepts

Antiderivative

The concept of an antiderivative is akin to "undoing" a derivative. When we take the derivative of a function, we find its rate of change. To revert this process, we find the antiderivative, or the original function up to a constant. In vector calculus, we often deal with functions that have multiple components. Here, each component requires its own antiderivative. For instance, in the given problem, the components of the vector function are integrated separately.

• For the component along the vector , we find the antiderivative of \( \cos(2t) \), resulting in \( \frac{1}{2}\sin(2t) \).
• For the component along the vector , the integration of \( -2\sin(t) \) yields \( 2\cos(t) \).
• The component along the vector , \( \frac{1}{1+t^2} \), results in \( \tan^{-1}(t) \) after integration.

To complete the antiderivative, we add a constant of integration, uniquely identifying each component in the final vector function.

Initial Condition

Initial conditions provide specific values that help uniquely determine the constants involved in antiderivatives. In the absence of initial conditions, more than one function can fulfill the criteria set by a derivative. Without initial values, each component of the vector function could have an infinite number of possibilities depending on the constants added during integration.
For the exercise, the initial condition is given as \( \mathbf{r}(0) = 3\mathbf{i} - 2\mathbf{j} + \mathbf{k} \). This means that at \( t = 0 \), the vector function must equal the stated vector.

• It ensures that \( C_1 = 3 \) in the component matching 3.
• For the component, \( C_2 = -4 \), aligning with \( -2 \).
• While the component mandates \( C_3 = 1 \).

Using these values, the antiderivative becomes a specific solution rather than a general one.

Vector Function

A vector function consists of multiple components, each corresponding to a particular axis in space. It maps real numbers to vectors, meaning each input of \( t \) gives a unique vector output. This is crucial in physics and engineering for representing paths or forces that change over time.
In the context of this exercise, the vector function \( \mathbf{r}(t) \) is derived from \( \mathbf{r}^{\prime}(t) \) by determining its antiderivative. Each component of \( \mathbf{r}(t) \) is formed by integrating the respective components of the derivative function:

• \( \frac{1}{2}\sin(2t) + C_1 \) in the component.
• \( 2\cos(t) + C_2 \) in the component.
• \( \tan^{-1}(t) + C_3 \) in the component.

Once the proper constants are found using the initial condition, the vector function accurately reflects its specified trajectory or configuration at \( t = 0 \).

Integration Constants

Integration constants appear as a result of integrating each component of a derivative. They represent a fundamental property of indefinite integrals—each integral solution is part of a family of functions, differing by a constant.
To solve for these constants, we use initial conditions, which provide the exact values needed to configure the antiderivative into a single, particular solution from the broader possible range. In our particular problem:

• \( C_1 \) determines the constant addition to the component.
• \( C_2 \) adjusts the component.
• \( C_3 \) finalizes the component.

The integration constants transform the general antiderivative back into the specific vector function originally differentiated to yield \( \mathbf{r}^{\prime}(t) \). This process ensures that the antiderivative satisfies given conditions and accurately describes the system or path being modeled.