Problem 15 - Calculus Early Transcendentals

Question

\(15-16\) Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position. $$\mathbf{a}(t)=\mathbf{i}+2 \mathbf{j}, \quad \mathbf{v}(0)=\mathbf{k}, \quad \mathbf{r}(0)=\mathbf{i}$$

Step-by-Step Solution

Answer

Velocity vector: \( \mathbf{v}(t) = t\mathbf{i} + 2t\mathbf{j} + \mathbf{k} \). Position vector: \( \mathbf{r}(t) = (\frac{t^2}{2} + 1)\mathbf{i} + t^2\mathbf{j} + t\mathbf{k} \).

Step 1: Integrate the acceleration vector to find the velocity vector

The given acceleration vector is \( \mathbf{a}(t) = \mathbf{i} + 2\mathbf{j} \). To find the velocity vector \( \mathbf{v}(t) \), integrate the acceleration with respect to time: \( \mathbf{v}(t) = \int (\mathbf{i} + 2\mathbf{j}) \, dt = (t + C_1)\mathbf{i} + (2t + C_2)\mathbf{j} + C_3\mathbf{k} \).

Step 2: Use initial conditions to find constants in velocity vector

We're given that \( \mathbf{v}(0) = \mathbf{k} \). Substituting \( t = 0 \) into the velocity equation gives us \( \mathbf{v}(0) = C_1\mathbf{i} + C_2\mathbf{j} + C_3\mathbf{k} = \mathbf{k} \), which implies \( C_1 = 0 \), \( C_2 = 0 \), and \( C_3 = 1 \). Therefore, the velocity vector is \( \mathbf{v}(t) = t\mathbf{i} + 2t\mathbf{j} + \mathbf{k} \).

Step 3: Integrate the velocity vector to find the position vector

To find the position vector \( \mathbf{r}(t) \), integrate the velocity vector: \( \mathbf{r}(t) = \int (t\mathbf{i} + 2t\mathbf{j} + \mathbf{k}) \, dt = (\frac{t^2}{2} + D_1)\mathbf{i} + (t^2 + D_2)\mathbf{j} + (t + D_3)\mathbf{k} \).

Step 4: Use initial conditions to find constants in position vector

We're given that \( \mathbf{r}(0) = \mathbf{i} \). Substituting \( t = 0 \) into the position equation gives \( \mathbf{r}(0) = D_1\mathbf{i} + D_2\mathbf{j} + D_3\mathbf{k} = \mathbf{i} \), which implies \( D_1 = 1 \), \( D_2 = 0 \), and \( D_3 = 0 \). Therefore, the position vector is \( \mathbf{r}(t) = (\frac{t^2}{2} + 1)\mathbf{i} + t^2\mathbf{j} + t\mathbf{k} \).

Key Concepts

Velocity Vector

A velocity vector is crucial in understanding how a particle moves in space over time. It provides us with information on both the direction and speed of the particle's movement.
In this exercise, to find the velocity vector, we start by integrating the given acceleration vector, \( \mathbf{a}(t) = \mathbf{i} + 2 \mathbf{j} \). This process involves finding the antiderivative of each component. After integration, we obtain the general form of the velocity vector:

• \(\mathbf{v}(t) = (t + C_1)\mathbf{i} + (2t + C_2)\mathbf{j} + C_3\mathbf{k}\).

Here, we introduce constants of integration, which we will later determine using the initial velocity conditions. This helps us precisely describe the particle's movement based on specific starting parameters.

Acceleration

Acceleration describes how the velocity of a particle changes over time, providing insight into changes in speed and direction. It is a vector quantity, meaning it has both magnitude and direction.
In this problem, the acceleration vector is given as \( \mathbf{a}(t) = \mathbf{i} + 2\mathbf{j} \), indicating a consistent change in velocity components:

• \(\mathbf{i}\) component increases linearly by 1 unit per time interval.
• \(\mathbf{j}\) component increases linearly by 2 units per time interval.

The absence of a \(\mathbf{k}\) component means there is no change in velocity along that axis, demonstrating how acceleration shapes the motion of a particle through modifying the velocity vector.

Integration

Integration, in calculus, is the process of finding the antiderivative or integral of a function. It is used to determine quantities such as area under a curve, but in this context, it helps us find velocity and position vectors from acceleration.
To find the velocity vector \(\mathbf{v}(t)\), we integrate the acceleration vector. Similarly, we obtain the position vector \(\mathbf{r}(t)\) by integrating the velocity vector. This involves finding integrals for each of the directional components separately. For example:

• \(\int \mathbf{i} \, dt = t\mathbf{i}\) results in the component \(t\mathbf{i}\).
• \(\int 2\mathbf{j} \, dt = 2t\mathbf{j}\) becomes \(2t\mathbf{j}\).

Each integration may include an undetermined constant, typically resolved using initial conditions.

Initial Conditions

Initial conditions serve as specific reference points that help us solve for unknowns in our equations, such as the constants arising from integration. They provide a snapshot of the system's state at the beginning of analysis.
In this exercise, we are provided with initial conditions for velocity and position:

• \(\mathbf{v}(0) = \mathbf{k} \)
• \(\mathbf{r}(0) = \mathbf{i} \)

Using these, we substitute \(t = 0\) into the respective integrated formulas to find values for constants \(C_1, C_2, C_3\) for velocity, and \(D_1, D_2, D_3\) for position.
This ensures the functions accurately match the system's initial state, crucial for making the equations practical and meaningful.