Problem 22 - Thomas Calculus

Question

In Exercises \(19-22, \mathbf{r}(t)\) is the position of a particle in space at time \(t .\) Find the angle between the velocity and acceleration vectors at time \(t=0\) $$ \mathbf{r}(t)=\frac{4}{9}(1+t)^{3 / 2} \mathbf{i}+\frac{4}{9}(1-t)^{3 / 2} \mathbf{j}+\frac{1}{3} t \mathbf{k} $$

Step-by-Step Solution

Answer

The angle between the velocity and acceleration vectors at \(t=0\) is \(90^\circ\).

Step 1: Find the Velocity Vector

To find the velocity vector \(\mathbf{v}(t)\), we need to take the derivative of the position vector \(\mathbf{r}(t)\) with respect to time \(t\). The derivative of \(\frac{4}{9}(1+t)^{3/2}\) with respect to \(t\) is \(\frac{4}{6}(1+t)^{1/2}\). Similarly, the derivative of \(\frac{4}{9}(1-t)^{3/2}\) is \(-\frac{4}{6}(1-t)^{1/2}\), and the derivative of \(\frac{1}{3}t\) is \(\frac{1}{3}\). Therefore, \(\mathbf{v}(t) = \frac{2}{3}(1+t)^{1/2}\mathbf{i} - \frac{2}{3}(1-t)^{1/2}\mathbf{j} + \frac{1}{3}\mathbf{k}\).

Step 2: Find the Acceleration Vector

The acceleration vector \(\mathbf{a}(t)\) is the derivative of the velocity vector \(\mathbf{v}(t)\) with respect to \(t\). The derivative of \(\frac{2}{3}(1+t)^{1/2}\) is \(\frac{1}{3}(1+t)^{-1/2}\), and the derivative of \(-\frac{2}{3}(1-t)^{1/2}\) is \(\frac{1}{3}(1-t)^{-1/2}\). Since the \(\mathbf{k}\)-component does not depend on \(t\), its derivative is \(0\). Thus, \(\mathbf{a}(t) = \frac{1}{3}(1+t)^{-1/2}\mathbf{i} + \frac{1}{3}(1-t)^{-1/2}\mathbf{j} + 0\mathbf{k}\).

Step 3: Evaluate at Time t=0

Substitute \(t=0\) into \(\mathbf{v}(t)\) and \(\mathbf{a}(t)\). For velocity, \(\mathbf{v}(0) = \frac{2}{3}(1)^{1/2}\mathbf{i} - \frac{2}{3}(1)^{1/2}\mathbf{j} + \frac{1}{3}\mathbf{k} = \frac{2}{3}\mathbf{i} - \frac{2}{3}\mathbf{j} + \frac{1}{3}\mathbf{k}\). For acceleration, \(\mathbf{a}(0) = \frac{1}{3}(1)^{-1/2}\mathbf{i} + \frac{1}{3}(1)^{-1/2}\mathbf{j} + 0\mathbf{k} = \frac{1}{3}\mathbf{i} + \frac{1}{3}\mathbf{j}\).

Step 4: Calculate the Dot Product

The dot product \(\mathbf{v}(0) \cdot \mathbf{a}(0)\) is calculated as follows: \([\frac{2}{3}\mathbf{i} - \frac{2}{3}\mathbf{j} + \frac{1}{3}\mathbf{k}] \cdot [\frac{1}{3}\mathbf{i} + \frac{1}{3}\mathbf{j} + 0\mathbf{k}] = \frac{2}{3} \cdot \frac{1}{3} + (-\frac{2}{3}) \cdot \frac{1}{3} + \frac{1}{3} \cdot 0 = \frac{2}{9} - \frac{2}{9} = 0.\) Thus, the dot product is zero.

Step 5: Determine the Angle

Since the dot product \(\mathbf{v}(0) \cdot \mathbf{a}(0) = 0\), the cosine of the angle \(\theta\) between the two vectors is \(\cos(\theta) = 0\). In vector mathematics, if \(\cos(\theta) = 0\), it implies that the angle \(\theta\) is \(90^\circ\).

Key Concepts

Velocity Vector

In vector calculus, when you want to understand how a particle's position changes over time, you look at the velocity vector. The velocity vector is simply the derivative of the position vector, denoted as \( \mathbf{v}(t) \). It's crucial for determining how fast and in what direction a particle is moving at any given moment.

To find the velocity vector from the given position vector \( \mathbf{r}(t) = \frac{4}{9}(1+t)^{3/2} \mathbf{i} + \frac{4}{9}(1-t)^{3/2} \mathbf{j} + \frac{1}{3} t \mathbf{k} \), we compute the derivative of each component with respect to \( t \):

• The \( \mathbf{i} \) component becomes \( \frac{2}{3} (1+t)^{1/2} \).
• The \( \mathbf{j} \) component simplifies to \(-\frac{2}{3} (1-t)^{1/2} \).
• The \( \mathbf{k} \) component remains constant at \( \frac{1}{3} \).

This yields the velocity vector \( \mathbf{v}(t) = \frac{2}{3}(1+t)^{1/2}\mathbf{i} - \frac{2}{3}(1-t)^{1/2}\mathbf{j} + \frac{1}{3}\mathbf{k} \). Understanding the velocity vector helps us predict movement and directions of particles.

Acceleration Vector

Acceleration is a measure of how quickly the velocity of a particle changes. The acceleration vector is obtained by taking the derivative of the velocity vector. This lets us see how both the speed and direction of a particle's motion are changing.

For the velocity vector \( \mathbf{v}(t) = \frac{2}{3}(1+t)^{1/2}\mathbf{i} - \frac{2}{3}(1-t)^{1/2}\mathbf{j} + \frac{1}{3}\mathbf{k} \), the calculation of the acceleration vector involves computing:

• The \( \mathbf{i} \) component to be \( \frac{1}{3}(1+t)^{-1/2} \).
• The \( \mathbf{j} \) component becomes \( \frac{1}{3}(1-t)^{-1/2} \).
• The \( \mathbf{k} \) component, since it is constant, has a derivative of \( 0 \).

Thus, the acceleration vector is represented as \( \mathbf{a}(t) = \frac{1}{3}(1+t)^{-1/2}\mathbf{i} + \frac{1}{3}(1-t)^{-1/2}\mathbf{j} + 0\mathbf{k} \). Analyzing acceleration provides insights into changing forces and variations in particle trajectory over time.

Dot Product

The dot product is an operation that gives a single number from two vectors. It highlights the degree to which two vectors, such as velocity and acceleration, go in the same direction. Calculating this product involves multiplying corresponding components and adding their results.

For example, the velocity vector at time \( t=0 \) is \( \mathbf{v}(0) = \frac{2}{3}\mathbf{i} - \frac{2}{3}\mathbf{j} + \frac{1}{3}\mathbf{k} \), and the acceleration vector is \( \mathbf{a}(0) = \frac{1}{3}\mathbf{i} + \frac{1}{3}\mathbf{j} + 0\mathbf{k} \). The dot product becomes:
\[ (\frac{2}{3} \cdot \frac{1}{3}) + (-\frac{2}{3} \cdot \frac{1}{3}) + (\frac{1}{3} \cdot 0) = \frac{2}{9} - \frac{2}{9} = 0.\]
Finding a zero dot product indicates that the velocity and acceleration vectors are perpendicular, a key insight into their relative orientations.

Angle Between Vectors

Understanding the angle between vectors in vector calculus, especially with velocity and acceleration vectors, helps illustrate their relationship. When two vectors are perpendicular, their dot product is zero, and this info helps find the angle between them.

Geometrically, the angle \( \theta \) between two vectors \( \mathbf{v} \) and \( \mathbf{a} \) is found using:
\[ \cos(\theta) = \frac{\mathbf{v} \cdot \mathbf{a}}{||\mathbf{v}|| \cdot ||\mathbf{a}||}.\]
Here, since their dot product \( \mathbf{v}(0) \cdot \mathbf{a}(0) = 0 \, \cos(\theta) = 0 \) implies that \( \theta = 90^\circ \). This perpendicular relationship means the velocity and acceleration vectors are orthogonal, and their movements involve no direct interference. These insights are beneficial for both basic and applied mathematics, including physics to solve real-world problems.