Problem 17 - Thomas Calculus

Question

\({r}(t)\) is the position of a particle in space at time \(t .\) Find the angle between the velocity and acceleration vectors at time \(t=0 .\) \begin{equation} \mathbf{r}(t)=\left(\ln \left(t^{2}+1\right) \mathbf{i}+\left(\tan ^{-1} t\right) \mathbf{j}+\sqrt{t^{2}+1} \mathbf{k}\right. \end{equation}

Step-by-Step Solution

Answer

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

Step 1: Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Differentiate \( \mathbf{r}(t) = \left( \ln(t^2 + 1) \right) \mathbf{i} + \left( \tan^{-1}t \right) \mathbf{j} + \sqrt{t^2 + 1} \mathbf{k} \) to obtain the velocity vector.\[\mathbf{v}(t) = \mathbf{r}'(t) = \left( \frac{2t}{t^2 + 1} \right) \mathbf{i} + \left( \frac{1}{1 + t^2} \right) \mathbf{j} + \left( \frac{t}{\sqrt{t^2 + 1}} \right) \mathbf{k}\]

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 time \( t \). Differentiate \( \mathbf{v}(t) \) to find \( \mathbf{a}(t) \).\[\mathbf{a}(t) = \mathbf{v}'(t) = \left( \frac{2(t^2 + 1) - 2t(2t)}{(t^2 + 1)^2} \right) \mathbf{i} + \left( \frac{-2t}{(1 + t^2)^2} \right) \mathbf{j} + \left( \frac{\sqrt{t^2 + 1} - t^2 / \sqrt{t^2 + 1}}{t^2 + 1} \right) \mathbf{k}\]

Step 3: Evaluate Vectors at t=0

Substitute \( t = 0 \) into \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \) to find the velocity and acceleration vectors at time \( t = 0 \).\[\mathbf{v}(0) = \left( 0 \right) \mathbf{i} + \left( 1 \right) \mathbf{j} + \left( 0 \right) \mathbf{k} = \mathbf{j}\]\[\mathbf{a}(0) = \left( 2 \right) \mathbf{i} + \left( 0 \right) \mathbf{j} + \left( 1 \right) \mathbf{k}\]

Step 4: Calculate Dot Product

Compute the dot product of \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \). Since \( \mathbf{v}(0) = \mathbf{j} \) and \( \mathbf{a}(0) = 2\mathbf{i} + \mathbf{k} \), the dot product is:\[\mathbf{v}(0) \cdot \mathbf{a}(0) = (0)(2) + (1)(0) + (0)(1) = 0\]

Step 5: Determine Magnitudes

Find the magnitudes of \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \). The magnitude of a vector \( \mathbf{x} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \) is given by \( \sqrt{a^2 + b^2 + c^2} \).\[\|\mathbf{v}(0)\| = \sqrt{0^2 + 1^2 + 0^2} = 1\]\[\|\mathbf{a}(0)\| = \sqrt{2^2 + 0^2 + 1^2} = \sqrt{5}\]

Step 6: Calculate the Angle Cosine

Use the dot product formula to find the cosine of the angle \( \theta \) between \( \mathbf{v}(0) \) and \( \mathbf{a}(0) \):\[\cos \theta = \frac{\mathbf{v}(0) \cdot \mathbf{a}(0)}{\|\mathbf{v}(0)\| \cdot \|\mathbf{a}(0)\|} = \frac{0}{1 \cdot \sqrt{5}} = 0\]This implies that the angle between them is 90°.

Key Concepts

Particle Motion

Particle motion is a vital concept in physics and calculus that describes how a particle moves through space over time. Imagine a particle moving through a three-dimensional space like a speck of dust in a room. Its position changes as time progresses. We can represent this motion mathematically using a position vector, often denoted as \( \mathbf{r}(t) \), where \( t \) represents time.

Understanding particle motion involves analyzing how this position vector changes. This can offer insights into the particle's speed, direction, and path taken through space. By differentiating the position vector with respect to time, we get the velocity vector, which describes how fast and in what direction the particle is moving at any given point in time.

The movement from one position to another over time is fundamental to many physics problems, helping predict where and how fast objects move. It's a foundational concept not just in physics but also in engineering, robotics, and animations.

Velocity Vector

In vector calculus, the velocity vector is derived from the position vector of a moving particle. It tells us how fast the position is changing and in what direction. Mathematically, the velocity vector \( \mathbf{v}(t) \) is the first derivative of the position vector \( \mathbf{r}(t) \) with respect to time \( t \).

For the given position vector \( \mathbf{r}(t) = \left( \ln(t^2 + 1) \right) \mathbf{i} + \left( \tan^{-1}t \right) \mathbf{j} + \sqrt{t^2 + 1} \mathbf{k} \), the velocity vector is:

• \( \mathbf{v}(t) = \frac{2t}{t^2 + 1} \mathbf{i} + \frac{1}{1 + t^2} \mathbf{j} + \frac{t}{\sqrt{t^2 + 1}} \mathbf{k} \)

Each component of the velocity vector corresponds to the rate of change of the respective components of the position vector.

Understanding the velocity vector helps us determine the direction of the particle's motion and how it changes with time. it's crucial for problems involving speed and trajectory and helps in visualizing the particle's path in space.

Acceleration Vector

Acceleration is the next step in understanding particle motion. The acceleration vector tells us how the velocity of the particle changes over time. It’s essentially the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to time.

To find the acceleration vector \( \mathbf{a}(t) \), differentiate the velocity vector once more. For our given problem, the acceleration vector is:

• \( \mathbf{a}(t) = \left( \frac{2(t^2 + 1) - 2t(2t)}{(t^2 + 1)^2} \right) \mathbf{i} + \left( \frac{-2t}{(1 + t^2)^2} \right) \mathbf{j} + \left( \frac{\sqrt{t^2 + 1} - t^2 / \sqrt{t^2 + 1}}{t^2 + 1} \right) \mathbf{k} \)

At any point in time, the acceleration vector reveals how quickly the particle is speeding up or slowing down and in which direction.

Calculating acceleration is key in many applications, such as vehicle motion analysis, physics simulations, and determining forces acting on objects.

Dot Product

The dot product, also known as the scalar product, is a fundamental operation in vector calculus. It takes two vectors and returns a single number (a scalar). It is especially useful in determining the angle between two vectors and in physics to find work done by or on a force.

The formula for the dot product of two vectors \( \mathbf{a} \) and \( \mathbf{b} \) is \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \).

In our exercise, the dot product was used to determine if the velocity and acceleration vectors at \( t = 0 \) were perpendicular. Given:

• \( \mathbf{v}(0) = \mathbf{j} \)
• \( \mathbf{a}(0) = 2 \mathbf{i} + \mathbf{k} \)

The dot product is zero, indicating that these vectors are orthogonal (at a 90-degree angle to each other).

This concept is vital for understanding geometry of vectors, and more broadly, predicting the behavior of systems in motion.

Angle Between Vectors

Finding the angle between two vectors is a common problem in vector calculus and physics. It helps us understand the spatial relationship between directions in which the vectors point. This is especially significant when examining forces and directions in physics.

The angle \( \theta \) between two vectors \( \mathbf{a} \) and \( \mathbf{b} \) can be found using the dot product formula:

• \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\| \|\mathbf{b}\|} \)

Here, \( \| \mathbf{a} \| \) and \( \| \mathbf{b} \| \) are the magnitudes of the vectors. In our example, the cosine of the angle was zero, which means the angle \( \theta \) is 90 degrees, confirming the vectors are perpendicular.

Understanding angles between vectors is crucial for determining relative directionality and impact of forces in physics and engineering. Knowing whether vectors are perpendicular informs us about independence in directions and helps optimize designs, routes, or solve motion-related problems efficiently.