Problem 19 - Thomas Calculus

Question

In Exercises 19 and \(20, \mathbf{r}(t)\) is the position vector of a particle in space at time \(t .\) Find the time or times in the given time interval when the velocity and acceleration vectors are orthogonal. $$ \mathbf{r}(t)=(t-\sin t) \mathbf{i}+(1-\cos t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi $$

Step-by-Step Solution

Answer

Velocity and acceleration are orthogonal at \(t = 0, \pi,\) and \(2\pi\).

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 each component of \(\mathbf{r}(t)\) to find:\[\mathbf{v}(t) = \frac{d}{dt}(t - \sin t)\mathbf{i} + \frac{d}{dt}(1 - \cos t)\mathbf{j}.\] Thus, \[\mathbf{v}(t) = (1 - \cos t)\mathbf{i} + (\sin t)\mathbf{j}.\]

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 each component of \(\mathbf{v}(t)\) to find: \[\mathbf{a}(t) = \frac{d}{dt}((1 - \cos t)\mathbf{i} + \sin t\mathbf{j}).\] Thus, \[\mathbf{a}(t) = (\sin t)\mathbf{i} + (\cos t)\mathbf{j}.\]

Step 3: Determine Orthogonality Condition

Velocity and acceleration vectors are orthogonal if their dot product is zero. Calculate the dot product of \(\mathbf{v}(t)\) and \(\mathbf{a}(t)\):\[(1 - \cos t)\sin t + \sin t\cos t = \sin t.\] Thus, the condition for orthogonality is \(\sin t = 0\).

Step 4: Solve the Orthogonality Equation

Solve \(\sin t = 0\) within the interval \(0 \leq t \leq 2\pi\). The solutions are \(t = 0\), \(t = \pi\), and \(t = 2\pi\).

Step 5: Verify the Solutions

Check these values of \(t\) in the original time interval. Since \(0 \leq t \leq 2\pi\) includes \(t = 0\), \(t = \pi\), and \(t = 2\pi\), these are valid solutions where the velocity and acceleration vectors are orthogonal.

Key Concepts

Orthogonal Vectors

Orthogonal vectors are vectors in space that are at a right angle to each other. When two vectors are orthogonal, their dot product equals zero. This is a fundamental concept in vector calculus that helps understand spatial relationships and interactions. In the context of the original exercise, when the velocity vector and acceleration vector are orthogonal, it indicates that there is no net component of acceleration in the direction of motion at that time. Mathematically, for vectors \( \mathbf{u} \) and \( \mathbf{v} \), they are orthogonal if their dot product \( \mathbf{u} \cdot \mathbf{v} = 0 \). This is a helpful tool in physics and engineering to analyze forces and movements.

Velocity Vector

A velocity vector is a vector that represents the rate of change of position of an object with respect to time. It provides both the speed and direction of the object's movement. To find the velocity vector, differentiate the position vector with respect to time. In the example from the exercise, the position vector is given by \( \mathbf{r}(t) = (t - \sin t) \mathbf{i} + (1 - \cos t) \mathbf{j} \). By differentiating it, the velocity vector \( \mathbf{v}(t) \) becomes \( (1 - \cos t) \mathbf{i} + (\sin t) \mathbf{j} \). This shows how the position of the particle changes with time, providing insights into its path and speed.

Acceleration Vector

An acceleration vector describes how the velocity of a particle changes over time. It is obtained by differentiating the velocity vector with respect to time. The acceleration vector gives us the direction and magnitude of the acceleration at any given point in time. From the exercise, the velocity vector \( \mathbf{v}(t) = (1 - \cos t) \mathbf{i} + (\sin t) \mathbf{j} \) is differentiated to find the acceleration vector \( \mathbf{a}(t) \), which is \( (\sin t) \mathbf{i} + (\cos t) \mathbf{j} \). This vector lets us understand how fast and in what manner the velocity is changing, essential for analyzing motion in physics and engineering.

Dot Product

The dot product of two vectors is a measure of the extent to which two vectors point in the same direction. It is calculated as the product of the magnitudes of the two vectors and the cosine of the angle between them. The dot product is a central calculation in determining orthogonality.For vectors \( \mathbf{u} = (u_1, u_2) \) and \( \mathbf{v} = (v_1, v_2) \), the dot product is given by \( \mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 \). In the supplied exercise, the orthogonality is checked through the dot product of the velocity vector \( \mathbf{v}(t) \) and the acceleration vector \( \mathbf{a}(t) \). With the equation \( (1 - \cos t)\sin t + \sin t\cos t = \sin t \), it highlights that when \( \sin t = 0 \), the vectors are orthogonal, thus making their dot product zero.