Problem 162 - Calculus Volume 3

Question

Find the velocity, acceleration, and speed of a particle with the given position function. \(\mathbf{r}(t)=\langle\sin t, t, \cos t\rangle\). The graph is shown here:

Step-by-Step Solution

Answer

Velocity is \( \langle \cos t, 1, -\sin t \rangle \), acceleration is \( \langle -\sin t, 0, -\cos t \rangle \), speed is \( \sqrt{2} \).

Step 1: Understand the Position Function

The position function of the particle is given by \( \mathbf{r}(t) = \langle \sin t, t, \cos t \rangle \). This function describes the position of the particle in 3-dimensional space as a function of time \( t \).

Step 2: Differentiate to Find Velocity

The velocity of the particle, \( \mathbf{v}(t) \), is the derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \). Differentiate each component: \( \mathbf{v}(t) = \frac{d}{dt} \langle \sin t, t, \cos t \rangle = \langle \cos t, 1, -\sin t \rangle \).

Step 3: Differentiate to Find Acceleration

The acceleration of the particle, \( \mathbf{a}(t) \), is the derivative of the velocity vector \( \mathbf{v}(t) \) with respect to \( t \). Differentiate each component: \( \mathbf{a}(t) = \frac{d}{dt} \langle \cos t, 1, -\sin t \rangle = \langle -\sin t, 0, -\cos t \rangle \).

Step 4: Calculate Speed

Speed is the magnitude of the velocity vector \( \mathbf{v}(t) \). Compute the magnitude using the formula for Euclidean norm: \( |\mathbf{v}(t)| = \sqrt{(\cos t)^2 + 1^2 + (-\sin t)^2} = \sqrt{\cos^2 t + 1 + \sin^2 t} = \sqrt{1 + 1} = \sqrt{2} \).

Step 5: Summarize the Results

The velocity \( \mathbf{v}(t) = \langle \cos t, 1, -\sin t \rangle \), the acceleration \( \mathbf{a}(t) = \langle -\sin t, 0, -\cos t \rangle \), and the constant speed is \( \sqrt{2} \).

Key Concepts

Understanding the Position Function

In vector calculus, the position function \( \mathbf{r}(t) = \langle \sin t, t, \cos t \rangle \) is crucial for describing the motion of a particle in three-dimensional space. This function encapsulates the particle's trajectory as it moves over time. Each component of the vector—\( \sin t \), \( t \), and \( \cos t \)—represents the particle's position along the x, y, and z axes, respectively.

By plugging in different values of \( t \), you can determine where the particle is located at specific times. This function acts as the starting point for more analyses, such as determining velocity and acceleration, which are derived from this position function.

Exploring Velocity and Acceleration

Velocity and acceleration are dynamic aspects of motion derived from the position function. The velocity function \( \mathbf{v}(t) \) is obtained by differentiating the position function \( \mathbf{r}(t) \) with respect to time \( t \). This reveals how fast the particle's position is changing. For the function given, \( \mathbf{v}(t) = \langle \cos t, 1, -\sin t \rangle \), showing the rate of change along each axis.

Similarly, acceleration \( \mathbf{a}(t) \) is found by differentiating the velocity function. It describes how the particle's velocity is changing over time, acting as a further derivative. For this example, \( \mathbf{a}(t) = \langle -\sin t, 0, -\cos t \rangle \), illustrating how changes in velocity occur along each axis. Understanding these concepts is essential for studying motion dynamics.

The Role of Euclidean Norm in Calculating Speed

The Euclidean norm is pivotal for determining the speed of an object in motion, essentially measuring the ''length'' or magnitude of a vector. In our context, the speed of a particle is the magnitude of its velocity vector \( \mathbf{v}(t) \). This is calculated by taking the square root of the sum of the squares of its components: \( \sqrt{(\cos t)^2 + 1^2 + (-\sin t)^2} \).

Regardless of time, this simplifies to \( \sqrt{2} \), indicating constant speed. This method showcases how fundamental the Euclidean norm is for converting a multidimensional description of movement into a singular, comprehensive measure of speed.

Differentiation in Calculus: A Powerful Tool for Analysis

Differentiation in calculus provides the foundation to analyze and interpret changing quantities within mathematical contexts. By applying differentiation to our position function \( \mathbf{r}(t) \), we can find both the velocity and acceleration, which tell us about the motion of a particle.

Differentiation entails computing derivatives, which are essentially rates of change. In our example, the velocity \( \mathbf{v}(t) = \langle \cos t, 1, -\sin t \rangle \) results from differentiating the position function. Further differentiating the velocity gives us the acceleration \( \mathbf{a}(t) = \langle -\sin t, 0, -\cos t \rangle \).

• This enables exploration of physical dynamics.
• Differentiation explains transitions in motion, force, or momentum.

This forms the backbone of calculus's application in physics and beyond.