Problem 162 - Calculus 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: \( \langle \cos t, 1, -\sin t \rangle \); Acceleration: \( \langle -\sin t, 0, -\cos t \rangle \); Speed: \( \sqrt{2} \).

Step 1: Understand the Problem

You are given the position function of a particle \( \mathbf{r}(t) = \langle \sin t, t, \cos t \rangle \). You need to find the velocity, acceleration, and speed of this particle as functions of time \( t \).

Step 2: Find the Velocity

The velocity function \( \mathbf{v}(t) \) is the derivative of the position function \( \mathbf{r}(t) \) with respect to \( t \). Differentiate each component of \( \mathbf{r}(t) \).\[\begin{align*}\frac{d}{dt}\sin t &= \cos t, \\frac{d}{dt}t &= 1, \\frac{d}{dt}\cos t &= -\sin t.\end{align*}\]Therefore, \( \mathbf{v}(t) = \langle \cos t, 1, -\sin t \rangle \).

Step 3: Calculate the Acceleration

Acceleration \( \mathbf{a}(t) \) is the derivative of the velocity function \( \mathbf{v}(t) \). Differentiate each component of \( \mathbf{v}(t) \).\[\begin{align*}\frac{d}{dt}\cos t &= -\sin t, \\frac{d}{dt}1 &= 0, \\frac{d}{dt}(-\sin t) &= -\cos t.\end{align*}\]Therefore, \( \mathbf{a}(t) = \langle -\sin t, 0, -\cos t \rangle \).

Step 4: Determine the Speed of the Particle

Speed is the magnitude of the velocity vector. Calculate it using the formula \( \text{speed} = \| \mathbf{v}(t) \| \).\[\begin{align*}\| \mathbf{v}(t) \| &= \sqrt{(\cos t)^2 + (1)^2 + (-\sin t)^2} \&= \sqrt{\cos^2 t + 1 + \sin^2 t}.\end{align*}\]Using the identity \( \cos^2 t + \sin^2 t = 1 \), we have:\[\| \mathbf{v}(t) \| = \sqrt{1+1} = \sqrt{2}.\]Thus, the speed of the particle is \( \sqrt{2} \).

Key Concepts

Understanding Velocity in Multivariable Calculus

In multivariable calculus, velocity is a key concept that describes the rate of change of a particle's position in space over time. Think of it as how fast and in which direction something is moving.
To determine the velocity of a particle given its position function, such as \( \mathbf{r}(t) = \langle \sin t, t, \cos t \rangle \), we take the derivative of each component with respect to time \( t \). This gives us the velocity function \( \mathbf{v}(t) \) as a vector that includes all components of motion.

• **Differentiating Components**: By differentiating \( \sin t, t, \cos t \), we find \([\cos t, 1, -\sin t]\). This vector represents how each axis (x, y, z) changes with time.
• **Direction and Magnitude**: The direction of \( \mathbf{v}(t) \) tells you where the particle is headed, while the magnitude (or length of the vector) relates to its speed.

Understanding these aspects can help you track and predict motion in a space defined by multiple variables. This same process applies to any position function you may encounter.

Exploring Acceleration in Multivariable Motion

Acceleration in multivariable calculus is the rate at which velocity changes over time. This is analogous to how quickly a car speeds up or slows down on a road. The particle's acceleration is derived by differentiating the velocity function \(\mathbf{v}(t) = \langle \cos t, 1, -\sin t \rangle \).
Calculating \(\mathbf{a}(t)\) involves taking the derivative of each component of the velocity vector. This yields \( \langle -\sin t, 0, -\cos t \rangle \), providing insight into how the particle's velocity changes in each spatial dimension.

• **Negative Components**: Here, both \(-\sin t\) and \(-\cos t\) indicate that the velocity decreases in those directions as \( t \) increases.
• **Constant Component**: The 0 in the middle indicates no change in the velocity component in the \( y \)-direction, signaling a constant velocity along that axis.

By understanding acceleration, you can predict how the path of a moving object will curve or shift, which is essential for modeling dynamic systems.

Grasping the Concept of Speed with Vectors

Speed, unlike velocity, doesn't involve direction. It simply indicates how quickly something moves in space. In multivariable calculus, speed is derived from finding the magnitude (or norm) of the velocity vector \( \mathbf{v}(t) \).
For the given function, the formula used is \( \| \mathbf{v}(t) \| = \sqrt{(\cos t)^2 + (1)^2 + (-\sin t)^2} \). Simplified, this results in \( \| \mathbf{v}(t) \| = \sqrt{2} \), meaning the particle's speed is constant over time at \( \sqrt{2} \).

• **Directionless Measure**: Speed solely accounts for how much ground is covered per unit of time, ignoring the direction of movement entirely.
• **Consistent Speed**: In this example, the constant speed shows uniform motion along the particle's path.

Thus, even when multiple variables affect a particle's motion, the speed represents a singular, stable measure of movement, crucial for evaluating dynamics. Multivariable calculus uses these concepts to solve real-world problems involving complex motion.