Problem 93 - Calculus 3

Question

Find the minimum speed of a particle traveling along the curve \(\mathbf{r}(t)=\langle t+\cos t, t-\sin t\rangle t \in[0,2 \pi)\)

Step-by-Step Solution

Answer

The minimum speed is \(\sqrt{3 - 2\sqrt{2}}\).

Step 1: Understand the Problem Statement

We need to find the minimum speed of a particle moving along the path described by the vector function \(\mathbf{r}(t) = \langle t + \cos t, t - \sin t \rangle\) for \(t \in [0, 2\pi)\). Speed is given by the magnitude of the velocity, which is the derivative of the position vector function with respect to time.

Step 2: Find the Velocity Function

To find the velocity of the particle, differentiate the position vector \(\mathbf{r}(t)\) with respect to \(t\):\[ \mathbf{v}(t) = \frac{d}{dt} \langle t + \cos t, t - \sin t \rangle = \langle 1 - \sin t, 1 - \cos t \rangle \]

Step 3: Calculate the Speed Function

The speed of the particle is the magnitude of the velocity vector. So, calculate:\[ \|\mathbf{v}(t)\| = \sqrt{(1 - \sin t)^2 + (1 - \cos t)^2} \]

Step 4: Simplify the Speed Expression

Simplify the expression for speed:\[ \|\mathbf{v}(t)\| = \sqrt{(1 - \sin t)^2 + (1 - \cos t)^2} = \sqrt{2 - 2\sin t - 2\cos t + \sin^2 t + \cos^2 t} \]Since \(\sin^2 t + \cos^2 t = 1\), we have:\[ \|\mathbf{v}(t)\| = \sqrt{2 + 1 - 2\sin t - 2\cos t} = \sqrt{3 - 2\sin t - 2\cos t} \]

Step 5: Simplify Further Using a Trigonometric Identity

We express \(3 - 2\sin t - 2\cos t\) in terms of a single trigonometric function.Use the identity: \(A\sin t + B\cos t = R\sin(t + \phi)\) where \(R = \sqrt{A^2 + B^2}\).Here, \(A = 1, B = 1\), so \[ R = \sqrt{1^2 + 1^2} = \sqrt{2} \]Hence,\[-2\sin t - 2\cos t = -2\sqrt{2}\sin(t + \frac{3\pi}{4})\]which means:\[ 3 - 2\sin t - 2\cos t = 3 + 2\sqrt{2}\sin(t + \frac{3\pi}{4}) \]

Step 6: Determine Minimum Speed

The expression for speed simplifies to:\[ \|\mathbf{v}(t)\| = \sqrt{3 + 2\sqrt{2}\sin(t + \frac{3\pi}{4})} \] The function inside the square root, \(3 + 2\sqrt{2}\sin(t + \frac{3\pi}{4})\), is minimized when \(\sin(t + \frac{3\pi}{4}) = -1\).This gives \[3 - 2\sqrt{2}\].

Step 7: Compute Minimum Speed

Substitute the minimum value back into the speed expression:\[ \min \|\mathbf{v}(t)\| = \sqrt{3 - 2\sqrt{2}} \] which is the minimum speed of the particle.

Key Concepts

Particle Motion

In vector calculus, understanding particle motion involves analyzing a particle's path over time. This movement is often represented through vector functions, which provide both the position and direction of the particle at any given moment. The exercise deals with a particle following a path described by the vector function \(\mathbf{r}(t) = \langle t+\cos t, t-\sin t \rangle \). This shows how the particle's position changes with respect to time \(t\).

For any vector function like this, each coordinate can change both along and perpendicular to its original path, creating motion that could curve, loop, or spiral. By plotting these points over a continuous interval here from 0 to \(2\pi\), we can visualize the complete trajectory of the particle. Understanding the function's form helps us predict how the particle will behave as time progresses. Thus, grasping particle motion involves breaking down the components—considering both changes in position (using trigonometric additions here for complexity) and aligning them to the interval over which time \(t\) changes.

• The vector function not only points out where the particle is but also where it might be at any subsequent moment.
• This movement translates complex physical phenomena into more digestible mathematical entities.

Speed of a Particle

Speed is a fundamental aspect of particle motion, representing how quickly a particle is moving along its path. Unlike velocity, which is a vector, speed is a scalar. This means it only measures magnitude and not direction.

To find the speed of the particle described by \(\mathbf{r}(t)\), we first determine the velocity by differentiating the position vector function with respect to time \(t\). Computing the speed requires calculating the magnitude of this velocity vector. It provides a single number that reflects the rate of movement, disregarding which way the particle is headed.

• Speed = |Velocity|
• Magnitude implies a focus on size, or rate of motion, giving us the speed at any specific point in time.

For complex paths like this one, involving trigonometric functions of \(t\), understanding speed as the magnitude lets us detect how quickly the curve is traced by the particle, without considering direction.

Velocity Function

The velocity function of a particle describes its rate and direction of travel at any given time. Here, the velocity function is determined by differentiating the position vector \(\mathbf{r}(t)\). This yields \(\mathbf{v}(t) = \langle 1 - \sin t, 1 - \cos t \rangle\), showing how the rate of change for each component (both x and y) varies with time.

The velocity vector reveals two key components:

• The change in position along the horizontal and vertical axes.
• Both components are influenced by sinusoidal functions, indicating periodic fluctuations in velocity.

Velocity being a vector explains how the movement of the particle has direction—it can change its course as defined by the differentiated function. Each component of the vector provides critical insights into how exactly the path changes with respect to time 🎯 emphasizing both the evolving position and timing properties of particle motion.

Trigonometric Identities

Trigonometric identities play a significant role in simplifying particle motion expressions. In our exercise, the calculation of speed involves simplifying expressions into manageable forms using identities. Particularly, the expression \(3 - 2\sin t - 2\cos t\) is transformed using the identity
\(A\sin t + B\cos t = R\sin(t + \phi)\), where \(R = \sqrt{A^2 + B^2}\).

Such identities allow expression reductions by consolidating them into a single trigonometric function, helping to solve for when specific conditions, like speed minimization, occur. Understanding and applying these identities:

• Simplifies complex trigonometric expressions making them more interpretable.
• Offers pathways to identify minima or maxima by focusing on sinusoidal peaks and troughs.

Without these identities, navigating problems involving trigonometric functions would be quite intricate. By allowing transformations such as \(-2\sin t - 2\cos t\) into \(-2\sqrt{2}\sin(t + \frac{3\pi}{4})\), calculations reduce to more straightforward values facilitating analysis.