Problem 2 - Thomas Calculus

Question

In Exercises \(1-8,\) find the curve's unit tangent vector. Also, find the length of the indicated portion of the curve. $$ \mathbf{r}(t)=(6 \sin 2 t) \mathbf{i}+(6 \cos 2 t) \mathbf{j}+5 t \mathbf{k}, \quad 0 \leq t \leq \pi $$

Step-by-Step Solution

Answer

The unit tangent vector is \( \mathbf{T}(t) = \frac{1}{13}(12 \cos 2t \mathbf{i} - 12 \sin 2t \mathbf{j} + 5 \mathbf{k}) \) and the arc length is \( 13\pi \).

Step 1: Find the Derivative of the Position Vector

First, we find the derivative of the position vector \( \mathbf{r}(t) \). This gives us the velocity vector \( \mathbf{v}(t) \). The derivative of each component is: \( \frac{d}{dt}(6 \sin 2t) = 12 \cos 2t \), \( \frac{d}{dt}(6 \cos 2t) = -12 \sin 2t \), and \( \frac{d}{dt}(5t) = 5 \). Therefore, \( \mathbf{v}(t) = 12 \cos 2t \mathbf{i} - 12 \sin 2t \mathbf{j} + 5 \mathbf{k} \).

Step 2: Calculate the Magnitude of the Velocity Vector

Next, compute the magnitude of \( \mathbf{v}(t) \) to find \( |\mathbf{v}(t)| \). The magnitude is calculated as \[ |\mathbf{v}(t)| = \sqrt{(12 \cos 2t)^2 + (-12 \sin 2t)^2 + 5^2} \]. Simplifying this, we get \[ |\mathbf{v}(t)| = \sqrt{144 \cos^2 2t + 144 \sin^2 2t + 25} = \sqrt{144 + 25} = 13 \].

Step 3: Find the Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is defined as \( \frac{\mathbf{v}(t)}{|\mathbf{v}(t)|} \). We have \( \mathbf{v}(t) = 12 \cos 2t \mathbf{i} - 12 \sin 2t \mathbf{j} + 5 \mathbf{k} \) and \( |\mathbf{v}(t)| = 13 \). Thus, \[ \mathbf{T}(t) = \frac{1}{13}(12 \cos 2t \mathbf{i} - 12 \sin 2t \mathbf{j} + 5 \mathbf{k}) \].

Step 4: Set Up the Integral for the Arc Length

The arc length \( S \) of the curve from \( t = 0 \) to \( t = \pi \) is given by \[ S = \int_{0}^{\pi} |\mathbf{v}(t)| \, dt \]. Since \( |\mathbf{v}(t)| = 13 \), this becomes \[ S = \int_{0}^{\pi} 13 \, dt \].

Step 5: Evaluate the Integral for Arc Length

Evaluate the integral \( \int_{0}^{\pi} 13 \, dt \), which simplifies to \[ S = 13 \cdot [t]_{0}^{\pi} = 13(\pi - 0) = 13\pi \].

Key Concepts

Position Vector

The position vector, often denoted as \( \mathbf{r}(t) \), is a crucial concept in vector calculus that describes the location of a point in space as it changes over time. It serves as a function of a variable, typically time (\( t \)), and maps out the path or trajectory of an object.
In this exercise, the position vector given is \( \mathbf{r}(t) = (6 \sin 2t) \mathbf{i} + (6 \cos 2t) \mathbf{j} + 5t \mathbf{k} \). This equation demonstrates how the object's position shifts in a 3D space along the \( x \), \( y \), and \( z \) axes.

• \( 6 \sin 2t \): Describes movement along the \( x \)-axis.
• \( 6 \cos 2t \): Describes movement along the \( y \)-axis.
• \( 5t \): Highlights alteration along the \( z \)-axis.

By analyzing each vector component, we can see the cyclical pattern in the \( xy \)-plane, forming a circular path, while the \( z \)-component linearly increases with time, giving the resulting trajectory a spiral-like form.

Velocity Vector

To understand motion, the velocity vector \( \mathbf{v}(t) \) is determined by taking the derivative of the position vector with respect to time. This represents both the speed and direction of the object's motion at any point.
In this case, differentiating each component of \( \mathbf{r}(t) \) leads to \( \mathbf{v}(t) = 12 \cos 2t \mathbf{i} - 12 \sin 2t \mathbf{j} + 5 \mathbf{k} \).

• \( 12 \cos 2t \): This component shows how the velocity switches direction in the \( x \)-axis, depending on \( t \).
• -12 \sin 2t: Demonstrates how the \( y \)-axis velocity also varies sinusoidally with time.
• 5: The constant \( z \)-component indicates a steady movement upwards along the \( z \)-axis.

Thus, the velocity vector helps in visualizing how swiftly and in which direction the point is traversing its path.

Magnitude

The magnitude of a vector quantifies its length, disregarding direction. For the velocity vector, its magnitude \( |\mathbf{v}(t)| \) signifies the object's speed at a given point in time.
Calculated as \( |\mathbf{v}(t)| = \sqrt{(12 \cos 2t)^2 + (-12 \sin 2t)^2 + 5^2} \), the simplification follows:

• \( 144 \cos^2 2t + 144 \sin^2 2t + 25 = 144 \cdot (\cos^2 2t + \sin^2 2t) + 25 \).
• Using the identity \( \cos^2 2t + \sin^2 2t = 1 \), this simplifies to \( 144 + 25 = 169 \).
• Hence, the magnitude is \( |\mathbf{v}(t)| = \sqrt{169} = 13 \).

Understanding the magnitude is essential as it precisely measures the rate of change of position in all three dimensions at once.

Arc Length

Arc length describes the total distance traveled along a curve between two points. It provides a broader understanding of the entire journey, rather than just a snapshot at a particular moment.
The arc length \( S \) for the curve from \( t = 0 \) to \( t = \pi \) is determined by the integral \( S = \int_{0}^{\pi} |\mathbf{v}(t)| \, dt \), with \( |\mathbf{v}(t)| = 13 \).

• Substituting into the integral, we have \( S = \int_{0}^{\pi} 13 \, dt \).
• This evaluates to \( 13[t]_{0}^{\pi} = 13(\pi - 0) \).
• Consequently, the arc length is \( S = 13\pi \).

The result, \( 13\pi \), reflects the entire scope of the path taken, capturing the full span of motion from start to finish.