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 \( \mathbf{T}(t) = \frac{12}{13} \cos 2t \, \mathbf{i} - \frac{12}{13} \sin 2t \, \mathbf{j} + \frac{5}{13} \mathbf{k} \). The length of the curve is \( 13\pi \).

Step 1: Calculate the derivative of \( \mathbf{r}(t) \)

The unit tangent vector requires finding the derivative of the position vector \( \mathbf{r}(t) \). The original vector is \( \mathbf{r}(t) = (6 \sin 2t) \mathbf{i} + (6 \cos 2t) \mathbf{j} + 5t \mathbf{k} \). Therefore, the derivative \( \mathbf{r}'(t) \) will be:\[ \mathbf{r}'(t) = \frac{d}{dt}(6 \sin 2t) \mathbf{i} + \frac{d}{dt}(6 \cos 2t) \mathbf{j} + \frac{d}{dt}(5t) \mathbf{k} \].Calculating each component:- \( \frac{d}{dt}(6 \sin 2t) = 12 \cos 2t \)- \( \frac{d}{dt}(6 \cos 2t) = -12 \sin 2t \)- \( \frac{d}{dt}(5t) = 5 \)Thus, \( \mathbf{r}'(t) = 12 \cos 2t \, \mathbf{i} - 12 \sin 2t \, \mathbf{j} + 5 \mathbf{k} \).

Step 2: Find the magnitude of \( \mathbf{r}'(t) \)

The magnitude, \( \| \mathbf{r}'(t) \| \), is necessary for finding the unit tangent vector. It is given by:\[ \| \mathbf{r}'(t) \| = \sqrt{(12 \cos 2t)^2 + (-12 \sin 2t)^2 + 5^2} \].Simplifying the expression inside the square root:\[= \sqrt{144 \cos^2 2t + 144 \sin^2 2t + 25} \]\[= \sqrt{144(\cos^2 2t + \sin^2 2t) + 25} \]\[= \sqrt{144 + 25} = \sqrt{169} = 13 \].

Step 3: Compute the unit tangent vector

The unit tangent vector \( \mathbf{T}(t) \) is obtained by dividing \( \mathbf{r}'(t) \) by its magnitude \( \| \mathbf{r}'(t) \| \):\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} \]\[ = \frac{12 \cos 2t \, \mathbf{i} - 12 \sin 2t \, \mathbf{j} + 5 \, \mathbf{k}}{13} \]\[ = \frac{12}{13} \cos 2t \, \mathbf{i} - \frac{12}{13} \sin 2t \, \mathbf{j} + \frac{5}{13} \mathbf{k} \].

Step 4: Find the length of the curve from \( t = 0 \) to \( t = \pi \)

The arc length \( L \) of the curve from \( t = 0 \) to \( t = \pi \) is calculated by\[ L = \int_{0}^{\pi} \| \mathbf{r}'(t) \| \, dt \]\[ = \int_{0}^{\pi} 13 \, dt \]\[ = 13 \times \left[ t \right]_{0}^{\pi} \]\[ = 13 \times (\pi - 0) \]\[ = 13 \pi \].

Key Concepts

Arc Length

The concept of arc length is a way to measure the distance along a curved path. In vector calculus, this means finding the length of a curve described by a vector function. It's the vector version of the curve's total distance traveled when moving from one point to another.
To find the arc length of a curve defined by a vector function \(\mathbf{r}(t)\), integrate the magnitude of the derivative of the vector function over the interval of interest. Essentially, you're summing up infinitely small line segments along the curve that each have a length of \(\| \mathbf{r}'(t) \| dt\). This integral is written as:

• \( L = \int_{a}^{b} \| \mathbf{r}'(t) \| dt \)

"Here, \( a \) and \( b \) are the start and end values of the parameter \( t \). For our specific example in the exercise, the arc length of the curve from \( t = 0 \) to \( t = \pi \) was found to be \( 13\pi \), calculated by integrating the constant magnitude \( 13 \) over the interval."
The calculation highlights the precision and care needed when working with vector-valued functions to determine a curve's length. Each tiny segment contributes to the full path length.

Vector Calculus

Vector calculus is a branch of mathematics that deals with vector fields and differentiating and integrating vector functions. It provides tools to handle fields like electromagnetism, fluid flow, and more geometric contexts in three dimensions.
When dealing with vector calculus:

• You often work with vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), which represent unit vectors along the x, y, and z axes.
• Vector functions allow description of curves and surfaces in three dimensions.

"In our exercise, we used vector calculus to determine both the unit tangent vector and the arc length. Vector calculus helps translate physical and geometric scenarios into mathematical language, enabling precise calculations of various properties like length, curvature, and direction of paths."
Vector functions can describe complex shapes and motions. This ability to encapsulate multi-dimensional behavior in vector format makes vector calculus a powerful tool in both mathematics and real-world applications.

Derivative of Vector Functions

The derivative of a vector function is crucial in finding the rate of change and the direction of curves in space. By differentiating a vector function, we get a new vector that points along the curve, representing its velocity or tangent direction at any point.
To differentiate a vector function \( \mathbf{r}(t) \), apply differentiation rules to each of its components:

• For example, if \( \mathbf{r}(t) = (x(t)) \mathbf{i} + (y(t)) \mathbf{j} + (z(t)) \mathbf{k} \), then \( \mathbf{r}'(t) = x'(t) \mathbf{i} + y'(t) \mathbf{j} + z'(t) \mathbf{k} \).

"In our exercise, the derivative \( \mathbf{r}'(t) \) was calculated as \( 12\cos 2t \mathbf{i} - 12\sin 2t \mathbf{j} + 5\mathbf{k} \), representing how the position changes with respect to our parameter \( t \)."
Once the derivative is found, it aids in determining the tangent and velocity vectors, essential for analyzing motion and geometry. This procedure is fundamental to understanding and applying vector calculus concepts effectively.