Problem 1 - 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. $$ \mathbf{r}(t)=(2 \cos t) \mathbf{i}+(2 \sin t \mathbf{j})+\sqrt{5} t \mathbf{k}, \quad 0 \leq t \leq \pi $$

Step-by-Step Solution

Answer

The unit tangent vector is \( \left(-\frac{2}{3} \sin t \right) \mathbf{i} + \left(\frac{2}{3} \cos t \right) \mathbf{j} + \left(\frac{\sqrt{5}}{3} \right) \mathbf{k} \). The curve length is \( 3\pi \).

Step 1: Determine the Derivative of the Vector Function

Given the vector function \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (2 \sin t) \mathbf{j} + \sqrt{5}t \mathbf{k} \), we need to find its derivative, \( \mathbf{r}'(t) \). Differentiate each component with respect to \( t \):- The derivative of \( 2 \cos t \) with respect to \( t \) is \( -2 \sin t \).- The derivative of \( 2 \sin t \) with respect to \( t \) is \( 2 \cos t \).- The derivative of \( \sqrt{5}t \) with respect to \( t \) is \( \sqrt{5} \).Thus, \( \mathbf{r}'(t) = (-2 \sin t) \mathbf{i} + (2 \cos t) \mathbf{j} + \sqrt{5} \mathbf{k} \).

Step 2: Calculate the Magnitude of the Velocity Vector

The magnitude of \( \mathbf{r}'(t) \), denoted as \( ||\mathbf{r}'(t)|| \), is calculated using the formula for the magnitude of a vector:\[||\mathbf{r}'(t)|| = \sqrt{(-2 \sin t)^2 + (2 \cos t)^2 + (\sqrt{5})^2}.\]Simplify the expression:\[||\mathbf{r}'(t)|| = \sqrt{4 \sin^2 t + 4 \cos^2 t + 5} = \sqrt{4(\sin^2 t + \cos^2 t) + 5} = \sqrt{9} = 3.\]Thus, \( ||\mathbf{r}'(t)|| = 3 \).

Step 3: Find the Unit Tangent Vector

The unit tangent vector \( \mathbf{T}(t) \) is given by:\[\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} = \frac{(-2 \sin t) \mathbf{i} + (2 \cos t) \mathbf{j} + \sqrt{5} \mathbf{k}}{3}.\]Thus, the unit tangent vector is:\[\mathbf{T}(t) = \left(-\frac{2}{3} \sin t \right) \mathbf{i} + \left(\frac{2}{3} \cos t \right) \mathbf{j} + \left(\frac{\sqrt{5}}{3} \right) \mathbf{k}.\]

Step 4: Calculate the Length of the Curve

The length \( L \) of the curve from \( t = 0 \) to \( t = \pi \) is given by the integral:\[L = \int_{0}^{\pi} ||\mathbf{r}'(t)|| \; dt = \int_{0}^{\pi} 3 \; dt.\]Evaluate the integral:\[L = 3 \left[ t \right]_{0}^{\pi} = 3(\pi - 0) = 3\pi.\]Thus, the length of the indicated portion of the curve is \( 3\pi \).

Key Concepts

Vector Calculus

Vector calculus is a powerful mathematical tool used to analyze vector fields and multi-dimensional functions. In this exercise, we deal with vector functions which define a vector in three-dimensional space in terms of a parameter, usually denoted as \( t \). Vectors help represent geometric paths and curves in space, as seen with our function \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (2 \sin t) \mathbf{j} + \sqrt{5} t \mathbf{k} \).
Understanding how these vectors behave is critical in physics and engineering, as it represents how an object moves in space. The first step in solving problems with vector calculus is often to find the derivative of the vector function, giving us information about the velocity or tangent to the curve at any point.
The unit tangent vector, found by normalizing the derivative, helps indicate the direction of the curve at each point. It's like pointing a compass along the path, showing the precise direction of travel at every instant.

Curve Length

The length of a curve is an important concept in vector calculus, especially in fields like engineering and physics, where you need to understand the distance traveled along a path. In this exercise, we are asked to find the distance along the curve described by the vector function \( \mathbf{r}(t) \) from \( t = 0 \) to \( t = \pi \).
First, you find the magnitude of the derivative \( ||\mathbf{r}'(t)|| \), which gives the speed or the rate of change of the position vector. In our solution, this was calculated to be 3 for all \( t \). This indicates that the speed of travel along the path is constant.
The integration then sums up these speeds over the interval from 0 to \( \pi \). The formula \( L = \int_{0}^{\pi} 3 \; dt \) represents adding this constant speed at every tiny interval, resulting in the total path length, \( 3\pi \). Knowing the length helps in understanding the total distance traversed by an object moving along the path described by the vector function.

Derivative of Vector Functions

Finding the derivative of a vector function is a core technique in vector calculus as it reveals how the vector changes with respect to a parameter. This change is not just in magnitude but also in direction, illustrating the dynamic nature of the curve.
Consider our vector function \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (2 \sin t) \mathbf{j} + \sqrt{5}t \mathbf{k} \). The derivative, \( \mathbf{r}'(t) = (-2 \sin t) \mathbf{i} + (2 \cos t) \mathbf{j} + \sqrt{5} \mathbf{k} \), is computed by differentiating each component. This derivative represents the instantaneous rate of change of the vector at any point \( t \), essentially the velocity of the curve.
The derivative lays the foundation for further calculations such as finding the unit tangent vector. It gives a snapshot of the curve's direction and movement at any given point, crucial when predicting the path of a moving object or analyzing steady-state conditions within the given parameter interval.