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 \( \left(-\frac{2}{3} \sin t\right) \mathbf{i} + \left(\frac{2}{3} \cos t\right) \mathbf{j} + \left(\frac{\sqrt{5}}{3}\right) \mathbf{k} \) and the length of the curve is \( 3\pi \).

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

First, differentiate the vector function \(\mathbf{r}(t) = (2 \cos t) \mathbf{i} + (2 \sin t) \mathbf{j} + \sqrt{5} t \, \mathbf{k} \) with respect to \(t\). This gives us the tangent vector \(\mathbf{r}'(t)\):\[ \mathbf{r}'(t) = \frac{d}{dt}[2 \cos t] \mathbf{i} + \frac{d}{dt}[2 \sin t] \mathbf{j} + \frac{d}{dt}[\sqrt{5} t] \mathbf{k} = -2 \sin t \, \mathbf{i} + 2 \cos t \, \mathbf{j} + \sqrt{5} \, \mathbf{k}. \]

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

Calculate the magnitude of the derivative \(\mathbf{r}'(t)\), which is needed for finding the unit tangent vector. The magnitude is given by: \[ |\mathbf{r}'(t)| = \sqrt{(-2 \sin t)^2 + (2 \cos t)^2 + (\sqrt{5})^2}. \] Simplifying, we have \[|\mathbf{r}'(t)| = \sqrt{4 \sin^2 t + 4 \cos^2 t + 5} = \sqrt{4(\sin^2 t + \cos^2 t) + 5} = \sqrt{9} = 3. \]

Step 3: Determine the Unit Tangent Vector

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

Step 4: Find the Length of the Curve Segment

To find the length of the curve from \(t = 0\) to \(t = \pi\), integrate the magnitude of \(\mathbf{r}'(t)\) over this interval: \[ L = \int_{0}^{\pi} |\mathbf{r}'(t)| \, dt = \int_{0}^{\pi} 3 \, dt. \] Calculating this, \[ L = 3 \times [t]_{0}^{\pi} = 3(\pi - 0) = 3\pi. \]

Key Concepts

Unit Tangent Vector

In vector calculus, the unit tangent vector plays a crucial role in understanding the direction of a curve at any given point. It essentially gives us the direction in which the curve is heading but does not account for how fast we are moving along it. To find the unit tangent vector, we follow a systematic process:

• First, take the derivative of the vector function \( \mathbf{r}(t) \). This derivative, \( \mathbf{r}'(t) \), gives us the tangent vector, which indicates the instantaneous direction of the curve.
• Next, determine the magnitude of \( \mathbf{r}'(t) \). This is necessary because to make a vector 'unit' means to scale it to have a length of one.
• Finally, divide the tangent vector by its magnitude to obtain the unit tangent vector, \( \mathbf{T}(t) \).

This process ensures that the unit tangent vector is a normalized vector pointing in the direction of the curve, encapsulated in the formula \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} \). This vector is especially useful in curvature studies and motion analysis along curves.

Curve Length

The concept of curve length in vector calculus allows us to quantify how long a path is between two points on a parametric curve. For a curve defined by a vector function \( \mathbf{r}(t) \), the length \( L \) over an interval can be calculated by integrating the magnitude of the derivative of the vector function.Here's how you do it:

• First, ensure you have correctly found \( \mathbf{r}'(t) \), the derivative of your vector function. Its magnitude, \( |\mathbf{r}'(t)| \), is crucial as it represents the speed along the curve.
• Integrate this magnitude over the given interval. The integral \( \int_{a}^{b} |\mathbf{r}'(t)| \, dt \) where \( a \) and \( b \) are the boundaries of the portion of the curve, provides the total length along the curve segment.

Through this integration approach, we smoothly sum up all the infinitesimal segments along the curve. This technique is pivotal not only in math but also in physical applications, like finding the distance traveled by a particle moving along the path.

Parametric Equations

Parametric equations play an integral role when it comes to describing curves in a more flexible and dynamic way than the standard \( y=f(x) \) form. These equations express the coordinates of the points on the curve as functions of a variable, typically denoted as \( t \). This approach is powerful for various reasons:

• It allows representation of curves that cannot be described with a single function of \( y \) in terms of \( x \). Examples include complex shapes like circles and ellipses.
• Parametric equations facilitate the calculation of derivative and integrals, making them well-suited for calculus operations, such as finding tangents, normals, or lengths.
• They are versatile in representing motion. For example, a position of a point moving through space over time can be neatly captured through parametric forms.

In the exercise, the parametric form \( \mathbf{r}(t) = (2 \cos t) \mathbf{i} + (2 \sin t) \mathbf{j} + \sqrt{5} t \mathbf{k} \) elegantly encapsulates a 3D helix path, highlighting its utility in representing spatial curves efficiently. This versatility makes parametric equations a favorite tool across mathematics and physics.