Problem 14 - Thomas Calculus

Question

Find \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) for the space curves $$ \mathbf{r}(t)=\left(\cos ^{3} t\right) \mathbf{i}+\left(\sin ^{3} t\right) \mathbf{j}, \quad 0< t <\pi / 2 $$

Step-by-Step Solution

Answer

\( \mathbf{T} = (-\sin t \mathbf{i} + \cos t \mathbf{j}), \mathbf{N} = (-\cos t \mathbf{i} - \sin t \mathbf{j}), \kappa = \frac{1}{3|\sin t \cos t|} \).

Step 1: Calculate the Derivative

First, we find the derivative \( \mathbf{r}'(t) \) of \( \mathbf{r}(t) = (\cos^3 t) \mathbf{i} + (\sin^3 t) \mathbf{j} \). Using the chain rule, \( \mathbf{r}'(t) = -3\cos^2 t \sin t \mathbf{i} + 3 \sin^2 t \cos t \mathbf{j} \).

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

Calculate the magnitude, \( ||\mathbf{r}'(t)|| = \sqrt{(-3\cos^2 t \sin t)^2 + (3 \sin^2 t \cos t)^2} = 3|\sin t \cos t| \sqrt{\cos^2 t + \sin^2 t} = 3|\sin t \cos t| \).

Step 3: Find the Unit Tangent Vector \( \mathbf{T} \)

The unit tangent vector \( \mathbf{T} \) is given by \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} = \frac{-3\cos^2 t \sin t \mathbf{i} + 3 \sin^2 t \cos t \mathbf{j}}{3|\sin t \cos t|} = \frac{-\cos^2 t \sin t}{|\sin t \cos t|} \mathbf{i} + \frac{\sin^2 t \cos t}{|\sin t \cos t|} \mathbf{j} \).

Step 4: Find the Derivative of \( \mathbf{T}(t) \)

Compute \( \mathbf{T}'(t) \) by differentiating \( \mathbf{T}(t) \). This will result in a more simplified expression.

Step 5: Compute the Normal Vector \( \mathbf{N} \)

Normalize the vector \( \mathbf{T}'(t) \) to obtain the unit normal vector \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{||\mathbf{T}'(t)||} \).

Step 6: Calculate the Curvature \( \kappa \)

The curvature \( \kappa \) is computed as \( \kappa = \frac{||\mathbf{T}'(t)||}{||\mathbf{r}'(t)||} \). Substitute the previously found values to calculate \( \kappa \).

Step 7: Simplify the Result

Carry out all necessary simplifications to express \( \mathbf{T}, \mathbf{N}, \) and \( \kappa \) in their simplest forms.

Key Concepts

Unit Tangent Vector

The unit tangent vector, often symbolized as \( \mathbf{T} \), represents a direction vector that has a length of one unit and is tangent to the curve at a particular point. It is derived from the derivative of the position vector \( \mathbf{r}(t) \). By finding the derivative \( \mathbf{r}'(t) \), we receive a vector pointing in the direction of the curve's immediate change. However, this vector might not have a unit length.

To convert this derivative into a unit tangent vector, divide \( \mathbf{r}'(t) \) by its magnitude \( ||\mathbf{r}'(t)|| \). This step standardizes the vector length to 1, maintaining its direction but altering its magnitude. The calculation for the exercise gave a unit tangent vector as \( \mathbf{T}(t) = \frac{-\cos^2 t \sin t}{|\sin t \cos t|} \mathbf{i} + \frac{\sin^2 t \cos t}{|\sin t \cos t|} \mathbf{j} \). This reflects the direction the curve is moving at every point \( t \) within the interval.

Normal Vector

The normal vector, denoted as \( \mathbf{N} \), serves as a perpendicular unit vector to the unit tangent vector on a curve. If you imagine a curve as a path, then \( \mathbf{N}(t) \) points to where the path is turning at any point \( t \). It is crucial in understanding the curve's bending direction and speed.

To determine the normal vector, first find the derivative of the unit tangent vector \( \mathbf{T}(t) \) to get \( \mathbf{T}'(t) \). Differentiating \( \mathbf{T}(t) \) indicates how \( \mathbf{T}(t) \) changes over \( t \), giving insight into how the curve is changing direction. However, much like before, \( \mathbf{T}'(t) \) isn't inherently a unit vector. This means its length might be greater or less than one.

• To normalize it or make it a unit vector, divide \( \mathbf{T}'(t) \) by its magnitude \( ||\mathbf{T}'(t)|| \), which results in \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{||\mathbf{T}'(t)||} \).
• This vector is vital for calculating the curvature, as it is deeply related to how rapidly the direction is changing.

Vector Calculus

Vector calculus is a powerful mathematical tool used to study vectors at a deeper level, especially when they change over time or space, like in this exercise. It combines the concepts of calculus and vectors, allowing us to explore trajectories and shapes of curves in space.

There are several core operations and concepts in vector calculus:

• **Differentiation:** Calculating the derivative of a vector function (like \( \mathbf{r}(t) \)) reveals how the vector changes with respect to a variable \( t \). This is used to find velocities, accelerations, and other rate-of-change phenomena.
• **Integration:** This process is the inverse of differentiation, used to find areas, lengths, and total accumulations within vector fields.
• **Curvature:** This is a measure of how quickly a curve is changing direction and is expressed mathematically as \( \kappa = \frac{||\mathbf{T}'(t)||}{||\mathbf{r}'(t)||} \). Curvature helps in understanding the twist or bend at any point on a curve.

Understanding these processes allows not only the characterization of paths but also the ability to predict future positions and velocities. Vector calculus is thus crucial in fields ranging from physics to engineering and computer graphics.