Problem 18 - Thomas Calculus

Question

Length is independent of parametrization To illustrate that the length of a smooth space curve does not depend on the parametrization you use to compute it, calculate the length of one turn of the helix in Example 1 with the following parametrizations. $$ \begin{array}{l}{\text { a. } \mathbf{r}(t)=(\cos 4 t) \mathbf{i}+(\sin 4 t) \mathbf{j}+4 t \mathbf{k}, \quad 0 \leq t \leq \pi / 2} \\ {\mathbf{b} . \mathbf{r}(t)=[\cos (t / 2)] \mathbf{i}+[\sin (t / 2)] \mathbf{j}+(t / 2) \mathbf{k}, \quad 0 \leq t \leq 4 \pi} \\ {\mathbf{c} \cdot \mathbf{r}(t)=(\cos t) \mathbf{i}-(\sin t) \mathbf{j}-t \mathbf{k}, \quad-2 \pi \leq t \leq 0}\end{array} $$

Step-by-Step Solution

Answer

The length of one turn of the helix is \( 2\pi\sqrt{2} \) for all parametrizations.

Step 1: Derivative of the Curve (Parametrization a)

Given \( \mathbf{r}(t) = (\cos 4t) \mathbf{i} + (\sin 4t) \mathbf{j} + 4t \mathbf{k} \), compute the derivative: \( \mathbf{r}'(t) = (-4\sin 4t) \mathbf{i} + (4\cos 4t) \mathbf{j} + 4 \mathbf{k} \).

Step 2: Magnitude of the Derivative (Parametrization a)

Calculate the magnitude: \[ ||\mathbf{r}'(t)|| = \sqrt{(-4\sin 4t)^2 + (4\cos 4t)^2 + 4^2} = \sqrt{16\sin^2 4t + 16\cos^2 4t + 16} = \sqrt{32} = 4\sqrt{2} \]

Step 3: Compute Length (Parametrization a)

The length \( L \) is given by:\[ L = \int_0^{\pi/2} ||\mathbf{r}'(t)|| \ dt = \int_0^{\pi/2} 4\sqrt{2} \ dt = 4\sqrt{2} \cdot \frac{\pi}{2} = 2\pi\sqrt{2} \]

Step 4: Derivative of the Curve (Parametrization b)

Given \( \mathbf{r}(t) = [\cos(t/2)] \mathbf{i} + [\sin(t/2)] \mathbf{j} + (t/2) \mathbf{k} \), compute the derivative: \( \mathbf{r}'(t) = \left(-\frac{1}{2}\sin\left(\frac{t}{2}\right)\right) \mathbf{i} + \left(\frac{1}{2}\cos\left(\frac{t}{2}\right)\right) \mathbf{j} + \frac{1}{2} \mathbf{k} \).

Step 5: Magnitude of the Derivative (Parametrization b)

Calculate the magnitude: \[ ||\mathbf{r}'(t)|| = \sqrt{\left(-\frac{1}{2}\sin\left(\frac{t}{2}\right)\right)^2 + \left(\frac{1}{2}\cos\left(\frac{t}{2}\right)\right)^2 + \frac{1}{2}^2} \]which simplifies to:\[ = \frac{1}{2} \sqrt{\sin^2(t/2) + \cos^2(t/2) + 1} = \frac{1}{2} \sqrt{2} \]

Step 6: Compute Length (Parametrization b)

The length \( L \) is given by:\[ L = \int_0^{4\pi} ||\mathbf{r}'(t)|| \ dt = \int_0^{4\pi} \frac{\sqrt{2}}{2} \ dt = \frac{\sqrt{2}}{2} \cdot 4\pi = 2\pi\sqrt{2} \]

Step 7: Derivative of the Curve (Parametrization c)

Given \( \mathbf{r}(t) = (\cos t) \mathbf{i} - (\sin t) \mathbf{j} - t \mathbf{k} \), compute the derivative: \( \mathbf{r}'(t) = (-\sin t) \mathbf{i} - (\cos t) \mathbf{j} - 1 \mathbf{k} \).

Step 8: Magnitude of the Derivative (Parametrization c)

Calculate the magnitude:\[ ||\mathbf{r}'(t)|| = \sqrt{(-\sin t)^2 + (-\cos t)^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2} \]

Step 9: Compute Length (Parametrization c)

The length \( L \) is given by:\[ L = \int_{-2\pi}^{0} ||\mathbf{r}'(t)|| \ dt = \int_{-2\pi}^{0} \sqrt{2} \ dt = \sqrt{2} \cdot 2\pi = 2\pi\sqrt{2} \]

Step 10: Conclusion: Consistent Results

The length of one turn of the helix, as computed using all three parametrizations, is \( 2\pi\sqrt{2} \). This shows that the length is independent of the choice of parametrization.

Key Concepts

Parametrization

Parametrization is a way of representing a curve by using a parameter that varies over a specified interval. In mathematical terms, it usually indicates expressing a set of coordinates as functions of one or more independent variables.
The main idea behind parametrization is that it allows us to describe a curve more flexibly, with a focus on characterized motion or other properties.

• A parameter often represents time. For instance, if \( t \) is the parameter, the curve could represent the path of a moving object over time.
• By using different parametrizations for the same curve, we aim to describe it in various ways, often highlighting certain characteristics.

Despite the different representations, an important property is that certain values, such as the total length of the curve, remain consistent. This invariance is the essence of the exercise and showcases an intrinsic property of the set geometric path.
Parametrization essentially grants us the flexibility to express complex curves using simpler mathematical expressions, even when applied in different contexts as seen with a helix.

Helix

A helix is a spiraled curve that is characterized by its consistent loop and rise in space. Imagine a slinky toy or the steps of a spiral staircase; a helix moves both in a circular pattern and vertically at the same time. This creates its iconic spiral form.
A helix can be mathematically defined using vector functions which denote its path in 3-dimensional space.

• The standard parametrization of a helix might be \( \mathbf{r}(t) = (\cos t) \mathbf{i} + (\sin t) \mathbf{j} + t \mathbf{k} \).
• Here, \( t \) serves as the parameter, \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors in the respective x, y, and z axes, and the terms \( \cos t \) and \( \sin t \) ensure the uniformity of the circular motion.

Helices are found in various natural phenomena and engineered structures, making understanding their mathematical properties crucial.
By exploring the helix through different parametrizations in vector calculus, we appreciate both its mathematical diversity and its physical applications.

Vector Calculus

Vector Calculus is a branch of mathematics focused on differentiation and integration of vector fields, mostly in 3-dimensional spaces. This subject combines the fundamentals of calculus and vector algebra to solve engineering and physics problems involving fields like electromagnetism.
One of the most fundamental uses of vector calculus is to compute the arc length of curves in space.

• To calculate it, the curve must first be expressed as a vector-valued function.
• The process involves deriving this function and then calculating its magnitude for integration.

Vector calculus not only allows for the computation of lengths but also helps to better understand relationships involving forces and velocities in space.
Its application extends to physics, engineering, and even computer graphics for rendering realistic environments.

Derivative of a Vector Function

The derivative of a vector function is a vital part of vector calculus and is key in understanding the behavior of curves. Similar to traditional calculus, it gives us the rate of change of the vector function with respect to its parameter, often time.
Consider a vector function \( \mathbf{r}(t) = [f(t)] \mathbf{i} + [g(t)] \mathbf{j} + [h(t)] \mathbf{k} \); the derivative gives us information about the curve's velocity or slope at a point.

• To find this derivative, differentiate each component function with respect to \( t \).
• The result is another vector function representing the tangent vector at each point along the curve.

The derivative not only helps in computing the arc length but also in analyzing how certain curves may interact or change over space and time.
As seen in the exercise, even though different parametrizations produce varying derivatives, the magnitude calculation ultimately yields consistent results, confirming the invariance of arc length across parametrizations.