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. $$ \begin{array}{l}{\text { a. } \mathbf{r}(t)=(\cos 4 t) \mathbf{i}+(\sin 4 t) \mathbf{j}+4 t \mathbf{k}, \quad 0 \leq t \leq \pi / 2} \\ {\text { b. } \mathbf{r}(t)=[\cos (t / 2)] \mathbf{i}+[\sin (t / 2)] \mathbf{j}+(t / 2) \mathbf{k}, \quad 0 \leq t \leq 4 \pi} \\ {\text { c. } \mathbf{r}(t)=(\cos t) \mathbf{i}-(\sin t) \mathbf{j}-t \mathbf{k}, \quad-2 \pi \leq t \leq 0}\end{array} $$

Step-by-Step Solution

Answer

The length of each parametrization is \( 2\pi\sqrt{2} \).

Step 1: Differentiate Parametrization a

Given the parametrization \( \mathbf{r}(t) = (\cos 4t) \mathbf{i} + (\sin 4t) \mathbf{j} + 4t \mathbf{k} \), we need to compute its derivative. Differentiate each component with respect to \( t \). The derivative is \( \mathbf{r}'(t) = (-4 \sin 4t) \mathbf{i} + (4 \cos 4t) \mathbf{j} + 4 \mathbf{k} \).

Step 2: Compute Arc Length for Parametrization a

The arc length formula is \( L = \int_{a}^{b} \| \mathbf{r}'(t) \| \, dt \). Here, \( \| \mathbf{r}'(t) \| = \sqrt{(-4 \sin 4t)^2 + (4 \cos 4t)^2 + 4^2} = \sqrt{16(\sin^2 4t + \cos^2 4t + 1)} = \sqrt{32} = 4\sqrt{2} \). Thus, \( L = \int_{0}^{\pi/2} 4\sqrt{2} \, dt = 4\sqrt{2} \times \frac{\pi}{2} = 2\pi\sqrt{2} \).

Step 3: Differentiate Parametrization b

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

Step 4: Compute Arc Length for Parametrization b

Calculate \( \| \mathbf{r}'(t) \| = \sqrt{\left(-\frac{1}{2} \sin (t/2)\right)^2 + \left(\frac{1}{2} \cos (t/2)\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{1}{4}(\sin^2(t/2) + \cos^2(t/2) + 1)} = \frac{1}{2}\sqrt{2} \). Then, \( L = \int_{0}^{4\pi} \frac{1}{2}\sqrt{2} \, dt = 2\pi \sqrt{2} \).

Step 5: Differentiate Parametrization c

The parametrization is \( \mathbf{r}(t) = (\cos t) \mathbf{i} - (\sin t) \mathbf{j} - t \mathbf{k} \). Its derivative is \( \mathbf{r}'(t) = (-\sin t) \mathbf{i} - (\cos t) \mathbf{j} - \mathbf{k} \).

Step 6: Compute Arc Length for Parametrization c

Find \( \| \mathbf{r}'(t) \| = \sqrt{(-\sin t)^2 + (-\cos t)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \). The arc length is \( L = \int_{-2\pi}^{0} \sqrt{2} \, dt = \sqrt{2} \times 2\pi = 2\pi\sqrt{2} \).

Step 7: Conclusion: Compare Results

Each parametrization gives the same arc length \( 2\pi\sqrt{2} \), illustrating that the length of the helix is independent of parametrization.

Key Concepts

Parametrization

Parametrization is a mathematical technique that helps us describe a path or curve using a set of equations. Think of it as assigning a set of coordinates to each point on a curve using a parameter, which is usually denoted by the letter \( t \). This parameter often represents time, allowing us to think of the path as traced out over time. In the exercise, parametrization helps to express the curves in space using different sets of equations, all of which ultimately represent the same path.

When dealing with a parametrized curve, each point on the curve can be given by\( \mathbf{r}(t) = x(t) \mathbf{i} + y(t) \mathbf{j} + z(t) \mathbf{k} \), where \( x(t) \), \( y(t) \), and \( z(t) \) are functions of the parameter \( t \). This description captures the curve's position in space at each moment in time \( t \). In our example exercise, different expressions for \( \mathbf{r}(t) \) reflect different ways of tracing the same helical curve. Despite the appearance, all such parametrizations describe the same physical path, which is intrinsic to the curve's geometric properties.

Derivatives

Derivatives are essential when working with parametrized curves as they help understand the curve's behavior at every point. Taking the derivative of a parametrization with respect to \( t \) gives us the tangent vector \( \mathbf{r}'(t) \) at each point, which symbolizes the curve's direction and how fast it moves at that point.

To find the arc length of the curve, the magnitude of this derivative, denoted \( \| \mathbf{r}'(t) \| \), is used. It gives us the speed, or rate of change, along the curve. In our particular exercise, calculating this derivative allows us to then integrate over the desired interval, which provides the exact length of the curve. By differentiating each component of the curve, you discover how it changes over time, which is crucial in space curve calculations.

• Expression of change: Differential form \( d \mathbf{r} = \mathbf{r}'(t) d t \).
• Magnitude: Finds rate of movement, \( \| \mathbf{r}'(t) \| \).

Calculating each derivative was an important step to derive the common arc length among the different parametrizations in the example.

Space Curves

Space curves are curves that exist in three-dimensional space, as opposed to flat, two-dimensional curves. They can twist and turn in a way that flat curves cannot. In our scenario, the helix is an example of a space curve, which spirals around an axis in three-dimensional space.

Represented through their distinct parametrizations, space curves illustrate complex shapes and forms. The mathematical description \( \mathbf{r}(t) = (x(t), y(t), z(t)) \) provides all the necessary data to understand the position and direction of such curves in space. We saw in the original exercise how a helix, existing in 3D, has several parametrizations, yet they all represent the same physical curve. This shows that although the mathematical expression can change, the inherent geometry of the space curve remains unchanged.

• Not just a line: Occupies 3D space.
• Parametrizations: Offer different perspectives, same curve.

These characteristics make space curves an interesting and useful subject in fields ranging from physics to engineering.