Problem 32 - Essential Calculus Early Transcendentals

Question

Plot the space curve and its curvature function \(\kappa(t)\). Comment on how the curvature reflects the shape of the curve. $$\mathbf{r}(t)=\langle t-\sin t, 1-\cos t, 4 \cos (t / 2)\rangle, \quad 0 \leqslant t \leqslant 8 \pi$$

Step-by-Step Solution

Answer

Plot the curve using \( \mathbf{r}(t) \) and compute \( \kappa(t) \). Higher \( \kappa(t) \) values show more bending.

Step 1: Understand the space curve

The given space curve is represented by \( \mathbf{r}(t)=\langle t-\sin t, 1-\cos t, 4 \cos (t / 2)\rangle \) where \( 0 \leq t \leq 8 \pi \). It defines a 3D parametric curve.

Step 2: Visualize the space curve

Plot the space curve described by the position vector \( \mathbf{r}(t) \) using software or graphing tools to better understand its shape. Observe how changes in \( t \) affect the path traced in 3D space based on each component of \( \mathbf{r}(t) \).

Step 3: Calculate the first derivative

Find the first derivative \( \mathbf{r}'(t) \) of \( \mathbf{r}(t) \) to get the velocity vector. This is calculated as:\[ \mathbf{r}'(t) = \langle 1-\cos t, \sin t, -2 \sin(t/2) \rangle \]

Step 4: Calculate the second derivative

Determine the second derivative \( \mathbf{r}''(t) \) which represents the acceleration vector of the curve. Compute the derivative as follows:\[ \mathbf{r}''(t) = \langle \sin t, \cos t, -\cos(t/2) \rangle \]

Step 5: Find the curvature function

The curvature function \( \kappa(t) \) is given by:\[ \kappa(t) = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \]Calculate \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) and \( \| \mathbf{r}'(t) \| \) to express \( \kappa(t) \) explicitly.

Step 6: Compute the cross product

Compute the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \):\[ \mathbf{r}'(t) \times \mathbf{r}''(t) = \langle \sin^2(t/2)(1-\cos t), -2(1-\cos t), \cos^2(t/2) + \sin^2(t/2) \sin t \rangle \]

Step 7: Calculate magnitudes needed for curvature

Determine the magnitudes: - \( \| \mathbf{r}'(t) \| = \sqrt{(1-\cos t)^2 + \sin^2 t + (-2\sin(t/2))^2} \)- Magnitude of the cross product \( \| \mathbf{r}'(t) \times \mathbf{r}''(t) \| \).

Step 8: Substitute and simplify the curvature formula

Using the magnitudes, calculate \( \kappa(t) \). This step involves complex algebraic and trigonometric manipulations to simplify the expression.

Step 9: Plot the curvature function

Graph \( \kappa(t) \) against \( t \) using graphing software to visually interpret how the curvature varies as the parameter \( t \) changes.

Step 10: Analyze the curvature and curve shape

Examine the curvature function \( \kappa(t) \): - High curvature indicates regions where the curve is bending sharply.- Low curvature corresponds to flatter or straighter regions of the curve.Relate the observed \( \kappa(t) \) to the shape of the space curve from the plot.

Key Concepts

Curvature Function

The curvature function, denoted as \( \kappa(t) \), is a crucial concept when studying the bending of curves in three-dimensional space. It quantifies how sharply a curve bends at any point along its path. To compute the curvature function, one typically uses the formula:\[ \kappa(t) = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \]Where:

• \( \mathbf{r}'(t) \) is the first derivative, representing the velocity vector.
• \( \mathbf{r}''(t) \) is the second derivative, indicative of the acceleration vector.
• The numerator involves the cross product of these derivatives, which gives a vector perpendicular to the plane formed by \( \mathbf{r}'(t) \) and \( \mathbf{r}''(t) \).
• The denominator scales with the cube of the magnitude of the velocity vector.

Essentially, larger values of \( \kappa(t) \) suggest a more pronounced bending at a point, providing insight into changes in direction along the curve. To visualize the curvature function, plotting \( \kappa(t) \) against \( t \) indicates how the curvature varies at different segments of the space curve.

3D Parametric Curve

A 3D parametric curve is a mathematical description of a curve in three-dimensional space using parametric equations. These equations provide a way to represent coordinates \((x, y, z)\) as functions of a parameter \( t \). For the given exercise, the space curve is:\[ \mathbf{r}(t)=\langle t-\sin t, 1-\cos t, 4 \cos (t / 2)\rangle \]This representation breaks down into the components:

• \(x(t) = t - \sin t\)
• \(y(t) = 1 - \cos t\)
• \(z(t) = 4 \cos(t/2)\)

The interval for \( t \) is \( 0 \leq t \leq 8\pi \), covering multiple oscillations in 3D space.By plotting these equations, the curve emerges in space, showing how \( x \), \( y \), and \( z \) coordinates change with \( t \). Observing these changes helps to understand the path traced by the curve, and its bending characteristics are analyzed using the curvature function.

First Derivative

The first derivative of a parametric curve is essential for understanding the velocity at any point along its path. It represents the rate of change of the position vector over the parameter \( t \). For the given curve \( \mathbf{r}(t) \), the first derivative is:\[ \mathbf{r}'(t) = \langle 1-\cos t, \sin t, -2 \sin(t/2) \rangle \]Here's what each component describes:

• \( 1-\cos t \) shows the rate of change in the x-direction.
• \( \sin t \) is the rate of change in the y-direction.
• \( -2 \sin(t/2) \) is the rate of change in the z-direction.

Effectively, this derivative gives the velocity vector, indicating how fast and in which direction the curve is moving at any particular point. Analyzing the velocity vector helps understand the curve's general behavior and can be used in calculating the curvature.

Second Derivative

The second derivative of a parametric curve provides insight into its acceleration, showing how the velocity changes at each point in time. For the parametric curve \( \mathbf{r}(t) \), the second derivative is:\[ \mathbf{r}''(t) = \langle \sin t, \cos t, -\cos(t/2) \rangle \]Each part of the second derivative indicates:

• \( \sin t \) reflects acceleration in the x-direction.
• \( \cos t \) exhibits acceleration in the y-direction.
• \( -\cos(t/2) \) demonstrates acceleration in the z-direction.

Studying this vector is key to understanding how the curve evolves over time. It pinpoints areas where transitions in movement occur, affecting the overall shape and bending tendencies of the curve as determined by the curvature function. The second derivative plays an instrumental role in calculating the cross product needed for the curvature interpretation.