Problem 84 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that for a straight line \(\mathbf{r}(t)=\mathbf{r}_{0}+a_{0} t \mathbf{i}+\) \(b_{0} t \mathbf{j}+c_{0} t \mathbf{k}\) both \(\kappa\) and \(\tau\) are zero.

Step-by-Step Solution

Answer

Both \( \kappa \) and \( \tau \) are zero because the second derivative is zero.

Step 1: Define the position vector

The position vector of the straight line is given by \( \mathbf{r}(t) = \mathbf{r}_0 + a_0 t \mathbf{i} + b_0 t \mathbf{j} + c_0 t \mathbf{k} \), where \( \mathbf{r}_0 \) is the initial point and \( a_0, b_0, c_0 \) are constants representing the direction of the line.

Step 2: Find the derivative with respect to t

The first derivative of the position vector gives us the velocity vector, which can be calculated as \( \mathbf{r}'(t) = \frac{d}{dt}(\mathbf{r}_0 + a_0 t \mathbf{i} + b_0 t \mathbf{j} + c_0 t \mathbf{k}) = a_0 \mathbf{i} + b_0 \mathbf{j} + c_0 \mathbf{k} \).

Step 3: Determine the second derivative

The second derivative of the position vector is the acceleration vector. Since the first derivative is constant, the second derivative is \( \mathbf{r}''(t) = \frac{d^2}{dt^2}(\mathbf{r}_0 + a_0 t \mathbf{i} + b_0 t \mathbf{j} + c_0 t \mathbf{k}) = \mathbf{0} \).

Step 4: Calculate the curvature \( \kappa \)

The curvature \( \kappa \) is defined as \( \kappa = \frac{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|}{\| \mathbf{r}'(t) \|^3} \). Since \( \mathbf{r}''(t) = \mathbf{0} \), the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) = \mathbf{0} \). Therefore, \( \kappa = 0 \).

Step 5: Calculate the torsion \( \tau \)

Torsion \( \tau \) is defined as \( \tau = \frac{(\mathbf{r}'(t) \times \mathbf{r}''(t)) \cdot \mathbf{r}'''(t)}{\| \mathbf{r}'(t) \times \mathbf{r}''(t) \|^2} \). Since both \( \mathbf{r}''(t) = \mathbf{0} \) and \( \mathbf{r}'''(t) = \mathbf{0} \), the numerator and denominator are zero, making \( \tau = 0 \).

Key Concepts

Curvature and Torsion

In calculus, the concepts of curvature and torsion help us understand how a curve or line behaves in space. Curvature, denoted as \( \kappa \), measures how fast a curve changes direction as you move along it. For a straight line, there's no bending or change in direction, br>so the curvature is zero. It shows that the straight line moves uniformly without any turns.
Torsion, denoted as \( \tau \), measures how much a curve twists out of the plane of curvature. Think of it like the twist in a ribbon. For a straight line, br>since it lies entirely in a plane without twisting out of it, the torsion is zero too. It's a bit like how a ruler stays flat on your desk without twisting around.

• Curvature helps us see how a path bends.
• Torsion tells us if the path twists in three-dimensional space.

For the exercise in question, when calculating both \( \kappa \) and \( \tau \), we find that they are zero, confirming that the line is perfectly straight with no bends or twists.

Position Vector

The position vector is a simple yet crucial concept. Think of it as a way to pinpoint exactly where you are in space at any given time.In mathematical terms, the position vector is often expressed as \( \mathbf{r}(t) = \mathbf{r}_0 + a_0 t \mathbf{i} + b_0 t \mathbf{j} + c_0 t \mathbf{k} \). Here, \( \mathbf{r}_0 \) represents the initial position, a fixed starting point.

The coefficients \( a_0, b_0, \) and \( c_0 \) describe the direction of movement along the x, y, and z axes respectively.For a straight line, these coefficients remain constant, keeping the line trajectory uniform as time \( t \) changes.

• Position vectors tell us where an object is at a specific time.
• They serve as a guide, showing us the path an object follows.

In the context of this exercise, understanding the position vector helps us confirm the nature of the straight line, as it directly affects calculations of curvature and torsion.

Velocity and Acceleration Vectors

Velocity and acceleration vectors are essential components when analyzing movement in calculus.They help describe how fast an object is moving and how its speed changes over time.

Velocity Vector

The velocity vector is derived from the first derivative of the position vector. In this context, it's shown as \( \mathbf{r}'(t) = a_0 \mathbf{i} + b_0 \mathbf{j} + c_0 \mathbf{k} \). It remains constant, indicating constant speed without any change in direction.The direction of this vector indicates the precise path of motion.

Acceleration Vector

Acceleration is about changes in velocity and is described by the second derivative of the position vector:
\( \mathbf{r}''(t) = \mathbf{0} \). In this exercise, the acceleration vector is zero, signifying no change in speed or direction.

• Velocity vectors show the direction and speed of travel at any moment.
• Acceleration vectors tell us if and how speed changes during movement.

For a straight line, a constant velocity and a zero acceleration confirm the uniform, straight-line motion, reinforcing the results of curvature and torsion calculations.