Problem 21 - Thomas Calculus

Question

Show that \(\kappa\) and \(\tau\) are both zero for the line $$\mathbf{r}(t)=\left(x_{0}+A t\right) \mathbf{i}+\left(y_{0}+B t\right) \mathbf{j}+\left(z_{0}+C t\right) \mathbf{k}$$

Step-by-Step Solution

Answer

Both \( \kappa \) and \( \tau \) are zero for the line because it is straight.

Step 1: Differentiate the Position Vector

First, differentiate the given position vector \( \mathbf{r}(t) = (x_0 + At) \mathbf{i} + (y_0 + Bt) \mathbf{j} + (z_0 + Ct) \mathbf{k} \) with respect to \( t \). This gives the velocity vector \( \mathbf{r'}(t) = A \mathbf{i} + B \mathbf{j} + C \mathbf{k} \).

Step 2: Calculate the Second Derivative

Differentiate the velocity vector \( \mathbf{r'}(t) = A \mathbf{i} + B \mathbf{j} + C \mathbf{k} \) with respect to \( t \) to get the acceleration vector \( \mathbf{r''}(t) = 0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k} \).

Step 3: Compute the Curvature \( \kappa \)

The curvature \( \kappa \) is given by \( \kappa = \frac{\|\mathbf{r'}(t) \times \mathbf{r''}(t)\|}{\|\mathbf{r'}(t)\|^3} \). Because \( \mathbf{r''}(t) \) is the zero vector, \( \mathbf{r'}(t) \times \mathbf{r''}(t) \) is also zero, so \( \kappa = 0 \).

Step 4: Compute the Torsion \( \tau \)

The torsion \( \tau \) is given by \( \tau = \frac{((\mathbf{r'}(t)\times \mathbf{r''}(t)) \cdot \mathbf{r'''}(t))}{\|\mathbf{r'}(t) \times \mathbf{r''}(t)\|^2} \). Since \( \mathbf{r''}(t) \) and consequently \( \mathbf{r'}(t) \times \mathbf{r''}(t) \) are zero, \( \tau \) is undefined due to division by zero, but conventionally considered zero for a straight line.

Key Concepts

Curvature

In the world of differential calculus, curvature helps us understand how sharply a curve bends. For a straight line, as seen with the equation for line in the exercise, the curvature is a concept used to measure this bend or sharpness. But since a line doesn’t bend, its curvature is zero.
Curvature (\(\kappa\)) is mathematically defined as the rate of change of the tangent vector with respect to arc length. For this specific line, the position vector \(\mathbf{r}(t)\), when differentiated gives a constant velocity vector. Further differentiation leads to a zero acceleration vector.

• The cross product of the velocity and acceleration vectors results in zero.
• The formula for \(\kappa\) includes this cross product in its numerator, causing the curvature to become zero.

Therefore, the line does not exhibit any curvature, which confirms its straightness.

Torsion

Torsion is another concept used to describe the twisting of a space curve. For a straight line, torsion is somewhat hypothetical but is understood to be zero due to convention.
For this problem, torsion (\(\tau\)) finds how the curve departs from being planar and is calculated with a complex formula involving third derivatives. However, the key here is that since both the second and third derivatives lead to zero values, the torsion reaches a state of being essentially zero.

• The primary reason torsion is often undefined is because of division by zero in the formula.
• For lines (straight lines, specifically), \(\tau = 0\) by convention as they do not twist or depart from their plane.

This reinforces the straight character of this line as it does not twist or turn in space.

Position Vector

A position vector describes the location of a point in space in relation to the origin. In the context of this exercise, \(\mathbf{r}(t) = (x_0 + At) \mathbf{i} + (y_0 + Bt) \mathbf{j} + (z_0 + Ct) \mathbf{k}\) represents the position vector for a line parameterized by \(t\).
The position vector changes over time due to the values \(A, B,\) and \(C\) which are multiples of \(t\), indicating motion along the line.

• The position vector essentially sets the path traced out by the variable \(t\).
• This tracing path is important in both geometric and physical contexts, providing a vector form of motion.

Position vectors serve as the initial building block for determining other quantities like velocity and acceleration, thus playing a crucial role in analyzing movement.

Velocity Vector

The velocity vector expresses the rate of change of the position vector with time. In differential calculus, velocity helps in understanding how fast something moves and in what direction. For our line equation, differentiating \(\mathbf{r}(t)\) gives us the velocity vector: \(\mathbf{r'}(t) = A \mathbf{i} + B \mathbf{j} + C \mathbf{k}\).
This vector remains constant as it does not change with time \(t\) because \(A, B,\) and \(C\) are constants. This reflects that the line has a consistent motion along a straight path.

• The magnitude of the velocity directly corresponds to the speed along this line.
• Understanding the velocity vector is key in defining the dynamic properties of curves and lines in space.

The velocity vector becomes vital in calculating other features like curvature and torsion, adding depth to our analysis of linear motion.