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.

Short answer

Both \( \kappa \) and \( \tau \) are zero for a straight line, confirming it is straight.

Step-by-step solution

Understand the problem

We are asked to show that both the curvature \( \kappa \) and torsion \( \tau \) for a straight line parameterized by \( \mathbf{r}(t) = \mathbf{r}_0 + a_0 t \mathbf{i} + b_0 t \mathbf{j} + c_0 t \mathbf{k} \) are zero. This involves calculating \( \kappa \) and \( \tau \) for the given curve.

Find the derivatives of the position vector

The position vector is \( \mathbf{r}(t) = \mathbf{r}_0 + a_0 t \mathbf{i} + b_0 t \mathbf{j} + c_0 t \mathbf{k} \). Compute the first derivative \( \mathbf{r}'(t) = a_0 \mathbf{i} + b_0 \mathbf{j} + c_0 \mathbf{k} \). Then compute the second derivative \( \mathbf{r}''(t) = \mathbf{0} \), since there are no terms involving \( t \).

Calculate the curvature \(\kappa\)

The curvature \( \kappa \) is given by the formula \[ \kappa = \frac{\| \mathbf{r}''(t) \times \mathbf{r}'(t) \|}{\| \mathbf{r}'(t) \|^3} \]. We found \( \mathbf{r}''(t) = \mathbf{0} \), so \( \mathbf{r}''(t) \times \mathbf{r}'(t) = \mathbf{0} \). Therefore, \( \kappa = \frac{0}{\| \mathbf{r}'(t) \|^3} = 0 \).

Calculate the torsion \(\tau\)

The torsion \( \tau \) is given by the formula \[ \tau = \frac{( \mathbf{r}'(t) \times \mathbf{r}''(t) ) \cdot \mathbf{r}'''(t)}{\| \mathbf{r}''(t) \times \mathbf{r}'(t) \|^2} \]. Since \( \mathbf{r}''(t) = \mathbf{0} \) and hence \( \mathbf{r}'''(t) = \mathbf{0} \), both the numerator and denominator become zero, but since \( \mathbf{r}''(t) \times \mathbf{r}'(t) \) is \( \mathbf{0} \), \( \tau = 0 \).

Conclusion

Both the curvature \( \kappa \) and torsion \( \tau \) for the straight line parameterized by \( \mathbf{r}(t) = \mathbf{r}_0 + a_0 t \mathbf{i} + b_0 t \mathbf{j} + c_0 t \mathbf{k} \) are zero, which confirms the line is straight.

Key concepts

Position Vector

In mathematics, a position vector is a concept used to represent the location of a point in space. It originates from the origin of the coordinate system to the point under consideration.
For a parameterized curve, the position vector is often represented as \( \mathbf{r}(t) = \mathbf{r}_0 + a_0 t \mathbf{i} + b_0 t \mathbf{j} + c_0 t \mathbf{k} \). Here, \( \mathbf{r}_0 \) denotes the initial position of the curve. The components \( a_0, b_0, \) and \( c_0 \) describe how the position changes over time.
The role of the position vector is crucial in determining the direction, position, and speed at which a curve is traversed. It essentially describes the path of a moving particle or object in space when a parameter \( t \) changes.

First Derivative

The first derivative of the position vector \( \mathbf{r}(t) \) is vital in understanding the velocity of a curve at any given point. It determines the rate of change of the position concerning the parameter \( t \).
In the context of our straight line equation, the first derivative is given as \( \mathbf{r}'(t) = a_0 \mathbf{i} + b_0 \mathbf{j} + c_0 \mathbf{k} \).
This derivative signifies a constant vector, meaning the velocity of the line is uniform in all parts, and it doesn't change over time. Essentially, the first derivative is parallel to the line itself, as it describes the direction and rate at which it moves.

Second Derivative

The second derivative refers to the acceleration of the position vector and provides insight into the curvature of the path the curve takes.
For straight lines, as in our exercise, the second derivative is \( \mathbf{r}''(t) = \mathbf{0} \). This zero vector indicates that there is no change in the velocity concerning time, which implies a steady, non-curving path.
This absence of acceleration is why the curvature \( \kappa \), computed from the second derivative, is zero for straight lines. A zero second derivative confirms that the trajectory is consistent without any bends or twists.

Straight Line Parameterization

Straight line parameterization is a way to represent straight lines using a parameter, typically \( t \), allowing it to be described in vector form. This allows us to express a line as \( \mathbf{r}(t) = \mathbf{r}_0 + a_0 t \mathbf{i} + b_0 t \mathbf{j} + c_0 t \mathbf{k} \).
Using parameterization, the line can be easily navigated through changes in parameter \( t \), depicting how the line extends in different directions as \( t \) changes. The coefficients \( a_0, b_0, \) and \( c_0 \) provide the direction proportions of the line.
This form highlights its inherent properties like constant speed and directionality. The consistent rate and direction illustrate why both the curvature \( \kappa \) and torsion \( \tau \) for such a parameterized line are zero, showcasing its linear, untwisted nature.