Problem 27 - Thomas Calculus

Question

Differentiable curves with zero torsion lie in planes That a sufficiently differentiable curve with zero torsion lies in a plane is a special case of the fact that a particle whose velocity remains perpendicular to a fixed vector \(\mathbf{C}\) moves in a plane perpendicular to \(\mathbf{C} .\) This, in turn, can be viewed as the solution of the following problem in calculus. Suppose \(\mathbf{r}(t)=f(t) \mathbf{i}+g(t) \mathbf{j}+h(t) \mathbf{k}\) is twice differentiable for all \(t\) in an interval \([a, b],\) that \(\mathbf{r}=0\) when \(t=a,\) and that \(\mathbf{v} \cdot \mathbf{k}=0\) for all \(t\) in \([a, b] .\) Then \(h(t)=0\) for all \(t\) in \([a, b] .\) Solve this problem. (Hint: Start with \(\mathbf{a}=d^{2} \mathbf{r} / d t^{2}\) and apply the initial conditions in reverse order.)

Step-by-Step Solution

Answer

\( h(t) = 0 \) for all \( t \) in \( [a, b] \), so the curve lies in a plane.

Step 1: Understand the Problem

We are given a curve \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \) and we need to show that \( h(t) = 0 \) for all \( t \) in \( [a, b] \) if the velocity \( \mathbf{v} \cdot \mathbf{k} = 0 \). This implies that the motion of the curve is restricted to a plane perpendicular to the \( \mathbf{k} \) vector.

Step 2: Express Acceleration

We start by calculating the acceleration, \( \mathbf{a} = \frac{d^2 \mathbf{r}}{dt^2} \). Thus, \( \mathbf{a}(t) = \frac{d^2 f(t)}{dt^2} \mathbf{i} + \frac{d^2 g(t)}{dt^2} \mathbf{j} + \frac{d^2 h(t)}{dt^2} \mathbf{k} \).

Step 3: Apply the Given Condition

Since \( \mathbf{v} \cdot \mathbf{k} = 0 \), the velocity vector \( \mathbf{v} = \frac{d \mathbf{r}}{dt} = \frac{d f(t)}{dt} \mathbf{i} + \frac{d g(t)}{dt} \mathbf{j} + \frac{d h(t)}{dt} \mathbf{k} \) implies that \( \frac{d h(t)}{dt} = 0 \).

Step 4: Integrate to Find \( h(t) \)

Given \( \frac{d h(t)}{dt} = 0 \), integrate with respect to \( t \) to find \( h(t) \). Thus, \( h(t) = C \), where \( C \) is a constant of integration.

Step 5: Apply Initial Conditions

Since \( \mathbf{r}(a) = 0 \), it follows \( h(a) = 0 \). Therefore, applying this condition to \( h(t) = C \), we find that \( C = 0 \).

Step 6: Conclude Result

Since \( C = 0 \), it follows that \( h(t) = 0 \) for all \( t \) in \( [a, b] \). This shows that the curve lies in the plane \( h(t) = 0 \).

Key Concepts

Zero Torsion

A curve is said to have zero torsion when, intuitively, it does not "twist" out of the plane in which it lies.
This means the curve remains flat and can be fully described by tracing it on a single plane.
Torsion, in the context of curves, measures how much and how quickly the curve deviates from lying entirely in a single plane.
For a curve to possess zero torsion, its Frenet frame — a coordinate system moving along the curve — remains fixed in its orientation relative to the curve.
With zero torsion, the binormal vector doesn’t "turn" or "twist," meaning it remains consistent throughout the curve.
This concept is critical in understanding the behavior of curves in space, particularly in showing that certain conditions force a curve to remain planar.
In our problem, zero torsion supports the finding that the given curve doesn't lift off its initial plane.

Plane Curve

A plane curve is simply a curve that is entirely contained in a single plane.
Think of drawing a curve on a sheet of paper; that curve will not rise or move outside the flat surface of the paper.
The mathematical condition involving zero torsion inherently implies that a curve with zero torsion must reside in some plane.
This is because any aspect of the curve that would cause it to leave the plane effectively introduces torsion.
Plane curves are significant because they are easier to visualize and analyze than curves that extend into three-dimensional space.
Identifying whether a curve is planar or not has important implications in both theoretical studies and practical applications, such as in engineering and computer graphics.

Velocity Perpendicular to a Vector

When the velocity of a moving point along a curve is always perpendicular to a fixed vector, this imposes a specific constraint on the curve.
In the problem, this condition leads us to deduce characteristics about the curve's trajectory — namely, its planarity.
The perpendicularity condition can be understood through the dot product, which is a measure of how much two vectors "overlap."
If the dot product of velocity with the fixed vector is zero, then the velocity is perpendicular to that vector at all moments in time.
In our case, when velocity is perpendicular to a fixed vector along the z-axis, the motion must be confined to the x-y plane.
This concept explains why the z-component of the curve i.e., \(h(t)\), remains zero throughout the interval.

Twice Differentiable Functions

Twice differentiable functions are smooth enough to allow the computation of both first and second derivatives.
These functions are critical in the study of curves because they allow us to define and analyze both velocity and acceleration.
A twice differentiable curve is one where its defining function has both smoothness and continuity for its first and second derivatives.
For our exercise, twice differentiability ensures that the acceleration vector can be computed as \(\mathbf{a}(t) = \frac{d^2 f(t)}{dt^2} \mathbf{i} + \frac{d^2 g(t)}{dt^2} \mathbf{j} + \frac{d^2 h(t)}{dt^2} \mathbf{k}\).
This capability provides essential information about the curvature and motion characteristics, ensuring we can fully determine the behavior of the curve.
Ensuring these functions are twice differentiable is crucial for applying calculus principles to solve and understand geometric problems.