Problem 25 - 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 following result. 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 \([a, b] .\) Show that \(h(t)=0\) for all \(t\) in \([a, b] .\) (Hint: Start with \(\mathbf{a}=d \mathbf{r} / d t^{2}\) and apply the initial conditions in reverse order.)

Step-by-Step Solution

Answer

The condition \( \mathbf{v} \cdot \mathbf{k} = 0 \) implies \( h(t) = 0 \) for all \( t \) in \([a, b]\). Thus, the curve lies in the \( xy \)-plane.

Step 1: Understand the Problem Statement

We have a vector function \( \mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k} \) which represents the position vector of a particle. This curve is twice differentiable and we are given that \( \mathbf{v} \cdot \mathbf{k} = 0 \) for all \( t \) in the interval \([a, b]\), where \( \mathbf{v} \) is the velocity vector \( \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} \). Additionally, the curve begins at the origin (\( \mathbf{r}(a) = 0 \)). We need to prove that \( h(t) = 0 \) for all \( t \) in \([a, b]\).

Step 2: Express Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \), given by \( \mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = f'(t)\mathbf{i} + g'(t)\mathbf{j} + h'(t)\mathbf{k} \). Since \( \mathbf{v} \cdot \mathbf{k} = 0 \), this implies \( h'(t) = 0 \) for all \( t \).

Step 3: Integrate Velocity Component

Since \( h'(t) = 0 \), integrating with respect to \( t \) gives \( h(t) = C \), where \( C \) is a constant. This means that \( h(t) \) remains constant over the interval \([a, b]\).

Step 4: Apply Initial Condition

The initial condition \( \mathbf{r}(a) = 0 \) implies \( h(a) = 0 \) because \( \mathbf{r}(a) = f(a)\mathbf{i} + g(a)\mathbf{j} + h(a)\mathbf{k} = 0 \). Substituting into the equation for \( h(t) \), we get \( C = 0 \).

Step 5: Conclude the Solution

With \( C = 0 \), we have \( h(t) = 0 \) for the entire interval \([a, b]\), showing that the position vector \( \mathbf{r}(t) \) lies in the \( xy \)-plane for all \( t \).

Key Concepts

Differentiable Curves

A differentiable curve is one that behaves smoothly, which means, at every point, it has a defined tangent. You might think of it as a path that a particle can travel along without encountering any sharp turns or breaks.
To be classified as differentiable, the function describing the curve must have derivatives of any order, or at least the order necessary to solve a particular problem.

• A smooth curve allows for continuous and well-defined changes along its path.
• The higher the number of derivatives a curve can have, the smoother its motion is.
• When the curve has zero torsion, it implies there's no "twist" in the curve.

For a curve that lies entirely in a plane, being differentiable ensures that it can be described by straightforward equations and analyzed easily. In the exercise, zero torsion indicates that the curve is flat and doesn't spiral out of the sheet of paper, staying within the plane completely.

Velocity Vector

The velocity vector represents how fast and in what direction a point on a curve is moving. In mathematical terms, it's the derivative of the position vector of the curve. When considering the motion of a particle, the velocity vector is essential for understanding how the particle's position changes over time.

• The components of the velocity vector in the directions of the coordinate axes show the rate of change of position in those directions.
• In this context, given \(\mathbf{v} \cdot \mathbf{k} = 0\), it means there's no motion along the \( z\)-axis.
• This condition simplifies the analysis as it indicates movement is confined to the \( xy \)-plane.

The velocity vector is fundamental not only for determining speed and direction but also for understanding whether the curve is bounded to a specific plane or if it might have more freedom of movement.

Curve in a Plane

A curve in a plane is a path or trajectory contained entirely within a two-dimensional surface, such as the \( xy \)-plane. This means that one of the coordinate variables remains constant on this path.
For a curve described by the function \(\mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j} + h(t)\mathbf{k}\), if \(h(t) = 0\) for all \(t\), it confirms the curve exists solely in the \( xy \)-plane.

• Being in a plane simplifies the analysis of many mathematical problems, as it confines the motion to two dimensions.
• Complex behaviors such as twisting or spiraling are not present in planar curves of zero torsion.
• This ensures that certain dynamics, like those seen in the velocity and acceleration, have straightforward interpretations.

Understanding whether a curve is in a plane aids in constructing and solving geometric and physical problems, providing clarity and reducing the complexity of spatial motion.