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 a \(=d^{2} \mathbf{r} / d t^{2}\) and apply the initial conditions in reverse order.)

Step-by-Step Solution

Answer

The curve lies in the \( xy \)-plane because \( h(t) = 0 \) for \( t \) in \([a, b]\).

Step 1: Write Down the Position Vector

The position vector is given as \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \). This vector represents the path of the curve in 3D space. We are to show that the curve lies in a plane, specifically that \( h(t) = 0 \) along the interval \([a, b]\).

Step 2: Understand Given Conditions

We know that \( \mathbf{v} \cdot \mathbf{k} = 0 \) for all \( t \) in \([a, b]\). The velocity vector \( \mathbf{v} = \frac{d\mathbf{r}}{dt} = f'(t) \mathbf{i} + g'(t) \mathbf{j} + h'(t) \mathbf{k} \). Thus, \( h'(t) \cdot 0 = 0 \) implies \( h'(t) = 0 \) for \( t \) in \([a, b]\).

Step 3: Integrate the Velocity Component Equation

Since \( h'(t) = 0 \), integrate this equation with respect to \( t \) across the interval \([a, b]\): \( \int h'(t) \, dt = \int 0 \, dt \), which simplifies to \( h(t) = C \), where \( C \) is a constant of integration.

Step 4: Apply Initial Condition

We know that \( \mathbf{r}(a) = 0 \), which means \( f(a) = 0 \), \( g(a) = 0 \), and \( h(a) = 0 \). From this, \( h(a) = C = 0 \). Therefore, \( h(t) = 0 \) for \( t \) in \([a, b]\).

Step 5: Conclusion

We have shown that \( h(t) = 0 \) for all \( t \) in \([a, b]\) using the given initial conditions and the constraint on the curve's velocity. Therefore, the curve lies in the \( xy \)-plane.

Key Concepts

Zero Torsion

Zero torsion is a mathematical property of curves used in calculus and physics. Torsion, in general, indicates how much a curve deviates from being planar, which means it quantifies the twisting of a curve in three-dimensional space.
When a curve has zero torsion, it does not twist out of the plane it lies in. It remains perfectly planar without any kind of spiral movement.

• This is why curves with zero torsion can easily be positioned in a 2D plane like the xy-plane or any other plane parallel to it.
• In mechanical and aerodynamic contexts, knowing that a structure has zero torsion can simplify analyzing the stresses it experiences because it remains flat over the examined interval.

Velocity Vector

The concept of the velocity vector is essential in understanding motion in three dimensions. It represents the speed and direction of a moving object.
The velocity vector, denoted here by \ \mathbf{v} = \frac{d\mathbf{r}}{dt} \, is derived from the position vector \( \mathbf{r}(t) \). Essentially, it gives us an insight into the instantaneous rate of change of the position with respect to time.

• For our specific problem, we know that the velocity vector remains perpendicular to the fixed vector \ \mathbf{k} \, prompting that its component along this vector is zero: \ \mathbf{v} \cdot \mathbf{k} = 0\.
• Because of this perpendicularity, we conclude that the curve's path doesn't rise or fall in the direction of \ \mathbf{k} \, which simplifies the study of its path.

Position Vector

The position vector is a fundamental concept in vector calculus. It specifies the location of a particle in space at any given time.
In the provided exercise, the position vector \( \mathbf{r}(t) = f(t) \mathbf{i} + g(t) \mathbf{j} + h(t) \mathbf{k} \) gives the coordinates of a particle moving in 3D space over time.

• This vector effectively tracks the particle's path by detailing how far it deviates along each axis (i, j, k_axes) over an interval \([a, b]\).
• Here, the goal was to demonstrate that the path lies entirely on the xy-plane, implying no movement along the z-axis, which \( h(t) \) represents.

Integration

Integration is a critical process used in mathematics to find a function from its derivative or a sum of elements that continuously change (like area under a curve).
In our scenario, integrating the velocity component, \( h'(t) = 0 \), is essential. Through integration, we discover the original function that, when differentiated, yields the velocity.

• Using the integral \( \int 0 \, dt \), we solve this to determine \( h(t) = C \), signifying a constant function.
• Thus, integration negates the derivative, allowing us to identify whether changes occur along the z-axis in this context. Here, integrating confirmed no z-component; hence, the curve stays in the plane.

Initial Condition

Initial conditions are specifications given for functions or equations, usually indicating where a function begins or what its initial value is.
They are vital in solving differential equations as they ensure that the particular solution fits the context of the problem.

• In the exercise, the condition \( \mathbf{r}(a) = 0 \) was instrumental, determining that \( f(a), g(a), h(a) = 0 \) at the start \( t = a \).
• This constraint allowed us to identify that the constant from integrating, \( h(t) = C \), resolved to zero since initially, the curve registers zero at the z-component, thus maintaining its planar nature.