Problem 11 - Thomas Calculus

Question

Find \(\mathbf{B}\) and \(\tau\) for these space curves. \(\mathbf{r}(t)=\left(e^{t} \cos t\right) \mathbf{i}+\left(e^{t} \sin t\right) \mathbf{j}+2 \mathbf{k}\)

Step-by-Step Solution

Answer

\( \mathbf{B}(t) = -e^{2t} \mathbf{k} \); \( \tau = 0 \).

Step 1: Compute the First Derivative

To find the velocity vector \( \mathbf{v}(t) \), take the derivative of \( \mathbf{r}(t) \). The components are:- Derivative of \(e^t \cos t\) is \( e^t \cos t - e^t \sin t \).- Derivative of \(e^t \sin t\) is \( e^t \sin t + e^t \cos t \).- Derivative of \(2\) is \(0\).So, \( \mathbf{v}(t) = (e^t \cos t - e^t \sin t) \mathbf{i} + (e^t \sin t + e^t \cos t) \mathbf{j} \).

Step 2: Compute the Second Derivative

To find the acceleration vector \( \mathbf{a}(t) \), take the derivative of \( \mathbf{v}(t) \):- The \( i \)-component's derivative: \( (e^t \cos t - e^t \sin t) = e^t \cos t - e^t \sin t - e^t \sin t - e^t \cos t = -2e^t \sin t \).- The \( j \)-component's derivative: \( (e^t \sin t + e^t \cos t) = e^t \sin t + e^t \cos t + e^t \cos t - e^t \sin t = 2e^t \cos t \).Thus, \( \mathbf{a}(t) = -2e^t \sin t \mathbf{i} + 2e^t \cos t \mathbf{j} \).

Step 3: Compute the Binormal Vector \( \mathbf{B}(t) \)

The binormal vector \( \mathbf{B}(t) \) is given by the cross product of \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \). Calculate:\[\mathbf{B}(t) = \mathbf{v}(t) \times \mathbf{a}(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ e^t \cos t - e^t \sin t & e^t \sin t + e^t \cos t & 0 \ -2e^t \sin t & 2e^t \cos t & 0 \end{vmatrix}\]- Simplified using the determinant: \( \mathbf{k}(-e^{2t}) = -e^{2t} \mathbf{k} \) (only non-zero component).So, \( \mathbf{B}(t) = -e^{2t} \mathbf{k} \).

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

The torsion \( \tau(t) \) is given by:\[ \tau(t) = \frac{(\mathbf{a}(t) \times \mathbf{v}(t)) \cdot \mathbf{B}(t)}{\|\mathbf{v}(t) \times \mathbf{a}(t)\|^2}\]We know \( \mathbf{B}(t) = -e^{2t} \mathbf{k} \) from previous steps.To find the cross product: \( \mathbf{a}(t) \times \mathbf{v}(t) \), it is the same as previously derived, \( \mathbf{B}(t) \).Thus, \( \tau(t) = 0 \) since it is a planar curve (all terms involving \( \mathbf{k} \) vanish).

Step 5: Provide Final Answer

Given the above calculations, the binormal vector \( \mathbf{B}(t) = -e^{2t} \mathbf{k} \) and the torsion \( \tau = 0 \).

Key Concepts

Binormal Vector

In the context of space curves, the binormal vector, represented as \( \mathbf{B}(t) \), plays an important role in understanding the geometry of a curve. It is perpendicular to both the tangent and normal vectors of the curve. To find the binormal vector, we take the cross product of the velocity vector \( \mathbf{v}(t) \) and the acceleration vector \( \mathbf{a}(t) \). Both vectors are derived from the original curve equation.

In the provided example, the velocity and acceleration vectors are cross-multiplied using determinants to yield:

• Velocity Vector, \( \mathbf{v}(t) \): \( (e^t \cos t - e^t \sin t) \mathbf{i} + (e^t \sin t + e^t \cos t) \mathbf{j} \)
• Acceleration Vector, \( \mathbf{a}(t) \): \( -2e^t \sin t \mathbf{i} + 2e^t \cos t \mathbf{j} \)

We compute their cross product as demonstrated, leading to the binormal vector: \[ \mathbf{B}(t) = -e^{2t} \mathbf{k} \]The result shows us that the binormal vector points in the direction of the \( \mathbf{k} \)-axis, indicating how the plane of the curve twists in 3D space.

Torsion

Torsion is a measure of how much a curve is deviating from being planar. It gives us insight into the twisting of the curve in three-dimensional space. The formula for torsion \( \tau(t) \) involves the binormal vector:\[ \tau(t) = \frac{(\mathbf{a}(t) \times \mathbf{v}(t)) \cdot \mathbf{B}(t)}{\|\mathbf{v}(t) \times \mathbf{a}(t)\|^2} \]To understand torsion without diving into complex math, visualize a slinky: torsion informs you about the twist.

In planar curves, like the example provided, the torsion is often zero because the plane in which the curve lies does not twist. We found that \( \tau = 0 \) indicating it is a planar curve meaning no twisting occurs. This aids significantly in simplifying calculations regarding the curve's structure.

Cross Product

The cross product is an operation on two vectors in three-dimensional space that yields another vector perpendicular to the plane containing the initial vectors. It is denoted by \( \mathbf{u} \times \mathbf{v} \).

To compute the cross product of two vectors, use the determinant of a matrix formed by unit vectors and the components of the two vectors. The result is a vector that is orthogonal to both original vectors.

• Formula: \( \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ u_1 & u_2 & u_3 \ v_1 & v_2 & v_3 \end{vmatrix} \)

For example, finding the cross product in our problem of \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \) gave the binormal vector.This operation is crucial in many aspects of vector calculus and physics, where it helps find perpendicular vectors and calculate torque, among other applications.

Velocity Vector

The velocity vector is the first derivative of a position vector function with respect to time. It represents the instantaneous rate of change of position with time, essentially describing the direction and speed of motion along a curve. In simpler terms, it's like the arrow that shows where and how fast a particle is moving at any moment.

• Expression for velocity: Taking the derivative of the space curve \( \mathbf{r}(t) \), we derive \( \mathbf{v}(t) \).
• Example: In the exercise, the velocity vector \( \mathbf{v}(t) \) is formed by differentiating each component of \( \mathbf{r}(t) \), giving: \( (e^t \cos t - e^t \sin t) \mathbf{i} + (e^t \sin t + e^t \cos t) \mathbf{j} \).

Understanding the velocity vector is key to analyzing motion along a curve, especially in physics and engineering contexts, as it directly affects the calculation of other quantities like acceleration and force.