Problem 81 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that the unit binormal vector \(\mathbf{B}=\mathbf{T} \times \mathbf{N}\) has the property that \(\frac{d \mathbf{B}}{d s}\) is perpendicular to \(\mathbf{T}\).

Step-by-Step Solution

Answer

\( \frac{d \mathbf{B}}{d s} \) is perpendicular to \( \mathbf{T} \).

Step 1: Understanding the Unit Binormal Vector

The unit binormal vector \( \mathbf{B} \) is defined as the cross product of the unit tangent vector \( \mathbf{T} \) and the unit normal vector \( \mathbf{N} \). Thus, \( \mathbf{B} = \mathbf{T} \times \mathbf{N} \). To solve the problem, we need to show that the derivative of \( \mathbf{B} \) with respect to the arc length \( s \), i.e., \( \frac{d \mathbf{B}}{d s} \), is perpendicular to \( \mathbf{T} \).

Step 2: Differentiate the Unit Binormal Vector

We differentiate \( \mathbf{B} = \mathbf{T} \times \mathbf{N} \) with respect to \( s \). Using the product rule for cross products, we get: \( \frac{d}{ds}(\mathbf{B}) = \frac{d}{ds}(\mathbf{T}) \times \mathbf{N} + \mathbf{T} \times \frac{d}{ds}(\mathbf{N}) \).

Step 3: Establish Perpendicularity Condition

We need to show that \( \frac{d \mathbf{B}}{d s} \cdot \mathbf{T} = 0 \). For the perpendicularity, we calculate the dot product of \( \frac{d \mathbf{B}}{ds} \) with \( \mathbf{T} \): \[\left(\frac{d}{ds}(\mathbf{T}) \times \mathbf{N} + \mathbf{T} \times \frac{d}{ds}(\mathbf{N})\right) \cdot \mathbf{T}\].

Step 4: Simplifying the Dot Product

The first term \( \left(\frac{d}{ds}(\mathbf{T}) \times \mathbf{N}\right) \cdot \mathbf{T} \) is zero because the dot product of a vector with a cross product involving the same vector is always zero.The second term \( \left(\mathbf{T} \times \frac{d}{ds}(\mathbf{N})\right) \cdot \mathbf{T} \) is also zero using the same property.Thus, \( \frac{d \mathbf{B}}{d s} \cdot \mathbf{T} = 0 \).

Step 5: Conclusion

The calculations confirm that \( \frac{d \mathbf{B}}{d s} \cdot \mathbf{T} = 0 \), meaning \( \frac{d \mathbf{B}}{d s} \) is indeed orthogonal to \( \mathbf{T} \). Therefore, we have shown that \( \frac{d \mathbf{B}}{d s} \) is perpendicular to \( \mathbf{T} \).

Key Concepts

Unit Tangent Vector

In differential geometry, the unit tangent vector plays a crucial role. Imagine a curve in three-dimensional space. As you travel along this curve, the direction you face changes. The unit tangent vector, often denoted as \( \mathbf{T} \), represents the direction of the curve at any given point. This vector is crucial for understanding the properties of the curve.The unit tangent vector is derived by taking the derivative of the position vector \( \mathbf{r}(t) \) with respect to the parameter \( t \), and then normalizing this vector:

• First, differentiate the position vector: \( \mathbf{r}'(t) = \frac{d\mathbf{r}}{dt} \).
• Next, normalize to get the unit tangent vector: \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{||\mathbf{r}'(t)||} \).

Where \( ||\mathbf{r}'(t)|| \) is the magnitude of \( \mathbf{r}'(t) \). The unit tangent vector always maintains a magnitude of 1, pointing in the direction of motion along the curve.

Unit Normal Vector

The unit normal vector, denoted \( \mathbf{N} \), is another key concept in understanding curves. This vector is orthogonal, or perpendicular, to the unit tangent vector. The unit normal vector expresses how the curve is bending at any given point.To find \( \mathbf{N} \), we follow these steps:

• First, differentiate the unit tangent vector: \( \mathbf{T}'(t) = \frac{d\mathbf{T}}{dt} \).
• Normalize the resulting vector to ensure it has a magnitude of 1: \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{||\mathbf{T}'(t)||} \).

The unit normal vector thus provides information about the rate and direction of the curve's change in direction, essential for understanding features like curvature.

Cross Product

The cross product is a fundamental operation in vector algebra used to find a vector that is perpendicular to two given vectors in three-dimensional space.Given vectors \( \mathbf{A} \) and \( \mathbf{B} \), the cross product \( \mathbf{A} \times \mathbf{B} \) is calculated as follows:

• The magnitude is given by \( ||\mathbf{A} \times \mathbf{B}|| = ||\mathbf{A}|| \cdot ||\mathbf{B}|| \cdot \sin\theta \), where \( \theta \) is the angle between \( \mathbf{A} \) and \( \mathbf{B} \).
• The resulting vector is orthogonal to both \( \mathbf{A} \) and \( \mathbf{B} \).

In the context of differential geometry, the unit binormal vector \( \mathbf{B} = \mathbf{T} \times \mathbf{N} \) uses this operation to find a vector perpendicular to both the unit tangent and unit normal vectors. This helps in constructing an orthonormal frame that can describe the geometry of a curve fully.

Orthogonality

Orthogonality is a concept where two vectors are perpendicular to each other. When vectors are orthogonal, their dot product equals zero.For vectors \( \mathbf{u} \) and \( \mathbf{v} \), this relationship is expressed as:

• If \( \mathbf{u} \cdot \mathbf{v} = 0 \), then \( \mathbf{u} \) and \( \mathbf{v} \) are orthogonal.

This feature is crucial in many areas of mathematics and engineering. In differential geometry, orthogonality is used to analyze curves and surfaces. For example, verifying that \( \frac{d \mathbf{B}}{d s} \) (the derivative of the unit binormal vector) is orthogonal to \( \mathbf{T} \) shows that they are perpendicular and therefore confirms the overall behavior of the curve.Such analyses help in formulating properties of curves through orthonormal frames, facilitating deeper geometric understandings.