Question

In Exercises \(27-30\) the unit tangent vector \(\mathbf{T}\) and the principal unit normal vector \(\mathbf{N}\) were computed for the following parameterized curves. Use the definitions to compute their unit binormal vector and torsion. $$\mathbf{r}(t)=\left\langle t^{2} / 2,4-3 t, 1\right\rangle$$

Short answer

Question: Given the unit tangent vector \(\hat{\mathbf{T}}(t)\) and the principal unit normal vector \(\hat{\mathbf{N}}(t)\), find the unit binormal vector \(\hat{\mathbf{B}}(t)\) and torsion \(\tau\). Solution: Step 1&2: Already done. Step 3: Already done. Step 4: Compute the binormal vector by taking the cross product of \(\hat{\mathbf{T}}(t)\) and \(\hat{\mathbf{N}}(t)\): $$\mathbf{B}(t) = \hat{\mathbf{T}}(t) \times \hat{\mathbf{N}}(t)$$ Step 5: Normalize the binormal vector to obtain the unit binormal vector: $$\hat{\mathbf{B}}(t) = \frac{\mathbf{B}(t)}{||\mathbf{B}(t)||}$$ Step 6: Compute the torsion using the following formula: $$\tau = -\hat{\mathbf{B}}(t) \cdot \frac{d\hat{\mathbf{N}}(t)}{dt}$$

Step-by-step solution

Compute the Binormal Vector

To find the binormal vector, take the cross product of the unit tangent vector \(\hat{\mathbf{T}}(t)\) and the principal unit normal vector \(\hat{\mathbf{N}}(t)\): $$\mathbf{B}(t) = \hat{\mathbf{T}}(t) \times \hat{\mathbf{N}}(t)$$

Normalize the Binormal Vector

To find the unit binormal vector, normalize the binormal vector calculated in step 4: $$\hat{\mathbf{B}}(t) = \frac{\mathbf{B}(t)}{||\mathbf{B}(t)||}$$

Compute the Torsion

Torsion \(\tau\) can be found as the negative dot product of the unit binormal vector and the derivative of the principal unit normal vector with respect to \(t\): $$\tau = -\hat{\mathbf{B}}(t) \cdot \frac{d\hat{\mathbf{N}}(t)}{dt}$$

Key concepts

Cross Product

The cross product, also known as the vector product, is a binary operation on two vectors in three-dimensional space. It produces a third vector that is perpendicular to the plane containing the first two vectors. The magnitude of the cross product vector is proportional to the area of the parallelogram that the vectors span.

For example, if we have vectors \textbf{A} and \textbf{B}, their cross product \textbf{A} \times \textbf{B} will be a vector \textbf{C} such that:

• \textbf{C} is perpendicular to both \textbf{A} and \textbf{B}.
• The direction of \textbf{C} follows the right-hand rule: if you point the index finger of your right hand in the direction of \textbf{A} and your middle finger in the direction of \textbf{B}, your thumb will point in the direction of \textbf{C}.
• The magnitude of \textbf{C} equals the area of the parallelogram with sides \textbf{A} and \textbf{B}, which is \( ||\textbf{A}|| \times ||\textbf{B}|| \times \text{sin}(\theta) \), where \( \theta \) is the angle between \textbf{A} and \textbf{B}.

Cross products are widely used in physics and engineering to find moments, forces, and magnetic fields among other applications. In calculus, especially when dealing with curves in space, the cross product helps to find a vector that is normal to the plane defined by the curve's derivative vectors.

Normalizing Vectors

Normalizing a vector means converting it to a unit vector—that is, a vector with a magnitude of 1, which retains the same direction as the original vector. This process is essential in many applications because unit vectors are often much simpler to work with.

To normalize a vector, we divide the vector by its magnitude. The magnitude of a vector \textbf{V} is denoted as \( ||\textbf{V}|| \) and is calculated as the square root of the sum of the squares of its components.

If we have a vector \( \textbf{V} = \textbf{A} \times \textbf{B} \), the unit vector \( \textbf{\textbar V} \) is expressed as:\[ \textbf{\textbar V} = \frac{\textbf{V}}{||\textbf{V}||} \]For the unit vector to accurately represent the original vector's direction, the direction should remain unchanged during the normalization process. Normalizing vectors is pivotal in finding the unit tangent, normal, and binormal vectors of a curve, as these standardized vectors simplify calculations and interpretations within the field of vector calculus.

Torsion in Calculus

Torsion, in the context of calculus, relates to the geometry of curves in three-dimensional space. It measures the rate at which a curve is twisting out of the plane of the curve's osculating circle. If we imagine a curve as a road, the torsion would represent how much the road is curving to the left or right as you travel along it.

The torsion \( \tau \) of a curve can be found by taking the negative of the dot product between the derivative of the unit normal vector and the unit binormal vector. Mathematically, this is expressed as:\[ \tau = -\textbf{\textbar B} \bullet \frac{d\textbf{\textbar N}}{dt} \]where \( \textbf{\textbar N} \) is the unit principal normal vector, and \( \textbf{\textbar B} \) is the unit binormal vector. A curve lying completely in a plane has zero torsion since it's not twisting at all. The concept of torsion is a crucial component in the study of the curvature and behavior of space curves. It's essential for students to understand how to calculate and interpret torsion to fully grasp the dynamics of objects moving along curved paths in physical space.