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Magazine Chapter 14: Problem 41
The unit tangent vector \(\mathbf{T}\) and the principal unit normal vector N were computed for the following parameterized curves. Use the definitions to compute their unit binormal vector and torsion. $$r(t)=\langle 2 \sin t, 2 \cos t\rangle$$
Short Answer
Expert verified
Answer: The unit binormal vector B is $$\langle 0, 0, 1\rangle$$, and the torsion 蟿 is 0.
Step by step solution
01
Find the parameterized curve derivatives
First, let's find the first and second derivatives of the parameterized curve $$r(t) = \langle 2 \sin t, 2 \cos t\rangle$$ :
$$r'(t) = \frac{d}{dt}\langle 2 \sin t, 2 \cos t\rangle = \langle 2 \cos t, -2\sin t\rangle$$
$$r''(t) = \frac{d^2}{dt^2}\langle 2 \sin t, 2 \cos t\rangle = \langle -2 \sin t, -2 \cos t \rangle$$
02
Calculate the unit tangent vector
Normalize $$r'(t)$$ to get the unit tangent vector T:
$$\mathbf{T}(t) = \frac{r'(t)}{\|r'(t)\|} = \frac{\langle 2 \cos t, -2\sin t\rangle}{\| \langle 2 \cos t, -2\sin t\rangle \|} = \langle \cos t, -\sin t\rangle$$
03
Calculate the principal unit normal vector
Normalize the derivative of the unit tangent vector T to get the principal unit normal vector N:
First, find the derivative of T:
$$\mathbf{T}'(t) = \frac{d}{dt}\langle \cos t, -\sin t\rangle = \langle -\sin t, -\cos t\rangle$$
Now find the magnitude:
$$\|\mathbf{T}'(t)\| = \| \langle -\sin t, -\cos t\rangle \| = \sqrt{(-\sin t)^2 + (-\cos t)^2} = \sqrt{\sin^2 t + \cos^2 t} = 1$$
So, the principal unit normal vector N is:
$$\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} = \langle -\sin t, -\cos t\rangle$$
04
Calculate the unit binormal vector
The binormal vector B can be found by taking the cross product of T and N. Since T and N are two-dimensional, we can add a dummy coordinate, calculate the cross product in 3D, and get rid of the dummy coordinate:
$$\mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) = \langle \cos t, -\sin t, 0\rangle \times \langle -\sin t, -\cos t, 0\rangle = \langle 0, 0, \cos^2 t + \sin^2 t\rangle = \langle 0, 0, 1\rangle$$
05
Calculate the torsion
Now, let's differentiate B and compute the torsion 蟿:
$$\mathbf{B}'(t) = \frac{d}{dt}\langle 0, 0, 1\rangle = \langle 0, 0, 0\rangle$$
$$\tau = -\frac{\langle \mathbf{B}'(t), \mathbf{N}(t) \rangle}{\|\mathbf{T}(t) \times \mathbf{N}(t)\|^2} = -\frac{\langle \langle 0, 0, 0\rangle, \langle -\sin t, -\cos t\rangle\rangle}{1^2} = 0$$
The unit binormal vector B is $$\langle 0, 0, 1\rangle$$, and the torsion 蟿 is 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Parameterization of curves
Parameterization of curves is a way to describe a curve by expressing its points in terms of a single variable, often denoted as \( t \). This variable typically represents a parameter, such as time.
- Each value of \( t \) corresponds to a unique point on the curve.
- For example, the curve \( r(t) = \langle 2 \sin t, 2 \cos t \rangle \) is parameterized by \( t \).
- This parameterization relates closely to the unit circle since \( \sin t \) and \( \cos t \) represent circular coordinates.
Unit tangent vector
The unit tangent vector of a curve represents the direction of the curve at any point. It tells us where the curve is heading.
- To find this vector, first determine the derivative of the parameterized curve, \( r'(t) \).
- Normalize this derivative by dividing by its magnitude to get \( \mathbf{T}(t) = \frac{r'(t)}{\|r'(t)\|} \).
- In our example, \( r'(t) = \langle 2 \cos t, -2 \sin t \rangle \), leading to a unit tangent vector \( \mathbf{T}(t) = \langle \cos t, -\sin t \rangle \).
Principal unit normal vector
The principal unit normal vector describes how the curve changes direction. It points towards the center of the curve's curvature.
- Start by finding the derivative of the unit tangent vector \( \mathbf{T}(t) \).
- Normalize this derivative to find the principal unit normal vector \( \mathbf{N}(t) \).
- For our curve: \( \mathbf{T}'(t) = \langle -\sin t, -\cos t \rangle \) with a magnitude of 1, gives \( \mathbf{N}(t) = \langle -\sin t, -\cos t \rangle \).
Binormal vector
The binormal vector adds a third dimension to the analysis of space curves, particularly useful in three-dimensional contexts. It is perpendicular to both the tangent and normal vectors.
- Calculate it using the cross product of the tangent and normal vectors: \( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \).
- In our 2D example, we introduce a third component (a zero), \( \mathbf{T}(t) = \langle \cos t, -\sin t, 0 \rangle \) and \( \mathbf{N}(t) = \langle -\sin t, -\cos t, 0 \rangle \).
- The result is \( \mathbf{B}(t) = \langle 0, 0, 1 \rangle \), signifying that the binormal vector points out of the plane.
