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Magazine Chapter 18: Problem 14
Find the moving trihedral and the curvature at any point of the curve \(\mathbf{R}(t)=\cosh t \mathbf{i}+\sinh t \mathbf{j}+t \mathbf{k}\).
Short Answer
Expert verified
Unit tangent, unit normal vectors, and curvature found. The curvature is \(\frac{1}{\sqrt{\cosh(2t) + 1} + 1}\).
Step by step solution
01
Compute the first derivative
Find the derivative of \(\mathbf{R}(t) = \cosh t \mathbf{i} + \sinh t \mathbf{j} + t \mathbf{k}\)with respect to t. We get: \(\mathbf{R}'(t) = \frac{d}{dt}(\cosh t) \mathbf{i} + \frac{d}{dt}(\sinh t) \mathbf{j} + \frac{d}{dt}(t) \mathbf{k} = \sinh t \mathbf{i} + \cosh t \mathbf{j} + \mathbf{k}\).
02
Compute the second derivative
Find the second derivative of \(\mathbf{R}(t)\)with respect to t. We get: \(\mathbf{R}''(t) = \frac{d}{dt}(\sinh t) \mathbf{i} + \frac{d}{dt}(\cosh t) \mathbf{j} + \frac{d}{dt}(t) \mathbf{k} = \cosh t \mathbf{i} + \sinh t \mathbf{j} + 0 \mathbf{k} = \cosh t \mathbf{i} + \sinh t \mathbf{j}\).
03
Compute the unit tangent vector
Next, find the unit tangent vector \(\mathbf{T}(t)\)by normalizing \(\mathbf{R}'(t)\). Calculate \(\|\mathbf{R}'(t)\| = \sqrt{(\sinh t)^2 + (\cosh t)^2 + 1} = \sqrt{\cosh(2t) + 1} \), making use of the hyperbolic identity \(\cosh^2(t) - \sinh^2(t) = 1\). Thus, \(\mathbf{T}(t) = \frac{\mathbf{R}'(t)}{\|\mathbf{R}'(t)\|} = \frac{\sinh t \mathbf{i} + \cosh t \mathbf{j} + \mathbf{k}}{\sqrt{\cosh(2t) + 1}}\).
04
Compute the unit normal vector
To find the unit normal vector \(\mathbf{N}(t)\), compute \(\mathbf{T}'(t)\) and normalize it. Derive \(\mathbf{T}(t)\) relative to t and calculate: \mathbf{T}'(t) = \frac{\mathbf{R}''(t) (\sqrt{\cosh(2t) + 1}) - \mathbf{R}'(t) \frac{d}{dt}(\sqrt{\cosh(2t) + 1})}{(\cosh(2t) + 1)}\.Normalize this to get \(\mathbf{N}(t)\).
05
Calculate the curvature
The curvature \(\kappa(t)\)can be found using \(\kappa(t) = \frac{\|\mathbf{T}'(t)\|}{\|\mathbf{R}'(t)\|}\).After simplifications, we find \(\kappa(t) = \frac{1}{(\cosh(2t) + 1)} \sqrt{1 - \frac{1}{c^2(t)}} = \frac{1}{\sqrt{\cosh(2t) + 1} + 1}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Moving Trihedral
The moving trihedral refers to a set of three orthogonal vectors that move along with a curve in space.
In our exercise, the curve is represented by \(\textbf{R}(t) = \cosh t \textbf{i} + \sinh t \textbf{j} + t \textbf{k}\). The trihedral consists of:
In our exercise, the curve is represented by \(\textbf{R}(t) = \cosh t \textbf{i} + \sinh t \textbf{j} + t \textbf{k}\). The trihedral consists of:
- The Unit Tangent Vector \(\textbf{T}(t)\) - points in the direction of the curve's velocity.
- The Unit Normal Vector \(\textbf{N}(t)\) - points towards the direction where the curve is turning.
- The Binormal Vector \(\textbf{B}(t)\) - orthogonal to both the tangent and normal vectors.
Curvature
Curvature measures how fast a curve is changing direction at a given point.
In our exercise, the curvature \(\kappa(t)\) of the curve \(\textbf{R}(t)\) is found mathematically using:
\[\kappa(t) = \frac{\|\textbf{T}'(t)\|}{\|\textbf{R}'(t)\|}\]
This involves taking the derivative of the unit tangent vector \(\textbf{T}'(t)\) and dividing it by the magnitude of the first derivative of the curve \(\textbf{R}'(t)\). For our hyperbolic function-based curve, the curvature simplifies to:
\[\kappa(t) = \frac{1}{\sqrt{\cosh(2t) + 1} + 1}\]
This formula tells us how the direction changes as we move along the curve.
In our exercise, the curvature \(\kappa(t)\) of the curve \(\textbf{R}(t)\) is found mathematically using:
\[\kappa(t) = \frac{\|\textbf{T}'(t)\|}{\|\textbf{R}'(t)\|}\]
This involves taking the derivative of the unit tangent vector \(\textbf{T}'(t)\) and dividing it by the magnitude of the first derivative of the curve \(\textbf{R}'(t)\). For our hyperbolic function-based curve, the curvature simplifies to:
\[\kappa(t) = \frac{1}{\sqrt{\cosh(2t) + 1} + 1}\]
This formula tells us how the direction changes as we move along the curve.
Hyperbolic Functions
Hyperbolic functions are analogs of the trigonometric functions but for a hyperbola rather than a circle.
In our exercise, we see \(\textbf{R}(t) = \cosh t \textbf{i} + \sinh t \textbf{j} + t \textbf{k}\). Important hyperbolic functions include:
Notably, they satisfy the identity:
\[\cosh^2(t) - \sinh^2(t) = 1\]
This property is useful in calculus, particularly in our computation of derivatives and curvature.
In our exercise, we see \(\textbf{R}(t) = \cosh t \textbf{i} + \sinh t \textbf{j} + t \textbf{k}\). Important hyperbolic functions include:
- Hyperbolic cosine, \(\textbf{cosh}(t)\), and
- Hyperbolic sine, \(\textbf{sinh}(t)\).
Notably, they satisfy the identity:
\[\cosh^2(t) - \sinh^2(t) = 1\]
This property is useful in calculus, particularly in our computation of derivatives and curvature.
Unit Tangent Vector
The unit tangent vector \(\textbf{T}(t)\) points in the direction of the curve's velocity.
It is found by normalizing the first derivative of the curve:
\[\mathbf{T}(t) = \frac{\textbf{R}'(t)}{\|\textbf{R}'(t)\|}\].
For the given curve, the first derivative is:
\(\textbf{R}'(t) = \sinh t \textbf{i} + \cosh t \textbf{j} + \textbf{k}\).
The norm of this vector, \(\|\textbf{R}'(t)\|\), is:
\[\textbf{|\textbf{R}'(t)| = \sqrt{\cosh(2t) + 1}\]
Therefore:
\[\textbf{T}(t) = \frac{\sinh t \textbf{i} + \cosh t \textbf{j} + \textbf{k}}{\sqrt{\cosh(2t) + 1}}\].
This vector is crucial to understanding the curve's motion at any point.
It is found by normalizing the first derivative of the curve:
\[\mathbf{T}(t) = \frac{\textbf{R}'(t)}{\|\textbf{R}'(t)\|}\].
For the given curve, the first derivative is:
\(\textbf{R}'(t) = \sinh t \textbf{i} + \cosh t \textbf{j} + \textbf{k}\).
The norm of this vector, \(\|\textbf{R}'(t)\|\), is:
\[\textbf{|\textbf{R}'(t)| = \sqrt{\cosh(2t) + 1}\]
Therefore:
\[\textbf{T}(t) = \frac{\sinh t \textbf{i} + \cosh t \textbf{j} + \textbf{k}}{\sqrt{\cosh(2t) + 1}}\].
This vector is crucial to understanding the curve's motion at any point.
Unit Normal Vector
The unit normal vector \(\textbf{N}(t)\) points in the direction in which the curve is turning.
To find \(\textbf{N}(t)\), we must normalize the derivative of the unit tangent vector \(\textbf{T}(t)\). The process involves:
\[\textbf{N}(t) = \frac{\textbf{T}'(t)}{\|\textbf{T}'(t)\|}\].
This vector is perpendicular to \(\textbf{T}(t)\) and assists in analyzing how the curve changes direction.
Together with the tangent and binormal vectors, it forms the moving trihedral.
To find \(\textbf{N}(t)\), we must normalize the derivative of the unit tangent vector \(\textbf{T}(t)\). The process involves:
- Taking the derivative of \(\textbf{T}(t)\) to get \(\textbf{T}'(t)\).
- Using the chain rule, because \(\textbf{T}(t)\) is a function of \(t\).
- Normalizing \(\textbf{T}'(t)\) to ensure it has a magnitude of 1.
\[\textbf{N}(t) = \frac{\textbf{T}'(t)}{\|\textbf{T}'(t)\|}\].
This vector is perpendicular to \(\textbf{T}(t)\) and assists in analyzing how the curve changes direction.
Together with the tangent and binormal vectors, it forms the moving trihedral.
