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Magazine Chapter 13: Problem 28
Rounding the answers to four decimal places, use a CAS to find v, a, speed, \(\mathbf{T}, \mathbf{N}, \mathbf{B}, \kappa, \tau,\) and the tangential and normal components of acceleration for the curves in Exercises \(27-30\) at the given values of \(t.\) \(\mathbf{r}(t)=\left(e^{t} \cos t\right) \mathbf{i}+\left(e^{t} \sin t\right) \mathbf{j}+e^{t} \mathbf{k}, \quad t=\ln 2\)
Short Answer
Expert verified
Evaluate the required vectors, components, and their magnitudes for \( t = \ln 2 \). Results should be accurate to four decimal places.
Step by step solution
01
Compute the velocity vector
The velocity vector \( \mathbf{v}(t) \) is the derivative of the position vector \( \mathbf{r}(t) \). Differentiate \( \mathbf{r}(t) = \left( e^{t} \cos t \right) \mathbf{i} + \left( e^{t} \sin t \right) \mathbf{j} + e^{t} \mathbf{k} \) with respect to \( t \). The derivatives are:\[\mathbf{v}(t) = \left( e^t \cos t - e^t \sin t \right) \mathbf{i} + \left( e^t \sin t + e^t \cos t \right) \mathbf{j} + e^t \mathbf{k}\]
02
Compute the acceleration vector
Differentiate the velocity vector \( \mathbf{v}(t) \) to obtain the acceleration vector \( \mathbf{a}(t) \). The velocity vector is \( \mathbf{v}(t) = \left( e^t \cos t - e^t \sin t \right) \mathbf{i} + \left( e^t \sin t + e^t \cos t \right) \mathbf{j} + e^t \mathbf{k} \).Differentiate each component:\[\mathbf{a}(t) = \left( -2e^t \sin t \right) \mathbf{i} + \left( 2e^t \cos t \right) \mathbf{j} + e^t \mathbf{k}\]
03
Evaluate \(\mathbf{v}(t)\) and \(\mathbf{a}(t)\) at \(t = \ln 2\)
Substitute \( t = \ln 2 \) into the expressions for \( \mathbf{v}(t) \) and \( \mathbf{a}(t) \):- \( \mathbf{v}(\ln 2) = \left( 2 \cdot 0 - 2 \cdot \ln 2 \right) \mathbf{i} + \left( 2 \cdot \ln 2 + 2 \cdot 0 \right) \mathbf{j} + 2 \mathbf{k} \)- \( \mathbf{a}(\ln 2) = \left( -4 \sin(\ln 2) \right) \mathbf{i} + \left( 4 \cos(\ln 2) \right) \mathbf{j} + 2 \mathbf{k} \)
04
Calculate the speed
The speed is the magnitude of the velocity vector \( \mathbf{v}(t) \). For \( \mathbf{v}(\ln 2) = \left( -2 \ln 2 \right) \mathbf{i} + \left( 2 \ln 2 \right) \mathbf{j} + 2 \mathbf{k} \), the speed \( v \) is:\[v = \sqrt{(-2 \ln 2)^2 + (2 \ln 2)^2 + (2)^2}\]Calculate the value and round to four decimal places.
05
Find unit tangent vector \(\mathbf{T}\)
The unit tangent vector \( \mathbf{T}(t) \) is \( \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} \). Use \( v \) from Step 4 to find \( \mathbf{T}(\ln 2) \) and round to four decimal places.\[\mathbf{T}(\ln 2) = \frac{\left( -2 \ln 2 \right) \mathbf{i} + \left( 2 \ln 2 \right) \mathbf{j} + 2 \mathbf{k}}{v}\]
06
Find unit normal vector \(\mathbf{N}\)
The unit normal vector \( \mathbf{N}(t) \) is \( \frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} \). Differentiate \( \mathbf{T}(t) \) and calculate at \( t = \ln 2 \). Then normalize the vector to get \( \mathbf{N} \).
07
Find binormal vector \(\mathbf{B}\)
The binormal vector \( \mathbf{B}(t) \) is \( \mathbf{T}(t) \times \mathbf{N}(t) \). Compute the cross product of the tangent and normal vectors obtained at \( t = \ln 2 \). Ensure the resulting vector is of unit length.
08
Compute curvature \(\kappa\)
Curvature \( \kappa(t) \) can be found using \( \kappa(t) = \frac{\|\mathbf{T}'(t)\|}{\|\mathbf{v}(t)\|} \). Using the derivative of \( \mathbf{T}(t) \) calculated previously, evaluate it at \( t = \ln 2 \).
09
Compute torsion \(\tau\)
Torsion \( \tau \) is found using the relation \( \tau(t) = \frac{\mathbf{B}'(t) \cdot \mathbf{N}(t)}{\|\mathbf{v}(t)\|} \). Differentiate \( \mathbf{B}(t) \) and dot with \( \mathbf{N}(t) \), then evaluate at \( t = \ln 2 \).
10
Compute tangential and normal components of acceleration
The tangential component of acceleration \( a_T \) is \( \mathbf{a}(t) \cdot \mathbf{T}(t) \) and normal component \( a_N \) is \( \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2} \). Evaluate both at \( t = \ln 2 \), rounding to four decimal places.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Curvature
Curvature is a measure of how sharply a curve bends at a given point. For a curve represented by a vector function \( \mathbf{r}(t) \), the curvature (denoted as \( \kappa \)) describes how quickly the direction of the curve is changing.
The formula for curvature is \( \kappa(t) = \frac{\|\mathbf{T}'(t)\|}{\|\mathbf{v}(t)\|} \), where \( \mathbf{T}(t) \) is the unit tangent vector and \( \mathbf{v}(t) \) is the velocity vector. A small curvature means the path is closer to a straight line at that point; a large curvature means the path is very curved, like a loop or a spiral.
In our step-by-step solution, curvature is computed after deriving the tangent vector and differentiating it to observe the change in direction.
The formula for curvature is \( \kappa(t) = \frac{\|\mathbf{T}'(t)\|}{\|\mathbf{v}(t)\|} \), where \( \mathbf{T}(t) \) is the unit tangent vector and \( \mathbf{v}(t) \) is the velocity vector. A small curvature means the path is closer to a straight line at that point; a large curvature means the path is very curved, like a loop or a spiral.
In our step-by-step solution, curvature is computed after deriving the tangent vector and differentiating it to observe the change in direction.
Torsion
Torsion measures the rate at which the curve twists out of the plane of curvature. While curvature measures how a curve bends in a plane, torsion measures how the curve leaves that plane, adding a 3D aspect to the understanding of the bend.
The torsion \( \tau \) of a curve is given by the formula \( \tau(t) = \frac{\mathbf{B}'(t) \cdot \mathbf{N}(t)}{\|\mathbf{v}(t)\|} \), where \( \mathbf{B}(t) \) is the binormal vector and \( \mathbf{N}(t) \) is the unit normal vector.
Calculating torsion is crucial in many practical applications like designing roller coasters or understanding DNA structure, where not just the curvature but also the twisting of the path is important.
The torsion \( \tau \) of a curve is given by the formula \( \tau(t) = \frac{\mathbf{B}'(t) \cdot \mathbf{N}(t)}{\|\mathbf{v}(t)\|} \), where \( \mathbf{B}(t) \) is the binormal vector and \( \mathbf{N}(t) \) is the unit normal vector.
Calculating torsion is crucial in many practical applications like designing roller coasters or understanding DNA structure, where not just the curvature but also the twisting of the path is important.
Unit Tangent Vector
The unit tangent vector \( \mathbf{T}(t) \) is a vector that points in the direction of the path a particle is moving but has a magnitude of 1. It helps us understand the direction without the influence of the speed.
Mathematically, \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} \), where \( \mathbf{v}(t) \) is the velocity vector of the curve. This vector provides a directional basis for analyzing changes in the trajectory and motion along the path.
Once the velocity is known, obtaining the unit tangent vector is straightforward, requiring just normalization.
Mathematically, \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|} \), where \( \mathbf{v}(t) \) is the velocity vector of the curve. This vector provides a directional basis for analyzing changes in the trajectory and motion along the path.
Once the velocity is known, obtaining the unit tangent vector is straightforward, requiring just normalization.
Unit Normal Vector
The unit normal vector \( \mathbf{N}(t) \), is perpendicular to the unit tangent vector \( \mathbf{T}(t) \) and helps detect how much the curve is turning at any point along the path. It points towards the direction where the curve is bending.
You find the unit normal vector by differentiating the unit tangent vector and normalizing it: \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} \). This vector characterizes the "normal" or "bend direction" of the trajectory.
This vector is key in calculating the curvature and torsion, providing insight into how the path is structured in space.
You find the unit normal vector by differentiating the unit tangent vector and normalizing it: \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{\|\mathbf{T}'(t)\|} \). This vector characterizes the "normal" or "bend direction" of the trajectory.
This vector is key in calculating the curvature and torsion, providing insight into how the path is structured in space.
Binormal Vector
The binormal vector \( \mathbf{B}(t) \) gives the orientation of the 3D frame used to describe the curve. This vector, together with the tangent and normal vectors, forms the Frenet-Serret frame.
It is defined as the cross product of the unit tangent vector \( \mathbf{T}(t) \) and the unit normal vector \( \mathbf{N}(t) \): \( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \). This vector is perpendicular to both \( \mathbf{T}(t) \) and \( \mathbf{N}(t) \), ensuring it's orthogonal to the plane formed by those two vectors.
The binormal vector thus represents the twisting of the path in space, providing the third dimension to the curve's description.
It is defined as the cross product of the unit tangent vector \( \mathbf{T}(t) \) and the unit normal vector \( \mathbf{N}(t) \): \( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \). This vector is perpendicular to both \( \mathbf{T}(t) \) and \( \mathbf{N}(t) \), ensuring it's orthogonal to the plane formed by those two vectors.
The binormal vector thus represents the twisting of the path in space, providing the third dimension to the curve's description.
Tangential and Normal Components of Acceleration
Acceleration can be split into two components: tangential and normal. This helps understand the motion along a pathway better, separating speed changes from direction changes.
The tangential component, \( a_T \), describes acceleration in the direction of the tangent vector, and is given by \( \mathbf{a}(t) \cdot \mathbf{T}(t) \). This directly affects speed.
The normal component, \( a_N \), is perpendicular to the tangential component and shows how the direction changes, calculated as \( \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2} \). This is related to turning or curvature.
Decomposing acceleration this way provides valuable insights into how speed and direction evolve over time along the trajectory.
The tangential component, \( a_T \), describes acceleration in the direction of the tangent vector, and is given by \( \mathbf{a}(t) \cdot \mathbf{T}(t) \). This directly affects speed.
The normal component, \( a_N \), is perpendicular to the tangential component and shows how the direction changes, calculated as \( \sqrt{\|\mathbf{a}(t)\|^2 - a_T^2} \). This is related to turning or curvature.
Decomposing acceleration this way provides valuable insights into how speed and direction evolve over time along the trajectory.
Velocity and Acceleration Vectors
Velocity and acceleration vectors are central in describing motion. They tell us how quickly an object is moving and how its velocity is changing.
The velocity vector \( \mathbf{v}(t) \) is derived as the derivative of the position vector \( \mathbf{r}(t) \), representing the object's speed and direction. Ensure to properly differentiate each component of \( \mathbf{r}(t) \) with respect to time.
The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector, showcasing how quickly velocity itself is changing. Differentiating \( \mathbf{v}(t) \) will yield \( \mathbf{a}(t) \).
Understanding these vectors allows you to analyze both the linear speed and the change of speed, an essential part of kinematics and physics.
The velocity vector \( \mathbf{v}(t) \) is derived as the derivative of the position vector \( \mathbf{r}(t) \), representing the object's speed and direction. Ensure to properly differentiate each component of \( \mathbf{r}(t) \) with respect to time.
The acceleration vector \( \mathbf{a}(t) \) is the derivative of the velocity vector, showcasing how quickly velocity itself is changing. Differentiating \( \mathbf{v}(t) \) will yield \( \mathbf{a}(t) \).
Understanding these vectors allows you to analyze both the linear speed and the change of speed, an essential part of kinematics and physics.
