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Chapter 13: Problem 12

Find \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) for the space curves in Exercises \(9-16\) $$ \mathbf{r}(t)=(6 \sin 2 t) \mathbf{i}+(6 \cos 2 t) \mathbf{j}+5 t \mathbf{k} $$

Short Answer

Expert verified
\( \mathbf{T}(t) = \frac{1}{13} ((12 \cos 2t) \mathbf{i} - (12 \sin 2t) \mathbf{j} + 5 \mathbf{k}) \), \( \mathbf{N}(t) = \cos 2t \mathbf{i} + \sin 2t \mathbf{j} \), and \( \kappa = \frac{24}{169} \).

Step by step solution

01

Find the derivative of the position vector

First, we find the derivative of the position vector \( \mathbf{r}(t) \) with respect to \( t \). This gives us the velocity vector, \( \mathbf{r}'(t) \).\[\mathbf{r}'(t) = \frac{d}{dt} [(6 \sin 2t) \mathbf{i} + (6 \cos 2t) \mathbf{j} + 5t \mathbf{k}]\]\[\mathbf{r}'(t) = (12 \cos 2t) \mathbf{i} - (12 \sin 2t) \mathbf{j} + 5 \mathbf{k}\]
02

Compute the unit tangent vector \( \mathbf{T}(t) \)

The unit tangent vector \( \mathbf{T}(t) \) is given by the normalized velocity vector.Compute the magnitude of \( \mathbf{r}'(t) \):\[|\mathbf{r}'(t)| = \sqrt{(12 \cos 2t)^2 + (-12 \sin 2t)^2 + 5^2} = \sqrt{144 + 25} = 13\]Thus, \( \mathbf{T}(t) \) is:\[\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} = \frac{1}{13} \left( (12 \cos 2t) \mathbf{i} - (12 \sin 2t) \mathbf{j} + 5 \mathbf{k} \right)\]
03

Find the derivative of the unit tangent vector \( \mathbf{T}(t) \)

Differentiate \( \mathbf{T}(t) \) with respect to \( t \) to find \( \mathbf{T}'(t) \):\[\mathbf{T}'(t) = \frac{d}{dt} \left( \frac{12}{13} \cos 2t \mathbf{i} - \frac{12}{13} \sin 2t \mathbf{j} + \frac{5}{13} \mathbf{k} \right)\]\[\mathbf{T}'(t) = \left(\frac{-24}{13} \sin 2t \right) \mathbf{i} - \left( \frac{24}{13} \cos 2t \right) \mathbf{j}\]
04

Compute the principal normal vector \( \mathbf{N}(t) \)

Normalize \( \mathbf{T}'(t) \) to find \( \mathbf{N}(t) \).Calculate the magnitude of \( \mathbf{T}'(t) \):\[|\mathbf{T}'(t)| = \sqrt{\left(\frac{-24}{13} \sin 2t\right)^2 + \left(\frac{-24}{13} \cos 2t\right)^2}\]\[= \frac{24}{13}\]Thus, the principal normal vector is:\[\mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|} = \cos 2t \mathbf{i} + \sin 2t \mathbf{j}\]
05

Calculate the curvature \( \kappa \)

The curvature \( \kappa \) is given by:\[\kappa = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|}\]Substituting the values:\[\kappa = \frac{\frac{24}{13}}{13} = \frac{24}{169}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Tangent Vector
The unit tangent vector, denoted as \( \mathbf{T}(t) \), is a crucial concept when studying curves in space. It points in the direction of the curve and has a magnitude (or length) of 1. To find \( \mathbf{T}(t) \), you need the velocity vector, which is the derivative of the position vector \( \mathbf{r}(t) \). The process involves:
  • Calculating the derivative \( \mathbf{r}'(t) \), which gives the velocity vector.
  • Finding the magnitude of this velocity vector, noted as \( |\mathbf{r}'(t)| \).
  • Dividing the velocity vector \( \mathbf{r}'(t) \) by its magnitude to get the unit tangent vector: \( \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{|\mathbf{r}'(t)|} \).
This ensures that \( \mathbf{T}(t) \) is a vector that follows the direction of the curve, providing a straightforward path to measure orientation along the path.
Principal Normal Vector
The principal normal vector, represented as \( \mathbf{N}(t) \), is perpendicular to the unit tangent vector \( \mathbf{T}(t) \). It points towards the center of curvature of the curve, offering insights into the way the curve bends.
To find \( \mathbf{N}(t) \), you:
  • Differentiate the unit tangent vector \( \mathbf{T}(t) \) with respect to time \( t \).
  • Calculate the magnitude of the new vector \( \mathbf{T}'(t) \).
  • Normalize this result by dividing \( \mathbf{T}'(t) \) by its magnitude: \( \mathbf{N}(t) = \frac{\mathbf{T}'(t)}{|\mathbf{T}'(t)|} \).
This results in a unit vector at each point along the curve, showing how sharply the curve turns at that point.
Curvature
Curvature \( \kappa \) is an integral measure of how a curve deviates from being a straight line. It is effectively the 'sharpness' of the curve at any given point. A larger curvature value implies a tighter turn.
To calculate the curvature, you:
  • Compute the magnitude of \( \mathbf{T}'(t) \), which is the derivative of the unit tangent vector.
  • Obtain the magnitude of the velocity vector \( |\mathbf{r}'(t)| \).
  • The formula for curvature \( \kappa \) is: \( \kappa = \frac{|\mathbf{T}'(t)|}{|\mathbf{r}'(t)|} \).
Curvature provides essential information on how the curve's path changes over time, helping visualize its twists and turns.
Position Vector
The position vector \( \mathbf{r}(t) \) represents the location of a point on a curve relative to an origin in space. It's like giving the exact address of each spot along a curve.
In the given exercise, the position vector is defined as:
  • \( \mathbf{r}(t)=(6 \sin 2 t) \mathbf{i}+(6 \cos 2 t) \mathbf{j}+5 t \mathbf{k} \)
This vector combines the three spatial dimensions \( x \), \( y \), and \( z \) coordinates using unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \). Understanding this helps visualize how each component contributes to the curve's path over time.
Velocity Vector
The velocity vector is derived from the derivative of the position vector \( \mathbf{r}(t) \). It reflects the rate of change of the position and tells you how fast and in what direction the point moves along the curve.
For the exercise, the velocity vector is:
  • \( \mathbf{r}'(t) = (12 \cos 2t) \mathbf{i} - (12 \sin 2t) \mathbf{j} + 5 \mathbf{k} \)
This demonstrates how each component \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) impacts movement along the spatial curve. This vector is foundational for other calculations like the unit tangent vector and curvature.

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Most popular questions from this chapter

Find \(\mathbf{B}\) and \(\tau\) for these space curves. \(\mathbf{r}(t)=(6 \sin 2 t) \mathbf{i}+(6 \cos 2 t) \mathbf{j}+5 t \mathbf{k}\)

Evaluate the integrals in Exercises \(1-10 .\) \(\int_{0}^{1}\left[\frac{2}{\sqrt{1-t^{2}}} \mathbf{i}+\frac{\sqrt{3}}{1+t^{2}} \mathbf{k}\right] d t\)

Find an equation for the circle of curvature of the curve \(\mathbf{r}(t)=t \mathbf{i}+(\sin t) \mathbf{j}\) at the point \((\pi / 2,1) .\) (The curve parametrizes the graph of \(y=\sin x\) in the \(x y\) -plane.)

Solve the initial value problems in Exercises \(11-20\) for \(\mathbf{r}\) as a vector function of \(t .\) $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d \mathbf{r}}{d t}=-t \mathbf{i}-t \mathbf{j}-t \mathbf{k}} \\ {\text { Initial condition: }} & {\mathbf{r}(0)=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}}\end{array} $$

Motion along a circle Show that the vector-valued function \(\mathbf{r}(t)=(2 \mathbf{i}+2 \mathbf{j}+\mathbf{k})\) $$ +\cos t\left(\frac{1}{\sqrt{2}} \mathbf{i}-\frac{1}{\sqrt{2}} \mathbf{j}\right)+\sin t\left(\frac{1}{\sqrt{3}} \mathbf{i}+\frac{1}{\sqrt{3}} \mathbf{j}+\frac{1}{\sqrt{3}} \mathbf{k}\right) $$ describes the motion of a particle moving in the circle of radius 1 centered at the point \((2,2,1)\) and lying in the plane \(x+y-2 z=2\) .
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