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Chapter 13: Problem 5

For Exercises \(1-8\) you found \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) in Section 13.4 (Exercises \(9-16\) ). Find now \(\mathbf{B}\) and \(\tau\) for these space curves. $$ \mathbf{r}(t)=\left(t^{3} / 3\right) \mathbf{i}+\left(t^{2} / 2\right) \mathbf{j}, \quad t>0 $$

Short Answer

Expert verified
The binormal vector \( \mathbf{B} \) is \( \mathbf{k} \) and the torsion \( \tau \) is 0.

Step by step solution

01

Find the Tangent Vector (T)

The tangent vector \( \mathbf{T} \) is found by taking the derivative of the position vector \( \mathbf{r}(t) \). First, differentiate each component of \( \mathbf{r}(t) \) with respect to \( t \). Given \( \mathbf{r}(t) = \left(t^3 / 3\right) \mathbf{i} + \left(t^2 / 2\right) \mathbf{j} \), we get:\[ \mathbf{r}'(t) = \left(t^2\right) \mathbf{i} + \left(t\right) \mathbf{j}. \]Normalize \( \mathbf{r}'(t) \) to find \( \mathbf{T}(t) \):\[ \mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\| \mathbf{r}'(t) \|} = \frac{t^2\mathbf{i} + t\mathbf{j}}{\sqrt{t^4 + t^2}} = \frac{t\mathbf{i} + \mathbf{j}}{\sqrt{t^2 + 1}}. \]
02

Find the Normal Vector (N)

The normal vector \( \mathbf{N} \) is obtained by differentiating \( \mathbf{T}(t) \) with respect to \( t \). Then normalize this derivative to get \( \mathbf{N}(t) \).Differentiate \( \mathbf{T}(t) \):\[ \frac{d}{dt}\left( \frac{t\mathbf{i} + \mathbf{j}}{\sqrt{t^2 + 1}} \right). \]Solving this yields:\[ \mathbf{T}'(t) = \frac{1}{(t^2 + 1)^{3/2}} \left(\mathbf{j} - t^2\mathbf{j} - 2t^3\mathbf{i}\right). \]Normalize \( \mathbf{T}'(t) \) to find \( \mathbf{N}(t) \):\[ \mathbf{N}(t) = \frac{1}{\sqrt{t^2 + 1}} \left( -t\mathbf{i} + \mathbf{j} \right). \]
03

Find the Binormal Vector (B)

The binormal vector \( \mathbf{B} \) is the cross product of \( \mathbf{T} \) and \( \mathbf{N} \):\[ \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t). \]Given:- \( \mathbf{T}(t) = \frac{t\mathbf{i} + \mathbf{j}}{\sqrt{t^2 + 1}} \)- \( \mathbf{N}(t) = \frac{-t\mathbf{i} + \mathbf{j}}{\sqrt{t^2 + 1}} \)Cross product:\[ \mathbf{B}(t) = \left( \frac{t\mathbf{i} + \mathbf{j}}{\sqrt{t^2 + 1}} \right) \times \left( \frac{-t\mathbf{i} + \mathbf{j}}{\sqrt{t^2 + 1}} \right) = \frac{1}{t^2 + 1} \mathbf{k}. \]
04

Find the Torsion (蟿)

Torsion \( \tau \) measures how much the curve is twisting out of the plane of the osculating circle. It is calculated using the formula:\[ \tau = -\frac{\left( \mathbf{B} \cdot \mathbf{T}' \right)}{\left\| \mathbf{r}' \right\|} \]From previous steps:- \( \mathbf{B}(t) = \mathbf{k} \)- \( \mathbf{T}'(t) \) calculated in Step 2Since there is no \( \mathbf{k} \)-component in \( \mathbf{T}'(t) \), we have:\[ \tau = 0. \]

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Key Concepts

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

Tangent Vector
The tangent vector, often represented by \( \mathbf{T}(t) \), is crucial in understanding the direction in which a space curve is heading at any given point. To find this vector, we start with the curve's position vector \( \mathbf{r}(t) \). By differentiating this position vector with respect to time \( t \), we obtain \( \mathbf{r}'(t) \). This derivative gives us a vector that is tangent to the curve.
  • For the given problem, \( \mathbf{r}(t) = \left(\frac{t^3}{3}$) \mathbf{i} + \frac{t^2}{2} \right) \mathbf{j} \), the derivative is \( \mathbf{r}'(t) = t^2 \mathbf{i} + t \mathbf{j} \).
  • To normalize, divide this vector by its magnitude which is \( \sqrt{t^4 + t^2} \).
  • The normalized tangent vector is \( \mathbf{T}(t) = \frac{t \mathbf{i} + \mathbf{j}}{\sqrt{t^2 + 1}} \).
Understanding the tangent vector helps us identify the instantaneous direction of the curve.
Normal Vector
The normal vector, denoted as \( \mathbf{N}(t) \), defines the direction in which a curve is turning. It's perpendicular to the tangent vector. To find \( \mathbf{N}(t) \), we differentiate the tangent vector \( \mathbf{T}(t) \) with respect to \( t \), and then normalize it.
  • Differentiate \( \mathbf{T}(t) = \frac{t \mathbf{i} + \mathbf{j}}{\sqrt{t^2 + 1}} \) to get \( \mathbf{T}'(t) = \frac{1}{(t^2 + 1)^{3/2}} (\mathbf{j} - t^2 \mathbf{j} - 2t^3 \mathbf{i}) \).
  • Then normalize \( \mathbf{T}'(t) \) to get \( \mathbf{N}(t) = \frac{-t \mathbf{i} + \mathbf{j}}{\sqrt{t^2 + 1}} \).
Finding the normal vector helps visualize how sharply a curve bends.
Binormal Vector
The binormal vector, \( \mathbf{B}(t) \), is orthogonal to both the tangent and normal vectors. It provides insight into the third dimension of a space curve, forming a right-handed coordinate system with \( \mathbf{T}(t) \) and \( \mathbf{N}(t) \). The binormal vector is derived by taking the cross product of \( \mathbf{T}(t) \) and \( \mathbf{N}(t) \).
  • For our problem, \( \mathbf{T}(t) = \frac{t \mathbf{i} + \mathbf{j}}{\sqrt{t^2 + 1}} \) and \( \mathbf{N}(t) = \frac{-t \mathbf{i} + \mathbf{j}}{\sqrt{t^2 + 1}} \).
  • The cross product \( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \) results in \( \mathbf{B}(t) = \frac{1}{t^2 + 1} \mathbf{k} \).
The binormal vector is essential for understanding how the curve moves in three-dimensional space.
Curvature
Curvature, symbolized as \( \kappa \), quantifies how much a curve deviates from being a straight line. It reflects how sharply a curve bends at any point.To find curvature, there is a dependence on the normal vector and the derivative of the tangent vector. These components help compute the rate of change of the tangent vector with respect to arc length.
  • It's often found using the formula \( \kappa(t) = \left| \frac{d \mathbf{T}}{ds} \right| \), where \( s \) is the arc length.
  • For practical calculations, another formula is used involving derivatives and magnitudes.
Curvature is important for predicting the behavior of the curve, such as how tight the path is turning.
Torsion
Torsion, denoted as \( \tau \), measures the rate of twist of a curve out of the plane formed by the tangent and normal vectors. Essentially, it identifies how a curve spirals in space.Torsion is calculated as: \[ \tau = -\frac{( \mathbf{B} \cdot \mathbf{T}' )}{\| \mathbf{r}' \|} \]For the given exercise:
  • \( \mathbf{B}(t) = \mathbf{k} \) was previously determined.
  • The derivative \( \mathbf{T}'(t) \) doesn't have a \( \mathbf{k} \)-component, meaning \( \mathbf{B} \cdot \mathbf{T}' \) results in 0.
  • Thus, \( \tau = 0 \), indicating no twisting out of the plane.
Torsion provides valuable insight when dealing with complex, spiraling curves in space.

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Most popular questions from this chapter

In Exercises 16 and \(17,\) two planets, planet \(A\) and planet \(B\) , are orbiting their sun in circular orbits with \(A\) being the inner planet and \(B\) being farther away from the sun. Suppose the positions of \(A\) and \(B\) at time \(t\) are $$ \mathbf{r}_{A}(t)=2 \cos (2 \pi t) \mathbf{i}+2 \sin (2 \pi t) \mathbf{j} $$ and $$ \mathbf{r}_{B}(t)=3 \cos (\pi t) \mathbf{i}+3 \sin (\pi t) \mathbf{j} $$ respectively, where the sun is assumed to be located at the origin and distance is measured in astronomical units. (Notice that planet \(A\) moves faster than planet \(B . )\) The people on planet \(A\) regard their planet, not the sun, as the center of their planetary system (their solar system). Using planet \(A\) as the origin of a new coordinate system, give parametric equations for the location of planet \(B\) at time \(t .\) Write your answer in terms of \(\cos (\pi t)\) and \(\sin (\pi t) .\)

Firing from \(\left(x_{0}, y_{0}\right)\) Derive the equations $$ \begin{aligned} x &=x_{0}+\left(v_{0} \cos \alpha\right) t \\ y &=y_{0}+\left(v_{0} \sin \alpha\right) t-\frac{1}{2} g t^{2} \end{aligned} $$ (see Equation ( 5\()\) in the text) by solving the following initial value problem for a vector \(\mathbf{r}\) in the plane. $$ \begin{array}{ll}{\text { Differential equation: }} & {\frac{d^{2} \mathbf{r}}{d t^{2}}=-g \mathbf{j}} \\ {\text { Initial conditions: }} & {\mathbf{r}(0)=x_{0} \mathbf{i}+y_{0} \mathbf{j}} \\ {} & {\frac{d \mathbf{r}}{d t}(0)=\left(v_{0} \cos \alpha\right) \mathbf{i}+\left(v_{0} \sin \alpha\right) \mathbf{j}}\end{array} $$

Motion along an ellipse A particle moves around the ellipse \((y / 3)^{2}+(z / 2)^{2}=1\) in the \(y z\) -plane in such a way that its position at time \(t\) is $$ \mathbf{r}(t)=(3 \cos t) \mathbf{j}+(2 \sin t) \mathbf{k} $$ Find the maximum and minimum values of \(|\mathbf{v}|\) and \(|\mathbf{a}| .\) (Hint: Find the extreme values of \(|\mathbf{v}|^{2}\) and \(|\mathbf{a}|^{2}\) first and take square roots later.)

Use a CAS to perform the following steps in Exercises \(58-61 .\) a. Plot the space curve traced out by the position vector \(\mathbf{r}\) . b. Find the components of the velocity vector \(d \mathbf{r} / d t\) . c. Evaluate \(d \mathbf{r} / d t\) at the given point \(t_{0}\) and determine the equation of the tangent line to the curve at \(\mathbf{r}\left(t_{0}\right) .\) d. Plot the tangent line together with the curve over the given interval. $$ \mathbf{r}(t)=\sqrt{2 t \mathbf{i}}+e^{t} \mathbf{j}+e^{-t} \mathbf{k}, \quad-2 \leq t \leq 3, \quad t_{0}=1 $$

In Exercises 19 and \(20, \mathbf{r}(t)\) is the position vector of a particle in space at time \(t .\) Find the time or times in the given time interval when the velocity and acceleration vectors are orthogonal. $$ \mathbf{r}(t)=(t-\sin t) \mathbf{i}+(1-\cos t) \mathbf{j}, \quad 0 \leq t \leq 2 \pi $$
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