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Chapter 11: Problem 30

, find the curvature \(\kappa\), the unit tangent vector \(\mathbf{T}\), the unit normal vector \(\mathbf{N}\), and the binormal vector \(\mathbf{B}\) at \(t=t_{1} .\) $$ \mathbf{r}(t)=\cos ^{3} t \mathbf{i}+\sin ^{3} t \mathbf{k} ; t_{1}=\pi / 2 $$

Short Answer

Expert verified
The curvature \( \kappa \), tangent \( \mathbf{T} \), normal \( \mathbf{N} \), and binormal \( \mathbf{B} \) need proper reevaluation or assumptions for \( t_1 \) due to vanishing constructs.

Step by step solution

01

Find the Velocity Vector

The velocity vector \( \mathbf{v}(t) \) is the derivative of \( \mathbf{r}(t) \) with respect to \( t \). Compute \( \mathbf{v}(t) \) as follows:\[\mathbf{v}(t) = \frac{d}{dt} [\cos^3 t \mathbf{i} + \sin^3 t \mathbf{k}] = -3\cos^2 t \sin t \mathbf{i} + 3\sin^2 t \cos t \mathbf{k}.\]
02

Evaluate the Velocity at \( t_1 \)

Substitute \( t = \frac{\pi}{2} \) into \( \mathbf{v}(t) \) to find \( \mathbf{v}(t_1) \):\[\mathbf{v}\left(\frac{\pi}{2}\right) = -3\cos^2\left(\frac{\pi}{2}\right)\sin\left(\frac{\pi}{2}\right)\mathbf{i} + 3\sin^2\left(\frac{\pi}{2}\right)\cos\left(\frac{\pi}{2}\right)\mathbf{k} = \mathbf{0}.\]
03

Find the Acceleration Vector

Compute the acceleration vector \( \mathbf{a}(t) \), which is the derivative of \( \mathbf{v}(t) \):\[\mathbf{a}(t) = \frac{d\mathbf{v}(t)}{dt} = (6\cos^3 t - 9\cos t \sin^2 t) \mathbf{i} + (-9\sin t \cos^2 t + 6\sin^3 t) \mathbf{k}.\]
04

Evaluate the Acceleration Vector at \( t_1 \)

Substitute \( t = \frac{\pi}{2} \) into \( \mathbf{a}(t) \) to find \( \mathbf{a}(t_1) \):\[\mathbf{a}\left(\frac{\pi}{2}\right) = (0) \mathbf{i} + (6\sin^3\left(\frac{\pi}{2}\right)) \mathbf{k} = 6\mathbf{k}.\]
05

Find the Unit Tangent Vector \( \mathbf{T} \)

Since \( \mathbf{v}(t_1) = \mathbf{0} \) (a technical issue to resolve involving tangent vectors), we ideally require rescaling of a non-zero vector. The next step comes naturally from functional dependency and calculations when necessary components are practically not zero if assumptions extend from derivations.
06

Clarify Results and Adjust

In typical methods with-fully-defined functions across calculations, results draw from vector magnitude and normalizing leading to further potential cross product (otherwise perturbed composite where necessary). Binormal outcomes \( \mathbf{B} \) normally express ratios within cross product sequence after further assumptions managed systematically sort out limits. Each such test stage reveals sensitivity given precise algebra requisite.

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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 \( \mathbf{T} \), provides a direction on a curve that shows where the curve is heading at any point. To find the unit tangent vector from a parametric representation of the curve, we first calculate the velocity vector \( \mathbf{v}(t) \), which is the derivative of the position vector \( \mathbf{r}(t) \). The velocity vector points in the direction of the curve's tangent.
Once we have the velocity vector, to get the unit tangent vector, we divide it by its magnitude, \( \| \mathbf{v}(t) \| \). This process normalizes the vector, making its length equal to one while maintaining its direction. Therefore, \( \mathbf{T}(t) = \frac{\mathbf{v}(t)}{\| \mathbf{v}(t) \|} \).
  • Ensures awareness of directionality as curves progress.
  • Essential in understanding path derivatives and behaviors.
In some cases, like the given exercise, where \( \mathbf{v}(t_1) = \mathbf{0} \), we face challenges requiring alternative definitions or points for precise before reaching critical evaluation points.
Unit Normal Vector
The unit normal vector, \( \mathbf{N} \), is perpendicular to the unit tangent vector and points towards the curve's local center of curvature. Calculating the unit normal vector involves differentiating the tangent vector, \( \mathbf{T}(t) \), with respect to \( t \) and then normalizing the resulting vector to get \( \mathbf{N}(t) \).
This complex interaction of instantaneous rate changes specifies:
  • Curvature indication and directional drift detection.
  • Locational differentials affecting arcs and angles into normals.
The length of \( \mathbf{N}(t) \) is usually one since it's normalized. It captures the tendency of the path to curve in its domain, defining how sharply paths turn. Even when \( \mathbf{v}(t_1) = \mathbf{0} \) presents positioning oddities like in our exercise, this component remains vital in contextual geographies.
Binormal Vector
The binormal vector, \( \mathbf{B} \), completes the right-handed orthonormal frame along a space curve created by the "TNB Frame." It's derived from the cross product of \( \mathbf{T}(t) \) and \( \mathbf{N}(t) \). Hence, \( \mathbf{B}(t) = \mathbf{T}(t) \times \mathbf{N}(t) \).
In characteristics of 3D space navigation:
  • Models path normality in axial-line correlations.
  • Provides rotational norms and helps underpin features.
Its direction is orthogonal to both \( \mathbf{T} \) and \( \mathbf{N} \). At points where the tangent vector might equal zero, careful assessment requires re-packaging visual interpretations to aid full insights, ensuring representation limitations are managed as necessary.
Vector Calculus
Vector calculus is fundamentally about understanding how vectors change, interact, and represent physical realities in multi-dimensional spaces. It involves differentiation and integration of vector fields, providing a framework for analyzing physical phenomena like electromagnetic waves, fluid flow, and more.
Through procedures such as:
  • Deriving curves and surfaces behavior in vector spaces.
  • Calculating arc speeds, momenta representation.
Applications of vector calculus stretch beyond simple curve studies into solving complex differential equations, evaluating line integrals, and more. It magnifies calculative dynamics and vector properties in spacepatterns, essential to mastering engineering and physics disciplines.

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Most popular questions from this chapter

EXPL 48. In this exercise you will derive Kepler's First Law, that planets travel in elliptical orbits. We begin with the notation. Place the coordinate system so that the sun is at the origin and the planet's closest approach to the sun (the perihelion) is on the positive \(x\) -axis and occurs at time \(t=0\). Let \(\mathbf{r}(t)\) denote the position vector and let \(r(t)=\|\mathbf{r}(t)\|\) denote the distance from the sun at time \(t\). Also, let \(\theta(t)\) denote the angle that the vector \(\mathbf{r}(t)\) makes with the positive \(x\) -axis at time \(t\). Thus, \((r(t), \theta(t))\) is the polar coordinate representation of the planet's position. Let \(\mathbf{u}_{1}=\mathbf{r} / r=(\cos \theta) \mathbf{i}+(\sin \theta) \mathbf{j} \quad\) and \(\quad \mathbf{u}_{2}=(-\sin \theta) \mathbf{i}+(\cos \theta) \mathbf{j}\) Vectors \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) are orthogonal unit vectors pointing in the directions of increasing \(r\) and increasing \(\theta\), respectively. Figure 12 summarizes this notation. We will often omit the argument \(t\), but keep in mind that \(\mathbf{r}, \theta, \mathbf{u}_{1}\), and \(\mathbf{u}_{2}\) are all functions of \(t .\) A prime in. dicates differentiation with respect to time \(t\). (a) Show that \(\mathbf{u}_{1}^{\prime}=\theta^{\prime} \mathbf{u}_{2}\) and \(\mathbf{u}_{2}^{\prime}=-\theta^{\prime} \mathbf{u}_{1}\). (b) Show that the velocity and acceleration vectors satisfy $$ \begin{array}{l} \mathbf{v}=r^{\prime} \mathbf{u}_{1}+r \theta^{\prime} \mathbf{u}_{2} \\ \mathbf{a}=\left(r^{\prime \prime}-r\left(\theta^{\prime}\right)^{2}\right) \mathbf{u}_{1}+\left(2 r^{\prime} \theta^{\prime}+r \theta^{\prime \prime}\right) \mathbf{u}_{2} \end{array} $$ (c) Use the fact that the only force acting on the planet is the gravity of the sun to express a as a multiple of \(\mathbf{u}_{1}\), then explain how we can conclude that $$ \begin{aligned} r^{\prime \prime}-r\left(\theta^{\prime}\right)^{2} &=\frac{-G M}{r^{2}} \\ 2 r^{\prime} \theta^{\prime}+r \theta^{\prime \prime} &=0 \end{aligned} $$ (d) Consider \(\mathbf{r} \times \mathbf{r}^{\prime}\), which we showed in Example 8 was a constant vector, say D. Use the result from (b) to show that \(\mathbf{D}=r^{2} \theta^{\prime} \mathbf{k} .\) (e) Substitute \(t=0\) to get \(\mathbf{D}=r_{0} v_{0} \mathbf{k}\), where \(r_{0}=r(0)\) and \(v_{0}=\|\mathbf{v}(0)\|\). Then argue that \(r^{2} \theta^{\prime}=r_{0} v_{0}\) for all \(t\). (f) Make the substitution \(q=r^{\prime}\) and use the result from (e) to obtain the first-order (nonlinear) differential equation in \(q\) : $$ q \frac{d q}{d r}=\frac{r_{0}^{2} v_{0}^{2}}{r^{3}}-\frac{G M}{r^{2}} $$ (g) Integrate with respect to \(r\) on both sides of the above equation and use an initial condition to obtain $$ q^{2}=2 G M\left(\frac{1}{r}-\frac{1}{r_{0}}\right)+v_{0}^{2}\left(1-\frac{r_{0}^{2}}{r^{2}}\right) $$ (h) Substitute \(p=1 / r\) into the above equation to obtain $$ \frac{r_{0}^{2} v_{0}^{2}}{\left(\theta^{\prime}\right)^{2}}\left(\frac{d p}{d t}\right)^{2}=2 G M\left(p-p_{0}\right)+v_{0}^{2}\left(1-\frac{p^{2}}{p_{0}^{2}}\right) $$

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