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Chapter 13: Problem 9

Find \(\mathbf{T}, \mathbf{N},\) and \(\kappa\) for the space curves in Exercises \(9-16\) $$ \mathbf{r}(t)=(3 \sin t) \mathbf{i}+(3 \cos t) \mathbf{j}+4 t \mathbf{k} $$

Short Answer

Expert verified
\(\mathbf{T}(t) = \frac{1}{5}[(3 \cos t) \mathbf{i} - (3 \sin t) \mathbf{j} + 4 \mathbf{k}],\, \mathbf{N}(t) = -\sin t \, \mathbf{i} - \cos t \, \mathbf{j},\, \kappa = \frac{3}{25}.\)

Step by step solution

01

Find the velocity vector

The velocity vector \(\mathbf{v}(t)\) is found by differentiating \(\mathbf{r}(t)\) with respect to \(t\). Thus, \[\mathbf{v}(t) = \frac{d}{dt}[(3 \sin t) \mathbf{i} + (3 \cos t) \mathbf{j} + 4t \mathbf{k}] = (3 \cos t) \mathbf{i} - (3 \sin t) \mathbf{j} + 4 \mathbf{k}.\]
02

Find the tangent vector

The tangent vector \(\mathbf{T}(t)\) is the unit vector of \(\mathbf{v}(t)\), which is \(\mathbf{T}(t) = \frac{\mathbf{v}(t)}{\|\mathbf{v}(t)\|}\). First find the magnitude of \(\mathbf{v}(t)\):\[\|\mathbf{v}(t)\| = \sqrt{(3 \cos t)^2 + (-3 \sin t)^2 + 4^2} = \sqrt{9 \cos^2 t + 9 \sin^2 t + 16} = \sqrt{25} = 5.\]Thus, \[\mathbf{T}(t) = \frac{1}{5}[(3 \cos t) \mathbf{i} - (3 \sin t) \mathbf{j} + 4 \mathbf{k}].\]
03

Find the acceleration vector

Differentiate the velocity vector \(\mathbf{v}(t)\) to find the acceleration vector \(\mathbf{a}(t)\). Therefore, \[\mathbf{a}(t) = \frac{d}{dt}[(3 \cos t) \mathbf{i} - (3 \sin t) \mathbf{j} + 4 \mathbf{k}] = (-3 \sin t) \mathbf{i} - (3 \cos t) \mathbf{j}.\]
04

Calculate the normal vector

The normal vector \(\mathbf{N}(t)\) is calculated as \(\mathbf{N}(t) = \frac{d\mathbf{T}/dt}{\|d\mathbf{T}/dt\|}\). First find the derivative \(\frac{d\mathbf{T}}{dt}\):\[\frac{d\mathbf{T}}{dt} = \frac{d}{dt}\left[\frac{1}{5}[(3 \cos t) \mathbf{i} - (3 \sin t) \mathbf{j} + 4 \mathbf{k}]\right] = \frac{1}{5}[ (-3 \sin t) \mathbf{i} - (3 \cos t) \mathbf{j} ].\]Then, find the magnitude: \[\|\frac{d\mathbf{T}}{dt}\| = \frac{1}{5}\sqrt{9 \sin^2 t + 9 \cos^2 t} = \frac{3}{5}.\]Thus, \[\mathbf{N}(t) = \frac{1}{\frac{3}{5}}\left(\frac{1}{5}(-3 \sin t) \mathbf{i} - (3 \cos t) \mathbf{j}\right) = (-\sin t) \mathbf{i} - (\cos t) \mathbf{j}.\]
05

Calculate curvature

Curvature \(\kappa\) is calculated using the formula \( \kappa = \frac{\|\mathbf{v}(t) \times \mathbf{a}(t)\|}{\|\mathbf{v}(t)\|^3} \). Evaluate the cross product: \[\mathbf{v}(t) \times \mathbf{a}(t) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 3 \cos t & -3 \sin t & 4 \ -3 \sin t & -3 \cos t & 0 \end{vmatrix} = \mathbf{i}(0 - (-12 \cos t)) - \mathbf{j}(0 - (-12 \sin t)) + \mathbf{k}((-9)).\]This simplifies to \( \mathbf{v}(t) \times \mathbf{a}(t) = 12 \cos t \mathbf{i} + 12 \sin t \mathbf{j} - 9 \mathbf{k} \). The magnitude is \(\sqrt{(12 \cos t)^2 + (12 \sin t)^2 + (-9)^2} = 15\). So, \[\kappa = \frac{15}{5^3} = \frac{15}{125} = \frac{3}{25}.\]

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Key Concepts

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

Tangent Vector
A tangent vector is a fundamental concept when analyzing curves in space. It provides us with the direction in which the curve is heading at any given point. To find the tangent vector \( \mathbf{T}(t) \), we first need to derive the velocity vector \( \mathbf{v}(t) \), which is the rate of change of the position vector \( \mathbf{r}(t) \) with respect to time \( t \). Once we have the velocity vector, we convert it into a unit vector (a vector with a magnitude of 1) to specifically denote the direction along the curve, which gives us the tangent vector.The tangent vector is essential because:
  • It helps understand the trajectory or path of the curve.
  • It's used in calculating other properties like curvature and normal vectors.
  • Provides clues about how rapidly or slowly the curve is turning.
Overall, the tangent vector offers a simplified glimpse into the motion along the curve.
Space Curves
Space curves are curves that exist in three-dimensional space and are represented by a vector function like \( \mathbf{r}(t) = (3 \sin t) \mathbf{i} + (3 \cos t) \mathbf{j} + 4t \mathbf{k} \). These curves illustrate how a point travels in 3D space as a function of time.Space curves are important because:
  • They allow us to model and analyze phenomena in physics and engineering that occur in three dimensions.
  • They can represent paths of moving objects, making them useful for kinematic studies.
  • Using vector calculus, we can derive useful information like tangents, normals, and curvature, which describe the curve鈥檚 behavior.
In studying space curves, the ability to break down motion into understandable components aids in applications ranging from robotics to computer graphics.
Normal Vector
The normal vector, denoted \( \mathbf{N}(t) \), is an important part of the framework when analyzing curves. It is always perpendicular, or orthogonal, to the tangent vector \( \mathbf{T}(t) \). To find \( \mathbf{N}(t) \), you must first differentiate the tangent vector and then normalize it (convert it to a unit vector).Why the normal vector matters:
  • It's crucial for understanding the lateral behavior of a curve鈥攈ow the curve 'bends'.
  • In physics, it鈥檚 used to analyze force directions, like in the physics of motion.
  • Plays a key role in calculating curvature, which measures the curve's deviation from being straight.
The normal vector is essential when one needs to exploring how curves interact with their surrounding space.
Velocity Vector
The velocity vector \( \mathbf{v}(t) \) tells us how fast and in which direction a point on the curve is moving, essentially describing the instantaneous motion at a point. It's calculated by differentiating the position vector \( \mathbf{r}(t) \) with respect to time \( t \).Key aspects of the velocity vector:
  • It provides both the speed and direction of the point on the curve at each instant.
  • By transforming it into a unit vector, it becomes the tangent vector, which helps in further calculations.
  • Essential in defining the dynamics of motion in three-dimensional space.
In practical terms, the velocity vector is the first step in dissecting any movement or change along space curves.

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Most popular questions from this chapter

Use a CAS to perform the following steps in Exercises \(49-52 .\) \(\begin{array}{l}{\text { a. Plot the space curve traced out by the position vector } \mathbf{r} \text { . }} \\ {\text { b. Find the components of the velocity vector } d \mathbf{r} / d t \text { . }} \\ {\text { c. Evaluate } d \mathbf{r} / d t \text { at the given point } t_{0} \text { and determine the equa- }} \\ {\text { tion of the tangent line to the curve at } \mathbf{r}\left(t_{0}\right) .} \\ {\text { d. Plot the tangent line together with the curve over the given }} \\ {\text { interval. }}\end{array}\) $$ \mathbf{r}(t)=\sqrt{2} t \mathbf{i}+e^{t} \mathbf{j}+e^{-1} \mathbf{k}, \quad-2 \leq t \leq 3, \quad t_{0}=1 $$

Osculating circle Find a parametrization of the osculating circle for the parabola \(y=x^{2}\) when \(x=1 .\)

The eccentricity of Earth's orbit is \(e=0.0167,\) so the orbit is nearly circular, with radius approximately \(150 \times 10^{6} \mathrm{km} .\) Find the rate \(d A / d t\) in units of \(\mathrm{km}^{2} /\) sec satisfying Kepler's second law.

Component test for continuity at a point Show that the vector function \(r\) defined by \(\mathbf{r}(t)=f(t) \mathbf{i}+g(t) \mathbf{j}+h(t) \mathbf{k}\) is continuous at \(t=t_{0}\) if and only if \(f, g,\) and \(h\) are continuous at \(t_{0}\) .

Volleyball A volleyball is hit when it is 4 ft above the ground and 12 ft from a 6 -ft-high net. It leaves the point of impact with an initial velocity of 35 \(\mathrm{ft} / \mathrm{sec}\) at an angle of \(27^{\circ}\) and slips by the opposing team untouched. a. Find a vector equation for the path of the volleyball. b. How high does the volleyball go, and when does it reach maximum height? c. Find its range and flight time. d. When is the volleyball 7 ft above the ground? How far e. Suppose that the net is raised to 8 ft. Does this change things? Explain.
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