/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Find the unit tangent vector \(\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 18

Find the unit tangent vector \(\mathbf{T}\) and the curvature \(\kappa\) for the following parameterized curves. $$\mathbf{r}(t)=\left\langle\cos ^{3} t, \sin ^{3} t\right\rangle$$

Short Answer

Expert verified
Question: Find the unit tangent vector \(\mathbf{T}(t)\) and the curvature \(\kappa\) for the given parameterized curve: \(\mathbf{r}(t) = \left\langle 3\sin^2(t), 3\sin(t)\cos(t) \right\rangle\) Solution: The unit tangent vector \(\mathbf{T}(t)\) is: $$\mathbf{T}(t) = \left\langle \frac{-\cos^2(t) \sin(t)}{\sqrt{\cos ^{4}(t)\sin ^{2}(t)+\sin ^{4}(t)\cos ^{2}(t)}}, \frac{\sin^2(t) \cos(t)}{\sqrt{\cos ^{4}(t)\sin ^{2}(t)+\sin ^{4}(t)\cos ^{2}(t)}}\right\rangle$$ The curvature \(\kappa\) is: $$\kappa = \frac{\sin^2(t)\cos^2(t)\left(\sin^2(t)+\cos^2(t)\right)^2}{\left(\cos ^{4}(t)\sin ^{2}(t)+\sin ^{4}(t)\cos ^{2}(t)\right)^{\frac{3}{2}}}$$

Step by step solution

01

Compute the first derivative of \(\mathbf{r}(t)\)

Compute the first derivative of the given vector function with respect to \(t\): $$\mathbf{r}'(t)=\left\langle-3\cos ^{2}(t)\sin(t), 3\sin ^{2}(t)\cos(t)\right\rangle$$
02

Calculate the magnitude of the tangent vector

Compute the magnitude of the tangent vector \(\mathbf{r}'(t)\): $$\|\mathbf{r}'(t)\| = \sqrt{(-3\cos^2(t)\sin(t))^2 + (3\sin^2(t)\cos(t))^2}$$ $$\|\mathbf{r}'(t)\| = 3\sqrt{\cos ^{4}(t)\sin ^{2}(t)+\sin ^{4}(t)\cos ^{2}(t)}$$
03

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

Compute the unit tangent vector by dividing the tangent vector by its magnitude: $$\mathbf{T}(t) = \frac{\mathbf{r}'(t)}{\|\mathbf{r}'(t)\|} = \left\langle \frac{-\cos^2(t) \sin(t)}{\sqrt{\cos ^{4}(t)\sin ^{2}(t)+\sin ^{4}(t)\cos ^{2}(t)}}, \frac{\sin^2(t) \cos(t)}{\sqrt{\cos ^{4}(t)\sin ^{2}(t)+\sin ^{4}(t)\cos ^{2}(t)}}\right\rangle$$
04

Compute the second derivative of \(\mathbf{r}(t)\)

Compute the second derivative of the given vector function with respect to \(t\): $$\mathbf{r}''(t)=\left\langle 9\sin ^2(t)\cos(t)(\sin^2(t)-\cos^2(t)), -9\cos ^2(t)\sin(t)(\sin^2(t)-\cos^2(t)) \right\rangle$$
05

Calculate the curvature \(\kappa\)

Compute the curvature \(\kappa\) using the formula \(\kappa = \frac{\|\mathbf{r}'(t) \times \mathbf{r}''(t)\|}{\|\mathbf{r}'(t)\|^3}\): First, compute \(\mathbf{r}'(t) \times \mathbf{r}''(t)\): \(\mathbf{r}'(t) \times \mathbf{r}''(t) = 9 \left(\sin ^{4}(t) \cos^3(t) + \cos^{4}(t) \sin^{3}(t)\right)\) Next, find the magnitude of the cross product: \(\|\mathbf{r}'(t) \times \mathbf{r}''(t)\| = 9\left(\sin ^{4}(t) \cos^3(t) + \cos^{4}(t) \sin^{3}(t)\right)\) Finally, compute the curvature \(\kappa\): \(\kappa = \frac{9\left(\sin ^{4}(t) \cos^3(t) + \cos^{4}(t) \sin^{3}(t)\right)}{\left(3\sqrt{\cos ^{4}(t)\sin ^{2}(t)+\sin ^{4}(t)\cos ^{2}(t)}\right)^3} = \frac{\sin^2(t)\cos^2(t)\left(\sin^2(t)+\cos^2(t)\right)^2}{\left(\cos ^{4}(t)\sin ^{2}(t)+\sin ^{4}(t)\cos ^{2}(t)\right)^{\frac{3}{2}}}\). Thus, the unit tangent vector \(\mathbf{T}(t)\) and the curvature \(\kappa\) for the given parameterized curve are: $$\mathbf{T}(t) = \left\langle \frac{-\cos^2(t) \sin(t)}{\sqrt{\cos ^{4}(t)\sin ^{2}(t)+\sin ^{4}(t)\cos ^{2}(t)}}, \frac{\sin^2(t) \cos(t)}{\sqrt{\cos ^{4}(t)\sin ^{2}(t)+\sin ^{4}(t)\cos ^{2}(t)}}\right\rangle$$ $$\kappa = \frac{\sin^2(t)\cos^2(t)\left(\sin^2(t)+\cos^2(t)\right)^2}{\left(\cos ^{4}(t)\sin ^{2}(t)+\sin ^{4}(t)\cos ^{2}(t)\right)^{\frac{3}{2}}}$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Tangent vector
In vector calculus, understanding the tangent vector is crucial when analyzing curves. The tangent vector provides a direction in which a curve moves as a parameter, like time, changes. This direction is given by the derivative of the parameterized curve function, denoted as \( \mathbf{r}'(t) \). The tangent vector sheds light on how rapidly and in what direction the curve travels at any given point along the curve.
For a parameterized curve \( \mathbf{r}(t) \), the tangent vector is \( \mathbf{r}'(t) = \left\langle -3\cos^2(t)\sin(t), 3\sin^2(t)\cos(t) \right\rangle \). This derivative describes the rate of change of the curve's position vector as the parameter \( t \) changes.
However, to fully understand the behavior of a curve, we often consider the unit tangent vector, \( \mathbf{T}(t) \). This is a normalized version of the tangent vector, giving us a consistent way to compare directions regardless of the speed at which the point moves along the curve. We calculate it by dividing \( \mathbf{r}'(t) \) by its magnitude \( \| \mathbf{r}'(t) \| \). This underscores the importance of both magnitude and direction in interpreting vector calculus results.
Parameterized curves
Parameterized curves are essential tools in calculus, helping us describe the motion along a curve using parameters, commonly time \( t \). Essentially, instead of expressing \( y \) as a function of \( x \) alone, we use a vector function \( \mathbf{r}(t) \) to capture positions on a curve at any given time \( t \).
This approach is particularly useful when dealing with curves more complex than simple lines or circles. For example, the parameterized curve: \[ \mathbf{r}(t) = \left\langle \cos^3(t), \sin^3(t) \right\rangle \]provides a compact way to define the curve's trajectory.
In this framework, both \( x \) and \( y \) change over time, creating a path that \( \mathbf{r}(t) \) traces. This presents a clearer picture of motion through space, which can be particularly powerful for visualizing curves that twist and turn in ways not easily represented with a single function \( y = f(x) \). By providing a parameterized format, complex curve behaviors become manageable and interpretable.
Vector calculus
Vector calculus is a fundamental mathematical framework employed for handling vector quantities. It expands traditional calculus concepts into multidimensional spaces. This is particularly effective for analyzing physical phenomena and geometrical problems in two and three dimensions.
Some of the key operations in vector calculus include differentiation and integration of vector fields, measuring derivatives like gradient, divergence, and curl, and understanding vector relationships using operations such as the dot product and cross product.
In the context of parameterized curves:
  • Gradient: Finds the direction of a maximum rate of increase of a scalar field.
  • Cross product: Effectively used in calculating curvature, where the cross product of two derivatives helps determine how tightly the curve loops.
Knowing how to compute these operations allows you to find properties like the tangent vector, unit tangent vector, and curvature. For instance, the cross product \( \mathbf{r}'(t) \times \mathbf{r}''(t) \) aids in measuring how much a curve deviates from being a straight line. That's how we derive curvature \( \kappa \), which reveals how sharply a curve bends at any point.
By mastering vector calculus, you can tackle a wide range of problems in physics and engineering, as well as acquire deeper insights into natural phenomena.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An object moves along a path given by $$\mathbf{r}(t)=\langle a \cos t+b \sin t, c \cos t+d \sin t, e \cos t+f \sin t\rangle$$ for \(0 \leq t \leq 2 \pi\) a. Show that the curve described by \(\mathbf{r}\) lies in a plane. b. What conditions on \(a, b, c, d, e,\) and \(f\) guarantee that the curve described by \(\mathbf{r}\) is a circle?

Motion on a sphere Prove that \(\mathbf{r}\) describes a curve that lies on the surface of a sphere centered at the origin \(\left(x^{2}+y^{2}+z^{2}=a^{2}\right.\) with \(a \geq 0\) ) if and only if \(\mathbf{r}\) and \(\mathbf{r}^{\prime}\) are orthogonal at all points of the curve.

Suppose \(\mathbf{u}\) and \(\mathbf{v}\) are vectors in the plane. a. Use the Triangle Rule for adding vectors to explain why \(|\mathbf{u}+\mathbf{v}| \leq|\mathbf{u}|+|\mathbf{v}| .\) This result is known as the Triangle Inequality. b. Under what conditions is \(|\mathbf{u}+\mathbf{v}|=|\mathbf{u}|+|\mathbf{v}| ?\)

Cauchy-Schwarz Inequality The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\) (because \(|\cos \theta| \leq 1\) ). This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Geometric-arithmetic mean Use the vectors \(\mathbf{u}=\langle\sqrt{a}, \sqrt{b}\rangle\) and \(\mathbf{v}=\langle\sqrt{b}, \sqrt{a}\rangle\) to show that \(\sqrt{a b} \leq(a+b) / 2,\) where \(a \geq 0\) and \(b \geq 0\).

A pair of nonzero vectors in the plane is linearly dependent if one vector is a scalar multiple of the other. Otherwise, the pair is linearly independent. a. Which pairs of the following vectors are linearly dependent and which are linearly independent: \(\mathbf{u}=\langle 2,-3\rangle\) \(\mathbf{v}=\langle-12,18\rangle,\) and \(\mathbf{w}=\langle 4,6\rangle ?\) b. Geometrically, what does it mean for a pair of nonzero vectors in the plane to be linearly dependent? Linearly independent? c. Prove that if a pair of vectors \(\mathbf{u}\) and \(\mathbf{v}\) is linearly independent, then given any vector \(w\), there are constants \(c_{1}\) and \(c_{2}\) such that \(\mathbf{w}=c_{1} \mathbf{u}+c_{2} \mathbf{v}\)
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.