/*! 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 58 Use the result of Exercise 55 to... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 58

Use the result of Exercise 55 to find the curvature function of the following curves. $$\mathbf{r}(t)=\left\langle a \cos ^{3} t, a \sin ^{3} t\right\rangle \quad \text { (astroid) }$$

Short Answer

Expert verified
Answer: The curvature function of the curve defined by $\mathbf{r}(t) = \langle a\cos^3t, a\sin^3t \rangle$ is given by: $$ \kappa(t) = \frac{\|\mathbf{r'}(t) \times \mathbf{r''}(t)\|}{\|\mathbf{r'}(t)\|^3} $$ where $$ \mathbf{r'}(t) = \left\langle -3a\cos^2t \sin t, 3a\sin^2t \cos t \right\rangle $$ $$ \mathbf{r''}(t) = \left\langle 3a(\sin^3t + 2\cos^3t\sin t - 3\cos^2t\sin^2t \cos t), 3a(-\cos^3t - 2\sin^3t\cos t + 3\cos^2t\sin^2t \sin t)\right\rangle $$ and the magnitudes of the first derivative and the cross product of the first and second derivatives are $\|\mathbf{r'}(t)\|^2 = 9a^2(\cos^4t\sin^2t + \sin^4t\cos^2t)$, and $\|\mathbf{r'}(t) \times \mathbf{r''}(t)\|$ is the square root of the expression given in the solution above.

Step by step solution

01

Compute the first derivative

To compute the first derivative of the curve, we need to differentiate both components with respect to the parameter t: $$ \mathbf{r'}(t) = \left\langle \frac{d}{dt}(a \cos^3t), \frac{d}{dt}(a \sin^3t) \right\rangle $$ Using the chain rule, we get: $$ \mathbf{r'}(t) = \left\langle -3a\cos^2t \sin t, 3a\sin^2t \cos t \right\rangle $$
02

Compute the second derivative

To compute the second derivative of the curve, we need to differentiate each component of the first derivative with respect to the parameter t: $$ \mathbf{r''}(t) = \left\langle \frac{d}{dt}(-3a\cos^2t \sin t), \frac{d}{dt}(3a\sin^2t \cos t) \right\rangle $$ Using the product rule and chain rule, we get: $$ \mathbf{r''}(t) = \left\langle 3a(\sin^3t + 2\cos^3t\sin t - 3\cos^2t\sin^2t \cos t), 3a(-\cos^3t - 2\sin^3t\cos t + 3\cos^2t\sin^2t \sin t)\right\rangle $$
03

Compute the magnitudes of first and second derivatives

To find the curvature, we need to compute the magnitudes of the first and second derivatives. For the first derivative: $$ \|\mathbf{r'}(t)\|^2 = (-3a\cos^2t\sin t)^2 + (3a\sin^2t\cos t)^2 = 9a^2(\cos^4t\sin^2t + \sin^4t\cos^2t) $$ For the second derivative: $$ \|\mathbf{r''}(t)\| = \sqrt{\left[3a(\sin^3t + 2\cos^3t\sin t - 3\cos^2t\sin^2t \cos t)\right]^2 + \left[3a(-\cos^3t - 2\sin^3t\cos t + 3\cos^2t\sin^2t \sin t)\right]^2} $$
04

Calculate the curvature function

Finally, we can calculate the curvature using the formula from Exercise 55: $$ \kappa(t) = \frac{\|\mathbf{r'}(t) \times \mathbf{r''}(t)\|}{\|\mathbf{r'}(t)\|^3} $$ Here, we need to compute the cross product of the first and second derivatives before plugging them into the formula above: $$ \mathbf{r'}(t) \times \mathbf{r''}(t) = (-3a\cos^2t \sin t)(3a(-\cos^3t - 2\sin^3t\cos t + 3\cos^2t\sin^2t \sin t)) - (3a\sin^2t \cos t)(3a(\sin^3t + 2\cos^3t\sin t - 3\cos^2t\sin^2t \cos t)) $$ $$ \|\mathbf{r'}(t) \times \mathbf{r''}(t)\| = \sqrt{\left[(-3a\cos^2t \sin t)(3a(-\cos^3t - 2\sin^3t\cos t + 3\cos^2t\sin^2t \sin t)) - (3a\sin^2t \cos t)(3a(\sin^3t + 2\cos^3t\sin t - 3\cos^2t\sin^2t \cos t))\right]^2} $$ By plugging the magnitudes of the first and second derivatives into the formula for curvature, we get the curvature function of the curve.

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.

Derivatives
Derivatives are fundamental in calculus for determining how a function changes. In the context of curves, the derivative helps us understand how the curve bends or twists as the parameter changes. For a vector function like \( \mathbf{r}(t) = \langle a \cos^3 t, a \sin^3 t \rangle \), we find the derivative by differentiating each component with respect to the parameter \( t \).

By using differentiation, we compute:
  • \( \frac{d}{dt}(a \cos^3 t) = -3a \cos^2 t \sin t \)
  • \( \frac{d}{dt}(a \sin^3 t) = 3a \sin^2 t \cos t \)
These derivatives describe how fast the curve moves in the direction of the cosine and sine components, respectively.
Chain Rule
The chain rule is a powerful tool in calculus for handling composite functions. It helps us break down the differentiation of functions that are nested within others. In this exercise, the functions \( a \cos^3 t \) and \( a \sin^3 t \) involve both trigonometric functions and an exponent.

To differentiate \( a \cos^3 t \), the chain rule provides:
  • First, differentiate the outer function: \( (\cos t)^3 \). The derivative is \( 3\cos^2 t \cdot (-\sin t) \).
  • Next, multiply by the derivative of the inner function: \(-\sin t\).
The chain rule is applied similarly to \( a \sin^3 t \), allowing us to effectively compute the first derivative \( \mathbf{r'}(t) \).
Cross Product
The cross product is a mathematical operation that helps to find a vector perpendicular to two given vectors. It is mostly used in contexts involving three-dimensional vectors. In this problem, we work with two-dimensional vectors, but we can imagine the vectors as lying in a plane within three-dimensional space.

The formula for calculating a cross product in the plane extends to using infinitesimally small components "out of the plane" to determine the vector perpendicular. This process is important for computing the curvature function, where the cross product of the first and second derivatives gives a measure of how the curve is twisting or bending in space.
Curvature Function
Curvature describes how sharply a curve bends at a point. It's crucial for understanding the curve's geometry and is expressed by the curvature function \( \kappa(t) \).

To find the curvature, we use the formula:\[\kappa(t) = \frac{\|\mathbf{r'}(t) \times \mathbf{r''}(t)\|}{\|\mathbf{r'}(t)\|^3}\]Here's what happens:
  • The numerator \( \|\mathbf{r'}(t) \times \mathbf{r''}(t)\| \) involves the cross product magnitude, indicating the twisting.
  • The denominator \( \|\mathbf{r'}(t)\|^3 \) involves the cube of the speed of the curve, standardizing the curvature measure.
Ultimately, the curvature function helps us measure the curve's bending at any point \( t \), providing insights into the curve's shape dynamics.

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

Alternative derivation of curvature Derive the computational formula for curvature using the following steps. a. Use the tangential and normal components of the acceleration to show that \(\mathbf{v} \times \mathbf{a}=\kappa|\mathbf{v}|^{3} \mathbf{B} .\) (Note that \(\mathbf{T} \times \mathbf{T}=\mathbf{0} .\) ) b. Solve the equation in part (a) for \(\kappa\) and conclude that \(\kappa=\frac{|\mathbf{v} \times \mathbf{a}|}{|\mathbf{v}|^{3}},\) as shown in the text.

Relationship between \(\mathbf{T}, \mathbf{N},\) and a Show that if an object accelerates in the sense that \(\frac{d^{2} s}{d t^{2}}>0\) and \(\kappa \neq 0,\) then the acceleration vector lies between \(\mathbf{T}\) and \(\mathbf{N}\) in the plane of \(\mathbf{T}\) and \(\mathbf{N}\). Show that if an object decelerates in the sense that \(\frac{d^{2} s}{d t^{2}}<0,\) then the acceleration vector lies in the plane of \(\mathbf{T}\) and \(\mathbf{N},\) but not between \(\mathbf{T}\) and \(\mathbf{N} .\)

Evaluate the following definite integrals. $$\int_{0}^{2} t e^{t}(\mathbf{i}+2 \mathbf{j}-\mathbf{k}) d t$$

Suppose the vector-valued function \(\mathbf{r}(t)=\langle f(t), g(t), h(t)\rangle\) is smooth on an interval containing the point \(t_{0}\) The line tangent to \(\mathbf{r}(t)\) at \(t=t_{0}\) is the line parallel to the tangent vector \(\mathbf{r}^{\prime}\left(t_{0}\right)\) that passes through \(\left(f\left(t_{0}\right), g\left(t_{0}\right), h\left(t_{0}\right)\right) .\) For each of the following functions, find an equation of the line tangent to the curve at \(t=t_{0} .\) Choose an orientation for the line that is the same as the direction of \(\mathbf{r}^{\prime}.\) $$\mathbf{r}(t)=\left\langle e^{t}, e^{2 t}, e^{3 t}\right\rangle ; t_{0}=0$$

Cusps and noncusps a. Graph the curve \(\mathbf{r}(t)=\left\langle t^{3}, t^{3}\right\rangle .\) Show that \(\mathbf{r}^{\prime}(0)=0\) and the curve does not have a cusp at \(t=0 .\) Explain. b. Graph the curve \(\mathbf{r}(t)=\left\langle t^{3}, t^{2}\right\rangle .\) Show that \(\mathbf{r}^{\prime}(0)=\mathbf{0}\) and the curve has a cusp at \(t=0 .\) Explain. c. The functions \(\mathbf{r}(t)=\left\langle t, t^{2}\right\rangle\) and \(\mathbf{p}(t)=\left\langle t^{2}, t^{4}\right\rangle\) both satisfy \(y=x^{2} .\) Explain how the curves they parameterize are different. d. Consider the curve \(\mathbf{r}(t)=\left\langle t^{m}, t^{n}\right\rangle,\) where \(m>1\) and \(n>1\) are integers with no common factors. Is it true that the curve has a cusp at \(t=0\) if one (not both) of \(m\) and \(n\) is even? Explain.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

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.