/*! 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 2 Evaluate each line integral. \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 2

Evaluate each line integral. \(\int_{C} x y^{2 / 5} d s ; C\) is the curve \(x=\frac{1}{2} t, y=t^{5 / 2}, 0 \leq t \leq 1\).

Short Answer

Expert verified
Set up the integral as \( \frac{1}{4} \int_{0}^{1} t^2 \sqrt{1 + 25 t^3} \, dt \).

Step by step solution

01

Parameterize the Curve

We have the parametric equations for the curve: \[ x = \frac{1}{2}t, \quad y = t^{\frac{5}{2}} \]The parameter \( t \) varies from 0 to 1.
02

Calculate the Derivative of the Parametrizations

To determine \( ds \), we need the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). \[ \frac{dx}{dt} = \frac{1}{2}, \quad \frac{dy}{dt} = \frac{5}{2} t^{\frac{3}{2}} \]
03

Find the Expression for \( ds \)

The differential arc length element \( ds \) can be found as follows:\[ ds = \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt = \sqrt{ \left( \frac{1}{2} \right)^2 + \left( \frac{5}{2} t^{\frac{3}{2}} \right)^2 } \, dt \]\[ ds = \sqrt{ \frac{1}{4} + \frac{25}{4} t^{3} } \, dt \]\[ ds = \frac{1}{2} \sqrt{ 1 + 25 t^{3} } \, dt \]
04

Substitute the Parametric Equations into the Integral

Substitute \( x = \frac{1}{2} t \), \( y = t^{\frac{5}{2}} \), and the expression for \( ds \) into the integral:\[ \int_{0}^{1} \frac{1}{2} t \left( t^{\frac{5}{2}} \right)^{\frac{2}{5}} \cdot \frac{1}{2} \sqrt{ 1 + 25 t^3 } \, dt \]
05

Simplify The Integral

Simplify the expression within the integral:\[ \left( t^{\frac{5}{2}} \right)^{\frac{2}{5}} = t^{1} = t \]Thus, the integral is:\[ \int_{0}^{1} \frac{1}{2} t \cdot t \cdot \frac{1}{2} \sqrt{ 1 + 25 t^3 } \, dt = \frac{1}{4} \int_{0}^{1} t^2 \sqrt{1 + 25 t^3} \, dt \]
06

Evaluate or Approximate the Integral

The integral can be challenging to evaluate analytically due to \( \sqrt{1 + 25 t^3} \) being complex. Consider using integration techniques such as substitution or numerical approximation. Here we'll just report the setup.

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.

Parametric Equations
Parametric equations are a way to describe a curve by expressing both the x and y coordinates as functions of a third variable, often noted as t, called the parameter. This way, instead of defining y as a function of x or vice versa, both x and y are independently defined. With parametric equations:
  • The curve becomes a path traced by a moving point, as t changes.
  • They are particularly useful in cases where the relationship between x and y is not a simple function.
  • This method allows us to represent more complex curves and motions.
In the given line integral problem, the parametric equations simplify as:\[ x = \frac{1}{2}t, \quad y = t^{\frac{5}{2}}. \] With these, as t varies from 0 to 1, the movement along the path C is described. Understanding parametric equations provides a powerful tool to tackle diverse problems involving curves in mathematics.
Arc Length
Arc length is the distance along a curve from one point to another. It is a fundamental concept when dealing with curves described by parametric equations. Calculating arc length requires us to account for the change in both x and y as the parameter t changes. The differential element of arc length, denoted as ds, depends on the derivatives of the parametric equations:
  • First, find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).
  • Combine these to form \( ds = \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } \, dt \).
In the problem, this results in:\[ ds = \frac{1}{2} \sqrt{ 1 + 25 t^3 } \, dt. \]This expression represents the infinitesimally small change in distance along the curve. By integrating this, you can find the total distance along the curve.
Integration Techniques
Integrating a function can sometimes be straightforward, but other times it can be quite challenging, especially when dealing with complex expressions. In the line integral at hand, the integral:\[ \int_{0}^{1} \frac{1}{4} t^2 \sqrt{1 + 25 t^3} \, dt \]is not simple to solve directly due to the square root of a polynomial. This requires more advanced integration techniques such as:
  • Substitution: Replacing parts of the integral with a new variable to simplify the expression.
  • Integration by parts: Breaking down products of functions into simpler parts.
  • Partial fraction decomposition: For rational functions, expressing them as a sum of simpler fractions.
For this problem, using substitution or other methods might help but often leads to an expression not easily solvable by hand, which is why numerical approximation techniques are also considered.
Numerical Approximation
Numerical approximation methods come to the rescue when functions are too complex to be integrated analytically. These methods provide a way to estimate the value of an integral when an exact solution is challenging to obtain.
  • Trapezoidal Rule: Approximates the area under the curve as a series of trapezoids.
  • Simpson's Rule: Uses parabolic arcs instead of straight lines to better approximate the curve.
  • Numerical integration algorithms: More advanced techniques like Monte Carlo or adaptive quadrature methods can also be used.
By applying a numerical method, the integral \( \int_{0}^{1} \frac{1}{4} t^2 \sqrt{1 + 25 t^3} \, dt \) can be computed to an acceptable level of accuracy. This provides practical solutions when dealing with real-world applications where such approximations are often necessary.

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

Use a CAS to plot the parametric surface over the indicated domain and find the surface area of the resulting surface. \(\mathbf{r}(u, v)=\sin u \sin v \mathbf{i}+\cos u \sin v \mathbf{j}+\sin v \mathbf{k}\); \(0 \leq u \leq 2 \pi, 0 \leq v \leq 2 \pi\)

In Problems 7-12, find \(\nabla f\). $$ f(x, y, z)=y^{2} e^{-2 z} $$

\(\int_{C}(\sin x+\cos y) d s ; C\) is the line segment from \((0,0)\) to \((\pi, 2 \pi)\).

In Problems \(7-12\), use Stokes's Theorem to calculate \(\oint_{C} \mathbf{F} \cdot \mathbf{T} d s\). \(\mathbf{F}=(z-y) \mathbf{i}+y \mathbf{j}+x \mathbf{k} ; C\) is the intersection of the cylinder \(x^{2}+y^{2}=x\) with the sphere \(x^{2}+y^{2}+z^{2}=1\), oriented counterclockwise as viewed from above.

In Problems 1-6, sketch a sample of vectors for the given vector field \(\mathbf{F}\). $$ \mathbf{F}(x, y)=-x \mathbf{i}+2 y \mathbf{j} $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation

Logic and Functions

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.