/*! 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 56 Let \(\mathscr{R}\) be the regio... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 56

Let \(\mathscr{R}\) be the region that lies between the curves \(y=x^{m}\) and \(y=x^{n}, 0 \leqslant x \leqslant 1,\) where \(m\) and \(n\) are integers with \(0 \leqslant n < m\). (a) Sketch the region \(\mathscr{R}\) . (b) Find the coordinates of the centroid of \(\mathscr{R}\) . (c) Try to find values of \(m\) and \(n\) such that the centroid lies outside \(\mathscr{R} .\)

Short Answer

Expert verified
The centroid of the region typically lies within it; test with extreme \( m \) and \( n \) values to find exceptions.

Step by step solution

01

Understand the Problem

The region \( \mathscr{R} \) is bounded by two power functions, \( y = x^m \) and \( y = x^n \), over the interval \( 0 \leq x \leq 1 \). Since \( 0 \leq n < m \), the curve \( y = x^n \) lies above \( y = x^m \). We need to sketch this region and find the centroid's coordinates.
02

Sketch the Region

Plot the curves \( y = x^m \) and \( y = x^n \) on the same axes. Note that both curves will start from \( (0,0) \) and end at \( (1,1) \). Since \( n < m \), the curve \( y = x^n \) will always lie above \( y = x^m \) for \( x \) in \([0,1]\). The region \( \mathscr{R} \) is the area between these two curves within the interval.
03

Calculate the Area of \( \mathscr{R} \)

The area \( A \) between the curves can be found using the integral: \( A = \int_{0}^{1} (x^n - x^m) \, dx \). Calculate this integral as follows:\[A = \int_{0}^{1} x^n \, dx - \int_{0}^{1} x^m \, dx = \left[ \frac{x^{n+1}}{n+1} \right]_0^1 - \left[ \frac{x^{m+1}}{m+1} \right]_0^1 = \frac{1}{n+1} - \frac{1}{m+1}\]
04

Determine the Coordinates of the Centroid

The coordinates of the centroid \( (\bar{x}, \bar{y}) \) can be determined by the formula:\( \bar{x} = \frac{1}{A} \int_{0}^{1} x (x^n - x^m) \, dx \),\( \bar{y} = \frac{1}{2A} \int_{0}^{1} (x^n)^2 - (x^m)^2 \, dx \).Calculate each:\[\bar{x} = \frac{1}{A} \left( \int_{0}^{1} x^{n+1} \, dx - \int_{0}^{1} x^{m+1} \, dx \right) = \frac{n+1}{m+n+2} - \frac{m+1}{m+n+2}\]\[\bar{y} = \frac{1}{2A} \left( \int_{0}^{1} x^{2n} \, dx - \int_{0}^{1} x^{2m} \, dx \right) = \frac{n+1}{2n+1} - \frac{m+1}{2m+1}\]Substitute \( A = \frac{1}{n+1} - \frac{1}{m+1} \) to get \( \bar{x} \) and \( \bar{y} \).
05

Analyze the Centroid's Position

To find \( m \) and \( n \) such that the centroid \( (\bar{x}, \bar{y}) \) lies outside \( \mathscr{R} \), consider values that push the centroid outside of the interval \( 0 \leq \bar{x} \leq 1 \). This scenario will occur when \( m \) and \( n \) are such that the integrals are significantly skewed. Testing out extreme cases where \( n \) is much smaller than \( m \) can show possible instances.
06

Verify with Examples

Try an example with \( n = 1 \) and \( m = 3 \). Calculate \( \bar{x} \) and \( \bar{y} \) to see if they fall outside by plugging into the centroid formulas derived earlier. If the results show these coordinates do not lie between the two curves over \( [0,1] \), the centroid is outside of \( \mathscr{R} \).

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Key Concepts

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

Centroid Calculation
Calculating the centroid of a region between curves is like finding the "average" position of the entire area. For the region \( \mathscr{R} \) between the curves \( y = x^m \) and \( y = x^n \), the centroid \((\bar{x}, \bar{y})\) gives us insight into the shape's balance point.
  • The x-coordinate of the centroid \( \bar{x} \) is the "average" x-value of the region, accounting for the area distribution.
  • The y-coordinate \( \bar{y} \) provides the "average" y-value considering the height differences between the curves.
To find \( \bar{x} \), use the formula:\[ \bar{x} = \frac{1}{A} \int_{0}^{1} x (x^n - x^m) \, dx \]For \( \bar{y} \), it's:\[ \bar{y} = \frac{1}{2A} \int_{0}^{1} ((x^n)^2 - (x^m)^2) \, dx \]The process involves evaluating the integrals for powers of x, finding the difference between the two, and then scaling them by the area \( A \). This reveals how the region's mass is distributed along the axis.
Area Between Curves
When finding the area between two curves, you're interested in the space enclosed between the curves across a specified interval. For the functions \( y=x^m \) and \( y=x^n \), determining this area involves integrals.
  • The formula to calculate the area is: \[ A = \int_{0}^{1} (x^n - x^m) \, dx \]
  • It's essential to subtract the lower function value, \( y=x^m \), from the higher \( y=x^n \), as \( x^n \) is greater than \( x^m \) for \( n < m \).
Integration results in finding the net "height" between the curves over the interval from 0 to 1. This calculation helps you understand how wide or narrow the region \( \mathscr{R} \) is, based on the domain specified.
Power Functions
Power functions are expressions where a variable is raised to a constant power. In this calculus problem, we've encountered \( y=x^m \) and \( y=x^n \), which are classic power functions. Here's why they're important:
  • They have predictable shapes based on the power: for \( m, n > 1 \), they curve upwards; for \( 0 < m, n < 1 \), they flatten.
  • Understanding how different power values affect the curves helps in solving area and centroid problems.
By grasping these functions' behavior, you recognize why one function is always above the other (\( y=x^n \) above \( y=x^m \)), crucial for setting up the integrals that calculate area and centroid.
Integrals in Calculus
Integrals are a fundamental concept in calculus, used here to find both the area and centroid of the region \( \mathscr{R} \). When working with curves:
  • The integral \( \int_{a}^{b} f(x) \, dx \) measures the area under the curve \( f(x) \) from \( x=a \) to \( x=b \).
  • For centroids and areas, you're often dealing with integrals that describe the difference between two curves, like \( \int_{0}^{1} (x^n - x^m) \, dx \).
Integrals offer a way to compute intricate properties of a region analytically. They help provide exact values for areas and points of balance (like the centroid) that otherwise would be cumbersome to calculate by geometric approximation.

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