/*! 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 28 A long needle of length \(L\) mu... [FREE SOLUTION] | 91Ó°ÊÓ

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A long needle of length \(L\) much bigger than 1 is dropped on a grid with horizontal and vertical lines one unit apart. Show that the average number \(a\) of lines crossed is approximately $$a=\frac{4 L}{\pi}$$

Short Answer

Expert verified
The average number of lines crossed is approximately \(a = \frac{4L}{\pi}\).

Step by step solution

01

Understand The Problem Context

A long needle of length \(L\) is dropped on an infinite grid with horizontal and vertical lines that are all one unit apart. We need to find out how many grid lines, on average, the needle crosses when it is dropped.
02

Consider The Geometry

Consider a circle centered around one endpoint of the needle. Let \(\theta\) be the angle between the needle and the vertical lines of the grid. As the needle rotates and moves within the grid, \(\theta\) ranges from 0 to \(\pi/2\) radians.
03

Define Parameters

Denote \(x\) as the perpendicular distance from the center of the needle to the nearest vertical line, and \(y\) as the perpendicular distance to the nearest horizontal line. Since the needle is very long, we will treat the grid as being indefinitely large.
04

Analyze Lines Crossed

The endpoint of the needle moves horizontally a distance of \(L\cos(\theta)\) and vertically \(L\sin(\theta)\) as it rotates. The number of vertical lines crossed is approximately \(\frac{L\cos(\theta)}{1}\) and horizontal lines as \(\frac{L\sin(\theta)}{1}\).
05

Calculate Expected Lines Crossed

Take the expected value of lines crossed over all values of \(\theta\) between 0 and \(\pi/2\), giving the average: \[ a = \frac{2}{\pi} \int_0^{\pi/2} L \left( \cos(\theta) + \sin(\theta) \right) \, d\theta. \]
06

Solve the Integral

The integral is two separate integrals: \[ a = \frac{2L}{\pi} \left( \int_0^{\pi/2} \cos(\theta) \, d\theta + \int_0^{\pi/2} \sin(\theta) \, d\theta \right). \] These integrals are standard calculation: \( \int_0^{\pi/2} \cos(\theta) \, d\theta = 1 \) and \( \int_0^{\pi/2} \sin(\theta) \, d\theta = 1 \).
07

Simplify and Conclude

Substitute back into the integral results: \[ a = \frac{2L}{\pi} (1 + 1) = \frac{4L}{\pi}. \] Thus, the average number of lines crossed is approximately \(\frac{4L}{\pi}\).

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Key Concepts

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

Understanding Integral Calculus
Integral calculus is a fundamental part of calculus that focuses on the accumulation of quantities and the areas under curves. It helps us find quantities where it isn't immediately obvious how they are changing. In this exercise, we use integral calculus to determine the average number of grid lines crossed by the needle. The integral component comes in when we calculate: \[ a = \frac{2}{\pi} \int_0^{\pi/2} L \left( \cos(\theta) + \sin(\theta) \right) \, d\theta. \]
This formula accumulates all the possible orientations and positions of the needle on the grid. Here's a simplified breakdown:
  • First, we recognize the function inside the integral, \(L (\cos(\theta) + \sin(\theta) )\), represents the horizontal and vertical distances the needle covers.
  • The integral of each trigonometric function, \(\cos(\theta)\) and \(\sin(\theta)\), is calculated over the interval from 0 to \(\pi/2\), which simplifies to 1 for each.
  • This result sums up across the possible angles \(\theta\), reflecting the needle's movement across the grid over multiple trials.
By understanding this component, it becomes clear how integral calculus allows us to model and solve real-world probability and geometry problems like Buffon's Needle.
Geometric Probability
Geometric probability deals with the likelihood of certain geometric patterns or shapes occurring, often using areas or lengths to express probability. In Buffon's Needle Problem, we're dealing with the probability of the needle crossing grid lines as it's randomly dropped.
  • The method relies on an understanding of the physical orientation and position of the needle relative to the grid.
  • Here, since the needle is infinitely long compared to the grid, this simplifies into computing the probability over a range of angles, \(\theta\), covered by the needle.
  • For each angle \(\theta\) from 0 to \(\pi/2\), the distances \(x\) and \(y\) that decide crossing possibilities are expressed through trigonometric functions.
Considering geometric probability in this problem, we evaluate how many times, on average, these geometric parameters cause the needle to intersect lines, fundamentally blending probability with geometry.
Calculating Expected Value
Expected value is a fundamental concept in probability describing the average outcome of a random event over time. In Buffon's Needle Problem, the expected value helps determine the average number of crossings by the needle. It's calculated by integrating over all possible needle orientations and applying weights based on probabilities.To approach this:
  • The expected value is computed by averaging the number of crossings over every possible angle \(\theta\).
  • This involves integrating each possible distance the needle can travel vertically and horizontally, summed over angle
    changes, weighted by the probability of these placements.
  • Here, each angle from 0 to \(\pi/2\) contributes equally to the expected crossings, given by the integrals of \(\cos(\theta)\) and \(\sin(\theta)\).
The resulting expected number, expressed as \(\frac{4L}{\pi}\), tells us the average behavior of the needle across countless trials, providing a probabilistic insight into this seemingly random but mathematically predictable process.

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Most popular questions from this chapter

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