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6800 | https://www.hirequotient.com/marginal-cost | Marginal Cost Calculator
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Use our free Marginal Cost Calculator to determine the cost of additional units produced
Marginal Cost Calculator
What is Marginal Cost?
Definition of Marginal Cost
Importance of Understanding Marginal Cost
The Formula for Marginal Cost
Practical Examples of Marginal Cost
Real-World Applications of Marginal Cost
The Role of Marginal Cost in Business Decisions
Marginal Cost Formula
Understanding the Marginal Cost Formula
The Marginal Cost Formula
Breaking Down the Formula
Practical Application of the Marginal Cost Formula
Importance of the Marginal Cost Formula
Marginal Cost Formula in Decision Making
How to Calculate Marginal Cost
Step-by-Step Guide to Calculating Marginal Cost
Step 1: Determine Total Cost
Step 2: Calculate the Change in Total Cost
Step 3: Calculate the Change in Quantity
Step 4: Apply the Marginal Cost Formula
Example Calculation
Why Calculating Marginal Cost is Important
Optimizing Production
Tools for Calculating Marginal Cost
Conclusion
Understanding Marginal Cost Through Examples
Example 1: Manufacturing Industry
Scenario: A company produces electronic gadgets. The total cost of producing 500 gadgets is $50,000. When the production increases to 501 gadgets, the total cost rises to $50,100.
Calculation:
Δ𝑇𝐶 = $50,100 − $50,000 = $100
Δ𝑄 = 501 − 500 = 1
Marginal Cost (MC) = $100 / 1 = $100
Explanation: The marginal cost of producing one additional gadget is $100. This helps the company decide if producing more gadgets will be profitable.
Example 2: Food and Beverage Industry
Scenario: A bakery produces 1,000 loaves of bread at a total cost of $5,000. If the total cost for producing 1,001 loaves is $5,005, the marginal cost can be determined.
Calculation:
Δ𝑇𝐶 = $5,005 − $5,000 = $5
Δ𝑄 = 1,001 − 1,000 = 1
MC = $5 / 1 = $5
Explanation: The marginal cost of producing an additional loaf of bread is $5. This information is crucial for pricing and production decisions.
Example 3: Software Development
Scenario: A software company incurs a total cost of $200,000 to develop 10 software licenses. If the cost increases to $205,000 for 11 licenses, the marginal cost needs to be calculated.
Calculation:
Δ𝑇𝐶 = $205,000 − $200,000 = $5,000
Δ𝑄 = 11 − 10 = 1
MC = $5,000 / 1 = $5,000
Explanation: The marginal cost of producing one additional software license is $5,000. This helps the company evaluate the cost-effectiveness of scaling up production.
Example 4: Automobile Industry
Scenario: An automobile manufacturer spends $1,000,000 to produce 100 cars. If producing 101 cars increases the total cost to $1,010,000, the marginal cost can be calculated.
Calculation:
Δ𝑇𝐶 = $1,010,000 − $1,000,000 = $10,000
Δ𝑄 = 101 − 100 = 1
MC = $10,000 / 1 = $10,000
Explanation: The marginal cost of producing an additional car is $10,000. This insight helps the manufacturer decide on optimal production levels.
Example 5: Pharmaceutical Industry
Scenario: A pharmaceutical company has a total cost of $10,000,000 for producing 1,000 batches of a drug. If the cost to produce 1,001 batches is $10,020,000, the marginal cost is calculated as follows:
Calculation:
Δ𝑇𝐶 = $10,020,000 − $10,000,000 = $20,000
Δ𝑄 = 1,001 − 1,000 = 1
MC = $20,000 / 1 = $20,000
Explanation: The marginal cost of producing an additional batch of the drug is $20,000. This helps the company in budgeting and pricing strategies.
Section 6: Finding Marginal Cost Curve
Understanding the Marginal Cost Curve
How to Plot the Marginal Cost Curve
Example of a Marginal Cost Curve
| Quantity (Q) | Total Cost (TC) | Marginal Cost (MC) |
---
| 10 | $1,000
| 20 | $1,800 | $80 |
| 30 | $2,400 | $60 |
| 40 | $3,200 | $80 |
| 50 | $4,500 | $130 |
Significance of the Marginal Cost Curve
Analyzing the Marginal Cost Curve
Conclusion
Section 7: Difference Between Marginal Cost and Marginal Revenue
Understanding Marginal Cost
Understanding Marginal Revenue
Key Differences Between Marginal Cost and Marginal Revenue
Example to Illustrate the Difference
Importance in Business Strategy
Conclusion
Frequently Asked Questions (FAQs)
1.What is marginal cost?
Answer: Marginal cost is the additional cost incurred by producing one more unit of a product or service. It helps businesses determine the cost-effectiveness of increasing production.
2.How to calculate marginal cost?
Answer: To calculate marginal cost, use the formula: Marginal Cost (MC) = ΔTC / ΔQ where ΔTC is the change in total cost and ΔQ is the change in quantity produced.
3.How to find marginal cost?
Answer: Marginal cost can be found by analyzing the change in total costs and output levels. Calculate the difference in total costs for two different production levels and divide by the change in output.
4.What is the best definition of marginal cost?
Answer: The best definition of marginal cost is the additional cost incurred by producing one more unit of a product. It helps businesses optimize production and pricing strategies.
5.What is the difference between marginal cost and marginal revenue?
Answer: Marginal cost is the cost of producing one additional unit, while marginal revenue is the additional revenue generated from selling one more unit. Profit maximization occurs when marginal cost equals marginal revenue.
6.What is marginal cost in economics?
Answer: In economics, marginal cost is the cost of producing one additional unit of output. It is crucial for analyzing production efficiency and cost management.
7.How is marginal cost (MC) calculated?
Answer: Marginal cost (MC) is calculated by dividing the change in total cost (ΔTC) by the change in quantity produced (ΔQ).
8.What is a marginal cost?
Answer: A marginal cost is the cost associated with producing one additional unit of a product or service. It helps in understanding the cost dynamics of production.
9.How to find marginal cost from total cost?
Answer: To find marginal cost from total cost, calculate the change in total cost for a given change in output and divide by the change in quantity.
10.How to calculate marginal cost from a table?
Answer: To calculate marginal cost from a table, find the change in total cost and the change in quantity produced between two data points and divide the former by the latter.
11.What does marginal cost mean?
Answer: Marginal cost means the additional cost incurred when producing one more unit of a product. It is a key concept in cost analysis and production planning.
12.When marginal cost is graphed, it creates which of the following?
Answer: When marginal cost is graphed, it typically creates a U-shaped curve, showing initially decreasing costs followed by increasing costs as production increases.
13.How does a firm calculate marginal cost?
Answer: A firm calculates marginal cost by dividing the change in total cost by the change in quantity produced, helping in decision-making and cost management.
14.Marginal cost includes which of the following?
Answer: Marginal cost includes variable costs such as materials, labor, and overheads directly associated with production.
15.What is marginal benefit and marginal cost?
Answer: Marginal benefit is the additional benefit from consuming one more unit, while marginal cost is the additional cost of producing one more unit.
16.How to calculate marginal cost in economics?
Answer: To calculate marginal cost in economics, use the formula: MC = ΔTC / ΔQ where ΔTC is the change in total cost and ΔQ is the change in quantity produced.
17.How to find marginal cost on a graph?
Answer: To find marginal cost on a graph, identify the slope of the total cost curve at a given output level, representing the additional cost of producing one more unit.
18.What is the marginal cost of producing the 200th pizza?
Answer: The marginal cost of producing the 200th pizza is found by calculating the difference in total cost before and after producing the 200th pizza and dividing by the change in quantity.
19.What is an example of marginal cost?
Answer: An example of marginal cost is a bakery producing 100 loaves of bread at a total cost of $500. If producing 101 loaves increases the total cost to $505, the marginal cost of the 101st loaf is $5.
20.How to find marginal cost calculator?
Answer: To find a marginal cost calculator, use online financial calculators designed for cost analysis. These tools allow you to input changes in total cost and quantity to quickly calculate marginal cost.
21.Why is it necessary to know fixed, variable, and total costs to determine marginal cost?
Answer: Knowing fixed, variable, and total costs is necessary to determine marginal cost because it helps in identifying the change in total cost resulting from a change in production level. Fixed costs remain constant, while variable costs change with production levels, impacting the marginal cost calculation.
22.Which curve(s) does the marginal cost curve intersect at their minimum point?
Answer: The marginal cost curve intersects both the average total cost (ATC) and average variable cost (AVC) curves at their respective minimum points, indicating the most efficient production levels.
23.Why does marginal cost decrease then increase?
Answer: Marginal cost initially decreases due to economies of scale but eventually increases due to diminishing returns, where each additional unit requires more resources, raising the cost.
24.How to graph marginal cost?
Answer: To graph marginal cost, plot the quantity of output on the horizontal axis and the marginal cost on the vertical axis. Connect the data points to form the marginal cost curve, typically U-shaped, indicating changes in cost efficiency.
25.Average total cost is falling when marginal cost is below it and rising when marginal cost is above it.
Answer: This relationship indicates that marginal cost influences the direction of average total cost.
26.What is the main difference between marginal revenue and marginal cost?
Answer: The main difference between marginal revenue and marginal cost is that marginal revenue is the additional income from selling one more unit, while marginal cost is the additional cost of producing one more unit. Profit maximization occurs when these two values are equal.
27.How to find marginal average cost?
Answer: To find marginal average cost, calculate the average total cost for different levels of output and determine the change in average cost as output increases.
28.Why is marginal cost important?
Answer: Marginal cost is important because it helps businesses determine the optimal production level, set pricing strategies, and allocate resources efficiently, ensuring profitability and cost control.
29.When marginal revenue equals marginal cost:
Answer: When marginal revenue equals marginal cost, the firm maximizes its profit, as each additional unit produced neither adds to nor subtracts from total profit.
30.Which best describes the graphical portrayal of marginal cost?
Answer: The graphical portrayal of marginal cost is typically a U-shaped curve, showing initially decreasing costs due to economies of scale, followed by increasing costs due to diminishing returns.
31.Why is it important to consider marginal benefits and costs when you do a cost-benefit analysis?
Answer: Considering marginal benefits and costs in a cost-benefit analysis ensures that decisions are made based on the additional benefits and costs of one more unit, leading to optimal resource allocation and efficiency.
32.What is the marginal cost of producing the fifth unit of output?
Answer: The marginal cost of producing the fifth unit of output is the additional cost incurred by producing that unit, calculated by finding the difference in total cost before and after its production.
33.How to find minimum marginal cost?
Answer: To find the minimum marginal cost, analyze the marginal cost curve and identify the lowest point, where producing additional units is most cost-efficient.
34.What does the marginal cost equal for any firm?
Answer: For any firm, the marginal cost equals the additional cost of producing one more unit of output. It is a crucial factor in production and pricing decisions.
35.What is marginal cost in economics?
Answer: In economics, marginal cost refers to the additional cost incurred by producing one more unit of a good or service. It is used to determine the optimal production level and pricing strategies.
36.Which of the following industries has a marginal cost that is close to zero?
Answer: Industries with digital products, such as software or online services, often have a marginal cost close to zero because producing additional units incurs negligible costs.
37.How to calculate marginal resource cost?
Answer: To calculate marginal resource cost, determine the change in total resource cost when one additional unit of the resource is employed and divide by the change in the quantity of the resource used.
38.What is the relationship between the marginal cost and the slope of the cost function?
Answer: The relationship between marginal cost and the slope of the cost function is that marginal cost represents the slope of the total cost function, indicating the rate at which costs change with production levels.
39.Why does marginal cost intersect the minimum of the ATC?
Answer: Marginal cost intersects the minimum of the average total cost (ATC) because, at this point, producing one more unit neither increases nor decreases the average cost, indicating optimal production efficiency.
40.Why would an already-successful business owner conduct a marginal cost analysis for their product?
Answer: An already-successful business owner would conduct a marginal cost analysis to identify opportunities for further cost optimization, ensuring continued profitability and competitive advantage.
41.What is marginal cost pricing?
Answer: Marginal cost pricing involves setting prices equal to the marginal cost to encourage efficient resource allocation and competitive pricing.
42.What is the relationship between marginal cost and marginal benefit?
Answer: The relationship between marginal cost and marginal benefit is that optimal production occurs when marginal cost equals marginal benefit, ensuring efficient resource use and maximum net benefit.
43.What is the marginal cost of producing the 24th car?
Answer: The marginal cost of producing the 24th car is the additional cost incurred by producing that unit, calculated by finding the difference in total cost before and after its production.
44.What is marginal resource cost?
Answer: Marginal resource cost is the additional cost incurred by employing one more unit of a resource, such as labor or raw materials, in the production process.
45.Which curve measures the marginal cost of production?
Answer: The marginal cost curve measures the marginal cost of production, showing how costs change with different levels of output.
46.What does marginal cost mean?
Answer: Marginal cost means the additional cost incurred when producing one more unit of a product. It is a critical concept in cost analysis and production planning.
47.What is marginal cost and benefit?
Answer: Marginal cost is the additional cost of producing one more unit, while marginal benefit is the additional benefit received from consuming one more unit. Optimal decisions are made when these values are equal.
48.Explain the concept of marginal cost. How does it relate to cost? How is it found?
Answer: Marginal cost is the cost of producing one more unit of a product. It is found by dividing the change in total cost by the change in quantity produced. It relates to overall cost by indicating the efficiency of production at different levels.
49.How do you get marginal cost?
Answer: To get marginal cost, calculate the change in total cost for an additional unit of production and divide by the change in quantity produced.
50.How to do marginal cost?
Answer: To calculate marginal cost, determine the change in total cost when production increases by one unit and divide by the change in quantity.
51.What is the relationship between marginal cost and marginal benefit?
Answer: The relationship between marginal cost and marginal benefit is that optimal production occurs when marginal cost equals marginal benefit, ensuring efficient resource use and maximum net benefit.
52.How to get the marginal cost?
Answer: To get the marginal cost, calculate the difference in total costs before and after producing one more unit and divide by the change in quantity.
53.How to calculate marginal benefit and marginal cost?
Answer: To calculate marginal benefit, determine the additional benefit from consuming one more unit. To calculate marginal cost, use the formula: MC = ΔTC / ΔQ where ΔTC is the change in total cost and ΔQ is the change in quantity produced.
54.Refer to the table below. What is the marginal cost of producing the 200th pizza?
Answer: The marginal cost of producing the 200th pizza is found by calculating the difference in total cost before and after producing the 200th pizza and dividing by the change in quantity.
55.How is marginal cost calculated?
Answer: Marginal cost is calculated by dividing the change in total cost (ΔTC) by the change in quantity produced (ΔQ).
56.What is an example of marginal cost?
Answer: An example of marginal cost is a factory producing 100 units at a total cost of $1,000. If producing 101 units increases the total cost to $1,020, the marginal cost of the 101st unit is $20.
57.What is marginal cost formula?
Answer: The marginal cost formula is: Marginal Cost (MC) = ΔTC / ΔQ where ΔTC is the change in total cost and ΔQ is the change in quantity produced.
58.How to find marginal cost calculator?
Answer: To find a marginal cost calculator, use online financial calculators designed for cost analysis. These tools allow you to input changes in total cost and quantity to quickly calculate marginal cost.
59.How to calculate marginal cost?
Answer: To calculate marginal cost, use the formula: MC = ΔTC / ΔQ where ΔTC represents the change in total cost, and ΔQ represents the change in quantity produced.
60.Why is it necessary to know fixed, variable, and total costs to determine marginal cost?
Answer: Knowing fixed, variable, and total costs is necessary to determine marginal cost because it helps in identifying the change in total cost resulting from a change in production level. Fixed costs remain constant, while variable costs change with production levels, impacting the marginal cost calculation.
61.How to get marginal cost from total cost?
Answer: To get marginal cost from total cost, calculate the difference in total costs before and after producing an additional unit and divide by the change in quantity.
62.Which curve(s) does the marginal cost curve intersect at their minimum point?
Answer: The marginal cost curve intersects both the average total cost (ATC) and average variable cost (AVC) curves at their respective minimum points, indicating the most efficient production levels.
63.How to find marginal cost economics?
Answer: In economics, marginal cost is found by calculating the change in total costs between two production levels and dividing by the change in quantity produced.
64.How to calculate marginal cost from total cost?
Answer: To calculate marginal cost from total cost, determine the change in total cost when output increases by one unit and divide by the change in quantity.
65.How to graph marginal cost?
Answer: To graph marginal cost, plot the quantity of output on the horizontal axis and the marginal cost on the vertical axis. Connect the data points to form the marginal cost curve, typically U-shaped, indicating changes in cost efficiency.
66.What is marginal cost equal to?
Answer: Marginal cost is equal to the additional cost incurred by producing one more unit of output. It is calculated by dividing the change in total cost by the change in quantity produced.
67.Average total cost is falling when marginal cost is below it and rising when marginal cost is above it. This relationship indicates that marginal cost influences the direction of average total cost.
Answer: Average total cost is falling when marginal cost is below it and rising when marginal cost is above it. This relationship indicates that marginal cost influences the direction of average total cost.
68.How to solve marginal cost?
Answer: To solve for marginal cost, use the formula: MC = ΔTC / ΔQ Calculate the change in total cost and divide by the change in quantity produced.
69.What is the main difference between marginal revenue and marginal cost?
Answer: The main difference between marginal revenue and marginal cost is that marginal revenue is the additional income from selling one more unit, while marginal cost is the additional cost of producing one more unit. Profit maximization occurs when these two values are equal.
70.How to find marginal average cost?
Answer: To find marginal average cost, calculate the average total cost for different levels of output and determine the change in average cost as output increases.
71.How to compute marginal cost?
Answer: To compute marginal cost, use the formula: MC = ΔTC / ΔQ Calculate the difference in total cost for an additional unit of production and divide by the change in quantity.
72.Why is marginal cost important?
Answer: Marginal cost is important because it helps businesses determine the optimal production level, set pricing strategies, and allocate resources efficiently, ensuring profitability and cost control.
73.When marginal revenue equals marginal cost:
Answer: When marginal revenue equals marginal cost, the firm maximizes its profit, as each additional unit produced neither adds to nor subtracts from total profit.
74.Which best describes the graphical portrayal of marginal cost?
Answer: The graphical portrayal of marginal cost is typically a U-shaped curve, showing initially decreasing costs due to economies of scale, followed by increasing costs due to diminishing returns.
75.What is the marginal cost of producing a fifth soccer net? $1.00 $1.50 $2.00 $2.50
Answer: The marginal cost of producing a fifth soccer net can be calculated by determining the change in total cost before and after producing the fifth net and dividing by the change in quantity.
76.Why is it important to consider marginal benefits and costs when you do a cost-benefit analysis?
Answer: Considering marginal benefits and costs in a cost-benefit analysis ensures that decisions are made based on the additional benefits and costs of one more unit, leading to optimal resource allocation and efficiency.
77.What is the marginal cost of producing the fifth unit of output?
Answer: The marginal cost of producing the fifth unit of output is the additional cost incurred by producing that unit, calculated by finding the difference in total cost before and after its production.
78.What are marginal cost?
Answer: Marginal cost refers to the additional cost incurred when producing one more unit of a product. It is a key metric for cost analysis and production planning.
79.How to find minimum marginal cost?
Answer: To find the minimum marginal cost, analyze the marginal cost curve and identify the lowest point, where producing additional units is most cost-efficient.
80.When the marginal benefit of an output exceeds the marginal cost, producing more will increase overall profit until marginal benefit equals marginal cost.
Answer: When the marginal benefit of an output exceeds the marginal cost, producing more will increase overall profit until marginal benefit equals marginal cost.
81.What does the marginal cost equal for any firm? For any firm, the marginal cost equals the additional cost of producing one more unit of output. It is a crucial factor in production and pricing decisions.
Answer: For any firm, the marginal cost equals the additional cost of producing one more unit of output. It is a crucial factor in production and pricing decisions.
82.How do I calculate marginal cost?
Answer: To calculate marginal cost, use the formula: MC = ΔTC / ΔQ Calculate the difference in total cost for an additional unit of production and divide by the change in quantity.
83.What does marginal cost equal?
Answer: Marginal cost equals the additional cost incurred when producing one more unit of a product. It helps businesses determine optimal production levels and pricing strategies.
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6801 | https://www.wyzant.com/resources/answers/448087/find_two_consecutive_integers_such_that_the_sum_of_their_squares_is_145 | find two consecutive integers such that the sum of their squares is 145 | Wyzant Ask An Expert
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Math Word Problem Help!
Stella G.
asked • 02/11/18
find two consecutive integers such that the sum of their squares is 145
i have a test tmr and cannot figure this problem out to save my life. ive looked at it and have tried to figure out #s that square and equal 145 but i honestly have no idea. HELP!!!
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Naomi S.answered • 02/11/18
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1 st integer = x
2nd integer = x+1
x 2+(x+1)2 =145
x 2 +x 2+2x+1 =145
2x 2+2x=144
2x 2+2x-144=0
x 2+x-72=0
(x+9)(x-8)=0
x=-9 and x=8
so
1 st integer = ±8
2nd integer = ±9
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Scott S.answered • 02/11/18
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Lets call our integers x and y.
How do we represent the "sum of their squares" with a mathematical expression?
This would be x 2+ y 2 and we are told this is equal to 145.
But I can't solve one equation with two variables.x 2 + y 2 = 145 has many solutions. The trick is in how you represent "consecutive integers." They come one after the other such as (4,5), (26,27) etc. How do we get from one number to the next consecutive one? We add one. So if my first number is x, the next consecutive number is x + 1. So lets replace the y in our equation with (x+1)
x 2 + (x+1)2 = 145
Now we have an equation with one variable that I can solve.
Foil first to get x 2+x 2+2x + 1 = 145
2x 2 + 2x +1 = 145
Since this is a quadratic equation we can use the quadratic formula or factor to solve it.
First get everything to one side 2x 2 + 2x - 144 = 0
and you will get that x = 8 or -9
These are the two possible answers for the first consecutive integer x. Then we can figure out the next consecutive integer (y) by adding one to each. So there are two sets of answers, 8 and 9, or -9 and -8. A simple check shows they both will give you 145.
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Scott S.
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The guessing approach may work faster in this particular case. I will just square two consecutive numbers to see what I get. Lets take 4 and 5. 4 2 + 5 2= 16 + 25 = 41. This is too low. Lets try two bigger numbers like 9 and 10. 9 2+ 10 2 - 190. This is too big but we can keep narrowing down until we try 8 and 9. But this won't necessarily give you the other set of answers like above.
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6802 | https://www.map.mathshell.org/lessons.php?taskid=620 | Formative Assessment Lessons
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Using Proportional Reasoning
Mathematical goals
This lesson unit is intended to help you assess how well students are able to reason proportionally when comparing the relationship between two quantities expressed as unit rates and/or part-to-part ratios. In particular, it will help you assess how well students are able to:
Describe a ratio relationship between two quantities.
Compare ratios expressed in different ways.
Use proportional reasoning to solve a real-world problem.
Introduction
The lesson unit is structured in the following way:
Before the lesson, students work individually on an assessment task designed to reveal their current understanding and difficulties. You then review their solutions and create questions for students to consider, in order to improve their work.
After a whole-class introduction, students work in groups, putting diagrams and descriptions of orange and soda mixtures into strength order. Students then compare their work with their peers.
Next, in a whole-class discussion, students critique some sample work stating reasons why two mixtures would or wouldn’t taste the same. Students then revise and correct any misplaced cards.
After a final whole-class discussion, students work individually either on a new assessment task, or return to the original task and try to improve their responses.
Materials Required
Each student will need a mini-whiteboard, pen, and eraser, and a copy of Mixing Drinks and Mixing Drinks (revisited).
Each small group of students will need the cut-up Card Set: Orange and Soda Mixtures and Card Set: Blank Cards, a sheet of poster paper and a glue stick.
You may wish to have some orange juice and soda for mixing/tasting but this is not essential.
Time needed
15 minutes before the lesson, a 100-minute lesson (or two 55-minute lessons), and 15 minutes in a follow-up lesson. Timings given are approximate and will depend on the needs of your class.
Lesson Type
C Concept Development
Mathematical Practices
This lesson involves a range of mathematical practices from the standards, with emphasis on:
MP1:Make sense of problems and persevere in solving them
MP2:Reason abstractly and quantitatively
MP3:Construct viable arguments and critique the reasoning of others
MP4:Model with mathematics
MP5:Use appropriate tools strategically
MP6:Attend to precision
MP7:Look for and make use of structure
MP8:Look for and express regularity in repeated reasoning
Mathematical Content Standards
This lesson asks students to select and apply mathematical content from across the grades, including the content standards:
6.RP:Understand ratio concepts and use ratio reasoning to solve problems.
Resources
Lesson (complete)
proportional reasoning r1.pdf(2008.6K PDF)(2008.6K PDF/Acrobat 10 Jul 2015)
Projector Resources
l080_slides using proportional reasoning - gamma.ppt(1701.5K PPT)(1701.5K MS PowerPoint 09 Apr 2015)
State, district and CCSSI standards appear courtesy of their respective authors.
All other material Copyright © 2007-2015 Mathematics Assessment Resource Service, University of Nottingham. |
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2023 AIME I Problems/Problem 11
Contents
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1 Problem
2 Solution 1 (Minimal Casework)
3 Solution 2
4 Solution 3 (Double Recursive Equations)
5 Solution 4 (Recursive)
6 Solution 5 (Similar to Solution 3)
7 Solution 6 (Stars and Bars)
8 Solution 7 (Fibonacci)
9 Solution 8 (Polyominoes)
10 Solution 9 (Dynamic Programming)
11 Video Solution
12 Video Solution (Mathematical Dexterity)
13 See also
Problem
Find the number of subsets of that contain exactly one pair of consecutive integers. Examples of such subsets are and
Solution 1 (Minimal Casework)
Define to be the number of subsets of that have consecutive element pairs, and to be the number of subsets that have consecutive pair.
Using casework on where the consecutive element pair is, there is a unique consecutive element pair that satisfies the conditions. It is easy to see that
We see that , , and . This is because if the element is included in our subset, then there are possibilities for the rest of the elements (because cannot be used), and otherwise there are possibilities. Thus, by induction, is the th Fibonacci number.
This means that .
~mathboy100
Solution 2
We can solve this problem using casework, with one case for each possible pair of consecutive numbers.
If we have (1,2) as our pair, we are left with the numbers from 3-10 as elements that can be added to our subset. So, we must compute how many ways we can pick these numbers so that the set has no consecutive numbers other than (1,2). Our first option is to pick no more numbers, giving us . We can also pick one number, giving us because 3 cannot be picked. Another choice is to pick two numbers and in order to make sure they are not consecutive we must fix one number in between them, giving us . This pattern continues for each amount of numbers, yielding for 3 numbers and for four numbers. Adding these up, we have + + + + = .
If we have (2,3) as our pair, everything works the same as with (1,2), because 1 is still unusable as it is consecutive with 2. The only difference is we now have only 4-10 to work with. Using the same pattern as before, we have + + + = .
This case remains pretty much the same except we now have an option of whether or not to include 1. If we want to represent this like we have with our other choices, we would say for choosing no numbers and for choosing 1, leaving us with + = 2 choices (either including the number 1 in our subset or not including it). As far as the numbers from 5-10, our pattern from previous cases still holds. We have + + + = 13. With 2 choices on one side and 13 choices on the other side, we have = combinations in all.
Following the patterns we have already created in our previous cases, for the numbers 1-3 we have + = 3 choices (1, 2, or neither) and for the numbers 6-10 we have + + = 8 choices. With 3 choices on one side and 8 choices on the other side, we have = combinations in all.
Again following the patterns we have already created in our previous cases, for the numbers 1-4 we have + + = 5 choices and for the numbers 5-10 we have the same + + = 5 choices. = combinations in all.
By symmetry, the case with (6,7) will act the same as case 4 with (4,5). This goes the same for (7,8) and case 3, (8.9) and case 2, and (9,10) and case 1.
Now, we simply add up all of the possibilities for each case to get our final answer. 34 + 21 + 26 + 24 + 25 + 24 + 26 + 21 + 34 =
-Algebraik
Solution 3 (Double Recursive Equations)
Denote by the number of subsets of a set that consists of consecutive integers, such that each subset contains exactly one pair of consecutive integers.
Denote by the number of subsets of a set that consists of consecutive integers, such that each subset does not contain any consecutive integers.
Denote by the smallest number in set .
First, we compute .
Consider . We do casework analysis.
Case 1: A subset does not contain .
The number of subsets that has exactly one pair of consecutive integers is .
Case 2: A subset contains but does not contain .
The number of subsets that has exactly one pair of consecutive integers is .
Case 3: A subset contains and .
To have exactly one pair of consecutive integers, this subset cannot have , and cannot have consecutive integers in .
Thus, the number of subsets that has exactly one pair of consecutive integers is .
Therefore, for ,
For , we have . For , we have .
Second, we compute .
Consider . We do casework analysis.
Case 1: A subset does not contain .
The number of subsets that has no consecutive integers is .
Case 2: A subset contains .
To avoid having consecutive integers, the subset cannot have .
Thus, the number of subsets that has no consecutive integers is .
Therefore, for ,
For , we have . For , we have .
By solving the recursive equations above, we get .
~ Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution 4 (Recursive)
Let be the number of cases from a subset ranging from - . We approach the subset from the start and the back to build up a equation. We figure out cases if and is in the subset.
If and are both not in the subset, We get .
If is in the subset, and is not, We get multiple cases from this. If is in the subset, would not be in the subset. Now we have {,,,,,} and we'll have to determine how many subsets which there is no consecutive integers. After we do some testing, We get = . From this case we get . But if is not in the subset, We can just get . The same goes for when is in the subset and is not, it is also + .
But when and are both in the subset, We'll have to find more cases. If and are both not in the subset, We simply get . When is in the subset and isn't, We know that is not in the subset since and forms a pair, The sum of this case will be . The same also goes for when is in the subset and isnt. There is no cases for if and are both in the subset, Since that forms pairs.
Summarizing we get: = + + + + .
+ = .
We get a equation of = + + +
We compute the basic cases. = = = =
We get our answer, .
~toub3490
Solution 5 (Similar to Solution 3)
Let be the number of subsets of the set such that there exists exactly 1 pair of consecutive elements. Let be the number of subsets of the set such that there doesn't exist any pair of consecutive elements. First, lets see how we can construct For each subset counted in either: 1. 2. , or 3. and The first case counts subsets (as cannot be included and the rest cannot have any consecutive elements), The second counts and the third counts Thus, Next, Lets try to construct For each subset counted in either: 1. or 2. The first case counts subsets and the second counts Thus, Since and we have that so (The is the th Fibonacci number). From here, we can construct a table of the values of until By listing out possibilities, we can solve for our first 3 values.
Our answer is ~AtharvNaphade
Solution 6 (Stars and Bars)
Note: This is a very common stars and bars application.
Casework on number of terms, let the number of terms be . We can come up with a generalized formula for the number of subsets with n terms.
Let be the differences between the n terms. For example, in the set {2, 3, 6},
Let the range of the set be k for now, . We select one pair of terms to be consecutive by selecting one of the (n-1) terms to be 1. WLOG, let . .
To ensure the other are greater than 1 such that no two other terms are consecutive or the same, let . where are positive integers.
Finally, we add in , the distance between 0 and the first term of the set, and , the distance between the last term and 11. This way, the "distance" from 0 to 11 is "bridged" by , k, and . There are n positive terms, by Balls and Urns, there are ways of doing this. However, recall that there were ways, and we had used a WLOG to choose which two digits are consecutive. The final formula for the number of valid n-element subsets is hence for .
Case 1: Two terms , so
Case 2: Three terms , so
Case 3: Four terms , so
Case 4: Five terms , so
Case 5: Six terms , so
We can check by the Pigeonhole principle that there cannot be more than six terms, so the answer is .
~Mathandski
Solution 7 (Fibonacci)
Note that there are subsets of a set of consecutive integers that contains no two consecutive integers. (This can be proven by induction.)
Now, notice that if we take and as the consecutive integers in our subset, we need to make a subset of the remaining integers such that it doesn't contain any two consecutive integers. Clearly, and cannot be chosen, and since and are sufficiently far apart, it is obvious we do not need to be concerned that an element of the set is consecutive with any element of the set
Thus, we can count the number of ways to choose a subset from the first set without any two elements being consecutive and multiply this by the number of ways to choose a subset from the second set without any two elements being consecutive. From above, and noting that the first set has consecutive integer elements and the second set has consecutive integer elements, we know that this is
Summing this over for all yields
~lpieleanu
Solution 8 (Polyominoes)
The problem is the same as laying out a line of polynomoes to cover spots : 1 triomino (), dominoes (), and monominoes (). The spots cover the members of the subset. The total number spots is 11, because one spot always covers the 0, and the other spots cover 1 through 10.
There are 5 ways to choose polyomino sets, and many ways to order each set:
Polyominoes Orderings
The sum is .
~BraveCobra22aops
Solution 9 (Dynamic Programming)
Let be the number of subsets of a set consecutive integers such that the maximum value in the set is and there exists exactly one pair of consecutive integers. Define similarly, but without any pair of consecutive integers. The base cases are , , and .
The transitions are:
Note that is the Fibonacci numbers.
Summing over yields
~Mathenthus
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution (Mathematical Dexterity)
See also
2023 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10Followed by
Problem 12
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
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The Circle of Parity
What do you think is common between this IPL and last year's ODI World Cup?
Mathematics Assoc, BITS Pilani
Apr 27, 2024
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19th November, 2023
The hearts of millions of cricket fans across India sank as their beloved team, riding high on a remarkable streak of 10 consecutive victories, suffered a loss to Australia in the all important finals of the ICC Men’s Cricket World Cup, 2023. With all factors seemingly aligned in our favor, and fostering hopes to continue the tradition of host countries winning the tournament, witnessing our star players teary eyed was undeniably heart wrenching.
But do you know, it led to the occurrence of an interesting Mathematical phenomenon. The results of the tournament became such that each team in the tournament could be mapped in this flow of a team losing to one and winning to another.
Often referred to as the “Circle of Parity” or the “Parity Problem”, this Circle is more frequently observed in Seasonal Leagues like NFL and UEFA Champions League which follow a round robin / double round robin format. In a round robin format every team plays every other team exactly once (and exactly twice in a double round robin).
Defining formally, Circle Of Parity is basically a circular arrangement of teams in which the teams arranged in the circle have won and lost from each other thus completing a circle. Suppose there are ‘k’ teams; T₁, T₂, T₃, ……, Tₖ ; if a CoP is possible between these teams the circle would start from T₁: T₁ beats T₂, T₂ beats T₃, T₃ beats T₄ , and so on up till Tₖ beats T₁, and a circle would thus complete.
So, coming back to the Cricket World Cup, the tournament structure in its entirety was not exactly Round Robin. Yes, the league stages were so, followed by the two semifinals and the heart wrenching finals. But it led to the creation of an Interesting Problem Statement in our minds.
If an n-team tournament is organized in a round robin format where every team plays every other team exactly once. What is the probability of the formation of at least one CoP ? In other words, out of the ⁿC₂ (n choose 2) matches being played, which can generate a total of 2^(ⁿC₂) combinations of unique tournament results. What is the probability that a CoP will be formed, assuming that for each match, either of the teams have an equal probability to win (also called a Competitive Parity), and assuming that each match does yield a result (we get a result for each match for sure where one team wins and the other losses) ?
To present in a different way, if we represent these teams by mere points on a sheet, then this actually is a Graph Theory problem where we want a complete cycle among all the points (vertices). And such a cycle which traverses all the vertices exactly once indeed has a special name, it is called the Hamiltonian Cycle. There can be a case where there are multiple Hamiltonian cycles in a graph, and a case where there is no Hamiltonian cycle. Also note that the Network we will obtain on plotting all the results of a tournament for our case will be a Directed Network in a complete Graph- since we are dealing with a Round Robin tournament where every team plays every other team once resulting in each vertex being connected to every other vertex, the Graph will be complete and the result of a particular match decides in what direction the edge connecting two nodes is, the Network will be directed. Hence, the probability of tracing a Circle of Parity in an n-team tournament is equivalent to looking at the formation of at least one Hamiltonian cycle in a complete graph in a directed network.
Let’s take a random result for a 5-team tournament, to put things in perspective. Representing the teams as vertices numbered 1, 2, 3, 4 and 5 respectively and the matches between them by the directed edges with the arrow pointing towards the winning team, let’s say the matches occurred in such a way that we could represent it by the above graph. A Hamiltonian cycle (1→3→5→2→4→1) traced by the yellow edges does get formed in this network. Hence, we do get a CoP in this particular 5-team tournament.
Do take a moment to ponder on this and share your approach with us in the comment section on Substack. And Stay Tuned for our upcoming series of articles where we delve into different methodologies to reach the said probability expression and how we use certain algorithms and simulations to substantiate our hypothesis.
Keywords:
ICC on X , Complete Graph, Directed Graph, Hamiltonian Path, Circle of Parity
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Chinmay Ravi Jha
Apr 28, 2024
Out of the ⁿC₂ (n choose 2) matches being played, which can generate a total of 2^(ⁿC₂) combinations of unique tournament results. What is the probability that a CoP will be formed?
Is the answer 50%?
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Glory419 is your go to store for inflation busting high quality brand name pre owned items!! We search EVERYWHERE looking for unique high quality brand name low price items so we can pass on the ...
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Glory419 is your go to store for inflation busting high quality brand name pre owned items!! We search EVERYWHERE looking for unique high quality brand name low price items so we can pass on the saving to YOU the CUSTOMER!! Customer Service is our #1 priority here at GLORY 419 so please feel free to contact us if your not satisfied and we guarantee a solution can be worked out. Glory419 is also open to offers and recommendations!!! If you see something that interests you PLEASE reach out and we will do our best to consider each reasonable offer. Our goal is to bust Inflation and destroy outlandish retail prices. We specialize in Jerseys, Hats, Sports Clothing, Shoes, Collectibles, Fashion, Media and much more. Also, Glory 419 cares about saving the environment. By repurposing items not only are we saving you the CUSTOMER $, but also saving items from the landfill!!! So Please help us bring the GLORY times back and look at our 1300+ plus items.
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id (4444)- Feedback left by buyer. Past month Verified purchase Shipping was very fast, with secure packaging, in excellent condition as described and a good value. Lot Of 5 Magazines From Baseball Hall Of Fame Memories And Dreams Excellent ⚾⚾⚾⚾ (#176273253343)
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6806 | https://micro.magnet.fsu.edu/cells/nucleus/nuclearenvelope.html | | | | | | | | | | | | | |
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6807 | https://www.sciencedirect.com/science/article/pii/0022510X69900100 | Muscular weakness in osteomalacia and hyperparathyroidism - ScienceDirect
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Cited by (69)
Journal of the Neurological Sciences
Volume 8, Issue 3, May–June 1969, Pages 511-520
Muscular weakness in osteomalacia and hyperparathyroidism
Author links open overlay panel R.Smith, G.Stern
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Abstract
A detailed prospective study has been made of the incidence and clinical and biochemical features of proximal muscular weakness in hyperparathyroidism and osteomalacia. In 41 patients with primary hyperparathyroidism, proximal weakness was found only in 1 patient, with a probable parathyroid carcinoma, whereas in 11 osteomalacic subjects there was weakness in at least 8. Investigations showed that the occurrence of weakness was independent of plasma calcium concentration, and no constant biochemical abnormality was found. This confirms the results of a previous study. Weakness always responded to vitamin D therapy, but pre-treatment plasma vitamin D levels were low or normal. It is suggested that myopathy is due to a disturbance of vitamin D metabolism and the reasons for this are discussed.
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References (20)
G.D. Kerr et al.
Hypercalcaemia and gastric hypersecretion in the familial endocrine-adenoma syndrome
Lancet (1967)
E. Kodicek
Turnover and distribution of vitamin D and the mode of action
J. Lund et al.
Biologically active metabolite of vitamin D 3 from bone, liver, and blood serum
J. Lipid Res. (1966)
P.D. Byers et al.
Trephine for full thickness iliac crest biopsy
Brit. med. J. (1967)
D.R. Davies et al.
Tertiary hyperparathyroidism
Brit. med. J. (1968)
C.E. Dent et al.
Hereditary forms of rickets and osteomalacia
J. Bone Jt. Surg. (1956)
C.E. Dent et al.
Radiological changes associated with certain metabolic bone diseases
Brit. J. Radiol. (1954)
K. Ekbom et al.
Weakness of proximal limb muscles probably due to myopathy after partial gastrectomy
Acta med. scand. (1964)
B. Frame et al.
Myopathy in primary hyperparathyroidism
Ann. intern. med. (1968)
F. Glisson
There are more references available in the full text version of this article.
Cited by (69)
MUSCLE WEAKNESS IN OSTEOMALACIA
1976, Lancet Show abstract The muscle weakness that frequently accompanies osteomalacia and rickets may arise from a variety of causes. Particularly in patients with muscle weakness, identification of the metabolic disorder is important, since effective treatment is often possible.
### Vitamin D and its role in skeletal muscle
2013, Calcified Tissue International
### A higher dose of vitamin D reduces the risk of falls in nursing home residents: A randomized, multiple-dose study
2007, Journal of the American Geriatrics Society
### Vitamin D deficiency, muscle function, and falls in elderly people
2002, American Journal of Clinical Nutrition
### Hypovitaminosis D myopathy without biochemical signs of osteomalacic bone involvement
2000, Calcified Tissue International
### Role of vitamin d in skeletal muscle function
1986, Endocrine Reviews
View all citing articles on Scopus
View full text
Copyright © 1969 Published by Elsevier B.V.
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6808 | https://www.khanacademy.org/math/trigonometry/trigonometry-right-triangles/sine-and-cosine-of-complementary-angles/v/pythagorean-trig-identity-from-soh-cah-toa | Intro to the Pythagorean trig identity (video) | Khan Academy
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Trigonometry
Course: Trigonometry>Unit 1
Lesson 5: Sine and cosine of complementary angles
Intro to the Pythagorean trig identity
Sine & cosine of complementary angles
Using complementary angles
Relate ratios in right triangles
Trig word problem: complementary angles
Trig challenge problem: trig values & side ratios
Trig ratios of special triangles
Math>
Trigonometry>
Right triangles & trigonometry>
Sine and cosine of complementary angles
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Intro to the Pythagorean trig identity
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Sal introduces and proves the identity (sinθ)^2+(cosθ)^2=1, which arises from the Pythagorean theorem!Created by Sal Khan.
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Marvin Cohen 11 years ago Posted 11 years ago. Direct link to Marvin Cohen's post “Are arccos and secant the...” more Are arccos and secant the same?
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Steven 11 years ago Posted 11 years ago. Direct link to Steven's post “No, they are completely d...” more No, they are completely different.
sec(x) = 1/cos(x)
y = cos(x) => arccos(y) = x
Arccos(x) is the inverse function of cos(x), whereas sec(x) is the reciprocal of cos(x).
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Leonardo Tenenbaum 12 years ago Posted 12 years ago. Direct link to Leonardo Tenenbaum's post “At 2:08, he says "sin^2(t...” more At 2:08 , he says "sin^2(theta)" why didn't he say "sin(theta)^2"?
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sanurakla 3 years ago Posted 3 years ago. Direct link to sanurakla's post “Does the theta have to al...” more Does the theta have to always bee in that exact place or can it be any unknown angle?
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llama 3 years ago Posted 3 years ago. Direct link to llama's post “It can be any angle. Thet...” more It can be any angle. Theta just means the angle we are evaluating
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Sam 11 years ago Posted 11 years ago. Direct link to Sam's post “Sal writes sin²Θ + cos²Θ ...” more Sal writes sin²Θ + cos²Θ = 1, shouldn't it be (sinΘ)² + (cosΘ)² = 1?
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C4LOwenZ 2 years ago Posted 2 years ago. Direct link to C4LOwenZ's post “At the near end of the vi...” more At the near end of the video, Sal says that this identity is part of the motivation for discovering the unit circle definitions. Why? I don't see a connection.
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Nicholas Bishop 2 years ago Posted 2 years ago. Direct link to Nicholas Bishop's post “According to the Pythagor...” more According to the Pythagorean theorem, the sum of the squares of the lengths of these two sides should equal the square of the length of the hypotenuse:
x² + y² = 1²
But because x = cosθ and y = sinθ for a point (x, y) on the unit circle, this becomes:
(cosθ)² + (sinθ)² = 1
or
cos²θ + sin²θ = 1
So the identity sin²θ + cos²θ = 1 is inherent in the definition of the sine and cosine functions in terms of the unit circle, and it provides a mathematical confirmation of the geometric relationship between the unit circle and right triangles
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redthumb.liberty 10 years ago Posted 10 years ago. Direct link to redthumb.liberty's post “I have a proof if you'd l...” more I have a proof if you'd like to see the complete evolution of the double angle identity:
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BlueLobster 3 years ago Posted 3 years ago. Direct link to BlueLobster's post “Theta is a variable for a...” more Theta is a variable for angles, but shouldn't Theta's real value equal >361?
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Hackcraft_ 2 years ago Posted 2 years ago. Direct link to Hackcraft_'s post “Theta is an angle, and an...” more Theta is an angle, and angles are constrained to 360deg. Any other value greater than 360 can be wrapped around to a value between 0 and 360. Thus, yes, any angle can have infinite ways to represent it. However there are no real or fake values for an angle. It only has countless representations.
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Video transcript
So we've got a right triangle drawn over here where this base's length is a, the height here is b, and the length of the hypotenuse is c. And we already know when we see something like this, we know from the Pythagorean theorem, the relationship between a, b, and c, we know there's a squared plus b squared is going to be equal to the hypotenuse squared, is going to be equal to c squared. What I want to do in this video is explore how we can relate trig functions to, essentially, the Pythagorean theorem. And to do that, let's pick one of these non-right angles. So let's pick this angle right over here as theta, and let's just think about this what the sine of theta is and what the cosine of theta is, and see if we can mess with them a little bit to somehow leverage the Pythagorean theorem. So before we do that, let's just write down sohcahtoa just so we remember the definitions of these trig functions. So sine is opposite over hypotenuse. Cah, cosine is adjacent over hypotenuse. And toa, tan is opposite over adjacent, we won't be using tan, at least in this video. So let's think about sine of theta. I will do it, I'll do it in this blue color. So sine of theta is what? It is opposite over hypotenuse, so it is equal to the length of b or it is equal to b-- b is the length-- b over the length of the hypotenuse, which is c. Now what is cosine of theta? Well, the adjacent side, the side of this angle that is not the hypotenuse, it has length a. So it's the length of the adjacent side over the length of the hypotenuse. Now how could I relate these things? Well it seems like, if I square sine of theta, then I'm going to have sine squared theta is equal to b squared over c squared, and cosine squared theta is going to be a squared over c squared. Seems like I might be able to add them to get something that's pretty close to the Pythagorean theorem here. So let's try that out. So sine squared theta is equal to b squared over c squared. I just squared both sides. Cosine squared theta is equal to a squared over c squared. So what's this sum? What's sine squared theta plus cosine squared theta? Is going to be equal to what? Sine squared theta is b squared over c squared, plus a squared over c squared, which is going to be equal to-- Well we have a common denominator of c squared. And the numerator, we have b squared plus a squared. Now, what is b squared plus a squared? Well, we have it right over here, Pythagorean theorem tells us, b squared plus a squared or a squared plus b squared is going to be equal to c squared. So this numerator simplifies to c squared. And the whole expression is c squared over c squared, which is just equal to 1. So using the sohcahtoa definition, in a future video, we'll use the unit circle definition. But you see just using the units, just even using the sohcahtoa definition of our trig functions, we see probably the most important of all the trig identities. That the sine squared theta, sine squared of an angle, plus the cosine squared of that same angle-- I'm introducing orange unnecessarily-- is going to be equal to 1. Now you might probably be saying, OK Sal, that's kind of cool, but what's the big deal about this? Why should I care about this? Well, the big deal is now you give me the sine of an angle and I can solve this equation for the cosine of that angle, or vice versa. So this is actually a pretty powerful, powerful thing. And this is also part of the motivation even for the unit circle definition of trig functions.
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6809 | http://www.360doc.com/content/18/0503/13/52760486_750765825.shtml | 二次函数最大值，最小值，有几种求法？
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二次函数最大值，最小值，有几种求法？
昵称vMN36 2018-05-03 来源|170阅读|1 转藏
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一般来说,如果这个一元二次函数的定义域是R的话:
(1)函数开口向上,即a>0时,则没有最大值,只有最小值,即函数的顶点,可用函数的顶点公式:(-b/2a,(4ac-b^2)/4a)来求.
(2)函数开口向上,即a<0时,则没有最小值,只有最大值,求法同上.
若该函数的定义域不是R的话:
(1)函数开口向上,即a>0时:
①当-b/2a在定义域内时,有最小值,再看定义域区间
假设是闭区间[m,n],若-b/2a>(n m)/2,则最大值是x=m时的函数值,若-b/2a<(n m)/2,则相反,若两者相同,则最大值即是端点值.
当定义域区间是开区间(m,n)时,则无最大值
还有就是区间是半开半闭的情况时,即[m,n)或(m,n]时,按上面闭区间的方法计算,但若x取不到,则没有最大值
②当-b/2a不在定义域内时,
假设是闭区间[m,n],则最小值和最小值就是两个端点值,算一下再比较大小就行
当定义域区间是开区间(m,n)时,则无最大最小值
当区间是半开半闭的情况,即[m,n)或(m,n]时,按上面闭区间的方法计算,关键是看能不能取到,但肯定是只有一个最值的
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Takeoff and Landing Performance: Key Calculations Explained
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AS 310 (04) - FALL 2023 HOMEWORK #6- Lesson 1 3, 14(Takeoff & Landing Performance) YOUR N AME: _________ DUE: Friday, N ovember 17 th @11:59 pm Symbols : a Acceleration or deceleration (ft/sec 2 ) F Rolling friction force (lb) F N Net accelerating or decelerating force (lb) D Total drag (lb) g acceleration due to gravity (ft/sec 2) = 32.2 ft/sec 2 L Lift (lb) S f Flare distance (feet) S gl Final glide distance (feet) S gr Ground roll distance (feet) S r Rotation distance (feet) t Time (sec) T Thrust (lb) T A Thrust available (lb) T R Thrust required (lb) V Velocity or Airspeed (knots), to convert to ft/sec multiply by 1.69 V ref Final Approach Airspeed (knots), to convert to ft/sec multiply by 1.69 V r Rotation Airspeed (knots), to convert to ft/sec multiply by 1.69 V s Stall Speed (knots), to convert to ft/sec multiply by 1.69 V t Transition Airspeed (knots), to convert to ft/sec multiply by 1.69 V TD Touch down Airspeed (knots), to convert to ft/sec multiply by 1.69 1) Ground roll 2) Rotation (nose wheel off the ground) 3) Transition (main gear off the ground) 4) Climb to clear 35-foot obstacle height (constant pitch attitude) 1) Final "glide" (50-foot obstacle cleared to flare) 2) Flare (beginning of level off to main gear touchdown) 3) Rotation (touchdown to nose gear down) 4) Ground roll (all gear down to stop) Stephanie Cook
W Aircraft Weight (lb) g Glide angle m Coefficient of friction for takeoff, or braking coefficient of friction for landing Equations : Takeoff forces: F n = T - D - F = ma F= m (W-L) Takeoff ground roll: Landing forces: F n= T - D - F = ma
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F= m (W-L) Final glide distance: Landing flare distance: Landing rotation distance: Landing ground roll distance: Total landing distance:
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Takeoff Segments:
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Landing Segments:
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Questions: 1. If the net accelerating force (F N ) on an airplane's takeoff roll is 5,000 lbs., calculate the rolling friction force (F) given the following information: Thrust = 7,000 lbs., Drag = 1,560 lbs. 2. An airplane's rotation speed (V r ) is 110 kts (multiply knots by 1.69 to convert speed to ft/sec). If the airplane's constant acceleration is 10 ft/sec 2 , calculate the airplane's ground roll distance (S gr ) in feet from a brakes-locked position. 3. A 10,000-lb aircraft accelerates down a runway for takeoff. Assuming Vr = 110 x 1.69 = 185.9 ft/sec S = V^2 / 2a = 185.9^2 / 2 x 10 = 34,558.81 / 20 = 1727.94 ft Fn = T - D - F 5000 = 7000 - 1560 - F 5000 = 5440 - F F = 440 lbs
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6811 | https://arxiv.org/html/2507.03208v1 | The Dependently Typed Higher-Order Form for the TPTP World
1 Introduction
2 Preliminaries
2.1 The TPTP World and Infrastructure
2.2 Dependently Typed Higher-Order Logic
Dependent Connectives
2.2.1 Polymorphic DHOL.
2.2.2 Choice.
2.2.3 Translation.
3 DTF
3.1 Syntax
3.2 Type Checking
3.3 Semantics
Standard Models.
General Models.
Models for Polymorphic DHOL.
4 Problem Dataset
5 Tools
5.1 The Logic Embedding Tool
5.2 DLash
5.3 MMT
5.4 TPTP Systems
6 Conclusion
1 1 institutetext: University of Innsbruck, Computational Logic, Austria
1 1 email: d.ranalter@gmail.com 2 2 institutetext: University of Melbourne, School of Computing, Australia
2 2 email: ckaliszyk@unimelb.edu.au 3 3 institutetext: University of Erlangen-Nuremberg, Computer Science, Germany
3 3 email: florian.rabe@fau.de 4 4 institutetext: University of Miami, Department of Computer Science, USA
4 4 email: geoff@cs.miami.edu
The Dependently Typed Higher-Order Form
for the TPTP World
Daniel Ranalter 11 0009-0006-2861-548XCezary Kaliszyk 2211 0000-0002-8273-6059Florian Rabe 33 0000-0003-3040-3655Geoff Sutcliffe 44 0000-0001-9120-3927
Abstract
Much of the current research and development in the field of automated reasoning builds on the infrastructure provided by the TPTP World. The TPTP language for logical formulae is central to the far-reaching adoption of the TPTP World. This paper introduces the Dependently Typed higher-order Form (DTF) of the TPTP language. It takes advantage of already established binders in the syntax, and is thus a minimally intrusive extension to the Typed Higher-order Form (THF). A starting set of over 100 problems is provided to exhibit the usefulness and incite interest in DTF. Some tools that are already able to reason about problems in the DTF language are discussed.
Keywords:
Automated Theorem Proving Dependent Types Higher-Order Logic.
1 Introduction
The TPTP World is a well-established infrastructure that supports research, development, and deployment of Automated Theorem Proving (ATP) systems. The TPTP language is one of the keys to the success of the TPTP World. It has variants that support uniform expression of logical formulae across a wide range of logics. The TPTP language is used for writing both problems and solutions, which enables convenient communication between ATP systems and tools. The majority of modern ATP systems accept input in TPTP syntax. The TPTP language variants that form the basis for this work are the monomorphic and polymorphic typed higher-order forms (TH0 and TH1)[32, 8] (see Section2.1 for the background and further variants).
All the existing typed TPTP language variants are simply typed. However, there is a steady increase of interest in dependently typed systems, such as Agda, Rocq[1, 37], and Lean. This interest extends to the SMT community, where the proposed version 3.0 of SMT-LIB is to include dependent types 1 1 1smt-lib.org/version3. Dependent types allow for the elegant formulation of complex data structures, possibly even a direct encoding of correctness properties. This paper introduces the Dependently Typed higher-order Form (DTF) of the TPTP language.
While dependent types are frequently used in interactive theorem proving, Automated Theorem Proving (ATP) has yet to embrace dependent types. Rothgang et al. made first steps towards bringing ATP and dependent types together, by introducing dependently typed higher-order logic (DHOL)[17, 18]. With only two minor extensions to the familiar syntax of Church-style HOL, DHOL makes dependent types easily accessible: HOL base types are extended into dependent base types that can take term arguments, and the function type A→B→𝐴 𝐵 A\rightarrow B italic_A → italic_B is changed into a dependent function type Π x:A.B\Pi x:A.B roman_Π italic_x : italic_A . italic_B. Originally DHOL did not allow quantifying over types or stating the equality of types, but a polymorphic version is in development.
As in FOL and HOL, DHOL allows arbitrary axioms that may constrain equality of terms in undecidable ways, and consequently DHOL’s type checking is undecidable (see Section3.2). To manage this complication Rothgang et al. provide an algorithm that reduces the well-formedness of a statement to a set of proof obligations. Thus theorem proving is needed to check the well-formedness of a problem’s formulae, not just to prove the conjecture. Happily, typically that does not make it harder to prove the conjecture. To increase ATP support for DHOL, Rothgang et al. define a translation from well-typed DHOL to HOL that preserves provability in both directions, thereby making DHOL available for regular HOL ATP systems, albeit without leveraging DHOL’s dependent types for more efficient proving. Furthermore, the translation introduces additional axioms capturing the constraints of the dependent types, thereby potentially complicating proof search. Several interactive theorem provers had previously employed the same idea, sacrificing decidable typing to gain the expressivity of dependent types, while keeping the general feel of the language simple. Most importantly, PVS essentially contains DHOL as a fragment, but extends it beyond the capabilities of current automated provers. Mizar, using soft typing on top of first-order set theory, can also capture DHOL-like features.
A detail missing from the original formulation of DHOL was the choice operator. Ranalter et al. investigated the effects of losing the non-emptiness constraint in DHOL on Hilbert’s choice in . To this end, they extended the – to the authors knowledge – first native implementation of DHOL into the ATP system Lash, by Niederhauser et al.. Their experiments strongly suggest that native reasoning in DHOL significantly outperforms reasoning on translated problems.
This work describes how DHOL is being integrated into the TPTP World, in a new TPTP language variant “Dependently Typed higher-order Form” (DTF), with monomorphic and polymorphic subvariants (DT0 and DT1). DTF requires only very minor changes to the familiar TPTP language syntax, mostly using existing notions for binders and application operators, thereby providing the ATP community with the necessary foundations on which research into dependently typed automated reasoning can thrive. A set of over 100 problems in DTF, taken from several different sources, has been curated as an initial contribution to the TPTP problem library. The problems provide a spread of interesting formulations focusing on a variety of difficulty levels in proving the conjecture as well as in type checking.
Section2 reviews the TPTP World and establishes the necessary background for DHOL, slightly generalizing the original DHOL definition to make it more suitable for TPTP. Section3 introduces the new DTF form. Section4 gives a short overview of the starting set of problems, and Section5 introduces tools that already support the new form. Finally, Section6 concludes and gives an outlook over future work.
2 Preliminaries
2.1 The TPTP World and Infrastructure
The TPTP World infrastructure includes the TPTP language, the TPTP problem library, the TSTP solution library, the SZS ontologies, the Specialist Problem Classes (SPCs) and problem difficulty ratings, SystemOnTPTP and StarExec, and the CADE ATP System Competition (CASC). The problem library is a large collection of Thousands of Problems for Theorem Proving – hence the name. The problem library release v9.1.0 contains over 26000 problems from over 50 different domains, written in the TPTP language. The problems are categorized into Specialist Problem Classes according to their syntactic and logical status. The TSTP solution library is the result of running numerous ATP systems on the problems in that library and collecting their output. The TPTP and TSTP libraries provide the basis for assigning a difficulty rating to each problem, according to which ATP systems are able to solve the problem.
The most salient feature of the TPTP World for this work is the TPTP language. Originally the TPTP language supported only first-order clause normal form (CNF). Over time, more complex logics were added, starting with first-order form (FOF) in TPTP release v2.0.0. Releases v3.0.0 and v4.0.0 added monomorphic typed higher-order (TH0) and monomorphic typed first-order (TF0) forms to the mix respectively. These got extended to their polymorphic variants TF1 and TH1 in releases v5.0.0 and v6.0.0. Release v7.0.0 of the TPTP started to include extended typed first-order form (TXF) which extends the typed first-order form with conditionals, let expressions, and boolean terms. All the listed extensions to the TPTP are classical in nature. This changed with the addition of non-classical typed first-order form (NTF) in release v9.0.0. A general principle of the TPTP language is: “We provide the syntax, you provide the semantics”. As such, there is no a priori commitment to any semantics for each of the language forms, although in almost all cases the intended logic and semantics are well known.
Problems and solutions are built from annotated formulae of the form
language(name,role,formula,source,useful_info)
The language s supported are cnf (clause normal form), fof (first-order form), tff (typed first-order form), and thf (typed higher-order form). The role, e.g., axiom, lemma, conjecture, defines the use of the formula. In a formula, terms and atoms follow Prolog conventions – functions and predicates start with a lowercase letter or are ’single quoted’, and variables start with an uppercase letter. The language also supports interpreted symbols that either start with a $, e.g., the truth constants $true and $false, or are composed of non-alphabetic characters, e.g., integer/rational/real numbers such as 27, 43/92, -99.66. The logical connectives in the TPTP language are !>, ?, @+, @-, !, ?, ~, |, &, =>, <=, <=>, and <~>, for the mathematical connectives Π Π\Pi roman_Π, Σ Σ\Sigma roman_Σ, choice (indefinite description), definite description, ∀for-all\forall∀, ∃\exists∃, ¬\neg¬, ∨\vee∨, ∧\wedge∧, ⇒⇒\Rightarrow⇒, ⇐⇐\Leftarrow⇐, ⇔⇔\Leftrightarrow⇔, and ⊕direct-sum\oplus⊕ respectively. Equality and inequality are expressed as the infix operators = and !=. The source and useful_info are optional.
2.2 Dependently Typed Higher-Order Logic
Dependently typed higher-order logic (DHOL) is an extension of Church’s higher-order logic (HOL) introduced by Rothgang et al.. It takes the widely supported HOL and equips it with dependent types, i.e., types that take term arguments. As such, it is a classical and extensional type theory, as opposed to the theory used in Rocq[1, 37], Lean, or others[3, 13] that rely on an intensional type theory. Notable exceptions to this trend are PVS, NuPRL, and F.
The extensionality of DHOL comes at the cost of making type checking undecidable because it must consider term equality, which may be subject to arbitrary axioms. Essentially, typing becomes undecidable if a type depends on a type for which equality is undecidable. This is because type checking t 𝑡 t italic_t against type an 𝑎 𝑛 a\ n italic_a italic_n must be done by inferring the type of t 𝑡 t italic_t, say am 𝑎 𝑚 a\ m italic_a italic_m, and then checking am=an 𝑎 𝑚 𝑎 𝑛 a\ m=a\ n italic_a italic_m = italic_a italic_n, and thus m=n 𝑚 𝑛 m=n italic_m = italic_n. If all dependent type symbols depend only on types for which equality is decidable (e.g., the examples below where we only use natural numbers with Presburger arithmetic), type checking is decidable. Otherwise, e.g., when using types depending on natural numbers with Peano arithmetic, type checking is undecidable.
The gain of having judgmental and provable equality coincide is significant: It positions DHOL much closer to how mathematics is usually done in the context of ATP. The availability of dependent types allows the elegant definition of data structures such as lists of fixed-length, intervals of numbers, or vector spaces over some field. It also allows encoding constraints in the types, which can remove the need for lengthy and error-prone guards in programming and track invariants useful for theorem proving. The cost – which might seem steep at first glance – is mitigated by the ever-increasing performance of ATP systems, and the fact that in many cases the proof obligations resulting from type checking are much simpler than the original proving problem.
The changes to the TPTP syntax to accommodate DTF are small: the definition of the simple base type is changed to a type that can accept term arguments, and the simple function type A→B→𝐴 𝐵 A\rightarrow B italic_A → italic_B is changed to Π x:A.B\Pi x:A.B roman_Π italic_x : italic_A . italic_B. This makes it possible to let the result type of the function depend on the specific term of the argument.
Figure1 gives the grammar of DHOL. A dependent base type a 𝑎 a italic_a with arity n 𝑛 n italic_n is written a:Π x 1:A 1,⋯,x n:A n.𝚝𝚢𝚙𝚎 a:\Pi x_{1}:A_{1},\cdots,x_{n}:A_{n}.\mathtt{type}italic_a : roman_Π italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT : italic_A start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , ⋯ , italic_x start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT : italic_A start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT . typewriter_type, and it is a simple base type if n=0 𝑛 0 n=0 italic_n = 0. Declarations of this form are part of the theory against which the type checking procedure is performed. In addition to base type declarations, theories may declare constant symbols c 𝑐 c italic_c and axioms ▷F▷𝐹\triangleright\;F▷ italic_F. A context specifies typed variables and assumptions. Contexts are superficially similar to theories, but denote local declarations, and as such, do not contain type declarations. ∘\circ∘ and ∙∙\bullet∙ denote the empty theory and context respectively. The order in a theory or context matters because the well-typedness of declarations might depend on preceding axioms. Types, as they appear in statements and typing judgements, are either fully applied base types, (dependent) function types, or classical booleans o 𝑜 o italic_o. Terms are built from variables/constants, lambda abstraction, application, and the usual connectives and quantifiers. Regular HOL can be recovered by omitting the highlighted elements – this is exactly the case when the arity of all base types is 0.
T,U 𝑇 𝑈 T,U italic_T , italic_U::=∘\circ∘ | T,a:(Π x:A.)∗𝚝𝚢𝚙𝚎 T,\;a:{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}(\Pi x% :A.)^{}}\mathtt{type}italic_T , italic_a : ( roman_Π italic_x : italic_A . ) start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT typewriter_type | T,c:A:𝑇 𝑐 𝐴 T,\;c:A italic_T , italic_c : italic_A | T,▷F 𝑇▷𝐹 T,\;\triangleright\;F italic_T , ▷ italic_F theories
Γ,Δ Γ Δ\Gamma,\Delta roman_Γ , roman_Δ::=∙∙\bullet∙ | Γ,x:A:Γ 𝑥 𝐴\Gamma,\,x:A roman_Γ , italic_x : italic_A | Γ,▷F Γ▷𝐹\Gamma,\,\triangleright\;F roman_Γ , ▷ italic_F context
A,B 𝐴 𝐵 A,B italic_A , italic_B::=at 1⋯t n 𝑎 subscript 𝑡 1⋯subscript 𝑡 𝑛 a\,{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}t_{1}\,% \cdots\,t_{n}}italic_a italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ⋯ italic_t start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT | Π x:A.B{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\Pi x:}A.B roman_Π italic_x : italic_A . italic_B | o 𝑜 o italic_o types
t,u,F,G 𝑡 𝑢 𝐹 𝐺 t,u,F,G italic_t , italic_u , italic_F , italic_G::=x 𝑥 x italic_x | c 𝑐 c italic_c | λ x:A.t\lambda x:A.t italic_λ italic_x : italic_A . italic_t | tu 𝑡 𝑢 t\,u italic_t italic_u | ∀x:A.F\forall x:A.F∀ italic_x : italic_A . italic_F | ∃x:A.F\exists x:A.F∃ italic_x : italic_A . italic_F | F⇒G⇒𝐹 𝐺 F\Rightarrow G italic_F ⇒ italic_G
|F∧G 𝐹 𝐺 F\land G italic_F ∧ italic_G | F∨G 𝐹 𝐺 F\lor G italic_F ∨ italic_G | ⊥bottom\bot⊥ | ⊤top\top⊤ | ¬F 𝐹\neg F¬ italic_F | t=A u subscript 𝐴 𝑡 𝑢 t=_{A}u italic_t = start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT italic_u terms (incl. formulae F,G 𝐹 𝐺 F,G italic_F , italic_G)
Figure 1: The grammar of DHOL
The following example encodes the familiar notion of fixed-length lists. As prerequisites, we give the usual notion of natural numbers in a simple type 𝚗𝚊𝚝 𝚗𝚊𝚝\mathtt{nat}typewriter_nat and a simple type 𝚌𝚑𝚊𝚛 𝚌𝚑𝚊𝚛\mathtt{char}typewriter_char of characters for the elements of the lists:
𝚗𝚊𝚝:𝚝𝚢𝚙𝚎 0:𝚗𝚊𝚝 𝚜𝚞𝚌:𝚗𝚊𝚝→𝚗𝚊𝚝+:𝚗𝚊𝚝→𝚗𝚊𝚝→𝚗𝚊𝚝\displaystyle\mathtt{nat}:\mathtt{type}\qquad 0:\mathtt{nat}\qquad\mathtt{suc}% :\mathtt{nat}\rightarrow\mathtt{nat}\qquad\mathtt{+}:\mathtt{nat}\rightarrow% \mathtt{nat}\rightarrow\mathtt{nat}typewriter_nat : typewriter_type 0 : typewriter_nat typewriter_suc : typewriter_nat → typewriter_nat + : typewriter_nat → typewriter_nat → typewriter_nat
▷∀n:𝚗𝚊𝚝.+ 0 n=𝚗𝚊𝚝 n▷∀n,m:𝚗𝚊𝚝.+(𝚜𝚞𝚌 n)m=nat 𝚜𝚞𝚌(+n m)\displaystyle\triangleright\;\forall n:\mathtt{nat}.\mathtt{+}\;0\;n={\mathtt% {nat}}n\qquad\triangleright\;\forall n,m:\mathtt{nat}.\mathtt{+}\;(\mathtt{suc% }\;n)\;m={nat}\mathtt{suc}\;(\mathtt{+}\;n\;m)▷ ∀ italic_n : typewriter_nat . + 0 italic_n = start_POSTSUBSCRIPT typewriter_nat end_POSTSUBSCRIPT italic_n ▷ ∀ italic_n , italic_m : typewriter_nat . + ( typewriter_suc italic_n ) italic_m = start_POSTSUBSCRIPT italic_n italic_a italic_t end_POSTSUBSCRIPT typewriter_suc ( + italic_n italic_m )
𝚌𝚑𝚊𝚛:𝚝𝚢𝚙𝚎 𝚊:𝚌𝚑𝚊𝚛 𝚋:𝚌𝚑𝚊𝚛…:𝚌𝚑𝚊𝚛 𝚝𝚢𝚙𝚎 𝚊:𝚌𝚑𝚊𝚛 𝚋:𝚌𝚑𝚊𝚛…\displaystyle\mathtt{char}:\mathtt{type}\qquad\mathtt{a}:\mathtt{char}\qquad% \mathtt{b}:\mathtt{char}\qquad...typewriter_char : typewriter_type typewriter_a : typewriter_char typewriter_b : typewriter_char …
Then 𝚟𝚎𝚌n 𝚟𝚎𝚌 𝑛\mathtt{vec}\,n typewriter_vec italic_n encodes the type of fixed-length lists of characters of length n 𝑛 n italic_n:
𝚟𝚎𝚌:Π n:𝚗𝚊𝚝.𝚝𝚢𝚙𝚎 𝚗𝚒𝚕:𝚟𝚎𝚌 0 𝚌𝚘𝚗𝚜:Π x:𝚗𝚊𝚝.𝚌𝚑𝚊𝚛→𝚟𝚎𝚌 n→𝚟𝚎𝚌(𝚜𝚞𝚌 n)\displaystyle\mathtt{vec}:\Pi n:\mathtt{nat}.\mathtt{type}\quad\mathtt{nil}:% \mathtt{vec}\;0\quad\mathtt{cons}:\Pi x:\mathtt{nat}.\mathtt{char}\rightarrow% \mathtt{vec}\;n\rightarrow\mathtt{vec}\;(\mathtt{suc}\ n)typewriter_vec : roman_Π italic_n : typewriter_nat . typewriter_type typewriter_nil : typewriter_vec 0 typewriter_cons : roman_Π italic_x : typewriter_nat . typewriter_char → typewriter_vec italic_n → typewriter_vec ( typewriter_suc italic_n )
++:Π n,m:𝚗𝚊𝚝.𝚟𝚎𝚌 n→𝚟𝚎𝚌 m→𝚟𝚎𝚌(+n m)\displaystyle\mathtt{++}:\Pi n,m:\mathtt{nat}.\mathtt{vec}\;n\rightarrow% \mathtt{vec}\;m\rightarrow\mathtt{vec}\;(\mathtt{+}\;n\;m)+ + : roman_Π italic_n , italic_m : typewriter_nat . typewriter_vec italic_n → typewriter_vec italic_m → typewriter_vec ( + italic_n italic_m )
Dependent Connectives
In DHOL it is desirable to make the binary connectives conjunction, implication, and disjunction dependent in the sense that the well-formedness of the second argument may assume the truth (for conjunction and implication) or the falsity (for disjunction) of the first argument. Consider the statement a=A b⇒fa=B(a)fb subscript 𝐴 𝑎 𝑏⇒𝑓 𝑎 subscript 𝐵 𝑎 𝑓 𝑏 a={A}b\Rightarrow f>a={B(a)}f>b italic_a = start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT italic_b ⇒ italic_f italic_a = start_POSTSUBSCRIPT italic_B ( italic_a ) end_POSTSUBSCRIPT italic_f italic_b. The well-formedness of the right-hand side requires the left-hand side as a premise. More precisely, Γ⊢F:o proves Γ 𝐹:𝑜\Gamma\vdash F:o roman_Γ ⊢ italic_F : italic_o resp. Γ⊢F proves Γ 𝐹\Gamma\vdash F roman_Γ ⊢ italic_F expresses that F 𝐹 F italic_F is a well-formed resp. provable formula in context Γ Γ\Gamma roman_Γ. The definition of well-formed formulae is:
Γ⊢F⇒G proves Γ⇒𝐹 𝐺\Gamma\vdash F\Rightarrow G roman_Γ ⊢ italic_F ⇒ italic_G if Γ⊢F proves Γ 𝐹\Gamma\vdash F roman_Γ ⊢ italic_F and Γ,▷F⊢G proves Γ▷𝐹 𝐺\Gamma,\,{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \triangleright\;F}\vdash G roman_Γ , ▷ italic_F ⊢ italic_G
Γ⊢F∧G proves Γ 𝐹 𝐺\Gamma\vdash F\wedge G roman_Γ ⊢ italic_F ∧ italic_G if Γ⊢F proves Γ 𝐹\Gamma\vdash F roman_Γ ⊢ italic_F and Γ,▷F⊢G proves Γ▷𝐹 𝐺\Gamma,\,{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \triangleright\;F}\vdash G roman_Γ , ▷ italic_F ⊢ italic_G
Γ⊢F∨G proves Γ 𝐹 𝐺\Gamma\vdash F\vee G roman_Γ ⊢ italic_F ∨ italic_G if Γ⊢F proves Γ 𝐹\Gamma\vdash F roman_Γ ⊢ italic_F and Γ,▷¬F⊢G proves Γ▷𝐹 𝐺\Gamma,\,{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}% \triangleright\;\neg F}\vdash G roman_Γ , ▷ ¬ italic_F ⊢ italic_G
where the marked parts make the connectives dependent. The usual natural deduction proof rules of implication and conjunction are the same as for the non-dependent versions. The proof rules for disjunction are adjusted as follows:
Γ⊢F Γ,▷¬F⊢G:o Γ⊢F∨G Γ,▷¬F⊢G Γ⊢F∨G Γ,▷F⊢C Γ,▷¬F,▷G⊢C Γ,▷F∨G⊢C proves Γ 𝐹 Γ▷𝐹 proves 𝐺:𝑜 proves Γ 𝐹 𝐺 proves Γ▷𝐹 𝐺 proves Γ 𝐹 𝐺 proves Γ▷𝐹 𝐶 Γ▷𝐹▷𝐺 proves 𝐶 proves Γ▷𝐹 𝐺 𝐶\frac{\Gamma\vdash F\quad\Gamma,\,{\color[rgb]{0,0,1}\definecolor[named]{% pgfstrokecolor}{rgb}{0,0,1}\triangleright\;\neg F}\vdash G:o}{\Gamma\vdash F% \vee G}\quad\frac{\Gamma,\,{\color[rgb]{0,0,1}\definecolor[named]{% pgfstrokecolor}{rgb}{0,0,1}\triangleright\;\neg F}\vdash G}{\Gamma\vdash F\vee G% }\quad\frac{\Gamma,\,\triangleright\;F\vdash C\quad\Gamma,\,{\color[rgb]{0,0,1% }\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}\triangleright\;\neg F},\,% \triangleright\;G\vdash C}{\Gamma,\,\triangleright\;F\vee G\vdash C}divide start_ARG roman_Γ ⊢ italic_F roman_Γ , ▷ ¬ italic_F ⊢ italic_G : italic_o end_ARG start_ARG roman_Γ ⊢ italic_F ∨ italic_G end_ARG divide start_ARG roman_Γ , ▷ ¬ italic_F ⊢ italic_G end_ARG start_ARG roman_Γ ⊢ italic_F ∨ italic_G end_ARG divide start_ARG roman_Γ , ▷ italic_F ⊢ italic_C roman_Γ , ▷ ¬ italic_F , ▷ italic_G ⊢ italic_C end_ARG start_ARG roman_Γ , ▷ italic_F ∨ italic_G ⊢ italic_C end_ARG
As usual, it is possible to choose some connectives as primitives, from which the others are defined. Rothgang et al. choose equality and implication. Contrary to HOL, they included implication because they could not define the dependent binary connectives solely from equality. For the TPTP World, it is better not to choose primitive connectives – that choice should be left to the ATP system developers. Therefore DTF extends the work by Rothgang et al. to make all connectives primitive. ATP systems can choose which connectives to treat as abbreviations, but in doing so must take the dependent nature of the connectives into account. Note that dependent connectives break the commutativity of conjunction and disjunction. While seemingly disruptive, sacrificing commutativity in this way is common practice, e.g., for short-circuit evaluation of Boolean terms in programming languages. To clarify the impact on theorem proving, Table1 summarizes typical proof rules for FOL and their status in DHOL. Roughly speaking, all rules that do not affect the order of subformulae remain sound, while the rest of the rules require the additional check to ensure the result remains well-formed. In particular, all rules needed to perform CNF or clause normal form transformations remain available.
Table 1: Typical proof rules for FOL and their status in DHOL
| Rule | Holds in DHOL |
| For disjunction and conjunction |
| associativity | ✓ |
| commutativity | Only if both sides are well-formed |
| idempotence, e.g., A∧X∧A⇔A∧X⇔𝐴 𝑋 𝐴 𝐴 𝑋 A\wedge X\wedge A\Leftrightarrow A\wedge X italic_A ∧ italic_X ∧ italic_A ⇔ italic_A ∧ italic_X | ✓(Drop the second occurrence) |
| de Morgan laws | ✓ |
| distributivity of one over the other | ✓ |
| absorption, e.g., A∧(A∨B)⇔A⇔𝐴 𝐴 𝐵 𝐴 A\wedge(A\vee B)\Leftrightarrow A italic_A ∧ ( italic_A ∨ italic_B ) ⇔ italic_A | ✓ |
| For implication |
| A⇒B⇔¬A∨B⇔⇒𝐴 𝐵 𝐴 𝐵 A\Rightarrow B\Leftrightarrow\neg A\vee B italic_A ⇒ italic_B ⇔ ¬ italic_A ∨ italic_B | ✓ |
| ¬(A⇒B)⇔A∧¬B⇔⇒𝐴 𝐵 𝐴 𝐵\neg(A\Rightarrow B)\Leftrightarrow A\wedge\neg B¬ ( italic_A ⇒ italic_B ) ⇔ italic_A ∧ ¬ italic_B | ✓ |
| ¬(A⇒B)⇔¬B⇒¬A⇔⇒𝐴 𝐵⇒𝐵 𝐴\neg(A\Rightarrow B)\Leftrightarrow\neg B\Rightarrow\neg A¬ ( italic_A ⇒ italic_B ) ⇔ ¬ italic_B ⇒ ¬ italic_A | Only if both sides are well-formed |
| For quantifiers and equality |
| all rules | ✓ |
| Common calculus rules |
| classical reasoning | ✓ |
| weakening | ✓ |
| contraction | ✓(Drop the second occurrence) |
| exchange | Only if still well-formed |
| cut | ✓ |
| resolution | Only if the clauses remain well-formed |
Developing advanced calculi for DHOL is beyond the scope of this paper. However, for example, one way to generalize resolution is to store clauses as lists [L 1,…,L n]subscript 𝐿 1…subscript 𝐿 𝑛[L_{1},\ldots,L_{n}][ italic_L start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_L start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ] where the well-formedness of each L i subscript 𝐿 𝑖 L_{i}italic_L start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT may depend on ¬L j subscript 𝐿 𝑗\neg L_{j}¬ italic_L start_POSTSUBSCRIPT italic_j end_POSTSUBSCRIPT for j<i 𝑗 𝑖 j<i italic_j < italic_i. Resolving [A,L→]𝐴→𝐿[A,\vec{L}][ italic_A , over→ start_ARG italic_L end_ARG ] and [¬A,M→]𝐴→𝑀[\neg A,\vec{M}][ ¬ italic_A , over→ start_ARG italic_M end_ARG ] to [L→,M→]→𝐿→𝑀[\vec{L},\vec{M}][ over→ start_ARG italic_L end_ARG , over→ start_ARG italic_M end_ARG ] is sound if the resolvent is well-formed, i.e., if the well-formedness of the L i subscript 𝐿 𝑖 L_{i}italic_L start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT resp. M i subscript 𝑀 𝑖 M_{i}italic_M start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT does not depend on ¬A 𝐴\neg A¬ italic_A resp. A 𝐴 A italic_A.
2.2.1 Polymorphic DHOL.
DHOL as presented in the previous section and is monomorphic. ATP for polymorphic DHOL, as well as proofs of properties for such an extension of the calculus, is ongoing parallel work. Polymorphic logics are already available in the TPTP language, so it is natural to offer polymorphic DTF. All the polymorphic example problems considered so far use only shallow/rank-1 polymorphism in line with the existing polymorphic first- and higher-order forms for TPTP.
2.2.2 Choice.
Hilbert’s choice operator has been part of HOL since its inception by Church. As such, it is natural to include it in DHOL. This introduces some complications: Due to the usual non-emptiness constraint on types, the semantics of choice are clear in HOL. However, DHOL no longer abides by this constraint, requiring a design decision that affects well-typedness and provability. Experiments done in suggest that the variant of choice dubbed “strong choice” results in more efficient automated reasoning. The eponymous characteristic of strong choice is the requirement that ∃x:A.t\exists x:A.t∃ italic_x : italic_A . italic_t needs to be true for (ε x:A.t):A(\varepsilon x:A.t):A( italic_ε italic_x : italic_A . italic_t ) : italic_A to be well-typed. Such a requirement for typing fits well with DHOL in general, and as ATP is the main concern this is the variant of choice, as it were. The problem set described in Section4 includes some examples supporting this variant.
2.2.3 Translation.
In order to take advantage of the ATP systems available for regular HOL, Rothgang et al. define a dependency-erasure, and thereby a translation from DHOL into regular HOL. They also prove that this translation is sound and complete for well-typed DHOL problems. Due to this result, and the implementation of the translation into the preprocessor of the Leo-III theorem prover, there existed reasoning support for DHOL even before native DHOL reasoning was implemented in the Lash ATP system by Niederhauser et al.. Information lost due to the erasure of term dependencies is captured in Partial Equivalence Relations (PERs) – symmetric and transitive relations on pairs of terms – with the idea that the relation is reflexive exactly for those terms that were previously of the same dependent type. The translation is shown in Figure2. The translation t¯¯𝑡\overline{t}over¯ start_ARG italic_t end_ARG of a term t 𝑡 t italic_t is defined inductively on the structure of the terms. The erasure of one type declaration results in three erased declarations: the erased type, the PER constant and an axioms stating it’s properties. The definition of the erasure on ∀for-all\forall∀- and ∃\exists∃-quantified terms is notable as it uses a PER as guard on the argument. To see why, note that, e.g., ∀x:A.t\forall x:A.t∀ italic_x : italic_A . italic_t can be defined in terms of equality as λ x:A.t=A→o λ x:A.⊤\lambda x:A.t=_{A\rightarrow o}\lambda x:A.\top italic_λ italic_x : italic_A . italic_t = start_POSTSUBSCRIPT italic_A → italic_o end_POSTSUBSCRIPT italic_λ italic_x : italic_A . ⊤. The erasure creates a PER from this typed equality with the guarded input in the premise, and the erased term in the consequence of the implication as seen in the erasure of ∀for-all\forall∀.
theories contexts
∘¯¯\displaystyle\overline{\circ}>over¯ start_ARG ∘ end_ARG=∘absent\displaystyle=>\circ= ∘∙¯¯∙\displaystyle\overline{\bullet}>over¯ start_ARG ∙ end_ARG=∙absent∙\displaystyle=>\bullet= ∙
T,D¯¯𝑇 𝐷\displaystyle\overline{T,D}>over¯ start_ARG italic_T , italic_D end_ARG=T¯,D¯absent¯𝑇¯𝐷\displaystyle=>\overline{T},\overline{D}= over¯ start_ARG italic_T end_ARG , over¯ start_ARG italic_D end_ARG Γ,D¯¯Γ 𝐷\displaystyle\overline{\Gamma,D}>over¯ start_ARG roman_Γ , italic_D end_ARG=Γ¯,D¯absent¯Γ¯𝐷\displaystyle=>\overline{\Gamma},\overline{D}= over¯ start_ARG roman_Γ end_ARG , over¯ start_ARG italic_D end_ARG
c:A¯¯:𝑐 𝐴\displaystyle\overline{c:A}>over¯ start_ARG italic_c : italic_A end_ARG=c:A¯,▷A∗cc:absent 𝑐¯𝐴▷superscript 𝐴∗𝑐 𝑐\displaystyle=>c:\overline{A},\;\triangleright\;{A}^{\ast}>c>c= italic_c : over¯ start_ARG italic_A end_ARG , ▷ italic_A start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT italic_c italic_c x:A¯¯:𝑥 𝐴\displaystyle\overline{x:A}>over¯ start_ARG italic_x : italic_A end_ARG=x:A¯,▷A∗xx:absent 𝑥¯𝐴▷superscript 𝐴∗𝑥 𝑥\displaystyle=>x:\overline{A},\;\triangleright\;{A}^{\ast}>x>x= italic_x : over¯ start_ARG italic_A end_ARG , ▷ italic_A start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT italic_x italic_x
▷F¯¯▷𝐹\displaystyle\overline{\triangleright\;F}>over¯ start_ARG ▷ italic_F end_ARG=▷F¯absent▷¯𝐹\displaystyle=>\triangleright\;\overline{F}= ▷ over¯ start_ARG italic_F end_ARG▷F¯¯▷𝐹\displaystyle\overline{\triangleright\;F}>over¯ start_ARG ▷ italic_F end_ARG=▷F¯absent▷¯𝐹\displaystyle=>\triangleright\;\overline{F}= ▷ over¯ start_ARG italic_F end_ARG
a:Π x 1:A 1.⋯Π x n:A n.𝚝𝚢𝚙𝚎¯=\overline{a:\Pi x_{1}:A_{1}.>\cdots>\Pi x_{n}:A_{n}.>\mathtt{type}}>=over¯ start_ARG italic_a : roman_Π italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT : italic_A start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT . ⋯ roman_Π italic_x start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT : italic_A start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT . typewriter_type end_ARG =
a𝚝𝚢𝚙𝚎 𝑎 𝚝𝚢𝚙𝚎 a>\mathtt{type}italic_a typewriter_type
a∗:A 1¯→⋯→A n¯→a→a→o:superscript 𝑎∗→¯subscript 𝐴 1⋯→¯subscript 𝐴 𝑛→𝑎→𝑎→𝑜{a}^{\ast}:\overline{A_{1}}\to\cdots\to\overline{A_{n}}\to a\to a\to o italic_a start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT : over¯ start_ARG italic_A start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_ARG → ⋯ → over¯ start_ARG italic_A start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT end_ARG → italic_a → italic_a → italic_o
▷∀x 1:A 1¯.⋯∀x n:A n¯.∀u,v:a.a∗x 1⋯x n u v⇒u=a v\triangleright\;\forall x_{1}:\overline{A_{1}}.>\cdots\forall x_{n}:\overline% {A_{n}}.>\forall u,v:a.>{a}^{\ast}\ x_{1}>\cdots>x_{n}\ u\ v\Rightarrow u=% {a}v▷ ∀ italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT : over¯ start_ARG italic_A start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_ARG . ⋯ ∀ italic_x start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT : over¯ start_ARG italic_A start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT end_ARG . ∀ italic_u , italic_v : italic_a . italic_a start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ⋯ italic_x start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT italic_u italic_v ⇒ italic_u = start_POSTSUBSCRIPT italic_a end_POSTSUBSCRIPT italic_v
c¯¯𝑐\displaystyle\overline{c}>over¯ start_ARG italic_c end_ARG=c absent 𝑐\displaystyle=>c= italic_c x¯¯𝑥\displaystyle\overline{x}>over¯ start_ARG italic_x end_ARG=x absent 𝑥\displaystyle=>x= italic_x
o¯¯𝑜\displaystyle\overline{o}>over¯ start_ARG italic_o end_ARG=o absent 𝑜\displaystyle=>o= italic_o at 1…t n¯¯𝑎 subscript 𝑡 1…subscript 𝑡 𝑛\displaystyle\overline{a\ t{1}\ \ldots\ t_{n}}>over¯ start_ARG italic_a italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT … italic_t start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT end_ARG=a absent 𝑎\displaystyle=>a= italic_a
Π x:A.B¯\displaystyle\overline{\Pi x:A.B}>over¯ start_ARG roman_Π italic_x : italic_A . italic_B end_ARG=A¯→B¯absent¯𝐴→¯𝐵\displaystyle=>\overline{A}\rightarrow\overline{B}= over¯ start_ARG italic_A end_ARG → over¯ start_ARG italic_B end_ARG λ x:A.t¯\displaystyle\overline{\lambda x:A.t}>over¯ start_ARG italic_λ italic_x : italic_A . italic_t end_ARG=λ x:A¯.t¯\displaystyle=>\lambda x:\overline{A}.>\overline{t}= italic_λ italic_x : over¯ start_ARG italic_A end_ARG . over¯ start_ARG italic_t end_ARG
¬t¯¯𝑡\displaystyle\overline{\neg t}>over¯ start_ARG ¬ italic_t end_ARG=¬t¯absent¯𝑡\displaystyle=>\neg\overline{t}= ¬ over¯ start_ARG italic_t end_ARG tu¯¯𝑡 𝑢\displaystyle\overline{t>u}>over¯ start_ARG italic_t italic_u end_ARG=t¯u¯absent¯𝑡¯𝑢\displaystyle=>\overline{t}>\overline{u}= over¯ start_ARG italic_t end_ARG over¯ start_ARG italic_u end_ARG
t⇒u¯¯⇒𝑡 𝑢\displaystyle\overline{t\Rightarrow u}>over¯ start_ARG italic_t ⇒ italic_u end_ARG=t¯⇒u¯absent¯𝑡⇒¯𝑢\displaystyle=>\overline{t}\Rightarrow\overline{u}= over¯ start_ARG italic_t end_ARG ⇒ over¯ start_ARG italic_u end_ARG t=A u¯¯subscript 𝐴 𝑡 𝑢\displaystyle\overline{t=_{A}u}>over¯ start_ARG italic_t = start_POSTSUBSCRIPT italic_A end_POSTSUBSCRIPT italic_u end_ARG=A∗t¯u¯absent superscript 𝐴∗¯𝑡¯𝑢\displaystyle=>{A}^{\ast}>\overline{t}>\overline{u}= italic_A start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT over¯ start_ARG italic_t end_ARG over¯ start_ARG italic_u end_ARG
t∧u¯¯𝑡 𝑢\displaystyle\overline{t\land u}>over¯ start_ARG italic_t ∧ italic_u end_ARG=t¯∧u¯absent¯𝑡¯𝑢\displaystyle=>\overline{t}\land\overline{u}= over¯ start_ARG italic_t end_ARG ∧ over¯ start_ARG italic_u end_ARG t∨u¯¯𝑡 𝑢\displaystyle\overline{t\lor u}>over¯ start_ARG italic_t ∨ italic_u end_ARG=t¯∨u¯absent¯𝑡¯𝑢\displaystyle=>\overline{t}\lor\overline{u}= over¯ start_ARG italic_t end_ARG ∨ over¯ start_ARG italic_u end_ARG
⊥¯¯bottom\displaystyle\overline{\bot}>over¯ start_ARG ⊥ end_ARG=⊥absent bottom\displaystyle=>\bot= ⊥⊤¯¯top\displaystyle\overline{\top}>over¯ start_ARG ⊤ end_ARG=⊤absent top\displaystyle=>\top= ⊤
∀x:A.t¯\displaystyle\overline{\forall x:A.t}>over¯ start_ARG ∀ italic_x : italic_A . italic_t end_ARG=∀x:A¯.A∗x x⇒t¯\displaystyle=>\forall x:\overline{A}.>{A}^{\ast}>x>x\Rightarrow\overline{t}= ∀ italic_x : over¯ start_ARG italic_A end_ARG . italic_A start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT italic_x italic_x ⇒ over¯ start_ARG italic_t end_ARG∃x:A.t¯\displaystyle\overline{\exists x:A.t}>over¯ start_ARG ∃ italic_x : italic_A . italic_t end_ARG=∃x:A¯.A∗x x∧t¯\displaystyle=>\exists x:\overline{A}.>{A}^{\ast}>x>x\land\overline{t}= ∃ italic_x : over¯ start_ARG italic_A end_ARG . italic_A start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT italic_x italic_x ∧ over¯ start_ARG italic_t end_ARG
PER for each type
o∗tu superscript 𝑜∗𝑡 𝑢\displaystyle{o}^{\ast}>t>u>italic_o start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT italic_t italic_u=t=o u absent 𝑡 subscript 𝑜 𝑢\displaystyle=>t={o}u= italic_t = start_POSTSUBSCRIPT italic_o end_POSTSUBSCRIPT italic_u
(at 1…t n)∗uv superscript 𝑎 subscript 𝑡 1…subscript 𝑡 𝑛∗𝑢 𝑣\displaystyle{(a\ t{1}\ \ldots\ t_{n})}^{\ast}>u>v>( italic_a italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT … italic_t start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ) start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT italic_u italic_v=a∗t 1¯⋯t n¯uv absent superscript 𝑎∗¯subscript 𝑡 1⋯¯subscript 𝑡 𝑛 𝑢 𝑣\displaystyle=>{a}^{\ast}\ \overline{t_{1}}>\cdots>\overline{t_{n}}\ u\ v= italic_a start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT over¯ start_ARG italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT end_ARG ⋯ over¯ start_ARG italic_t start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT end_ARG italic_u italic_v
(Π x:A.B)∗t u\displaystyle{(\Pi x:A.B)}^{\ast}>t>u>( roman_Π italic_x : italic_A . italic_B ) start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT italic_t italic_u=∀x,y:A¯.A∗x y⇒B∗(t x)(u y)\displaystyle=>\forall x,y:\overline{A}.>{A}^{\ast}>x>y\Rightarrow{B}^{% \ast}(t>x)(u>y)= ∀ italic_x , italic_y : over¯ start_ARG italic_A end_ARG . italic_A start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT italic_x italic_y ⇒ italic_B start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT ( italic_t italic_x ) ( italic_u italic_y )
Figure 2: The translation from DHOL to HOL
As an example of erasure, consider the list of char s [a, b], represented by a term 𝚌𝚘𝚗𝚜 1𝚊(𝚌𝚘𝚗𝚜 0𝚋𝚗𝚒𝚕)𝚌𝚘𝚗𝚜 1 𝚊 𝚌𝚘𝚗𝚜 0 𝚋 𝚗𝚒𝚕\mathtt{cons}\;1\;\mathtt{a}\;(\mathtt{cons}\;0\;\mathtt{b}\;\mathtt{nil})typewriter_cons 1 typewriter_a ( typewriter_cons 0 typewriter_b typewriter_nil ) of type 𝚟𝚎𝚌 2 𝚟𝚎𝚌 2\mathtt{vec}\;2 typewriter_vec 2, where 0,𝚜𝚞𝚌 0,𝚜𝚞𝚌(𝚜𝚞𝚌 0),…0 𝚜𝚞𝚌 0 𝚜𝚞𝚌 𝚜𝚞𝚌 0…0,\mathtt{suc}\;0,\mathtt{suc}\;(\mathtt{suc}\;0),...0 , typewriter_suc 0 , typewriter_suc ( typewriter_suc 0 ) , … is abbreviated as 0,1,2,…0 1 2…0,1,2,...0 , 1 , 2 , …. Applying the erasure gives 𝚌𝚘𝚗𝚜𝚊(𝚌𝚘𝚗𝚜𝚋𝚗𝚒𝚕)𝚌𝚘𝚗𝚜 𝚊 𝚌𝚘𝚗𝚜 𝚋 𝚗𝚒𝚕\mathtt{cons}\;\mathtt{a}\;(\mathtt{cons}\;\mathtt{b}\;\mathtt{nil})typewriter_cons typewriter_a ( typewriter_cons typewriter_b typewriter_nil ) of type 𝚟𝚎𝚌 𝚟𝚎𝚌\mathtt{vec}typewriter_vec. A predicate would be generated, establishing that this particular list is in the PER of vectors of length 2: 𝚟𝚎𝚌∗ 2tt superscript 𝚟𝚎𝚌∗2 𝑡 𝑡{\mathtt{vec}}^{\ast}\;2\;t\;t typewriter_vec start_POSTSUPERSCRIPT ∗ end_POSTSUPERSCRIPT 2 italic_t italic_t where t stands for 𝚌𝚘𝚗𝚜𝚊(𝚌𝚘𝚗𝚜𝚋𝚗𝚒𝚕)𝚌𝚘𝚗𝚜 𝚊 𝚌𝚘𝚗𝚜 𝚋 𝚗𝚒𝚕\mathtt{cons}\;\mathtt{a}\;(\mathtt{cons}\;\mathtt{b}\;\mathtt{nil})typewriter_cons typewriter_a ( typewriter_cons typewriter_b typewriter_nil ). While one might think that unary predicates would be sufficient as a type guard, PERs becomes necessary to express the typing and equality of higher-order functions: functions are well-typed if they map well-typed inputs to well-typed outputs, and they are equal if they agree on well-typed inputs.
3 DTF
After establishing the theoretic background, this section presents the realization of DHOL in the TPTP language. Syntax and semantics are given, as well as an exposition to the problem of type checking.
3.1 Syntax
The syntax of DTF requires almost no change to the existing TPTP syntax. The TPTP language already defines the !> binder for types. In the typed TPTP language variants it is currently used for only polymorphism, e.g.,
cons : !>[A: $tType]: ( A > ( list @ A ) > ( list @ A ) )
is a type declaration for a polymorphic cons. The TPTP syntax does not forbid listing terms in the types of such variable lists. This fact is used to unobtrusively extend TPTP by dependent types. A dependent type symbol declaration is written with m 𝑚 m italic_m terms of n 𝑛 n italic_n types as
a : !>[x 1:A 1:subscript 𝑥 1 subscript 𝐴 1 x_{1}:A_{1}italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT : italic_A start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT, ..., x m:A n:subscript 𝑥 𝑚 subscript 𝐴 𝑛 x_{m}:A_{n}italic_x start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT : italic_A start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT]:$tType
or alternatively
a : A 1 subscript 𝐴 1 A_{1}italic_A start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT> ... >A n subscript 𝐴 𝑛 A_{n}italic_A start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT> $tType.
Such types use the application operator @, to instantiate the terms to the dependent type:
a @ t 1 subscript 𝑡 1 t_{1}italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT @ ... @ t m subscript 𝑡 𝑚 t_{m}italic_t start_POSTSUBSCRIPT italic_m end_POSTSUBSCRIPT.
In polymorphic problems, the variable list is prepended with the type variables, which may appear in the same binder. An example of a problem in DTF is shown in Figure3.
thf(elem_type,type, elem: $tType ).
thf(nat_type,type, nat: $tType ).
thf(zero_type,type, zero: nat ).
thf(suc_type,type, suc: nat > nat ).
thf(plus_type,type, plus: nat > nat > nat ).
thf(list_type,type, list: nat > $tType ).
thf(nil_type,type, nil: list @ zero ).
thf(cons_type,type, cons:
!>[N: nat] : (elem > (list @ N) > (list @ (suc @ N))) ).
thf(app_type,type, app:
!>[N: nat,M: nat] : ((list @ N) > (list @ M) >
(list @ (plus @ N @ M))) ).
thf(ax1,axiom,
! [N: nat] : ((plus @ zero @ N) = N) ).
thf(ax2,axiom,
! [N: nat,X: list @ N] : ((app @ zero @ N @ nil @ X) = X) ).
thf(plus_assoc,axiom,
! [M1: nat,M2: nat,M3: nat] :
( (plus @ M1 @ (plus @ M2 @ M3))
= (plus @ ( plus @ M1 @ M2) @ M3)) ).
thf(list_app_assoc_base,conjecture,
! [M2: nat,L2: list @ M2,M3: nat,L3: list @ M3] :
( (app @ zero @ (plus @ M2 @ M3) @ nil @
(app @ M2 @ M3 @ L2 @ L3))
= (app @ (plus @ zero @ M2) @ M3 @
(app @ zero @ M2 @ nil @ L2) @ L3)) ).
Figure 3: The base case of associativity of append on fixed-length lists.
3.2 Type Checking
Due to equality reflection, type checking for DHOL is, in general, undecidable. Nevertheless, problems need to be well-typed, otherwise the translation outlined in Section2.2.3 might not be sound. Type checking in DTF thus takes on a larger role than in other logics in the TPTP World.
While performing the usual type checking procedure in DHOL, obligations of the form at 1⋯t n≡au 1⋯u n 𝑎 subscript 𝑡 1⋯subscript 𝑡 𝑛 𝑎 subscript 𝑢 1⋯subscript 𝑢 𝑛 a\ t_{1}\ \cdots\ t_{n}\equiv a\ u_{1}\cdots\ u_{n}italic_a italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ⋯ italic_t start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT ≡ italic_a italic_u start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ⋯ italic_u start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT, are generated. These establish equality of the dependent base types applied to arguments t 1⋯t n,u 1⋯u n subscript 𝑡 1⋯subscript 𝑡 𝑛 subscript 𝑢 1⋯subscript 𝑢 𝑛 t_{1}\cdots t_{n},u_{1}\cdots u_{n}italic_t start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ⋯ italic_t start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT , italic_u start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT ⋯ italic_u start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT of appropriate types. The type equality holds if all pairs t i,u i subscript 𝑡 𝑖 subscript 𝑢 𝑖 t_{i},u_{i}italic_t start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT , italic_u start_POSTSUBSCRIPT italic_i end_POSTSUBSCRIPT are equal, which depends on the available axioms. This can create interesting situations where a problem must include axioms that are not necessary for proving the conjecture itself, but are necessary for type checking it. The common example of fixed-length lists is one such example: the statement of the associativity of append is well-typed only if addition on 𝚗𝚊𝚝 𝚗𝚊𝚝\mathtt{nat}typewriter_nat is associative, and thus requires including the defining equations of addition. To prove the problem only the defining equations of appending lists are needed.
The undecidability of type checking can lead to compromises. One such compromise is “shallow type checking”. When a problem file is shallowly checked, only the simply typed skeleton of the problem is considered, i.e., term arguments to types as well as dependent functions are ignored. This collapses to type checking as is done on non-dependently typed problems, and is decidable. This form of type checking is sufficient to catch many careless mistakes in the formulation of problems, and provides a basic check of issues often found in human-written DHOL problems. Examples are: mismatches in the number of arguments of a base type or function, and egregious type mismatches. Shallow type checking provides a valuable sanity check for users, especially considering the complexity that problems in DHOL forms can reach.
3.3 Semantics
As for HOL, there are two kinds of semantics for DHOL: standard models are intuitive and are the ones that are usually used; non-standard (Henkin) models are a generalization that is needed for completeness. A full account is given in the forthcoming , which is summarized below. The rules of DHOL, as given by Rothgang et al., already define which formulae are theorems.
Standard Models.
Given a theory T 𝑇 T italic_T, a standard model M∈⟦T⟧M\in\llbracket T\rrbracket italic_M ∈ ⟦ italic_T ⟧ is a tuple providing an interpretation for every declaration in T 𝑇 T italic_T. Similarly, given a context Γ Γ\Gamma roman_Γ, an assignment α∈⟦Γ⟧M\alpha\in\llbracket\Gamma\rrbracket^{M}italic_α ∈ ⟦ roman_Γ ⟧ start_POSTSUPERSCRIPT italic_M end_POSTSUPERSCRIPT for Γ Γ\Gamma roman_Γ is a tuple providing an interpretation for every declaration in Γ Γ\Gamma roman_Γ. These induce the interpretation function ⟦−⟧α M\llbracket-\rrbracket^{M}_{\alpha}⟦ - ⟧ start_POSTSUPERSCRIPT italic_M end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_α end_POSTSUBSCRIPT (with α 𝛼\alpha italic_α omitted if the context is empty), which is defined inductively for all the syntax. In particular, the possible components of a model are defined by induction on declarations:
•For a type symbol with arguments Γ=x 1:A 1,…,x n:A n:Γ subscript 𝑥 1 subscript 𝐴 1…subscript 𝑥 𝑛:subscript 𝐴 𝑛\Gamma=x_{1}:A_{1},\ldots,x_{n}:A_{n}roman_Γ = italic_x start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT : italic_A start_POSTSUBSCRIPT 1 end_POSTSUBSCRIPT , … , italic_x start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT : italic_A start_POSTSUBSCRIPT italic_n end_POSTSUBSCRIPT, a function ⟦Γ⟧M→𝒮 ℰ 𝒯\llbracket\Gamma\rrbracket^{M}\to\mathcal{SET}⟦ roman_Γ ⟧ start_POSTSUPERSCRIPT italic_M end_POSTSUPERSCRIPT → caligraphic_S caligraphic_E caligraphic_T
•For a term symbol c:A:𝑐 𝐴 c:A italic_c : italic_A, a value from ⟦A⟧M\llbracket A\rrbracket^{M}⟦ italic_A ⟧ start_POSTSUPERSCRIPT italic_M end_POSTSUPERSCRIPT
•For an axiom ▷F▷𝐹\triangleright\;F▷ italic_F, a unique choice ✓✓\checkmark✓ if ⟦M⟧F=1\llbracket M\rrbracket^{F}=1⟦ italic_M ⟧ start_POSTSUPERSCRIPT italic_F end_POSTSUPERSCRIPT = 1, and no choice otherwise
For the components of an assignment:
•For a term variable x:A:𝑥 𝐴 x:A italic_x : italic_A, a value from ⟦A⟧α M\llbracket A\rrbracket^{M}_{\alpha}⟦ italic_A ⟧ start_POSTSUPERSCRIPT italic_M end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_α end_POSTSUBSCRIPT
•For an assumption ▷F▷𝐹\triangleright\;F▷ italic_F, a unique choice ✓✓\checkmark✓ if ⟦M⟧α F=1\llbracket M\rrbracket^{F}_{\alpha}=1⟦ italic_M ⟧ start_POSTSUPERSCRIPT italic_F end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_α end_POSTSUBSCRIPT = 1, and no choice otherwise
For types and terms, the model is defined by induction in the usual way, in particular
•⟦o⟧α M={0,1}\llbracket o\rrbracket^{M}_{\alpha}={0,1}⟦ italic_o ⟧ start_POSTSUPERSCRIPT italic_M end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_α end_POSTSUBSCRIPT = { 0 , 1 }
•⟦Π x:A.B⟧α M\llbracket\Pi x:A.B\rrbracket^{M}{\alpha}⟦ roman_Π italic_x : italic_A . italic_B ⟧ start_POSTSUPERSCRIPT italic_M end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_α end_POSTSUBSCRIPT is the set of functions f 𝑓 f italic_f mapping every u∈⟦A⟧α M u\in\llbracket A\rrbracket^{M}{\alpha}italic_u ∈ ⟦ italic_A ⟧ start_POSTSUPERSCRIPT italic_M end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_α end_POSTSUBSCRIPT to some f(u)∈⟦B⟧α u M f(u)\in\llbracket B\rrbracket^{M}_{\alpha^{u}}italic_f ( italic_u ) ∈ ⟦ italic_B ⟧ start_POSTSUPERSCRIPT italic_M end_POSTSUPERSCRIPT start_POSTSUBSCRIPT italic_α start_POSTSUPERSCRIPT italic_u end_POSTSUPERSCRIPT end_POSTSUBSCRIPT where α u superscript 𝛼 𝑢\alpha^{u}italic_α start_POSTSUPERSCRIPT italic_u end_POSTSUPERSCRIPT extends α 𝛼\alpha italic_α with the value u 𝑢 u italic_u for x 𝑥 x italic_x
General Models.
The definition of general models generalizes the Henkin models from HOL by applying methods from categorical models of type theory. First, akin to assignments for Γ Γ\Gamma roman_Γ, substitutions γ:Γ→Δ:𝛾→Γ Δ\gamma:\Gamma\to\Delta italic_γ : roman_Γ → roman_Δ as lists of terms or ✓✓\checkmark✓ by induction on Γ Γ\Gamma roman_Γ are defined:
•For a term variable x:A:𝑥 𝐴 x:A italic_x : italic_A, a term of type Δ⊢A[γ]proves Δ 𝐴 delimited-[]𝛾\Delta\vdash A[\gamma]roman_Δ ⊢ italic_A [ italic_γ ]
•For an assumption ▷F▷𝐹\triangleright\;F▷ italic_F, the unique choice ✓✓\checkmark✓ if Δ⊢F[γ]proves Δ 𝐹 delimited-[]𝛾\Delta\vdash F[\gamma]roman_Δ ⊢ italic_F [ italic_γ ], and no choice otherwise
Equality of contexts and substitutions is defined by applying the existing equality judgments for types and terms component-wise. For every theory T 𝑇 T italic_T, this yields the syntactic category ⟦T⟧¯¯delimited-⟦⟧𝑇\overline{\llbracket T\rrbracket}over¯ start_ARG ⟦ italic_T ⟧ end_ARG of T 𝑇 T italic_T-contexts and substitutions. A general model is then defined as any pushout-preserving contravariant functor Φ:⟦T⟧¯→𝒮ℰ𝒯:Φ→¯delimited-⟦⟧𝑇 𝒮 ℰ 𝒯\Phi:\overline{\llbracket T\rrbracket}\to\mathcal{SET}roman_Φ : over¯ start_ARG ⟦ italic_T ⟧ end_ARG → caligraphic_S caligraphic_E caligraphic_T. From such a Φ Φ\Phi roman_Φ, an interpretation function is extracted using Φ(x:A)\Phi(x:A)roman_Φ ( italic_x : italic_A ) as the interpretation of the type A 𝐴 A italic_A and Φ(t)Φ 𝑡\Phi(t)roman_Φ ( italic_t ) as the interpretation of the term t:A:𝑡 𝐴 t:A italic_t : italic_A (seen as a substitution x:A→∙:𝑥→𝐴∙x:A\to\bullet italic_x : italic_A → ∙). These general models must further satisfy Φ(o)={0,1}Φ 𝑜 0 1\Phi(o)={0,1}roman_Φ ( italic_o ) = { 0 , 1 }, and Φ⊧F models Φ 𝐹\Phi\models F roman_Φ ⊧ italic_F is defined as Φ(F)=1 Φ 𝐹 1\Phi(F)=1 roman_Φ ( italic_F ) = 1. Here the pushout-preservation essentially corresponds to the preservation of substitution, i.e., interpretation and substitution commute. The lack of any preservation of exponentials allows for a non-compositional interpretation of function types. This approach can be seen as a generalization of Henkin models, which also preserve substitution but do not need to interpret function types compositionally. Contrary to Henkin models, the interpretation of λ 𝜆\lambda italic_λ and application terms can also be non-compositional in these general models as long as substitution is preserved.
Models for Polymorphic DHOL.
As mentioned above, a rank-1 polymorphic variant of DHOL is being developed in parallel work. It is straightforward to extend standard models to polymorphic DHOL. The syntax of binding a type variable corresponds to abstracting over an arbitrary set on the semantic side. In particular, the interpretation of a polymorphic term/type symbol with n 𝑛 n italic_n type variables takes n 𝑛 n italic_n sets as arguments. Polymorphic axioms correspond to universal quantification over sets. The definition of syntactic category and general models is expected to carry over to polymorphic DHOL as well. This has not been investigated in detail.
4 Problem Dataset
Over 100 problems in DTF format have been collected for addition to the TPTP problem library. Their classification is presented in Table2 and discussed here (with 36 problems just for testing DHOL prover features omitted). The number of problems in each class is given in the last column. The problems concern several domains that can benefit from dependent types. While shows DHOL to be sound and complete, the strength of the existing automation for this foundation (discussed in Section5) still needs to be improved. For this reason, some of the harder problems were broken down into simpler subproblems that can be proven independently. Some list properties that require both induction and reasoning with dependent types are an instance of this. For example, the fact that list append is associative, ListAppAssoc, is split into three subproblems, showing the particular induction scheme, the proof of the base case, and the step case. These three subproblems are easier to prove than their combined version, which is also included. Some problems benefit from intermediate lemmas, e.g. the instantiation of the inductive step case. These are found in the “Lemmas” categories of Table2.
One of the simplest classes of examples are lists that depend on their length (also called vectors, for example in the Rocq library). As the list libraries of most interactive theorem provers are substantial, it is relatively easy to experiment with many properties of dependently typed lists. Such properties include the aforementioned associativity of append, corollaries of this statement, or involution statements about the reverse function. Some of these list examples are extended to their polymorphic generalizations, which are in the “Polymorphic” categories.
The idea of expressing well-known but sometimes challenging properties extends to several other algebraic data types, such as matrices that have fixed dimensions, and lists of lists. Red-black trees are a well-known data structure for balanced trees where the invariant can be expressed using dependent types, and again several problems concerning this type are included. The Fin type present in several proof libraries has been manually recreated, and some problems about these are in the ROCQ category of Table2. The collection includes the five examples from category theory that were originally presented in , slightly reformatted to match the TPTP syntax. To make use of the choice operator, several problems about dependent higher-order Skolemization are included. Choice is also used in a function definition with no fixed point, and conjectures establishing this are presented in the “no FP” category. Finally, several simple tests to evaluate the ability of provers to perform native DHOL inferences are provided.
Some of the dependent HOL problems are more interesting from a proof perspective – the deep type checking is there only to make sure the problem is well-formed. For example, for all the dependent list problems, the type checking obligations are there mostly to make sure no incorrect calls are being made, but they are relatively straightforward to discharge. It is the proof that requires more logical reasoning. Other problems, while relatively straightforward in terms of proving, are harder to type check. This is because it is possible to use dependent types to encode important properties and invariants in the type system.
| Problem Type | Problem Category | Problem Count |
---
| Monomorphic Complete | Category theory | 5 |
| | Choice basic | 11 |
| | Choice list | 3 |
| | Choice no fixed point | 10 |
| | List app assoc | 3 |
| | List app assoc corollary | 1 |
| | List app nil | 4 |
| | List of lists | 1 |
| | List reversal involution | 1 |
| | List reversal inv lemma | 3 |
| | Matrices | 5 |
| | ROCQ | 3 |
| Monomorphic Lemmas | Choice no fixed point | 10 |
| | List app assoc | 5 |
| | List app assoc corollary | 5 |
| | List reversal involution | 5 |
| | List reversal inv lemma | 11 |
| Polymorphic Complete | List app assoc poly | 3 |
| | List app nil poly | 4 |
| | List reversal involution poly | 1 |
| | Red-black tree | 3 |
| Polymorphic Lemmas | List app assoc poly | 14 |
| | List reversal involution poly | 13 |
| | Red-black tree | 9 |
Table 2: The categories of the DTF problems.
5 Tools
This section discusses the tools capable of processing problems in DTF format.
5.1 The Logic Embedding Tool
The Leo-III prover includes the Logic Embedding Tool, which has been extended to support polymorphic DTF. The tool implements the erasure presented in Section2.2.3, and incorporates the polymorphic extension. The tool can generate both the type checking obligations and the translated problem separately. This makes it possible to translate DTF problems into THF problems (that do not have dependent types). The embedding tool is available as NTFLET in SystemB4TPTP 2 2 2tptp.org/cgi-bin/SystemB4TPTP. The embedding tool enables the use of existing higher-order ATP systems for solving DTF problems, by pipelining the output from NTFLET to a THF ATP system of the user’s choosing. This has been implemented as the DT2H2X ATP systems, available in SystemOnTPTP 3 3 3tptp.org/cgi-bin/SystemOnTPTP.
5.2 DLash
The Lash prover is a partial reimplementation of the tableaux calculus of Satallax, using a central term representation with perfect sharing. This design facilitated the implementation of the DLash extension of Lash, which handles DTF. In addition to the erasure implementation, DLash can process monomorphic dependently typed higher-order logic with choice. As with the Logic Embedding Tool, type checking and proving can be requested separately. DLash, like Satallax, includes a strategy language used to build so-called modes. The current version includes 36 dedicated modes for dependent types, tailored to specific problem types. DLash is available in SystemOnTPTP 4 4 4tptp.org/cgi-bin/SystemOnTPTP.
5.3 MMT
MMT is a logical framework designed to formalize and manage large collections of interconnected formal systems and their libraries, using modular theory graphs. A particular application of MMT is rapid prototyping, and it was the tool originally used to develop and prototype DHOL. The MMT/DHOL implementation offers reconstruction of omitted types and implicit arguments as well as parsing against user-defined notations. It can be used to interactively author and type check DHOL problems and export them in TPTP format. It uses the PER translation, and calls the Leo-III prover to discharge the resulting proof obligations. MMT is mostly useful for developing formalizations, rather than proving TPTP conjectures. Therefore, it does not provide a TPTP import at this point, but provides additional evidence of the well-typedness DTF problems.
5.4 TPTP Systems
As discussed in Section2.1, TPTP includes several generic tools capable of processing problems and solutions. For DTF problems:
•TPTP4X pretty-prints DTF problems and solutions, and offers various transformations/augmentations of problems.
•BNFParser produces the abstract syntax tree from parsing a DTF problem.
•Leo-III-STC validates the syntax and types of DTF problems.
•ProblemStats outputs various syntactic measures for problems.
All these tools are available in SystemB4TPTP 5 5 5tptp.org/cgi-bin/SystemB4TPTP. For DTF proofs:
•ProofStats outputs various syntactic measures for DAG-structured proofs.
•IDV provides interactive viewing of proofs from DTF problems.
All these tools are available in SystemOnTSTP 6 6 6tptp.org/cgi-bin/SystemOnTSTP.
6 Conclusion
This paper has described DTF, the dependently typed higher-order form of the TPTP language. It responds to the growing interest in dependently typed automated reasoning as exemplified by the number of TPTP problems and tools that have cropped up in the short time since DHOL was first described. It can be seen as pushing the boundary of automated theorem proving towards language features that have previously been found only in interactive provers.
DHOL problems sometimes used differing standards, which defeated the uniformity advantage that the TPTP language provides. This work unifies them, and provides over a 100 problems from different domains, benefiting from the use of dependent types. We hope that the availability of dependent types in the TPTP will stimulate research into dependently typed automated theorem proving, by making it easier to exchange and compare results. Extending existing systems with support for DTF, and improving the performance of the systems that already exist, will be important next steps. In particular, the extension of superposition-based theorem proving to dependent types is a tantalizing goal.
Acknowledgements:
The authors thank Johannes Niederhauser and Colin Rothgang for granting access to their DHOL problems that are in the problem dataset. This work was supported by the ERC PoC grant no. 101156734 “FormalWeb3”.
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6812 | https://arxiv.org/pdf/2212.10116 | arXiv:2212.10116v3 [math.NT] 18 Jul 2023
On the representability of sequences as constant terms
Alin Bostan 1, Armin Straub 2, and Sergey Yurkevich 1,3
1
Inria, Univ. Paris-Saclay, France
alin.bostan@inria.fr
2
University of South Alabama, USA
straub@southalabama.edu
3
University of Vienna, Austria
sergey.yurkevich@univie.ac.at
Abstract
A constant term sequence is a sequence of rational numbers whose
n-th term is the constant term of P n(x)Q(x), where P (x) and Q(x)are multivariate Laurent polynomials. While the generating functions of such sequences are invariably diagonals of multivariate rational func-tions, and hence special period functions, it is a famous open question, raised by Don Zagier, to classify diagonals that are constant terms. In this paper, we provide such a classification in the case of sequences sat-isfying linear recurrences with constant coefficients. We also consider the case of hypergeometric sequences and, for a simple illustrative fam-ily of hypergeometric sequences, classify those that are constant terms.
Keywords: Integer sequences, C-finite sequences, hypergeometric sequences, constant term sequences, P-finite sequences, Laurent polynomials, Gauss congruences, diagonals of rational functions.
1 Introduction
Recognizing and interpreting integrality of sequences defined by recursions is at the same time an extensively studied and a difficult topic in number theory. Even in the case of P-finite sequences A(n) (also called P-recursive ,or holonomic ), defined by linear recurrences with polynomial coefficients
pr(n)A(n + r) = pr−1(n)A(n + r − 1) + · · · + p0(n)A(n), pi(n) ∈ Z[n],
1neither a criterion nor even an algorithm is known for classifying/deciding integrality. An attempt for such a classification is the famous and widely open conjecture by Christol [ Chr90 , Conjecture 4, p. 55]. Roughly speaking, it states that a P-finite sequence ( A(n)) n≥0 with (at most) geometric growth is integral if and only if ( A(n)) n≥0 is the coefficient sequence of the diagonal of a rational function R(x) ∈ Z(x1, . . . , x d) ∩ Z for some d ≥ 1. Recall that the diagonal of a multivariate power series
R(x) = ∑
n1,n 2,...,n d≥0
c(n1, n 2, . . . , n d)xn1
1
xn2
2
· · · xnd
d
(1) is the univariate power series Diag( R) whose coefficient sequence is given by
A(n) = c(n, n, . . . , n ). For a precise statement of Christol’s conjecture see Conjecture 6.3 below. Often integrality of sequences can be explained by the underlying com-binatorial nature. For example, the Catalan numbers C(n) satisfying (n + 2) C(n + 1) = 2(2 n + 1) C(n), C(0) = 1 ,
are clearly integers because they count triangulations of convex polygons with n + 2 vertices. On the other hand, for many other integral and P-finite sequences, combinatorial interpretations are not a priori known; this is the case, for instance, for the Ap´ ery numbers A(n) (associated with the irrationality proof of ζ(3)) defined by (n + 1) 3A(n + 1) = (2 n + 1)(17 n2 + 17 n + 5) A(n) − n3A(n − 1) ,A(0) = 1 , A (1) = 5 .
In both examples above, integrality can be seen from the explicit formulas
C(n) =
(2nn
)
−
( 2nn + 1
)
and A(n) =
n
∑
k=0
(nk
)2(n + kk
)2
.
Putting Christol’s conjecture in practice gives a different justification for the integrality of these two examples. It namely holds that
∑
n≥0
C(n)tn = Diag
( 1 − y
1 − x(y + 1) 2
)
and
∑
n≥0
A(n)tn = Diag
( 1
1 − (xy + x + y)( zw + z + w)
)
,
2and the integrality of C(n) and A(n) follows from that of the coefficients in the Taylor expansions of the corresponding multivariate rational functions. In the context of the current text, however, we would like to empha-size a slightly different viewpoint, which does not only justify integrality of the two examples, but also implies some interesting arithmetic prop-erties. Writing ct[ P (x)] for the constant term of a Laurent polynomial
P (x) ∈ Q[x±11 , . . . , x ±1
d
], one can prove that [ Str14 , Rem. 1.4]
C(n) = ct [(x−1 + 2 + x)n(1 − x)] and
A(n) = ct
[( (x + y)( z + 1)( x + y + z)( y + x + 1)
xyz
)n]
.
Similar identities as in the examples of the Catalan and Ap´ ery numbers can be deduced for many other integral P-finite sequences. This motivates the following definition and the subsequent natural question.
Definition 1.1. A sequence A(n) is a constant term if it can be represented as
A(n) = ct[ P (x)nQ(x)] , (2) where P, Q ∈ Q[x±1] are Laurent polynomials in x = ( x1, . . . , x d). Using the geometric series it is easy to see that generating functions of constant term sequences can be expressed as diagonals of rational functions. The converse is, however, not true in general. This leads to the follow-ing question which was raised by Zagier [ Zag18 , p. 769, Question 2] and Gorodetsky [ Gor21 ] in the case Q = 1 (see Proposition 5.1 below for an indication why this case is of particular arithmetic significance).
Question 1.2. Which P-finite sequences are constant terms?
To our knowledge, Question 1.2 is widely open. In fact, the initial mo-tivation for the present text was the goal of answering the following very particular sub-question asked by the second author in [ Str22 , Question 5.1]:
Question 1.3. Is the Fibonacci sequence (F (n)) n≥0 a constant term se-quence?
Recall that the Fibonacci sequence is the coefficient sequence in the Tay-lor expansion of the univariate rational function x/ (1 − x − x2), or equiva-lently the P-finite sequence ( F (n)) n≥0 defined by F (n+2) = F (n+1)+ F (n)and F (0) = 0 , F (1) = 1. 3Already in [ Str22 ] the second author noted that a representation of the Fibonacci numbers as constant terms with Q = 1 is impossible since (F (n)) n≥0 does not satisfy the so-called Gauss congruences (see ( 9)). Ex-ploiting the fact that for any prime p, the value F (p) (mod p) depends on
p (mod 5), we can show (see Example 3.2 ) that the answer to Question 1.3
is negative. The reason for this is that, as we will prove, for any constant term sequence A(n), the sequence A(p) (mod p) must be constant for large enough primes p. Note that this is not a sufficient criterion, since already the Lucas numbers L(n) (defined by the same recursion as the Fibonacci num-bers, but with different initial terms L(0) = 2 , L (1) = 1, see ( 8)) do satisfy the Gauss congruences but are not constant terms (see Example 4.2 ). In the present text, we are able to answer Question 1.2 in the case of diagonals of rational functions F (x) ∈ Q(x) in a single variable. Such sequences are precisely the (rational) C-finite sequences (also known as C-recursive sequences), and are characterized by the fact that they satisfy a linear recursion with constant rational coefficients. More explicitly, we define a sequence A(n) of rational numbers to be C-finite if there exists a polynomial P (x) ∈ Q[x] such that for every n ≥ 0 we have
P (N )A(n) = 0 , (3) where N denotes the shift operator N ℓ(A(n)) := A(n + ℓ) for all ℓ ≥ 0. Equivalently, there exist integers r > 0 and n0 ≥ 0, and complex numbers
c0, . . . , c r−1 with c0 6 = 0 such that
A(n + r) = cr−1A(n + r − 1) + · · · + c0A(n) for all n ≥ n0. (4) We recall that associated to the recursion ( 4), the characteristic roots are usually defined as the roots of
χ(λ) := λr − cr−1λr−1 − · · · − c0.
For our purpose, however, it is useful to define the characteristic roots of a C-finite sequence A(n) as the roots of P (x), where P (x) is chosen with minimal degree such that ( 3) holds. Note that the only difference between considering roots of χ and P is that 0 can be a root of the latter. Equivalently, 0 is defined to be a characteristic root of A(n) of multiplicity m0 if the minimal
n0 in ( 4) (chosen so that r is minimal) equals m0. With these definitions we obtain the following:
Proposition 1.4. Let A(n) be a C-finite sequence. A(n) is a constant term if and only if it has a single characteristic root λ and λ ∈ Q.
4This proposition immediately answers Question 1.3 but also shows that, for example, the sequence A(n) = 2 n + 1 is not a constant term sequence either (in both of these cases, there are two different characteristic roots). Evidently, however, it is the sum of two constant terms: we see that the class of constant term sequences is not a ring. Therefore, to fix this issue, it is natural to consider the class of sequences given as Q-linear combinations of constant terms:
Question 1.5. Which P-finite sequences are finite Q-linear combinations of constant terms?
Again in the case of C-finite sequences, we can answer this question completely with the main result of the present work:
Theorem 1.6. Let A(n) be a C-finite sequence. Then A(n) is an r-term Q-linear combination of constant terms if and only if it has at most r distinct characteristic roots, all of which are rational.
Having completed the classification of C-finite sequences that can be written as (sums of) constant terms, there are two most natural directions for further work. On the one hand, it is reasonable to go from diagonals in one variable to diagonals in two variables. By the combination of results due to P´ olya [ P´ o22 ] and Furstenberg [ Fur67 ] this is known to be exactly the class of algebraic generating functions. One is then lead to the following question which we leave for future work:
Question 1.7. Which sequences A(n) with algebraic generating function are constant terms?
Another reasonable direction is to try to classify those hypergeometric sequences which are constant terms. Recall that a P-finite sequence A(n) is called hypergeometric if it satisfies a recursion of order one, i.e. α(n)A(n +1) = β(n)A(n) for some polynomials α(n), β (n) ∈ Q[n]. In this sense, this class of sequences is arguably the simplest (and best understood) among P-finite ones. Still, Christol’s conjecture remains open even in this very special case. In fact, it is still an open question whether the generating function of the sequence
A(n) =
( 1
9
)
n
( 4
9
)
n
( 5
9
)
n
n!2 ( 1
3
)
n
can be represented as the diagonal of a rational function. Here and later, (x)n := x(x + 1) · · · (x + n − 1) denotes the rising factorial. We can use the same methods as in the C-finite case to prove that A(n) is not a constant 5term sequence (see Lemma 6.6 ). By classifying when the family ( 21 ) of hypergeometric sequences is a constant term, we are further able to conclude that not all hypergeometric diagonals are constant terms. The following question, however, remains open in general:
Question 1.8. Which hypergeometric sequences are constant terms?
The organization of the paper is as follows: In Section 2, we review properties of C-finite sequences that will be important for our purposes. In particular, we state Theorem 2.2 which is due to Minton [ Min14 ] and which is a crucial ingredient of our approach. In Section 3, we derive certain congruences that are satisfied by any constant term sequence; these are already enough to answer Question 1.3 . By combining these congruences with Minton’s Theorem, we prove in Section 4 our main Theorem 1.6 , thus answering Question 1.2 and Question 1.5 in the case of C-finite sequences. In the short Section 5 we prove a statement which is pleasingly similar to Minton’s theorem and which allows to classify the constant terms with Q = 1 among all constant terms. Finally, in Section 6, we turn our attention to hypergeometric sequences and discuss Question 1.8 .Throughout the article, p denotes a prime number, Fp the finite field with p elements and Zp the ring of p-adic integers.
2 Trace sequences
Let A(n) be a C-finite sequence. Denote by λ1, λ 2, . . . , λ d ∈ Q the char-acteristic roots, and let mj be the multiplicity of the root λj . Recall that
λ0 = 0 is defined to be a characteristic root of A(n) of multiplicity m0 if the minimal n0 in ( 4) equals m0. A(n) can be written as a linear combination
A(n) = A0(n) +
d
∑
j=1
mj−1
∑
r=0
cj,r nrλnj (5) for certain coefficients cj,r ∈ Q (more precisely, cj,r ∈ Q(λ1, . . . , λ d)) and
A0(n) a sequence of finite support {0, 1, . . . , m 0−1}. We refer to [ EvdPSW03 ]or [ KP11 , Chapter 4] for introductions to C-finite sequences. Note that al-lowing 0 as a characteristic root is equivalent to not restricting the numera-tor of the rational generating function of A(n) to have degree less than the degree of its denominator. In the following, we will refer to
Asep (n) = A0(0) +
d
∑
j=1
cj, 0λnj (6) 6as the separable part of A(n). We note that, if A(n) ∈ Q, then Asep (n) ∈ Q.A sequence A(n) is said to be a trace sequence if it is a Q-linear combi-nation of traces Tr( θn) = θn
1
· · · + θnr of algebraic numbers θ with Galois conjugates θ1 = θ, θ 2, . . . , θ r (with the understanding that Tr(0 n) is 1 for
n = 0 and 0 otherwise). Equivalently, a trace sequence is a C-finite se-quence for which the multiplicity of each characteristic root is mj = 1 and for which ci, 0 = cj, 0 in ( 5) whenever λi and λj have the same minimal poly-nomial. We further note as in [ BHS18 ] that the condition to be a trace sequence is equivalent to the property that the generating function F (x)is F (0) plus a Q-linear combination of functions of the form xu ′(x)/u (x), where u ∈ Q[x] is irreducible and u(0) = 1.
Example 2.1. For the Fibonacci numbers F (n), the representation ( 5)takes the form
F (n) = ϕn
+
− ϕn
−
√5 , ϕ± = 1 ± √5
2 . (7) Because the coefficients of ϕn
+
and ϕn
−
differ in sign, the Fibonacci numbers
F (n) are not a trace sequence. On the other hand, the Lucas numbers
L(n) = ϕn
+
ϕn
−
= tr[ M n], M =
[0 11 1
]
, (8) which satisfy the same recurrence as the Fibonacci numbers, are a trace sequence. In particular, it follows from Theorem 2.2 that the Lucas numbers
L(n) satisfy the Gauss congruences ( 9). Minton [ Min14 ] classified those C-finite sequences that satisfy the Gauss congruences ( 9) (see [ BHS18 ] for another proof of Minton’s result).
Theorem 2.2 (Minton, 2014) . Let A(n) be C-finite. Then the following are equivalent: (a) For all large enough primes p and for all r ≥ 1, A(n) satisfies the Gauss congruences
A(prn) ≡ A(pr−1n) (mod pr). (9)
(b) For all large enough primes p, A(n) satisfies the congruences
A(pn ) ≡ A(n) (mod p). (10)
(c) A(n) is a trace sequence.
7We conclude from Minton’s Theorem 2.2 the following result, which we employ in the proof of our main result (Theorem 1.6 ). To see the impor-tance of Lemma 2.3 , we note that, as we will show later (in Corollary 3.4 ), the sequences A(n) which are linear combinations of constant terms satisfy the congruences A(prn) ≡ A(pn ) (mod p) for all r ≥ 1 and large enough primes p.
Lemma 2.3. Let A(n) be C-finite. If A(n) satisfies the congruences
A(prn) ≡ A(pn ) (mod p) (11)
for all r ≥ 1 and for all large enough primes p, then the separable part
Asep (n) is a trace sequence. Proof. It follows from comparing ( 5) with ( 6) that for n large enough
A(n) = Asep (n) + n ˜A(n),
where Asep (n) and ˜A(n) are rational and satisfy the minimal recurrence for
A(n). In particular, each of these sequences is in Zp for large enough p, since denominators can only arise from the coefficients of the recurrence and the initial conditions. It follows that
Asep (pn ) ≡ A(pn ) (mod p)for all large enough p. Consequently, the congruences ( 11 ) are also satisfied by the C-finite sequence Asep (n). That is, for all r ≥ 1 and large enough pAsep (prn) ≡ Asep (pn ) (mod p). (12) On the other hand, let us consider the C-finite sequence Asep (n) in Fp. To avoid confusion, we denote this reduced sequence by asep
p
(n). Since the characteristic polynomial of Asep (n) over Q is separable, it is also separable for all large enough primes p (this can be seen by looking at the discriminant which, if nonzero over Q, can only vanish modulo finitely many primes). Consequently, we have a version of ( 6) with coefficients and roots in Fp.Namely,
asep
p
(n) =
d
∑
j=1
dj μnj , dj , μ j ∈ Fp.
Denoting with ϕp : Fp → Fp the Frobenius automorphism defined by ϕp(z) =
zp, we therefore have
asep
p
(psn) =
d
∑
j=1
dj μpsnj =
d
∑
j=1
dj (ϕsp(μj )) n
8for each s ∈ Z>0. Note that ϕp acts as a permutation on the roots μj .Writing m for the order of this permutation, we have ϕmp (μj ) = μj and thus
asep
p
(pmn) = asep
p
(n).
Consequently, the corresponding sequence Asep (n) satisfies
Asep (pmn) ≡ Asep (n) (mod p).
Combined with the congruences ( 12 ), this implies that
Asep (pn ) ≡ Asep (n) (mod p)for all large enough primes p. Theorem 2.2 therefore implies that Asep (n) is a trace sequence.
3 Congruences for constant terms
In this section we will show that if A(n) is a constant term sequence then it must satisfy certain congruences for large enough primes p. As a con-sequence, this allows us to conclude that the Fibonacci numbers are not a constant term sequence, thus answering Question 1.3 from the introduction. For a Laurent polynomial P ∈ Q[x±1], let deg( P ) denote the maximal degree with which any variable or its inverse appears in P .
Lemma 3.1. Let A(n) = ct[ P (x)nQ(x)] with P, Q ∈ Zp[x±1]. Then
A(prn + k) ≡ A(k) ct[ P (x)pr−1n] (mod pr)
for all integers n, k ≥ 0 and r ≥ 1, provided that p > deg( P kQ).Proof. Recall that (see, for instance, [ RY15 , Proposition 1.9]), for any Lau-rent polynomial F ∈ Zp[x±1],
F (x)pr
≡ F (xp)pr−1
(mod pr). (13) As in [ Str22 ], it follows from ( 13 ) that
A(prn + k) = ct[ P (x)pr nP (x)kQ(x)]
≡ ct[ P (xp)pr−1nP (x)kQ(x)] (mod pr)= ct[ P (x)pr−1nΛp[P (x)kQ(x)]] ,
9where Λ p denotes the section operator Λp
∑
k∈Zd
akxk
= ∑
k∈Zd
apkxk.
If p > deg( P kQ), then Λp[P (x)kQ(x)] = ct[ P (x)kQ(x)] = A(k)and the claim follows.
Example 3.2. For the Fibonacci numbers F (n), it is a well-known conse-quence of ( 7) that, modulo any prime p, we have the congruences
F (p) ≡
{
1, if p ≡ 1, 4 mod 5 ,
−1, if p ≡ 2, 3 mod 5 , (mod p).
Since this is incompatible with Lemma 3.1 (setting r = n = 1 and k = 0 implies that A(p) ≡ A(0) · c (mod p) for some c ∈ Q that is independent of p), we see that F (n) is not a constant term sequence. On the other hand, by Theorem 2.2 , the Lucas numbers L(n) (from ( 8)) satisfy the congruences L(prn) ≡ L(pn ) (mod p) for r ≥ 1 and p large enough. As such, Lemma 3.1 is not sufficient to conclude that L(n) is not a constant term sequence. However, we will be able to conclude in Example 4.2
the stronger result that both the Fibonacci numbers and the Lucas numbers cannot be expressed as a Q-linear combination of constant terms.
Corollary 3.3. Let A(n) = ct[ P (x)nQ(x)] with P, Q ∈ Zp[x±1]. Then
A(psn + k) ≡ A(prn + k) (mod pr)
for all integers n, k ≥ 0 and s ≥ r ≥ 1, provided that p > deg( P kQ).Proof. It follows from Lemma 3.1 and ( 13 ) that
A(psn + k) ≡ A(k) ct[ P (x)ps−1n] (mod ps)
≡ A(k) ct[ P (xps−r
)pr−1n] (mod pr)= A(k) ct[ P (x)pr−1n],
as claimed.
10 The simple but useful special case r = 1 and k = 0 of the corollary above takes the following form. Here, p is large enough if p > deg( Q) and
P, Q ∈ Zp[x±1].
Corollary 3.4. Let A(n) = ct[ P (x)nQ(x)] with P, Q ∈ Q[x±1]. If p is a large enough prime, then, for all integers n ≥ 0 and r ≥ 1,
A(pr n) ≡ A(pn ) (mod p).
4 C-finite sequences that are constant terms
In this section, we prove our main result, Theorem 1.6 stated in the in-troduction. We thus classify those C-finite sequences that are constant terms or linear combinations of such. We start by proving the following weaker version, since it illustrates well our approach and the usefulness of the congruences proved in Section 3. We then extend the argument to prove Theorem 1.6 in full generality.
Proposition 4.1. Let A(n) be a C-finite sequence. A(n) is a Q-linear com-bination of constant terms if and only if all characteristic roots are rational. Proof. For one direction, note that ct[( x + λ)n(λ/x )r ] =
(nr
)
λn = n(n − 1) · · · (n − r + 1)
r! λn. (14) Varying r, the right-hand side forms a basis for the span of the sequences (nrλn)n≥0. A sequence A0(n) of finite support can be represented as
A0(n) = ct[ xn(A(0) + A(1) x−1 + · · · + A(N )x−N )] ,
where N is the largest integer for which A0(N ) is non-zero. It therefore follows with ( 5) that, if all characteristic roots λ are rational, then A(n) can be represented as a linear combination of constant terms. On the other hand, suppose that A(n) is a linear combination of constant terms. Note that this implies that any shift A(n + k), where k ∈ Z≥0, is a linear combination of constant terms as well. These shifts generate the space VA of rational solutions of the minimal constant-coefficient recursion satisfied by A(n). Thus, any sequence in VA is a linear combination of constant terms. Assume, for contradiction, that there is a characteristic root λ that is not rational. Then among the sequences in VA there is always 11 a sequence B(n) of the form ( 6) (that is, B(n) equals its separable part) which is not a trace sequence. For instance, if λ1, . . . , λ d are the roots of the minimal polynomial of λ,then the space Vλ of rational sequences of the form b(n) = c1λn
1
· · · +
cdλnd , with c1, . . . , c d ∈ Q, is a d-dimensional subspace of VA. Clearly, each sequence in Vλ is of the form ( 6). Note that λn
1
· · · + λnd and its multiples are the only trace sequences in Vλ. Since d ≥ 2, we can therefore choose a sequence B(n) in Vλ that is not a trace sequence. It follows from Corollary 3.4 that B(n) satisfies the congruences
B(prn) ≡ B(pn ) (mod p)for all r ≥ 1 and all large enough primes p. Lemma 2.3 therefore implies that Bsep (n) = B(n) is a trace sequence. This is a contradiction, and we conclude that all characteristic roots must be rational.
Example 4.2. Recall from ( 7) that the Fibonacci numbers F (n) are C-finite with characteristic roots (1 ± √5) /2. Since these are not rational, it follows from Proposition 4.1 that F (n) cannot be expressed as a linear combination of constant terms. The same argument applied to ( 8) shows that the Lucas numbers L(n)cannot be expressed as a linear combination of constant terms either. Al-ternatively, this can also be concluded from the relationship 2L(n + 1) − L(n) = 5 F (n)combined with the fact that Fibonacci numbers are not a sum of constant terms. We next prove the case r = 1 of Theorem 1.6 , that is Proposition 1.4 ,stating that a C-finite sequence A(n) is a single constant term if and only if it has a single characteristic root λ and λ ∈ Q.
Proof of Proposition 1.4 . It follows from ( 5) and ( 14 ) that if A(n) is a C-finite sequence with the single characteristic root λ ∈ Q (possibly repeated or possibly 0), then A(n) is a constant term, namely A(n) = ct[( x+λ)nQ(x−1)] for a suitable polynomial Q(x). On the other hand, suppose that A(n) = ct[ P (x)nQ(x)] is a single con-stant term. Since A(n) is a C-finite sequence, it has a representation of the form ( 5) or, equivalently,
A(n) = A0(n) +
d
∑
j=1
λnj pj (n) (15) 12 for pairwise distinct λj ∈ Q× and nonzero pj (n) ∈ Q[n]. As before, A0(n)is a sequence with finite support, corresponding to the characteristic root 0. It follows from Proposition 4.1 that all characteristic roots λj are rational, and this further implies that pj (n) ∈ Q[n]. Let c0 = ct[ P (x)] ∈ Q. From Lemma 3.1 (with r = 1 and n = 1) it follows that
A(p + n) ≡ A(n) · c0 (mod p)for all n ≥ 0 and all large enough primes p (namely, p > deg( P nQ) and large enough so that c0 ∈ Zp). Combining this congruence with ( 15 ) and applying Fermat’s little theorem to reduce λp+nj and pj (p + n) modulo p to
λn+1
j
and pj (n) respectively, we find that
d
∑
j=1
λn+1
j
pj (n) ≡ c0
A0(n) +
d
∑
j=1
λnj pj (n)
(mod p) (16) for all large enough p (in particular, so that p is larger than any denominator occuring in the pj (n) and so that A0(p+n) = 0). Note that both sides of ( 16 )are independent of p. Since they agree modulo any large enough p, it follows that they must be equal (for each fixed value of n). Accordingly, we have the identity
d
∑
j=1
λn+1
j
pj (n) = c0
A0(n) +
d
∑
j=1
λnj pj (n)
for all n ≥ 0. (17) Note that both sides of ( 17 ) are C-finite sequences so that, because the representation ( 15 ) is unique, we must have, in particular, c0A0(n) = 0. If
c0 = 0 then it follows by comparison with the left-hand side of ( 17 ) that
d = 0 so that A(n) = A0(n) with the single characteristic root λ = 0. In the other case, that is if c0 6 = 0, we have A0(n) = 0, so 0 is not a characteristic root. Further comparing both sides of ( 17 ), we find that λj = c0 for all j.Since the λj are distinct, we conclude that d = 1 so that A(n) = λn
1
p1(n)with the single characteristic root λ1 ∈ Q×.
We now extend Proposition 1.4 to the case of r-term Q-linear combi-nations of constant terms, thus proving our main result, Theorem 1.6 . We recall that its statement is that a C-finite A(n) sequence is an r-term Q-linear combination of constant terms if and only if it has at most r characteristic roots, all of which are rational. 13 Proof of Theorem 1.6 . The case r = 1 is proved by Proposition 1.4 . With the same argument as in ( 14 ) it follows that any C-finite sequence with r
characteristic roots, all of which are rational, can be represented as a linear combination of r constant terms. Therefore, suppose that r > 1 and that
A(n) = ct[ P1(x)nQ1(x)] + · · · + ct[ Pr (x)nQr(x)] is an r-term Q-linear combination of constant terms with Pj , Q j ∈ Q[x±1]. We need to show that A(n) has at most r characteristic roots, all of which are rational. As in the proof of Proposition 1.4 , we find that all characteristic roots of A(n) are rational and that A(n) can be represented in the form ( 15 )with pj (n) ∈ Q[n]. Let cj = ct[ Pj (x)] ∈ Q. It follows from Lemma 3.1 that
A(p + n) ≡ c1 ct[ P1(x)nQ1(x)] + · · · + cr ct[ Pr (x)nQr(x)] (mod p)for all n ≥ 0 and all large enough primes p. On the other hand, for large p,by Fermat’s little theorem,
A(p + n) ≡
d
∑
j=1
λn+1
j
pj (n) (mod p).
Note that the right-hand sides of the last two congruences are independent of p. Since the congruences hold modulo all large enough primes, we conclude that
d
∑
j=1
λn+1
j
pj (n) = c1 ct[ P1(x)nQ1(x)] + · · · + cr ct[ Pr (x)nQr(x)] .
Note that the sequence
B(n) :=
d
∑
j=1
λn+1
j
pj (n) − c1A(n) =
d
∑
j=1
(λj − c1)λnj pj (n) − c1A0(n)is C-finite and is an ( r − 1)-term Q-linear combination of constant terms. By induction, we may conclude that B(n) has at most r − 1 characteristic roots, all of which are rational. By comparison with ( 15 ), we see that A(n)has at most one more characteristic root than B(n). Thus A(n) has at most
r characteristic roots, which is what we had to show.
14 Theorem 1.6 classifies those rational recursive sequences with constant coefficients which can be represented as a linear combination of r constant terms. In particular, a rational C-finite sequence A(n) is a linear combina-tion of constant terms if and only if all of its characteristic roots are rational. It is natural to wonder whether we can restrict to integer sequences and con-clude that all characteristic roots must be integral. This can be achieved by using the following proposition 1, that we could not locate in the vast literature on C-finite sequences.
Proposition 4.3. Let A(n) be a C-finite sequence with characteristic roots
λ1, . . . , λ d ∈ Q. If A(n) is an integer sequence, then λ1, . . . , λ d ∈ Z.Proof. By assumption, A(n) is equal to ∑di=1 pi(n)λni , where the λi’s are mutually distinct rational numbers and the pi(x)’s are polynomials in Q[x]. We will prove that if for some N ∈ Z \ { 0} we have that A(n) ∈ 1
N
Z for all n, then all the λi’s are integers. Let us start with the observation that this is true if d = 1; indeed, if
λ1 = a/b with coprime integers a, b , and p1(n) = q(n)/c with q(x) ∈ Z[x]and c ∈ Z, then the assumption “ A(n) = p1(n)λn
1
∈ 1
N
Z for all n” implies that bn divides N q (n) for all n, hence b = 1. Let us now treat the general case. Denote by V the d × d Vandermonde matrix attached with the λi’s, that is V = ( λj−1
i
)1≤i,j ≤d. Since the λi’s are mutually distinct, V is in GL d(Q). Therefore, the equality [p1(n)λn
1
, . . . , p d(n)λnd ] · V = [ A(n), . . . , A (n + d − 1)] implies that each term pi(n)λni is equal to 1 / det( V ) times a linear combi-nation of the integers A(n), . . . , A (n + d − 1) with fixed rational coefficients. In other terms, each pi(n)λni is in 1
Ni
Z for some Ni ∈ Z \ { 0} independent of n. By the case d = 1, this implies that all the λi’s are integers.
Remark 4.4. Proposition 4.3 also follows 2 from “Fatou’s lemma” (not to be confused with Fatou’s famous result in Lebesgue integration theory) which states that if f (x) ∈ Z is a rational function, then one can write it as
f (x) = P (x)/Q (x) with P, Q ∈ Z[x] and Q(0) = 1. This lemma is stated in Fatou’s 1904 short communication [ Fat04 , p. 313] and proved in his PhD thesis [ Fat06 , p. 369] To see how Proposition 4.3 follows from Fatou’s lemma, note that since the generating function of the sequence A(n) is rational with integer co-efficients, we may represent it as in the lemma. The characteristic roots
1The proof of Prop. 4.3 was communicated to us by Carlo Sanna (Politecnico di Torino).
2We are indebted to the anonymous referee for pointing out this connection.
15 λ1, . . . , λ d are the zeros of the reversal of Q(x) which is monic because
Q(0) = 1. By Gauss’ lemma it follows that the rational numbers λ1, . . . , λ d
are integers. We conclude this section with the following immediate consequence of Proposition 1.4 , Theorem 1.6 and Proposition 4.3 :
Corollary 4.5. Let A(n) ∈ Z be a C-finite sequence. A(n) is a constant term if and only if it has a single characteristic root λ and λ ∈ Z. More generally, A(n) is an r-term Q-linear combination of constant terms if and only if it has at most r distinct characteristic roots, all of which are integral.
5 An analog of Minton’s theorem
In this section, we record the following result which, though having a much simpler proof, is pleasingly similar to Theorem 2.2 due to Minton [ Min14 ]. Moreover, this result gives a classification of constant term sequences of the form A(n) = ct[ P (x)n] among all constant term sequences ct[ P (x)nQ(x)].
Proposition 5.1. Suppose A(n) = ct[ P (x)nQ(x)] with P, Q ∈ Q[x±1].Then the following are equivalent: (a) For all large enough primes p and for all r ≥ 1, A(n) satisfies the Gauss congruences (9).(b) For all large enough primes p, A(n) satisfies the congruences (10 ).(c) A(n) = A(0) ct[ P (x)n].Proof. We conclude from Lemma 3.1 with r = 1 and k = 0 that
A(pn ) ≡ A(0) ct[ P (x)n] (mod p)for large enough p (namely, if p > deg( Q)). If A(n) satisfies the congru-ences ( 10 ), we find that, for large enough p,
A(n) ≡ A(0) ct[ P (x)n] (mod p).
In that case, since this congruence holds modulo infinitely many p, we con-clude the equality A(n) = A(0) ct[ P (x)n]. Thus the third condition follows from the second. To complete the proof, we need to show that the third condition implies the first. This follows from Lemma 3.1 with k = 0 and Q = 1.
16 Remark 5.2. Note that Proposition 5.1 does not imply that if A(n) = ct[ P (x)nQ(x)] satisfies the Gauss congruences ( 9) for large enough primes, then Q must be constant. For instance, for any P (x) ∈ Z[x±1], the con-stant terms ct[ P (x2)n(1 + x)] = ct[ P (x2)n] = ct[ P (x)n] satisfy the Gauss congruences for all primes p, even though the first constant term has a non-constant Q. Proposition 5.1 rather shows that if (a) or (b) are fulfilled, then
Q can be replaced by ct[ Q].
6 Hypergeometric constant terms
Exiting the class of C-finite sequences, we find it natural to ask (Question 1.8
in the introduction): Which hypergeometric sequences 3 A(n) are constant term sequences? The reason for the specialization to hypergeometric sequences is three-fold. First, it can be argued that it is the easiest P-finite case. Second, similar to constant terms, hypergeometric sequences are not stable under addition. Finally, as we will see below in Lemma 6.1 , the congruences proven in Section 3 behave nicely with the hypergeometric assumption. It follows from Lemma 3.1 (specialized to n = 1 and r = 1) that if
A(n) = ct[ P (x)nQ(x)] with P, Q ∈ Q[x±1], then
A(p + k) ≡ A(k) ct[ P (x)] (mod p)for all integers k ≥ 0, provided that p > deg( P kQ) and P, Q ∈ Zp[x±1]. In other words, if A(n) is a constant term, then there exists a constant c ∈ Q
such that, for each k ∈ Z≥0, the congruences
A(p + k) ≡ A(k) · c (mod p) (18) hold for all large enough primes p. We shall now show that, for hypergeo-metric sequences, the congruences ( 18 ) follow from the base case k = 0.
Lemma 6.1. Let A(n) be a hypergeometric sequence. Suppose that there exists a constant c ∈ Q such that
A(p) ≡ c (mod p) (19)
for all large enough primes p. Then, for each k ∈ Z≥0, the congruence (18 )
holds for sufficiently large primes p.
3Recall that a sequence A(n) is hypergeometric if it satisfies a first-order recurrence
α(n)A(n+ 1) = β(n)A(n) for some polynomials α(n), β (n)∈Q[n]. For our purposes we will assume α(n)6= 0 for all n≥0.
17 Proof. Since A(n) is hypergeometric, we have A(n + 1) = ρ(n)A(n) for a rational function ρ(n) = β(n)/α (n) with α(n), β (n) ∈ Z[n]. Fix k ∈ Z≥0
and suppose that the congruence ( 18 ) holds for all large enough primes p.By applying the hypergeometric recurrence twice, we obtain
A(p + k + 1) = ρ(p + k)A(p + k) ≡ ρ(k)cA (k) = cA (k + 1) (mod p),
which is ( 18 ) with k + 1 in place of k. Here we used that ρ(p + k) ≡ ρ(k)(mod p), which holds true provided that α(k) 6 ≡ 0 (mod p). The latter is true for all sufficiently large primes p since, by assumption, α(k) 6 = 0. The claim therefore follows by induction on k.
Remark 6.2. Note that Lemma 6.1 does not hold for non-hypergeometric sequences in general. For instance, it does not hold for the Lucas numbers
L(n) as defined in ( 8). These form a trace sequence so that, by Minton’s Theorem 2.2 , the Gauss congruences ( 9) are satisfied. In the case n = 1, these imply the congruences ( 19 ). However, the Lucas numbers do not satisfy the congruences ( 18 ) for k > 0. Lemma 3.1 gives a necessary condition for A(n) to be a constant term sequence. It is natural to wonder whether, or to what extent, this condition is sufficient: Is any integer hypergeometric sequence A(n) that satisfies the congruences ( 19 ) a constant term? Natural sources of potential counterex-amples to this question are families of integer sequences that are quotients of binomial coefficients but cannot be written as products of those, for example
A(n) = (8n
4n
)( 4nn
)( 2nn
)−1 (see [ Bob09 , Thm. 1.2]). We recall that the corresponding question for diagonals ( 1) (namely, to classify which hypergeometric sequences A(n) are coefficients of diagonals) also remains open. The following conjecture due to Christol [ Chr90 , Con-jecture 4, p. 55] attempts such a classification. In its statement, we call a sequence ( A(n)) n≥0 almost integral if there exists a positive integer K such that Kn+1 A(n) ∈ Z for all integers n ≥ 0. An almost integral sequence with (at most) geometric growth is called globally bounded .
Conjecture 6.3 ([ Chr90 ]) . Let (A(n)) n≥0 be a sequence of rational num-bers. The generating function ∑
n≥0
A(n)tn is the diagonal of a rational function if and only if (A(n)) n≥0 is P-finite and globally bounded.
Any hypergeometric sequence is P-finite since it satisfies, by definition, a recurrence with polynomial coefficients of order one. Moreover, thanks to a result of Christol [ Chr87 , Chr90 ] it is easy to check when a hypergeometric sequence is integral (in the case when α(n) and β(n) in the definition split 18 in Q[n]). This makes hypergeometric sequences a natural source of potential counterexamples to Conjecture 6.3 . We refer to [ BBC +13 ], [ AKM20 ] and [BY22 ] for recent progress in this area. Here, we only mention that even for
A(n) =
( 1
9
)
n
( 4
9
)
n
( 5
9
)
n
n!2 ( 1
3
)
n
(20) the conjecture is open. In other words, it is an open question whether the sequence ( 20 ) is the diagonal of a rational function. On the other hand, we will show in this section that ( 20 ) is not a constant term. Before doing so, we first prove the following result answering Question 1.8 for a special family of hypergeometric sequences.
Lemma 6.4. Let m ≥ 2 be an integer and consider the sequence
Am(n) =
( 1
m
)
n
(1 − 1
m
)
n
n!2 . (21)
(a) Am(n) is a diagonal for all m ≥ 2.(b) Am(n) is a constant term if and only if m ∈ { 2, 3, 4, 6}.
Note that the classification in Lemma 6.4 suggests that constant term sequences are special among diagonals and often have significant additional arithmetic properties. Indeed, the cases m ∈ { 2, 3, 4, 6} (see A002894 ,
A006480 , A000897 and A113424 in the on-line encyclopedia of integer se-quences [ ST22 ]) correspond precisely to those special hypergeometric func-tions underlying Ramanujan’s theory of elliptic functions ( m = 2 being the classical case and m = 3 , 4, 6 corresponding to the alternative bases). We refer to [ BBG95 ] for more information.
Example 6.5. The hypergeometric sequence
B(n) = 5 3n
( 1
5
)
n
( 4
5
)
n
n!2 = 1 , 20 , 1350 , 115500 , 10972500 , . . . (22) is an integer sequence and grows at most exponentially. As suggested by Christol’s Conjecture 6.3 and stated in Lemma 6.4 , the sequence B(n) is a diagonal. However, B(n) is not a constant term. The proof of Lemma 6.4
in this case proceeds by showing that we have the congruences
B(p) ≡
{
20 , if p ≡ ± 1 mod 5 ,
30 , otherwise , (mod p),
which contradict Lemma 6.1 .19 Proof of Lemma 6.4 . Part (a) follows from the fact that the generating func-tion of Am(n) is the Hadamard (term-wise) product of (1 − x)−1/m and (1 − x)1/m −1. The latter are algebraic functions and hence diagonals by a result of Furstenberg [ Fur67 ]. Since diagonals are closed under Hadamard products [ Chr88 ], it follows that Am(n) is a diagonal. That Am(n) is a constant term if m ∈ { 2, 3, 4, 6} follows from the fol-lowing alternative representations as products of binomial coefficients: 24nA2(n) = (2 n)! 2
n!4 =
(2nn
)2
,
33nA3(n) = (3 n)!
n!3 =
(3n
2n
)( 2nn
)
,
43nA4(n) = (4 n)!
(2 n)! n!2 =
(4n
2n
)( 2nn
)
,
24n33nA6(n) = (6 n)!
(3 n)!(2 n)! n! =
(6n
3n
)( 3nn
)
.
In the remainder, we will show that Am(n) is not a constant term if m 6 ∈{2, 3, 4, 6}. If m is coprime to p (as it is for large enough primes p), then the right-hand side of
mp ( 1
m
)
p
= 1 · (m + 1)(2 m + 1) · · · (( p − 1) m + 1) is a product of all the residues modulo p. In particular, exactly one factor is of the form ap where a ∈ { 1, 2, . . . , m − 1} is characterized by ap ≡ 1(mod m). By Wilson’s theorem, we therefore have
mp ( 1
m
)
p
≡ − ap (mod p2)or, equivalently,
mp ( 1
m
)
p
p! ≡ a (mod p).
Similarly,
mp (1 − 1
m
)
p
= ( m − 1)(2 m − 1) · · · (pm − 1) and, again, the right-hand side features a product of all residues modulo p.Exactly one factor is of the form bp where b ∈ { 1, 2, . . . , m − 1} is charac-terized by bp ≡ − 1 (mod m). It follows that b = m − a. Combined, we conclude that
m2pAm(p) = m2p ( 1
m
)
p
(1 − 1
m
)
p
p!2 ≡ a(m − a) (mod p). (23) 20 Since a ∈ { 1, 2, . . . , m − 1} is characterized by a ≡ 1/p (mod m) it, in particular, depends only on the residue class of p modulo m. As p ranges through all primes, it follows from Dirichlet’s theorem on primes in arith-metic progressions, that each value a ∈ { 1, 2, . . . , m − 1} with a coprime to m appears infinitely many times. There are φ(m) many such values of a,where φ is Euler’s totient function. Consequently, the quantity a(m − a) on the right-hand side of ( 23 ) takes φ(m)/2 many different values as p ranges through all primes p > m .On the other hand, if Am(n) is a constant term sequence, then by ( 19 )there exists a constant c ∈ Q such that m2pAm(p) ≡ c (mod p) for all large enough primes p. If ( 23 ) holds for infinitely many p, we necessarily have
c = a(m − a), which is only possible if φ(m)/2 = 1. Thus, if φ(m) > 2 then m2pAm(p) cannot satisfy the congruences ( 19 )for all large enough primes and, hence, the sequences m2nAm(n) and Am(n)cannot be constant terms. Since φ(m) > 2 for all integers m ≥ 2 except for
m ∈ { 2, 3, 4, 6}, the claim follows.
For hypergeometric sequences, we therefore have the following inclusions
{constant terms } ( {diagonals } ⊆ { P-finite & globally bounded seq’s } .
We note that these inclusions are also true for C-finite as well as for P-finite sequences. An example for the strictness of the first inclusion in the realm of hypergeometric sequences is given by the sequence ( 22 ) and in the class of C-finite sequences by the Fibonacci numbers. The second inclusion is a consequence of a result due to Lipshitz [ Lip88 ] and it is strict if and only if Christol’s Conjecture 6.3 (restricted to hypergeometric sequences) is false. A potential candidate of a globally bounded hypergeometric sequence that is not a diagonal is sequence ( 20 ). We now show that this sequence is not a constant term.
Lemma 6.6. The hypergeometric sequence A(n) defined in (20 ) is not a constant term sequence. Proof. Proceeding as in the proof of Lemma 6.4 , we find
mp ( r
m
)
p
= r(m + r) · · · (( p − 1) m + r),
where the right-hand side is a product over all residues modulo p. Exactly one factor is of the form ap where a ∈ { 1, 2, . . . , m − 1} is characterized by
ap ≡ r (mod m). In that case,
mp ( r
m
)
p
p! ≡ a (mod p).
21 If p ≡ 1 (mod 9), we therefore find 9p ( 1
9
)
p
p! ≡ 3p ( 1
3
)
p
p! ≡ 1, 9p ( 4
9
)
p
p! ≡ 4, 9p ( 5
9
)
p
p! ≡ 5 (mod p),
which combine to 35pA(p) = 3 5p
( 1
9
)
p
( 4
9
)
p
( 5
9
)
p
p!2 ( 1
3
)
p
≡ 1 · 4 · 5
1 ≡ 20 (mod p).
On the other hand, if p ≡ − 1 (mod 9), then 9p ( 1
9
)
p
p! ≡ 8, 3p ( 1
3
)
p
p! ≡ 2, 9p ( 4
9
)
p
p! ≡ 5, 9p ( 5
9
)
p
p! ≡ 4 (mod p),
which combine to 35pA(p) = 3 5p
( 1
9
)
p
( 4
9
)
p
( 5
9
)
p
p!2 ( 1
3
)
p
≡ 8 · 4 · 5
2 ≡ 80 (mod p).
As in the proof of Lemma 6.4 we conclude that 3 5nA(n) and, hence, A(n)cannot be a constant term.
Acknowledgments. Our warm thanks go to the PolSys team of the Sor-bonne University for having provided a very nice atmosphere for the three authors to collaborate during summer 2022. We are grateful to Carlo Sanna
who provided the elegant and elementary proof of Proposition 4.3 and per-mitted us to include it in our article. We also thank the anonymous referee for the careful reading and for bringing Proposition 4.3 in connection to Fatou’s lemma (see Remark 4.4 ). The first and third authors were supported by ANR-19-CE40-0018 De Rerum Natura
and the WTZ collaboration /Amadeus project FR-09/2021 (46411YJ). The second author gratefully acknowledges support through a Collaboration Grant (#514645) awarded by the Simons Foundation. The third author is funded by DOC scholarship P-26101 of the ¨OAW .
References
[AKM20] Y. Abdelaziz, C. Koutschan, and J.-M. Maillard. On Christol’s conjecture. Journal of Physics A: Mathematical and Theoret-ical , 53(20):205201, 2020. 22 [BBC +13] A. Bostan, S. Boukraa, G. Christol, S. Hassani, and J.-M. Maillard. Ising n-fold integrals as diagonals of rational func-tions and integrality of series expansions. Journal of Physics A: Mathematical and Theoretical , 46(18):185202, May 2013. [BBG95] B. C. Berndt, S. Bhargava, and F. G. Garvan. Ramanujan’s Theories of Elliptic Functions to Alternative Bases. Trans-actions of the American Mathematical Society , 347(11):4163, November 1995. [BHS18] F. Beukers, M. Houben, and A. Straub. Gauss congruences for rational functions in several variables. Acta Arithmetica ,184:341–362, 2018. [Bob09] J. W. Bober. Factorial ratios, hypergeometric series, and a family of step functions. Journal of the London Mathematical Society. Second Series , 79(2):422–444, 2009. [BY22] A. Bostan and S. Yurkevich. On a class of hypergeometric diagonals. Proceedings of the American Mathematical Society ,150(3):1071–1087, 2022. [Chr87] G. Christol. Fonctions hyperg´ eom´ etriques born´ ees. Groupe de travail d’analyse ultram´ etrique , 14:1–16, 1986-1987. Talk no. 8. [Chr88] G. Christol. Diagonales de fractions rationnelles. In S´ eminaire de Th´ eorie des Nombres, Paris, 1986–87 , volume 7 of Progr. Math. , pages 65–90. Birkh¨ auser Boston, Boston, MA, 1988. [Chr90] G. Christol. Globally bounded solutions of differential equa-tions. In K. Nagasaka and E. Fouvry, editors, Analytic Number Theory , number 1434 in Lecture Notes in Mathematics, pages 45–64. Springer Berlin Heidelberg, January 1990. [EvdPSW03] G. Everest, A. van der Poorten, I. Shparlinski, and T. Ward.
Recurrence Sequences , volume 104 of Mathematical Surveys and Monographs . American Mathematical Society, Provi-dence, RI, 2003. [Fat04] P. Fatou. Sur les s´ eries entieres a coefficients entiers. Comptes-Rendus de l’Acad´ emie des Sciences , 138(A):342–344, 1904. 23 [Fat06] P. Fatou. S´ eries trigonom´ etriques et s´ eries de Taylor. Acta Math. , 30(1):335–400, 1906. [Fur67] H. Furstenberg. Algebraic functions over finite fields. Journal of Algebra , 7(2):271–277, October 1967. [Gor21] O. Gorodetsky. New representations for all sporadic Ap´ ery-like sequences, with applications to congruences. Experimental Mathematics , 2021. [KP11] M. Kauers and P. Paule. The Concrete Tetrahedron . Springer-Verlag, 2011. [Lip88] L. Lipshitz. The diagonal of a D-finite power series is D-finite.
Journal of Algebra , 113(2):373–378, 1988. [Min14] G. T. Minton. Linear recurrence sequences satisfying congru-ence conditions. Proceedings of the American Mathematical Society , 142(7):2337–2352, April 2014. [P´ o22] G. P´ olya. Sur les s´ eries enti` eres, dont la somme est une fonction alg´ ebrique. L’Enseignement Math´ ematique , 22:38– 47, 1921/1922. [RY15] E. Rowland and R. Yassawi. Automatic congruences for diag-onals of rational functions. Journal de Th´ eorie des Nombres de Bordeaux , 27(1):245–288, 2015. [ST22] N. J. A. Sloane and The OEIS Foundation Inc. The On-Line Encyclopedia of Integer Sequences, 2022. Published electron-ically at .[Str14] A. Straub. Multivariate Ap´ ery numbers and supercongruences of rational functions. Algebra & Number Theory , 8(8):1985– 2007, 2014. [Str22] A. Straub. On congruence schemes for constant terms and their applications. Research in Number Theory , 8(3), May 2022. [Zag18] D. B. Zagier. The arithmetic and topology of differential equa-tions. In V. Mehrmann and M. Skutella, editors, Proceedings of the European Congress of Mathematics, Berlin, 18-22 July, 2016 , pages 717–776. European Mathematical Society, 2018. 24 |
6813 | https://en.wiktionary.org/wiki/echelon | Jump to content
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Contents
Beginning
1 English
1.1 Etymology
1.2 Pronunciation
1.3 Noun
1.3.1 Alternative forms
1.3.2 Derived terms
1.3.3 Related terms
1.3.4 Translations
1.3.5 See also
1.4 Verb
1.4.1 Translations
1.5 Adjective
1.6 Further reading
1.7 Anagrams
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From Wiktionary, the free dictionary
See also: Echelon and échelon
English
[edit]
WOTD – 10 February 2018
Etymology
[edit]
Borrowed from French échelon (“rung; echelon”), from échelle (“ladder”) + -on (diminutive suffix). Échelle is derived from Latin scāla (“ladder”), from scandō (“to ascend, climb”), from Proto-Indo-European skend- (“to jump”).
Pronunciation
[edit]
IPA(key): (Received Pronunciation) /ˈɛʃ.ə.lɒn/, /ˈeɪ-/, /ˈɛk.ə.lɒn/
(General American) IPA(key): /ˈɛʃ.əˌlɑn/
| | |
--- |
| Audio (General Australian): | (file) |
Hyphenation: ech‧e‧lon
Noun
[edit]
echelon (plural echelons)
A level or rank in an organization, profession, or society.
1987, “Scenario A: The Reference Scenario”, in C. F. Hollander, H. A. Becker, editors, Growing Old in the Future: Scenarios on Health and Ageing 1984–2000: Scenario-report, Commissioned by the Steering Committee on Future Health Scenarios, Dordrecht: Martinus Nijhoff Publishers, →DOI, →ISBN, section 2.5.3 (GP Care), page 85:
: Other important functions performed by the GP [general practitioner] are those of referring patients to other (health) care facilities and acting as contact person for other providers of aid, both for other facilities in first echelon care and with respect to second echelon care (outpatient care and treatment in hospital).
2004, Alfred Kuo-liang Ho, “Deng’s Political Reforms”, in China’s Reforms and Reformers, Westport, Conn.: Praeger Publishers, Greenwood Publishing Group, →ISBN, page 120:
: Officials in China are divided into three echelons, or generations. The first echelon joined the Communist Party in the 1920s, soon after the party was founded. […] The second echelon joined the Communist Party before the founding of the People's Republic of China in 1949. The party was then still struggling, and many of the second echelon died during the wars. They were again true patriots. The third echelon joined the Communist Party after the founding of the People's Republic.
2018, “Mad I Got It”, performed by Denzel Curry:
: Still in the hood and I still sip teaWhen I go in, I'ma go for the winOn point like syringe and it's like thatLiving life full of sin, gotta cup full of Hen'Put my tears in a pen and I write backIce pack cause' my heart lostI'm so cold that I Jack FrostI'm 'bout be on the upper echelonNiggas gonna hate on the man that I become
2. (cycling) A line of riders seeking maximum drafting in a crosswind, resulting in a diagonal line across the road.
1972, Journal of the American Association for Health, Physical Education, and Recreation, volume 43, [Washington, D.C.]: American Association for Health, Physical Education, and Recreation, →OCLC, page 22, column 2:
: In an echelon, in which several cyclists are sitting in on one another, each rider takes his turn of about 200 meters at the front before dropping to the rear.
1989, Fred Matheny, Bicycling Magazine’s Complete Guide to Riding and Racing Techniques, Emmaus, Pa.: Rodale Press, →ISBN, page 99:
: Cyclists in an echelon take up more room in the traffic lane than they do when riding single or double file.
3. (military) A formation of troops, ships, aircraft, etc., in diagonal parallel rows. [from late 18th c.]
1833 October, “a royalist”, “Sketches of the War of the French in Spain in the Year 1823”, in The United Service Journal and Naval and Military Magazine, part III, number 59, London: Published for Henry Colburn, by Richard Bentley, New Burlington Street, →OCLC, page 183:
: The troops selected by his Royal Highness for this daring exploit, consisted of the war battalions of the 3d, 6th, and 7th regiments of the Royal Guard, forming the first echelon, […]
1863, Ed. [Édouard] De La Barre Duparcq, “Infantry Formation and Tactics”, in George W[ashington] Cullum, transl., Elements of Military Art and History: Comprising the History and Tactics of the Separate Arms; the Combination of the Arms; and the Minor Operations of War, New York, N.Y.: D[avid] Van Nostrand, 192 Broadway, page 84:
: The order in echelons is favorable for attack, because it readily conforms to the nature of the ground, and does not necessitate engaging more than a part of the forces; it is adopted for the purpose of attacking a particular point of the enemy's line.
1899, Winston Spencer Churchill, “The Battle of Omdurman: September 2, 1898”, in F[rancis] Rhodes, editor, The River War: An Historical Account of the Reconquest of the Soudan [...] In Two Volumes, volume II, London: Longmans, Green, and Co., 89 Paternoster Row, →OCLC, pages 145–146:
: As soon as the infantry had replenished their ammunition, they wheeled to the left in échelon of brigades, and began to march towards Surgham ridge. The movements of a great force are slow. It was not desirable that the British division, which led the échelon, should remain in the low ground north of Surgham—where it was commanded, had no field of fire, and could see nothing—and accordingly both these brigades moved forward almost together to occupy the crest of the ridge.
2022 March 28, Julian Borger, quoting Yaroslav Honchar, “The drone operators who halted Russian convoy headed for Kyiv”, in The Guardian:
: “The first echelon of the Russian force was stuck without heat, without oil, without bombs and without gas. And it all happened because of the work of 30 people,” Honchar said.
Alternative forms
[edit]
échelon
Derived terms
[edit]
echelon form
echelonic
echelon lens
echelonment
echelon parking
multiechelon
rear echelon
Related terms
[edit]
en echelon
Translations
[edit]
level or rank
| |
| Czech: stupeň (cs) m, úroveň (cs) f, vrstva (cs) f Danish: grad (da) c, trin n, lag (da) n Finnish: taso (fi) French: échelon (fr) m German: Stufe (de), Rang (de) m, Skala (de) f or n Indonesian: eselon (id) Norwegian: Bokmål: grad (no) m, nivå (no) n Persian: پله (fa) (pelle) Portuguese: escalão (pt) Russian: звено́ (ru) n (zvenó), эшело́н (ru) m (ešelón) Spanish: escalón (es) m, esfera (es) f, cuadro (es) m, nivel (es) m, rango (es) m, peldaño (es) m, estrato (es) m, estamento (es) m Swedish: grad (sv), nivå (sv), skikt (sv) |
(cycling) line of riders seeking maximum drafting in a crosswind
| |
| Czech: terezín (cs) m Dutch: waaier (nl) m Spanish: abanico (es) m |
(military) formation of troops
| |
| Belarusian: эшэло́н m (ešelón) Bulgarian: ешело́н m (ešelón) Czech: (vojenský) útvar (cs) m Danish: led (da) n, niveau (da) n Finnish: vinorivi French: échelon (fr) m German: Echelon m Indonesian: eselon (id) Italian: scaglione (it) m Polish: eszelon (pl) m Portuguese: escalão (pt) m Russian: эшело́н (ru) m (ešelón) Spanish: escalón (es) Ukrainian: ешело́н m (ešelón) |
See also
[edit]
chevron
Verb
[edit]
echelon (third-person singular simple present echelons, present participle echeloning, simple past and past participle echeloned)
(transitive, military) To form troops into an echelon.
1860, François-Xavier Garneau, “Battle of Carillon (Ticonderoga). 1758.”, in Andrew Bell, transl., History of Canada, from the Time of Its Discovery till the Union Year (1840–1): Translated from “L’Histoire du Canada” of F. X. Garneau, Esq., and Accompanied with Illustrative Notes, etc., etc. [...] In Three Volumes, volume II, Montreal, Que.: Printed and published by John Lovell, St. Nicholas Street, →OCLC, page 205:
: July 1, Montcalm made a movement in advance, echeloning his troops from Fort Carillon to the foot of Lake George, to curb the enemy, and obstruct their landing.
1968, Earl F[rederick] Ziemke, “Offensives on Both Flanks—the South Flank”, in Stalingrad to Berlin: The German Defeat in the East (Army Historical Series), Washington, D.C.: Office of the Chief of Military History, U.S. Army, →OCLC; republished Washington, D.C.: Center of Military History, United States Army, 2002, →OCLC, page 225:
: Behind the 17th Panzer Division the corps echeloned the main force of its other division, the 16th Panzer Division.
Translations
[edit]
to form troops into an echelon
| |
| Belarusian: эшалані́раваць impf or pf (ešalaníravacʹ) Bulgarian: ешелонирам (ešeloniram) French: échelonner (fr) Italian: scaglionare (it) Polish: eszelonować impf Portuguese: escalonar Russian: эшелони́ровать (ru) impf or pf (ešelonírovatʹ) Spanish: escalonar (es) Ukrainian: ешелонува́ти impf or pf (ešelonuváty) |
Adjective
[edit]
| Examples |
| An example of a 4×6 matrix in row echelon form (the indicate the nonzero pivots, and the indicate entries which may be zero or nonzero): |
echelon (not comparable)
(linear algebra) Of a matrix: having undergone Gaussian elimination with the result that the leading coefficient or pivot (that is, the first nonzero number from the left) of a nonzero row is to the right of the pivot of the row above it, giving rise to a stepped appearance in the matrix.
2001, Malcolm Pemberton, Nicholas Rau, “Systems of Linear Equations”, in Mathematics for Economists: An Introductory Textbook, Manchester; New York, N.Y.: Manchester University Press, →ISBN, section 12.1 (Echelon Matrices), page 204:
: An echelon matrix is a matrix, not necessarily square, with the following two properties: (i) There is at least one non-zero entry; rows consisting entirely of zeros, if any, lie below rows with at least one non-zero entry. (ii) In each non-zero row after the first, the left-most non-zero entry lies to the right of the left-most non-zero entry in the preceding row. […] In each of the non-zero rows of an echelon matrix, the left-most non-zero entry is called the pivot, […]
Further reading
[edit]
echelon (disambiguation) on Wikipedia.Wikipedia
echelon formation on Wikipedia.Wikipedia
row echelon form on Wikipedia.Wikipedia
Anagrams
[edit]
Chelone, chelone
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6814 | https://www.youtube.com/watch?v=HXee18A4JY0 | Introducing Electrochemistry; Redox Reactions Revisited | OpenStax Chemistry 2e 4.2, 17.1
Michael Evans
29900 subscribers
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Description
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Posted: 19 Nov 2022
00:00 The Lemon Battery
01:44 Chapter Outline
03:39 Oxidation-Reduction (Redox) Reactions
05:11 Defining Oxidation Number
07:00 Guidelines for Oxidation Number
Transcript:
The Lemon Battery hi guys to start off this video I wanted to show off one of my favorite chemistry demonstrations the lemon battery so what we're going to do is actually measure the fact that there is a voltage associated with this little linen construction that I've created here so I've got a piece of copper metal stuck into one side of the lemon right here and a piece of zinc metal stuck into the other side and I've squeezed it a little bit to get some of that lemon juice out and flowing around the copper and zinc metal and what I'm going to do is connect these voltmeter leads to the copper and zinc and we're going to see what happens and I've got it set to millivolts here for dramatic effects so I want to connect the leads here what we can see is that there's actually a voltage generated and we can actually get this pretty high I had it up at 900 millivolts earlier see if I can get it back up there see it's it's topping out at about seven eight hundred there we go there we go now we're up at about 900. in this unit we're going to talk about electrochemistry and we're going to learn all about how and why this situation gives rise to a voltage this is comparable to about two-thirds of a AAA battery so a good amount of voltage out of this thing we're going to understand where this comes from on a chemical level what it is chemically about the copper and zinc that gives rise to an electrical potential difference that has to do with reox chemistry electron transfer and more broadly the field of electrochemistry which concerns extracting electrical energy from chemical energy or converting electrical energy into chemical energy going the other way that's what this unit is all Chapter Outline about in this video series we're going to talk all about electrochemistry which concerns the interplay between electrical and chemical energy in redox or electron transfer reactions and so we're going to start by reviewing redox reactions and redox chemistry defining oxidation number oxidation and reduction and then we're going to see how we can specially engineer a cell called a galvanic cell to take advantage of the energy built into a spontaneous redox reaction to produce a voltage and a cell like this that uses a spontaneous redox reaction to produce an electrical potential difference and drive electric current is known as a galvanic cell we'll then learn how to calculate the potential of a galavanic cell given the components of the cell and the concentrations of any aqueous species any species in solution in electrode and cell Potentials in the fourth section we're going to relate cell potential to free energy and chemical equilibrium and this makes an important point about this unit as a whole electrochemistry as we'll discuss it here is really an application of chemical equilibrium and chemical thermodynamics at its root so we'll see equations for example that are going to be reminiscent of equations we've seen previously for example the nernst equation is going to look a lot like a relation between free energy and the equilibrium constant that we've come across previously from there we're going to move into applications of electrochemistry starting with batteries and fuel cells and then talking about corrosion and then finally electrolysis which in a sense is the opposite of a galivantic cell and an electrolysis process we use an external circuit to send electrons into a chemical system and drive an otherwise non-spontaneous or unfavorable redox reaction Oxidation-Reduction (Redox) Reactions let's start by reviewing redox chemistry and redox reactions so a redox reaction or the more long-winded term here is oxidation reduction reaction involves the transfer of one or more electrons from one species to another and in thinking about transferring electrons we can talk about a species that loses electrons that's what we call oxidation and a species that gains electrons and that's what we call reduction and at the bottom of this slide we see an example of a redox reaction between metallic Elemental sodium and a solid and chlorine gas cl2 gas two NAS react with the cl2 to produce two NaCl solids two sodium chlorides notice here that sodium metal becomes sodium cation this is an oxidation process sodium starts out in zero oxidation state we'll talk about how to determine this on the next slide and in NaCl sodium is in the plus one oxidation state and so the oxidation number has increased this is oxidation if we look at cl2 on the other hand cl2 chlorine is in the zero oxidation state and in NaCl it goes to the negative one oxidation states of its oxidation number has decreased or been reduced therefore it is undergoing reduction Defining Oxidation Number now let's formalize these terms a little bit more by defining oxidation number or oxidation state so oxidation number is a way of defining the charge on an atom similar in spirit to formal charge but with a different strategy for thinking about how we distribute electrons to assign a charge oxidation number is specifically defined as the charge that an atom would possess if it were involved in only ionic bonds so what we do with any covalent bonds is we give both electrons in the bond say it's just a single Bond although this works for double and triple we give all of the electrons in that bond to the more electronegative atom that's why you're prompted here to consider electronegativity the charges that result after we do that are the oxidation numbers and on the rest of this line here you see some rules for assigning oxidation number that are going to be very helpful for becoming more efficient so that you don't have to kind of mentally split bonds every time you see a compound right really quickly before we dig into the rules if we go back to NaCl in an ionic compound all you need to do is split up the ionic compound into its component ions with their appropriate charges and you can if we're talking about monatomic cation monatomic anion you can easily infer what the oxidation numbers are so for example in NaCl N A has a charge of plus one that's an n a plus cation that's why it's in the plus one oxidation state and cl has a charge of negative one that's the chloride anion and that's chlorine in the oxidation state of negative one so for monatomic ionic compounds or binary ionic compounds that don't involve polyatomic ions this is pretty straightforward Guidelines for Oxidation Number rule number one is also very straightforward the oxidation number of an atom in its Elemental form is zero and this includes any monatomic metallic element and the diatomic elements like br2 cl2 I2 H2 those atoms are all in the zero oxidation state since electrons if they're involved in a bond are shared equally right those covalent bonds in H2 and cl2 for example are nonpolar covalent bonds as we just mentioned if we're talking about a monatomic n a plus Li plus mg2 plus al3 plus the oxidation number of a monatomic ion is simply equal to its charge rule number three lists some common oxidation numbers for for common elements hydrogen most commonly it's going to be linked to a non-metal in which I'm most commonly yeah it's going to be linked to a non-metal in which case its oxidation number is plus one when the non-metal is more electronegative than hydrogen which is typical but when hydrogen is linked to a metallic element well then hydrogen itself is more electronegative than the metal and so its oxidation number is negative one and again you can infer this from the definition at the top of the slide by taking the bond that hydrogen is involved in and giving those electrons to the more electronegative atom in the bond oxygen is typically more electronegative than whatever it's connected to and so most typically it's in the oxidation number of negative two there are very very few exceptions to this the halogens well if we're talking about fluorine which is the most electronegative element on the periodic table fluorine always is in the oxidation state of negative one for our purposes for the other halogens these are negative one when they're linked to less electronegative atoms via single bonds for instance but when they show up in polyatomic anions polyatomic oxyanions like perchlorate hypochlorite Etc they're going to have a positive oxidation number since they're linked to oxygen which is more electronegative than those halogens below fluorine so they'll have some positive oxidation number value it's actually not necessarily equal to the number of oxygens that the halogen is linked to this is a little bit misleading on the slide but you can work it out by applying rule 4 as well as Oxygen's typical oxidation state of negative two rule 4 says that the sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge of the molecule or ion and this makes sense it's just like formal charge right the total charge on all of the atoms needs to add up to the net charge of the molecular ion |
6815 | https://www.jogc.com/article/S1701-2163(15)30390-X/fulltext | Parvovirus B19 Infection in Pregnancy - Journal of Obstetrics and Gynaecology Canada
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SOGC CLINICAL PRACTICE GUIDELINEVolume 36, Issue 12p1107-1116 December 2014
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Parvovirus B19 Infection in Pregnancy
Joan Crane, MD
Joan Crane, MD
Affiliations
St. John’s NL
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∙ William Mundle, MD
William Mundle, MD
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Windsor ON
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∙ Isabelle Boucoiran, MD
Isabelle Boucoiran, MD
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Vancouver BC
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∙ MATERNAL FETAL MEDICINE COMMITTEE
MATERNAL FETAL MEDICINE COMMITTEE
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Robert Gagnon, MD,
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Robert Gagnon, MD
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Emmanuel Bujold, MD,
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Emmanuel Bujold, MD
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Melanie Basso, RN,
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Melanie Basso, RN
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Hayley Bos, MD,
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Hayley Bos, MD
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Richard Brown, MD,
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Richard Brown, MD
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Stephanie Cooper, MD,
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Stephanie Cooper, MD
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Katy Gouin, MD,
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Katy Gouin, MD
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N. Lynne McLeod, MD,
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N. Lynne McLeod, MD
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Savas Menticoglou, MD,
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William Mundle, MD,
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Christy Pylypjuk, MD,
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Anne Roggensack, MD,
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Frank Sanderson, MD
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Affiliations & Notes Article Info
1 St. John’s NL
2 Windsor ON
3 Vancouver BC
4 Verdun QC
5 Quebec QC
6 Vancouver BC
7 Victoria BC
8 Montreal QC
9 Calgary AB
10 Quebec QC
11 Halifax NS
12 Winnipeg MB
13 Windsor ON
14 Winnipeg MB
15 Calgary AB
16 Saint John NB
Footnotes:
This Clinical Practice Guideline has been prepared by the Maternal Fetal Medicine committee, reviewed by Infectious Disease and Family Physician Advisory Committees, and approved by the Executive and Council of the Society of Obstetricians and Gynaecologists of Canada.
Disclosure statements have been received from all contributors.
This document reflects emerging clinical and scientific advances on the date issued and is subject to change. The information should not be construed as dictating an exclusive course of treatment or procedure to be followed. Local institutions can dictate amendments to these opinions. They should be well documented if modified at the local level. None of these contents may be reproduced in any form without prior written permission of the SOGC.
DOI: 10.1016/S1701-2163(15)30390-X External LinkAlso available on ScienceDirect External Link
Copyright: © 2014 Society of Obstetricians and Gynaecologists of Canada.
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Abstract
Key Words
INTRODUCTION
CLINICAL PRESENTATION
PARVOVIRUS B19 INFECTION IN PREGNANCY
MANAGEMENT OF PARVOVIRUS B19
DIAGNOSIS OF FETAL INFECTION
MANAGEMENT OF FETAL HYDROPS AND ANEMIA
REFERENCES
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Abstract
Key Words
INTRODUCTION
CLINICAL PRESENTATION
PARVOVIRUS B19 INFECTION IN PREGNANCY
MANAGEMENT OF PARVOVIRUS B19
DIAGNOSIS OF FETAL INFECTION
MANAGEMENT OF FETAL HYDROPS AND ANEMIA
REFERENCES
Article metrics
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Abstract
Objectives
This guideline reviews the evidence relating to the effects of parvovirus B19 on the pregnant woman and fetus, and discusses the management of women who are exposed to, who are at risk of developing, or who develop parvovirus B19 infection in pregnancy.
Outcomes
The outcomes evaluated were maternal outcomes including erythema infectiosum, arthropathy, anemia, and myocarditis, and fetal outcomes including spontaneous abortion, congenital anomalies, hydrops fetalis, stillbirth, and long-term effects.
Evidence
Published literature was retrieved through searches of PubMed and The Cochrane Library on July 8, 2013, using appropriate controlled vocabulary (MeSH terms “parvovirus” and “pregnancy”) and key words (parvovirus, infection, pregnancy, hydrops). Results were restricted to systematic reviews, randomized control trials/controlled clinical trials, and observational studies. There were no date restrictions but results were limited to English or French language materials. Grey (unpublished) literature was identified through searching the websites of health technology assessment and health technology assessment-related agencies, clinical practice guideline collections, and national and international medical specialty.
Values
The quality of evidence in this document was rated using the criteria described in the Report of the Canadian Task Force on Preventive Health Care (Table 1).
Recommendations
1.
Investigation for parvovirus B19 infection is recommended as part of the standard workup for fetal hydrops or intrauterine fetal death. (II-2A)
2.
Routine screening for parvovirus immunity in low-risk pregnancies is not recommended. (II-2E)
3.
Pregnant women who are exposed to, or who develop symptoms of, parvovirus B19 infection should be assessed to determine whether they are susceptible to infection (non-immune) or have a current infection by determining their parvovirus B19 immunoglobulin G and immunoglobulin M status. (II-2A)
4.
If parvovirus B19 immunoglobulin G is present and immunoglobulin M is negative, the woman is immune and should be reassured that she will not develop infection and that the virus will not adversely affect her pregnancy. (II-2A)
5.
If both parvovirus B19 immunoglobulin G and immunoglobulin M are negative (and the incubation period has passed), the woman is not immune and has not developed the infection. She should be advised to minimize exposure at work and at home. Absence from work should be considered on a case-by-case basis. (II-2C) Further studies are recommended to address ways to lessen exposure including the risk of occupational exposure. (III-A)
6.
If a recent parvovirus B19 infection has been diagnosed in the woman, referral to an obstetrician or a maternal–fetal medicine specialist should be considered. (III-B) The woman should be counselled regarding risks of fetal transmission, fetal loss, and hydrops and serial ultrasounds should be performed every 1 to 2 weeks, up to 12 weeks after infection, to detect the development of anemia (using Doppler measurement of the middle cerebral artery peak systolic velocity) and hydrops. (III-B) If hydrops or evidence of fetal anemia develops, referral should be made to a specialist capable of fetal blood sampling and intravascular transfusion. (II-2B)
Key Words
parvovirus
infection
pregnancy
hydrops
ABBREVIATIONS IgG
immunoglobulin G
IgM
immunoglobulin M
MCA
middle cerebral artery
MSAFP
maternal serum alpha fetoprotein
PCR
polymerase chain reaction
INTRODUCTION
Parvovirus B19 is a single-stranded DNA virus that is responsible for erythema infectiosum, a common person is usually no longer infectious. Current data suggest childhood illness.1
1.
Levy, R. ∙ Weissman, A. ∙ Blomberg, G. ...
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The virus was identified in 1975 during routine blood screening for hepatitis B surface antigen,2
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and was identified as the cause of erythema infectiosum in 1983.3
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It was subsequently linked to cases of non-immune hydrops and fetal death.4–7
4.
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Intrauterine parvovirus infection associated with hydrops fetalis
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7.
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The B19 parvovirus strain infects only humans and animal strains infect only animals, not humans.1
1.
Levy, R. ∙ Weissman, A. ∙ Blomberg, G. ...
Infection by parvovirus B19 during pregnancy: a review
Obstet Gynecol Surv. 1997; 52:254-259
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Scopus (91)
PubMed
Google Scholar
Parvovirus B19 is most commonly spread by respiratory secretions or from hand to mouth contact.8
8.
Adler, S. ∙ Koch, W.C.
Human parvovirus B19
Remington, J.S. ∙ Klein, J.O. (Editors)
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Google Scholar
Other modes of transmission include blood product infusion and transplacental transfer. As the main mode of transmission is respiratory, epidemics of parvovirus B19 infection can occur. Outbreaks usually happen in spring (but can occur any time of the year), and mainly affect children aged 4 to 11. Outbreaks usually occur yearly, with larger epidemics every four to five years, and may last up to six months.9–11
9.
Rodis, J.F.
Parvovirus infection
Clin Obstet Gynecol. 1999; 42:107-120
quiz 174–5
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de Jong, E.P. ∙ Walther, F.J. ∙ Kroes, A.C. ...
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11.
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Most cases in pregnant women seem to occur in late spring and summer.12
12.
Harger, J.H. ∙ Adler, S.P. ∙ Koch, W.C. ...
Prospective evaluation of 618 pregnant women exposed to parvovirus B19: risks and symptoms
Obstet Gynecol. 1998; 91:413-420
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Viremia occurs 4 to 14 days after exposure and may last up to 20 days.13
13.
Anderson, L.J.
Role of parvovirus B19 in human disease
Pediatr Infect Dis J. 1987; 6:711-718
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PubMed
Google Scholar
Fever and prodromal symptoms may develop in the last few days of the incubation period,14
14.
Markenson, G.R. ∙ Yancey, M.K.
Parvovirus B19 infections in pregnancy
Semin Perinatol. 1998; 22:309-317
Abstract
Full Text (PDF)
PubMed
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but many people remain asymptomatic. A rash and arthralgia may begin around day 15, by which time the person is usually no longer infectious. Current data suggest that infection with parvovirus B19 usually confers lifelong immunity.14
14.
Markenson, G.R. ∙ Yancey, M.K.
Parvovirus B19 infections in pregnancy
Semin Perinatol. 1998; 22:309-317
Abstract
Full Text (PDF)
PubMed
Google Scholar
Because outbreaks can be frequent and many infectious people are asymptomatic, encounters that risk exposure to parvovirus infection are often unrecognized.
Approximately 50% to 75% of women of reproductive age have developed immunity to parvovirus B19.11,15–18
11.
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Parvovirus B19 in pregnancy: prenatal diagnosis and management of fetal complications
Curr Opin Obstet Gynecol. 2012; 24:95-101
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18.
Lamont, R.F. ∙ Sobel, J.D. ∙ Vaisbuch, E. ...
Parvovirus B19 infection in human pregnancy
BJOG. 2011; 118:175-186
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Without known exposure, about 1% to 3% of susceptible pregnant women will develop serologic evidence of infection in pregnancy,16,19
16.
Valeur-Jensen, A.K. ∙ Pedersen, C.B. ∙ Westergaard, T. ...
Risk factors for parvovirus B19 infection in pregnancy
JAMA. 1999; 281:1099-1105
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19.
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rising to over 10% in epidemic periods.10
10.
de Jong, E.P. ∙ Walther, F.J. ∙ Kroes, A.C. ...
Parvovirus B19 infection in pregnancy: new insights and management
Prenat Diagn. 2011; 31:419-425
Crossref
Scopus (88)
PubMed
Google Scholar
Where there is extensive opportunity for exposure to parvovirus B19, such as in a daycare centre or school, it is estimated that 20% to 30% of susceptible women19,20
19.
Centers for Disease Control (CDC)
Risks associated with human parvovirus B19 infection
MMWR Morb Mortal Wkly Rep. 1989; 38:81-88
93–7.
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20.
Gillespie, S.M. ∙ Cartter, M.L. ∙ Asch, S. ...
Occupational risk of human parvovirus B19 infection for school and day-care personnel during an outbreak of erythema infectiosum
JAMA. 1990; 263:2061-2065
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will develop infection, while 50% of susceptible women exposed through household contacts will become infected.19,21
19.
Centers for Disease Control (CDC)
Risks associated with human parvovirus B19 infection
MMWR Morb Mortal Wkly Rep. 1989; 38:81-88
93–7.
PubMed
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21.
Chorba, T. ∙ Coccia, P. ∙ Holman, R.C. ...
The role of parvovirus B19 in aplastic crisis and erythema infectiosum (fifth disease)
J Infect Dis. 1986; 154:383
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Nursery school teachers have a 3-fold higher risk of acute infection than other pregnant women, and other school teachers have a 1.6-fold increased risk.16
16.
Valeur-Jensen, A.K. ∙ Pedersen, C.B. ∙ Westergaard, T. ...
Risk factors for parvovirus B19 infection in pregnancy
JAMA. 1999; 281:1099-1105
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Scopus (173)
PubMed
Google Scholar
The population-attributable risk of infection in susceptible pregnant women is about 55% from their own children and 6% for occupational exposure.16
16.
Valeur-Jensen, A.K. ∙ Pedersen, C.B. ∙ Westergaard, T. ...
Risk factors for parvovirus B19 infection in pregnancy
JAMA. 1999; 281:1099-1105
Crossref
Scopus (173)
PubMed
Google Scholar
Women at increased risk include mothers of preschool and school-age children, workers at daycare centres, and school teachers. Assessment of parvovirus B19 immunity at the beginning of the pregnancy can be considered in this population.
Since the publication of the 2002 guideline, there have been publications of the natural history, outcomes, diagnosis, and management of parvovirus in pregnancy. This updated guideline provides a review of this literature. The quality of evidence reported in these guidelines has been described using the Evaluation of Evidence criteria outlined in the Report of the Canadian Task Force on Preventive Health Care (Table 1).
| Quality of evidence assessment | Classification of recommendations† |
--- |
| I: Evidence obtained from at least one properly randomized controlled trial II-1: Evidence from well-designed controlled trials without randomization II-2: Evidence from well-designed cohort (prospective or retrospective) or case-control studies, preferably from more than one centre or research group II-3: Evidence obtained from comparisons between times or places with or without the intervention. Dramatic results in uncontrolled experiments (such as the results of treatment with penicillin in the 1940s) could also be included in this category III: Opinions of respected authorities, based on clinical experience, descriptive studies, or reports of expert committees | A. There is good evidence to recommend the clinical preventive action B. There is fair evidence to recommend the clinical preventive action C. The existing evidence is conflicting and does not allow to make a recommendation for or against use of the clinical preventive action; however, other factors may influence decision-making D. There is fair evidence to recommend against the clinical preventive action E. There is good evidence to recommend against the clinical preventive action L. There is insufficient evidence (in quantity or quality) to make a recommendation; however, other factors may influence decision-making |
Table 1
Key to evidence statements and grading of recommendations, using the ranking of the Canadian Task Force on Preventive Health Care
The quality of evidence reported in these guidelines has been adapted from The Evaluation of Evidence criteria described in the Canadian Task Force on Preventive Health Care.87
87.
Woolf, S.H. ∙ Battista, R.N. ∙ Angerson, G.M. ...
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†
Recommendations included in these guidelines have been adapted from the Classification of Recommendations criteria described in the Canadian Task Force on Preventive Health Care.87
87.
Woolf, S.H. ∙ Battista, R.N. ∙ Angerson, G.M. ...
Canadian Task Force on Preventive Health Care. New grades for recommendations from the Canadian Task Force on Preventive Health Care
CMAJ. 2003; 169:207-208
PubMed
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Open table in a new tab
CLINICAL PRESENTATION
The multiple ways parvovirus B19 may present are described below and summarized in Table 2.
1.
Asymptomatic: Up to 50% of non-pregnant women who develop parvovirus B19 infection, and up to 70% of infected pregnant women, will be asymptomatic.9,18–23
9.
Rodis, J.F.
Parvovirus infection
Clin Obstet Gynecol. 1999; 42:107-120
quiz 174–5
Crossref
Scopus (37)
PubMed
Google Scholar
18.
Lamont, R.F. ∙ Sobel, J.D. ∙ Vaisbuch, E. ...
Parvovirus B19 infection in human pregnancy
BJOG. 2011; 118:175-186
Crossref
Scopus (132)
PubMed
Google Scholar
19.
Centers for Disease Control (CDC)
Risks associated with human parvovirus B19 infection
MMWR Morb Mortal Wkly Rep. 1989; 38:81-88
93–7.
PubMed
Google Scholar
20.
Gillespie, S.M. ∙ Cartter, M.L. ∙ Asch, S. ...
Occupational risk of human parvovirus B19 infection for school and day-care personnel during an outbreak of erythema infectiosum
JAMA. 1990; 263:2061-2065
Crossref
Scopus (141)
PubMed
Google Scholar
21.
Chorba, T. ∙ Coccia, P. ∙ Holman, R.C. ...
The role of parvovirus B19 in aplastic crisis and erythema infectiosum (fifth disease)
J Infect Dis. 1986; 154:383
383
Crossref
Scopus (202)
PubMed
Google Scholar
22.
Plummer, F.A. ∙ Hammond, G.W. ∙ Forward, K. ...
An erythema infectiosum-like illness caused by human parvovirus infection
N Engl J Med. 1985; 313:74-79
Crossref
Scopus (139)
PubMed
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23.
Chisaka, H. ∙ Ito, K. ∙ Niikura, H. ...
Clinical manifestations and outcomes of parvovirus B19 infection during pregnancy in Japan
Tohoku J Exp Med. 2006; 209:277-283
Crossref
Scopus (41)
PubMed
Google Scholar
2.
Erythema infectiosum (fifth disease): Children with parvovirus B19 infection most commonly develop erythema infectiosum, initially presenting with flu-like symptoms, fever, and headache, followed 1 to 4 days later by a “slapped cheek” rash that becomes lacy in appearance, and after about 1 week may spread to the trunk and limbs.13
13.
Anderson, L.J.
Role of parvovirus B19 in human disease
Pediatr Infect Dis J. 1987; 6:711-718
Crossref
Scopus (158)
PubMed
Google Scholar
Adults with parvovirus B19 infection usually do not have an extensive rash. The onset of the rash usually coincides with the appearance of parvovirus B19 antibodies (IgM), suggesting that this symptom is immune-mediated.14
14.
Markenson, G.R. ∙ Yancey, M.K.
Parvovirus B19 infections in pregnancy
Semin Perinatol. 1998; 22:309-317
Abstract
Full Text (PDF)
PubMed
Google Scholar
Other dermatologic syndromes associated with parvovirus infection in adults include papular-purpuric “gloves and socks” syndrome.
3.
Arthropathy: For those adults with symptoms, the most common symptom is arthropathy. It affects up to 50% of pregnant women with parvovirus infection,12
12.
Harger, J.H. ∙ Adler, S.P. ∙ Koch, W.C. ...
Prospective evaluation of 618 pregnant women exposed to parvovirus B19: risks and symptoms
Obstet Gynecol. 1998; 91:413-420
Crossref
Scopus (146)
PubMed
Google Scholar
and may last several weeks to months. The arthropathy usually presents as symmetric polyarthralgia, affecting the hands, wrists, ankles, and knees.12,19,24,25
12.
Harger, J.H. ∙ Adler, S.P. ∙ Koch, W.C. ...
Prospective evaluation of 618 pregnant women exposed to parvovirus B19: risks and symptoms
Obstet Gynecol. 1998; 91:413-420
Crossref
Scopus (146)
PubMed
Google Scholar
19.
Centers for Disease Control (CDC)
Risks associated with human parvovirus B19 infection
MMWR Morb Mortal Wkly Rep. 1989; 38:81-88
93–7.
PubMed
Google Scholar
24.
White, D.G. ∙ Woolf, A.D. ∙ Mortimer, P.P. ...
Human parvovirus arthropathy
Lancet. 1985; 1:419-421
Abstract
Scopus (362)
PubMed
Google Scholar
25.
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1.
Levy, R. ∙ Weissman, A. ∙ Blomberg, G. ...
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Maternal:
•
Asymptomatic
•
Erythema infectiosum/rash
•
Arthopathy
•
Anemia
•
Myocarditis
Fetal:
•
Fetal loss
•
•
• Myocarditis
Table 2
Presentation of parvovirus B19 infection
Open table in a new tab
PARVOVIRUS B19 INFECTION IN PREGNANCY
Pregnancy does not appear to affect the course of the infection, but infection may affect the pregnancy.27
27.
Alger, L.S.
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The transmission rate of maternal parvovirus B19 infection to the fetus is 17% to 33%.12,38,39
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Most fetuses infected with parvovirus B19 have spontaneous resolution with no adverse outcomes.1,14
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Levy, R. ∙ Weissman, A. ∙ Blomberg, G. ...
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(Table 3)
| Author | Cases (N) | Fetal loss | Hydrops |
--- --- |
| Public Health Laboratory Service Working Party on Fifth Disease38 38. Public Health Laboratory Service Working Party on Fifth Disease Prospective study of human parvovirus (B19) infection in pregnancy BMJ. 1990; 300:1166-1170 Crossref PubMed Google Scholar | 186 | | 1 |
| Rodis et al.43 43. Rodis, J.F. ∙ Quinn, D.L. ∙ Gary, Jr., G.W. ... Management and outcomes of pregnancies complicated by human B19 parvovirus infection: a prospective study Am J Obstet Gynecol. 1990; 163:1168-1171 Abstract Full Text (PDF) Scopus (148) PubMed Google Scholar | 39 | | 0 |
| Gratacós et al.39 39. Gratacós, E. ∙ Torres, P.J. ∙ Vidal, J. ... The incidence of human parvovirus B19 infection during pregnancy and its impact on perinatal outcome J Infect Dis. 1995; 171:1360-1363 Crossref Scopus (185) PubMed Google Scholar | 60 | 5 | 0 |
| Harger et al.12 12. Harger, J.H. ∙ Adler, S.P. ∙ Koch, W.C. ... Prospective evaluation of 618 pregnant women exposed to parvovirus B19: risks and symptoms Obstet Gynecol. 1998; 91:413-420 Crossref Scopus (146) PubMed Google Scholar | 52 | 2 | 0 |
| Miller et al.42 42. Miller, E. ∙ Fairley, C.K. ∙ Cohen, B.J. ... Immediate and long term outcome of human parvovirus B19 infection in pregnancy Br J Obstet Gynaecol. 1998; 105:174-178 Crossref Scopus (275) PubMed Google Scholar | 427 | | 7 |
| Guidozzi et al.44 44. Guidozzi, F. ∙ Ballot, D. ∙ Rothberg, A.D. Human B19 parvovirus infection in an obstetric population. A prospective study determining fetal outcome J Reprod Med. 1994; 39:36-38 PubMed Google Scholar | 64 | 1 | 0 |
| Rodis et al.47 47. Rodis, J.F. ∙ Rodner, C. ∙ Hansen, A.A. ... Long-term outcome of children following maternal human parvovirus B19 infection Obstet Gynecol. 1998; 91:125-128 Crossref Scopus (87) PubMed Google Scholar | 113 (115 fetuses) | | 2 |
| Koch et al.46 46. Koch, W.C. ∙ Harger, J.H. ∙ Barnstein, B. ... Serologic and virologic evidence for frequent intrauterine transmission of human parvovirus B19 with a primary maternal infection during pregnancy Pediatr Infect Dis J. 1998; 17:489-494 Crossref Scopus (91) PubMed Google Scholar | 43 | 0 | 0 |
| Enders et al.55 55. Enders, M. ∙ Weidner, A. ∙ Zoellner, I. ... Fetal morbidity and mortality after acute human parvovirus B19 infection in pregnancy: prospective evaluation of 1018 cases Prenat Diagn. 2004; 24:513-518 Crossref Scopus (269) PubMed Google Scholar | 1018 | | 40 |
| Schwarz et al.26 26. Schwarz, T.F. ∙ Roggendorf, M. ∙ Hottenträger, B. ... Human parvovirus B19 infection in pregnancy Lancet. 1988; 2:566-567 Crossref Scopus (91) PubMed Google Scholar | 39 | 7 | 10 |
| Simms et al.48 48. Simms, R.A. ∙ Liebling, R.E. ∙ Patel, R.R. ... Management and outcome of pregnancies with parvovirus B19 infection over seven years in a tertiary fetal medicine unit Fetal Diagn Ther. 2009; 25:373-378 Crossref Scopus (45) PubMed Google Scholar | 47 | | 8 |
| Total | 2090 fetuses | | 68 (2.9%) |
Table 3
Risk of hydrops and fetal death with parvovirus B19 maternal infection
Does not include data of Gratacós et al.,39
39.
Gratacós, E. ∙ Torres, P.J. ∙ Vidal, J. ...
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Harger et al.,12 Giudozzi et al.,44 Koch et al.,46 or Schwarz et al.26 because gestational age was not indicated for all cases of infection.
Open table in a new tab
Fetal Effects of Parvovirus B19 Infection
Parvovirus infection can lead to spontaneous miscarriage and stillbirth.40,41
40.
Leduc, L. ∙ SOGC Maternal-Fetal Medicine Committee
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The spontaneous loss rate of fetuses affected with parvovirus B19 before 20 weeks’ gestation is 13.0% and after 20 weeks’ gestation is 0.5%.12,26,38,42–49
12.
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26.
Schwarz, T.F. ∙ Roggendorf, M. ∙ Hottenträger, B. ...
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Miller, E. ∙ Fairley, C.K. ∙ Cohen, B.J. ...
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(Table 3). The reason for this difference is uncertain, but the largest study suggests it may be related to multisystem organ damage, which is possible even without anemia or hydrops.10
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Currently, there does not appear to be any evidence that parvovirus B19 infection increases the risk of congenital anomalies in humans,1,14
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Parvovirus B19 has been associated with hydrops fetalis.12,19,26,38,39,42–44,46,49,54–56
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39.
Gratacós, E. ∙ Torres, P.J. ∙ Vidal, J. ...
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44.
Guidozzi, F. ∙ Ballot, D. ∙ Rothberg, A.D.
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46.
Koch, W.C. ∙ Harger, J.H. ∙ Barnstein, B. ...
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49.
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The overall incidence in fetuses whose mothers have been infected by parvovirus during pregnancy is 2.9% (Table 3). The risk of fetal hydrops appears to be greater when infection occurs earlier in pregnancy. Enders et al. noted the rate of hydrops to be 4.7% if maternal infection occurred before 25 weeks’ gestation compared with 2.3% after this gestation.55
55.
Enders, M. ∙ Weidner, A. ∙ Zoellner, I. ...
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Possible mechanisms for hydrops include fetal anemia due to the virus crossing the placenta, combined with the shorter half-life of fetal red blood cells (especially during the hepatic stage of hematopoiesis), leading to the severe anemia, hypoxia, and high output cardiac failure that are associated with fetal hydrops. Other possible causes include fetal viral myocarditis leading to cardiac failure, and impaired hepatic function caused by direct damage to hepatocytes and indirect damage due to hemosiderin deposits.12,19,38,39,42–44
12.
Harger, J.H. ∙ Adler, S.P. ∙ Koch, W.C. ...
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Obstet Gynecol. 1998; 91:413-420
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PubMed
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Centers for Disease Control (CDC)
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93–7.
PubMed
Google Scholar
38.
Public Health Laboratory Service Working Party on Fifth Disease
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PubMed
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39.
Gratacós, E. ∙ Torres, P.J. ∙ Vidal, J. ...
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42.
Miller, E. ∙ Fairley, C.K. ∙ Cohen, B.J. ...
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43.
Rodis, J.F. ∙ Quinn, D.L. ∙ Gary, Jr., G.W. ...
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PubMed
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44.
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If a fetus develops hydrops, ultrasound signs include ascites, skin edema, pleural and pericardial effusions, and placental edema.1
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Levy, R. ∙ Weissman, A. ∙ Blomberg, G. ...
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although some studies found molecular evidence of parvovirus B19 in 18% to 27% of cases of non-immune hydrops.14
14.
Markenson, G.R. ∙ Yancey, M.K.
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Semin Perinatol. 1998; 22:309-317
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Investigation for parvovirus B19 infection is recommended as part of the standard workup for fetal hydrops or intrauterine fetal death. (II-2A)
MANAGEMENT OF PARVOVIRUS B19
Exposure/Infection in Pregnancy
Systematic screening for parvovirus immunity in low-risk pregnancies is not currently recommended.66
66.
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10.
de Jong, E.P. ∙ Walther, F.J. ∙ Kroes, A.C. ...
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The presence of parvovirus B19 IgM antibodies with no evidence of parvovirus B19 IgG antibodies suggests either a very recent infection or a false-positive result.9,69
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In this situation, it is recommended that testing for parvovirus B19 IgG and IgM be repeated in 1 to 2 weeks. If recent infection has occurred, then the IgG should also be positive at that time.9–11,18
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If both parvovirus B19 IgG and IgM are negative, the woman is not immune and is therefore susceptible to infection.9,69
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If she has had a recent exposure to the virus, and may be incubating the infection, it is suggested that the IgG and IgM tests be repeated 2 to 4 weeks later. If exposure is ongoing, serology may be repeated every 2 to 4 weeks. Occasionally maternal IgM levels in acute infection may be below detection. In these cases PCR can be used in maternal serum for the diagnosis of acute infection.10,11
10.
de Jong, E.P. ∙ Walther, F.J. ∙ Kroes, A.C. ...
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However, the interpretation of this result is complicated by the possible persistence of low parvovirus B19 DNA levels in the blood for several months after acute infection.
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According to the limited published data, this measurement has a sensitivity of 83% to 100%, and a specificity of 93% to 100% for diagnosis of anemia in parvovirus B19 infected fetuses.76,77,79
76.
Cosmi, E. ∙ Mari, G. ∙ Delle Chiaie, L. ...
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77.
Delle Chiaie, L. ∙ Buck, G. ∙ Grab, D. ...
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Chauvet, A. ∙ Dewilde, A. ∙ Thomas, D. ...
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Other ultrasound signs of parvovirus B19 infection include increased placenta thickness, echogenic bowel/ meconium peritonitis, first trimester increased nuchal translucency, and amniotic fluid abnormalities.10,80
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As fetuses with hydrops tend to move less, women should also be instructed to monitor fetal movement daily.9
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If there is a delay in establishing the woman’s immunity status, serial ultrasounds for the detection of hydrops and anemia may be obtained until information regarding immunity is available.81
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Recommendations
2.
Routine screening for parvovirus immunity in low-risk pregnancies is not recommended. (II-2E)
3.
Pregnant women exposed to, or who develop symptoms of, parvovirus B19 infection should be assessed to determine whether they are susceptible to infection (non-immune) or have a current infection by determining their parvovirus B19 immunoglobulin G and immunoglobulin M status. (II-2A)
4.
If parvovirus B19 immunoglobulin G is present and immunoglobulin M is negative, the woman is immune and should be reassured that she will not develop infection and that the virus will not adversely affect her pregnancy. (II-2A)
5.
If both parvovirus B19 immunoglobulin G and immunoglobulin M are negative (and the incubation period has passed), the woman is not immune and has not developed the infection. She should be advised to minimize exposure at work and at home. Absence from work should be considered on a case-by-case basis. (II-2C) Further studies are recommended to address ways to lessen exposure including the risk of occupational exposure. (III-A)
DIAGNOSIS OF FETAL INFECTION
Parvovirus B19 cannot usually be cultured in regular culture media.1
1.
Levy, R. ∙ Weissman, A. ∙ Blomberg, G. ...
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It can be identified histologically by characteristic intranuclear inclusions or by the presence of viral particles by electron microscopy.1
1.
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Fetal infection can be detected with amniotic fluid or fetal serum using the most sensitive molecular methods available (nested PCR or reverse transcription PCR).49
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Although there is the possibility of diagnosing parvovirus B19 infection with amniotic fluid obtained by amniocentesis, invasive diagnosis of this condition is not required for all suspected or confirmed maternal infections. If amniocentesis is performed for a fetal indication, a PCR for parvovirus B19 should be requested as part of the workup. The presence of viral particles, however, can only be seen during the viremic stage. The presence of parvovirus B19 IgM in fetal blood cannot be depended upon to make the diagnosis of fetal infection,7
7.
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because the fetus does not begin to make its own IgM until 22 weeks’ gestation. There have been false-negative results even when the fetus is beyond 22 weeks.82
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the authors judged it to be weak, and thus it cannot be used as a reliable marker of fetal parvovirus B19 infection.14
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MANAGEMENT OF FETAL HYDROPS AND ANEMIA
Every pregnancy identified with fetal anemia or hydrops should be referred to a tertiary care centre with a maternal–fetal medicine specialist. The current management of fetuses with hydrops or anemia due to parvovirus B19 infection is to consider cordocentesis, to assess fetal hemoglobin and reticulocyte count, and intrauterine transfusion, if necessary.14
14.
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If the fetus is at or near term, delivery should be considered.14
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If delivery of a hydropic or anemic infant is planned this should occur in a tertiary care centre with staff and resources to manage these neonates. The use of corticosteroids to accelerate lung maturity is not contraindicated. For fetuses at younger gestational ages, the options of expectant management or intravascular transfusion have been proposed.9,14
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No randomized trials to date have evaluated the best management for fetal hydrops or anemia caused by parvovirus B19 infection. A summary of 14 studies involving a total of 1436 cases of fetal parvovirus infection found a survival rate of 82% with transfusion compared with 55% in those who were not transfused.80
80.
von Kaisenberg, C.S. ∙ Jonat, W.
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The upper limit of gestational age for transfusion is case- and centre-dependent. Two to three transfusions may be required before resolution of the fetal hydrops or anemia, which usually occurs 3 to 6 weeks after the first transfusion.8
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Adler, S. ∙ Koch, W.C.
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The degree of hydrops may not correlate with fetal hemoglobin because of myocarditis. The role of fetal echocardiography should be explored.
The role of Doppler measurement of the MCA peak systolic flow in the management of hydropic fetuses needs further research, but cohort studies suggest it helps to determine the likelihood of anemia as the cause of the hydrops and to measure its severity.75–77,79
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Delle Chiaie, L. ∙ Buck, G. ∙ Grab, D. ...
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If a recent parvovirus B19 infection has been diagnosed in the woman, referral to an obstetrician or a maternal–fetal medicine specialist should be considered. (III-B) The woman should be counselled regarding risks of fetal transmission, fetal loss, and hydrops and serial ultrasounds should be performed every 1 to 2 weeks, up to 12 weeks after infection, to detect the development of anemia (using Doppler measurement of the middle cerebral artery peak systolic velocity) and hydrops. (III-B) If hydrops or evidence of fetal anemia develops, referral should be made to a specialist capable of fetal blood sampling and intravascular transfusion. (II-2B)
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6816 | https://www.geeksforgeeks.org/maths/why-do-we-use-plus-or-minus-in-square-root/ | Why do we use plus or minus in square root? - GeeksforGeeks
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Why do we use plus or minus in square root?
Last Updated : 23 Jul, 2025
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The arithmetic value which is used for representing the quantity and used in making calculations are defined as Numbers. A symbol like “4,5,6” which represents a number is known as a numeral. Without numbers, we can't do counting of things, dates, time, money, etc., these numbers are also used for measurement and used for labelling.
The properties of numbers make them helpful in performing arithmetic operations on them. These numbers can be written in numeric forms and also in words.
For example,3 is written as three in words, 35 is written as thirty-five in words, etc. Students can write the numbers from 1 to 100 in words to learn more.
There are different types of numbers, which we can learn. They are whole and natural numbers, odd and even numbers, rational and irrational numbers, etc.
What is a Number System?
A Number System is a method of showing numbers by writing, which is a mathematical way of representing the numbers of a given set, by mathematically using the numbers or symbols. The writing system for denoting numbers logically using digits or symbols is defined as the Number System.
We can use the digits from 0 to 9 to form all the numbers. With these digits, anyone can create infinite numbers.
For example, 156, 3907, 3456, 1298, 784859, etc.
What is a Square Root?
Square roots of a number are defined as a number which on multiplication by itself gives the original number. Suppose a is the square root of b, then it is represented as
a = √b
We can express the same equation as a2= b. Here, ’√’ this symbol we used to represent the root of numbers is termed as are.
The positive number when it is to be multiplied by itself represents the square of the number. The square root of the square of any positive number gives the original number.
For example, the square of 4 is 16, 4 2 = 16, and the square root of 16, √16 = ±4.Since 4 is a perfect square, hence it is easy to find the square root of such numbers, but for an imperfect square, it's really tricky.
To represent a number ‘a' as a square root using this symbol can be written as: ‘√a‘, where a is the number.
The number here under the radical symbol is called the radicand. For example, the square root of 4 is also represented as a radical of 4. Both represent the same value.
Formula to find the square root is: a = √b
Properties of Square Roots
It is defined as a one-to-one function that takes a positive number as an input and returns the square root of the given input number.
f(x) = √x
For example, here if x = 9, then the function returns the output value as 3.
Properties of the square root are as follows:
If a number is a perfect square number, then there definitely exists a perfect square root.
If a number ends with an even number of zeros (0’s), then we can have a square root.
The two square root values can be multiplied. For example, √3 can be multiplied by √2, then the result will be √6.
When two same square roots are multiplied, then the result must be a radical number. It shows that the result is a non-square root number. For example, when √7 is multiplied by √7, the result obtained is 7.
The square root of negative numbers is undefined. Hence the perfect square cannot be negative.
Some of the numbers end with 2, 3, 7, or 8 (in the unit digit), then the perfect square root does not exist.
Some of the numbers end with 1, 4, 5, 6, or 9 in the unit digit, then the number will have a square root.
It is easy to find the square root of a number that is a perfect square.
Perfect squares are those positive numbers that can be written as the multiplication of a number by itself, or you can say that a perfect square is a number which is the value of power 2 of any integer.
Number that can be expressed as the product of two equal integers.
For example,16 is a perfect square because it is the product of two equal integers, 4 × 4 = 16. However, 24 is not a perfect square because it cannot be expressed as the product of two equal integers. (8 × 3 = 24).
Number which is obtained by squaring a whole number is termed as a perfect square. If we assume N is a perfect square of a whole number y, this can be written as
N = Product of y and y = y 2.
So, the perfect square formula can be expressed as:
N = Y2
Let's Use the formula with values.
If y = 9, and N = y 2.
This means, N = 9 2 = 81.
Here, 81 is a perfect square of 9 because it is the square of a whole number.
So real square roots of 81 is +9, -9
With the help of square roots, we can identify whether a number is a perfect square or not, if we calculate the square root of the given number.
If the square root is a whole number, then the given number will be a perfect square, and if the square root value is not a whole number, then the given number is not a perfect square.
For example, to check whether 24 is a perfect square or not, we will calculate its square root. √24 = 4.898979. As we can see, 4.898979 is not a whole number, so, 24 is not a perfect square.
Let's take another example of
The number 49. √49 = ±7. We can see that 7 is a whole number, therefore, 49 is a perfect square.
Why do we use plus or minus in square root?
If we want both the positive and the negative square root of a radicand then we put the symbol ± (read as plus minus) in front of the root.
The numbers that are not a perfect square are members of the irrational numbers. This means that numbers or square root can't be written as the quotient of two integers.
Related Article
If the square root of a number plus two is the same number, then find the number?
Why do we use numbers?
Square Root
Sample Problems
Question 1: What are the two square roots of 100?
Solution:
Here 100 is the perfect square of 10, so this can have two roots one negative and one positive
or we can say real square root of 100 is ±10
or 10 2 = 10 × 10 = 100
(-10)2 = - 10 × - 10 = 100
Hence, the two square roots of 100 are +10 and -10.
Question 2: What are the square roots of 12?
Solution:
Square root of 12
Here 12 is not a perfect square so this number doesn't have two square roots we can't write it as √12 = ±3.464
Therefore √12 = 3.464 is an irrational number, the numbers that are not a perfect square are members of the irrational numbers. This means that numbers or square roots can't be written as the quotient of two integers.
Question 3: What are the two square roots of 144?
Solution:
square root of 144
Here square root of 144 is perfect square of 12, i.e a whole number this has two square roots +12, -12
Therefore √144 = ± 12
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下面小编来给大家介绍一下SAT数学题的解题方法，希望能给同学们带来帮助。
当然这些内容中的绝大部分在中国大陆的数学教学中高二之前就已经全部学习完毕了。
在一次完整的New SAT考试里，数学部分一共包括3个Section ，其中2个Section各需在25分钟内完成，1个Section需要在20分钟内完成。题型则分为两种：选择题和填写空格题。所以说大家还是需要多做SAT数学练习题来备考SAT数学考试的。
给大家介绍几种常见的数学解题方法：
代入法
例题：If x and y are two different integers and the product 35xy is the square of an integer, which of the following could be equal to xy?
A. 5 B.70 C. 105 D. 140 E. 350解答：将答案依次带入，发现只有D选项35xy=35×140=35×35×4=(35×2)2。
特殊值法
例题：xy=x+y. If y>2, what are all possible values of x that satisfy the equation above?
A. x<0, B. 0
解答：y>2,不妨假设它为3，3x=x+3,x=3/2,一目了然D即为正确选项。
巧解法
例题：At a certain diner, Joe orders 3 doughnuts and a cup of coffee and is charged $2.25. Stella orders 2 doughnuts and a cup of coffee and is charged $1.70. What is the price of 2 doughnuts?
A. $ 0.55 B. $ 0.60 C. $ 1.10 D. $ 1.30 E. $ 1.80
解答：按照常规的解题思路，我们可以用字母x和y分别指代“炸面包圈”和“咖啡”的价格，然后建立2个方程式3 x+y= $ 2.25 , 2 x+y= $ 1.70，最后求出方程组的解x=$ 0.55,
y =$ 0.6。
但是只要稍作思考，我们不难发现解题的捷径—— 3个炸面包圈加1杯咖啡的价格与2个炸面包圈加1 杯咖啡的价格相减，余额就是1个炸面包圈的价格：$ 2.25 -$ 1.70 =$ 0.55 。此时，我们还要注意的是，千万不能掉入命题者的陷阱而选择A,，因为A选项只是1个炸面包圈的价格。所以，正确答案应该是C选项。
以上就是SAT数学中有效的解题方法，更多SAT干货敬请关注新航道SAT考试频道。
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6818 | https://mecheng.iisc.ac.in/suresh/me256/notes3_2007.pdf | Lecture notes #3 of ME 256: Variational Methods and Structural Optimization Jan.-May, 2007 1 of 3 Ananthasuresh, IISc III. Fundamental lemmas of calculus of variations We are now familiar with the notions of a functional, vector spaces (of which function spaces are one type), Gâteaux variation and Fréchet differential. We also know operationally useful definition of Gâteaux variation of a functional. We did all this because we want to derive the necessary conditions for a minimum of the given functional. But then, Gâteaux variation depends on an arbitrary function h . In contrast, the gradient (i.e., the derivative) of an ordinary function does not have such an arbitrary entity. Of course, we noted that h exists in the definition of Gâteaux variation just as a direction is there in the definition of a directional derivative of an ordinary function. In any case, h is there and we have to deal with it. At this point, h is there in between us and the necessary conditions for a minimum of a functional. This is where the fundamental lemma of calculus of variations helps us. So, let us look at it. Lemma 1 If ( ) F x is continuous in [ ] , a b and if ( ) ( ) 0 b a F x h x dx = ∫ for every function ( ) ( ) 0 , h x c a b ∈ such that ( ) ( ) 0 h a h b = = , then ( ) 0 F x = for all [ ] , x a b ∈ . It is a simple but profound statement. It is simple in that one can easily see why this is true. It is profound because many results of calculus of variations rest on this. It is interesting to note that its proof was attempted in 1854 by Stegmann before Du Bois-Raymond proved it in 1879. So, we can perhaps assume that Euler, Lagrange and others who dealt with necessary and sufficient conditions for a minimum of a functional tacitly assumed that it is true. For the same of completeness, let us look at a proof of this lemma. It will be proved by contradiction—a legitimate method of proving things although it is simply a process of verifying what you know as truth. Proof of lemma 1 by contradiction Let us say that ( ) F x is not zero over its entire domain. Let us assume that it is positive for some interval 1 2 [ , ] x x contained within [ , ] a b . Let ( ) ( )( ) 1 2 h x x x x x = − − for 1 2 [ , ] x x x ∈ and zero outside of 1 2 [ , ] x x . Note that ( )( ) 1 2 x x x x − − is positive for 1 2 [ , ] x x x ∈ . Now, consider: Lecture notes #3 of ME 256: Variational Methods and Structural Optimization Jan.-May, 2007 2 of 3 Ananthasuresh, IISc ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( )( ) 1 2 1 2 2 1 2 1 1 2 0 ( ) ( ) 0 0 x x b b a a x x x x x x F x h x dx F x h x dx F x h x dx F x h x dx F x h x dx F x x x x x dx = + + = + + = − − > ∫ ∫ ∫ ∫ ∫ ∫ Thus, we get a contradiction to what is given the lemma. So, we can conclude that ( ) F x cannot be non-zero anywhere in the domain [ , ] a b . This proves the lemma. Lemma 2 If ( ) F x is continuous in [ , ] a b and if ( ) ( ) 0 b a F x h x dx ′ = ∫ for every ( ) ( ) 1 , h x c a b ∈ such that ( ) ( ) 0 h a h b = = , then ( ) F x = constant for all [ ] , x a b ∈ . To prove this, let c be defined as in ( ) ( ) 0 b a F x c dx − = ∫ and let ( ) ( ) ( ) x a h x F c d ξ ξ = − ∫ so that ( ) h x satisfies the conditions laid out in the statement of the lemma. Now, consider: ( ) { } ( ) ( ) ( ) ( ) ( ) ( ) 0 b b a a F x c h x dx F x h x dx c h b h a ′ ′ − = − − = ∫ ∫ (Why is this true? The reason lies in the statement of the lemma.) and ( ) ( ) 2 ( ) ( ) ( ) b b a a F x c h x dx F x c dx ′ − = − ∫ ∫ (Why is this true? The reason lies in our assume ( ) h x .) Therefore, ( ) ( ) 2 ( ) ( ) ( ) 0 ( ) 0 ( ) b b a a F x c h x dx F x c dx F x c F x c ′ − = − = ⇒ − = ⇒ = ∫ ∫ for all [ ] , x a b ∈ . This proved this second lemma. In calculus of variations, two more lemmas are also stated. Lemma 3 Lecture notes #3 of ME 256: Variational Methods and Structural Optimization Jan.-May, 2007 3 of 3 Ananthasuresh, IISc If ( ) F x is continuous in [ , ] a b and if ( ) ( ) 0 b a F x h x dx ′′ = ∫ for every ( ) ( ) 2 , h x c a b ∈ such that ( ) ( ) 0 h a h b = = and ( ) ( ) 0 h a h b ′ ′ = = , then ( ) 0 1 F x c c x = + for all [ ] , x a b ∈ where 0 c and 1 c are constants. Lemma 4 If ( ) 1 F x and ( ) 2 F x are continuous in [ , ] a b and if ( ) ( ) 1 2( ) ( ) 0 b a F x h x F x h x dx ′ + = ⎡ ⎤ ⎣ ⎦ ∫ for every ( ) ( ) 1 , h x c a b ∈ such that ( ) ( ) 0 h a h b = = , then ( ) 2 F x is differentiable and ( ) 2 1( ) F x F x ′ = for all [ , ] x a b ∈ . Lemmas 3 and 4 can also be proved by contradiction in the same way as the first two by assuming certain functions for ( ) h x . It is also interesting that lemmas 2-4 can also be derived from lemma 1 using simple rule of integration by parts. In fact, as we will see later in the course, the rule of integration by parts is an essential tool of calculus of variations. We must also recall that Green’s theorem and divergence theorem are essentially integration by parts in higher dimensions. |
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Glomus Jugulare Tumors Treatment & Management
Medical Therapy
Some cases require no treatment. Often, glomus jugulare tumors are diagnosed within the sixth or seventh decade of life and can be followed by imaging only and may not need surgical intervention.
A study from Vanderbilt University found that in the absence of brainstem compression or concern for malignancy, observation of glomus jugulare tumors can be a viable initial management approach for elderly patients. Of 15 patients studied (80% female; median age, 69.6 yr), radiologic growth occurred in 5 patients. The median growth rate of the 5 enlarging tumors was 0.8 mm/yr (range, 0.6-1.6 mm/yr) using maximum linear dimension, or 0.4 cm3/yr (0.1-0.9 cm3/yr) with volumetric analysis. No deaths were attributable to tumor progression or treatment.
Medical therapy may be indicated in some cases. Alpha-blockers and beta-blockers are useful for tumors secreting catecholamines. They are usually administered for 2-3 weeks before embolization and/or surgery to avoid potentially lethal blood pressure lability and arrhythmias. Successful treatment of pulmonary metastases with etoposide (VP-16) and cisplatin has been described. In a preliminary report, a somatostatin analogue (octreotide) has been successfully used for growth control of somatostatin receptor–positive tumors.
Surgical Therapy
Surgery is the treatment of choice for glomus jugulare tumors. However, radiation therapy, particularly stereotactic radiosurgery (eg, Gamma Knife surgery), has been shown to provide good tumor growth control with a low risk of treatment-related cranial nerve injury. [1, 17, 18, 19, 20, 31, 32, 33, 34] Numerous studies of stereotactic radiosurgery have shown reduction or stabilization of tumor size and improvement in overall neurologic deficit. [1, 17, 18, 19, 20]
The short- and intermediate-term risk of progression to nonserviceable hearing following stereotactic radiosurgery for jugular paraganglioma is low. In a series of 85 patients who underwent SRA for the treatment of jugular paraganglioma, the Kaplan-Meier estimated rates of serviceable hearing at 1, 3, and 5 years following SRS were 91%, 80%, and 80%, respectively. Sixty percent of patients with pulsatile tinnitus who underwent SRS experienced varying levels of symptomatic improvement following treatment.
A large retrospective, multicenter, international study analyzed the long-term outcome in 132 patients with primary radiation treatment or radiation after partial resection of a glomus tumor. The study found long-term successful control of the tumor growthi and mprovement of tinnitus and overall neurological status, as well as cranial nerve function.
Of 22 patients with glomus jugulare tumors who underwent Gamma Knife surgery, neurologic status improved in 12 patients, 7 showed stable clinical condition, and 3 patients developed new moderate deficits. The average tumor volume was 7.26 cm3. Tumor volume following surgery was unchanged in 13 patients and was decreased in 8; tumor regrowth occurred in 1 patient. Tumor progression-free survival was 95.5% at 5 and 7 years.
A German study of 32 patients who underwent stereotactic radiosurgery for glomus jugulare tumors showed that stereotactic linear accelerator (LINAC) radiosurgery achieved excellent long-term tumor control, along with a low rate of morbidity. According to the study, following LINAC stereotactic radiosurgery, 10 of 27 patients showed a significant improvement of their previous neurologic complaints, whereas 12 patients remained unchanged. No tumor progression was observed. Five patients died due to unrelated causes. Overall survival rates after 5, 10, and 20 years were 100%, 95.2% and 79.4%, respectively.
In a study of 28 patients treated with radiosurgery and 2 patients with stereotactic radiosurgery, crude overall survival, tumor control, clinical control, and long-term grade 1 toxicity rates were 97%, 97%, 97%, and 13% (4/30), respectively. No statistically significant risk factor was associated with lower tumor control in the series. Univariate analysis showed a statistically significant association between patients having 1 cranial nerve (CN) involvement before radiosurgery and a higher risk of lack of improvement of symptoms (odds ratio 5.24, 95% confidence interval 1.06–25.97, P=0.043).
Because resection of glomus jugulare tumors can be challenging due to their inherent vascularity, preoperative embolization of these tumors with ethylene vinyl alcohol (Onyx) has been proposed. A study by Gaynor et al showed a dramatic reduction of blood loss and facilitation of surgical resection, but these results came at the price of a higher incidence of cranial nerve neuropathy.
Because this tumor is rare and may present with various symptoms, surgery may be contraindicated for various reasons, including age and general physical condition. Surgical resection of the glomus tumor is relatively simple and complication free for type I tumors. Large tumors that affect the lower cranial nerves and extend beyond the petrous apex carry a significant risk of postoperative complications, especially in older patients. In these cases, other modalities of treatment should be considered (eg, embolization, radiation, Gamma Knife rardiosurgery, intratumoral injection of cyanoacrylate glue).
Ehret et al evaluated the effectiveness and safety of image-guided robotic radiosurgery (RRS) and reported overall local control of 99% after a median follow-up of 35 months and stated that 56% of patients experienced symptom improvement or recovered entirely after undergoing RRS. Another study reported that quality of life analyses after RRS revealed no significant decline, while bodily pain significantly decreased.
The surgical approach depends on the localization and extension of the tumor. Intraoperative monitoring including EEGs and somatosensory-evoked potentials (SSEPs) are routinely used.
Fisch type A tumors can be excised by a transmeatal or perimeatal approach. Type B tumors require an extended posterior tympanotomy.
Type C tumors require radical resection via a standard combined transmastoid-infratemporal or transtemporal-infratemporal approach with or without internal carotid artery (ICA) trapping, preceded by external carotid artery (ECA) embolization or superselective embolization. Intraoperatively, temporarily occlude the transverse or sigmoid sinus with EEG monitoring to determine whether vein bypass should be performed for total resection. Surgery leads to therapeutic success in about 90% of patients. Intratumoral injection of cyanoacrylate glue has been proposed to control bleeding.
Large type D tumors need to be treated with a combined otologic and neurosurgical approach. An infratemporal approach with a skull base resection and a posterior fossa exploration is the most advisable in attempting to remove the entire tumor. Partial resection of the tumor needs to be followed by radiation and follow-up MRI/CT scanning.
Radiation therapy and radiosurgery may be indicated. Both classic fractionated radiation therapy (40-50 Gy) and stereotactic radiosurgery (eg, Gamma Knife surgery) are successful in long-term control of tumor growth and in decrease of catecholamine excretion in functional tumors; however, the short duration of observation after stereotactic radiosurgery does not allow for definite conclusions. Radiation treatment is advised as the sole treatment modality for elderly or infirm patients who are symptomatic, especially those with extensive or growing tumors.
Gross total resection of some extensive tumors may be extremely difficult and may carry unwarranted risk. In such cases, radiotherapy may be indicated to treat residual tumor following subtotal resection. [9, 10] However, a study by Prahbu showed that even complex glomus tumors can be managed surgically.
In a study of 51 patients with jugular foramen tumors who underwent less-aggressive surgical interventions to preserve neurovascular structures, overall tumor recurrence-regrowth-free survival, symptom-progression-free survival, and overall survival at 15 years were 78.9%, 86.8%, and 80.6%, respectively. The tumor recurrence-regrowth rate was 11.8%, swallowing function improved or stabilized in 96.1%, and facial function improved or stabilized in 94.1%. Overall neurologic status improved or stabilized in 90% of patients. In the study patients, the mean lesion size was 3.8 cm, and 43 cases (84.3%) were Fisch type D, including 37 cases (72.5%) of type Di1 and Di2. Thirty-seven cases (72.5%) were Glasscock-Jackson type III-IV. Gross-total resection and subtotal resection were achieved in 26 (51.0%) and 22 (43.1%) cases, respectively.
(See the images below.)
Content.
Preoperative, Intraoperative, and Posoperative Details
If routine screening for catecholamine is positive (3 times the reference range), alpha-blockers and beta-blockers are administered for 2-3 weeks before surgery and embolization. This helps to avoid blood pressure lability and arrhythmias. In emergent cases, 3 days of treatment is adequate.
Surgical approach depends on the localization and extent of the tumor. Fisch type A tumors can be excised by a transmeatal or perimeatal approach. Type B tumors require an extended posterior tympanotomy. Type C tumors require radical resection via a standard combined transmastoid-infratemporal or transtemporal-infratemporal approach with or without ICA trapping, preceded by external carotid artery embolization or superselective embolization. Surgery leads to therapeutic success in about 90% of patients. Treat large type D tumors with a combined otologic and neurosurgical approach. An infratemporal approach with a skull base resection and a posterior fossa exploration is advisable in the attempt to remove the entire tumor.
Patients are usually in the sixth decade of life; therefore, careful monitoring of cardiac function is advisable, especially if a catecholamine secreting tumor was only partially resected.
Postoperative lower cranial nerve deficits need to be carefully diagnosed, and, when present, early rehabilitation is advocated.
Radiologic and, when indicated, endocrinologic monitoring for tumor growth or regrowth is indicated every 6 months to 1 year for 2 years and then, depending on the dynamics of the tumor behavior, every 2 years.
(See the images below.)
Complications of surgery include death, cranial nerve palsies, bleeding, cerebrospinal fluid (CSF) leak, meningitis, uncontrollable hypotension/hypertension, and tumor regrowth.
Complications of radiation include ICA thrombosis, sigmoid sinus thrombosis, secondary tumor development, pituitary-hypothalamic insufficiency, CSF leak, tumor growth, and radiation necrosis of bone, brain, or dura.
Toxicities of single-fraction stereotactic radiosurgery (SRS) include vertigo, nausea, and headache along with lower cranial neuropathies.
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Sharma M, Meola A, Bellamkonda S, Jia X, Montgomery J, Chao ST, et al. Long-Term Outcome Following Stereotactic Radiosurgery for Glomus Jugulare Tumors: A Single Institution Experience of 20 Years. Neurosurgery. 2018 Nov 1. 83 (5):1007-1014. [QxMD MEDLINE Link].
Patel NS, Link MJ, Driscoll CLW, Pollock BE, Lohse CM, Carlson ML. Hearing Outcomes After Stereotactic Radiosurgery for Jugular Paraganglioma. Otol Neurotol. 2018 Jan. 39 (1):99-105. [QxMD MEDLINE Link].
Sheehan JP, Tanaka S, Link MJ, et al. Gamma Knife surgery for the management of glomus tumors: a multicenter study. J Neurosurg. 2012 Aug. 117(2):246-54. [QxMD MEDLINE Link].
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Tables
Contributor Information and Disclosures
Ryszard M Pluta, MD, PhD (Retired) Associate Professor, Neurosurgical Department Medical Research Center, Polish Academy of Sciences, Poland; (Retired) Clinical Staff Scientist, Surgical Neurology Branch, National Institute of Neurological Disorders and Stroke, National Institutes of Health (NIH); (Retired) Fishbein Fellow, JAMA
Ryszard M Pluta, MD, PhD is a member of the following medical societies: Congress of Neurological Surgeons, Polish Society of Neurosurgeons
Disclosure: Nothing to disclose.
Brian A Iuliano, MD Attending Neurosurgeon, Central Maryland Neurosurgical Associates
Disclosure: Nothing to disclose.
Francisco Talavera, PharmD, PhD Adjunct Assistant Professor, University of Nebraska Medical Center College of Pharmacy; Editor-in-Chief, Medscape Drug Reference
Disclosure: Received salary from Medscape for employment.
Brian H Kopell, MD Associate Professor, Department of Neurosurgery, Icahn School of Medicine at Mount Sinai; Director, Center for Neuromodulation, Co-Director, The Bonnie and Tom Strauss Movement Disorders Center, Department of Neurosurgery, Mount Sinai Health System
Brian H Kopell, MD is a member of the following medical societies: Alpha Omega Alpha, American Association of Neurological Surgeons, American Society for Stereotactic and Functional Neurosurgery, Congress of Neurological Surgeons, International Parkinson and Movement Disorder Society, North American Neuromodulation Society
Disclosure: Received income in an amount equal to or greater than $250 from: Medtronic; Abbott Neuromodulation; Turing Medical.
Duc Hoang Duong, MD Professor, Chief Physician, Departments of Neurological Surgery and Neuroscience, Epilepsy Center, Charles Drew University of Medicine and Science
Duc Hoang Duong, MD is a member of the following medical societies: American Neurological Association, Congress of Neurological Surgeons, North American Skull Base Society
Disclosure: Nothing to disclose.
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6821 | https://www.sciencedirect.com/science/article/pii/S2346809216300812 | Homogeneous functions: New characterization and applications - ScienceDirect
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1. Introduction and preliminaries
2. The q-analogues of the Euler theorem
3. Applications of the q-Euler operator
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Transactions of A. Razmadze Mathematical Institute
Volume 171, Issue 2, August 2017, Pages 171-181
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Homogeneous functions: New characterization and applications
Author links open overlay panel Moncef Elghribi a, Hakeem A.Othman b c, Al-Hossain Ahmed Al-Nashri b
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Abstract
Positive homogeneous functions on R of a negative degree are characterized by a new counterpart of the Euler’s homogeneous function theorem using quantum calculus and replacing the classical derivative operator by Jackson derivative. As application we start by characterizing the harmonic functions associated to Jackson derivative. Then, the solution of the Cauchy problem associated to the analogue of the Euler operator is given. Using this solution we study the associated ν-potential. Its Markovianity property is treated.
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Keywords
Homogeneous functions
Euler’s theorem
Quantum calculus
Cauchy problem
Markovian semigroups
1. Introduction and preliminaries
The notion of a homogeneous function arises in connection with the spherical harmonic functions. The solid harmonic also can be defined as homogeneous functions that obey Laplace’s equation. The Euler theorem is used in proving that the Hamiltonian is equal to the total energy. In thermodynamics, extensive thermodynamic functions are homogeneous functions. In this context, Euler’s theorem is applied in thermodynamics by taking Gibbs free energy. Also, Euler’s theorem is of value in analytical mechanics and has been widely implemented as a theoretical basis for the reversal of wide magnetic and gravity data sets in terms of single sources, see, , . In mathematics, a homogeneous function is a function f with multiplicative scaling behavior, i.e, if the argument is multiplied by a factor α, then the result is multiplied by some power λ of this factor. Positive homogeneous functions are characterized by Euler’s homogeneous function theorem which consists of: f is positive homogeneous of degree λ∈R if and only if ∑j=1 n x j∂∂x j f(x)=λ f(x). The operator ∑j=1 n x j∂∂x j is called the Euler operator (see). In microeconomics, they use homogeneous production functions, including the function of Cobb–Douglas, developed in 1928, the degree of such homogeneous functions can be negative which was interpreted as decreasing returns to scale. The concept of homogeneity in methods for enforcement finds modeling physical phenomena and, in particular, for directly solving inverse problems for potential fields, see, where the gravity field of a mass point has the potential V. The tests of homogeneity for this potential can be implemented using equation of Euler’s theorem to study the homogeneity for the gravity potential V with a negative degree of homogeneity (λ=−1), see. Also, Gel’fand and Shilov studied the homogeneous distributions of negative integer degree λ on R.
At the last quarter of the XX century, q-calculus appeared as a connection between mathematics and physics (see for more details, , , , ). It has a lot of applications in different basic hyper-geometric functions and other sciences as quantum theory, mechanics and theory of relativity. We shall briefly recall some of the concepts, notations and known results on q-calculus as given in, , , , , , . Let q∈(0,1). A q-number [n]q is defined by [n]q≔1+q++⋯+q n,n∈N. Generally a q-complex number is given by [a]q is [a]q=1−q a 1−q,a∈C. The factorial of a number [n]q is defined by q!=1,[n]q!=[n]q[n−1]q...q,n∈N. The q-derivative also referred to as Jackson derivative is defined as follows (1)D q f(x)={f(x)−f(q x)(1−q)x,q∈(0,1),x≠0 f′(0),x=0, such that, lim q⟶1 D q f(x)=d f(x)d x, if f is differentiable at x. This derivative (2.1) verifies the following q-derivation property (2)D q(f.g)(x)=g(q x)D q f(x)+f(x)D q g(x)=f(q x)D q g(x)+g(x)D q f(x). And the high q-derivatives are D q 0 f≔f,D q n f≔D q(D q n−1 f),n∈N. Notice, that a continuous function on an interval, which does not include 0 is continuous q-differentiable. The q-derivative operator D q and the operator X defined by X f(x)=x f(x) give a bounded representation of a a∗−q a∗a=1 on H 2(B q,μ q) which is the completion of the analytic functions on B q={z∈C:‖z‖2<1 1−q} with respect to the inner product defined by a measure μ q on the complex plane that replaces the Lebesgue measure on the unit circle, for more details see. As q tends to 1, μ q tends to the Gauss measure on the complex plane. This representation generalizes the Bargmann representation of analytic functions on the complex plane. The operator D q+X, viewed as a non-commutative (or quantum) random variable, has a q-Gaussian distribution in the vacuum state. The operator X D q will be called the q-Euler operator. This operator has a q-Poisson distribution in the vacuum state. It is obvious that X D q is the q-deformation of the operator X d d x verifying: as q tends to 1, X D q tends to X d d x. Now, if we replace X d d x by X D q, what are the q-analogues of the Euler’s theorem?
In this paper we give a response to this above question as follows: positive homogeneous functions f on R of a negative degree λ are characterized by a new counterpart of the Euler’s homogeneous function theorem using the q-Euler operator, i.e, f is homogeneous of degree λ if and only if Δ E,q f(x)=[λ]q f(x), for q∈(0,1) and x∈R+. As application we start by characterizing the q-harmonic functions. Then, the solution of the Cauchy problem associated to the q-Euler operator is given. Using this solution we study the associated ν-potential. Its Markovianity property is treated.
2. The q-analogues of the Euler theorem
Let f:R+⟶R. We say that f is homogeneous of degree λ∈R, if for all x∈R and for all α>0 f(α x)=α λ f(x).
Definition 2.1
For q∈(0,1), we define the q-Euler operator by Δ E,q=X.D q.
Proposition 2.1
Let f,g:R+⟶R , then for q∈(0,1), we have(3)Δ E,q(f.g)(x)=g(x)Δ E,q f(x)+f(q x)Δ E,q g(x).
Proof
By definition, we have, for x≠0 Δ E,q(f.g)(x)=x D q(f.g)(x)=x f(x)g(x)−f(q x)g(q x)(1−q)x=x f(x)−f(q x)(1−q)x g(x)+x f(q x)g(x)−g(q x)(1−q)x=x(D q f)(x)g(x)+x f(q x)(D q g)(x)=g(x)Δ E,q f(x)+f(q x)Δ E,q g(x). Now, for x=0, we know that D q(f.g)(x)|x=0=(f.g)′(0)=f′(0)g(0)+g′(0)f(0). Then, applying the multiplication operator x, we obtain Δ E,q(f.g)(0)=g(0)(Δ E,q f)(0)+f(q.0)(Δ E,q g)(0), which completes the proof.□
Theorem 2.2
If f is homogeneous of degree λ∈R , then we have Δ E,q f(x)=[λ]q f(x),q∈(0,1).
Proof
Let f be a homogeneous function of degree λ∈R. Then for x≠0, we have D q(f)(x)=f(x)−f(q x)(1−q)x=f(x)−q λ f(x)(1−q)x=(1−q λ)(1−q)f(x)x=[λ]q f(x)x. Therefore, we get Δ E,q f(x)=[λ]q f(x). Now, for x=0 and λ∈R∖{0}, we have (X.D q f)(0)=0 and f′(0)=0. But, we know that f is homogeneous of degree λ then f(α x)=α λ f(x). In particular for x=0,f(0)=α λ f(0) for α>0, which implies that f(0)=0 for λ∈R∗. Then, we get (X.D q f)(0)=0=[λ]q f(0). Now, for λ=0 and x=0(X.D q f)(0)=0=q f(0). Hence, for x∈R+ and λ∈R, we obtain Δ E,q f(x)=[λ]q f(x), which gives the desired statement, which completes the proof.□
Theorem 2.3
Let q∈(0,1),x∈R and λ<0 . If we have Δ E,q f(x)=[λ]q f(x).Then, f is homogeneous of degree λ.
Proof
Let f:R+⟶R such that Δ E,q f(x)=[λ]q f(x). For all x∈R+, define g:(0,∞)⟶R by g(α)=f(α x)−α λ f(x). Then, for α>0, we get D q(g)(α)=f(α x)−f(q α x)(1−q)α−[λ]q α λ−1 f(x)=x α f(α x)−f(q α x)(1−q)x−[λ]q α λ−1 f(x)=1 α Δ E,q f(α x)−[λ]q α λ−1 f(x). Then, we can obtain α D q g(α)=Δ E,q f(α x)−[λ]q α λ f(x)=[λ]q(f(α x)−α λ f(x))=[λ]q g(α). Since α is arbitrary, g satisfies the following q-differential equation (4)D q g(α)−[λ]q α g(α)=0.Eq. (4) is equivalent to g(α)−g(q α)(1−q)α=[λ]q α g(α). Then, we get g(α)=q−λ g(q α). By a simple iteration, we obtain g(α)=q−λ n g(q n α). Then, for n⟶∞, we get g(α)=0, which is equivalent to f(α x)=α λ f(x). This completes the proof.□
Combining Theorem 2.2 with Theorem 2.3 above, we obtain the following theorem, which will be called q-analogues of Euler’s Theorem:
Theorem 2.4
Let λ<0 . Then, f is homogeneous of degree λ if and only if Δ E,q f(x)=[λ]q f(x),for q∈(0,1)and x∈R+.
Remark 1
It is obvious from Theorem 2.4 that, as q tends to 1, we refind the classical Euler’s theorem for λ<0.
3. Applications of the q-Euler operator
3.1. q-harmonic function
This subsection deals with the study of a link taken homogeneous function and a new notion of q-harmonic functions.
Definition 3.1
f is called q-harmonic function if D q 2(f)=0.
Theorem 3.1
Let f be λ -homogeneous. Then f is q -harmonic if and only if λ=0 .
Proof
“⇒”Let f be λ-homogeneous and q-harmonic. Then from Theorem 2.2, we have D q 2(f)(x)=0 and from Theorem 2.2(5)X.D q(f)(x)=[λ]q f(x). Then we get (X.D q)2(f)(x)=[λ]q 2 f(x). But we know that, D q X−q X D q=1. Then, D q X=1+q X D q from which we get X(1+q X D q)2 D q(f)(x)=[λ]q 2 f(x)X.D q(f)(x)+q X 2 D q 2 f(x)=[λ]q 2 f(x). Thus [λ]q f(x)=[λ]q 2 f(x). This gives λ=0. “⇐”Let f be 0-homogeneous. Then for λ=0, we have f(α x)=f(x). Therefore, we get D q f(x)=f(x)−f(q x)(1−q)x=f(x)−f(x)(1−q)x=0. Then f is q-harmonic, which completes the proof.□
3.2. Cauchy problem associated to the q-Euler operator
Let f:R+⟶R be homogeneous function of degree λ where 0<λ≤1. Consider, the following Cauchy problem (6){∂∂t U(t,x)=Δ E,q U(t,x),x∈R+U(0,x)=f(x).
Theorem 3.2
The Cauchy problem(6)admits a unique solution given by(7)U(t,x)=f(x)+∑n=1∞∑k=0 n−1(−1)k t n[f(q k x)−f(q k+1 x)]n(n−1−k)!k!(1−q)n.
Proof
We start by verifying that U(t,x)≔f(x)+∑n=1∞∑k=0 n−1(−1)k t n[f(q k x)−f(q k+1 x)]n(n−1−k)!k!(1−q)n is a solution of the system (6). On the one hand, we have ∂U(t,x)∂t=∑n=1∞∑k=0 n−1(−1)k t n−1f(q k x)−f(q k+1 x)!k!(1−q)n. On the other hand, we have Δ E,q U(t,x)=x D q U(t,x)=U(t,x)−U(t,q x)1−q=f(x)−f(q x)1−q+∑n=1∞∑k=0 n−1(−1)k t n[f(q k x)−f(q k+1 x)]n(n−1−k)!k!(1−q)n+1−∑n=1∞∑k=0 n−1(−1)k t n[f(q k+1 x)−f(q k+2 x)]n(n−1−k)!k!(1−q)n+1. By indices change in the right sums of the above equation, we obtain Δ E,q U(t,x)=f(x)−f(q x)1−q+∑n=1∞∑k=0 n−1(−1)k t n[f(q k x)−f(q k+1 x)]n(n−1−k)!k!(1−q)n+1+∑n=1∞∑k=1 n(−1)k t n[f(q k x)−f(q k+1 x)]n(n−k)!(k−1)!(1−q)n+1=x D q f(x)+∑n=1∞∑k=0 n(−1)k t nf(q k x)−f(q k+1 x)!k!(1−q)n+1=∑n=0∞∑k=0 n(−1)k t nf(q k x)−f(q k+1 x)!k!(1−q)n+1=∑n=1∞∑k=0 n−1(−1)k t n−1f(q k x)−f(q k+1 x)!k!(1−q)n=∂U(t,x)∂t,which shows that U(t,x) is a solution of (6). Let us show by recursion on n∈N⋆ that (8)Δ E,q n f(x)=1(1−q)n∑k=0 n−1(k n−1)(−1)k(f(q k x)−f(q k+1 x)).
We have for, n=1 Δ E,q f(x)≔x D q f(x)=1 1−q(f(x)−f(q x)),x∈R+. Now, suppose that Eq. (7) is verified, then we get Δ E,q n+1 f(x)=Δ E,q(Δ E,q n f(x))=x D q(Δ E,q n f(x))=1 1−q[Δ E,q n f(x)−Δ E,q n f(q x)]=1(1−q)n∑k=0 n−1(k n−1)(−1)k q k x D q f(q k x)−1(1−q)n∑k=0 n−1(k n−1)(−1)k q k+1 x D q f(q k+1 x). By indices change in the right sums of the above equation, we obtain (9)Δ E,q n+1 f(x)=1(1−q)n∑k=0 n−1(k n−1)(−1)k q k x D q f(q k x)+1(1−q)n∑k=1 n(k−1 n−1)(−1)k q k x D q f(q k x)=1(1−q)n(x D q f(x)+(−1)n q n x D q f(q n x))+1(1−q)n∑k=1 n−1{(k n−1)+(k−1 n−1)}(−1)k q k x D q f(q k x)=1(1−q)n(x D q f(x)+(−1)n q n x D q f(q n x))+1(1−q)n∑k=1 n−1(k n)(−1)k q k x D q f(q k x)=1(1−q)n∑k=0 n(k n)(−1)k q k x D q f(q k x)=1(1−q)n+1∑k=0 n(k n)(−1)k(f(q k x)−f(q k+1 x)). This shows Eq. (6) for all n∈N. Then, using Eq. (6) we get (10)Q t f(x)≔∑n=0∞t k n!Δ E,q n f(x)=f(x)+∑n=1∞t k n!Δ E,q n f(x)=f(x)+∑n=1∞∑k=0 n−1(−1)k t n(f(q k x)−f(q k+1 x))n(n−1−k)!k!(1−q)n=U(t,x). Finally, we show the uniqueness of the above solution. Let V(t,x) be another solution of Eq. (6), we set W(t,x)=Q−t V(t,x). Then∂W(t,x)∂t=−Δ E,q W(t,x)+Q−t(Δ E,q V(t,x))=−Δ E,q W(t,x)+Δ E,q Q−t V(t,x)=0 from which, we deduce that W(t,x)=W(0,x)=V(0,x)=f(x). This implies that V(t,x)=Q t f(x)=U(t,x), which completes the proof.□
3.3. ν–q-potential
Using the semigroup {Q t}t we come to the following.
Definition 3.2
For ν>0, we define the ν–q-potential by: H ν,q f(x)=∫0∞e−ν t(Q t(f)(x)−f(x))d t.
Theorem 3.3
The ν–q -potential is the unique solution of the following Poisson equation :(ν I−Δ E,q)F=1 ν D q.
Proof
By Definition 3.2 and Eq. (10) we have H ν,q f(x)=∫0∞e−ν t∑n=1∞∑k=0 n−1(−1)k t n(f(q k x)−f(q k+1 x))n(n−1−k)!k!(1−q)n d t=∑n=1∞∑k=0 n−1(−1)k(f(q k x)−f(q k+1 x))n(n−1−k)!k!(1−q)n∫0∞e−ν t t n d t. One can show easily that ∫0∞e−ν t t n d t=n!ν n+1. Then, (11)H ν,q f(x)=∑n=1∞∑k=0 n−1(−1)k t n(f(q k x)−f(q k+1 x))(n−k)!k!(1−q)n((n−1)!ν n+1). On the other hand we have X.D q Q t f(x)=Δ E,q e t Δ E,q f(x).=∑n=0∞t n n!Δ E,q n+1 f(x).=Δ E,q f(x)+∑n=1∞t n n!Δ E,q n+1 f(x). Using Eq. (9), we get Δ E,q Q t f(x)=Δ E,q f(x)+∑n=1∞t n n!1(1−q)n+1∑k=0 n(k n)(−1)k(f(q k x)−f(q k+1 x))from which we obtain Δ E,q(Q t f−f)(x)=∑n=1∞∑k=0 n t n(−1)k k!(n−k)!(1−q)n+1(f(q k x)−f(q k+1 x)). Then, from Definition 3.2, we get (12)Δ E,q H ν,q f(x)=∑n=1∞∑k=0 n n!(−1)k ν n+1 k!(n−k)!(1−q)n+1(f(q k x)−f(q k+1 x)). By the change indices (n−1=j) in the right sums of Eq. (11), we have H ν,q f(x)=∑j=0∞∑k=0 j j!(−1)k ν n+2 k!(j−k)!(1−q)j+1(f(q k x)−f(q k+1 x)). Using Eq. (12) we obtain H ν,q f(x)=1 ν 2(1−q)(f(x)−f(q x))+∑j=0∞∑k=0 j j!(−1)k ν n+2 k!(j−k)!(1−q)j+1(f(q k x)−f(q k+1 x))=1 ν 2(1−q)(f(x)−f(q x)+1 ν Δ E,q H ν,q f(x)),which is equivalent to ν H ν,q f(x)−Δ E,q H ν,q f(x)=1 ν D q f(x). This implies that(ν I−Δ E,q)H ν,q=1 ν D q,which completes the proof.□
3.4. Markovianity property
Recall that from {P t}t≥0 is called a Markov semigroup if it satisfies
(a)P 0=I d
(b)P t+s=P t P s for s,t≥0
(c)Strong continuity : P t f→f as t→0 for all f.
(d)P t f≥0 whenever f≥0
(e)P t 1=1 for all t≥0.
Note that conditions (d) and (e) imply Contraction Property: ‖P t f‖≤‖f‖ for all f and t.
Theorem 3.4
The family{Q t}t≥0 is Markov semigroup.
Proof
(a)It is obvious that Q 0=I d.
(b)Let s,t≥0, then Q t+s=e(s+t)Δ E,q=e s Δ E,q e t Δ E,q=Q t Q s.
(c)Let t≥0, then ‖Q t f−f‖≤∑n=1∞t n‖Δ E,q‖n‖f‖n!=(e t‖Δ E,q‖−1)‖f‖→0 as t→0.
(d)Using Theorem 2.2, we get Δ E,q f(x)=[λ]q f(x). Similarly using Theorem 2.2, we obtain (Δ E,q)n f(x)=([λ]q)n f(x). Then, Q t f(x)=∑n=o∞t n n![λ]q n f(x)=e t[λ]q f(x). Hence, when f≥0, we obtain Q t f≥0.
(e)Using Eq. (10), we get Q t 1=1,for all t≥0.
This completes the proof.□
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References
D.T. Thompson EULDPH - A new technique for making computer assisted depth estimates from magnetic data Geophysics, 47 (1982), pp. 31-37 View in ScopusGoogle Scholar
I. Marson, E.E. Klingele Advantages of using the vertical gradient of gravity for 3-D interpretation Geophysics, 58 (1993), pp. 1588-1595 View in ScopusGoogle Scholar
P. Stavrev, A. Reid Degrees of homogeneity of potential fields and structural indexes of Euler deconvolution Geophysics, 72 (1) (2007), pp. 1-12 CrossrefGoogle Scholar
I.M. Gel’fand, G.E. Shilov Generalized Functions, vol. I, Academic Press, Inc., New York (1968)
H.F. Jackson q-Difference equations Amer. J. Math., 32 (1910), pp. 305-314 CrossrefGoogle Scholar
T.E. Mason On properties of the solution of linear q-difference equations with entire function coefficients Amer. J. Math., 37 (1915), pp. 439-444 CrossrefGoogle Scholar
C.R. Adams On the linear ordinary q-difference equation Am. Math. Ser. II, 30 (1929), pp. 195-205 Google Scholar
W.H. Abdi On q-Laplace transforms Proc. Acad. Sci. India, 29 A (1960), pp. 389-408 Google Scholar
W.H. Abdi On certain q-difference equations and q-Laplace transforms Proc. Nat. Inst. Sci. India Acad, 28 A (1962), pp. 1-15 Google Scholar
G. Gasper, M. Rahman Basic hypergeometric series Encyclopedia of Mathematics and its Application, vol. 35, Cambridge University Press, Cambridge (1990) Google Scholar
G. Bangerezako Variational q-calculus J. Math. Anal. Appl., 289 (2004), pp. 650-665 View PDFView articleView in ScopusGoogle Scholar
H.V. Leeuwen, H. Maassen A q-deformation of the Gauss distribution J. Math. Phys., 36 (9) (1995), pp. 4743-4756 Google Scholar
R. Rundnicki, K. Pichor, M. Tyran-Kaminska Markov semigroups and their applications Dynamics of Dissipation, Lecture Notes in Physics, vol. 597, Springer, Berlin, Heidelberg (2002), pp. 215-238 Google Scholar
Further reading
M. Bozejko, R. Speicher An example of a generalized Brownian motion Comm. Math. Phys., 137 (1991), pp. 519-531 View in ScopusGoogle Scholar
L. Euler Differential Calculus Gostechizdat (1949) (Russian translation) Google Scholar
R.O. Hansen, L. Suciu Multiple-source Euler deconvolution Geophysics, 67 (2002), pp. 525-535 View in ScopusGoogle Scholar
P. Protter Stochastic integration and differential equations A New Approach, Springer, Berlin (1992) Google Scholar
H. Rguigui Quantum Ornstein–Uhlenbeck semigroups Quantum Stud. Math. Found., 2 (2) (2015), pp. 159-175 CrossrefView in ScopusGoogle Scholar
H. Rguigui Quantum λ-potentials associated to quantum Ornstein–Uhlenbeck semigroups Chaos Solitons Fractals, 73 (2015), pp. 80-89 View PDFView articleView in ScopusGoogle Scholar
H. Rguigui Characterization of the QWN-conservation operator and applications Chaos Solitons Fractals, 84 (2016), pp. 41-48 View PDFView articleView in ScopusGoogle Scholar
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6822 | https://dermnetnz.org/topics/connective-tissue-naevi | Connective tissue naevi (nevi)
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Connective tissue naevus
Connective tissue naevus — extra information
Categories:
Lesions (benign)
ICD-10:
D23.9, D21.9
ICD-11:
LC20, LC2Y, LC02, 2F7C
SNOMED CT:
400091006, 399926001, 277812008, 23914407, 56525001, 254146000
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Lesions (benign)
Connective tissue naevus
Author: Vanessa Ngan, Staff Writer, 2003.
IntroductionCollagenomasElastomasOther connective tissue naeviTreatment
What is a connective tissue naevus?
A connective tissue naevus (American spelling nevus) is an uncommon skin lesion that occurs when the deeper layers of the skin do not develop correctly or the components of these layers occur in the wrong proportion. There may be too much collagen; this is called a collagenoma. Or there may be too much elastic tissue; this is called an elastoma. The naevi may also include excess glycosaminoglycans (ground substance of the dermis), excess smooth muscle or excess fat.
Connective tissue naevi are sometimes found in other diseases including tuberous sclerosis, chronic myelocytic leukaemia, syphilis and encephalocraniocutaneous lipomatosis.
Connective tissue naevi are thought to be due to genetic defects in the skin cells.
Collagenomas
Collagenomas include:
Familial cutaneous collagenoma (FCC)
Isolated collagenomas
Shagreen patch (associated with tuberous sclerosis)
Connective tissue naevus
Connective tissue naevus
Connective tissue naevus
Familial cutaneous collagenomalesions usually appear during adolescence. They are characterised by multiple hard nodules of varying sizes over the upper two thirds of the back. They are sometimes associated with cardiac disease.
Shagreen patch is commonly associated with tuberous sclerosis. There are flesh coloured ‘orange-peel’ textured lesions of varying sizes, which are usually found on the lower back and nape of the neck.
Elastomas
Elastomas or elastin naevi include:
Isolated elastomas
Buschke-Ollendorf syndrome
Elastosis perforans serpiginosa
Buschke-Ollendorf syndrome is a rare hereditary disorder where there is an increased accumulation of elastin in the dermis (elastoma). Lesions may be present at birth but more usually appear within the first year of life. They are firm yellowish wrinkled nodules often group together to form plaques. The abdomen, back, buttock, thighs or arms are commonly affected. Other manifestations of the syndrome appear with time and may include osteopoikilosis (inherited bone disorder identified on X-Ray), eye disorders and spinal problems.
Elastosis perforans serpiginosa (EPS) is a perforating disorder. In this case, abnormal elastic fibres are extruded through the epidermis. It presents in adolescence and typically affects one or more sites on the face, neck and/or arms. Groups of scaly papules are generally arranged in an arc or ring shape. EPS is associated with other disorders of connective tissue such as Marfan syndrome, Ehler Danlos syndrome, osteogenesis imperfecta, pseudoxanthoma elasticum and Down syndrome.
Elastosis perforans serpiginosa
Other connective tissue naevi
Congenital smooth muscle hamartoma is first noted at birth or during the first few weeks of life. It is usually a single irregularly shaped lesion that once established remains unchanged.
Fat naevus (naevus lipomatosus superficialis) occurring at birth includes the very rare Michelin tyre baby disorder that is characterised by generalised folding of redundant skin.
Congenital fibromatosis (infantile myofibromatosis) is characterised by single or multiple benign tumours that appear to be derived from connective tissue and smooth muscle cells. These tumours may involve the skin and underlying tissues, bones, and/or certain internal organs. They are present at birth or develop within the first few weeks of life. Lesions usually resolve spontaneously, however severe or widespread involvement of internal organs may cause potentially life-threatening complications.
Mucinous naevus is formed by a deposit of mucin and can be present at birth or develop later in life.
Treatment of connective tissue naevi
Connective tissue naevi that are not associated with other diseases do not require any treatment. Patients with naevi should undergo full examination to rule out any associated conditions.
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References
Book: Textbook of Dermatology. Ed Rook A, Wilkinson DS, Ebling FJB, Champion RH, Burton JL. Fourth edition. Blackwell Scientific Publications.
On DermNet
Naevi (birthmarks)
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Naevus lipomatosus superficialis – pathology
Elastosis
Fibroblastic connective tissue naevus pathology
Nonmelanocytic congenital naevi— common skin lesions course
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Connective tissue nevus— Medscape Reference
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6823 | https://www.cut-the-knot.org/Curriculum/Geometry/GeoGebra/PPC.shtml | PPC: Apollonius' Problem with Two Points and a Circle
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PPC: Apollonius' Problem with Two Points and a Circle<
Below I give an Euclidean construction to a variant of the Problem of Apollonius:
Find a circle through two given points C and D and tangent to a given circle A(B), with center at A and passing through B:
Rather obviously the problem has no solutions if the given circle separates the given points (i.e., if one of the points is inside, the other outside the circle.)
(The applet below illustrates the proof. Points A,B,C,D,E are draggable. This permits changing the configuration, but within certain limitations.)
Solution
|Contact||Front page||Content||Geometry|
Copyright © 1996-2018 Alexander Bogomolny
Construction
Find a circle through two given points C and L and tangent to a given circle A(B), with center at A and passing through B:
Start with finding any circle that passes through the given points and intersects with the given circle in two point, say, F and G:
Let H be the intersection of C D and F G. Find H S and H T - the tangents from H to A(B). Find the intersections K and L of the perpendicular bisector of C D with A S and A T, respectively. Circles K(S) and L(T) solve the problem.
Why does the construction work?
Assuming the problem solved, let (O) denote the auxiliary circle. Then F G is the radical axis of (O) and A(B), the common tangent at S is the radical axis of K(S) and A(B), while the tangent at T is the radical axis of L(T) and A(B). Lastly, C D is the radical axis of any pair of (O), K(S), or L(T). It follows that H is the radical center of the three circles K(S),A(B), and (O), as well as the radical center of L(T),C(A,B), and (O). You can notice by dragging point E on the circle (O) in the applet that H does not depend on the position of (O). Thus H is uniquely determined as the intersection of F G and C D.
The problem has two solutions.
References
N. Altshiller-Court, College Geometry, Dover, 1980, #470
J. L. Coolidge, A Treatise On the Circle and the Sphere, AMS - Chelsea Publishing, 1971, p. 168
R. A. Johnson, Advanced Euclidean Geometry (Modern Geometry), Dover, 2007, #169
|Contact||Front page||Content||Geometry|
Copyright © 1996-2018 Alexander Bogomolny
73255532 |
6824 | https://www.mountsinai.org/health-library/discharge-instructions/bronchiolitis-discharge | Bronchiolitis - discharge Information | Mount Sinai - New York
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Bronchiolitis - discharge
RSV bronchiolitis - discharge; Respiratory syncytial virus bronchiolitis - discharge
Your child has bronchiolitis, which causes swelling and mucus to build up in the smallest air passages of the lungs.
Now that your child is going home from the hospital, follow your health care provider's instructions on how to care for your child. Use the information below as a reminder.
Bronchiolitis is an inflammation of the bronchioles (smaller airways that branch off the main airway) usually caused by a viral infection.
When You're in the Hospital
In the hospital, your provider helped your child breathe better. They also made sure your child received enough fluids.
What to Expect at Home
Your child will likely still have symptoms of bronchiolitis after leaving the hospital.
Wheezing may last for up to 5 days.
Coughing and stuffy nose will slowly get better over 7 to 14 days.
Sleeping and eating may take up to 1 week to return to normal.
You may need to take time off work to care for your child.
Home Care
Breathing moist (wet) air helps loosen the sticky mucus that may be choking your child. You can use a humidifier to make the air moist. Follow the directions that came with the humidifier.
Do not use steam vaporizers because they can cause burns. Use cool mist humidifiers instead.
If your child's nose is stuffy, your child will not be able to drink or sleep easily. You can use warm tap water or saline nose drops to loosen the mucus. Both of these work better than any medicine you can buy.
Place 3 drops of warm water or saline in each nostril.
Wait 10 seconds, then use a soft rubber suction bulb to suck out the mucus from each nostril.
Repeat several times until your child is able to breathe through the nose quietly and easily.
Before anyone touches your child, they must wash their hands with warm water and soap or use an alcohol-based hand cleanser before doing so. Try to keep other children away from your child.
Do not let anyone smoke in the house, car, or anywhere near your child.
Eating and Drinking
It is very important for your child to drink enough fluids.
Offer breast milk or formula if your child is younger than 12 months.
Offer regular milk if your child is older than 12 months.
Eating or drinking may make your child tired. Feed small amounts, but more often than usual.
If your child throws up because of coughing, wait a few minutes and try to feed your child again.
Medicines
Some asthma medicines help children with bronchiolitis. Your provider may prescribe such medicines for your child.
Do not give your child decongestant nose drops, antihistamines, or any other cold medicines unless your provider tells you to.
When to Call the Doctor
Contact your provider right away if your child has any of the following:
Hard time breathing
Chest muscles are pulling in with each breath
Breathing faster than 50 to 60 breaths per minute (when not crying)
Making a grunting noise
Sitting with shoulders hunched over
Wheezing becomes more intense
Skin, nails, gums, lips, or area around the eyes is bluish or grayish
Extremely tired
Not moving around very much
Limp or floppy body
Nostrils are flaring out when breathing
References
Greenland JR, Jones KD, Singer JP. Bronchiolitis. In: Broaddus VC, Ernst JD, King TE, et al, eds. Murray and Nadel's Textbook of Respiratory Medicine. 7th ed. Philadelphia, PA: Elsevier; 2022:chap 72.
Kliegman RM, St. Geme JW, Blum NJ, et al. Wheezing, bronchiolitis, and bronchitis. In: Kliegman RM, St. Geme JW, Blum NJ, et al, eds. Nelson Textbook of Pediatrics. 22nd ed. Philadelphia, PA: Elsevier; 2025:chap 439.
Scarfone RJ, Seiden JA. Pediatric lower airway obstruction. In: Walls RM, ed. Rosen's Emergency Medicine: Concepts and Clinical Practice. 10th ed. Philadelphia, PA: Elsevier; 2023:chap 163.
Version Info
Last reviewed on: 4/1/2024
Reviewed by: Charles I. Schwartz, MD, FAAP, Clinical Assistant Professor of Pediatrics, Perelman School of Medicine at the University of Pennsylvania, General Pediatrician at PennCare for Kids, Phoenixville, PA. Also reviewed by David C. Dugdale, MD, Medical Director, Brenda Conaway, Editorial Director, and the A.D.A.M. Editorial team.
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6825 | https://mathguy.us/Handbooks/CalculusHandbook.pdf | Copyright 2008-23, Earl Whitney, Reno NV. All Rights Reserved Math Handbook of Formulas, Processes and Tricks (www.mathguy.us) Calculus Prepared by: Earl L. Whitney, FSA, MAAA Version 5.6 April 8, 2023 Note to Students This Calculus Handbook was developed primarily through work with a number of AP Calculus classes, so it contains what most students need to prepare for the AP Calculus Exam (AB or BC) or a first‐year college Calculus course. In addition, a number of more advanced topics have been added to the handbook to whet the student’s appetite for higher level study. It is important to note that some of the tips and tricks noted in this handbook, while generating valid solutions, may not be acceptable to the College Board or to the student’s instructor. The student should always check with their instructor to determine if a particular technique that they find useful is acceptable. Why Make this Handbook? One of my main purposes for writing this handbook is to encourage the student to wonder, to ask “what about … ?” or “what if … ?” I find that students are so busy today that they don’t have the time, or don’t take the time, to find the beauty and majesty that exists within Mathematics. And, it is there, just below the surface. So be curious and seek it out. The answers to all of the questions below are inside this handbook, but are seldom taught. What is oscillating behavior and how does it affect a limit? Is there a generalized rule for the derivative of a product of multiple functions? What’s the partial derivative shortcut to implicit differentiation? What are the hyperbolic functions and how do they relate to the trigonometric functions? When can I simplify a difficult definite integral by breaking it into its even and odd components? What is Vector Calculus? Additionally, ask yourself: Why … ? Always ask “why?” Can I come up with a simpler method of doing things than I am being taught? What problems can I come up with to stump my friends? Those who approach math in this manner will be tomorrow’s leaders. Are you one of them? Please feel free to contact me at mathguy.us@gmail.com if you have any comments. Thank you and best wishes! Earl Cover art by Rebecca Williams, Twitter handle: @jolteonkitty Version 5.6 Page 2 of 242 April 8, 2023 Page Description Chapter 1: Functions and Limits 8 Functions 10 Continuity Examples 11 Limits 12 Techniques for Finding Limits 14 Indeterminate Forms 16 When Limits Fail to Exist Chapter 2: Differentiation 17 Definition, Basic Rules, Product Rule 18 Quotient, Chain and Power Rules; Exponential and Logarithmic Functions 19 Trigonometric and Inverse Trigonometric Functions 23 Generalized Product Rule 25 Inverse Function Rule 26 Partial Differentiation 27 Implicit Differentiation 30 Logarithmic Differentiation Chapter 3: Applications of Derivatives 31 Maxima and Minima (i.e., Extrema) 33 Inflection Points 34 Special Case: Extrema and Inflection Points of Polynomials 35 Relationship Among f(x), f'(x) and f''(x) 39 Curve Sketching 44 Determining the Shape of a Curve Based On Its Derivatives 45 Rolles's Theorem and the Mean Value Theorem (MVT) 46 Related Rates 49 Kinematics (Particle Motion) 52 Differentials 53 Curvature 54 Newton's Method Chapter 4: Integration 56 Indefinite Integration (Antiderivatives) 57 Exponential and Logarithmic Functions 57 Trigonometric Functions 60 Inverse Trigonometric Functions 62 Selecting the Right Function for an Intergral Calculus Handbook Table of Contents Version 5.6 Page 3 of 242 April 8, 2023 Calculus Handbook Table of Contents Page Description Chapter 5: Techniques of Integration 63 u -Substitution 65 Integration by Partial Fractions 68 Integration by Parts 72 Integration by Trigonometric Substitution 74 Gamma Function 76 Beta Function 78 Impossible Integrals Chapter 6: Hyperbolic Functions 79 Definitions 80 Identities 81 Relationship to Trigonometric Functions 82 Inverse Hyperbolic Functions 83 Graphs of Hyperbolic Functions and Their Inverses 84 Derivatives 85 Integrals Chapter 7: Definite Integrals 87 Riemann Sums 92 Rules of Definite Integration 92 Fundamental Theorems of Calculus 94 Properties of Definite Integrals 95 Solving Definite Integrals with Directed Line Segments 96 u -Subsitution 98 Special Techniques for Evaluation 100 Derivative of an Integral Chapter 8: Applications of Integration 101 Area Under a Curve 102 Area Between Curves 103 Area in Polar Form 105 Areas of Limacons 107 Arc Length 110 Comparison of Formulas for Rectangular, Polar and Parametric Forms 111 Area of a Surface of Revolution 112 Volumes of Solids of Revolution Chapter 9: Improper Integrals 118 Definite Integrals with Infinite Limits of Integration 119 Definite Integrals with Discontinuous Integrands Version 5.6 Page 4 of 242 April 8, 2023 Calculus Handbook Table of Contents Page Description Chapter 10: Differential Equations 120 Definitions 121 Separable First Order Differential Equations 123 Slope Fields 124 Logistic Function 125 Numerical Methods Chapter 11: Vector Calculus 129 Introduction 129 Special Unit Vectors 129 Vector Components 130 Properties of Vectors 131 Dot Product 132 Cross Product 134 Triple Products 135 Kinematics (Particle Motion) 136 Gradient 137 Divergence 138 Curl 139 Laplacian Chapter 12: Sequences 140 Definitions and Types of Sequences 141 More Definitions and Theorems 142 Limits (Convergence and Divergence) 143 Basic Recursive Sequence Theory Chapter 13: Series 147 Introduction 148 Key Properties 148 n-th Term Convergence Theorems 148 Power Series 149 Telescoping Series 150 Geometric Series 151 Estimating the Value of Series with Positive Terms 152 Riemann Zeta Function (p -Series) 156 Bernoulli Numbers 158 Convergence Tests 163 Alternating Series 165 Radius and Interval of Convergence of Power Series 168 Summary of Convergence/Divergence Tests Version 5.6 Page 5 of 242 April 8, 2023 Calculus Handbook Table of Contents Page Description Chapter 14: Taylor and MacLaurin Series 169 Taylor Series 169 MacLaurin Series 171 LaGrange Remainder Chapter 15: Miscellaneous Cool Stuff 172 e 173 Derivation of Euler's Formula 175 Logarithms of Negative Real Numbers and Complex Numbers 176 What Is i i 177 Derivative of e to a Complex Power (ez) 178 Derivatives of a Circle 179 Derivatives of a Ellipse 180 Derivatives of a Hyperbola 181 Derivative of: (x+y)3=x3+y3 182 Inflection Points of the PDF of the Normal Distribution Appendices 183 Appendix A: Key Definitions 203 Appendix B: Key Theorems 207 Appendix C: List of Key Derivatives and Integrals 214 Appendix D: Key Functions and Their Derivatives 218 Appendix E: Geometry and Trigonometry Formulas 223 Appendix F: Polar and Parametric Equations 234 Appendix G: Interesting Series 235 Index Useful Websites www.mathguy.us mathworld.wolfram.com Wolfram Math World – A premier site for mathematics on the Web. This site contains definitions, explanations and examples for elementary and advanced math topics. Mathguy.us – Developed specifically for math students from Middle School to College, based on the author's extensive experience in professional mathematics in a business setting and in math tutoring. Contains free downloadable handbooks, PC Apps, sample tests, and more.
Version 5.6 Page 6 of 242 April 8, 2023 Calculus Handbook Table of Contents Schaum’s Outlines Other Useful Books An important student resource for any high school math student is a Schaum’s Outline. Each book in this series provides explanations of the various topics in the course and a substantial number of problems for the student to try. Many of the problems are worked out in the book, so the student can see how they can be solved. Schaum’s Outlines are available at Amazon.com, Barnes & Noble and other booksellers.
Version 5.6 Page 7 of 242 April 8, 2023 Chapter 1 Functions and Limits Functions Definitions Expression: A meaningful arrangement of mathematical values, variables and operations. Relation: An expression that defines a connection between a set of inputs and a set of outputs. The set of inputs is called the Domain of the relation. The set of outputs is called the Range of the relation. Function: A relation in which each element in the domain corresponds to exactly one element in the range. One‐to‐One Function: A function in which each element in the range is produced by exactly one element in the domain. Continuity: A function, 𝑓, is continuous at 𝑥ൌ𝑎 iff: o 𝑓ሺ𝑎ሻ is defined, o lim ௫ →𝑓ሺ𝑥ሻ exists, and o lim ௫ →𝑓ሺ𝑥ሻൌ𝑓ሺ𝑎ሻ o If 𝑥ൌ𝑎 is an endpoint, then the limit need only exist from the left or the right. Continuity Rules If 𝑓ሺ𝑥ሻ and 𝑔ሺ𝑥ሻ are continuous functions at a point ሺ𝑥, 𝑦ሻ, and if 𝑐 is a constant, then the following are also true at ሺ𝑥, 𝑦ሻ: 𝑓ሺ𝑥ሻ𝑔ሺ𝑥ሻ is continuous. Addition 𝑓ሺ𝑥ሻെ𝑔ሺ𝑥ሻ is continuous. Subtraction 𝑐∙𝑓ሺ𝑥ሻ is continuous. Scalar Multiplication 𝑓ሺ𝑥ሻ∙𝑔ሺ𝑥ሻ is continuous. Multiplication ሺ௫ሻ ሺ௫ሻ is continuous if 𝑔ሺ𝑥ሻ്0. Division 𝑓ሺ𝑥ሻ is continuous if 𝑓ሺ𝑥ሻ exists. Exponents ඥ𝑓ሺ𝑥ሻ is continuous if ඥ𝑓ሺ𝑥ሻ exists. Roots Note: All polynomial functions are continuous on the interval ሺെ∞, ∞ሻ. Note: lim ௫→𝑓ሺ𝑥ሻ exists if and only if: lim ௫→ି𝑓ሺ𝑥ሻൌlim ௫→ା𝑓ሺ𝑥ሻ. Version 5.6 Page 8 of 242 April 8, 2023 Chapter 1 Functions and Limits Types of Discontinuities A Discontinuity occurs at a location where the graph of a relation or function is not connected. Removable Discontinuity. A discontinuity that can be “repaired” by adding a single point to the graph. Typically, this will show up as a hole in a graph. In the function 𝒇ሺ𝒙ሻൌ 𝒙𝟑ା𝟏 𝒙ା𝟏 , a removable discontinuity exists at 𝑥ൌെ1. Mathematically, a removable discontinuity is a point at which the limit of 𝑓ሺ𝑥ሻ at 𝑐 exists but does not equal 𝑓ሺ𝑐ሻ. That is, lim ௫ →ି𝑓ሺ𝑥ሻ ൌlim ௫ →ା𝑓ሺ𝑥ሻ ്𝑓ሺ𝑐ሻ Note: a removable discontinuity exists at 𝑥ൌ𝑐 whether or not 𝑓ሺ𝑐ሻ exists. Essential Discontinuity. A discontinuity that is not removable. Mathematically, an essential discontinuity is a point at which the limit of 𝑓ሺ𝑥ሻ at 𝑐 does not exist. This includes: o Jump Discontinuity. A discontinuity at which the limit from the left does not equal the limit from the right. That is, lim ௫ →ି𝑓ሺ𝑥ሻ ്lim ௫ →ା𝑓ሺ𝑥ሻ In the function 𝒇ሺ𝒙ሻൌ ห𝒙𝟑ି𝟏ห 𝟒ሺ𝒙ି𝟏ሻ , a jump discontinuity exists at 𝑥ൌ1. o Infinite Discontinuity. These occur at vertical asymptotes. In the function 𝒇ሺ𝒙ሻൌ 𝟏 𝒙𝟐ା𝟓𝒙ା𝟔 , infinite discontinuities exist at 𝑥ൌሼെ3, െ2ሽ. Version 5.6 Page 9 of 242 April 8, 2023 Chapter 1 Functions and Limits Continuity Examples Case 1 Case 2 Case 3 Case 4 Jump Discontinuity Not continuous Limit does not exist ݂ሺ5ሻ may or may not exist (it does not exist in the graph shown) Removable Discontinuity Not continuous Limit exists ݂ሺ5ሻ does not exist Removable Discontinuity Not continuous Limit exists ݂ሺ5ሻ exists but does not equal the limit No Discontinuity Continuous Limit exists ݂ሺ5ሻ exists and is equal the limit Version 5.6 Page 10 of 242 April 8, 2023 Chapter 1 Functions and Limits Limits Definitions Formal Definition: Let 𝒇 be a function defined on an open interval containing 𝒂, except possibly at 𝒙ൌ𝒂, and let 𝑳 be a real number. Then, the statement: lim ௫ →𝑓ሺ𝑥ሻൌ𝐿 means that for each 𝜀0, there exists a 𝛿0 such that: 0 ൏|𝑥െ𝑎| ൏𝛿 implies |𝑓ሺ𝑥ሻെ𝐿| ൏𝜀. Written using math symbols: ∀ 𝜀0 ∃ 𝛿0 ∋ 0 ൏|𝑥െ𝑎| ൏𝛿 ⇒ |𝑓ሺ𝑥ሻെ𝐿| ൏𝜀. Informal Definition: The limit is the value 𝑳 that a function approaches as the value of the input variable 𝒙 approaches the desired value 𝒂. Limits may exist approaching 𝑥ൌ𝑎 from either the left ቀlim 𝒙 →𝒂ష𝑓ሺ𝑥ሻቁ or the right ሺlim 𝒙 →𝒂శ𝑓ሺ𝑥ሻሻ. If the limits from the left and right are the same (e.g., they are both equal to 𝐿), then the limit exists at 𝑥ൌ𝑎 and we say lim ௫ →𝑓ሺ𝑥ሻൌ𝐿. Limit Rules Assuming that each of the requisite limits exist, the following rules apply: lim ௫ →ሾ𝑓ሺ𝑥ሻ𝑔ሺ𝑥ሻሿൌlim ௫ →𝑓ሺ𝑥ሻlim ௫ →𝑔ሺ𝑥ሻ Addition of Limits lim ௫ →ሾ𝑓ሺ𝑥ሻെ𝑔ሺ𝑥ሻሿൌlim ௫ →𝑓ሺ𝑥ሻെlim ௫ →𝑔ሺ𝑥ሻ Subtraction of Limits lim ௫ →ሾ𝑐∙𝑓ሺ𝑥ሻሿൌ𝑐∙lim ௫ →𝑓ሺ𝑥ሻ Scalar Multiplication lim ௫ →ሾ𝑓ሺ𝑥ሻ∙𝑔ሺ𝑥ሻሿൌlim ௫ →𝑓ሺ𝑥ሻ∙lim ௫ →𝑔ሺ𝑥ሻ Multiplication of Limits lim ௫ →ቂ ሺ௫ሻ ሺ௫ሻቃൌ ୪୧୫ ೣ →ೌሺ௫ሻ ୪୧୫ ೣ →ೌሺ௫ሻ Division of Limits lim ௫ →𝑓ሺ𝑥ሻൌቂlim ௫ →𝑓ሺ𝑥ሻቃ Powers lim ௫ →ൣඥ𝑓ሺ𝑥ሻ ൧ൌටlim ௫ →𝑓ሺ𝑥ሻ Roots Also, assuming that each of the requisite limits exists, the typical properties of addition and multiplication (e.g., commutative property, associative property, distributive property, inverse property, etc.) apply to limits. Version 5.6 Page 11 of 242 April 8, 2023 Chapter 1 Functions and Limits Techniques for Finding Limits Substitution The easiest method, when it works, for determining a limit is substitution. Using this method, simply substitute the value of ݔ into the limit expression to see if it can be calculated directly. Example 1.1: lim ௫ →ଷ൬ݔ2 ݔെ2൰ൌ3 2 3 െ2 ൌ Simplification When substitution fails, other methods must be considered. With rational functions (and some others), simplification may produce a satisfactory solution. Example 1.2: lim ௫ →ହቆݔଶെ25 ݔെ5 ቇൌlim ௫ →ହ൬ሺݔ5ሻሺݔെ5ሻ ሺݔെ5ሻ ൰ൌݔ5 ൌ Rationalization Rationalizing a portion of the limit expression is often useful in situations where a limit is indeterminate. In the example below the limit expression has the indeterminate form ሺെ∞∞ሻ. Other indeterminate forms are discussed later in this chapter. Example 1.3: lim ௫ →ିஶቀݔඥݔଶെ8ݔቁ First, notice that this limit is taken to െ∞, which can often cause confusion. So, let’s modify it so that we are taking the limit to ∞. We do this using the substitution ݔൌെݕ. lim ௫ →ିஶቀݔඥݔଶെ8ݔቁൌlim ௬ →ାஶቀെݕඥݕଶ8ݕቁ Next, let’s rationalize the expression in the limit by multiplying by a name for one, using its conjugate. (cont’d) Version 5.6 Page 12 of 242 April 8, 2023 Chapter 1 Functions and Limits lim ௬ →ାஶቀെݕඥݕଶ8ݕቁൌlim ௬ →ାஶ൭െݕඥݕଶ8ݕ 1 ∙ ݕඥݕଶ8ݕ ݕඥݕଶ8ݕ ൱ ൌlim ௬ →ାஶቆെݕଶݕଶ8ݕ ݕඥݕଶ8ݕ ቇൌlim ௬ →ାஶቆ 8ݕ ݕඥݕଶ8ݕ ቇ ൌlim ௬ →ାஶቆ 8ݕ ݕඥݕଶ8ݕ ൊݕ ݕቇൌlim ௬ →ାஶ ۉ ۇ 8 1 ට1 8 ݕی ۊ ൌ 8 1 √1 ൌ4 L’Hospital’s Rule If ݂ and ݃ are differentiable functions and ݃ᇱሺݔሻ്0 near ܽ and if: lim ௫→݂ሺݔሻൌ0 and lim ௫→݃ሺݔሻൌ0 ۽܀ lim ௫→݂ሺݔሻൌേ∞ and lim ௫→݃ሺݔሻൌേ∞ Then, ܔܑܕ ࢞→ࢇ ࢌሺ࢞ሻ ࢍሺ࢞ሻ ൌ ܔܑܕ ࢞→ࢇ ࢌ′ሺ࢞ሻ ࢍ′ሺ࢞ሻ Note: L’Hospital’s rule can be repeated as many times as necessary as long as the result of each step is an indeterminate form. If a step produces a form that is not indeterminate, the limit should be calculated at that point. Example 1.4: lim ௫ → sin ݔ ݔ ൌ lim ௫ → d dx sin ݔ ݀ ݀ݔݔ ൌ lim ௫ → cos ݔ 1 ൌ 1 1 ൌ Example 1.5: lim ௫ → ݔ ݁ଷ௫െ1 ൌ lim ௫ → d dx ݔ d dx ሺ݁ଷ௫െ1ሻ ൌ 1 3݁ଷ௫ൌ 1 3 ∙1 ൌ Example 1.6: (involving successive applications of L’Hospital’s Rule) lim ௫ →ஶ 3ݔଷ2ݔ1 4ݔଷെ5ݔଶെ2 ൌ lim ௫ →ஶ 9ݔଶ2 12ݔଶെ10ݔ ൌ lim ௫ →ஶ 18ݔ 24ݔെ10 ൌlim ௫ →ஶ 18 24 ൌ Version 5.6 Page 13 of 242 April 8, 2023 Chapter 1 Functions and Limits L’Hospital’s Rule Indeterminate Forms of Limits The following table presents some types of indeterminate forms that may be encountered and suggested methods for evaluating limits in those forms. Form Steps to Determine the Limit 0 0 or ∞ ∞ Use L’Hospital’s Rule 0 ∙∞ ∞െ∞ For either of these forms: 1. Convert to or ஶ ஶ 2. Use L’Hospital’s Rule 0 ∞ 1ஶ For any of these forms: 1. Take ݈݊ of the term or write the term in exponential form 2. Convert to or ஶ ஶ 3. Use L’Hospital’s Rule For ݕൌሾ݂ሺݔሻሿሺ௫ሻ, convert to: ܔܖ࢟ൌࢍሺ࢞ሻ∙ܔܖࢌሺ࢞ሻ or ࢟ൌࢋࢍሺ࢞ሻ∙ܔܖࢌሺ࢞ሻ Example 1.7: Form ∙∞ ܔܑܕ ࢞→ିஶ࢞ࢋ࢞ൌlim ௫→ିஶቀݔ ݁ି௫ቁൌlim ௫→ିஶ൬ 1 െ݁ି௫൰ൌlim ௫→ିஶെ݁௫ൌ Example 1.8: Form ∞െ∞ ܔܑܕ ࢞→ሺ࣊ ⁄ ሻష ሺܛ܍܋࢞െܜ܉ܖ࢞ሻ ൌ lim ௫→ሺగଶ ⁄ ሻష൬1 cos ݔെsin ݔ cos ݔ൰ ൌ lim ௫→ሺగଶ ⁄ ሻష ሺ1 െsin ݔሻ cos ݔ ൌ lim ௫→ሺగଶ ⁄ ሻష െcos ݔ െsin ݔൌ L’Hospital’s Rule Version 5.6 Page 14 of 242 April 8, 2023 Chapter 1 Functions and Limits L’Hospital’s Rule L’Hospital’s Rule L’Hospital’s Rule Example 1.9: Form 𝟎𝟎 𝐥𝐢𝐦 𝒙→𝟎శ𝒙𝒙 let: 𝑦ൌlim ௫→శ𝑥௫ ln 𝑦ൌlim ௫→శ𝑥 ln 𝑥ൌlim ௫→శ ൬ln 𝑥 𝑥ିଵ൰ൌlim ௫→శ ቆ𝑥ିଵ െ𝑥ିଶቇ ൌlim ௫→శ ሺെ𝑥ሻൌ0 Then, since ln 𝑦ൌ0, we get 𝑦ൌ𝑒ൌ𝟏 Example 1.10: Form ∞𝟎 𝐥𝐢𝐦 𝒙→ஶ𝒙𝟏/𝒙 let: 𝑦ൌlim ௫→ஶ𝑥ଵ/௫ ln 𝑦ൌlim ௫→ஶ ln 𝑥 𝑥 ൌlim ௫→ஶ ቆ𝑥ିଵ 1 ቇൌlim ௫→ஶ ൬1 𝑥൰ൌ0 Then, since ln 𝑦ൌ0, we get 𝑦ൌ𝑒ൌ𝟏 Example 1.11: Form 𝟏ஶ 𝐥𝐢𝐦 𝒙→𝟎శሺ𝟏𝐬𝐢𝐧𝟒𝒙ሻ𝐜𝐨𝐭𝒙 let: 𝑦ൌlim ௫→శሺ1 sin 4𝑥ሻୡ୭୲௫ ln 𝑦ൌlim ௫→శሾሺcot 𝑥ሻ∙lnሺ1 sin 4𝑥ሻሿൌlim ௫→శ lnሺ1 sin 4𝑥ሻ tan 𝑥 ln 𝑦 ൌlim ௫→శ ቀ4 cos 4𝑥 1 sin 4𝑥ቁ secଶ𝑥 ൌ ቀ4 ∙1 1 0ቁ 1ଶ ൌ4 Then, since ln 𝑦ൌ4, we get 𝑦ൌ𝒆𝟒 Version 5.6 Page 15 of 242 April 8, 2023 Chapter 1 Functions and Limits When Limits Fail to Exist There are several circumstances when limits fail to exist: When taken separately, limits from the left and right are different. This generally occurs at a jump discontinuity. In the graph of ሺ࢞ሻൌ |࢞| ࢞, a jump discontinuity exists at ݔൌ0, so Oscillating behavior at the limit point. Consider the function ࢌሺ࢞ሻൌ܋ܗܛ ࢞ , as ݔ →0. In any neighborhood δ around ݔൌ0, the value of the function varies from െ1 to 1. Therefore, This function is also discontinuous at ݔൌ0, though it is difficult to see this on the graph. Unbounded behavior at the limit point. Typically, this will happen at a vertical asymptote. In the graph of ࢌሺ࢞ሻൌܔܖ|࢞|, an infinite discontinuity exists at ݔൌ0 because the logarithms of positive real numbers that approach zero become large negative numbers without bound. Therefore, Note: in this case, we may write: lim ௫→ln|ݔ| ൌെ∞ lim ௫ →൬cos 1 ݔ൰ does not exist. lim ௫ → |ݔ| ݔ does not exist. lim ௫ →ln|ݔ| does not exist. Version 5.6 Page 16 of 242 April 8, 2023 Chapter 2 Differentiation Basic Rules of Differentiation Definition of a Derivative ݀ ݀ݔ ݂ሺݔሻൌlim → ݂ሺݔ݄ሻെ݂ሺݔሻ ݄ ݀ ݀ݔ ݂ሺݔሻൌlim ௫ → ݂ሺݔሻെ݂ሺܽሻ ݔെܽ Note: In these rules, ࢉ is a constant, and ࢛ and ࢜ are functions differentiable in ࢞. Basic Derivative Rules ݀ ݀ݔሺܿሻൌ0 ݀ ݀ݔሺܿ∙ݑሻൌܿ∙݀ ݀ݔሺݑሻ ݀ ݀ݔሺݑݒሻൌ݀ ݀ݔሺݑሻ݀ ݀ݔሺݒሻ ݀ ݀ݔሺݑെݒሻൌ݀ ݀ݔሺݑሻെ݀ ݀ݔሺݒሻ ݀ݕ ݀ݔൌ 1 ݀ݔ ݀ݕ ݀ݕ ݀ݔൌ ݀ݕ ݀ݑ ݀ݔ ݀ݑ The Product, Quotient and Chain Rules are shown in Leibnitz, Lagrange, and differential forms. Product Rule (two terms) ݀ ݀ݔ ሾ ݂ሺݔሻ∙݃ሺݔሻ ሿൌ݂ሺݔሻ∙݀ ݀ݔሾ ݃ሺݔሻ ሿ݃ሺݔሻ∙݀ ݀ݔሾ ݂ሺݔሻ ሿ ሺ݂∙݃ሻᇱൌ݂∙݃ᇱ݃∙݂′ ݀ ሺݑݒሻൌݑ ݀ݒݒ ݀ݑ Product Rule (three terms) ݀ ݀ݔ ሾ ݂ሺݔሻ∙݃ሺݔሻ ሿൌ݀ ݀ݔሾ ݂ሺݔሻ ሿ∙݃ሺݔሻ∙݄ሺݔሻ݂ሺݔሻ∙݀ ݀ݔሾ ݃ሺݔሻ ሿ∙݄ሺݔሻ݂ሺݔሻ∙݃ሺݔሻ∙݀ ݀ݔሾ ݄ሺݔሻ ሿ ሺ݂∙݃∙݄ሻᇱൌሺ݂ᇱ∙݃∙݄ሻሺ݂∙݃ᇱ∙݄ሻሺ݂∙݃∙݄′ሻ ݀ ሺݑݒݓሻൌݒݓ ݀ݑݑݓ ݀ݒݑݒ ݀ݓ Version 5.6 Page 17 of 242 April 8, 2023 Chapter 2 Differentiation Quotient Rule ݀ ݀ݔ ݂ሺݔሻ ݃ሺݔሻ ൨ൌ ݃ሺݔሻ∙ ݀ ݀ݔሾ݂ሺݔሻሿെ݂ሺݔሻ∙݀ ݀ݔሾ ݃ሺݔሻ ሿ ሾ݃ሺݔሻሿଶ ൬݂ ݃൰ ᇱ ൌ݃∙݂ᇱെ݂∙݃ᇱ ݃ଶ ݀ ቀݑ ݒቁൌݒ ݀ݑെݑ ݀ݒ ݒଶ Chain Rule ݀ݕ ݀ݔൌ݀ݕ ݀ݑ∙݀ݑ ݀ݔ ݄ᇱሺݔሻൌ݂ᇱ൫݃ሺݔሻ൯ ∙ ݃′ሺݔሻ, where: ݄ൌ݂∘݃ ݀ݕൌ݀ݕ ݀ݑ ∙ ݀ݑ Power Rule ݀ ݀ݔሺݔሻൌ݊∙ݔିଵ ݀ ݀ݔሺݑሻൌ݊∙ݑିଵ݀ݑ ݀ݔ Exponential and Logarithmic Functions ሺܽ0, ്ܽ1ሻ ݀ ݀ݔ݁௫ൌ ݁௫ ݀ ݀ݔ݁௨ൌ݁௨∙݀ݑ ݀ݔ ݀ ݀ݔܽ௫ൌ ܽ௫ ∙ ln ܽ ݀ ݀ݔܽ௨ൌܽ௨∙ln ܽ∙݀ݑ ݀ݔ ݀ ݀ݔln ݔൌ 1 ݔ ݀ ݀ݔln ݑൌ1 ݑ∙݀ݑ ݀ݔ ݀ ݀ݔlogݔൌ 1 ݔ ln ܽ ݀ ݀ݔlogݑൌ 1 ݑln ܽ∙݀ݑ ݀ݔ Version 5.6 Page 18 of 242 April 8, 2023 Chapter 2 Differentiation Derivatives of Special Functions Trigonometric and Inverse Trigonometric Functions Trigonometric Functions ݀ ݀ݔsin ݔൌcos ݔ ݀ ݀ݔsin ݑൌcos ݑ∙݀ݑ ݀ݔ ݀ ݀ݔcos ݔൌെsin ݔ ݀ ݀ݔcos ݑൌെsin ݑ∙݀ݑ ݀ݔ ݀ ݀ݔtan ݔൌ secଶݔ ݀ ݀ݔtan ݑൌsecଶݑ∙݀ݑ ݀ݔ ݀ ݀ݔcot ݔൌെ cscଶݔ ݀ ݀ݔcot ݑൌെcscଶݑ∙݀ݑ ݀ݔ ݀ ݀ݔsec ݔൌ sec ݔ tan ݔ ݀ ݀ݔsec ݑൌsec ݑtan ݑ∙ ݀ݑ ݀ݔ ݀ ݀ݔcsc ݔൌെcsc ݔcot ݔ ݀ ݀ݔcsc ݑൌെcsc ݑcot ݑ∙ ݀ݑ ݀ݔ Inverse Trigonometric Functions (Basic Formulas) ݀ ݀ݔsinିଵݔൌ 1 √1 െݔଶ ݀ ݀ݔsinିଵݑൌ 1 √1 െݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q IV ݀ ݀ݔcosିଵݔൌ െ1 √1 െݔଶ ݀ ݀ݔcosିଵݑൌ െ1 √1 െݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q II ݀ ݀ݔtanିଵݔൌ 1 1 ݔଶ ݀ ݀ݔtanିଵݑൌ 1 1 ݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q IV ݀ ݀ݔcotିଵݔൌ െ1 1 ݔଶ ݀ ݀ݔcotିଵݑൌ െ1 1 ݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q II ݀ ݀ݔsecିଵݔൌ 1 |ݔ| √ݔଶെ1 ݀ ݀ݔsecିଵݑൌ 1 |ݑ| √ݑଶെ1 ∙݀ݑ ݀ݔ Angle in Q I or Q II ݀ ݀ݔcscିଵݔൌ െ1 |ݔ| √ݔଶെ1 ݀ ݀ݔcscିଵݑൌ െ1 |ݑ| √ݑଶെ1 ∙݀ݑ ݀ݔ Angle in Q I or Q IV Version 5.6 Page 19 of 242 April 8, 2023 Chapter 2 Differentiation Development of Basic Inverse Trig Derivatives Inverse Sine If ݕൌsinିଵݔ, then ݔൌsin ݕ. Take the derivative of both sides of this equation, and consider the result in conjunction with the triangle at right. sin ݕൌݔ cos ݕ ݀ݕ ݀ݔൌ1 ݀ݕ ݀ݔൌ 1 cos ݕൌ 1 √1 െݔଶ Inverse Tangent If ݕൌtanିଵݔ, then ݔൌtan ݕ. Take the derivative of both sides of this equation, and consider the result in conjunction with the triangle at right. tan ݕൌݔ secଶݕ ݀ݕ ݀ݔൌ1 ݀ݕ ݀ݔൌ 1 secଶݕൌcosଶݕൌ൬ 1 √1 ݔଶ൰ ଶ ൌ 1 1 ݔଶ Inverse Secant If ݕൌsecିଵݔ, then ݔൌsec ݕ. Take the derivative of both sides of this equation, and consider the result in conjunction with the triangle at right. sec ݕൌݔ sec ݕtan ݕ ݀ݕ ݀ݔൌ1 ݀ݕ ݀ݔൌ 1 sec ݕtan ݕൌcosଶݕ sin ݕൌ൬1 ݔ൰ ଶ ൊ√ݔଶെ1 |ݔ| ൌ |ݔ| ݔଶ√ݔଶെ1 ൌ 1 |ݔ|√ݔଶെ1 Note the use of the absolute value sign in this derivative. This occurs because the ݏ݁ܿିଵ function is defined only in quadrants 1 and 2, and the sine function is always positive in these two quadrants. The student may verify that the slope of the ݏ݁ܿିଵ function is always positive. Version 5.6 Page 20 of 242 April 8, 2023 Chapter 2 Differentiation Graphs of Inverse Trig Functions Graphs of the Inverse Trigonometric (IT) Functions over their principal ranges are provided below. Asymptotes are shown as dotted lines. Notice the following about these graphs: The graphs of sinିଵݔ, tanିଵݔ, secିଵݔ have positive slopes over their entire domains. So, their derivatives are always positive. The graphs of cosିଵݔ, cotିଵݔ, cscିଵݔ have negative slopes over their entire domains. So, their derivatives are always negative. Each IT function has a principal range of length ߨ radians, i.e., two quadrants. In one of these quadrants, the corresponding trigonometric function value is negative, and in the other it is positive. For example, cosିଵݔ has range ሾ0, ߨሿ, Quadrants I and II. In Quadrant I, cos ݔ is positive and in Quadrant II, cos ݔ is negative. At each ݔ‐value, cofunction pairs (e.g., sinିଵݔ and cosିଵݔ) have slopes with opposite values, i.e., the same absolute value but one slope is positive while the other slope is negative. Cofunction pairs (e.g., sinିଵݔ and cosିଵݔ) are reflections of each other over the horizontal line that contains their intersection. There is not universal agreement on the principal range of cotିଵݔ. Some sources, including the TI nSpire and a number of Calculus textbooks, set the range to ሺ0, ߨሻ, as shown on this page. Others, including Wolfram MathWorld and the US National Institute of Standards and Technology, set the range to ቀെ గ ଶ, గ ଶቃ. Version 5.6 Page 21 of 242 April 8, 2023 Chapter 2 Differentiation Generalized Inverse Trig Derivatives Derivatives Note that “ܽ” is defined to be positive in these formulas in order to meet the domain restrictions of inverse Trigonometric functions. ݀ ݀ݔsinିଵቀݔ ܽቁൌ 1 √ܽଶെݔଶ ݀ ݀ݔsinିଵቀݑ ܽቁൌ 1 √ܽଶെݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q IV ݀ ݀ݔcosିଵቀݔ ܽቁൌ െ1 √ܽଶെݔଶ ݀ ݀ݔcosିଵቀݑ ܽቁൌ െ1 √ܽଶെݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q II ݀ ݀ݔtanିଵቀݔ ܽቁൌ ܽ ܽଶݔଶ ݀ ݀ݔtanିଵቀݑ ܽቁൌ ܽ ܽଶݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q IV ݀ ݀ݔcotିଵቀݔ ܽቁൌ െܽ ܽଶݔଶ ݀ ݀ݔcotିଵቀݑ ܽቁൌ െܽ ܽଶݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q II ݀ ݀ݔsecିଵቀݔ ܽቁൌ ܽ |ݔ| √ݔଶെܽଶ ݀ ݀ݔsecିଵቀݑ ܽቁൌ ܽ |ݑ| √ݑଶെܽଶ∙݀ݑ ݀ݔ Angle in Q I or Q II ݀ ݀ݔcscିଵቀݔ ܽቁൌ െܽ |ݔ| √ݔଶെܽଶ ݀ ݀ݔcscିଵቀݑ ܽቁൌ െܽ |ݑ| √ݑଶെܽଶ∙݀ݑ ݀ݔ Angle in Q I or Q IV Sample Developments of Generalized Formulas from Basic Formulas ݀ ݀ݔsinିଵቀݔ ܽቁ ൌ 1 ට1 െቀݔ ܽቁ ଶ ∙ ൬1 ܽ൰ ൌ 1 ට1 െݔଶ ܽଶ ∙ √ܽଶ ൌ 1 √ܽଶെݔଶ ݀ ݀ݔtanିଵቀݔ ܽቁ ൌ 1 1 ቀݔ ܽቁ ଶ ∙ ൬1 ܽ൰ ൌ 1 ܽݔଶ ܽ ൌ 1 ܽଶ ܽݔଶ ܽ ൌ 1 ܽଶݔଶ ܽ ൌ ܽ ܽଶݔଶ ݀ ݀ݔsecିଵቀݔ ܽቁ ൌ 1 ቚݔ ܽቚ ටቀݔ ܽቁ ଶ െ1 ∙ ൬1 ܽ൰ ൌ 1 |ݔ| ܽ ටݔଶ ܽଶെܽଶ ܽଶ ∙√ܽଶ ൌ ܽ |ݔ| √ݔଶെܽଶ Version 5.6 Page 22 of 242 April 8, 2023 Chapter 2 Differentiation Generalized Product Rule Product Rule (three terms) ݀ ݀ݔ ሺݑݒݓሻൌ ݀ ݀ݔ ሾݑሺݒݓሻሿൌݑ ݀ ݀ݔሺݒݓሻݒݓ ݀ݑ ݀ݔ ൌݑ ∙ ൬ݒ ݀ݓ ݀ݔݓ ݀ݒ ݀ݔ൰ݒݓ ݀ݑ ݀ݔ ࢊ ࢊ࢞ ሺ࢛࢜࢝ሻൌ࢛࢜ ࢊ࢝ ࢊ࢛࢞࢝ ࢊ࢜ ࢊ࢞࢜࢝ ࢊ࢛ ࢊ࢞ Product Rule (four terms) ݀ ݀ݔ ሺݑݒݓݐሻൌ ݀ ݀ݔ ሾሺݑݒሻሺݓݐሻሿൌݑݒ ݀ ݀ݔሺݓݐሻݓݐ ݀ ݀ݔሺݑݒሻ ൌݑݒ ∙ ൬ݓ ݀ݐ ݀ݔݐ ݀ݓ ݀ݔ൰ݓݐ ∙ ൬ݑ ݀ݒ ݀ݔݒ ݀ݑ ݀ݔ൰ ࢊ ࢊ࢞ ሺ࢛࢚࢜࢝ሻൌ࢛࢜࢝ ࢊ࢚ ࢊ࢛࢚࢞࢜ ࢊ࢝ ࢊ࢛࢚࢞࢝ ࢊ࢜ ࢊ࢚࢞࢜࢝ ࢊ࢛ ࢊ࢞ Generalized Product Rule (n terms) ࢊ ࢊ࢞ ൭ෑ࢛ ୀ ൱ൌ ቌෑ࢛ ஷ ቍ ∙ ࢊ ࢊ࢞ ൫࢛ ൯ ୀ Example 2.1: Product Rule (six terms) – from Generalized Product Rule ࢊ ࢊ࢞ ሺ࢛࢙࢚࢜࢝࢘ሻൌ࢛࢙࢜࢝࢘ ࢊ࢚ ࢊ࢛࢚࢞࢜࢝࢘ ࢊ࢙ ࢊ࢛࢙࢚࢞࢜࢝ ࢊ࢘ ࢊ࢛࢙࢚࢞࢜࢘ ࢊ࢝ ࢊ࢛࢙࢚࢞࢝࢘ ࢊ࢜ ࢊ࢙࢚࢞࢜࢝࢘ ࢊ࢛ ࢊ࢞ In words: 1. Take the derivative of each function in the product. 2. Multiply it by all of the other functions in the product. 3. Add all of the resulting terms. Version 5.6 Page 23 of 242 April 8, 2023 Chapter 2 Differentiation Generalized Product Rule Example Generalized Product Rule (n terms) ࢊ ࢊ࢞ ൭ෑ࢛ ୀ ൱ൌ ቌෑ࢛ ஷ ቍ ∙ ࢊ ࢊ࢞ ൫࢛ ൯ ୀ Example 2.2: Find the derivative of: ࢌሺ࢞ሻൌሺ࢞࢞ሻ∙ࢋ࢞∙ܛܑܖ࢞∙܋ܗܛ࢞ Let: ࢛ൌሺ࢞࢞ሻ ࢜ൌࢋ࢞ ࢝ൌܛܑܖ࢞ ࢚ൌ܋ܗܛ࢞ Then, build the derivative based on the four components of the function: Original Function Term Derivative of Original Function Term Remaining Functions ࢛ൌሺ࢞࢞ሻ ૡ࢞ ࢋ࢞∙ܛܑܖ࢞∙܋ܗܛ࢞ ࢜ൌࢋ࢞ ࢋ࢞ ሺ࢞࢞ሻ∙ܛܑܖ࢞∙܋ܗܛ࢞ ࢝ൌܛܑܖ࢞ ܋ܗܛ࢞ ሺ࢞࢞ሻ∙ࢋ࢞∙܋ܗܛ࢞ ࢚ൌ܋ܗܛ࢞ െ࢞ܛܑܖ࢞ ሺ࢞࢞ሻ∙ࢋ࢞∙ܛܑܖ࢞ The resulting derivative is: ࢌ′ሺ࢞ሻൌሺૡ࢞ሻሺࢋ࢞∙ܛܑܖ࢞∙܋ܗܛ࢞ሻሺ࢞࢞ሻ∙ࢋ࢞∙ܛܑܖ࢞∙܋ܗܛ࢞+ ∙ሺ࢞࢞ሻ∙ࢋ࢞∙܋ܗܛ࢞∙܋ܗܛ࢞െሺૡ࢞࢞ሻ∙ࢋ࢞∙ܛܑܖ࢞∙ܛܑܖ࢞ In words: 1. Take the derivative of each function in the product. 2. Multiply it by all of the other functions in the product. 3. Add all of the resulting terms. Version 5.6 Page 24 of 242 April 8, 2023 Chapter 2 Differentiation Inverse Function Rule The Inverse Function Rule states the following: If ݂ሺݔሻ and ݃ሺݔሻ are inverse functions and ݂ᇱ൫݃ሺݔሻ൯്0, then ݃ᇱሺݔሻൌ ଵ ᇲ൫ሺ௫ሻ൯ To understand what this means, it may be best to look at what it says graphically and create an Inverse Function Diagram. Example 2.3: Let ݂ሺݔሻൌݔଶ3. Find the slope of ݃ሺݔሻൌ݂ିଵሺݔሻ at the point ሺ7, 2ሻ. To solve this, let’s look at the graph of ࢌሺ࢞ሻൌ࢞ and its inverse ࢍሺ࢞ሻൌേ√࢞െ. The figure at right shows these two plots, along with the axis of reflection and the lines tangent to the two curves at the desired points. Notice the following: ࢍሺૠሻൌ, so ࢌሺሻൌૠ ࢌᇱሺ࢞ሻൌ࢞, so ࢌᇱሺሻൌ ࢍᇱሺૠሻൌ ࢌᇲሺሻൌ (the answer) An Inverse Function Diagram (IFD) organizes this information as follows: IFD for Example 2.3 General IFD ࢌሺሻൌૠ ⇔ ࢍሺૠሻൌ ࢌሺ࢞ሻൌ࢟ ⇔ ࢍሺ࢟ሻൌ࢞ ࢌᇱሺሻൌ ࢟ࢋࢊ࢙ ሱ ۛ ۛ ۛ ሮ ࢍᇱሺૠሻൌ ࢌᇱሺ࢞ሻൌ ࢟ࢋࢊ࢙ ሱ ۛ ۛ ۛ ሮ ࢍᇱሺ࢟ሻൌ Version 5.6 Page 25 of 242 April 8, 2023 Chapter 2 Differentiation Partial Differentiation Partial differentiation is differentiation with respect to a single variable, with all other variables being treated as constants. Example 2.4: Consider the function ݂ሺݔ, ݕሻൌݔݕ2ݔ3ݕ. Notice in the partial derivative panels above, that the “off‐variable” is treated as a constant. In the left‐hand panel, the derivative is taken in its normal manner, including using the product rule on the ݔݕ‐term. In the middle panel, which takes the partial derivative with respect to ݔ, ݕ is considered to be the coefficient of ݔ in the ݔݕ‐term. In the same panel, the 3ݕ term is considered to be a constant, so its partial derivative with respect to ݔ is 0. In the right‐hand panel, which takes the partial derivative with respect to ݕ, ݔ is considered to be the coefficient of ݕ in the ݔݕ‐term. In the same panel, the 2ݔ term is considered to be a constant, so its partial derivative with respect to ݕ is 0. Partial derivatives provide measures of rates of change in the direction of the variable. So, for a 3‐dimensional curve, ߲ݖ ߲ݔ provides the rate of change in the ݔ‐direction and ߲ݖ ߲ݕ provides the rate of change in the ݕ‐direction. Partial derivatives are especially useful in physics and engineering. Example 2.5: Let ࢝ൌ࢞ࢋ࢟ܔܖࢠࢋ࢞ܛܑܖሺ࢟ࢠሻെ܋ܗܛሺ࢞࢟ࢠሻ. Then, ࣔ࢝ ࣔ࢞ൌ࢞ࢋ࢟ܔܖࢠࢋ࢞ܛܑܖሺ࢟ࢠሻܡܢ ܛܑܖሺ࢞࢟ࢠሻ ࣔ࢝ ࣔ࢟ൌ࢞ࢋ࢟ܔܖࢠࢋ࢞܋ܗܛሺ࢟ࢠሻ࢞ࢠ ܛܑܖሺ࢞࢟ࢠሻ ࣔ࢝ ࣔࢠൌ࢞ࢋ࢟ ࢠ ࢋ࢞܋ܗܛሺ࢟ࢠሻ࢞࢟ܛܑܖሺ࢞࢟ࢠሻ Full derivative: ݀ ݀ݔሺݔݕ2ݔ3ݕሻൌ ݔ݀ݕ ݀ݔݕ2 3 ݀ݕ ݀ݔ Partial derivative: ߲ ߲ݔሺݔݕ2ݔ3ݕሻൌ ݕ2 Partial derivative: ߲ ߲ݕሺݔݕ2ݔ3ݕሻൌ ݔ3 Version 5.6 Page 26 of 242 April 8, 2023 Chapter 2 Differentiation Implicit Differentiation Implicit differentiation is typically used when it is too difficult to differentiate a function directly. The entire expression is differentiated with respect to one of the variables in the expression, and algebra is used to simplify the expression for the desired derivative. Example 2.6: Find ௗ௬ ௗ௫ for the ellipse ௫మ ସ ௬మ ଽ ൌ36. We could begin by manipulating the equation to obtain a value for ݕ: ࢟ൌേටെ ࢞ . However, this is a fairly ugly expression for ݕ, and the process of developing ௗ௬ ௗ௫ is also ugly. It is many times easier to differentiate implicitly as follows: 1. Start with the given equation: ௫మ ସ ௬మ ଽ ൌ36 2. Multiply both sides by 36 to get rid of the denominators: 9ݔଶ4ݕଶൌ1296 3. Differentiate with respect to ݔ: 18ݔ8ݕ∙ݕᇱൌ0 4. Subtract 18ݔ: 8ݕ∙ݕᇱൌെ18ݔ 5. Divide by 8ݕ: ݕᇱൌ ିଵ଼௫ ଼௬ൌെ ଽ௫ ସ௬ 6. Sometimes you will want to substitute in the value of ݕ to get the expression solely in terms of ݔ: The result is still ugly and, in fact, it must be ugly. However, the algebra required to get the result may be cleaner and easier using implicit differentiation. In some cases, it is either extremely difficult or impossible to develop an expression for ݕ in terms of ݔ because the variables are so intertwined; see the example on the next page. ݕᇱൌ െ ଽ௫ ସ௬ൌ െ ଽ௫ ቆേටെ ࢞ ቇ ࢟ᇱൌ േ ࢞ ටെ ࢞ (ݔ്േ12) Version 5.6 Page 27 of 242 April 8, 2023 Chapter 2 Differentiation Implicit Differentiation (cont’d) Example 2.7: Find ௗ௬ ௗ௫ for the equation: ݔ∙sin ݕݕ∙cos ݔൌ0. Manipulating this equation to find ݕ as a function of ݔ is out of the question. So, we use implicit differentiation as follows: 1. Start with the given equation: ࢞∙ܛܑܖ࢟࢟∙܋ܗܛ࢞ൌ0 2. Differentiate with respect to ݔ using the product rule and the chain rule: ࢞∙ ࢊ ࢊ࢞ሺܛܑܖ࢟ሻܛܑܖ࢟∙ ࢊ ࢊ࢞ሺ࢞ሻ࢟∙ ࢊ ࢊ࢞ሺ܋ܗܛ࢞ሻ܋ܗܛ࢞∙ ࢊ ࢊ࢞ሺ࢟ሻൌ0 3. Simplify: ࢞∙ሺ܋ܗܛ࢟ሻ∙ ࢊ࢟ ࢊ࢞ܛܑܖ࢟࢟∙ሺെܛܑܖ࢞ሻ܋ܗܛ࢞∙ ࢊ࢟ ࢊ࢞ൌ0 4. Combine like terms and simplify: ࢞∙ሺ܋ܗܛ࢟ሻ∙ ࢊ࢟ ࢊ࢞܋ܗܛ࢞∙ ࢊ࢟ ࢊ࢞ܛܑܖ࢟࢟∙ሺെܛܑܖ࢞ሻൌ0 ሾݔ∙ሺcos ݕሻcos ݔሿ∙ ௗ௬ ௗ௫ ሾsin ݕെݕ∙ሺsin ݔሻሿൌ0 ሾݔ∙ሺcos ݕሻcos ݔሿ∙ ௗ௬ ௗ௫ ൌ ሾݕ∙ሺsin ݔሻെsin ݕሿ ࢊ࢟ ࢊ࢞ ൌ ࢟ ∙ ሺܛܑܖ࢞ሻ ି ܛܑܖ࢟ ࢞ ∙ ሺ܋ܗܛ࢟ሻ ା ܋ܗܛ࢞ (as long as: ݔ∙ሺcos ݕሻcos ݔ്0ሻ That’s as good as we can do. Notice that the derivative is a function of both ݔ and ݕ. Even though we cannot develop an expression for ݕ as a function of ݔ, we can still calculate a derivative of the function in terms of ݔ and ݕ. Viva implicit differentiation! Version 5.6 Page 28 of 242 April 8, 2023 Chapter 2 Differentiation Implicit Differentiation (cont’d) Implicit Differentiation Using Partial Derivatives Let ݖൌ݂ሺݔ, ݕሻ. Then, the following formula is often a shortcut to calculating ௗ௬ ௗ௫. ࢊ࢟ ࢊ࢞ ൌ െ ࣔࢠ ࣔ࢞ ࣔࢠ ࣔ࢟ Let’s re‐do the examples from the previous pages using the partial derivative method. Example 2.8: Find ௗ௬ ௗ௫ for the ellipse ௫మ ସ ௬మ ଽ ൌ36. Let: ࢠൌ࢞ ࢟ ૢ െ. Then, ࣔࢠ ࣔ࢞ൌ࢞ ࣔࢠ ࣔ࢟ൌ࢟ ૢ ࢊ࢟ ࢊ࢞ ൌ െ ࣔࢠ ࣔ࢞ ࣔࢠ ࣔ࢟ ൌ െ ࢞ ࢟ ૢ ൌ െૢ࢞ ࢟ Example 2.9: Find ௗ௬ ௗ௫ for the equation: ݔ∙sin ݕݕ∙cos ݔൌ0. Let: ݖൌ࢞ܛܑܖ࢟࢟܋ܗܛ࢞. Then, ࣔࢠ ࣔ࢞ൌܛܑܖ࢟െ࢟ܛܑܖ࢞ ࣔࢠ ࣔ࢟ൌ࢞܋ܗܛ࢟܋ܗܛ࢞ ࢊ࢟ ࢊ࢞ ൌ െ ࣔࢠ ࣔ࢞ ࣔࢠ ࣔ࢟ ൌ െ ܛܑܖ࢟െ࢟ܛܑܖ࢞ ࢞܋ܗܛ࢟܋ܗܛ࢞ ൌ ࢟ܛܑܖ࢞െܛܑܖ࢟ ࢞܋ܗܛ࢟܋ܗܛ࢞ Contrast the work required here with the lengthy efforts required to calculate these results on the two prior pages. So, implicit differentiation using partial derivatives can be fast and, because fewer steps are involved, improve accuracy. Just be careful how you handle each variable. This method is different and takes some getting used to. Version 5.6 Page 29 of 242 April 8, 2023 Chapter 2 Differentiation Logarithmic Differentiation Logarithmic differentiation is typically used when functions exist in both the base and the exponent of an exponential expression. Without this approach, the differentiation of the function would be much more difficult. The process involves several steps, as follows: 1. If possible, put the function in the form: ݕൌ݂ሺݔሻ 2. Take natural logarithms of both sides of the expression. 3. Take the derivatives of both sides of the expression. 4. Solve for ௗ௬ ௗ௫. Example 2.10: Calculate the derivative of the general case ࢟ൌ࢛࢜, where ݑ and ݒ are functions of ݔ, and are differentiable at ݔ. 1. Original equation ࢟ൌ࢛࢜ 2. Take natural logarithms of both sides ࢟ൌ࢛࢜ 3. Simplify right side ࢟ൌ࢜∙࢛ 4. Take derivatives of both sides ࢟∙ࢊ࢟ ࢊ࢞ൌ࢜∙ࢊ ࢊ࢞ሺ࢛ሻሺ࢛ሻࢊ࢜ ࢊ࢞ 5. Apply Product Rule and Chain Rule to right side ࢟∙ࢊ࢟ ࢊ࢞ൌ࢜∙ ࢛∙ࢊ࢛ ࢊ࢞ሺ࢛ሻࢊ࢜ ࢊ࢞൨ 6. Multiply both sides by ݕ ࢊ࢟ ࢊ࢞ൌ࢟∙࢜ ࢛∙ࢊ࢛ ࢊ࢞ሺ࢛ሻࢊ࢜ ࢊ࢞൨ 7. Substitute value of ݕ ࢊ࢟ ࢊ࢞ൌ࢛࢜∙࢜ ࢛∙ࢊ࢛ ࢊ࢞ሺܔܖ࢛ሻࢊ࢜ ࢊ࢞൨ 8. Simplify ࢊ ࢊ࢛࢞࢜ൌ࢛࢜࢜ି∙ࢊ࢛ ࢊ࢛࢞࢜ሺܔܖ࢛ሻࢊ࢜ ࢊ࢞ Version 5.6 Page 30 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Maxima and Minima Relative Extrema Relative maxima and minima (also called relative extrema) may exist wherever the derivative of a function is either equal to zero or undefined. However, these conditions are not sufficient to establish that an extreme exists; we must also have a change in the direction of the curve, i.e., from increasing to decreasing or from decreasing to increasing. Note: relative extrema cannot exist at the endpoints of a closed interval. First Derivative Test If a function, ݂, is continuous on the open interval ሺܽ, ܾሻ, and ܿ is a critical number ∈ሺܽ, ܾሻ (i.e., ݂ᇱሺܿሻ is either zero or does not exist), ݂ is differentiable on the open interval ሺܽ, ܾሻ, except possibly at c, Then If ݂ᇱሺݔሻ changes from positive to negative at ܿ, then ݂ሺܿሻ is a relative maximum. If ݂ᇱሺݔሻ changes from negative to positive at ܿ, then ݂ሺܿሻ is a relative minimum. The conclusions of this theorem are summarized in the table below: First Derivative Sign of ࢊ࢟ ࢊ࢞ left of ࢞ൌࢉ Sign of ࢊ࢟ ࢊ࢞ right of ࢞ൌࢉ Type of Extreme Case 1 ௗ௬ ௗ௫ ൌ0 or ௗ௬ ௗ௫ does not exist. െ െ None Case 2 െ Minimum Case 3 None Case 4 െ Maximum Illustration of First Derivative Test for Cases 1 to 4: Version 5.6 Page 31 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Second Derivative Test If a function, 𝑓, is continuous on the open interval ሺ𝑎, 𝑏ሻ, and 𝑐∈ሺ𝑎, 𝑏ሻ, and 𝑓ᇱሺ𝑐ሻൌ0 and 𝑓′ᇱሺ𝑐ሻ exists, Then If 𝑓ᇱᇱሺ𝑐ሻ൏0, then 𝑓ሺ𝑐ሻ is a relative maximum. If 𝑓ᇱᇱሺ𝑐ሻ0, then 𝑓ሺ𝑐ሻ is a relative minimum. The conclusions of the theorem are summarized in the table below: First Derivative Second Derivative Type of Extreme Case 1 ௗ௬ ௗ௫ ൌ0 or ௗ௬ ௗ௫ does not exist. ௗమ௬ ௗ௫మ൏0 Maximum Case 2 ௗమ௬ ௗ௫మ0 Minimum Case 3 ௗమ௬ ௗ௫మൌ0 or does not exist Test Fails Absolute Extrema Absolute extrema (also called “global extrema” or simply “extrema”) are the highest or lowest values of the function on the interval in question. If a function is continuous, its absolute extrema exist at the locations of either its relative extrema or the endpoints of the interval. Note that if an interval is open, the endpoint does not exist and so it cannot be an absolute extreme. This means that in some cases, a function will not have an absolute maximum or minimum on the interval in question. Discontinuities in a function can also cause a function to not have a relative maximum or minimum. A function may have 0, 1 or multiple absolute maxima and/or absolute minima on an interval. In the illustration to the right, the function has: Two absolute minima, at ሺെ1, െ1ሻ and ሺ2, െ1ሻ. No absolute maximum (due to the discontinuity). One relative maximum, at ሺ0, 3ሻ. One relative minimum – The point located at ሺ2, െ1ሻ is both a relative minimum and an absolute minimum. In the event that the second derivative is zero or does not exist (Case 3), we cannot conclude whether or not an extreme exists. In this case, it may be a good idea to use the First Derivative Test at the point in question. Version 5.6 Page 32 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Inflection Points Definition An inflection point is a location on a curve where concavity changes from upward to downward or from downward to upward. At an inflection point, ݂′ᇱሺݔሻൌ0 or ݂′ᇱሺݔሻ does not exist. However, it is not necessarily true that if ݂′ᇱሺݔሻൌ0, then there is an inflection point at ݔൌܿ. Testing for an Inflection Point To find the inflection points of a curve in a specified interval, Determine all ݔ‐values (ݔൌܿ) for which ݂′ᇱሺܿሻൌ0 or ݂′ᇱሺܿሻ does not exist. Consider only ܿ‐values where the function has a tangent line. Test the sign of ݂′ᇱሺݔሻ to the left and to the right of ݔൌܿ. If the sign of ݂′ᇱሺݔሻ changes from positive to negative or from negative to positive at ݔൌܿ, then ሺܿ, ݂ሺܿሻሻ is an inflection point of the function. Note: inflection points cannot exist at the endpoints of a closed interval. Concavity A function, ݂, is concave upward on an interval if ݂’ሺݔሻ is increasing on the interval, i.e., if ݂ᇱᇱሺݔሻ0. A function, ݂, is concave downward on an interval if ݂’ሺܿሻ is decreasing on the interval, i.e., if ݂ᇱᇱሺݔሻ൏0. Concavity changes at inflection points, from upward to downward or from downward to upward. In the illustration at right, an inflection point exists at the point ሺ2, 3ሻ. Second Derivative Sign of ௗమ௬ ௗ௫మ left of ࢞ൌࢉ Sign of ௗమ௬ ௗ௫మ right of ࢞ൌࢉ Inflection Point? Case 1 ௗమ௬ ௗ௫మൌ0 or ௗమ௬ ௗ௫మ does not exist െ െ No Case 2 െ Yes Case 3 No Case 4 െ Yes Version 5.6 Page 33 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Special Case: Extrema and Inflection Points of Polynomials For a polynomial, ݂ሺݔሻ, critical values of exist at all ݔ‐values for which ݂ᇱሺݔሻൌ0. However, critical values do not necessarily produce extrema. Possible inflection points exist at all ݔ‐ values for which ݂ᇱᇱሺݔሻൌ0. However, not all of these ݔ‐values produce inflection points. To find the extrema and inflection points of a polynomial we can look at the factored forms of ݂ᇱሺݔሻ and ݂ᇱᇱሺݔሻ, respectively. Every polynomial can be factored into linear terms with real roots and quadratic terms with complex roots as follows: ܲሺݔሻൌ݇ሺݔെݎ ଵሻభ∙ሺݔെݎ ଶሻమ… ሺݔെݎ ሻ∙ܳଵሺݔሻ∙ܳଶሺݔሻ… ܳሺݔሻ where, k is a scalar (constant), each ݎ is a real root of ݂ሺݔሻ, each exponent ܽ is an integer, and each ܳ is a quadratic term with complex roots. Extrema The exponents (ܽ) of the linear factors of ݂ᇱሺݔሻ determine the existence of extrema. An odd exponent on a linear term of ݂ᇱሺݔሻ indicates that ݂ᇱሺݔሻ crosses the ݔ‐axis at the root of the term, so ݂ሺݔሻ has an extreme at that root. Further analysis is required to determine whether the extreme is a maximum or a minimum. An even exponent on a linear term of ݂ᇱሺݔሻ indicates that ݂ᇱሺݔሻ bounces off the ݔ‐axis at the root of the term, so ݂ሺݔሻ does not have an extreme at that root. Example 3.1: Consider ݂ᇱሺݔሻൌሺݔ3ሻଷሺݔ2ሻଶ൫ݔ√3൯ ଷ൫ݔെ√3൯ ଷ ሺݔെ4ሻଶሺݔെ7ሻ. The original polynomial, ݂ሺݔሻ, has critical values for each term: ܥܸൌሼെ3, െ2, െ√3, √3, 4, 7ሽ. However, extrema exist only for the terms with odd exponents: ܧݔݐݎ݁݉ܽൌሼെ3, െ√3, √3, 7ሽ. Inflection Points (PI) The exponents (ܽ) of the linear factors of ݂ᇱᇱሺݔሻ determine the existence of inflection points. An odd exponent on a linear term of ݂ᇱᇱሺݔሻ indicates that ݂ሺݔሻ has an inflection point at the root of that term. An even exponent on a linear term of ݂ᇱᇱሺݔሻ indicates that ݂ሺݔሻ does not have an inflection point at the root of that term. Example 3.2: Consider ݂ᇱᇱሺݔሻൌሺݔ3ሻଷሺݔ2ሻଶ൫ݔ√3൯ ଷ൫ݔെ√3൯ ଷ ሺݔെ4ሻଶሺݔെ7ሻ. Inflection points exist only for the terms with odd exponents: ܲܫൌሼെ3, െ√3, √3, 7ሽ. Version 5.6 Page 34 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Key Points on ሺܠሻ, ᇱሺܠሻ and ᇱᇱሺܠሻ – Alauria Diagram An Alauria Diagram shows a single curve as ݂ሺݔሻ, ݂ᇱሺݔሻ or ݂ᇱᇱሺݔሻ on a single page. The purpose of the diagram is to answer the question: If the given curve is ݂ሺݔሻ, ݂ᇱሺݔሻ or ݂ᇱᇱሺݔሻ, where are the key points on the graph. If the curve represents ࢌሺ࢞ሻ: ݂ሺݔሻ’s ݔ‐intercepts (green and one yellow) exist where the curve touches the x‐axis. Relative maxima and minima (yellow) exist at the tops and bottoms of humps. Inflection points (orange) exist where concavity changes from up to down or from down to up. If the curve represents ࢌ′ሺ࢞ሻ (1st derivative): ݂ሺݔሻ’s ݔ‐intercepts cannot be seen. Relative maxima and minima of ݂ሺݔሻ (yellow) exist where the curve crosses the ݔ‐axis. If the curve bounces off the ݔ‐axis, there is no extreme at that location. Inflection points of ݂ሺݔሻ (orange) exist at the tops and bottoms of humps. If the curve represents ࢌ′′ሺ࢞ሻ (2nd derivative): ݂ሺݔሻ’s ݔ‐intercepts cannot be seen. Relative maxima and minima of ݂ሺݔሻ cannot be seen. Inflection points of ݂ሺݔሻ (orange) exist where the curve crosses the ݔ‐axis. If the curve bounces off the ݔ‐axis, there is no inflection point at that location. Version 5.6 Page 35 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Key Points on ሺܠሻ, ᇱሺܠሻ and ᇱᇱሺܠሻ The graphs below show ݂ሺݔሻ, ݂ᇱሺݔሻ or ݂ᇱᇱሺݔሻ for the same 5th degree polynomial function. The dotted blue vertical line identifies one location of an extreme (there are four, but only one is illustrated) The dashed dark red vertical line identifies one location of a point of inflection (there are three, but only one is illustrated). ࢌሺ࢞ሻ ࢌᇱሺ࢞ሻ ࢌᇱᇱሺ࢞ሻ In a graph of ࢌሺ࢞ሻ: Relative extrema exist at the tops and bottom of humps. Inflection points exist at locations where concavity changes from up to down or from down to up. In a graph of ࢌ′ሺ࢞ሻ: Relative extrema of ݂ሺݔሻ exist where the curve crosses the ݔ‐axis. If the curve bounces off the ݔ‐axis, there is no extreme at that location. Inflection points of ݂ሺݔሻ exist at the tops and bottoms of humps. In a graph of ࢌ′′ሺ࢞ሻ: Relative extrema of ݂ሺݔሻ cannot be seen. Inflection points of ݂ሺݔሻ exist where the curve crosses the ݔ‐axis. If the curve bounces off the ݔ‐axis, there is no inflection point at that location. Version 5.6 Page 36 of 242 April 8, 2023 Chapter 3 Applications of Differentiation What Does the Graph of 𝒇′ሺ𝒙ሻ Tell Us about 𝒇ሺ𝒙ሻ? Short answer: a lot! Consider the graph of the derivative of 𝑓ሺ𝑥ሻ when 𝒇ᇱሺ𝒙ሻൌ𝒆𝒙𝟏𝟎 ⁄ ∙𝐜𝐨𝐬𝒙 on the interval ቂെ 𝟑 𝟐𝝅, 𝟓 𝟐𝝅ቃ. Increasing vs. Decreasing We can tell if 𝑓ሺ𝑥ሻ is increasing or decreasing based on whether 𝑓ᇱሺ𝑥ሻ is positive or negative. Critical values exist where 𝑓ᇱሺ𝑥ሻ is zero or does not exist. Relative maxima and minima exist at critical values if the graph of 𝑓ᇱሺ𝑥ሻ crosses the 𝑥-axis. See the graph and chart below. Note that critical values, relative maxima and relative minima do not exist at endpoints of an interval. Concavity We can tell if 𝑓ሺ𝑥ሻ is concave up or concave down based on whether 𝑓ᇱሺ𝑥ሻ is increasing or decreasing. Inflection Points exist at the extrema of 𝑓ᇱሺ𝑥ሻ, i.e. at the top and bottom of any humps on the graph of 𝑓ᇱሺ𝑥ሻ. See the graph and chart below. Note that inflection points do not exist at endpoints of an interval. Increasing vs. Decreasing 𝒇ሺ𝒙ሻ Increasing Decreasing 𝒇ᇱሺ𝒙ሻ Positive Negative Concavity 𝒇ሺ𝒙ሻ Concave up Concave down 𝒇ᇱሺ𝒙ሻ Increasing Decreasing 𝒇ᇱᇱሺ𝒙ሻ Positive Negative All items in a column occur simultaneously. All items in a column occur simultaneously. Version 5.6 Page 37 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Simultaneous Behavior of 𝒇ሺ𝒙ሻ, 𝒇′ሺ𝒙ሻ, and 𝒇′′ሺ𝒙ሻ A question often faced by Calculus students is: Given information about one of 𝑓ሺ𝑥ሻ, 𝑓′ሺ𝑥ሻ, or 𝑓′′ሺ𝑥ሻ on a specific interval, what can be said about the behavior of the others on the same interval? For this purpose, we can use the Natalie Chart shown below: Using the Natalie Chart The Natalie Chart extends the information from the previous page into an expandible format. Information relating to the simultaneous behavior of 𝑓ሺ𝑥ሻ, 𝑓′ሺ𝑥ሻ, and 𝑓′′ሺ𝑥ሻ is provided in a single column in the chart. For example: If we are told that 𝑓′ሺ𝑥ሻ is increasing on a given interval, the first magenta column in the chart tells us that 𝑓ሺ𝑥ሻ is concave up and 𝑓′′ሺ𝑥ሻ is positive on the same interval. If we are told that 𝑓ᇱሺ𝑥ሻ൏0 on a given interval, the second blue column in the chart tells us that 𝑓ሺ𝑥ሻ is decreasing. We cannot determine any information about the behavior of 𝑓′′ሺ𝑥ሻ in this case, so those cells in the table are blank. Expanding the Natalie Chart Note that the information in the Natalie Chart is expandible to any set of three consecutive derivatives of a function by adding rows and columns. In this expanded chart, notice that knowing whether 𝑓ᇱᇱሺ𝑥ሻ is increasing or decreasing on an interval (green text) provides information about the simultaneous behavior of 𝑓ᇱሺ𝑥ሻ and 𝑓ᇱᇱᇱሺ𝑥ሻ on the same interval. Adding additional rows and columns can provide information about the simultaneous behavior of any three consecutive derivatives of any given function. Natalie Chart 𝒇ሺ𝒙ሻ Increasing Decreasing Concave Up Concave Down 𝒇ᇱሺ𝒙ሻ െ Increasing Decreasing 𝒇ᇱᇱሺ𝒙ሻ െ Expanded Natalie Chart 𝒇ሺ𝒙ሻ Increasing Decreasing Concave Up Concave Down 𝒇ᇱሺ𝒙ሻ െ Increasing Decreasing Concave Up Concave Down 𝒇ᇱᇱሺ𝒙ሻ െ Increasing Decreasing 𝒇′′′ሺ𝒙ሻ െ Version 5.6 Page 38 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Curve Sketching Curve Sketching is much easier with the tools of Calculus. In particular, the calculation of derivatives allows the student to identify critical values (relative maxima and minima) and inflection points for a curve. A curve can then be broken into intervals for which the various characteristics (e.g., increasing or decreasing, concave up or down) can be determined. The acronym DIACIDE may help the student recall the things that should be considered in sketching curves. DIACIDE: Derivatives: generally, the student should develop the first and second derivatives of the curve, and evaluate those derivatives at each key value (e.g., critical points, inflection points) of ݔ. Intercepts: to the extent possible, the student should develop both ݔ‐ and ݕ‐intercepts for the curve. ݔ‐intercepts occur where ݂ሺݔሻൌ0. ݕ‐ intercepts occur at ݔൌ0. Asymptotes: vertical asymptotes should be identified so that the curve can be split into continuous sub‐segments. Vertical asymptotes occur at values of ݔ where the curve approaches െ∞ or ∞; ݂ᇱሺݔሻ does not exist at these values of ݔ. Horizontal asymptotes are covered below under the category “End Behavior.” Critical Values: relative maxima and minima are locations where the curve changes from increasing to decreasing or from decreasing to increasing. They occur at “critical” ݔ‐values, where ݂ᇱሺݔሻൌ0 or where ݂ᇱሺݔሻ does not exist. Concavity: concavity is determined by the value of the second derivative: ݂′ᇱሺݔሻ൏0 implies downward concavity ݂′ᇱሺݔሻ0 implies upward concavity Inflection Points: an inflection point is a location on the curve where concavity changes from upward to downward or from downward to upward. At an inflection point, ݂′ᇱሺݔሻൌ0 or where ݂′ᇱሺݔሻ does not exist. Domain: the domain of a function is the set of all x‐values for which a y‐value exists. If the domain of a function is other than “all real numbers,” care should be taken to graph only those values of the function included in the domain. End Behavior: end behavior is the behavior of a curve on the left and the right, i.e., as ݔ tends toward െ∞ and ∞. The curve may increase or decrease unbounded at its ends, or it may tend toward a horizontal asymptote. Version 5.6 Page 39 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Example 3.3: Sketch the graph of f (x) = x3 – 5x2 + 3x + 6. DIACIDE: Derivatives, Intercepts, Asymptotes, Critical Values, Concavity, Inflection Points, Domain, End Behavior Derivatives: ݂ሺݔሻൌݔଷ – 5ݔଶ3ݔ6 ݂ᇱሺݔሻൌ3ݔଶെ10ݔ3 ݂ᇱᇱሺݔሻൌ6ݔെ10 Intercepts: Use synthetic division to find: ݔൌ2, so: ݂ሺݔሻൌሺݔെ2ሻ∙ሺ ݔଶെ3ݔെ3ሻ Then, use the quadratic formula to find: ݔൌ ଷേ√ଶଵ ଶ ൌሼെ0.791, 3.791ሽ ݔ‐intercepts, then, are: ሼെ0.791, 2, 3.791ሽ ݕ‐intercepts: ݂ሺ0ሻൌ6 Asymptotes: None for a polynomial Critical Values: ݂ᇱሺݔሻൌ3ݔଶെ10ݔ3 ൌ0 at ݔൌቄ ଵ ଷ, 3ቅ Critical Points are: ሼሺ. 333, 6.481ሻ, ሺ3, െ3ሻሽ ݂ᇱᇱሺ. 333ሻ൏0, so ሺ. 333, 6.481ሻ is a relative maximum ݂ᇱᇱሺ3ሻ0, so ሺ3, െ3ሻ is a relative minimum Concavity: ݂ᇱᇱሺݔሻ൏0 for ݔ൏1.667 (concave downward) ݂ᇱᇱሺݔሻ0 for ݔ1.667 (concave upward) Inflection Points: ݂ᇱᇱሺݔሻൌ6ݔെ10 ൌ0 at ݔൌ ହ ଷ ~ 1.667 Inflection Point is: ሺ1.667, 1.741ሻ Domain: All real values of ݔ for a polynomial End Behavior: Positive lead coefficient on a cubic equation implies that: lim ௫ →ିஶ݂ሺݔሻൌെ∞, and lim ௫ →ஶ݂ሺݔሻൌ∞ Note the two C’s.
Version 5.6 Page 40 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Example 3.4: Sketch the graph of 𝑓ሺ𝑥ሻൌ ଵ ∙ ୱ୧୬௫ ೣ DIACIDE: Derivatives: 𝑓ሺ𝑥ሻൌ ଵ ∙ ୱ୧୬௫ ೣ 𝑓ᇱሺ𝑥ሻൌ ଵ ∙ ሺୡ୭ୱ௫ ି ୱ୧୬௫ሻ ೣ 𝑓ᇱᇱሺ𝑥ሻൌ ିଶ ∙ ୡ୭ୱ௫ ೣ Intercepts: 𝑥‐intercept where sin 𝑥ൌ0, so, 𝑥ൌ𝑘∙𝜋, with 𝑘 being any integer 𝑦‐intercept at 𝑓ሺ0ሻൌ0 Asymptotes: No vertical asymptotes. Horizontal asymptote at 𝑦ൌ0. Critical Values: 𝑓ᇱሺ𝑥ሻൌ0 where cos 𝑥ൌsin 𝑥. Critical Points exist at 𝑥ൌቄ గ ସ𝑘∙𝜋, 𝑘∈𝑍ቅ ሺ. 707, 3.224ሻ is a relative maximum; ሺ3.927, െ0.139ሻ is a relative minimum There are an infinite number of relative maxima and minima, alternating at 𝑥‐ values that are 𝜋 apart. Concavity: The function is concave up where cos 𝑥൏0, i.e., Quadrants II and III and is concave down where cos 𝑥0, i.e., Quadrants I and IV. Inflection Points: 𝑓ᇱᇱሺ𝑥ሻൌ0 where cos 𝑥ൌ0 Inflection Points exist at: 𝑥ൌቄ గ ଶ𝑘∙𝜋, 𝑘∈𝑍ቅ Domain: All real values of 𝑥 End Behavior: lim ௫ →ିஶ𝑓ሺ𝑥ሻ does not exist, as the function oscillates up and down with each period lim ௫ →ஶ𝑓ሺ𝑥ሻൌ0 Version 5.6 Page 41 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Example 3.5: Sketch the graph of ݂ሺݔሻൌ ௫మିସ ௫మିଽ DIACIDE: Derivatives: ݂ሺݔሻൌ ௫మିସ ௫మିଽ ݂ᇱሺݔሻൌ ିଵ௫ ሺ௫మିଽሻమ ݂ᇱᇱሺݔሻൌ ଷ ∙ ሺ௫మାଷሻ ሺ௫మିଽሻయ Intercepts: ݔ‐intercept where ݔଶെ4 ൌ0, so, ࢞ൌേ ݕ‐intercept at ࢌሺሻൌ ૢ Asymptotes: Vertical asymptotes where: ݔଶെ9 ൌ0, so ࢞ൌേ. Horizontal asymptote at: ݕ ൌ lim ௫ →ஶ ݔ2െ4 ݔ2െ9 ൌ lim ௫ →ିஶ ݔ2െ4 ݔ2െ9 ൌ 1 Critical Values: ݂ᇱሺݔሻൌ0 where ݔൌ0; so ݂ᇱሺ0ሻൌ0 Since ݂ᇱᇱሺ0ሻൌെ ଵ ଼ଵ൏0, ቀ, ૢቁ is a relative maximum Concavity: The concavity of the various intervals are shown in the table on the next page Inflection Points: ݂ᇱᇱሺݔሻൌ0 where ݔଶ3 ൌ0 Therefore, there are no real inflection points Domain: All real values of ݔ, except at the vertical asymptotes So, the domain is: All Real ݔ്ሼെ3, 3ሽ End Behavior: lim ௫ →ିஶ ݔ2െ4 ݔ2െ9 ൌ 1 lim ௫ →ஶ ݔ2െ4 ݔ2െ9 ൌ 1 These imply the existence of a horizontal asymptote at ݕൌ1. Plot these intercepts on the graph. Plot the asymptotes on the graph. Plot the critical values on the graph. If there are inflection points, plot them on the graph. Version 5.6 Page 42 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Example 3.5 (cont’d) In some cases, it is useful to set up a table of intervals which are defined by the key values identified in green above: ࢞ൌሼെ, , ሽ. The key values are made up of: Vertical asymptotes Relative maxima and minima Inflection Points ࢞‐values ࢌሺ࢞ሻ ࢌᇱሺ࢞ሻ ࢌᇱᇱሺ࢞ሻ Graph Characteristics ሺെ∞, െ3ሻ curve increasing, concave up െ3 undefined undefined undefined vertical asymptote ሺെ3, 0ሻ െ curve increasing, concave down 0 . 444 0 െ relative maximum ሺ0, 3ሻ െ െ curve decreasing, concave down 3 undefined undefined undefined vertical asymptote ሺ3, ∞ሻ െ curve decreasing, concave up Version 5.6 Page 43 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Determining the Shape of a Curve Based On Its Derivatives The possible shapes of a curve, based on its first and second derivatives are: Increasing function Decreasing function Increasing function Decreasing function ݂ᇱሺݔሻ0 ݂ᇱሺݔሻ൏0 ݂ᇱሺݔሻ0 ݂ᇱሺݔሻ൏0 Concave up Concave up Concave down Concave down ݂ᇱᇱሺݔሻ0 ݂ᇱᇱሺݔሻ0 ݂ᇱᇱሺݔሻ൏0 ݂ᇱᇱሺݔሻ൏0 So, given a differentiable function with first and second derivatives identified, we need only match the shapes above to the intervals of the function and then join them together. If we are given points on the curve, we must also fit the shape through the given points. Example 3.6: Suppose we want to determine the approximate shape of the curve of the differentiable function defined by the following table. ݔ 1 ൏ݔ൏3 ݔൌ3 3 ൏ݔ൏5 ݔൌ5 5 ൏ݔ൏7 '( ) f x Positive 0 Negative Negative Negative "( ) f x Negative Negative Negative 0 Positive Curve Shape Flat – Relative Maximum Point of Inflection To get the shape of the function over the given interval, join the shapes for each subinterval together as shown at right. Note: If we are given points on the curve, we must also stretch or compress the various parts of the resulting shape to fit through the given points. ݔൌ3 ݔൌ5 Version 5.6 Page 44 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Rolle’s Theorem and the Mean Value Theorem Note: If the conditions for Rolle’s Theorem are satisfied, then either Rolle’s Theorem or the MVT can be applied. Rolle’s Theorem concludes that there is a value ܿ such that: ݂ᇱሺܿሻൌ0 The MVT concludes that there is a value ܿ such that: ݂ᇱሺܿሻൌ ሺሻିሺሻ ି ൌ ି ൌ0 These two conclusions are identical. Rolle’s Theorem If 1. ݂ሺݔሻ is continuous on ሾܽ, ܾሿ. 2. ݂ሺݔሻ is differentiable in ሺܽ, ܾሻ. 3. ݂ሺܽሻൌ݂ሺܾሻ. Then There is at least one value ܿ in ሺܽ, ܾሻ such that ݂ᇱሺܿሻൌ0. Conclusion in Words: There is at least one point in ሺܽ, ܾሻ with a horizontal tangent line. Mean Value Theorem (MVT) If 1. ݂ሺݔሻ is continuous on ሾܽ, ܾሿ. 2. ݂ሺݔሻ is differentiable in ሺܽ, ܾሻ. Then There is at least one value ܿ in ሺܽ, ܾሻ such that ݂ᇱሺܿሻൌ ሺሻିሺሻ ି . Conclusion in Words: There is at least one point in ሺܽ, ܾሻ where the slope of the tangent line has the same slope as the secant line over ሾܽ, ܾሿ. Version 5.6 Page 45 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Related Rates Related Rates Problems To solve problems that involve rates of change of two or more related variables, each with respect to a third variable, we must take derivatives with respect to the third variable (often, time) and remember to use the chain rule at each step. There are numerous methods that can be used to solve these problems; one that students have found particularly helpful is described and illustrated below. The General‐Specific Method This method breaks up the solution into the General and Specific Cases described in the problem, as follows: The General Case Deal with all variables in the abstract, without any numbers substituted for the variables. Set up any formulas required to solve the problem (e.g., volume of a cone). Take any derivatives (based on the above formulas) required to solve the problem. The Specific Case Record any values of variables for the specific situation described in the problem. Calculate any additional values required based on those provided in the problem (e.g., the length of the third side of a right triangle). After any derivatives are developed in the General Case, substitute values for the variables in the derivative equation. Derive the solution to the problem by solving the resulting equation. Notes: For some problems, you may need to draw a picture of the situation described in the problem. In these problems, you should draw a picture for the General Case and a second picture for the Specific Case. See Example 3.9, below. In the examples that follow, the General Case is shown on the left and the Specific Case is shown on the right. Version 5.6 Page 46 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Example 3.7: The volume of a cylinder is changing by 48 cm3 per second when the radius of the cylinder is 2 cm. If the height is twice the radius, find the rate of change of the radius when r = 2 cm. Note: ܸൌߨݎଶ݄. Example 3.8: The SA of a sphere is changing by 36 cm2 per second when the radius of the cylinder is 3 cm. Find the rate of change of the radius when r = 3 cm. Note: ܵܣൌ4ߨݎଶ. General Case We are asked to find ௗ ௗ௧ : ݄ൌ2ݎ ܸൌߨݎଶ݄ൌߨݎଶሺ2ݎሻൌ2ߨݎଷ Take the derivatives of both sides with respect to t: ௗ ௗ௧ ൌ6ߨݎଶ ௗ ௗ௧ After this part is done, move to the Specific Case. Specific Case Information given: ௗ ௗ௧ ൌ48 ݎൌ2 Substitute values into the equation derived in the General Case: ௗ ௗ௧ ൌ6ߨݎଶ ௗ ௗ௧ 48 ൌ6ߨ∙2ଶ∙݀ݎ ݀ݐ Do some algebra to calculate: ࢊ࢘ ࢊ࢚ൌ ସ଼ ଶସగൌ ࣊ cm/sec General Case We are asked to find ௗ ௗ௧ : ܵܣൌ4ߨݎଶ Take the derivatives of both sides with respect to t: ௗௌ ௗ௧ ൌ8ߨݎ ௗ ௗ௧ After this part is done, move to the Specific Case. Specific Case Information given: ௗௌ ௗ௧ ൌ36 ݎൌ3 Substitute values into the equation derived in the General Case: ௗௌ ௗ௧ ൌ8ߨݎ ௗ ௗ௧ 36 ൌ8ߨ∙3 ∙݀ݎ ݀ݐ Do some algebra to calculate: ࢊ࢘ ࢊ࢚ൌ ଷ ଶସగൌ ࣊ cm/sec Version 5.6 Page 47 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Example 3.9: A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second. How fast is the top of the ladder moving down the wall when its base is 7 feet from the wall? Example 3.10: The radius r of a circle is increasing at a rate of 3 cm/minute. Find the rate of change of the area when the circumference ܥൌ12ߨ cm. General Case We are asked to find ௗ௬ ௗ௧ Based on the drawing: ݔଶݕଶൌ625 Take the derivatives of both sides with respect to t: 2ݔ∙ ௗ௫ ௗ௧ 2ݕ∙ ௗ௬ ௗ௧ ൌ0 After this part is done, move to the Specific Case. ࢟ ࢞ 25 Specific Case Information given: ݔൌ7 Calculate: ݕൌ24 ௗ௫ ௗ௧ൌ 2 Substitute values into the equation derived in the General Case: 2ݔ∙ ௗ௫ ௗ௧ 2ݕ∙ ௗ௬ ௗ௧ ൌ0 2 ∙7 ∙ 2 2 ∙24 ∙ ௗ௬ ௗ௧ ൌ0 Do some algebra to calculate: ࢊ࢟ ࢊ࢚ൌെ ଶ଼ ସ଼ൌെ ૠ ൌെ. ૡ ܜ ܛ܍܋ ૠ 25 General Case We are asked to find ௗ ௗ௧ : ܣൌߨݎଶ Take the derivatives of both sides with respect to t: ௗ ௗ௧ ൌ2ߨݎ ௗ ௗ௧ After this part is done, move to the Specific Case. Specific Case Information given: ܥൌ2ߨݎൌ12ߨ ௗ ௗ௧ ൌ3 Substitute values into the equation derived in the General Case: ௗ ௗ௧ ൌ2ߨݎ ௗ ௗ௧ ௗ ௗ௧ ൌܥ∙ ௗ ௗ௧ ൌ12ߨ∙3 ൌ࣊ ܋ܕ ܕܑܖ Version 5.6 Page 48 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Kinematics (Particle Motion) Position Position is the location of a particle at a point in time. It is typically represented by the functions ݏሺݐሻ or ݔሺݐሻ. Displacement Displacement is a measure of the difference between a particle’s starting point and its ending point. It may be either positive or negative. A formula for displacement is: ∆ݏൌݏെݏ, where ݏ is the position at any point in time, and ݏ is the starting position. Distance Distance is a measure of the total movement of a particle; it is always a positive value. Total distance is the sum of the absolute values of the displacements of a particle in its various directions. Example 3.11: A particle moves from ݔൌ0 to ݔൌ6 to ݔൌ2. Displacement ൌ݁݊݀െݏݐܽݎݐൌ2 െ0 ൌ2 units Distance ൌ sum of absolute values of individual displacements ൌ |6 െ0| |2 െ6| ൌ10 units Velocity Velocity measures the rate of change in position. Instantaneous velocity is generally shown using the variable ݒ and average velocity is generally shown as ݒ̅. Velocity may also be shown as a vector ܞ ሬ റ, which has both magnitude and direction. The following formulas apply to velocity: Instantaneous velocity: ݒൌ ௗ௦ ௗ௧ (i.e, the derivative of the position function) Velocity at time ݐ: ݒൌݒܽݐ (where, ݒ is initial velocity and ܽ is a constant acceleration) Average velocity: ݒ̅ ൌ ୲୭୲ୟ୪ ୢ୧ୱ୮୪ୟୡୣ୫ୣ୬୲ ୲୭୲ୟ୪ ୲୧୫ୣ ൌ ∆௦ ∆௧ൌ ௦ሺ௧మሻି௦ሺ௧భሻ ௧మି௧భ Velocity may be either positive or negative. Version 5.6 Page 49 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Speed Speed, like velocity, measures the rate of change in position. However, unlike velocity, speed is always positive (it does not have direction). Instantaneous speed is the absolute value of velocity |ݒ| at a point in time. Average speed is based on distance instead of displacement. The following formulas apply to speed: Instantaneous speed: |ݒ| ൌ ቚ ௗ௦ ௗ௧ቚ (i.e, the absolute value of the velocity function) Average speed: ୲୭୲ୟ୪ ୢ୧ୱ୲ୟ୬ୡୣ ୲୭୲ୟ୪ ୲୧୫ୣ A note about speed: Speed is increasing when velocity and acceleration have the same sign (either or െ). Speed is decreasing when velocity and acceleration have different signs (one , one െ). Acceleration Acceleration measures the rate of change in velocity. Instantaneous acceleration is generally shown using the variable ܽ and average acceleration is generally shown as ܽ ത. Acceleration may also be shown as a vector ܉ ሬ റ, which has both magnitude and direction. The following formulas apply to acceleration: Instantaneous acceleration: ܽൌ ௗ௩ ௗ௧ ൌ ௗమ௦ ௗ௧మ Average acceleration: ܽ തൌ ୲୭୲ୟ୪ ୡ୦ୟ୬ୣ ୧୬ ୴ୣ୪୭ୡ୧୲୷ ୲୭୲ୟ୪ ୲୧୫ୣ ൌ ∆௩ ∆௧ ൌ ௩ሺ௧మሻି௩ሺ௧భሻ ௧మି௧భ Moving Among Functions The following diagram describes how to move back and forth among the position, velocity and acceleration functions. (Note: integration is handled in a subsequent chapter.) ௧௧ ሱۛۛۛۛۛۛۛۛۛሮ ௧௧ ሱۛۛۛۛۛۛۛۛۛሮ ܲݏ݅ݐ݅݊ ܸ݈݁ܿ݅ݐݕ ܣ݈ܿܿ݁݁ݎܽݐ݅݊ ூ௧௧ ር ۛ ۛ ۛ ۛ ۛ ۛ ሲ ூ௧௧ ር ۛ ۛ ۛ ۛ ۛ ۛ ሲ Version 5.6 Page 50 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Summary of Kinematics Terms Consider the following function definitions in relation to a particle in motion: 𝑠ሺ𝑡ሻ or 𝑥ሺ𝑡ሻ is the position function along the 𝑥-axis. 𝑣ሺ𝑡ሻൌ𝑠′ሺ𝑡ሻ is the velocity function along the 𝑥-axis. 𝑠𝑝𝑒𝑒𝑑ሺ𝑡ሻൌ|𝑣ሺ𝑡ሻ| is the speed function along the 𝑥-axis. 𝑎ሺ𝑡ሻൌ𝑣ᇱሺ𝑡ሻൌ𝑠′′ሺ𝑡ሻ is the acceleration function along the 𝑥-axis. Then the following terms relate to the functions defined above: Term/Description Meaning Initially 𝑡ൌ0 At the origin 𝑠ሺ𝑡ሻൌ0 At rest (i.e., zero velocity) 𝑣ሺ𝑡ሻൌ0 Positive velocity (moving to the right) 𝑣ሺ𝑡ሻ0 Negative velocity (moving to the left) 𝑣ሺ𝑡ሻ൏0 Average velocity (or the approximation of velocity over an interval ሾ𝑎, 𝑏ሿ) 𝑣̅ ൌ∆position ∆time ൌ𝑠ሺ𝑏ሻെ𝑠ሺ𝑎ሻ 𝑏െ𝑎 Instantaneous velocity at time 𝑡ൌ𝑐 𝑣ሺ𝑐ሻൌ𝑠′ሺ𝑐ሻ Particle changes directions 𝑣ሺ𝑡ሻൌ0, changes signs at time 𝑡 Speed is increasing (particle is accelerating) 𝑣ሺ𝑡ሻ, 𝑎ሺ𝑡ሻ have the same sign (+ or -) Speed is decreasing (particle is decelerating) 𝑣ሺ𝑡ሻ, 𝑎ሺ𝑡ሻ have different signs Positive acceleration 𝑎ሺ𝑡ሻ0 Negative acceleration 𝑎ሺ𝑡ሻ൏0 Average acceleration (or the approximation of acceleration over an interval ሾ𝑎, 𝑏ሿ) 𝑎 തൌ∆velocity ∆time ൌ𝑣ሺ𝑏ሻെ𝑣ሺ𝑎ሻ 𝑏െ𝑎 Instantaneous acceleration at time 𝑡ൌ𝑐 𝑎ሺ𝑐ሻൌ𝑣ᇱሺ𝑐ሻൌ𝑠′′ሺ𝑐ሻ acceleration speed velocity Version 5.6 Page 51 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Differentials Finding the Tangent Line Most problems that use differential to find the tangent line deal with three issues: Developing the equation of a tangent line at a point on a curve Estimating the value of a function using the tangent line. Estimating the change in the values of a function between two points, using the tangent line. In each case, the tangent line is involved, so let’s take a look at it. The key equation is: ݕൌ݂ሺܿሻ݂ᇱሺܿሻ∙ሺݔെܿሻ How does this equation come about? Let’s look at a curve and find the equation of the tangent line to that curve, in the general case. See the diagram below: Let our point on the curve be ൫ܿ, ݂ሺܿሻ൯. The slope of the tangent line at ൫ܿ, ݂ሺܿሻ൯ is ݂ᇱሺܿሻ. Use the point‐slope form of a line to calculate the equation of the line: ݕെݕଵൌ݉ሺݔെݔଵሻ ⇒ ݕെ݂ሺܿሻൌ݂ᇱሺܿሻ∙ሺݔെܿሻ Add ݂ሺܿሻ to both sides of the equation to obtain the form shown above Let’s take a closer look at the pieces of the equation: ݕൌ݂ሺܿሻ݂ᇱሺܿሻ∙ሺݔെܿሻ This is the “change part”. So, when you are asked about the change in ݂ሺݔሻ between two points or the potential error in measuring something, this is the part to focus on. ሺݔെܿሻis also shown as ∆ݔ. It is the difference between the x‐value you are evaluating and your anchor to the curve, which is the tangent point ൫ܿ, ݂ሺܿሻ൯. First, define your anchor, ܿ, and calculate ݂ሺܿሻ and ݂ᇱሺܿሻ. Substitute these into the equation and you are well on your way to a solution to the problem. Version 5.6 Page 52 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Curvature Curvature is the rate of change of the direction of a curve at a point, P (i.e., how fast the curve is turning at point P). Direction is based on ࣂ, the angle between the x‐axis and the tangent to the curve at P. The rate of change is taken with respect to ࢙, the length of an arbitrary arc on the curve near point P. We use the Greek letter kappa, ࣄ, for the measure of curvature. This is illustrated for the function ݕൌlnሺݔെ4ሻ3 at right. ߢൌ݀ߠ ݀ݏൌlim ௦ → Δߠ Δݏ This results in the following equations for ࣄ: ߢൌ ݀ଶݕ ݀ݔଶ ቈ1 ൬݀ݕ ݀ݔ൰ ଶ ଷଶ ൗ or ߢൌ െ݀ଶݔ ݀ݕଶ ቈ1 ൬݀ݔ ݀ݕ൰ ଶ ଷଶ ൗ Polar Form: Let ݎሺߠሻ be a function in polar form. Then, the polar form of curvature is given by: ߢൌݎଶ2ሺݎ′ሻଶെݎ ݎ′′ ሾݎଶሺݎ′ሻଶሿଷଶ ⁄ where, ݎᇱൌ݀ݎ ݀ߠ, ݎᇱᇱൌ݀ଶݎ ݀ߠଶ The Osculating Circle of a curve at Point P is the circle which is: Tangent to the curve at point P. Lies on the concave side of the curve at point P. Has the same curvature as the curve at point P. The Radius of Curvature of a curve at Point P is the radius of the osculating circle at point P. ܴൌ ଵ || The Center of Curvature of a curve at Point P is the center of the osculating circle at Point P. Version 5.6 Page 53 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Newton’s Method Sometimes it is difficult or impossible to find the exact roots of an equation. In such cases, approximate values may be found using numerical methods. Newton’s Method is a popular approach for determining roots this way, primarily because it is simple and easily programmed for use with a computer. Newton’s Method Use the following steps to identify a root of a function ݂ሺݔሻ using Newton’s Method. 1. Select an estimate of the root you are looking for. Call this estimate ࢞. It may be useful to graph the function for this purpose. 2. Use the differential formula (see above) to refine your estimate of the root: ݕൌ݂ሺݔሻ݂ᇱሺݔሻ∙ሺݔെݔሻ We want an estimate of ݔ when ݕൌ0. Setting ݕൌ0, the differential formula can be manipulated algebraically to get: ݔ ൌݔ െ ݂ሺݔሻ ݂ᇱሺݔሻ Let this value of ݔ be our next estimate, ࢞, of the value of the root we seek. Then, ࢞ ൌ࢞ െ ࢌሺ࢞ሻ ࢌᇱሺ࢞ሻ 3. Repeat the process to get subsequent values of ࢞, i.e., ࢞ ൌ࢞ି െ ࢌሺ࢞ିሻ ࢌᇱሺ࢞ିሻ 4. Continue Step 3 until the sequence ሼ࢞ሽ converges; that is, until successive estimates round to the same value based on a predetermined level of accuracy. When Newton’s Method Diverges Newton’s Method diverges under certain conditions. That is, for some functions and/or starting ࢞ values, successive values of ࢞ may not exist, may fluctuate back and forth between values, or may grow further and further away from the initial estimate of the root. When this occurs, you may want to select a different starting value of ࢞ and try again. However, the student should be aware that there are situations where Newton’s Method fails altogether. Version 5.6 Page 54 of 242 April 8, 2023 Chapter 3 Applications of Differentiation Example 3.12: Estimate the root of ࢌሺ࢞ሻൌࢋܛܑܖ࢞െ܋ܗܛ࢞ near ࢞ൌ to six decimal places. Let’s graph the function. In the graph, it is clear that there is a root close to ࢞ൌ. So, we are hopeful that Newton’s Method will converge quickly. We begin with the following: ݔൌ5 ݂ሺݔሻൌ݁ୱ୧୬௫െcos ݔ ݂ᇱሺݔሻൌ൫݁ୱ୧୬௫∙cos ݔ൯sin ݔ Now, let’s develop successive values of ݔ. Note: Microsoft Excel is useful for this purpose. ݔଵ ൌݔ െ ݂ሺݔሻ ݂ᇱሺݔሻൌ5 െ ݁ୱ୧୬ହെcos 5 ሺ݁ୱ୧୬ହ∙cos 5ሻsin 5 ൌ5 െ0.099643 െ0.8502 ൌ 5.1172 ݔଶ ൌݔଵ െ ݂ሺݔଵሻ ݂ᇱሺݔଵሻൌ5.1172 െ ݁ୱ୧୬ହ.ଵଵଶെcos 5.1172 ሺ݁ୱ୧୬ହ.ଵଵଶ∙cos 5.1172ሻsin 5.1172 ൌ 5.123764 ݔଷൌݔଶ െ ݂ሺݔଶሻ ݂ᇱሺݔଶሻൌ5.123764 െ ݁ୱ୧୬ହ.ଵଶଷସെcos 5.123764 ሺ݁ୱ୧୬ହ.ଵଶଷସ∙cos 5.123764ሻsin 5.123764 ൌ 5.123787 ݔସ ൌݔଷ െ ݂ሺݔଷሻ ݂ᇱሺݔଷሻൌ5.123787 െ ݁ୱ୧୬ହ.ଵଶଷ଼െcos 5.123787 ሺ݁ୱ୧୬ହ.ଵଶଷ଼∙cos 5.123787ሻsin 5.123787 ൌ . ૠૡૠ At this point, we stop the process because ݔସൌݔଷ when rounded to six decimals. The sequence of {ݔ} appears to have converged to 5.123787, which is our solution. If you like, you can use a modern graphing calculator to verify that this is in fact a good estimate of the desired root of ݂ሺݔሻ. Note: While the use of modern handheld graphing calculators makes Newton’s Method unnecessary in the Calculus classroom, its use in mathematical computer applications is essential. It is very useful in Microsoft Excel, Visual Basic, Python, Java and other applications in which the determination of a root is automated. Version 5.6 Page 55 of 242 April 8, 2023 Chapter 4 Integration Rules of Indefinite Integration Note: the rules presented in this chapter omit the “ C ” term that must be added to all indefinite integrals in order to save space and avoid clutter. Please remember to add the “ C ” term on all work you perform with indefinite integrals. Basic Rules නܿ݀ݑ ൌܿݑ න݂ܿሺݑሻ ݀ݑ ൌܿ න݂ሺݑሻ݀ݑ න݂ሺݑሻ݃ሺݑሻ݀ݑ ൌන݂ሺݑሻ݀ݑ න݃ሺݑሻ݀ݑ Integration by Parts නݑ݀ݒൌݑݒെනݒ݀ݑ Power Rule නሺݑሻ ݀ݑൌ 1 ݊1 ∙ݑାଵ ሺ്݊1ሻ න1 ݑ݀ݑൌln|ݑ| Exponential and Logarithmic Functions ሺܽ0, ്ܽ1ሻ න݁௨݀ݑൌ ݁௨ නܽ௨݀ݑൌ1 ln ܽ ܽ௨ න 1 ݑln ݑ݀ݑൌlnሺln ݑሻ Version 5.6 Page 56 of 242 April 8, 2023 Chapter 4 Integration Integrals of Special Functions Exponential and Logarithmic Functions න݁௫݀ݔൌ ݁௫ න݁௨݀ݑൌ݁௨ නܽ௫݀ݔൌܽ௫ ln ܽ නܽ௨݀ݑൌܽ௨ ln ܽ න1 ݔ݀ݔൌln|ݔ| න1 ݑ݀ݑൌln|ݑ| නln ݔ݀ݔൌݔln ݔെݔ නln ݑ݀ݔൌݑln ݑെݑ න 1 ݔln ݔ݀ݔൌlnሺln ݔሻ න 1 ݑln ݑ݀ݑൌlnሺln ݑሻ Trigonometric Functions නsin ݑ݀ݑൌെcos ݑ නcos ݑ݀ݑൌsin ݑ නtan ݑ݀ݑൌln |sec ݑ| ൌെln |cos ݑ| නsecଶݑ݀ݑൌtan ݑ නcot ݑ݀ݑൌെln |csc ݑ| ൌln |sin ݑ| නcscଶݑ݀ݑൌെcot ݑ නsec ݑ݀ݑൌln |sec ݑtan ݑ| නsec ݑtan ݑ݀ݑൌsec ݑ නcsc ݑ݀ݑൌെ ln |csc ݑcot ݑ| නcsc ݑcot ݑ݀ݑൌെcsc ݑ Version 5.6 Page 57 of 242 April 8, 2023 Chapter 4 Integration Derivations of the Integrals of Trigonometric Functions නܜ܉ܖ࢞ࢊ࢞ නtan ݔ݀ݔൌන sin ݔ cos ݔ݀ݔ Let: ݑ ൌ cos ݔ so that: ݀ݑ ൌ െsin ݔ ݀ݔ Then, නtan ݔ݀ݔൌെන1 ݑ ݀ݑ ൌ െln|ݑ| ܥ ൌ െln| cos ݔ| ܥ න܋ܗܜ࢞ࢊ࢞ නcot ݔ݀ݔൌන cos ݔ sin ݔ݀ݔ Let: ݑ ൌ sin ݔ so that: ݀ݑ ൌ cos ݔ ݀ݔ Then, නcot ݔ݀ݔൌන1 ݑ ݀ݑ ൌ ln|ݑ| ܥ ൌ െln| sin ݔ| ܥ නܛ܍܋࢞ࢊ࢞ Multiply the numerator and denominator by: ሺsec ݔtan ݔሻ Then, නsec ݔ݀ݔൌනsec ݔ∙ሺsec ݔtan ݔሻ ሺsec ݔtan ݔሻ݀ݔൌනሺsecଶݔsec ݔtan ݔሻ ሺsec ݔtan ݔሻ ݀ݔ Let: ݑ ൌ ሺsec ݔtan ݔሻ so that: ݀ݑ ൌ ሺsec ݔtan ݔsecଶݔሻ݀ݔ Then, නsec ݔ݀ݔൌන1 ݑ ݀ݑ ൌ ln|ݑ| ܥ ൌ ln| sec ݔtan ݔ| ܥ Version 5.6 Page 58 of 242 April 8, 2023 Chapter 4 Integration Derivations of the Integrals of Trig Functions (cont’d) න܋ܛ܋࢞ࢊ࢞ Multiply the numerator and denominator by: ሺcsc ݔcot ݔሻ Then, නcsc ݔ݀ݔൌනcsc ݔ∙ሺcsc ݔcot ݔሻ ሺcsc ݔcot ݔሻ݀ݔൌනሺcscଶݔcsc ݔcot ݔሻ ሺcsc ݔcot ݔሻ ݀ݔ Let: ݑ ൌ ሺcsc ݔcot ݔሻ so that: ݀ݑ ൌ ሺെcsc ݔtan ݔെcscଶݔሻ݀ݔ Then, නcsc ݔ݀ݔൌെන1 ݑ ݀ݑ ൌെ ln|ݑ| ܥ ൌ െln| csc ݔcot ݔ| ܥ Version 5.6 Page 59 of 242 April 8, 2023 Chapter 4 Integration Integration Involving Inverse Trig Functions Key Formulas: Base Formulas General Formulas න 1 √1 െ𝑢ଶ 𝑑𝑢ൌsinିଵ𝑢𝐶 න 1 √𝑎ଶെ𝑢ଶ𝑑𝑢ൌsinିଵቀ𝑢 𝑎ቁ𝐶 න 1 1 𝑢ଶ 𝑑𝑢ൌtanିଵ𝑢𝐶 න 1 𝑎ଶ𝑢ଶ𝑑𝑢ൌ1 𝑎tanିଵቀ𝑢 𝑎ቁ𝐶 න 1 𝑢 √𝑢ଶെ1 𝑑𝑢ൌsecିଵ|𝑢| 𝐶 න 1 𝑢√𝑢ଶെ𝑎ଶ𝑑𝑢ൌ1 𝑎secିଵቆ|𝑢| 𝑎ቇ𝐶 About Inverse Trig Functions As an example, sinିଵ𝑥 asks the question, what angle (in radians) has a sine value of 𝑥? So, sinିଵቀ ଵ ଶቁൌ గ sinିଵቀ√ଶ ଶቁൌ గ ସ sinିଵቀെ ଵ ଶቁൌെ గ It is important, for these purposes, to understand the quadrants in which each inverse Trig function is defined, as shown in the following charts. Solutions to problems involving inverse Trig functions may be expressed multiple ways. For example, in the triangle at right with sides 𝑎, 𝑏 and 𝑐, the measure of angle A can be expressed as follows: 𝑚∠A ൌsinିଵቀ𝑎 𝑐ቁൌtanିଵቀ𝑎 𝑏ቁൌsecିଵቀ𝑐 𝑏ቁ Some calculators will never give results using the secିଵ function, preferring to use the tanିଵ function instead; the answers are equivalent. For example, secିଵ|2𝑥| ൌtanିଵ√4𝑥ଶെ1. Ranges of Inverse Trigonometric Functions Function Gives a Result In: sinିଵ𝜃 െ𝜋 2 𝜃𝜋 2 cosିଵ𝜃 0 𝜃𝜋 tanିଵ𝜃 െ𝜋 2 𝜃𝜋 2 Version 5.6 Page 60 of 242 April 8, 2023 Chapter 4 Integration Indefinite Integrals of Inverse Trigonometric Functions Inverse Trigonometric Functions නsinିଵݑ݀ݑൌݑ sinିଵݑ ඥ1 െݑଶ නcosିଵݑ݀ݑൌݑ cosିଵݑെ ඥ1 െݑଶ නtanିଵݑ݀ݑൌݑ tanିଵݑെ1 2 lnሺݑଶ1ሻ නcotିଵݑ݀ݑൌݑ cotିଵݑ1 2 lnሺݑଶ1ሻ නsecିଵݑ݀ݑൌݑ secିଵݑെ ln ቀݑඥݑଶെ1ቁ secିଵݑ∈ቄቀ0, ߨ 2ቁቅ ൌݑ secିଵݑ ln ቀݑඥݑଶെ1ቁ secିଵݑ∈ቄቀߨ 2 , ߨቁቅ නcscିଵݑ݀ݑൌݑ cscିଵݑ ln ቀݑඥݑଶെ1ቁ cscିଵݑ∈ቄቀ0, ߨ 2ቁቅ ൌݑ cscିଵݑെ ln ቀݑඥݑଶെ1ቁ cscିଵݑ∈ቄቀെߨ 2 , 0ቁቅ Involving Inverse Trigonometric Functions න 1 √1 െݑଶ ݀ݑൌsinିଵݑ න 1 √ܽଶെݑଶ݀ݑൌsinିଵቀݑ ܽቁ න 1 1 ݑଶ ݀ݑൌtanିଵݑ න 1 ܽଶݑଶ݀ݑൌ1 ܽtanିଵቀݑ ܽቁ න 1 ݑ √ݑଶെ1 ݀ݔൌsecିଵ|ݑ| න 1 ݑ√ݑଶെܽଶ݀ݔൌ1 ܽsecିଵቆ|ݑ| ܽቇ Version 5.6 Page 61 of 242 April 8, 2023 Chapter 4 Integration Integrals of Special Functions Selecting the Right Function for an Integral Form Function Integral න 1 √ܽଶെݑଶ ݀ݑ sinିଵݑ න 1 √ܽଶെݑଶ݀ݑൌsinିଵቀݑ ܽቁ න 1 ܽଶݑଶ ݀ݑ tanିଵݑ න 1 ܽଶݑଶ݀ݑൌ1 ܽtanିଵቀݑ ܽቁ න 1 ݑ √ݑଶെܽଶ ݀ݔ secିଵݑ න 1 ݑ√ݑଶെܽଶ݀ݔൌ1 ܽ secିଵቆ|ݑ| ܽቇ න 1 √ݑଶܽଶ ݀ݑ sinhିଵݑ න 1 √ݑଶܽଶ݀ݑൌln ቀݑඥݑଶܽଶቁ න 1 √ݑଶെܽଶ ݀ݑ coshିଵݑ න 1 √ݑଶെܽଶ݀ݑൌln ቀݑඥݑଶെܽଶቁ න 1 ܽଶെݑଶ ݀ݑቚ ܽݑ tanhିଵݑ න 1 ܽଶെݑଶ ݀ݑൌ1 2ܽln ฬܽݑ ܽെݑฬ න 1 ݑଶെܽଶ ݀ݑቚ ݑܽ cothିଵݑ න 1 ݑ √ܽଶെݑଶ ݀ݑ sechିଵݑ න 1 ݑ√ܽଶെݑଶ݀ݑൌെ1 ܽln ቆܽ√ܽଶെݑଶ |ݑ| ቇ න 1 ݑ √ܽଶݑଶ݀ݑ cschିଵݑ න 1 ݑ√ܽଶݑଶ݀ݑൌെ1 ܽln ቆܽ√ܽଶݑଶ |ݑ| ቇ This is an inverse hyperbolic function. For more information, see Chapter 6. Note that you do not need to know about inverse hyperbolic functions to use the formulas on this page. Version 5.6 Page 62 of 242 April 8, 2023 Chapter 5 Techniques of Integration 𝒖–Substitution Often, an integrand will contain a function within a function. For example, in the integral √௫ ௫ 𝑑𝑥, we have the function √𝑥 within the ln function. When this happens, it is often useful to substitute another variable for the internal function. Typically the variable u is used to represent the inner function, so the process is called 𝒖–substitution. The typical process used for 𝑢–substitution is described in steps below. When trying this approach, note the following: 𝑢–substitution will work for all integrals, even ones that look ripe for it, though it does work often. If one attempted substitution does not work, the student should try another one. It takes practice to train the eye to identify what functions work well in this process. It is possible that the student will be faced with an integral than simply cannot be integrated by any elementary method (e.g., 𝑒ି௫మ 𝑑𝑥). Process Following are the steps for the general solution to an integral using 𝑢–substitution. 1. Set a portion of the integrand equal to a new variable, e.g., 𝑢. Look to the rest of the integrand in deciding what to set equal to 𝑢. You will need to have 𝑑𝑢 in the integrand as well, if this technique is to find success. 2. Find 𝑑𝑢 in terms of 𝑑𝑥. 3. Rearrange the integrand so that the integral exists in terms of 𝑢 instead of 𝑥. 4. Perform the integration. 5. Substitute the expression for 𝑢 back into the result of the integration. 6. If you are uncomfortable with the result, integrate it to see if you get the integrand as a result. If so, you have achieved your goal. And, don’t forget the 𝐶 for an indefinite integration. Example 5.1: Find: ୪୬√௫ ௫ 𝑑𝑥 නln √𝑥 𝑥 𝑑𝑥ൌන 1 2 ln 𝑥 𝑥 𝑑𝑥ൌ1 2 නln 𝑥1 𝑥𝑑𝑥 ൌ1 2 න𝑢 𝑑𝑢 ൌ1 2 ∙1 2 𝑢ଶൌ 𝟏 𝟒 ሺ𝐥𝐧𝒙ሻ𝟐𝑪 𝑢ൌln 𝑥 𝑑𝑢ൌ1 𝑥𝑑𝑥 Version 5.6 Page 63 of 242 April 8, 2023 Chapter 5 Techniques of Integration Example 5.2: Find: ሺଵି୪୬ ௧ሻమ ௧ 𝑑𝑡 Example 5.3: Find: ௗ௫ √ଽି ௫మ Recall: ଵ √ଵି௨మ 𝑑𝑢ൌsinିଵ𝑢 with െ గ ଶsinିଵ𝑢 గ ଶ Example 5.4: Find: ೣ ଵା మೣ 𝑑𝑥 Recall: ଵ ଵା௨మ 𝑑𝑢ൌtanିଵ𝑢 with െ గ ଶtanିଵ𝑢 గ ଶ නሺ1 െln 𝑡ሻଶ 𝑡 𝑑𝑡 ൌെනሺ1 െln 𝑡ሻଶ ൬െ1 𝑡 𝑑𝑡൰ ൌെ න𝑢ଶ 𝑑𝑢 ൌെ1 3 𝑢ଷൌ െ𝟏 𝟑ሺ𝟏െ𝐥𝐧𝒕ሻ𝟑𝑪 𝑢ൌ1 െln 𝑡 𝑑𝑢ൌെ1 𝑡𝑑𝑡 න 𝑑𝑥 √9 െ𝑥ଶ ൌන 1 3ට1 െቀ𝑥 3ቁ ଶ 𝑑𝑥 ൌන 1 ට1 െቀ𝑥 3ቁ ଶ 1 3 𝑑𝑥 ൌsinିଵ𝑢𝐶ൌ 𝐬𝐢𝐧ି𝟏ቀ𝒙 𝟑ቁ𝑪 𝑢ൌ𝑥 3 𝑑𝑢ൌ1 3 𝑑𝑥 න 𝑒௫ 1 𝑒ଶ௫ 𝑑𝑥 ൌන 1 1 ሺ𝑒௫ሻଶ 𝑒௫𝑑𝑥 ൌන 1 1 𝑢ଶ 𝑑𝑢 ൌtanିଵ𝑢ൌ 𝐭𝐚𝐧ି𝟏ሺ𝒆𝒙ሻ𝑪 𝑢ൌ𝑒𝑥 𝑑𝑢ൌ𝑒𝑥𝑑𝑥 Version 5.6 Page 64 of 242 April 8, 2023 Chapter 5 Techniques of Integration Partial Fractions Partial Fractions Every rational function of the form 𝑅ሺ𝑥ሻൌ ேሺ௫ሻ ሺ௫ሻ can be expressed as a sum of fractions with linear and quadratic forms in their denominators. For example: 𝑥ସ2𝑥ଷെ3𝑥4 ሺ𝑥െ4ሻଷሺ𝑥ଶ2𝑥4ሻଶൌ 𝑎ଵ ሺ𝑥െ4ሻ 𝑎ଶ ሺ𝑥െ4ሻଶ 𝑎ଷ ሺ𝑥െ4ሻଷ 𝑏ଵ𝑥𝑐ଵ ሺ𝑥ଶ2𝑥4ሻ 𝑏ଶ𝑥𝑐ଶ ሺ𝑥ଶ2𝑥4ሻଶ Our task is to determine the appropriate fractions, including the values of the 𝑎’s, 𝑏’s and 𝑐’s, so we can integrate the function. The result of integration tends to contain a number of natural logarithm terms and inverse tangent terms, as well as others. The following process can be used to determine the set of fractions (including the 𝑎’s, 𝑏’s and 𝑐’s) whose sum is equal to 𝑅ሺ𝑥ሻ. Process 1. If 𝑁ሺ𝑥ሻ has the same degree or higher degree than 𝐷ሺ𝑥ሻ, divide 𝑁ሺ𝑥ሻ by 𝐷ሺ𝑥ሻ to obtain the non-fractional (polynomial) component of the rational function. Proceed in the next steps with the fractional component of the rational function. Example 5.5: 𝑅ሺ𝑥ሻൌ ௫మାଶ௫ିହ ௫ିଶ ൌ𝑥4 ଷ ௫ିଶ . Since it is easy to integrate the polynomial portion of this result, (i.e., 𝑥4), it remains to integrate the fractional portion (i.e., ଷ ௫ିଶ). 2. To determine the denominators of the fractions on the right side of the equal sign, we must first factor the denominator of 𝑅ሺ𝑥ሻ, i.e., 𝐷ሺ𝑥ሻ. Note that every polynomial can be expressed as the product of linear terms and quadratic terms, so that: 𝐷ሺ𝑥ሻൌ𝑘ሺ𝑥െ𝑟 ଵሻሺ𝑥െ𝑟 ଶሻ… ሺ𝑥െ𝑟 ሻ ∙ ሺ𝑥ଶ𝑝ଵ𝑥𝑞ଵሻሺ𝑥ଶ𝑝ଶ𝑥𝑞ଶሻ… ሺ𝑥ଶ𝑝𝑥𝑞ሻ Where 𝑘 is the lead coefficient, the ሺ𝑥െ𝑟ሻ terms are the linear factors and the ሺ𝑥ଶ𝑝𝑥 𝑞ሻ are the quadratic terms of 𝐷ሺ𝑥ሻ. Version 5.6 Page 65 of 242 April 8, 2023 Chapter 5 Techniques of Integration We solve these equations to obtain: 3. Every rational function can be expressed as the sum of fractions of the following types: ሺ௫ିሻೞ or ௫ା ሺ௫మା௫ାሻ Where the exponents in the denominators, 𝑠 and 𝑡, take all values from 1 to the multiplicity of the factor in 𝐷ሺ𝑥ሻ. Examples 5.6 – 5.8: 2𝑥ଶ5𝑥െ3 ሺ𝑥2ሻଷ ൌ 𝑎ଵ ሺ𝑥2ሻ 𝑎ଶ ሺ𝑥2ሻଶ 𝑎ଷ ሺ𝑥2ሻଷ 𝑥ଷെ𝑥ଶ6𝑥െ2 ሺ𝑥ଶെ3𝑥7ሻଶൌ 𝑏ଵ𝑥𝑐ଵ ሺ𝑥ଶെ3𝑥7ሻ 𝑏ଶ𝑥𝑐ଶ ሺ𝑥ଶെ3𝑥7ሻଶ 𝑥ସ2𝑥ଷെ3𝑥4 ሺ𝑥െ1ሻଶሺ𝑥3ሻሺ𝑥ଶെ4𝑥1ሻൌ 𝑎ଵ ሺ𝑥െ1ሻ 𝑎ଶ ሺ𝑥െ1ሻଶ 𝑎ଷ ሺ𝑥3ሻ 𝑏ଵ𝑥𝑐ଵ ሺ𝑥ଶെ4𝑥1ሻ We must solve for the values of the 𝑎’s, 𝑏’s and 𝑐’s. This is accomplished by obtaining a common denominator and then equating the coefficients of each term in the numerator. This will generate a number of equations with the same number of unknown values of 𝑎, 𝑏 and 𝑐. Example 5.6a (using the first expression above): 2𝑥ଶ5𝑥െ3 ሺ𝑥2ሻଷ ൌ 𝑎ଵ ሺ𝑥2ሻ 𝑎ଶ ሺ𝑥2ሻଶ 𝑎ଷ ሺ𝑥2ሻଷ ൌ𝑎ଵሺ𝑥2ሻଶ ሺ𝑥2ሻଷ𝑎ଶሺ𝑥2ሻ ሺ𝑥2ሻଷ 𝑎ଷ ሺ𝑥2ሻଷൌ𝑎ଵሺ𝑥2ሻଶ𝑎ଶሺ𝑥2ሻ𝑎ଷ ሺ𝑥2ሻଷ Equating the numerators, then, 2𝑥ଶ5𝑥െ3 ൌ 𝑎ଵ𝑥ଶሺ4𝑎ଵ𝑎ଶሻ𝑥ሺ4𝑎ଵ2𝑎ଶ𝑎ଷሻ So that: 𝑎ଵൌ2 𝑎ଵൌ2 4𝑎ଵ𝑎ଶൌ5 𝑎ଶൌെ3 4𝑎ଵ2𝑎ଶ𝑎ଷൌെ3 𝑎ଷൌെ5 Finally concluding that: 2𝑥ଶ5𝑥െ3 ሺ𝑥2ሻଷ ൌ 2 ሺ𝑥2ሻ െ3 ሺ𝑥2ሻଶ െ5 ሺ𝑥2ሻଷ ൌ 2 ሺ𝑥2ሻെ 3 ሺ𝑥2ሻଶെ 5 ሺ𝑥2ሻଷ Version 5.6 Page 66 of 242 April 8, 2023 Chapter 5 Techniques of Integration 4. The final step is to integrate the resulting fractions. Example 5.6b (continuing from Step 3): න2𝑥ଶ5𝑥െ3 ሺ𝑥2ሻଷ 𝑑𝑥ൌන 2 ሺ𝑥2ሻെ 3 ሺ𝑥2ሻଶെ 5 ሺ𝑥2ሻଷ 𝑑𝑥 ൌ2 ln|𝑥2| 3 ሺ𝑥2ሻ 5 2ሺ𝑥2ሻ2 Version 5.6 Page 67 of 242 April 8, 2023 Chapter 5 Techniques of Integration Integration by Parts General From the product rule of derivatives we have: 𝑑 𝑢𝑣 ൌ 𝑢 𝑑𝑣𝑣 𝑑𝑢 Rearranging terms we get: 𝑢 𝑑𝑣 ൌ 𝑑 𝑢𝑣 െ 𝑣 𝑑𝑢 Finally, integrating both sides gives us: න𝑢 𝑑𝑣 ൌ න𝑑 𝑢𝑣 െ න𝑣 𝑑𝑢 න𝑢 𝑑𝑣 ൌ 𝑢𝑣 െ න𝑣 𝑑𝑢 This last formula is the one for integration by parts and is extremely useful in solving integrals. When performing an integration by parts, first define 𝑢 and 𝑑𝑣. L I A T E When integrating by parts, students often struggle with how to break up the original integrand into 𝑢 and 𝑑𝑣. LIATE is an acronym that is often used to determine which part of the integrand should become 𝑢. Here’s how it works: let 𝑢 be the function from the original integrand that shows up first on the list below. Logarithmic functions (e.g., ln 𝑥) Inverse trigonometric functions (e.g., tanିଵ𝑥) Algebraic functions (e.g., 𝑥ଷ𝑥െ2) Trigonometric functions (e.g., cos 𝑥) Exponential functions (e.g., 𝑒௫) In general, we want to let 𝑢 be a function whose derivative 𝑑𝑢 is both relatively simple and compatible with 𝑣. Logarithmic and inverse trigonometric functions appear first in the list because their derivatives are algebraic; so if 𝑣 is algebraic, 𝑣 𝑑𝑢 is algebraic and an integration with “weird” functions is transformed into one that is completely algebraic. Note that the LIATE approach does not always work, but in many cases it can be helpful. Version 5.6 Page 68 of 242 April 8, 2023 Chapter 5 Techniques of Integration Example 5.9: Find cosଶ𝑥𝑑𝑥 (note: ignore the 𝐶 until the end) නcosଶ𝑥𝑑𝑥ൌsin 𝑥cos 𝑥െනሺെsinଶ𝑥ሻ𝑑𝑥 ൌsin 𝑥cos 𝑥නሺsinଶ𝑥ሻ𝑑𝑥 ൌsin 𝑥cos 𝑥නሺ1 െcosଶ𝑥ሻ𝑑𝑥 ൌsin 𝑥cos 𝑥නሺ1ሻ𝑑𝑥െනሺcosଶ𝑥ሻ𝑑𝑥 නcosଶ𝑥𝑑𝑥ൌsin 𝑥cos 𝑥𝑥െනሺcosଶ𝑥ሻ𝑑𝑥 2 නcosଶ𝑥𝑑𝑥ൌsin 𝑥cos 𝑥𝑥 නcosଶ𝑥𝑑𝑥ൌ1 2 ሺsin 𝑥cos 𝑥𝑥ሻ𝐶 Example 5.9A: Find cosଶ𝑥𝑑𝑥 without using integration by parts Let’s use the Trig identity: cosଶ𝑥ൌ ଵାୡ୭ୱଶ௫ ଶ නcosଶ𝑥𝑑𝑥ൌන൬1 cos 2𝑥 2 ൰𝑑𝑥 ൌ1 2 නሺ1 cos 2𝑥ሻ𝑑𝑥 ൌ1 2 ൬𝑥1 2 sin 2𝑥൰𝐶 Next, recall that sin 2𝑥ൌ2 sin 𝑥cos 𝑥. So, නcosଶ𝑥𝑑𝑥ൌ1 2 ൬𝑥1 2 ∙2 sin 𝑥cos 𝑥൰𝐶 ൌ1 2 ሺsin 𝑥cos 𝑥𝑥ሻ𝐶 Let: 𝑢ൌcos 𝑥 𝑣ൌsin 𝑥 𝑑𝑢ൌെsin 𝑥𝑑𝑥 𝑑𝑣ൌcos 𝑥𝑑𝑥 Version 5.6 Page 69 of 242 April 8, 2023 Chapter 5 Techniques of Integration නሺ𝑥ଶ𝑒௫ሻ𝑑𝑥ൌ𝑥ଶ𝑒௫െන2𝑥𝑒𝑥𝑑𝑥 ൌ𝑥ଶ𝑒௫െ2 න𝑥 𝑒𝑥𝑑𝑥 ൌ𝑥ଶ𝑒௫െ2 ൬𝑥 𝑒𝑥െන𝑒𝑥𝑑𝑥൰ ൌ𝑥ଶ𝑒௫െ2ሺ𝑥 𝑒𝑥െ𝑒𝑥ሻ𝐶 ൌሺ𝑥ଶെ2𝑥2ሻ𝑒௫𝐶 Example 5.10: Find ln 𝑥𝑑𝑥 නln 𝑥𝑑𝑥ൌ𝑥ln 𝑥െන𝑥 𝑥𝑑𝑥 ൌ𝑥ln 𝑥െන1 𝑑𝑥 ൌ𝑥ln 𝑥െ𝑥𝐶 Example 5.11: Find ሺ𝑥ଶ𝑒௫ሻ𝑑𝑥 Example 5.12: Find tanିଵ𝑥𝑑𝑥 නtanିଵ𝑥𝑑𝑥ൌ𝑥tanିଵ𝑥െන 𝑥 1 𝑥2 𝑑𝑥 ൌ𝑥tanିଵ𝑥െ1 2 න 1 1 𝑥2 2𝑥 𝑑𝑥 ൌ𝑥tanିଵ𝑥െ1 2 lnሺ1 𝑥2ሻ𝐶 Let: 𝑢ൌln 𝑥 𝑣ൌ𝑥 𝑑𝑢ൌ1 𝑥𝑑𝑥 𝑑𝑣ൌ𝑑𝑥 Let: 𝑢ൌ𝑥ଶ 𝑣ൌ𝑒௫ 𝑑𝑢ൌ2𝑥 𝑑𝑥 𝑑𝑣ൌ𝑒௫𝑑𝑥 Let: 𝑢ൌ𝑥 𝑣ൌ𝑒௫ 𝑑𝑢ൌ𝑑𝑥 𝑑𝑣ൌ𝑒௫𝑑𝑥 Let: 𝑢ൌtanିଵ𝑥 𝑣ൌ𝑥 𝑑𝑢ൌ 1 1 𝑥ଶ𝑑𝑥 𝑑𝑣ൌ𝑑𝑥 Version 5.6 Page 70 of 242 April 8, 2023 Chapter 5 Techniques of Integration Integration by Parts – Tabular Method If the use of Integration by Parts results in another integral that must be solved using integration by parts, the Tabular Method can be used to simplify repeating the process and save time. This method is particularly useful when one of the terms of the integrand is a polynomial. Description of the Method Create a table like the one below, starting with the 𝑢- and 𝑑𝑣-substitutions to be used in the initial integration by parts. Start the 𝑑𝑣-column one line higher than the 𝑢-column. In the 𝑢-column, take consecutive derivatives until the derivative equals zero. In the 𝑑𝑣-column, take consecutive integrals until the derivative column equals zero. In the sign column, begin with a sign and alternate and – signs. Multiply the sign and the expressions in the 𝑢- and 𝑑𝑣 columns to obtain each term of the solution. Add all of the terms obtained as described above to obtain the complete solution. Example 5.13: Tabular Method to determine 𝒙𝟑𝐬𝐢𝐧𝟐𝒙𝒅𝒙 Terms Sign 𝒖, 𝒅𝒖, 𝒅𝟐𝒖… 𝒅𝒗, 𝒗, න𝒗, ඵ𝒗, … 𝑑𝑣 sin 2𝑥𝑑𝑥 𝑢, 𝑣 𝑥ଷ െ1 2 cos 2𝑥 𝑑𝑢, න𝑣 െ 3𝑥ଶ െ1 4 sin 2𝑥 𝑑ଶ𝑢, ඵ𝑣 6𝑥 1 8 cos 2𝑥 𝑑ଷ𝑢, ම𝑣 െ 6 1 16 sin 2𝑥 Solution: න𝑥ଷsin 2𝑥𝑑𝑥ൌሺ𝑥ଷሻ൬െ1 2 cos 2𝑥൰െሺ3𝑥ଶሻ൬െ1 4 sin 2𝑥൰ሺ6𝑥ሻ൬1 8 cos 2𝑥൰െ6 ൬1 16 sin 2𝑥൰𝐶 ൌെ1 2 𝑥ଷcos 2𝑥3𝑥ଶ 4 sin 2𝑥3𝑥 4 cos 2𝑥െ3 8 sin 2𝑥𝐶 take consecutive derivatives take consecutive integrals Version 5.6 Page 71 of 242 April 8, 2023 Chapter 5 Techniques of Integration න 𝑑𝑥 𝑥 √𝑥ଶ16 ൌන 4 secଶ𝜃𝑑𝜃 4 tan 𝜃 ඥሺ4 tan 𝜃ሻଶ16 ൌන 4 secଶ𝜃𝑑𝜃 4 tan 𝜃∙4 sec 𝜃 ൌ1 4 නsec 𝜃 tan 𝜃𝑑𝜃ൌ1 4 නcsc 𝜃𝑑𝜃 Let: 𝑥ൌ4 tan 𝜃 𝑑𝑥ൌ4 secଶ𝜃𝑑𝜃 Trigonometric Substitution Certain integrands are best handled with a trigonometric substitution. Three common forms are shown in the table below: Integral Contains this Form Try this Substitution ඥ𝑥ଶ𝑎ଶ 𝑜𝑟 ඥ𝑎ଶ𝑥ଶ 𝑥ൌ𝑎tan 𝜃 ඥ𝑥ଶെ𝑎ଶ 𝑥ൌ𝑎sec 𝜃 ඥ𝑎ଶെ𝑥ଶ 𝑥ൌ𝑎sin 𝜃 or 𝑥ൌ𝑎cos 𝜃 Why are these helpful? Quite simply because they eliminate what is often the most difficult part of the problem – the square root sign. Let’s look at each of the substitutions in the table. Using the substitution 𝑥ൌ𝑎tan 𝜃, we have: ඥ𝑥ଶ𝑎ଶൌඥሺ𝑎tan 𝜃ሻଶ𝑎ଶൌඥ𝑎ଶ ሺtanଶ𝜃1ሻൌඥ𝑎ଶ secଶ𝜃ൌ𝑎sec 𝜃 Using the substitution 𝑥ൌ𝑎sec 𝜃, we have: ඥ𝑥ଶെ𝑎ଶൌඥሺ𝑎sec 𝜃ሻଶെ𝑎ଶൌඥ𝑎ଶ ሺsecଶ𝜃െ1ሻൌඥ𝑎ଶ tanଶ𝜃ൌ𝑎tan 𝜃 Using the substitution 𝑥ൌ𝑎sin 𝜃, we have: ඥ𝑎ଶെ𝑥ଶൌඥ𝑎ଶെሺ𝑎sin 𝜃ሻଶൌඥ𝑎ଶ ሺ1 െsinଶ𝜃ሻൌඥ𝑎ଶ cosଶ𝜃ൌ𝑎cos 𝜃 Using the substitution 𝑥ൌ𝑎cos 𝜃, we have: ඥ𝑎ଶെ𝑥ଶൌඥ𝑎ଶെሺ𝑎cos 𝜃ሻଶൌඥ𝑎ଶ ሺ1 െcosଶ𝜃ሻൌඥ𝑎ଶ sinଶ𝜃ൌ𝑎sin 𝜃 Example 5.14: ൌെ1 4 ln|csc 𝜃cot 𝜃| 𝐶ൌെ1 4 ln ቤ√𝑥ଶ16 4 𝑥 ቤ𝐶 Version 5.6 Page 72 of 242 April 8, 2023 Chapter 5 Techniques of Integration Example 5.15: නඥ𝑥ଶ1 ଵ 𝑑𝑥 Let: 𝑥ൌtan 𝜃, so 𝜃ൌtanିଵ𝑥 and 𝑑𝑥ൌsecଶ𝜃𝑑𝜃 Then the limits of integration become: 𝜃ൌtanିଵ0 ൌ0 and 𝜃ൌtanିଵ1 ൌ గ ସ. නඥ𝑥ଶ1 ଵ 𝑑𝑥ൌනඥtanଶ𝜃1 గ ସ ∙ secଶ𝜃𝑑𝜃ൌනඥsecଶ𝜃 గ ସ ∙ secଶ𝜃𝑑𝜃ൌනsecଷ𝜃𝑑𝜃 గ ସ We need integration by parts to integrate ሺsecଷ𝜃𝑑𝜃ሻ. නsecଷ𝜃𝑑𝜃 గ ସ ൌsec 𝜃tan 𝜃อ 𝜋 4 0 െනsec 𝜃tan2 𝜃𝑑𝜃 గ ସ නsecଷ𝜃𝑑𝜃 గ ସ ൌsec 𝜃tan 𝜃อ 𝜋 4 0 െනsec 𝜃ሺsec2 𝜃െ1ሻ𝑑𝜃 గ ସ නsecଷ𝜃𝑑𝜃 గ ସ ൌsec 𝜃tan 𝜃อ 𝜋 4 0 െනsec3 𝜃𝑑𝜃 గ ସ නsec 𝜃𝑑𝜃 గ ସ Next, notice that ൬ secଷ𝜃𝑑𝜃 ഏ ర ൰ is on both sides of the equation. So, add it to both sides: 2 නsecଷ𝜃𝑑𝜃 గ ସ ൌsec 𝜃tan 𝜃อ 𝜋 4 0 නsec 𝜃𝑑𝜃 గ ସ 2 නsecଷ𝜃𝑑𝜃 గ ସ ൌsec 𝜃tan 𝜃อ 𝜋 4 0 ln|sec 𝜃tan 𝜃| อ 𝜋 4 0 2 නsecଷ𝜃𝑑𝜃 గ ସ ൌቀsec 𝜋 4 tan 𝜋 4 െsec 0 tan 0ቁቀln ቚsec 𝜋 4 tan 𝜋 4ቚെln|sec 0 tan 0|ቁ 2 නsecଷ𝜃𝑑𝜃 గ ସ ൌ൫√2 ∙1 െ1 ∙0൯ൣ ln൫√2 1൯െlnሺ1 0ሻ ൧ 2 නsecଷ𝜃𝑑𝜃 గ ସ ൌ√2 ln൫√2 1൯ Finally, divide both sides by 2: නඥ𝒙𝟐𝟏 𝟏 𝟎 𝒅𝒙 ൌ න𝐬𝐞𝐜𝟑𝜽𝒅𝜽 𝝅 𝟒 𝟎 ൌ √𝟐𝐥𝐧൫√𝟐𝟏൯ 𝟐 ~ 𝟏. 𝟏𝟒𝟕𝟕𝟗 Let: 𝑢ൌsec 𝜃 𝑣ൌtan 𝜃 𝑑𝑢ൌsec 𝜃tan 𝜃𝑑𝜃 𝑑𝑣ൌsecଶ𝜃𝑑𝜃 Version 5.6 Page 73 of 242 April 8, 2023 Chapter 5 Techniques of Integration Γሺ1ሻ ൌනሺ𝑒ି௧ሻ ஶ 𝑑𝑡 ൌ െ𝑒ି௧ቚ∞ 0 ൌെሺ0 െ1ሻ Gamma Function The Gamma Function is defined by the following definite integral: Γሺ𝑥ሻൌනሺ𝑡௫ିଵ𝑒ି௧ሻ ஶ 𝑑𝑡 In this context, 𝑥 is a constant and 𝑡 is the variable in the integrand. Using integration by parts: Γሺ𝑥1ሻൌනሺ𝑡௫𝑒ି௧ሻ ஶ 𝑑𝑡 ൌെ𝑡௫𝑒ି௧ቚ∞ 0 െනሺെ𝑥 𝑡𝑥െ1𝑒െ𝑡ሻ ஶ 𝑑𝑡 ൌቂlim ௧ →ஶሺെ𝑡௫𝑒ି௧ሻെ0ቃ𝑥නሺ 𝑡𝑥െ1𝑒െ𝑡ሻ ஶ 𝑑𝑡 ൌሾ0ሿ𝑥 Γሺ𝑥ሻ So, we obtain one of the key properties of the Gamma Function: Next, let’s compute: Γሺ1ሻൌ ሺ𝑒ି௧ሻ ஶ 𝑑𝑡 Now for something especially cool. Based on these two results, we have the following: Γሺ1ሻ ൌ 1 Γሺ2ሻ ൌ 1 ∙Γሺ1ሻ ൌ 1 ∙1 ൌ 1 ൌ 1! Γሺ3ሻ ൌ 2 ∙Γሺ2ሻ ൌ 2 ∙1 ൌ 2 ൌ 2! Γሺ4ሻ ൌ 3 ∙Γሺ3ሻ ൌ 3 ∙2 ൌ 6 ൌ 3! Γሺ5ሻ ൌ 4 ∙Γሺ4ሻ ൌ 4 ∙6 ൌ 24 ൌ 4! … Let: 𝑢ൌ𝑡௫ 𝑣ൌെ𝑒ି௧ 𝑑𝑢ൌ𝑥𝑡௫ିଵ 𝑑𝑡 𝑑𝑣ൌ𝑒ି௧𝑑𝑡 𝚪ሺ𝒙𝟏ሻൌ𝒙 𝚪ሺ𝒙ሻ 𝚪ሺ𝟏ሻൌ𝟏 𝚪ሺ𝒙𝟏ሻൌ𝒙! Version 5.6 Page 74 of 242 April 8, 2023 Chapter 5 Techniques of Integration Graph from: mathworld.wolfram.com/GammaFunction.html Using the Gamma Function to Solve a Definite Integral Example 5.16: න𝑒ି௫య𝑑𝑥 ஶ Let: 𝑡ൌ𝑥ଷ, so 𝑥ൌ𝑡 ଵଷ ൗ, 𝑑𝑥ൌ ଵ ଷ𝑡 ିଶଷ ൗ 𝑑𝑡 The limits of integration do not change. Then: න𝑒ି௫య𝑑𝑥 ஶ ൌන 𝑒ି௧∙1 3 𝑡 ିଶଷ ൗ 𝑑𝑡 ஶ ൌ1 3 න𝑡 ିଶଷ ൗ 𝑒ି௧𝑑𝑡 ஶ Compare this to the Gamma Function: Γሺ𝑥ሻൌන𝑡௫ିଵ ஶ 𝑒ି௧𝑑𝑡 If: 𝑥ൌ ଵ ଷ, we get: න𝑒ି௫య𝑑𝑥 ஶ ൌ1 3 න𝑡 ିଶଷ ൗ 𝑒ି௧𝑑𝑡 ஶ ൌ1 3 න𝑡൫ଵଷ ൗିଵ൯ 𝑒ି௧𝑑𝑡 ஶ ൌ1 3 Γ ൬1 3൰ ൌ 𝚪൬𝟒 𝟑൰ ~ 𝟎. 𝟖𝟗𝟐𝟗𝟖 Some values of 𝚪ሺ𝒙ሻ: Γሺ1ሻൌ0 Γ ൬1 2൰ൌ√𝜋 Γሺ2ሻൌ1 Γ ൬3 2൰ൌ√𝜋 2 Γሺ3ሻൌ2 Γ ൬5 2൰ൌ3√𝜋 4 Γሺ4ሻൌ6 Γ ൬7 2൰ൌ15√𝜋 8 Γ ൬1 3൰ ~ 2.67894 Note: There are a number of online Gamma Function calculators that provide values of Γሺ𝑥ሻ. The following properties of the Gamma Function are of particular interest: 𝚪ሺ𝒙𝟏ሻൌ𝒙 ! for integer values of 𝑥 𝚪ሺ𝒙𝟏ሻൌ𝒙∙𝚪ሺ𝒙ሻ for values of 𝑥 where Γሺ𝑥ሻ exists 𝚪ቀ 𝟏 𝟐ቁൌ√𝝅 𝚪ሺ𝒙ሻ∙𝚪ሺ𝟏െ𝒙ሻൌ 𝝅 𝐬𝐢𝐧 ሺ𝝅𝒙ሻ for ሼ0 ൏𝑥൏1ሽ 𝚪ሺ𝒙ሻ്𝟎 for any value of 𝑥 𝚪ሺ𝒙ሻ does not exist for 𝑥ൌ0, nor for negative integer values of 𝑥. Version 5.6 Page 75 of 242 April 8, 2023 Chapter 5 Techniques of Integration Beta Function The Beta Function is defined by the following equivalent definite integrals: Βሺ𝑥, 𝑦ሻൌන𝑡௫ିଵ ଵ ሺ1 െ𝑡ሻ௬ିଵ𝑑𝑡 Βሺ𝑥, 𝑦ሻൌන 𝑡௫ିଵ ሺ1 𝑡ሻ௫ା௬ ஶ 𝑑𝑡 Relation between Beta and Gamma Function Βሺ𝑥, 𝑦ሻൌΒሺ𝑦, 𝑥ሻ That is, the Beta function is symmetric. Βሺ𝑥, 𝑦ሻൌΓሺ𝑥ሻΓሺ𝑦ሻ Γሺ𝑥𝑦ሻൌሺ𝑥െ1ሻ! ሺ𝑦െ1ሻ!
ሺ𝑥𝑦െ1ሻ! Note: mathworld.wolfram.com/BetaFunction.html lists many other properties of the Beta Function. Example 5.17: න𝑡ସ ଵ ሺ1 െ𝑡ሻ𝑑𝑡ൌΒሺ5, 7ሻൌΓሺ5ሻΓሺ7ሻ Γሺ5 7ሻൌ4! ∙6!
11! ൌ 𝟏 𝟐𝟑𝟏𝟎 Trigonometric Form Rewrite the Beta Function integral with dummy variable 𝑢 instead of 𝑡. Βሺ𝑥, 𝑦ሻൌන𝑢௫ିଵ ଵ ሺ1 െ𝑢ሻ௬ିଵ𝑑𝑢 Substitute: 𝑢ൌsinଶ𝑡, so 𝑑𝑢ൌ2 sin 𝑡cos 𝑡𝑑𝑡, 𝑢หଵ ⇒ 𝑡ቚ గଶ ൗ to get: Βሺ𝑥, 𝑦ሻൌ2 න sinଶ௫ିଵሺ𝑡ሻ గଶ ൗ cosଶ௬ିଵሺ𝑡ሻ𝑑𝑡 Example 5.18: න sinଽሺ𝑡ሻ గଶ ൗ cosହሺ𝑡ሻ𝑑𝑡ൌ1 2 Βሺ5, 3ሻൌ1 2 ∙Γሺ5ሻΓሺ3ሻ Γሺ5 3ሻൌ4! ∙2!
2 ∙7! ൌ 𝟏 𝟐𝟏𝟎 Note, in this example: 2𝑥െ1 ൌ9, so 𝑥ൌ5, and 2𝑦െ1 ൌ5, so 𝑦ൌ3. Very useful! Very useful! Version 5.6 Page 76 of 242 April 8, 2023 Chapter 5 Techniques of Integration Example 5.19: Find the value of: න √tan 𝑥 ሺcos 𝑥sin 𝑥ሻଶ గଶ ൗ 𝑑𝑥ൌන √tan 𝑥 ሺcosଶ𝑥ሻሺ1 tan 𝑥ሻଶ గଶ ൗ 𝑑𝑥 Let: 𝑡ൌtan 𝑥, so 𝑑𝑡ൌsecଶ𝑥ൌ 1 cosଶ𝑥𝑑𝑥, 𝑥ቤ 𝜋2 ൗ 0 ⇒ 𝑡ቚ∞ 0 න √tan 𝑥 ሺcos 𝑥sin 𝑥ሻଶ గଶ ൗ 𝑑𝑥ൌන 𝑡 ଵଶ ൗ ሺ1 𝑡ሻଶ ஶ 𝑑𝑡 This integral is now in the second Beta Function form shown above: Βሺ𝑥, 𝑦ሻൌන 𝑡௫ିଵ ሺ1 𝑡ሻ௫ା௬ ஶ 𝑑𝑡, 𝑥െ1 ൌ1 2 , 𝑥𝑦ൌ2 ⇒ 𝑥ൌ3 2 , 𝑦ൌ1 2 න √tan 𝑥 ሺcos 𝑥sin 𝑥ሻଶ గଶ ൗ 𝑑𝑥ൌන 𝑡൫ଷଶ ൗିଵ൯ ሺ1 𝑡ሻ൫ଷଶ ൗାଵଶ ൗ൯ ஶ 𝑑𝑡 ൌΒ ൬3 2 , 1 2൰ൌ Γ ቀ3 2ቁΓ ቀ1 2ቁ Γ ቀ3 2 1 2ቁ ൌ √𝜋 2 ∙√𝜋 1!
ൌ𝝅 𝟐 Version 5.6 Page 77 of 242 April 8, 2023 Chapter 5 Techniques of Integration Impossible Integrals Some expressions are impossible to integrate using elementary methods. Examples are provided below. Error Function This integral may be encountered in exercises related to the normal probability distribution. It is important enough that tables of value associated with its definite integral form have been developed. erfሺ𝑥ሻൌන𝑒ି௧మ𝑑𝑡 ௫ Other Functions with Tables of Values A number of other integrals are important enough to have tables of values developed for them: Function Name Indefinite Form Definite Form Logarithmic Integral න1 ln 𝑥𝑑𝑥 𝑙𝑖ሺ𝑥ሻൌන 1 ln 𝑡𝑑𝑡 ௫ Sine Integral නsin 𝑥 𝑥 𝑑𝑥 𝑆𝑖ሺ𝑥ሻൌනsin 𝑡 𝑡 𝑑𝑡 ௫ Cosine Integral නcos 𝑥 𝑥 𝑑𝑥 𝐶𝑖ሺ𝑥ሻൌെන cos 𝑡 𝑡 𝑑𝑡 ஶ ௫ Exponential Integral න𝑒ି௫ 𝑥𝑑𝑥 𝐸𝑖ሺ𝑥ሻൌെන 𝑒ି௧ 𝑡𝑑𝑡 ஶ ି௫ Other Impossible Integrals නsin ൬1 𝑥൰𝑑𝑥, නcos ൬1 𝑥൰𝑑𝑥, නtan ൬1 𝑥൰𝑑𝑥, නsin൫√𝑥൯𝑑𝑥, න√𝑥sin 𝑥𝑑𝑥, නsinሺ𝑥ଶሻ𝑑𝑥 න𝑒൫௫మ൯𝑑𝑥, න𝑒൫ଵ௫ ൗ൯𝑑𝑥, න𝑒௫ 𝑥𝑑𝑥, නlnሺln 𝑥ሻ𝑑𝑥, නlnሺsin 𝑥ሻ𝑑𝑥, න𝑥 ln 𝑥𝑑𝑥 Many more functions that cannot be integrated using elementary methods can be found at: න𝑒ି௫మ𝑑𝑥 Version 5.6 Page 78 of 242 April 8, 2023 Chapter 6 Hyperbolic Functions Hyperbolic Functions Definitions ܛܑܖܐ࢞ൌࢋ࢞െࢋି࢞ ܜ܉ܖܐ࢞ൌࢋ࢞െࢋି࢞ ࢋ࢞ࢋି࢞ ܛ܍܋ܐ࢞ൌ ࢋ࢞ࢋି࢞ ܋ܗܛܐ࢞ൌࢋ࢞ࢋି࢞ ܋ܗܜܐ࢞ൌࢋ࢞ࢋି࢞ ࢋ࢞െࢋି࢞ ܋ܛ܋ܐ࢞ൌ ࢋ࢞െࢋି࢞ Geometric Representation The illustration at right provides a geometric representation of a value "z" and its hyperbolic function values relative to the unit hyperbola. The hyperbolic cosine "ݕൌcoshሺݔሻ", is the equation of the Catenary, the shape of hanging chain that is supported at both ends. Many of the properties of hyperbolic functions bear a striking resemblance to the corresponding properties of trigonometric functions (see next page). Graphs of Hyperbolic Functions Version 5.6 Page 79 of 242 April 8, 2023 Chapter 6 Hyperbolic Functions Hyperbolic Function Identities Comparison of Trigonometric and Hyperbolic Identities Hyperbolic Function Identity Trigonometric Function Identity sinhሺെݔሻൌെsinh ݔ sinሺെݔሻൌെsin ݔ coshሺെݔሻൌcosh ݔ cosሺെݔሻൌcos ݔ tanhሺെݔሻൌെtanh ݔ tanሺെݔሻൌെtan ݔ coshଶݔെsinhଶݔൌ1 sinଶݔcosଶݔൌ1 sechଶݔൌ1 െtanhଶݔ secଶݔൌ1 tanଶݔ cschଶݔൌcothଶݔെ1 cscଶݔൌ1 cotଶݔ sinhሺݔݕሻൌsinh ݔcosh ݕcosh ݔsinh ݕ sinሺݔݕሻൌsin ݔcos ݕcos ݔsin ݕ sinhሺݔെݕሻൌsinh ݔcosh ݕെcosh ݔsinh ݕ sinሺݔെݕሻൌsin ݔcos ݕെcos ݔsin ݕ sinh 2ݔൌ2 sinh ݔcosh ݔ sin 2ݔൌ2 sin ݔcos ݔ coshሺݔݕሻൌcosh ݔcosh ݕsinh ݔsinh ݕ cosሺݔݕሻൌcos ݔcos ݕെsin ݔsin ݕ coshሺݔെݕሻൌcosh ݔcosh ݕെsinh ݔsinh ݕ cosሺݔെݕሻൌcos ݔcos ݕsin ݔsin ݕ cosh 2ݔൌcoshଶݔsinhଶݔ cos 2ݔൌcosଶݔെsinଶݔ tanhሺݔݕሻൌtanh ݔtanh ݕ 1 tanh ݔtanh ݕ tanሺݔݕሻൌtan ݔtan ݕ 1 െtan ݔtan ݕ tanhሺݔെݕሻൌtanh ݔെtanh ݕ 1 െtanh ݔtanh ݕ tanሺݔെݕሻൌtan ݔെtan ݕ 1 tan ݔtan ݕ sinhଶݔൌെ1 cosh 2ݔ 2 sinଶݔൌ1 െcos 2ݔ 2 coshଶݔൌ1 cosh 2ݔ 2 cosଶݔൌ1 cos 2ݔ 2 Version 5.6 Page 80 of 242 April 8, 2023 Chapter 6 Hyperbolic Functions Hyperbolic Function Identities Relationship between Trigonometric and Hyperbolic Functions sinh ݔൌെ݅sinሺ݅ݔሻ cosh ݔൌcosሺ݅ݔሻ tanh ݔൌsinh ݔ cosh ݔൌെ݅tanሺ݅ݔሻ coth ݔൌcosh ݔ sinh ݔൌ݅cotሺ݅ݔሻ sech ݔൌ 1 cosh ݔൌsecሺ݅ݔሻ csch ݔൌ 1 sinh ݔൌ݅cscሺ݅ݔሻ Series Expansions Appendix G provides series expansions for the trigonometric functions sin ݔ and cos ݔ. Those are repeated here, along with the series expansions for the corresponding hyperbolic functions sinh ݔ and cosh ݔ. sin ݔൌݔെݔଷ 3! ݔହ 5! െݔ 7! ⋯ cos ݔൌ1 െݔଶ 2! ݔସ 4! െݔ 6! ⋯ sinh ݔൌݔݔଷ 3! ݔହ 5! ݔ 7! ⋯ cosh ݔൌ1 ݔଶ 2! ݔସ 4! ݔ 6! ⋯ It is possible to develop series expansions for the other four hyperbolic functions, but they involve the more esoteric Bernoulli numbers and Euler numbers. Instead, the student may wish to develop values for the other four hyperbolic functions from the expansions of sinh ݔ and cosh ݔ. Example 6.1: tanh ݔൌ ୱ୧୬୦௫ ୡ୭ୱ୦௫ ൌ ௫ ା ೣయ య! ା ೣఱ ఱ! ା ೣళ ళ! ା … ଵ ା ೣమ మ! ା ೣర ర! ା ೣల ల! ା … From these two relationships, the other four may be determined. Version 5.6 Page 81 of 242 April 8, 2023 Chapter 6 Hyperbolic Functions Inverse Hyperbolic Functions Logarithmic Forms of Inverse Hyperbolic Functions Principal Values Function Domain Function Range sinhିଵݔൌln ቀݔඥݔଶ1ቁ ሺെ∞, ∞ሻ ሺെ∞, ∞ሻ coshିଵݔൌln ቀݔඥݔଶെ1ቁ ሾ1, ∞ሻ ሾ0, ∞ሻ tanhିଵݔൌ1 2 ln ൬1 ݔ 1 െݔ൰ ሺെ1, 1ሻ ሺെ∞, ∞ሻ cothିଵݔൌtanhିଵ൬1 ݔ൰ൌ1 2 ln ൬ݔ1 ݔെ1൰ ሺെ∞, െ1ሻ∪ሺ1, ∞ሻ ሺെ∞, ∞ሻ sechିଵݔൌcoshିଵ൬1 ݔ൰ൌln ቆ1 √1 െݔଶ ݔ ቇ ሺ0, 1ሿ ሾ0, ∞ሻ cschିଵݔൌsinhିଵ൬1 ݔ൰ൌln ቆ1 ݔ√1 ݔଶ |ݔ| ቇ ሺെ∞, ∞ሻ ሺെ∞, ∞ሻ Graphs of Inverse Hyperbolic Functions Version 5.6 Page 82 of 242 April 8, 2023 Chapter 6 Hyperbolic Functions Graphs of Hyperbolic Functions and Their Inverses Version 5.6 Page 83 of 242 April 8, 2023 Chapter 6 Hyperbolic Functions Derivatives of Hyperbolic Functions and Their Inverses Hyperbolic Functions ݀ ݀ݔsinh ݔൌ cosh ݔ ݀ ݀ݔsinh ݑൌcosh ݑ∙݀ݑ ݀ݔ ݀ ݀ݔcosh ݔൌ sinh ݔ ݀ ݀ݔcosh ݑൌsinh ݑ∙݀ݑ ݀ݔ ݀ ݀ݔtanh ݔൌ sechଶݔ ݀ ݀ݔtanh ݑൌsechଶݑ∙݀ݑ ݀ݔ ݀ ݀ݔcoth ݔൌെcschଶݔ ݀ ݀ݔcoth ݑൌെcschଶݑ∙݀ݑ ݀ݔ ݀ ݀ݔsech ݔൌ െsech ݔ tanh ݔ ݀ ݀ݔsech ݑൌെsech ݑtanh ݑ ∙ ݀ݑ ݀ݔ ݀ ݀ݔcsch ݔൌെcsch ݔcoth ݔ ݀ ݀ݔcsch ݑൌെcsch ݑcoth ݑ ∙ ݀ݑ ݀ݔ Inverse Hyperbolic Functions ݀ ݀ݔsinhିଵݔൌ 1 √ݔଶ1 ݀ ݀ݔsinhିଵݑൌ 1 √ݑଶ1 ∙ ݀ݑ ݀ݔ ݀ ݀ݔcoshିଵݔൌ 1 √ݔଶെ1 ݀ ݀ݔcoshିଵݑൌ 1 √ݑଶെ1 ∙ ݀ݑ ݀ݔ ݀ ݀ݔtanhିଵݔൌ 1 1 െݔଶ ݀ ݀ݔtanhିଵݑൌ 1 1 െݑଶ∙݀ݑ ݀ݔ ݀ ݀ݔcothିଵݔൌ 1 1 െݔଶ ݀ ݀ݔcothିଵݑൌ 1 1 െݑଶ∙݀ݑ ݀ݔ ݀ ݀ݔsechିଵݔൌ െ1 ݔ √1 െݔଶ ݀ ݀ݔsechିଵݑൌ െ1 ݑ√1 െݑଶ ∙ ݀ݑ ݀ݔ ݀ ݀ݔcschିଵݔൌ െ1 |ݔ| √1 ݔଶ ݀ ݀ݔcschିଵݑൌ െ1 |ݑ| √1 ݑଶ ∙ ݀ݑ ݀ݔ Version 5.6 Page 84 of 242 April 8, 2023 Chapter 6 Hyperbolic Functions Integrals of Hyperbolic Functions and Their Inverses Hyperbolic Functions නsinh ݑ݀ݑൌcosh ݑ නcosh ݑ݀ݑൌsinh ݑ නtanh ݑ݀ݑൌlnሺcosh ݑሻ නsechଶݑ݀ݑൌtanh ݑ නcoth ݑ݀ݔൌln|sinh ݑ| නcschଶݑ݀ݑൌെcoth ݑ නsech ݑ݀ݑൌ2 tanିଵሺ݁௨ሻ නsech ݑtanh ݑ݀ݑൌെsech ݑ නcsch ݑ݀ݑൌln ቚtanh ቀݑ 2ቁቚ නcsch ݑcoth ݑ݀ݑൌെcoth ݑ Inverse Hyperbolic Functions නsinhିଵݑ݀ݑൌݑsinhିଵݑെඥݑଶ1 නcoshିଵݑ݀ݑൌݑcoshିଵݑඥݑଶെ1 නtanhିଵݑ݀ݑൌݑtanhିଵݑ1 2 lnሺ1 െݑଶሻ නcothିଵݑ݀ݑൌݑcothିଵݑ1 2 lnሺݑଶെ1ሻ නsechିଵݑ݀ݑൌݑsechିଵݑsinିଵݑ නcschିଵݑ݀ݑൌݑcschିଵݑsinhିଵݑ if ݑ0 ൌݑcschିଵݑെsinhିଵݑ if ݑ൏0 Note: the integration rules presented in this chapter omit the “ ܥ” term that must be added to all indefinite integrals in order to save space and avoid clutter. Please remember to add the “ ܥ” term on all work you perform with indefinite integrals. Be careful with these integrals. A couple of them have inverse trigonometric functions in the formulas. These are highlighted in blue. Version 5.6 Page 85 of 242 April 8, 2023 Chapter 6 Hyperbolic Functions Other Integrals Relating to Hyperbolic Functions න 1 √ݔଶܽଶ ݀ݔൌsinhିଵݔ ܽ න 1 √ݔଶܽଶ݀ݔൌln ቀݔඥݔଶܽଶቁ න 1 √ݔଶെܽଶ ݀ݔൌcoshିଵݔ ܽ න 1 √ݔଶെܽଶ݀ݔൌln ቚݔඥݔଶെܽଶቚ න 1 ܽଶെݔଶ ݀ݔቚ ܽݔൌ1 ܽtanhିଵݔ ܽ න 1 ܽଶെݔଶ ݀ݔൌ1 2ܽln ฬܽݔ ܽെݔฬ න 1 ܽଶെݔଶ ݀ݔቚ ݔܽൌ1 ܽcothିଵݔ ܽ න 1 ݔ √ܽଶെݔଶ ݀ݔൌെ1 ܽsechିଵ|ݔ| ܽ න 1 ݔ√ܽଶെݔଶ݀ݔൌെ1 ܽln ቆܽ√ܽଶെݔଶ |ݔ| ቇ න 1 ݔ √ܽଶݔଶ݀ݔൌെ1 ܽcschିଵ|ݔ| ܽ න 1 ݔ√ܽଶݔଶ݀ݑൌെ1 ܽln ቆܽ√ܽଶݔଶ |ݔ| ቇ Note: The results above are shown without their constant term (ܥ). When more than one result is shown, the results may differ by a constant, meaning that the constants in the formulas may be different. Example 6.2: From the first row above: න 1 √ݔଶܽଶ ݀ݔൌsinhିଵݔ ܽܥଵ and න 1 √ݔଶܽଶ ݀ݔൌln ቀݔඥݔଶܽଶቁܥଶ From earlier in this chapter, we know that the logarithmic form of sinhିଵݔ is: sinhିଵݔൌln ቀݔඥݔଶ1ቁ Then: න 1 √ݔଶܽଶ ݀ݔൌsinhିଵݔ ܽܥଵൌln ቌݔ ܽඨቀݔ ܽቁ ଶ 1ቍܥଵ ൌln ቆݔ√ݔଶܽଶ ܽ ቇܥଵൌln ቀݔඥݔଶܽଶቁെln ܽܥଵ So we see that ܥଶൌെln ܽܥଵ and so the formulas both work, but have different constant terms. Version 5.6 Page 86 of 242 April 8, 2023 Chapter 7 Definite Integrals Definite Integrals as Riemann Sums Riemann Sum A Riemann Sum is the sum of the areas of a set of rectangles that can be used to approximate the area under the curve over a closed interval. Consider a closed interval ሾ𝑎, 𝑏ሿ on 𝑥 that is partitioned into 𝑛 sub‐intervals of lengths ∆𝑥ଵ, ∆𝑥ଶ, ∆𝑥ଷ, … ∆𝑥. Let 𝑥 ∗ be any value of 𝑥 on the 𝑖‐th sub‐interval. Then, the Riemann Sum is given by: A graphical representation of a Riemann sum on the interval ሾ2, 5ሿ is provided at right. Note that the area under the curve from 𝑥ൌ𝑎 to 𝑥ൌ𝑏 is: The largest ∆𝑥 is called the mesh size of the partition. A typical Riemann Sum is developed with all ∆𝑥 the same (i.e., constant mesh size), but this is not required. The resulting definite integral, 𝑓ሺ𝑥ሻ 𝑑𝑥 is called the Riemann Integral of 𝑓ሺ𝑥ሻ on the interval ሾ𝑎, 𝑏ሿ. With constant mesh size, the Riemann Integral of 𝑓ሺ𝑥ሻ on the interval ሾ𝑎, 𝑏ሿ can be expressed: lim 𝑛 →∞𝑓ሺ𝑥𝑖 ∗ሻ∙∆𝑥 𝑖ୀଵ ൌන𝑓ሺ𝑥ሻ 𝑑𝑥 where, ∆𝑥ൌ interval length number of subintervals . Note: the term “under the curve” is generally used to refer to the area between a curve and the 𝑥‐axis. Some of this area may not be strictly under the curve if the curve is below the 𝑥‐axis. 𝑆ൌ𝑓ሺ𝑥 ∗ሻ∙ ୀଵ ∆𝑥 lim ୫ୟ୶ ∆௫ →𝑓ሺ𝑥 ∗ሻ∙∆𝑥 ୀଵ ൌන𝑓ሺ𝑥ሻ 𝑑𝑥 Version 5.6 Page 87 of 242 April 8, 2023 Chapter 7 Definite Integrals Methods for Calculating Riemann Sums Riemann Sums are often calculated using equal sub‐intervals over the interval specified. Below are examples of 4 commonly used approaches. Although some methods provide better answers than others under various conditions, the limits under each method as max ∆𝑥 →0 are the same, and are equal to the integral they are intended to approximate. Example 7.1: Given: 𝑓ሺ𝑥ሻൌ 8 2 2 x x dx . Using 3 n , estimate the area under the curve. ∆𝒙ൌ 𝟖ି𝟐 𝟑ൌ𝟐. The three intervals in question are: ሾ𝟐, 𝟒ሿ, ሾ𝟒, 𝟔ሿ, ሾ𝟔, 𝟖ሿ. Then, 𝐴ൌ𝑓ሺ𝑥ሻ ∙ ∆𝒙𝒊ൌ∆𝒙 ∙𝑓ሺ𝑥ሻ ଷ ୀଵ ଷ ୀଵ Left‐Endpoint Rectangles (use rectangles with left endpoints on the curve) 𝑳ൌ𝟐∙ሾ𝑓ሺ2ሻ𝑓ሺ4ሻ𝑓ሺ6ሻሿൌ𝟐∙ሺ2 12 30ሻൌ𝟖𝟖 units2 Right‐Endpoint Rectangles (use rectangles with right endpoints on the curve) 𝑹ൌ𝟐∙ሾ𝑓ሺ4ሻ𝑓ሺ6ሻ𝑓ሺ8ሻሿൌ𝟐∙ሺ12 30 56ሻൌ𝟏𝟗𝟔 units2 Trapezoid Rule (use trapezoids with all endpoints on the curve) 𝑻ൌ 𝑳 ା 𝑹 𝟐 ൌ 𝟖𝟖 ା 𝟏𝟗𝟔 𝟐 ൌ𝟏𝟒𝟐 units2 Midpoint Rule (use rectangles with midpoints on the curve) 𝑴ൌ𝟐∙ሾ𝑓ሺ3ሻ𝑓ሺ5ሻ𝑓ሺ7ሻሿൌ𝟐∙ሺ6 20 42ሻൌ𝟏𝟑𝟔 units2 Note: the actual value of the area under the curve is: නሺ𝑥ଶെ𝑥ሻ ଼ ଶ 𝑑𝑥ൌ138 Version 5.6 Page 88 of 242 April 8, 2023 Chapter 7 Definite Integrals Riemann Sums of Tables Using the TI‐84 Consider the following problem: Use a right Riemann Sum to approximate the area under the curve on the interval ሾ2, 13ሿ. 𝑥 2 4 5.5 8 9.2 10.3 11.8 13 𝑓ሺ𝑥ሻ 4 ‐1 ‐2 1 5 11 13 9 There are eight columns and, therefore, seven intervals in the table. The formula for the required Riemann Sum, then, is: 𝐴ൌ𝑓ሺ𝑥ሻ ∙ ∆𝑥 ୀଵ where the ∆𝑥 are the widths of the intervals and the 𝑓ሺ𝑥ሻ are the values of the function at the right side of each interval (because we are calculating a right Riemann Sum). The student can calculate this directly as: 𝐴ൌሺെ1ሻሺ4 െ2ሻሺെ2ሻሺ5.5 െ4ሻ1ሺ8 െ.5.5ሻ5ሺ9.2 െ8ሻ11ሺ10.3 െ9.2ሻ13ሺ11.8 െ10.3ሻ9ሺ13 െ11.8ሻൌ𝟒𝟓. 𝟗 Alternatively, the student can use the TI‐84 calculator as follows: Step 1: STAT – EDIT – L1 – enter the values of ∆𝑥 in the column for L1. Step 2: STAT – EDIT – L2 – enter the appropriate values of𝑓ሺ𝑥ሻ in the column for L2. Step 3: 2ND – QUIT – this will take you back to the TI‐84’s home screen. Step 3: L1 x L2 STO> L3 – this will put the product of columns L1 and L2 in column L3. Note that L3 will contain the areas of each of the rectangles in the Riemann Sum. Step 4: 2ND – LIST – MATH – SUM( – L3 – this will add the values in column L3, giving the value of 𝐴, which, for this problem, matches the sum of 𝟒𝟓. 𝟗 shown above. Note: entering L1, L2 or L3 requires use of the 2ND key. The student can review the contents of the lists L1, L2, and L3 using STAT – EDIT. For this problem, the display will look something like the image at right. The advantages of this are: It allows the student to check their work quickly. If the student is asked for some other kind of Riemann Sum, a portion of the required input is already in the TI‐84. Each student should use whichever method of calculating Riemann Sums works best for them. Version 5.6 Page 89 of 242 April 8, 2023 Chapter 7 Definite Integrals න𝑥ଶ ସ ଶ 𝑑𝑥 Riemann Integrals with Constant Mesh Size With constant mesh size, the Riemann Integral of 𝑓ሺ𝑥ሻ on the interval ሾ𝑎, 𝑏ሿ can be expressed: lim 𝑛 →∞𝑓ሺ𝑥 ∗ሻ∙∆𝑥 𝑖ୀଵ ൌන𝑓ሺ𝑥ሻ𝑑𝑥 where, ∆𝑥ൌ interval length number of intervals . A formula for a right‐endpoint Riemann Sum with 𝑛 sub‐intervals, then, can be developed using: 𝑥𝑖 ∗ൌright endpoint ൌleft endpoint 𝑖∙∆𝑥, 𝑖ൌ1, 2, … , 𝑛 Example 7.2: Provide the exact value of by expressing it as the limit of a Riemann sum. Start with the definition of a Riemann Sum with constant mesh size (see above): න𝑓ሺ𝑥ሻ𝑑𝑥 ൌ lim 𝑛 →∞𝑓ሺ𝑥𝑖 ∗ሻ∙∆𝑥 𝑖ୀଵ Using 𝑛 sub‐intervals and a right‐endpoint Riemann Sum, ∆𝑥ൌ4 െ2 𝑛 ൌ2 𝑛 𝑥 ∗ൌ2 𝑖ቀ ଶ ቁൌ2 ଶ , 𝑖ൌ1, 2, … , 𝑛 𝑓ሺ𝑥 ∗ሻൌ𝑓൬2 2𝑖 𝑛൰ ൌ ൬2 2𝑖 𝑛൰ ଶ ൌ 4 8𝑖 𝑛4𝑖ଶ 𝑛ଶ So, න𝑥ଶ𝑑𝑥 ସ ଶ ൌ lim →ஶቈ൬2 2𝑖 𝑛൰ ଶ ∙൬2 𝑛൰ ୀଵ ൌ 𝐥𝐢𝐦 𝒏→ஶቆ𝟒𝟖𝒊 𝒏𝟒𝒊𝟐 𝒏𝟐ቇ∙൬𝟐 𝒏൰ 𝒏 𝒊ୀ𝟏 Version 5.6 Page 90 of 242 April 8, 2023 Chapter 7 Definite Integrals Riemann Sum Methods – Over‐ or Under‐Estimates Left and Right Endpoint Methods Over‐ or Under‐estimates for the Left and Right Endpoint Methods depend on whether a function is increasing or decreasing over the interval used. Increasing Decreasing Method Over‐ or Under‐ Estimate Method Over‐ or Under‐ Estimate Left Endpoint Under Left Endpoint Over Right Endpoint Over Right Endpoint Under Midpoint and Trapezoid Methods Over‐ or Under‐estimates for the Midpoint and Trapezoidal Methods depend on whether a function is concave up or concave down over the interval used. Concave Up Concave Down Method Over‐ or Under‐ Estimate Method Over‐ or Under‐ Estimate Midpoint Under Midpoint Over Trapezoidal Over Trapezoidal Under Version 5.6 Page 91 of 242 April 8, 2023 Chapter 7 Definite Integrals Rules of Definite Integration First Fundamental Theorem of Calculus If 𝑓ሺ𝑥ሻ is a continuous function on ሾ𝑎, 𝑏ሿ, and 𝐹ሺ𝑥ሻ is any antiderivative of 𝑓ሺ𝑥ሻ, then න𝑓ሺ𝑥ሻ𝑑𝑥ൌ𝐹ሺ𝑏ሻെ𝐹ሺ𝑎ሻ Second Fundamental Theorem of Calculus If 𝑓ሺ𝑥ሻ is a continuous function on ሾ𝑎, 𝑏ሿ, then for every 𝑥∈ሾ𝑎, 𝑏ሿ 𝑑 𝑑𝑥 න𝑓ሺ𝑡ሻ ௫ 𝑑𝑡ൌ𝑓ሺ𝑥ሻ Chain Rule of Definite Integration If 𝑓ሺ𝑥ሻ is a continuous function on ሾ𝑎, 𝑏ሿ, then for every 𝑥∈ሾ𝑎, 𝑏ሿ 𝑑 𝑑𝑥 න 𝑓ሺ𝑡ሻ ሺ௫ሻ 𝑑𝑡ൌ𝑓൫𝑔ሺ𝑥ሻ൯ ∙ 𝑑 𝑑𝑥𝑔ሺ𝑥ሻ Mean Value Theorem for Integrals If 𝑓ሺ𝑥ሻ is a continuous function on ሾ𝑎, 𝑏ሿ, then there is a value 𝑐∈ሾ𝑎, 𝑏ሿ, such that න𝑓ሺ𝑥ሻ 𝑑𝑥ൌሺ𝑏െ𝑎ሻ∙𝑓ሺ𝑐ሻ The value 𝑓ሺ𝑐ሻ is called the Average Value of the function 𝑓ሺ𝑥ሻ on the interval ሾ𝑎, 𝑏ሿ. A formula for the average value is: Average Value ൌ 1 𝑏െ𝑎න𝑓ሺ𝑥ሻ 𝑑𝑥 Version 5.6 Page 92 of 242 April 8, 2023 Chapter 7 Definite Integrals Average Rate of Change vs. Average Value Average Rate of Change Average Value Math Level Algebra 1 Integral Calculus Requirement 𝑓ሺ𝑥ሻ is continuous on ሾ𝑎, 𝑏ሿ 𝑓ሺ𝑥ሻ is continuous on ሾ𝑎, 𝑏ሿ Description Slope of the secant line connecting the endpoints of the curve on the interval ሾ𝑎, 𝑏ሿ Height of the rectangle with area equal to the area under the curve on the interval ሾ𝑎, 𝑏ሿ Formula 𝑓ሺ𝑏ሻെ𝑓ሺ𝑎ሻ 𝑏െ𝑎 1 𝑏െ𝑎න𝑓ሺ𝑥ሻ 𝑑𝑥 Illustration Average rate of change of 𝑓ሺ𝑥ሻൌ ଵ 𝑥ଶ over the interval ሾ1, 5ሿ. 1 6 ሺ5ሻଶെ1 6 ሺ1ሻଶ 5 െ1 ൌ1 Average value of 𝑓ሺ𝑥ሻൌ ଵ 𝑥ଶ on the interval ሾ1, 5ሿ. 1 5 െ1 න൬1 6 𝑥ଶ൰ ହ ଵ 𝑑𝑥ൌ1.722 Version 5.6 Page 93 of 242 April 8, 2023 Chapter 7 Definite Integrals Properties of Definite Integrals Same Upper and Lower Limits න𝑓ሺ𝑥ሻ 𝑑𝑥ൌ0 Reversed Limits න𝑓ሺ𝑥ሻ 𝑑𝑥 ൌ െන𝑓ሺ𝑥ሻ 𝑑𝑥 Multiplication by a Scalar න𝑘 𝑓ሺ𝑥ሻ 𝑑𝑥 ൌ 𝑘∙න𝑓ሺ𝑥ሻ 𝑑𝑥 Telescoping Limits න 𝑓ሺ𝑥ሻ 𝑑𝑥 ൌ න𝑓ሺ𝑥ሻ 𝑑𝑥 න𝑓ሺ𝑥ሻ 𝑑𝑥 Sum or Difference න ሾ𝑓ሺ𝑥ሻ𝑔ሺ𝑥ሻሿ 𝑑𝑥 ൌ න𝑓ሺ𝑥ሻ 𝑑𝑥 න𝑔ሺ𝑥ሻ 𝑑𝑥 Linear Combination න ሾ𝑘∙𝑓ሺ𝑥ሻ𝑚∙𝑔ሺ𝑥ሻሿ 𝑑𝑥 ൌ 𝑘∙න𝑓ሺ𝑥ሻ 𝑑𝑥𝑚∙න𝑔ሺ𝑥ሻ 𝑑𝑥 The integral of a linear combination of functions is the linear combination of the integrals of the functions.
If the upper and lower limits of the integral are the same, its value is zero. Reversing the limits of an integral negates its value. The integral of the product of a scalar and a function is the product of the scalar and the integral of the function. The integral over the interval ሾ𝑎, 𝑏ሿ is equal to the integral over the interval ሾ𝑎, 𝑐ሿ, plus the integral over the interval ሾ𝑐, 𝑏ሿ. The integral of a sum (or difference) of functions is the sum (or difference) of the integrals of the functions. Version 5.6 Page 94 of 242 April 8, 2023 Chapter 7 Definite Integrals Solving Definite Integrals with Directed Line Segments A common problem in elementary Calculus is to use the values of definite integrals of a given function 𝑓ሺ𝑥ሻ over two or more intervals to obtain the value of a definite integral of 𝑓ሺ𝑥ሻ over a related interval. The illustration below shows how directed line segments can be used to simplify the calculations required for this kind of problem. Example 7.3: Given that 3𝑓ሺ𝑥ሻ𝑑𝑥ൌ84 ଼ ିଷ and 5𝑓ሺ𝑥ሻ ଼ ସ 𝑑𝑥ൌെ75, find 𝑓ሺ𝑥ሻ ସ ିଷ 𝑑𝑥. Step 1: Remove any scalar multipliers by dividing the values given by the scalar multipliers. Divide: 3𝑓ሺ𝑥ሻ𝑑𝑥ൌ84 ଼ ିଷ by 3 to get 𝑓ሺ𝑥ሻ ଼ ିଷ 𝑑𝑥ൌ28. Divide: 5𝑓ሺ𝑥ሻ ଼ ସ 𝑑𝑥ൌെ75 by 5 to get 𝑓ሺ𝑥ሻ ଼ ସ 𝑑𝑥ൌെ15. Step 2: Draw directed line segments for each of the definite integrals in the problem. Label each segment with its magnitude. The starting and ending points of each segment reflect the limits in the integral. Known values are shown in blue and the target value is in green. Notice that the first segment stretches over the interval ሾെ3, 8ሿ and has magnitude 28, reflecting 𝑓ሺ𝑥ሻ ଼ ିଷ 𝑑𝑥ൌ28. The other segments are constructed similarly. We want to find the magnitude of the third (green) segment. We could subtract the second segment from the first to obtain the solution segment. Its magnitude would be: 𝑓ሺ𝑥ሻ ସ ିଷ 𝑑𝑥ൌ28 െሺെ15ሻൌ4𝟑. If we do this, we are done; we have our solution. Alternatively, we could take a more fluid approach to this problem as in Step 3. Step 3 (if desired): Reorient segments as needed so we can follow the known directed segments from the beginning to the end of the interval required for the solution (i.e., from 𝑥ൌെ3 to 𝑥ൌ4). If we reorient the middle segment so it is pointing to the left, the magnitude of the new second segment becomes 15, reflecting the fact that we are moving to the left instead of to the right. Using Calculus, this reflects the fact that 𝑓ሺ𝑥ሻ ସ ଼ 𝑑𝑥ൌെ 𝑓ሺ𝑥ሻ ଼ ସ 𝑑𝑥ൌ15. We are now able to get to 𝑥ൌ4 by following the known segments in the directions shown. Then, we simply add the magnitudes of the known segments to get our solution: 𝑓ሺ𝑥ሻ ସ ିଷ 𝑑𝑥ൌ28 15 ൌ4𝟑.
Version 5.6 Page 95 of 242 April 8, 2023 Chapter 7 Definite Integrals Definite Integrals – 𝒖‐Substitution 𝒖‐substitution may be used in the evaluation of definite integrals as well as indefinite integrals (note: using 𝑢‐substitution with indefinite integrals is covered in Chapter 5). The process with definite integrals is slightly different and may even be a bit easier. Process Following are the steps for the general solution to a definite integral using 𝑢–substitution. 1. Set a portion of the integrand equal to a new variable, e.g., 𝑢. Look to the rest of the integrand in deciding what to set equal to 𝑢. You will need to have 𝑑𝑢 in the integrand as well, if this technique is to find success. 2. Find 𝑑𝑢 in terms of 𝑑𝑥. 3. Rearrange the integrand so that the integral exists in terms of 𝑢 instead of 𝑥. 4. Perform the integration. 5. Evaluate the values of the limits of integration in terms of the new variable and substitute these into the definite integral in terms of u. 6. Evaluate the result. Note that by calculating the limits of integration in terms of the new variable, 𝑢, we are able to avoid the step where we must substitute the expression for 𝑢 back into the result of the integration. This saves time and reduces the likelihood of error in the calculation. Example 7.4: Evaluate: ଶ ௗ௫ ሺଶ௫ିଵሻమ ିଵ න 2 𝑑𝑥 ሺ2𝑥െ1ሻଶ ିଵ ൌන 1 ሺ2𝑥െ1ሻଶ 2 𝑑𝑥 ିଵ ൌන 𝑢ିଶ 𝑑𝑢 ିଵ ିଷ ൌ െ𝑢ିଵฬ െ1 െ3 ൌ െ1 𝑢ฬ െ1 െ3 ൌ െ൬ 1 െ1 െ 1 െ3൰ൌ െ൬െ1 1 3൰ൌ 𝟐 𝟑 𝑢ൌ2𝑥െ1 𝑑𝑢ൌ2 𝑑𝑥 𝑥ൌ0 ⇒𝑢ൌെ1 𝑥ൌെ1 ⇒𝑢ൌെ3 Version 5.6 Page 96 of 242 April 8, 2023 Chapter 7 Definite Integrals Example 7.5: Evaluate: sin 2𝑥 గସ ⁄ ିగ 𝑑𝑥 Example 7.6: Evaluate: tan 𝑥secଶ𝑥 𝑑𝑥 గସ ⁄ For trig functions other than sine and cosine, we need to make sure the denominators of the functions are not zero within our interval. If they are zero, the function is not continuous on the interval and so the Fundamental Theorem of Calculus does not apply. For the current problem, we need to make sure cos 𝑥്0 over the interval ቂ0, గ ସቃ in order to use the Fundamental Theorem of Calculus. Since cos 𝑥ൌ0 at 𝑥ൌቄെ గ ଶ, గ ଶቅ in this neighborhood, we are okay to proceed. න tan 𝑥secଶ𝑥 𝑑𝑥 గସ ⁄ ൌන𝑢 𝑑𝑢 ଵ ൌ1 2 𝑢ଶฬ 1 0 ൌ 1 2 ሺ12 െ02ሻൌ 𝟏 𝟐 ALTERNATIVE APPROACH: setting 𝑢ൌsec 𝑥 න tan 𝑥secଶ𝑥 𝑑𝑥 గସ ⁄ ൌන sec 𝑥 ሺsec 𝑥 tan 𝑥 𝑑𝑥ሻ గସ ⁄ ൌන 𝑢 𝑑𝑢 √2 ଵ ൌ1 2 𝑢ଶቤ√2 1 ൌ 1 2 ቀ√2 2 െ12ቁൌ 𝟏 𝟐 න sin 2𝑥 𝑑𝑥 గସ ⁄ ିగ ൌ1 2 න sin 2𝑥 2 𝑑𝑥 గସ ⁄ ିగ ൌ1 2 න sin 𝑢 𝑑𝑢 𝜋2 ൗ െ2𝜋 ൌ1 2 ሺെcos 𝑢ሻቤ 𝜋2 ⁄ െ2𝜋 ൌ െ1 2 cos 𝑢ቤ 𝜋2 ⁄ െ2𝜋ൌെ 1 2 ሺ0 െ1ሻൌ 𝟏 𝟐 𝑢ൌ2𝑥 𝑑𝑢ൌ2 𝑑𝑥 𝑥ൌ𝜋 4 ⇒ 𝑢ൌ𝜋 2 𝑥ൌെ𝜋 ⇒𝑢ൌെ2𝜋 𝑢ൌtan 𝑥 𝑑𝑢ൌsecଶ𝑥𝑑𝑥 𝑥ൌ𝜋 4 ⇒ 𝑢ൌ1 𝑥ൌ0 ⇒𝑢ൌ0 𝑢ൌsec 𝑥 𝑑𝑢ൌsec 𝑥tan 𝑥 𝑑𝑥 𝑥ൌ𝜋 4 ⇒ 𝑢ൌ√2 𝑥ൌ0 ⇒𝑢ൌ1 Version 5.6 Page 97 of 242 April 8, 2023 Chapter 7 Definite Integrals Definite Integrals – Special Techniques Sometimes it is difficult or impossible to take an antiderivative of an integrand. In such cases, it may still be possible to evaluate a definite integral, but special techniques and creativity may be required. This section presents a few techniques that the student may find helpful. Even and Odd Functions The following technique can sometimes be used to solve a definite integral that has limits that are additive inverses (i.e, െ𝑎 and 𝑎). Every function can be split into even and odd components. The even and odd components of a given function, 𝑓ሺ𝑥ሻ, are: 𝑓 ୣ୴ୣ୬ሺ𝑥ሻൌ𝑓ሺ𝑥ሻ𝑓ሺെ𝑥ሻ 2 𝑓 ୭ୢୢሺ𝑥ሻൌ𝑓ሺ𝑥ሻെ𝑓ሺെ𝑥ሻ 2 Notice that: 𝑓 ୣ୴ୣ୬ሺ𝑥ሻൌ𝑓 ୣ୴ୣ୬ሺെ𝑥ሻ, so that 𝑓 ୣ୴ୣ୬ሺ𝑥ሻ is an even function. 𝑓 ୭ୢୢሺ𝑥ሻൌെ𝑓 ୭ୢୢሺെ𝑥ሻ, so that 𝑓 ୭ୢୢሺ𝑥ሻ is an odd function. 𝑓ሺ𝑥ሻൌ𝑓 ୣ୴ୣ୬ሺ𝑥ሻ𝑓 ୭ୢୢሺ𝑥ሻ Further recall that, for an odd function with limits that are additive inverses, any negative areas “under” the curve are exactly offset by corresponding positive areas under the curve. That is: න𝑓 ୭ୢୢሺ𝑥ሻ ି 𝑑𝑥ൌ0 Additionally, for an even function with limits that are additive inverses, the area under the curve to the left of the 𝑦‐axis is the same as the area under the curve to the right of the 𝑦‐axis. That is: න𝑓 ୣ୴ୣ୬ሺ𝑥ሻ ି 𝑑𝑥ൌන𝑓 ୣ୴ୣ୬ሺ𝑥ሻ ି 𝑑𝑥 න𝑓 ୣ୴ୣ୬ሺ𝑥ሻ 𝑑𝑥ൌ2 න𝑓 ୣ୴ୣ୬ሺ𝑥ሻ 𝑑𝑥 Therefore, we have: න𝑓ሺ𝑥ሻ ି 𝑑𝑥ൌන ሾ 𝑓 ୣ୴ୣ୬ሺ𝑥ሻ𝑓 ୭ୢୢሺ𝑥ሻ ሿ ି 𝑑𝑥 ൌන 𝑓 ୣ୴ୣ୬ሺ𝑥ሻ ି 𝑑𝑥 න 𝑓 ୭ୢୢሺ𝑥ሻ ି 𝑑𝑥 And, finally, substituting from the above equations: න𝒇ሺ𝒙ሻ 𝒂 ି𝒂 𝒅𝒙 ൌ 𝟐න 𝒇𝐞𝐯𝐞𝐧ሺ𝒙ሻ 𝒂 𝟎 𝒅𝒙 Let’s look at an example of how this can be used to evaluate a difficult definite integral on the next page.
Version 5.6 Page 98 of 242 April 8, 2023 Chapter 7 Definite Integrals Example 7.7: Evaluate First, define: 𝑓ሺ𝑥ሻൌ ୡ୭ୱ௫ ଵ ା ೣ . Notice that there are no singularities for this integral. That is, there are no points between the limits (i.e., െ గ ଶ൏𝑥൏ గ ଶ) at which 𝑓ሺ𝑥ሻ does not exist. So we may proceed in a normal fashion. Next, let’s look at the even and odd components of 𝑓ሺ𝑥ሻ. 𝑓 ୣ୴ୣ୬ሺ𝑥ሻൌ𝑓ሺ𝑥ሻ𝑓ሺെ𝑥ሻ 2 ൌ1 2 cos 𝑥 1 𝑒௫cosሺെ𝑥ሻ 1 𝑒ି௫൨ Noting that cosሺെ𝑥ሻൌcos 𝑥, we get: 𝑓 ୣ୴ୣ୬ሺ𝑥ሻൌ1 2 ቂcos 𝑥 1 𝑒௫ cos 𝑥 1 𝑒ି௫ቃൌcos 𝑥 2 1 1 𝑒௫ 1 1 𝑒ି௫൨ ൌcos 𝑥 2 ቈ ሺ1 𝑒ି௫ሻሺ1 𝑒௫ሻ ሺ1 𝑒௫ሻሺ1 𝑒ି௫ሻ ൌcos 𝑥 2 2 𝑒ି௫𝑒௫ 2 𝑒ି௫𝑒௫൨ൌcos 𝑥 2 The odd component of 𝑓ሺ𝑥ሻ is (note: this work is not necessary to evaluate the integral): 𝑓 ୭ୢୢሺ𝑥ሻൌ𝑓ሺ𝑥ሻെ𝑓ሺെ𝑥ሻ 2 ൌ1 2 cos 𝑥 1 𝑒௫െcosሺെ𝑥ሻ 1 𝑒ି௫൨ ൌ1 2 ቂcos 𝑥 1 𝑒௫െ cos 𝑥 1 𝑒ି௫ቃൌcos 𝑥 2 1 1 𝑒௫െ 1 1 𝑒ି௫൨ ൌcos 𝑥 2 ቈ ሺ1 𝑒ି௫ሻെሺ1 𝑒௫ሻ ሺ1 𝑒௫ሻሺ1 𝑒ି௫ሻ ൌcos 𝑥 2 𝑒ି௫െ𝑒௫ 2 𝑒ି௫𝑒௫൨ Since the value of the odd component of the definite integral is zero, we need only evaluate the even component of the definite integral using the formula on the previous page: න ቀ 𝐜𝐨𝐬𝒙 𝟏𝒆𝒙 ቁ 𝝅 𝟐 ି 𝝅 𝟐 𝒅𝒙 ൌ 2 න ቀcos 𝑥 2 ቁ గ ଶ 𝑑𝑥 ൌ sin 𝑥ቤ 𝜋2 ൗ 0 ൌ sin ቀ𝜋 2ቁെsinሺ0ሻ ൌ 1 െ0 ൌ 𝟏 න ቀ 𝐜𝐨𝐬𝒙 𝟏𝒆𝒙ቁ 𝝅 𝟐 ି 𝝅 𝟐 𝒅𝒙 f(x) = cos(x) 1 + ex feven(x) = cos(x) 2 fodd(x) = cos(x) 2 e -x - ex 2 + e -x + ex Version 5.6 Page 99 of 242 April 8, 2023 Chapter 7 Definite Integrals Derivative of an Integral The Second Fundamental Theorem of Calculus states that if 𝑓ሺ𝑥ሻ is a continuous function on the interval ሾ𝑎, 𝑏ሿ, then for every 𝑥∈ሾ𝑎, 𝑏ሿ, ௗ ௗ௫ 𝑓ሺ𝑥ሻ𝑑𝑥 ௫ ൌ𝑓ሺ𝑥ሻ. Essentially, this is a statement that integration and differentiation are inverses. But, there is more. If the upper limit is a function of 𝑥, say 𝑢ሺ𝑥ሻ, then we must apply the chain rule to get: 𝑑 𝑑𝑥න𝑓ሺ𝑡ሻ𝑑𝑡 ௨ ൌ 𝑓ሺ𝑢ሻ∙𝑑𝑢 𝑑𝑥 Note that 𝑎 is a constant and 𝑢 is a function in 𝑥. Also note that the value of the constant 𝑎 is irrelevant in this expression, as long as 𝑓ሺ𝑥ሻ is continuous on the required interval. If both of the limits in the integral are functions of 𝑥, we can take advantage of a property of definite integrals to develop a solution. Let 𝑢 and 𝑣 both be functions in 𝑥, and let 𝑎 be an arbitrary constant in the interval where 𝑓ሺ𝑥ሻ is continuous. Then, 𝑑 𝑑𝑥න𝑓ሺ𝑡ሻ𝑑𝑡 ௨ ௩ ൌ 𝑑 𝑑𝑥න𝑓ሺ𝑡ሻ𝑑𝑡 ௨ െ 𝑑 𝑑𝑥න𝑓ሺ𝑡ሻ𝑑𝑡 ௩ So, 𝑑 𝑑𝑥න𝑓ሺ𝑡ሻ𝑑𝑡 ௨ ௩ ൌ 𝑓ሺ𝑢ሻ∙𝑑𝑢 𝑑𝑥 െ 𝑓ሺ𝑣ሻ∙𝑑𝑣 𝑑𝑥 Example 7.8: 𝑑 𝑑𝑥න 𝑡ଶ𝑑𝑡 ଷୱ୧୬ଶ௫ ൌ ሺ3 sin 2𝑥ሻଶ∙ሺ6 cos 2𝑥ሻ ൌ 𝟓𝟒 𝐬𝐢𝐧𝟐𝟐𝒙 𝐜𝐨𝐬𝟐𝒙 Example 7.9: 𝑑 𝑑𝑥න 𝑒௧𝑑𝑡 ୲ୟ୬௫ ௫మ ൌ ሾ𝑒୲ୟ୬௫∙secଶ𝑥ሿ െ ൣ𝑒௫మ∙2𝑥൧ ൌ 𝒆𝐭𝐚𝐧𝒙𝐬𝐞𝐜𝟐𝒙 െ 𝟐𝒙𝒆𝒙𝟐 Version 5.6 Page 100 of 242 April 8, 2023 Chapter 8 Applications of Integration Area Under a Curve The area under a curve can be calculated directly by integrating the curve over the desired interval. Note the following: The area “under” a curve is actually the area between the axis and the curve. In this sense, the word “under” may be a bit of a misnomer. The area under a curve may be positive (if above the ݔ‐axis) or negative (if below the ݔ‐axis). Example 8.1: Find the area under the curve ݕൌݐଵଷ ⁄ െ2 on the interval ሾെ1, 1ሿ. නቀݐ ଵଷ ൗെ2ቁ ݀ݐൌ൬3 4 ݐ ସଷ ൗെ2ݐ൰ฬ1 െ1 ଵ ିଵ ൌ3 4 ∙ቀ1 ସଷ ൗቁെ2 ∙1൨െ 3 4 ∙ሺെ1ሻ ସଷ ൗെ2 ∙ሺെ1ሻ൨ ൌെ5 4 െ11 4 ൌെ Example 8.2: Find the area under the curve ݕൌ4 sec ߠtan ߠ on the interval ቂെ గ ଷ, గ ଷቃ. න ሺ4 sec ߠ tan ߠሻ ݀ߠ గଷ ൗ ିగଷ ൗ ൌሺ4 sec ߠሻቤ ߨ3 ൗ െߨ3 ൗ ൌ ൬ 4 cos ߠ൰ቤ ߨ3 ൗ െߨ3 ൗ ൌቌ 4 1 2 ቍെቌ4 1 2 ቍ ൌ 0 Note: this interesting result means that the negative area under the curve of ݂ሺߠሻൌ4 sec ߠ tan ߠ on the interval ቂെ గ ଷ, 0ቃ is exactly offset exactly by the positive area above the curve on the interval ቂ0, గ ଷቃ. Negative Area Positive Area Version 5.6 Page 101 of 242 April 8, 2023 Chapter 8 Applications of Integration Area Between Curves The area between two curves is the difference of the areas under the curves. It is always positive, so if the curves switch position in terms of which one is superior (on top or to the right), the integration must take that into account. Example 8.3: Find the area of the region is bounded by the ݕ‐axis and the curves ݕൌsin ݔ and ݕൌcos ݔ (i.e., inside the green lines in the illustration). First, we must find the point of intersection in Quadrant 1. sin ݔൌcos ݔ at ݔൌ గ ସ, so our interval of integration is ቂ0, గ ସቃ Next, consider which curve is superior to the other (i.e., which one is higher if the form of the equations is ݕൌ݂ሺݔሻ, or more to the right if the form of the equations is ݔൌ݃ሺݕሻ). The other curve is inferior. The inferior curve is subtracted from the superior curve in the integrand. On the interval ቂ0, గ ସቃ, ݕൌcos ݔ is the higher of the two curves. Finally, calculate the area by integrating the difference between the curves. ܣൌන ሺcos ݔെsin ݔሻ గସ ൗ ݀ݔൌሺsin ݔcos ݔሻቤ ߨ4 ൗ 0 ൌቀsin ߨ 4 cos ߨ 4ቁെሺsin 0 cos 0ሻ ൌቆ√2 2 √2 2 ቇെሺ0 1ሻ ൌ √െ Example 8.4: Find the area of the region between ݕൌ݁௫ and ݕൌݔଶെ1 on the interval ሾെ1, 1ሿ (i.e., inside the green lines in the illustration). On the interval ሾെ1, 1ሿ, the highest curve is ݕൌ݁௫. Calculate the area by integrating the difference between the curves. නሾ݁௫െሺݔଶെ1ሻሿ ଵ ିଵ ݀ݔൌනሺ݁௫െݔଶ1ሻ ଵ ିଵ ݀ݔ ൌ൬݁௫െ1 3 ݔଷݔ൰ฬ1 െ1 ൌ൬݁െ1 3 1൰െ൬1 ݁1 3 െ1൰ ൌ ݁െ1 ݁െ2 3 2 ൌ ࢋെ ࢋ Version 5.6 Page 102 of 242 April 8, 2023 Chapter 8 Applications of Integration Area in Polar Form Area in Polar Form is given by: Let: ݎൌ݂ሺߠሻ Then, ܣൌ1 2 නݎଶ ݀ߠ Why? The diagram at right illustrates the reason that we use the above formula for area. The integral adds all of the slices (see the color slices in the diagram) inside the curve in question. Each slice is a sector of a circle with radius ݎ and angle ݀ߠ (an infinitesimally small angle). The area of a single slice, then, is ௗఏ ଶగ times the area of the circle containing it. That is: ܣ௦ൌ݀ߠ 2ߨ∙ߨݎଶൌ1 2 ݎଶ݀ߠ Integrating this over the desired interval of ߠ results in the above formula for area. Example 8.5: Find the area in the first quadrant inside the lemniscate ݎଶൌ4 sin 2ߠ shown in the above diagram. First, we need to determine the limits of integration. Consider that the loop in Quadrant 1 begins and ends at locations where ݎൌ0. So, we need to find two values of the variable ߠ that make ݎൌ0. We do this by setting ݎൌ0 in the equation of the lemniscate. 0ଶൌ4 sin 2ߠ, which occurs when sin 2ߠൌ0, which occurs at ߠൌቄ0, గ ଶ, ߨ, ଷగ ଶ, … ቅ For our limits of integration, we will use 0 and గ ଶ because these two values define the loop in Quadrant 1. We can check this by evaluating ݎ for a value in the interval ቂ0, గ ଶቃ and making sure the resulting point is in Quadrant 1. Let’s find ݎ when ߠൌ గ ସ. ߠൌ గ ସ ⇒ ݎଶൌ4 sin ቀ2 ∙ గ ସቁൌ4 ∙1 ൌ4 ⇒ ݎൌ2 (in Quadrant 1) The area of the lemniscate above in Quadrant 1, then, is calculated as: ܣ ൌ 1 2 නݎଶ ݀ߠ ൌ 1 2 න ሺ4 sin 2ߠሻ గଶ ൗ ݀ߠ ൌන 2 sin 2ߠ గଶ ൗ ݀ߠൌെcos 2ߠቤ ߨ2 ൗ 0 ൌ Version 5.6 Page 103 of 242 April 8, 2023 Chapter 8 Applications of Integration Example 8.6: Calculate the area of the general lemniscate of the form ݎଶൌܽଶsin 2ߠ. Note that the area of the entire lemniscate is double that of the loop in Quadrant 1. Then, ܣൌ2 ቆ1 2 නݎଶ ݀ߠቇൌ න ሺܽଶsin 2ߠሻ గଶ ൗ ݀ߠൌܽଶන sin 2ߠ గଶ ൗ ݀ߠൌെ1 2 ܽଶcos 2ߠቤ ߨ2 ൗ 0 ൌࢇ Example 8.7: Find the area within the inner loop of the limaçon ݎൌ1 2 cos ߠ. First, we need to determine the limits of integration. Consider that the loop begins and ends at locations where ݎൌ0. So, we need to find the values of the variable ߠ that make ݎൌ0 and define the inner loop. We do this by setting ݎൌ0 in the equation of the lemniscate. 0 ൌ1 2 cos ߠ, which occurs when cos ߠൌെ ଵ ଶ, which occurs at ߠൌቄ ଶగ ଷ, ସగ ଷቅ Next, we need to make sure that the inner loop is defined as ߠ progresses from ଶగ ଷ to ସగ ଷ. We can do this by evaluating ݎ for a value of ߠ in the interval ቂ ଶగ ଷ, ସగ ଷቃ and making sure the resulting point is on the inner loop. Let’s find ݎ when ߠൌߨ. ߠൌߨ ⇒ ݎൌ1 2 cos ߨൌെ1 We check the polar point ሺെ1, ߨሻ on the curve and note that it is on the inner loop. Therefore, our limits of integration are the values ߠൌቄ ଶగ ଷ, ସగ ଷቅ. The area of the inner loop of the limaçon ݎൌ1 2 cos ߠ, then, is calculated as: ܣൌ1 2 නݎଶ ݀ߠൌ1 2 න ሺ1 2 cos ߠሻଶ ସగଷ ൗ ଶగଷ ൗ ݀ߠ ൌ 1 2 න ሺ1 4 cos ߠ4 cosଶߠሻ ସగଷ ൗ ଶగଷ ൗ ݀ߠ ൌ1 2 න ൬1 4 cos ߠ4 ∙1 cos 2ߠ 2 ൰ ସగଷ ൗ ଶగଷ ൗ ݀ߠ ൌ න ൬3 2 2 cos ߠcos 2ߠ൰ ସగଷ ൗ ଶగଷ ൗ ݀ߠ ൌ3 2 ߠ2 sin ߠ1 2 sin 2ߠอ 4ߨ3 ൗ 2ߨ3 ൗ ൌ ࣊െ√ Version 5.6 Page 104 of 242 April 8, 2023 Chapter 8 Applications of Integration Areas of Limaçons Limaçons that have both inner and outer loops present a challenge when calculating area. The general form of a limaçon is: ݎൌܾܽcos ߠ or ݎൌܾܽsin ߠ When |ܽ| ൏|ܾ|, the limaçon has an inner loop that covers part of its outer loop, so we must be careful calculating areas in this kind of limaçon. Example 8.8: Find the area between the loops (i.e., inside the outer loop but outside the inner loop) of the limaçon: ݎൌ1 െ2 sin ߠ. First, we need to find where ݎൌ1 െ2 sin ߠൌ0 so we can identify the starting and ending ߠ‐ values for the inner loop. After finding these values to be ߠൌ గ , ହగ , we can look at the curve over various intervals on ሾ0,2ߨሿ and calculate the areas associated with those intervals. ቂ0, ߨ 6ቃ: 1 2 න ሺ1 െ2 sin ߠሻଶ గ/ ݀ߠൌߨ3√3 െ8 4 ~ 0.0844 ߨ 6 , 5ߨ 6 ൨: 1 2 න ሺ1 െ2 sin ߠሻଶ ହగ/ గ/ ݀ߠൌ2ߨെ3√3 2 ~ 0.5435 5ߨ 6 , ߨ൨: 1 2 න ሺ1 െ2 sin ߠሻଶ గ ହగ/ ݀ߠൌߨ3√3 െ8 4 ~ 0.0844 ሾߨ, 2ߨሿ: 1 2 න ሺ1 െ2 sin ߠሻଶ ଶగ గ ݀ߠൌ3ߨ8 2 ~ 8.7124 The total area of the limaçon, including both the outer and inner loops, is the sum of these: ሾ0,2ߨሿ: 1 2 න ሺ1 െ2 sin ߠሻଶ ଶగ ݀ߠ ൌ 3ߨ ~ 9.4248 Version 5.6 Page 105 of 242 April 8, 2023 Chapter 8 Applications of Integration A sketch of the complete limaçon ݎൌ1 െ2 sin ߠ is shown in Figure 1 below. Since taking the area from 0 to 2ߨ includes the area completely inside the outer loop plus the area inside the inner loop, the total area can be thought of as shown in Figure 2. This illustrates that the area within the inner loop is included in ܣൌ ଵ ଶ ሺ1 െ2 sin ߠሻଶ ଶగ ݀ߠ twice, and therefore, must be subtracted twice when looking for the area between the loops. Subtracting it once leaves all of the area inside the outer loop (Figure 3). A second subtraction is required to obtain the area between the loops. Given all of the above, let’s calculate the key areas of the limaçon ݎൌ1 െ2 sin ߠ: The total area of the limaçon, including both the outer loop and the inner loop, is: Interval ሾ0,2ߨሿ: 1 2 න ሺ1 െ2 sin ߠሻଶ ଶగ ݀ߠ ൌ 3ߨ ~ 9.4248 The area inside the inner loop is calculated as: Interval ߨ 6 , 5ߨ 6 ൨: 1 2 න ሺ1 െ2 sin ߠሻଶ ହగ/ గ/ ݀ߠ ൌ 2ߨെ3√3 2 ~ 0.5435 The area between the loops (i.e., the solution to this example) is calculated as: 1 2 න ሺ1 െ2 sin ߠሻଶ ଶగ ݀ߠ െ 2 ∙1 2 න ሺ1 െ2 sin ߠሻଶ ହగ గ ݀ߠ ൌ ߨ3√3 ~ 8.3377 Figure 1 ݎൌ1 െ2 sin ߠ Graphed on ሾ0, 2ߨሿ Figure 2 ݎൌ|1 െ2 sin ߠ| Graphed on ሾ0, 2ߨሿ Figure 3 ݎൌ1 െ2 sin ߠ Graphed on ቂ0, గ ቃ∪ቂ ହగ , 2ߨቃ Version 5.6 Page 106 of 242 April 8, 2023 Chapter 8 Applications of Integration Arc Length The arc length, ܮ, of a curve, in its various forms, is discussed below: Rectangular Form: For a function of the form: ݕൌ݂ሺݔሻ, from ݔൌܽ to ݔൌܾ. ܮൌනඨ1 ൬݀ݕ ݀ݔ൰ ଶ ݀ݔ For a function of the form: ݔൌ݃ሺݕሻ, from ݕൌܿ to ݕൌ݀. ܮൌනඨ1 ൬݀ݔ ݀ݕ൰ ଶ ௗ ݀ݕ Example 8.9: Find the length of the arc on the hyperbolic curve ݕൌcosh ݔൌ ೣାషೣ ଶ on the ݔ‐ interval ሾ0, 2ሿ. Using the above formula, and noting that ௗ௬ ௗ௫ൌ ೣିషೣ ଶ ൌsinh ݔ: ܮൌනඨ1 ൬݀ݕ ݀ݔ൰ ଶ ݀ݔ ൌ නඨ1 ൬݁௫െ݁ି௫ 2 ൰ ଶ ଶ ݀ݔ ൌනඨ1 1 4 ሺ݁ଶ௫െ2 ݁ିଶ௫ሻ ଶ ݀ݔ ൌනඨ1 4 ሺ݁ଶ௫2 ݁ିଶ௫ሻ ଶ ݀ݔ ൌනඨ൬݁௫݁ି௫ 2 ൰ ଶ ଶ ݀ݔ ൌ න൬݁௫݁ି௫ 2 ൰ ଶ ݀ݔ ൌ݁௫െ݁ି௫ 2 ฬ2 0 ൌ ݁ଶെ݁ିଶ 2 െ1 െ1 2 ൌ ࢋെࢋି ൌ ܛܑܖܐ Version 5.6 Page 107 of 242 April 8, 2023 Chapter 8 Applications of Integration Polar Form: For a function of the form: ݎൌ݂ሺߠሻ, ܮൌනඨݎଶ൬݀ݎ ݀ߠ൰ ଶ ݀ߠ Example 8.10: Find the length of the arc of one petal on the rose ݎൌ2 cos 3ߠ. To find the interval which defines one petal, we set ݎൌ0. 0 ൌ2 cos 3ߠ, which occurs when cos 3ߠൌ0, which occurs at ߠൌቄ గ , గ ଶ, ହగ , ݁ݐܿቅ. A little investigation reveals we can define a full petal over the interval ߠ∈ቂ గ , గ ଶቃ. Next find: ௗ ௗఏൌെ6 sin 3ߠ. Then, the arc length of a single petal is: ܮൌනඨݎଶ൬݀ݎ ݀ߠ൰ ଶ ݀ߠ ൌ න ඥሺ2 cos 3ߠሻଶሺെ6 sin 3ߠሻଶ గଶ ൗ గ ൗ ݀ߠ ൌ න ඥ4 cosଶ3ߠ36 sinଶ3ߠ గଶ ൗ గ ൗ ݀ߠ ൌ 2 න ඥcosଶ3ߠ9 sinଶ3ߠ గଶ ൗ గ ൗ ݀ߠ ൌ 2 න ඥሺcosଶ3ߠsinଶ3ߠሻ8 sinଶ3ߠ గଶ ൗ గ ൗ ݀ߠ ൌ න ඥૡܛܑܖࣂ ࣊ ൗ ࣊ ൗ ࢊࣂ This expression is quite ugly but can be handled by a modern calculator. Its value is approximately . as calculated on both the TI‐84 Plus and the TI nSpire. Version 5.6 Page 108 of 242 April 8, 2023 Chapter 8 Applications of Integration Parametric Form: For a function of the form: ݔൌ݂ሺݐሻ, ݕൌ݃ሺݐሻ ܮൌනඨ൬݀ݔ ݀ݐ൰ ଶ ൬݀ݕ ݀ݐ൰ ଶ ݀ݐ Example 8.11: Find the length of the arc of one petal on the rose defined by the parametric equations ݔൌ2 cos 3ߠcos ߠ and ݕൌ2 cos 3ߠsin ߠ. This is the same curve defined in the example above. So we will integrate over the same interval: ߠ∈ቂ గ , గ ଶቃ. To integrate in parametric form, we need ௗ௫ ௗఏ and ௗ௬ ௗఏ. Let’s calculate them: ݀ݔ ݀ߠൌ2ሾሺcos 3ߠሻሺെsin ߠሻሺcos ߠሻሺെ3 sin 3ߠሻሿ ݀ݕ ݀ߠൌ2ሾሺcos 3ߠሻሺcos ߠሻሺsin ߠሻሺെ3 sin 3ߠሻሿ Then, ܮൌනඨ൬݀ݔ ݀ݐ൰ ଶ ൬݀ݕ ݀ݐ൰ ଶ ݀ݐ ൌන ඥ4ሾെሺcos 3ߠሻሺsin ߠሻെሺcos ߠሻሺ3 sin 3ߠሻሿଶ4ሾሺcos 3ߠሻሺcos ߠሻെሺsin ߠሻሺ3 sin 3ߠሻሿଶ గଶ ൗ గ ൗ ݀ߠ ൌ2 න ඨሺcosଶ3ߠሻሺsinଶߠሻ6ሺcos ߠሻሺcos 3ߠሻሺsin ߠሻሺsin 3ߠሻ9ሺcosଶߠሻሺsinଶ3ߠሻ ሺcosଶ3ߠሻሺcosଶߠሻെ6ሺcos ߠሻሺcos 3ߠሻሺsin ߠሻሺsin 3ߠሻ9ሺsinଶߠሻሺsinଶ3ߠሻ൨ గଶ ൗ గ ൗ ݀ߠ Notice in this expression that terms above and below each other can be combined to get: ܮൌ2 න ඥሾሺcosଶ3ߠሻሺsinଶߠcosଶߠሻ9ሺsinଶߠcosଶߠሻሺsinଶ3ߠሻሿ గଶ ൗ గ ൗ ݀ߠ ൌ2 න ඥሾሺcosଶ3ߠሻ9ሺsinଶ3ߠሻሿ గଶ ൗ గ ൗ ݀ߠ ൌ 2 න ඥૡܛܑܖࣂ ࣊ ൗ ࣊ ൗ ࢊࣂ This is exactly the same expression that was derived on the previous page in polar form.
Version 5.6 Page 109 of 242 April 8, 2023 Chapter 8 Applications of Integration Comparison of Formulas for Rectangular, Polar and Parametric Forms Rectangular Form Polar Form Parametric Form Form 𝑦ൌ𝑓ሺ𝑥ሻ position ൌ𝑠ሺ𝑡ሻ 𝑟ൌ𝑓ሺ𝜃ሻ 𝑥ൌ𝑓ሺ𝑡ሻ 𝑦ൌ𝑔ሺ𝑡ሻ Conversion 𝑥ଶ𝑦ଶൌ𝑟ଶ tan 𝜃ൌ𝑦 𝑥 𝑥ൌ𝑟cos 𝜃 𝑦ൌ𝑟sin 𝜃 𝑥ൌ𝑟cos 𝑡 𝑦ൌ𝑟sin 𝑡 Area Under Curve න𝑓ሺ𝑥ሻ 𝑑𝑥 1 2 න𝑟ଶ ఉ ఈ 𝑑𝜃 නሾ𝑔ሺ𝑡ሻ∙𝑓ᇱሺ𝑡ሻሿ ௗ 𝑑𝑡 Area Between Curves නሾ𝑓ሺ𝑥ሻെ𝑔ሺ𝑥ሻሿ 𝑑𝑥 1 2 නሾ𝑟 ௨௧ ଶ െ𝑟 ଶ ሿ ఉ ఈ 𝑑𝜃 Arc Length (𝑳) නඨ1 ൬𝑑𝑦 𝑑𝑥൰ ଶ 𝑑𝑥 නඥ𝑟ଶሺ𝑟′ሻଶ ఉ ఈ 𝑑𝜃 නඨ൬𝑑𝑥 𝑑𝑡൰ ଶ ൬𝑑𝑦 𝑑𝑡൰ ଶ ௗ 𝑑𝑡 Magnitude of Speed (2D) |𝑣ሺ𝑡ሻ| ൌฬ𝑑 𝑑𝑡ሾ𝑠ሺ𝑡ሻሿฬ ඥ𝑟ଶሺ𝑟′ሻଶ ඨ൬𝑑𝑥 𝑑𝑡൰ ଶ ൬𝑑𝑦 𝑑𝑡൰ ଶ Slope of Tangent Line 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥ൌ𝑑𝑦/𝑑𝜃 𝑑𝑥/𝑑𝜃ൌ𝑟ᇱsin 𝜃𝑟cos 𝜃 𝑟ᇱcos 𝜃െ𝑟sin 𝜃 𝑑𝑦 𝑑𝑥ൌ𝑑𝑦/𝑑𝑡 𝑑𝑥/𝑑𝑡ൌ𝑟ᇱsin 𝑡𝑟cos 𝑡 𝑟ᇱcos 𝑡െ𝑟sin 𝑡 Second Derivative 𝑑ଶ𝑦 𝑑𝑥ଶ 𝑑 𝑑𝜃൬𝑑𝑦 𝑑𝑥൰ 𝑑𝑥 𝑑𝜃 𝑑 𝑑𝑡൬𝑑𝑦 𝑑𝑥൰ 𝑑𝑥 𝑑𝑡 Horizontal Tangents 𝑑𝑦 𝑑𝑥ൌ0 𝑑𝑦 𝑑𝜃ൌ𝑟ᇱsin 𝜃𝑟cos 𝜃ൌ0 𝑑𝑦 𝑑𝑡ൌ𝑟ᇱsin 𝑡𝑟cos 𝑡ൌ0 Vertical Tangents ௗ௬ ௗ௫ undefined 𝑑𝑥 𝑑𝜃ൌ𝑟ᇱcos 𝜃െ𝑟sin 𝜃ൌ0 𝑑𝑥 𝑑𝑡ൌ𝑟ᇱcos 𝑡െ𝑟sin 𝑡ൌ0 Version 5.6 Page 110 of 242 April 8, 2023 Chapter 8 Applications of Integration Area of a Surface of Revolution Rotation about the ࢞‐Axis Rotation of a curve ݕൌ݂ሺݔሻ from ݔൌܽ to ݔൌܾ. ܵൌන2ߨ ݕ ݀ݏ ݎ ܵൌ2ߨනݕ ඨ1 ൬݀ݕ ݀ݔ൰ ଶ ݀ݔ ݀ݏൌඨ1 ൬݀ݕ ݀ݔ൰ ଶ ݀ݔ ݏ is the arc length of the curve on ሾܽ, ܾሿ. If the curve is defined by parametric equations, ݔൌ݂ሺݐሻ, ݕൌ݃ሺݐሻ: ܵൌ2ߨන ݕ ௧ୀ௧మ ௧ୀ௧భ ඨ ൬݀ݔ ݀ݐ൰ ଶ ൬݀ݕ ݀ݐ൰ ଶ ݀ݐ Rotation about the ࢟‐Axis Rotation of a curve ݔൌ݃ሺݕሻ from ݕൌܿ to ݕൌ݀. ܵൌන2ߨ ݔ ௗ ݀ݏ ݎ ܵൌ2ߨනݔ ௗ ඨ1 ൬݀ݔ ݀ݕ൰ ଶ ݀ݕ ݀ݏൌඨ1 ൬݀ݔ ݀ݕ൰ ଶ ݀ݕ ݏ is the arc length of the curve on ሾܿ, ݀ሿ. If the curve is defined by parametric equations, ݔൌ݂ሺݐሻ, ݕൌ݃ሺݐሻ: ܵൌ2ߨන ݔ ௧ୀ௧మ ௧ୀ௧భ ඨ ൬݀ݔ ݀ݐ൰ ଶ ൬݀ݕ ݀ݐ൰ ଶ ݀ݐ Version 5.6 Page 111 of 242 April 8, 2023 Chapter 8 Applications of Integration Volumes of Solids of Revolution Solids of Revolution Rotation about: x‐axis y‐axis Disk Method ܸൌߨනሺ݂ሺݔሻሻଶ݀ݔ ܸൌߨනሺ݂ሺݕሻሻଶ݀ݕ ௗ Washer Method(1) ܸൌߨනቂሺ݂ሺݔሻሻଶെ൫݃ሺݔሻ൯ ଶቃ݀ݔ ܸൌߨනሾሺ݂ሺݕሻሻଶെሺ݃ሺݕሻሻଶሿ݀ݕ ௗ Cylindrical Shell Method(2) ܸൌ2ߨනݕ݂ሺݕሻ݀ݕ ௗ or ܸൌ2ߨනݎ݂ሺݕሻ݀ݕ ௗ ܸൌ2ߨනݔ ݂ሺݔሻ݀ݔ or ܸൌ2ߨනݎ ݂ሺݔሻ݀ݔ Difference of Shells Method(2)(3) ܸൌ2ߨනݕሺ݂ሺݕሻെ݃ሺݕሻሻ݀ݕ ௗ or ܸൌ2ߨනݎሺ݂ሺݕሻെ݃ሺݕሻሻ݀ݕ ௗ ܸൌ2ߨනݔ ൫݂ሺݔሻെ݃ሺݔሻ൯݀ݔ or ܸൌ2ߨනݎ ൫݂ሺݔሻെ݃ሺݔሻ൯݀ݔ Area Cross Section Method(4) ܸൌනܣሺݔሻ݀ݔ ܸൌනܣሺݕሻ݀ݕ ௗ Notes: 1. The Washer Method is an extension of the Disk Method. 2. ݎ is the radius of the cylindrical shell. In cases where there is a gap between the axis of revolution and the functions being revolved, ݎ is the distance between the axis of revolution and either ݔ or ݕ, as appropriate. 3. The Difference of Shells Method is an extension of the Cylindrical Shell Method. 4. The function ܣ is the area of the cross section being integrated. Version 5.6 Page 112 of 242 April 8, 2023 Chapter 8 Applications of Integration Disk and Washer Methods The formulas for the Disk Method and Washer Method for calculating volumes of revolution are provided above. Below, we present an approach that can be used to calculate volumes of revolution using these methods. Under the Disk Method, we integrate the area of the region between a curve and its axis of revolution to obtain volume. Since each cross‐section of the resulting object will be a circle, we use the formula ܽݎ݁ܽൌߨݎଶ as our starting point. The resulting formula is: ܸൌ ߨනሺܿݎݏݏ ݏ݁ܿݐ݅݊ ݎܽ݀݅ݑݏሻଶ ݀ݔ or ܸൌ ߨනሺܿݎݏݏ ݏ݁ܿݐ݅݊ ݎܽ݀݅ݑݏሻଶ ௗ ݀ݕ The Washer Method is simply a dual application of the Disk Method. Consider the illustration at right. If we want the area of the shaded region, we subtract the area of the smaller circle from the area of the larger circle. The same occurs with the Washer Method; since we integrate cross‐sectional area to find volume, so to obtain the volume of revolution of a region between the two curves we integrate the difference in the areas between the two curves. Below is a set of steps that can be used to determine the volume of revolution of a region between two curves. The approach is illustrated based on the following example: Example 8.12: Find the volume that results from revolving the region between the curves ݕൌ 2√ݔ and ݕൌ ଵ ସݔଶ about the line ݕൌ6. Steps 1. Graph the equations provided and any other information given in the problem (illustrated below). Then, isolate the section of the graph that we want to work with (illustrated at right). The disks we will use are shown as green and orange vertical lines. The dashed objects are reflections of the curves and disks over the axis of revolution; these give us an idea of what the central cross‐section of the 3ܦ shape will look like after revolution. You do not need to draw these. Integration Interval Version 5.6 Page 113 of 242 April 8, 2023 Chapter 8 Applications of Integration 2. Identify whether there is a gap between the region to be revolved and the axis of revolution. In the example, the axis of revolution is ݕൌ6, so there is clearly a gap between a) the red and blue curves, and b) the axis of revolution. Therefore, we will use the Washer Method. 3. Set up the integral form to be used. a. Disk Method: ܸൌߨ ሺradiusሻଶ ݀ݔ or ܸൌߨ ሺradiusሻଶ ௗ ݀ݕ b. Washer Method: ܸൌߨ ሾሺbig radiusሻଶെሺsmall radiusሻଶሿ ݀ݔ or ܸൌ ߨ ሾሺbig radiusሻଶെሺsmall radiusሻଶሿ ௗ ݀ݕ 4. Identify the variable of integration (i.e., are we using ݀ݔ or ݀ݕ?). The disks used must be perpendicular to the axis of revolution. a. If we are revolving around an axis, use the variable of that axis. b. If the axis of revolution is a line of the form, ݔൌܽ or ݕൌܾ, use the opposite variable from the one that occurs in the equation of the axis. In the example, the axis of revolution is ݕൌ6, so we will integrate with respect to ݔ. Note: The expressions used in the integration must be in terms of the variable of integration. So, for example, if the variable of integration is ݕ and the equation of a curve is given as ݕൌ݂ሺݔሻ, we must invert this to the form ݔൌ݃ሺݕሻ before integrating. 5. Identify the limits of integration. In the example, the curves intersect at ݔൌ0 and ݔൌ4. This results in an equation for volume in the form: ܸ ൌ ߨනሾሺbig radiusሻଶെሺsmall radiusሻଶሿ ସ ݀ݔ 6. Substitute the expressions for the big and small radii inside the integral. In the example, we have the following: a. big radius ൌ6 െ ଵ ସݔଶ b. small radius ൌ6 െ2√ݔ This results in the following: ࢂ ൌ ࣊නቈ൬െ ࢞൰ െ൫െ√࢞൯ ࢊ࢞ ~ . ૠ Note that this matches the value calculated using the Difference of Shells Method below. Version 5.6 Page 114 of 242 April 8, 2023 Chapter 8 Applications of Integration Cylindrical Shell Methods The formulas for the Cylindrical Shell Method and Difference of Shells Method for calculating volumes of revolution are provided above. Below, we present an approach that can be used to calculate volumes of revolution using these methods. Under the Cylindrical Shell Method, we integrate the volume of a shell across the appropriate values of ݔ or ݕ. We use the formula for the volume of a cylinder as our starting point (i.e., ܸ݈ݑ݉݁ൌ2ߨݎ݄, where ݄ is typically the function provided). The resulting formula is: ܸൌ 2ߨනݎ ሺ݄݄݁݅݃ݐ ݂ ݏ݄݈݈݁ሻ ݀ݔ or ܸൌ 2ߨනݎ ሺ݄݄݁݅݃ݐ ݂ ݏ݄݈݈݁ሻ ௗ ݀ݕ The Difference of Shells Method is essentially a dual application of the Cylindrical Shell Method. We want the volume of the cylinder whose height is the difference between two functions (see illustration at right). Below is a set of steps that can be used to determine the volume of revolution of a region between two curves. The approach is illustrated based on the following example: Example 8.13: Find the volume that results from revolving the region between the curves ݕൌ 2√ݔ and ݕൌ ଵ ସݔଶ about the line ݕൌ6. Steps 1. Graph the equations provided and any other information given in the problem (illustrated below left). Then, isolate the section of the graph that we want to work with (illustrated below right). Also shown are reflections of the curves over the axis of revolution (dashed curves); this allows us to see the “other side” of the cylindrical shells we will use. A typical shell is shown as a green cylinder. Integration Interval ࢌሺ࢞ሻെࢍሺ࢞ሻ Version 5.6 Page 115 of 242 April 8, 2023 Chapter 8 Applications of Integration 2. Identify whether the integration involves one or two curves. a. One curve: Use the Cylindrical Shell Method. b. Two curves: Use the Difference of Shells Method. This is the case in the example. 3. Set up the integral form to be used. Let ݎ be the radius of the shell. a. Cylindrical Shell Method: ܸൌ2ߨ ݎ ݂ሺݔሻ ݀ݔ or ܸൌ2ߨ ݎ ݃ሺݕሻ ௗ ݀ݕ. b. Difference of Shells Method: ܸൌ2ߨ ݎ ሺdifference of shell heightsሻ ݀ݔ or ܸൌ2ߨ ݎ ሺdifference of shell heightsሻ ௗ ݀ݕ. 4. Identify the variable of integration (i.e., are we using ݀ݔ or ݀ݕ?). The shells used must be parallel to the axis of revolution. a. If we are revolving around an axis, consider the equation of that axis (i.e., the ݔ‐ axis has equation: ݕൌ0). b. The axis of revolution is a line of the form, ݔൌܽ or ݕൌܾ, use the same variable as the one that occurs in the equation of the axis of revolution. In the example, the axis of revolution is ݕൌ6, so we will integrate with respect to ݕ. ܸൌ2ߨනݎ ሺdifference of shell heightsሻ ௗ ݀ݕ 5. Identify the limits of integration. In the example, the curves intersect at ݕൌ0 and ݕൌ4. This results in an equation for volume in the form: ܸൌ2ߨනݎ ሺdifference of shell heightsሻ ସ ݀ݕ 6. Substitute the expressions for ࢘ and the difference of shell heights into the integral. In the example, we need to convert each equation to the form ݔൌ݃ሺݕሻ because ݕ is the variable of integration: a. ݕൌ ଵ ସݔଶ so ݔൌ2ඥݕ ݕൌ2√ݔ so ݔൌ ଵ ସݕଶ The difference of shell heights, then, is ቀ2ඥݕ െ 1 4 ݕ2ቁ. b. The radius of a shell is the difference between the line ݕൌ6 and the value of ݕ in the interval, so the radius is 6 െݕ. This results in the following: ࢂ ൌ ࣊නሺെ࢟ሻ ൬ඥ࢟ െ ࢟൰ ࢊ࢟ ~ . ૠ Note that this matches the value calculated using the Washer Method above. Version 5.6 Page 116 of 242 April 8, 2023 Chapter 8 Applications of Integration Volume by Area of a Cross‐Section Some problems require us to determine volume of a solid using its base and cross‐sectional area across that base. These are not problems based on revolution of a shape, so we use a more basic formula (that does not involve π): ܸൌ නሺܽݎ݁ܽ ܿݎݏݏ ݏ݁ܿݐ݅݊ሻ ݀ݔ or ܸൌ නሺܽݎ݁ܽ ܿݎݏݏ ݏ݁ܿݐ݅݊ሻ ௗ ݀ݕ Below is a set of steps that can be used to determine volume for this type of problem. The approach is illustrated using the following example: Example 8.14: Find the volume of a solid with a base of ݕൌ2√sin ݔ over the interval ሾ0, ߨሿ if the cross‐sections perpendicular to the ݔ‐axis are equilateral triangles whose bases stretch from the ݔ‐axis to the curve. Steps 1. Graph the curve of the base over the interval specified. 2. Determine the variable of integration. This will always be the variable whose axis is perpendicular to the cross‐sections specified. In the example, the variable of integration is ݔ. 3. Determine the limits of integration. This is typically the interval provided in the problem. In the example, this is the interval ሾ0, ߨሿ. 4. Draw the cross‐section you are provided in the problem. In the example, we are working with equilateral triangles with base equal to the function ݕൌ2√sin ݔ. 5. Determine the area of the cross‐section in terms of the appropriate variable. We need the area of an equilateral triangle for this example. This area can be developed from basic principles using the illustration at right, or from the formula: ܣൌ√ଷ ସܾଶ, where ܾ is the length of the base of the triangle. In the example: ܣൌ√ଷ ସܾଶൌ√ଷ ସ൫2√sinݔ൯ ଶൌ√3 sin ݔ 6. Integrate the area of the cross‐section using the limits determined in Step 3. ܸ ൌ න√3 sin ݔ గ ݀ݔ ൌ െ √3 cos ݔቚߨ 0 ൌ √ ~ . Version 5.6 Page 117 of 242 April 8, 2023 Chapter 9 Improper Integrals Improper Integration Improper integration refers to integration where the interval of integration contains one or more points where the integrand is not defined. Infinite Limits When either or both of the limits of integration are infinite, we replace the infinite limit by a variable and take the limit of the integral as the variable approaches infinity. න݂ሺݔሻ ஶ ݀ݔൌlim →ஶන݂ሺݔሻ ݀ݔ න݂ሺݔሻ ିஶ ݀ݔൌlim → ିஶන݂ሺݔሻ ݀ݔ න ݂ሺݔሻ ஶ ିஶ ݀ݔൌlim → ିஶන݂ሺݔሻ ݀ݔlim → ஶන݂ሺݔሻ ݀ݔ Note: in this third formula, you can select the value of ܾ to be any convenient value that produces convergent intervals. Example 9.1: න ൬1 ݔଶ൰ ஶ ଵ ݀ݔൌlim → ஶන൬1 ݔଶ൰ ଵ ݀ݔ ൌlim → ஶනሺݔିଶሻ ଵ ݀ݔൌlim → ஶሺെݔିଵሻቚܽ 1 ൌlim → ஶ൬െ1 ݔ൰ቚܽ 1 ൌlim → ஶ൬െ1 ܽ1 1൰ൌ0 1 ൌ1 Example 9.2: න ൬ 1 ݔଶ9൰ ିஶ ݀ݔൌlim → ିஶන൬ 1 ݔଶ9൰ ݀ݔ ൌ 1 3 lim → ିஶනቌ 1 ቀݔ 3ቁ ଶ 1 ቍ 1 3 ݀ݔ ൌ1 3 lim → ିஶቀtanିଵݔ 3ቁฬ0 ܽ ൌ1 3 lim → ିஶቀtanିଵ0 െtanିଵܽ 3ቁ ൌ 1 3 ቂ0 െቀെߨ 2ቁቃ ൌ ߨ 6 Version 5.6 Page 118 of 242 April 8, 2023 Chapter 9 Improper Integrals Discontinuous Integrand Limits are also required in cases where the function in an integrand is discontinuous over the interval of its limits. If there is a discontinuity at ݔൌܽ, If there is a discontinuity at ݔൌܾ, න݂ሺݔሻ ݀ݔൌlim ௧ → శන݂ሺݔሻ ௧ ݀ݔ න݂ሺݔሻ ݀ݔൌlim ௧ → షන݂ሺݔሻ ௧ ݀ݔ If there is a discontinuity at ݔൌܿ where ܽ൏ܿ൏ܾ, න݂ሺݔሻ ݀ݔൌlim ௧ → షන݂ሺݔሻ ௧ ݀ݔlim ௧ → శන݂ሺݔሻ ௧ ݀ݔ Example 9.3: න൬ 1 4 െݔ൰ ସ ݀ݔൌlim ௧ → ସషන൬ 1 4 െݔ൰ ௧ ݀ݔ ൌെlim ௧ → ସషሾlnሺ4 െݔሻሿฬݐ 0 ൌ lim ௧ → ସషሾlnሺ4 െݔሻሿฬ0 ݐ ൌlim ௧ → ସషሾlnሺ4 െ0ሻെlnሺ4 െݐሻሿ ൌln 4 െ ሺെ∞ሻ ൌ ∞ Example 9.4: න൬1 √ݔ൰ ଵ ݀ݔൌනቀݔିଵଶ ൗቁ ଵ ݀ݔൌlim ௧ →శනቀݔିଵଶ ൗቁ ଵ ௧ ݀ݔ ൌlim ௧ →శ ቀ2ݔ ଵଶ ൗቁฬ1 ݐൌlim ௧ →శ ൫2√ݔ൯ฬ1 ݐ ൌ lim ௧ →శ൫2√1 െ2√ݐ൯ ൌ 2 െ0 ൌ 2 Example 9.5: න ሺsec ݔtan ݔሻ గ గଶ ൗ ݀ݔൌ lim ௧ → గ/ଶశනሺsec ݔtan ݔሻ గ ௧ ݀ݔ ൌlim ௧ →గ ଶ శsec ݔቚߨ ݐൌlim ௧ →గ ଶ శሺsec ߨെsec ݐሻ ൌ െ1 ∞ ൌ ∞ Version 5.6 Page 119 of 242 April 8, 2023 Chapter 10 Differential Equations Differential Equations Definitions A Differential Equation is an equation that contains an independent variable, one or more dependent variables, and full or partial derivatives of the dependent variables. An Ordinary Differential Equation (ODE) is a differential equation that contains ordinary (not partial) derivatives. Generally, an ODE is expressed in one of the following forms: 𝐹൫𝑥, 𝑦, 𝑦ᇱ, 𝑦ᇱᇱ, … , 𝑦ሺሻ൯ൌ0 or 𝐹൫𝑥, 𝑦, 𝑦ᇱ, 𝑦ᇱᇱ, … , 𝑦ሺିଵሻ൯ൌ𝑦ሺሻ A Partial Differential Equation (PDE) is a differential equation that contains partial derivatives. The Order of a differential equation is the highest derivative of a dependent variable in the equation. A Linear ODE of Order 𝒏 is an equation of the form: 𝑎ሺ𝑥ሻ∙𝑦ሺሻ𝑎ିଵሺ𝑥ሻ∙𝑦ሺିଵሻ … 𝑎ଵሺ𝑥ሻ∙𝑦ᇱ𝑎ሺ𝑥ሻ∙𝑦ൌ𝑓ሺ𝑥ሻ where each of the 𝑎ሺ𝑥ሻ is a function in 𝑥 only, (i.e., not in 𝑦 or any of its derivatives). The 𝑎ሺ𝑥ሻ need not be linear functions. The label “Linear” refers to 𝑦 and its derivatives; that is, there are no powers of 𝑦 and its derivatives and no products of 𝑦 and/or any of its derivatives. For example, there are no terms like ൫𝑦ሺሻ൯ ଶ, ሺ𝑦∙𝑦ᇱᇱሻ, etc. A Separable first order ODE is one that can be written in the form: 𝑑𝑦 𝑑𝑥ൌ𝑓ሺ𝑥ሻ∙𝑔ሺ𝑦ሻ A Solution to a differential equation is any function that satisfies the differential equation in the interval specified. Initial Conditions are those that allow us to determine which of a possible set of solutions to a differential equation we seek. In essence, these allow us to determine the value of any constants that turn up in the integrations required to solve the differential equations. An Initial Value Problem is a differential equation whose solution depends on the initial conditions provided. The Actual Solution to a differential equation is the specific solution that satisfies both the differential equation and the initial conditions. An Explicit Solution is a solution that can be expressed in the form 𝑦ൌ𝑓ሺ𝑥ሻ. An Implicit Solution is a solution that cannot be expressed in the form 𝑦ൌ𝑓ሺ𝑥ሻ. Version 5.6 Page 120 of 242 April 8, 2023 Chapter 10 Differential Equations Separable First Order ODEs Most of the differentiable equations that will be encountered in first year Calculus will be separable first order differential equations. Typically, we will use Algebra to identify 𝑓ሺ𝑥ሻ and 𝑔ሺ𝑥ሻ to get the equation into the form ௗ௬ ௗ௫ ൌ𝑓ሺ𝑥ሻ∙𝑔ሺ𝑦ሻ. Next, we treat 𝑑𝑦 and 𝑑𝑥 as separate entities, and convert the equation to the form: 𝑑𝑦 𝑔ሺ𝑦ሻൌ𝑓ሺ𝑥ሻ 𝑑𝑥 Finally, we integrate both sides to obtain a solution: න𝑑𝑦 𝑔ሺ𝑦ሻൌන𝑓ሺ𝑥ሻ 𝑑𝑥 The final result will have a 𝐶 term. Typically, you need only one 𝐶 term since the constants from each integral can be subtracted to get a single constant term. Often, there is an initial condition provided which allows us to calculate the value of 𝐶. Example 10.1: Find the explicit actual solution to ௗ௬ ௗ௫ ൌ𝑒௬ if ሺ1, 0ሻ is a point on the curve. An explicit solution is one of the form “𝑦ൌ𝑓ሺ𝑥ሻ”. An actual solution is one in which we have solved for any constants that pop up. Let’s begin by separating the variables. 𝑑𝑦 𝑑𝑥ൌ𝑒௬ 𝑒ି௬𝑑𝑦ൌ𝑑𝑥 න𝑒ି௬𝑑𝑦ൌන𝑑𝑥 െ𝑒ି௬ൌ𝑥𝐶 Substituting ሺ1, 0ሻ for ሺ𝑥, 𝑦ሻ gives െ1 ൌ1 𝐶 so, 𝐶ൌെ2 െ𝑒ି௬ൌ𝑥െ2 𝑒ି௬ൌ2 െ𝑥 െ𝑦ൌlnሺ2 െ𝑥ሻ 𝒚ൌെ𝐥𝐧ሺ𝟐െ𝒙ሻ Note the resulting domain restriction: 𝑥൏2. Version 5.6 Page 121 of 242 April 8, 2023 Chapter 10 Differential Equations Example 10.2: Find the explicit actual solution to ௗ௬ ௗ௫ൌ ௫ √ଽା௫మ if ሺ4, 5ሻ is a point on the curve. An explicit solution is one of the form “𝑦ൌ𝑓ሺ𝑥ሻ”. An actual solution is one in which we have solved for any constants that pop up. Let’s begin by separating the variables. Note that since there is an 𝑥 in the numerator, we do not need to use inverse trig functions. Then, substituting ሺ25, 5ሻ for ሺ𝑢, 𝑦ሻ gives: 5 ൌ√25 𝐶 so, 𝐶ൌ0 𝑦ൌ√𝑢 ⇒ 𝒚ൌඥ𝟗𝒙𝟐 An alternative way to develop a solution, involving 𝑥 more directly, would be to replace the three lines immediately above with these: 𝑦ൌ√𝑢𝐶 ⇒ 𝑦ൌඥ9 𝑥ଶ𝐶 Then, substituting ሺ4, 5ሻ for ሺ𝑥, 𝑦ሻ gives: 5 ൌ√9 4ଶ𝐶 ⇒ 5 ൌ√25 𝐶 so, 𝐶ൌ0 𝒚ൌඥ𝟗𝒙𝟐 𝑢ൌ9 𝑥ଶ 𝑑𝑢ൌ2𝑥𝑑𝑥 𝑥ൌ4 ⇒𝑢ൌ25 and 𝑦ൌ5 𝑑𝑦 𝑑𝑥ൌ 𝑥 √9 𝑥ଶ 𝑑𝑦ൌ 𝑥 √9 𝑥ଶ 𝑑𝑥 න𝑑𝑦ൌ1 2 න 2𝑥 √9 𝑥ଶ 𝑑𝑥 𝑦ൌ1 2 න𝑢ିଵଶ ൗ 𝑑𝑢ൌ1 2 ∙2𝑢 ଵଶ ൗ𝐶 𝑦ൌ√𝑢𝐶 Version 5.6 Page 122 of 242 April 8, 2023 Chapter 10 Differential Equations Slope Fields A Slope Field (also called a Direction Field) is a graphical representation of the slopes of a curve at various points that are defined by a differential equation. Each position in the graph (i.e., each point ሺ𝑥, 𝑦ሻ) is represented by a line segment indicating the slope of the curve at that point. Example 10.3: 𝒅𝒚 𝒅𝒙 ൌ𝒆𝐬𝐢𝐧𝒙𝐜𝐨𝐬𝒙𝐬𝐢𝐧𝒙 Example 10.4: 𝒅𝒚 𝒅𝒙 ൌ𝒙𝟐െ𝒚𝟐 If you know a point on a curve and if you have its corresponding slope field diagram, you can plot your point and then follow the slope lines to determine the curve. Example 10.5: Find the explicit actual solution to 𝑓ᇱሺ𝑥ሻൌ ௫ ௬ if ሺ1, െ2ሻ is a point on the curve. ௗ௬ ௗ௫ൌ ௫ ௬ 𝑦 𝑑𝑦ൌ𝑥 𝑑𝑥 න𝑦𝑑𝑦ൌන𝑥 𝑑𝑥 1 2 𝑦ଶൌ1 2 𝑥ଶ𝐶ଵ 𝑦ଶൌ𝑥ଶ𝐶 Substituting ሺ1, െ2ሻ for ሺ𝑥, 𝑦ሻ gives: 𝐶ൌ3 𝑦ଶൌ𝑥ଶ3 𝑦ൌേ ඥ𝑥ଶ3 Finally, noting that ሺ1, െ2ሻ is a solution, we can narrow the solution down to: 𝒚ൌെ √𝒙𝟐𝟑 Slope Field for: 𝑑𝑦 𝑑𝑥ൌ𝑥 𝑦 Slope Field generator available at: /calculus/differential‐ equations/slope‐field‐ generator.php Version 5.6 Page 123 of 242 April 8, 2023 Chapter 10 Differential Equations Logistic Function A Logistic Function describes the growth of a population over time. Early in its growth phase, the model describes near‐exponential population growth. As the population grows larger, it eventually faces limits that reduce its growth rate. Late in its growth phase, a population approaches a maximum value, called the carrying capacity. Several forms of the Logistic Function for a population 𝑃ሺ𝑡ሻ, over time, are common: 𝑃ሺ𝑡ሻൌ 𝐾 1 ቀ𝐾െ𝑃 𝑃 ቁ𝑒ି௧ or 𝑃ሺ𝑡ሻൌ 𝑃 𝐾 𝑃ሺ𝐾െ𝑃ሻ𝑒ି௧ or 𝑃ሺ𝑡ሻൌ 𝐾 𝑃 𝑒௧ 𝐾𝑃ሺ𝑒௧െ1ሻ The symbols in these equations have the following meanings: 𝑃ሺ𝑡ሻ is the population at time 𝑡. 𝐾 is the carrying capacity of the population. It is the maximum population sustainable in the system 𝑃ൌ𝑃ሺ0ሻ is the initial population. 𝑟 is the rate of growth of the population, and is called the growth parameter. 𝑡 is the variable for time. The differential equation that leads to the Logistic Function is: 𝑑𝑃 𝑑𝑡ൌ𝑟𝑃൬1 െ𝑃 𝐾൰ Characteristics of the Logistic Function ௗ ௗ௧0 for all 𝑡 lim ௧ →ஶ𝑃ሺ𝑡ሻൌ𝐾 𝑃ሺ𝑡ሻ has an inflection point at 𝑡ൌ ଵ ln ቀ బെ1ቁ, when 𝑃ሺ𝑡ሻൌ ଶ. Therefore, the maximum rate of growth for the population occurs when 𝑃ሺ𝑡ሻൌ ଶ. Version 5.6 Page 124 of 242 April 8, 2023 Chapter 10 Differential Equations Numerical Methods If we know a point on a curve and the slope of the curve at each point, but do not know the equation of the curve, it is possible to estimate the value of another point on the same curve using numerical methods. Several of these numerical methods are presented below. Euler’s Method Euler’s Method estimates the location of the new point based on the position of the first point and the slope of the curve at intervals between the two points. Any number of intervals, 𝑛, can be used. Each interval is called a time step. The formulas involved are as follows. Let: ሺ𝑥, 𝑦ሻ be the initial (known) point. ሺ𝑥, 𝑦ሻ be the intermediate points, for 𝑘ൌ1, 2, … . ሺ𝑥, 𝑦ሻ be the desired point. Note that 𝑛 is the number of time steps and 𝑥 is known. ℎ be the distance between successive 𝑥‐values. That is, ℎൌ ௫ି௫బ . Then, Euler’s Method estimates each 𝑦ାଵ based on 𝑦 and the slope of the function at ሺ𝑥, 𝑦ሻ, using the formulas: 𝑥ାଵൌ𝑥ℎ 𝑦ାଵൌ𝑦𝑦ᇱሺ𝑥ሻ∙ℎ Example 10.6: Let ௗ௬ ௗ௫ ൌ𝑦ᇱሺ𝑥ሻൌ2𝑦െ𝑥. Estimate 𝑦ሺ2ሻ using 4 time steps if we know ሺ1, 2ሻ is a point on the curve. We start at point ሺ𝑥, 𝑦ሻൌሺ1,2ሻ, using a time step of ℎൌ ଶିଵ ସൌ0.25. The following table shows the iterations required to estimate 𝑦ሺ2ሻ. Values in the table are rounded to 2 decimals for display, but the exact values are used in all calculations. 𝑘 𝑥 𝑦 𝑦ᇱሺ𝑥ሻൌ2𝑦െ𝑥 𝑦ାଵ 0 1.00 2.00 2 ሺ2.00ሻെ1.00 ൌ3.00 2.00 3.00 ሺ0.25ሻൌ2.75 1 1.25 2.75 2 ሺ2.75ሻെ1.25 ൌ4.25 2.75 4.25 ሺ0.25ሻൌ3.81 2 1.50 3.81 2 ሺ3.81ሻെ1.50 ൌ6.13 3.81 6.13 ሺ0.25ሻൌ5.34 3 1.75 5.34 2 ሺ5.34ሻെ1.75 ൌ8.94 5.34 8.94 ሺ0.25ሻൌ7.58 4 2.00 7.58 Since it is natural to develop Euler’s Method in table form, it is relatively easy to adapt it to a spreadsheet program such as Microsoft Excel. A plot of successive values of 𝑥 is shown in the graph at right. Version 5.6 Page 125 of 242 April 8, 2023 Chapter 10 Differential Equations Modified Euler’s Method The Modified Euler’s Method is like Euler’s Method, but develops the slope at each point as the average of the slopes at the beginning and end of each interval. Using the same notation as on the previous page, the Modified Euler’s Method uses a two‐step formula: Predictor step: 𝑥ାଵൌ𝑥ℎ 𝑦ାଵൌ𝑦𝑦ᇱሺ𝑥ሻ∙ℎ Corrector step: 𝑦ାଵൌ𝑦 ଵ ଶሾ𝑦ᇱሺ𝑥ሻ𝑦ᇱሺ𝑥ାଵሻሿ∙ℎ In the corrector step, the estimate of 𝑦ᇱሺ𝑥ାଵሻ is based on the value of 𝑦ାଵ generated in the predictor step. Example 10.7: Let ௗ௬ ௗ௫ൌ𝑦ᇱሺ𝑥ሻൌ2𝑦െ𝑥. Estimate 𝑦ሺ2ሻ using 4 time steps if we know ሺ1, 2ሻ is a point on the curve. We start at point ሺ𝑥, 𝑦ሻൌሺ1,2ሻ, using a time step of ℎൌ ଶିଵ ସൌ0.25. The following table shows the iterations required to estimate 𝑦ሺ2ሻ. Values in the table are rounded to 2 decimals for display, but the exact values are used in all calculations. 𝑘 𝑥 𝑦 𝑦ᇱሺ𝑥ሻ or 𝑦ᇱሺ𝑥ାଵሻ 𝑦ାଵ 0 1.00 2.00 2 ሺ2.00ሻെ1.00 ൌ3.00 2.00 3.00 ሺ0.25ሻൌ2.75 Corrector 2 ሺ2.75ሻെ1.25 ൌ4.25 2.00 ሺ3.00 4.25ሻ/2 ∙ ሺ0.25ሻൌ2.91 1 1.25 2.91 2 ሺ2.91ሻെ1.25 ൌ4.56 2.91 4.56 ሺ0.25ሻൌ4.05 Corrector 2 ሺ4.05ሻെ1.50 ൌ6.59 2.91 ሺ4.56 6.59ሻ/2 ∙ ሺ0.25ሻൌ4.30 2 1.50 4.30 2 ሺ4.30ሻെ1.50 ൌ7.10 4.30 7.10 ሺ0.25ሻൌ6.08 Corrector 2 ሺ6.08ሻെ1.75 ൌ10.40 4.30 ሺ7.10 10.40ሻ/2 ∙ ሺ0.25ሻൌ6.49 3 1.75 6.49 2 ሺ6.49ሻെ1.75 ൌ11.23 6.49 11.23 ሺ0.25ሻൌ9.30 Corrector 2 ሺ9.30ሻെ2.00 ൌ16.59 6.49 ሺ11.23 16.59ሻ/2 ∙ ሺ0.25ሻൌ9.97 4 2.00 9.97 A plot of successive values of 𝑥 is shown in the graph at right. The Modified Euler’s Method is more complex than Euler’s Method, but it tends to be more accurate because it uses a better estimate of the slope in each interval. Though complex, this method is also relatively easy to adapt to a spreadsheet program such as Microsoft Excel. Order: A numerical method is said to be of order 𝑛 if it produces exact results for polynomials of degree 𝑛 or less. Euler’s method is of order 1. Modified Euler’s Method is of order 2. The Runge‐Kutta Method, described on the next page, is of order 4. Version 5.6 Page 126 of 242 April 8, 2023 Chapter 10 Differential Equations Runge‐Kutta Method Runge‐Kutta Method an order 4 numerical method for estimating points on a curve using an initial point and slopes of the curve at various locations. Using similar notation to that on the previous pages, the Runge‐Kutta Method uses the following formulas: 𝑥ାଵൌ𝑥ℎ 𝑦ାଵൌ𝑦1 6 ሺ𝑘ଵ2𝑘ଶ2𝑘ଷ𝑘ସሻ where the following 𝑘‐values are weighted together to obtain incremental values of 𝑦. 𝑓ሺ𝑥, 𝑦ሻ is the derivative of the function at 𝑥, i.e., 𝑓ሺ𝑥, 𝑦ሻൌ𝑦’ሺ𝑥ሻ. 𝑘ଵൌℎ∙𝑓ሺ𝑥, 𝑦ሻ 𝑘ଶൌℎ∙𝑓ሺ𝑥 ଵ ଶℎ, 𝑦 ଵ ଶ𝑘ଵሻ 𝑘ଷൌℎ∙𝑓ሺ𝑥 ଵ ଶℎ, 𝑦 ଵ ଶ𝑘ଶሻ 𝑘ସൌℎ∙𝑓ሺ𝑥ℎ, 𝑦𝑘ଷሻ Note that the slope, 𝑓ሺ𝑥, 𝑦ሻ, used in defining each successive 𝑘 value builds on the slope determined in the previous 𝑘 value. Example 10.8: Let ௗ௬ ௗ௫ൌ𝑦ᇱሺ𝑥ሻൌ2𝑦െ𝑥. Estimate 𝑦ሺ2ሻ using 4 time steps if we know ሺ1, 2ሻ is a point on the curve. Time Step 1: Once again, we start at point ሺ𝑥, 𝑦ሻൌሺ1,2ሻ, and ℎൌ0.25. The following steps show the calculation of 𝑦ሺ1.25ሻ: ሺ𝑥, 𝑦ሻൌሺ1, 2ሻ 𝑦ᇱൌ𝑓ሺ𝑥, 𝑦ሻൌ2𝑦െ𝑥 𝑘ଵൌℎ∙𝑓ሺ𝑥, 𝑦ሻൌሺ0.25ሻ𝑓ሺ1, 2ሻൌሺ0.25ሻሺ2 ∙2 െ1ሻൌ0.75 𝑘ଶൌℎ∙𝑓൬𝑥1 2 ℎ, 𝑦1 2 𝑘ଵ൰ൌሺ0.25ሻ𝑓ሺ1.125, 2.375ሻ ൌሺ0.25ሻሺ2 ∙2.375 െ1.125ሻൌ0.90625 𝑘ଷൌℎ∙𝑓൬𝑥1 2 ℎ, 𝑦1 2 𝑘ଶ൰ ൌሺ0.25ሻ𝑓ሺ1.125, 2.453125ሻ ൌሺ0.25ሻሺ2 ∙2.453125 െ1.125ሻൌ0.9453125 𝑘ସൌℎ∙𝑓ሺ𝑥ℎ, 𝑦𝑘ଷሻൌሺ0.25ሻ𝑓ሺ1.25, 2.9453125ሻ ൌሺ0.25ሻሺ2 ∙2.9453125 െ1.25ሻൌ1.16015625 𝑦ሺ1.25ሻൌ𝑦ଵൌ𝑦1 6 ሺ𝑘ଵ2𝑘ଶ2𝑘ଷ𝑘ସሻ ൌ2 1 6 ሺ0.75 2 ∙0.90625 2 ∙0.9453125 1.16015625ሻൌ 𝟐. 𝟗𝟒 Note: Since 𝑘‐ values have a specific meaning in this method, we have switched our index variable from 𝑘 to 𝑛. Version 5.6 Page 127 of 242 April 8, 2023 Chapter 10 Differential Equations Time Steps 2 to 4: Performing the same set of calculations for three more steps gives the following values, all rounded to two decimals: 𝑦ሺ1.50ሻൌ4.40 𝑦ሺ1.75ሻൌ6.72 𝑦ሺ2.00ሻൌ10.48 To nine decimal places, with 4 time steps, our calculated value of 𝑦ሺ2.00ሻ is 10.479962905. Changing the number of time steps produces the results in the following table. Number of Time Steps Value of 𝑦ሺ2.00ሻ 4 10.479962905 10 10.486111552 20 10.486305959 50 10.486319742 100 10.486320099 200 10.486320122 500 10.486320124 Actual 10.486320124 In summary, let’s compare the results under the three methods above to the true values for the function defined by our conditions: 𝑦ൌ ቀ ହ ସమ𝑒ଶ௫ ଵ ଶ𝑥 ଵ ସቁ. Estimates of 𝑦 at Each Time Step Under Four Numerical Methods Time Step 𝑥‐value Euler’s Method Modified Euler’s Method Runge‐ Kutta (4‐steps) Actual Value 1 1.25 2.75 2.90625 2.935546875 2.935901588 2 1.50 3.8125 4.30078125 4.396682739 4.397852286 3 1.75 5.34375 6.488769531 6.724219203 6.727111338 4 2.00 7.578125 9.966125488 10.479962905 10.486320124 Clearly, the higher the order, the more accurate the estimates were for the function defined in the example. This will tend to be true, but will not be true in every case. Increasing the number of steps, and correspondingly decreasing the value of ℎ, will also tend to increase the accuracy of the estimates. Even though there are a significant number of steps and calculations involved in developing Runge‐Kutta estimates, their accuracy may warrant the effort, especially if a spreadsheet proram is readily available to the student. Notice how the increasing the number of time steps in the calculation improves the accuracy of the results. With 500 time steps the result is accurate to 9 decimal places. Version 5.6 Page 128 of 242 April 8, 2023 Chapter 11 Vector Calculus Vectors A vector is a quantity that has both magnitude and direction. An example would be wind blowing toward the east at 30 miles per hour. Another example would be the force of 10 kg weight being pulled toward the earth (a force you can feel if you are holding the weight). Special Unit Vectors We define unit vectors to be vectors of length 1. Unit vectors having the direction of the positive axes will be quite useful to us. They are described in the chart and graphic below. Unit Vector Direction ܑ positive ݔ‐axis ܒ positive ݕ‐axis ܓ positive ݖ‐axis Vector Components The length of a vector, ܞ, is called its magnitude and is represented by the symbol ‖ܞ‖. If a vector’s initial point (starting position) is ሺݔଵ, ݕଵ, ݖଵሻ, and its terminal point (ending position) is ሺݔଶ, ݕଶ, ݖଶሻ, then the vector displaces ࢇൌ࢞െ࢞ in the ݔ‐direction, ࢈ൌ࢟െ࢟ in the ݕ‐ direction, and ࢉൌࢠെࢠ in the ݖ‐direction. We can, then, represent the vector as follows: ܞൌܑܾܽܒܿܓ The magnitude of the vector, ܞ, is calculated as: ‖ܞ‖ ൌ√ܽଶܾଶܿଶ If this looks familiar, it should. The magnitude of a vector in three dimesnsions is determined as the length of the space diagonal of a rectangular prism with sides ܽ, ܾ and ܿ. In two dimensions, these concepts contract to the following: ܞൌܑܾܽܒ ‖ܞ‖ ൌ√ܽଶܾଶ In two dimensions, the magnitude of the vector is the length of the hypotenuse of a right triangle with sides ܽ and ܾ. Graphical representation of unit vectors ܑ and j in two dimensions. Version 5.6 Page 129 of 242 April 8, 2023 Chapter 11 Vector Calculus Vector Properties Vectors have a number of nice properties that make working with them both useful and relatively simple. Let ݉ and ݊ be scalars, and let u, v and w be vectors. Then, If ܞൌܑܾܽܒ, then ܽൌ‖ܞ‖ cos ߠ and ܾൌ‖ܞ‖ sin ߠ Then, ܞൌ‖ܞ‖ cos ߠ ܑ ‖ܞ‖ sin ߠ ܒ (note: this formula is used in Force calculations) If ܝൌܽଵܑܾଵܒ and ܞൌܽଶܑܾଶܒ, then ܝܞൌሺܽଵܽଶሻܑሺܾଵܾଶሻܒ If ܞൌܑܾܽܒ, then ݉ܞൌሺ݉ܽሻܑሺܾ݉ሻܒ Define to be the zero vector (i.e., it has zero length, so that ܽൌܾൌ0). Note: the zero vector is also called the null vector. Note: ܞൌܑܾܽܒ can also be shown with the following notation: ܞൌ〈ܽ, ܾ〉. This notation is useful in calculating dot products and performing operations with vectors. Properties of Vectors ܞൌܞൌܞ Additive Identity ܞሺെܞሻൌሺെܞሻܞൌ Additive Inverse ܝܞൌܞܝ Commutative Property ܝሺܞܟሻൌሺܝܞሻܟ Associative Property ݉ሺ࢛݊ሻൌሺ݉݊ሻ࢛ Associative Property ݉ሺ࢛࢜ሻൌ࢛݉݉࢜ Distributive Property ሺ݉݊ሻ࢛ൌ࢛݉݊ܝ Distributive Property 1ሺܞሻൌܞ Multiplicative Identity Also, note that: ‖݉ܞ‖ ൌ|݉| ‖ܞ‖ Magnitude Property ܞ ‖ܞ‖ Unit vector in the direction of ܞ Version 5.6 Page 130 of 242 April 8, 2023 Chapter 11 Vector Calculus General Example 11.1 〈ܽଵ, ܾଵ, ܿଵ〉 〈4, െ3, 2〉 ∘〈ܽଶ, ܾଶ, ܿଶ〉 ∘〈2, െ2, 5〉 ܽଵܽଶܾଵܾଶܿଵܿଶ 8 6 10 ൌ24 alternative vector notation Vector Dot Product The Dot Product of two vectors, ܝൌܽଵܑܾଵܒܿଵܓ and ܞൌܽଶܑܾଶܒܿଶܓ, is defined as follows: ܝ∙ܞൌሺܽଵ∙ܽଶሻሺܾଵ∙ܾଶሻሺܿଵ∙ܿଶሻ It is important to note that the dot product is a scalar, not a vector. It describes something about the relationship between two vectors, but is not a vector itself. A useful approach to calculating the dot product of two vectors is illustrated here: ܝൌܽଵܑܾଵܒ܋ܓൌ〈ܽଵ, ܾଵ, ܿଵ〉 ܞൌܽଶܑܾଶܒ܋ܓൌ〈ܽଶ, ܾଶ, ܿଶ〉 In the example at right the vectors are lined up vertically. The numbers in the each column are multiplied and the results are added to get the dot product. In the example, 〈4, െ3, 2〉∘〈2, െ2, 5〉ൌ8 6 10 ൌ24. Properties of the Dot Product Let ݉ be a scalar, and let u, v and w be vectors. Then, ∘ܝൌܝ∘ൌ0 Zero Property ܑ∘ܒൌܒ∘ܓൌܓ∘ܑൌ0 ܑ, ܒ and ܓ are orthogonal to each other. ܝ∘ܞൌܞ∘ܝ Commutative Property ܝ∘ܝൌ‖ܝ‖ Magnitude Square Property ܝ∘ሺܞܟሻൌሺܝ∘ܞሻሺܝ∘ܟሻ Distributive Property ݉ሺܝ∘ܞሻൌሺ࢛݉ሻ∘ܞൌ࢛∘ሺ݉ܞሻ Multiplication by a Scalar Property More properties: If ܝ∘ܞൌ0 and ܝ് and ܞ്, then ܝ and ܞ are orthogonal (perpendicular). If there is a scalar ݉ such that ࢛݉ൌܞ, then ܝ and ܞ are parallel. If ߠ is the angle between ܝ and ܞ, then cos ߠൌ ܝ ∘ ܞ ‖ܝ‖ ‖ܞ‖ Version 5.6 Page 131 of 242 April 8, 2023 Chapter 11 Vector Calculus Vector Cross Product Cross Product In three dimensions, Let: ܝൌuଵܑ uଶܒ uଷܓ and ܞൌvଵܑ vଶܒ vଷܓ Then, the Cross Product is given by: ܝ x ܞ ൌ อ ܑ ܒ ܓ uଵ uଶ uଷ vଵ vଶ vଷ อ ൌ ሺuଶvଷെuଷvଶሻ ܑ ሺuଷvଵെuଵvଷሻ ܒ ሺuଵvଶെuଶvଵሻ ܓ ܝ x ܞൌ‖ܝ‖ ‖ܞ‖ sin ߠ ܖ The cross product of two nonzero vectors in three dimensions produces a third vector that is orthogonal to each of the first two. This resulting vector ܝ x ܞ is, therefore, normal to the plane containing the first two vectors (assuming ܝ and ܞ are not parallel). In the second formula above, ܖ is the unit vector normal to the plane containing the first two vectors. Its orientation (direction) is determined using the right hand rule. Right Hand Rule Using your right hand: Point your forefinger in the direction of ܝ, and Point your middle finger in the direction of ܞ. Then: Your thumb will point in the direction of ܝ x ܞ. In two dimensions, Let: ܝൌuଵܑ uଶܒ and ܞൌvଵܑ vଶܒ Then, ܝ x ܞ ൌ ቚuଵ uଶ vଵ vଶቚ ൌ ሺuଵvଶെuଶvଵሻ which is a scalar (in two dimensions). The cross product of two nonzero vectors in two dimensions is zero if the vectors are parallel. That is, vectors ܝ and ܞ are parallel if ܝ x ܞ ൌ0. The area of a parallelogram having ܝ and ܞ as adjacent sides and angle θ between them: ܣݎ݁ܽൌܝ x ܞൌ‖ܝ‖ ‖ܞ‖ sin θ. ܝ ܞ ܝx ܞ Version 5.6 Page 132 of 242 April 8, 2023 Chapter 11 Vector Calculus Properties of the Cross Product Let ݉ be a scalar, and let u, v and w be vectors. Then, x ܝൌܝ x ൌ Zero Property ܑ x ܒൌܓ, ܒ x ܓൌܑ, ܓ x ܑൌܒ ܑ, ܒ and ܓ are orthogonal to each other ܒ x ܑൌെܓ, ܓ x ܒൌെܑ, ܑ x ܓൌെܒ Reverse orientation orthogonality ܝ x ܝൌ Every non‐zero vector is parallel to itself ܝ x ܞൌെܞ x ܝ Anti‐commutative Property ܝ x ሺܞܟሻൌሺܝ x ܞሻሺܝ x ܟሻ Distributive Property ሺܝܞሻ x ܟൌሺܝ x ܟሻሺܞ x ܟሻ Distributive Property ሺmܝሻ x ܞ ൌ ܝ x ሺmܞሻ ൌ mሺܝ x ܞሻ Scalar Multiplication More properties: If ܝ x ܞൌ, then ܝ and ܞ are parallel. If ߠ is the angle between ܝ and ܞ, then o ‖ܝ x ܞ‖ ൌ‖ܝ‖ ‖ܞ‖ sin ߠ o sin ߠൌ ‖ܝ ୶ ܞ‖ ‖ܝ‖ ‖ܞ‖ Version 5.6 Page 133 of 242 April 8, 2023 Chapter 11 Vector Calculus Vector Triple Products Scalar Triple Product Let: ܝൌuଵܑ uଶܒ uଷܓ. Then the triple product ܝ∘ሺܞ x ܟሻ gives a scalar representing the volume of a parallelepiped with ܝ, ܞ, and ܟ as edges: ܝ∘ሺܞ x ܟሻ ൌ อ uଵ uଶ uଷ vଵ vଶ vଷ wଵ wଶ wଷ อ ܝ∘ሺܞ x ܟሻ ൌ ሺܝ x ܞሻ ∘ ܟ Other Triple Products ܝ ∘ ሺܝ x ܞሻൌܞ ∘ሺܝ x ܞሻൌ Duplicating a vector results in a product of ܝ x ሺܞ x ܟሻ ൌ ሺܝ ∘ ܟሻ ܞ െ ሺܝ ∘ ܞሻ ܟ ሺܝ x ܞሻ x ܟ ൌ ሺܝ ∘ ܟሻ ܞ െ ሺܞ ∘ ܟሻ ܝ ܝ∘ሺܞ x ܟሻൌ ܞ∘ሺܟ x ܠሻൌ ܟ∘ሺܠ x ܞሻ Note: vectors ܝ, ܞ, and ܟ are coplanar if and only if ܝ∘ሺܞ x ܟሻൌ0. No Associative Property The associative property of real numbers does not translate to triple products. In particular, ሺܝ∘ܞሻ ∙ ܟ ് ܝ ∙ ሺܞ∘ܟሻ No associative property of dot products/multiplication ܝ x ሺܞ x ܟሻ ് ሺܝ x ܞሻ x ܟ No associative property of cross products Version 5.6 Page 134 of 242 April 8, 2023 Chapter 11 Vector Calculus Kinematics (Particle Motion) – Vectors This page is an extension of the Kinematics pages in Chapter 3, adapted to 3‐dimensional space. The corresponding application to 2‐dimensional space would remove the third (i.e., ݖ) component of each vector presented. On this page, 〈 〉 notation is used for the vectors rather than ܑ, ܒ, ܓ notation. Position Position is the location of a particle at a point in time. It may be represented by the vector ܛൌ〈ݔሺݐሻ, ݕሺݐሻ, ݖሺݐሻ〉. Velocity Velocity measures the rate of change in position. Instantaneous velocity is the vector of first derivatives of the position vector ܞൌ〈ݔᇱሺݐሻ, ݕᇱሺݐሻ, ݖᇱሺݐሻ〉. Velocity vector components may be either positive or negative. Speed Speed is the magnitude of the velocity vector; it is always positive. The formula for speed is: ‖ܞ‖ ൌඥሾݔᇱሺݐሻሿଶሾݕᇱሺݐሻሿଶሾݖᇱሺݐሻሿଶ Acceleration Acceleration measures the rate of change in velocity. Instantaneous acceleration is the vector of second derivatives of the position vector ܉ൌ〈ݔᇱᇱሺݐሻ, ݕᇱᇱሺݐሻ, ݖᇱᇱሺݐሻ〉. Moving Among Vectors The following diagram describes how to move back and forth among the position, velocity and acceleration vectors. ௧௧ ሱۛۛۛۛۛۛۛۛۛሮ ௧௧ ሱۛۛۛۛۛۛۛۛۛሮ ܲݏ݅ݐ݅݊ ܸ݈݁ܿ݅ݐݕ ܣ݈ܿܿ݁݁ݎܽݐ݅݊ ூ௧௧ ር ۛۛ ۛ ۛ ۛ ۛ ሲ ூ௧௧ ር ۛۛ ۛ ۛ ۛ ۛ ሲ Displacement Displacement is a measure of the distance between a particle’s starting position and its ending position. The displacement vector from ݐൌܽ to ݐൌܾ may be calculated as: ∆ܛ ൌ 〈නݔᇱሺݐሻ݀ݐ , නݕᇱሺݐሻ݀ݐ , නݖᇱሺݐሻ݀ݐ 〉 Version 5.6 Page 135 of 242 April 8, 2023 Chapter 11 Vector Calculus Gradient Scalar Fields and Vector Fields A Scalar Field in three dimensions provides a value at each point in space. For example, we can measure the temperature at each point within an object. The temperature can be expressed as ܶൌ߶ሺݔ, ݕ, ݖሻ. (note: ߶ is the Greek letter phi, corresponding to the English letter “݂”.) A Vector Field in three dimensions provides a vector at each point in space. For example, we can measure a magnetic field (magnitude and direction of the magnetic force) at each point in space around a charged particle. The magnetic field can be expressed as ܯ ሬ ሬറൌܸ ሬറሺݔ, ݕ, ݖሻ. Note that the half‐arrows over the letters ܯ and ܸ indicate that the function generates a vector field. Del Operator When looking a scalar field it is often useful to know the rates of change (i.e., slopes) at each point in the ݔ‐, ݕ‐ and ݖ‐directions. To obtain this information, we use the Del Operator: સ ൌ ܑ߲ ߲ݔ ܒ߲ ߲ݕ ܓ߲ ߲ݖ Gradient The Gradient of a scalar field ߶ describes the rates of change in the ݔ, ݕ and ݖ directions at each point in the field in vector form. Therefore, the gradient generates a vector field from the points in the scalar field. The gradient is obtained by applying the del operator to ߶. ݃ݎܽ݀ ߶ ൌ સ߶ ൌ ܑ߲߶ ߲ݔ ܒ߲߶ ߲ݕ ܓ߲߶ ߲ݖ డథ డ௫, డథ డ௬ and డథ డ௭ are called directional derivatives of the scalar field ߶. Example 11.2: Suppose: ߶ሺݔ, ݕ, ݖሻൌsin ݔln ݕ݁ି௭ Then: డథ డ௫ ൌcos ݔ, డథ డ௬ ൌ ଵ ௬ and డథ డ௭ ൌെ݁ି௭ So, સ߶ ൌ ܑcos ݔ ܒ1 ݕ െ ܓ݁ି௭ ; providing all three directional derivatives in a single vector. Over a set of points in space, this results in a vector field. At point ܲൌሺ2, 0.5, െ1ሻ, સ߶ ൌ ሺcos 2ሻ ܑ 2 ܒ െ ݁ ܓ ~ െ0.416 ܑ 2 ܒ െ 2.718 ܓ Version 5.6 Page 136 of 242 April 8, 2023 Chapter 11 Vector Calculus Divergence Divergence The Divergence of a vector field describes the flow of material, like water or electrical charge, away from (if positive) or into (if negative) each point in space. The divergence maps the vector at each point in the material to a scalar at that same point (i.e., the dot product of the vector in ܄ and its associated rates of change in the ݔ, ݕ and ݖ directions), thereby producing a scalar field. Let ܄ൌܑV௫ ܒV௬ ܓV௭ where V௫, V௬, V௭ are each functions in ݔ, ݕ and ݖ. Then, ݀݅ݒ ܄ ൌ સ∘܄ ൌ ൬ܑ߲ ߲ݔ ܒ߲ ߲ݕ ܓ߲ ߲ݖ൰ ∘ ൫ܑV௫ ܒV௬ ܓV௭൯ ൌ ߲V௫ ߲ݔ ߲V௬ ߲ݕ ߲V௭ ߲ݖ Points of positive divergence are referred to as sources, while points of negative divergence are referred to as sinks. The divergence at each point is the net outflow of material at that point, so that if there is both inflow and outflow at a point, these flows are netted in determining the divergence (net outflow) at the point. Example 11.3: Let’s start with the vector field created by taking the gradient of ߶ on the prior page. Let: ܄ൌܑcos ݔ ܒ1 ݕ െ ܓ݁ି௭ In this expression, notice that: V௫ൌcos ݔ, V௬ൌ ଵ ௬, and V௭ൌെ݁ି௭. Then: ݀݅ݒ ܄ ൌ સ∘܄ ൌ ߲V௫ ߲ݔ ߲V௬ ߲ݕ ߲V௭ ߲ݖൌ െsin ݔെ1 ݕଶ݁ି௭ Let’s find the value of the divergence at a couple of points, and see what it tells us. At ܲ ଵൌሺെ1, 1, 0ሻ, we have: ݀݅ݒ ܞൌെsinሺെ1ሻെ ଵ ଵమ݁ିൌ 0.841. This value is greater than zero, indicating that ܲ ଵ is a “source”, and that the vector ܞ at ܲ ଵ produces an outflow. At ܲଶൌሺ3, െ1, 2ሻ, we have: ݀݅ݒ ܞൌെsinሺ3ሻെ ଵ ሺିଵሻమ݁ିଶൌ െ1.006. This value is less than zero, indicating that ܲଶ is a “sink”, and that the vector ܞ at ܲଶ produces an inflow. Version 5.6 Page 137 of 242 April 8, 2023 Chapter 11 Vector Calculus Curl Curl The Curl of a vector field describes the circulation of material, like water or electrical charge, about each point in the material. The curl maps the vector at each point in the original vector field to another vector (i.e., the cross product of the original vector and its associated rates of change in the ݔ, ݕ and ݖ directions) at that same point, thereby producing a new vector field. ܿݑݎ݈ ܄ ൌ સ x ܄ ൌ ൬ܑ߲ ߲ݔ ܒ߲ ߲ݕ ܓ߲ ߲ݖ൰ x ൫ܑV௫ ܒV௬ ܓV௭൯ ൌ ተ ተ ܑ ܒ ܓ ߲ ߲ݔ ߲ ߲ݕ ߲ ߲ݖ V௫ V௬ V௭ ተ ተൌ ܑ ቆ߲V௭ ߲ݕെ߲V௬ ߲ݖቇ ܒ ൬߲V௫ ߲ݖെ߲V௭ ߲ݔ൰ ܓ ቆ߲V௬ ߲ݔെ߲V௫ ߲ݕቇ The curl gives the direction of the axis of circulation of material at a point ܲ. The magnitude of the curl gives the strength of the circulation. If the curl at a point is equal to the zero vector (i.e., ), its magnitude is zero and the material is said to be irrotational at that point. Example 11.4: We need to use a more complex vector field for the curl to produce meaningful results. Let: ܄ൌܑ ሺݕݖcos ݔሻ ܒቀݔݕ ݖቁ െ ܓሺ݁ି௫௬௭ሻ In this expression, notice that: V௫ൌݕݖcos ݔ, V௬ൌ ௫௬ ௭, and V௭ൌെ݁ି௫௬௭. Then: ܿݑݎ݈ ܄ ൌ સ x ܄ ൌ ܑ ቆ߲V௭ ߲ݕെ߲V௬ ߲ݖቇ ܒ ൬߲V௫ ߲ݖെ߲V௭ ߲ݔ൰ ܓ ቆ߲V௬ ߲ݔെ߲V௫ ߲ݕቇ ൌܑ ቀݔݖ݁ି௫௬௭ݔݕ ݖଶቁ ܒ ሺݕcos ݔെݕݖ݁ି௫௬௭ሻ ܓ ቀݕ ݖെݖcos ݔቁ Let’s find the value of the curl at a point, and see what it tells us. Let ܲൌሺെ1, 1, 2ሻ. Then, ܿݑݎ݈ ܞൌሺെ2݁ଶെ0.25ሻ ܑሺcosሾെ1ሿെ2݁ଶሻ ܒሺ0.5 െ2 cosሾെ1ሿሻ ܓ ~ െ15.0ܑെ14.2ܒെ0.6ܓ The circulation, then, at Point P is around an axis in the direction of: െ15.0ܑെ14.2ܒെ0.6ܓ The strength of the circulation is given by the magnitude of the curl: ‖ܿݑݎ݈ ܞ‖ ൌඥሺെ15.0ሻଶሺെ14.2ሻଶሺെ0.6ሻଶൌ20.7 Version 5.6 Page 138 of 242 April 8, 2023 Chapter 11 Vector Calculus Laplacian Laplacian The Laplacian Operator is similar to the Del Operator, but involves second partial derivatives. સ ൌ ܑ߲ଶ ߲ݔଶ ܒ߲ଶ ߲ݕଶ ܓ߲ଶ ߲ݖଶ The Laplacian of a scalar field ߶ is the divergence of the gradient of the field. It is used extensively in the sciences. સ߶ ൌ સ ∘ સ߶ ൌ ߲ଶ߶ ߲ݔଶ ߲ଶ߶ ߲ݕଶ ߲ଶ߶ ߲ݖଶ Example 11.5: For the scalar field ߶ሺݔ, ݕ, ݖሻൌsin ݔln ݕ݁ି௭, we already calculated the Laplacian in the example for divergence above (but we did not call it that). It is repeated here with Laplacian notation for ease of reference. Gradient: સ߶ ൌ ܑ߲߶ ߲ݔ ܒ߲߶ ߲ݕ ܓ߲߶ ߲ݖ For the scalar field defined above: డథ డ௫ ൌcos ݔ, డథ డ௬ ൌ ଵ ௬ and డథ డ௭ ൌെ݁ି௭ So, સ߶ ൌ ܑcos ݔ ܒ1 ݕ െ ܓ݁ି௭ Laplacian (Divergence of the Gradient): સ߶ ൌ સ∘સ߶ ൌ ߲ଶ߶ ߲ݔଶ ߲ଶ߶ ߲ݕଶ ߲ଶ߶ ߲ݖଶ ൌ െsin ݔെ1 ݕଶ݁ି௭ Let’s then find the value of the Laplacian at a couple of points. At ܲ ଵൌሺെ1, 1, 0ሻ, we have: સ߶ ൌെsinሺെ1ሻെ ଵ ଵమ݁ିൌ 0.841. At ܲଶൌሺ3, െ1, 2ሻ, we have: સ߶ ൌെsinሺ3ሻെ ଵ ሺିଵሻమ݁ିଶൌ െ1.006. Version 5.6 Page 139 of 242 April 8, 2023 Chapter 12 Sequences Sequences Definitions A Sequence is an ordered set of numbers. A Term is an element in the ordered set of numbers. An Infinite Sequence has no end. A Finite Sequence has a final term. An Explicit Sequence is one that defines the terms of the sequence based on the number of the term. By convention, the number of the term is usually expressed in terms of the variables ݊ or ݇. We talk of the nth term or the kth term of the sequence or series. A Recursive Sequence is one that defines its terms based on one or more previous terms. Types of Sequences A term of a sequence is denoted ࢇ and an entire sequence of terms ሼࢇሽ. Generally (unless otherwise specified), ݊ൌ1 for the first term of a sequence, ݊ൌ2 for the second term, etc. Explicit Sequence: terms of the sequence ሼܽሽ are defined by an Explicit Formula. Example 12.1: ቄ ଶ ଵାቅ ൌ ቄ ଶ ଶ, ସ ଷ, ସ, ଼ ହ, … ቅ Example 12.2: ቄ ଵ ቅ ൌ ቄ1, ଵ ଶ, ଵ ଷ, ଵ ସ, … ቅ Example 12.3: ቄ3 ∙ቀ ଵ ଶቁ ቅ ൌ ቄ ଷ ଶ, ଷ ସ, ଷ ଼, ଷ ଵ, … ቅ Example 12.4: ሼሺെ1ሻሽ ൌ ሼെ1, 1, െ1, 1, … ሽ Example 12.5: ሼܤሽ ൌ ቄ1, െ ଵ ଶ, ଵ , 0, െ ଵ ଷ, 0, ଵ ସଶ, … ቅ Note: this is the sequence of Bernoulli Numbers; it begins with ܤ. Recursive Sequence: Usually, one or more initial terms are defined with values in a recursive sequence. Each subsequent term is defined in terms of previous terms. Example 12.6: ሼ݂ வଶൌ݂ ିଵ݂ ିଶ, ݂ ଵൌ݂ ଶൌ1ሽ ൌ ሼ1, 1, 2, 3, 5, 8, 13, … ሽ Example 12.7: ሼ݂ வଶൌ݂ ିଶെ݂ ିଵ, ݂ ଵൌ3, ݂ ଶൌ1ሽ ൌ ሼ3, 1, 2, െ1, 3, െ4, 7, … ሽ Version 5.6 Page 140 of 242 April 8, 2023 Chapter 12 Sequences More Definitions for Sequences Monotonic Sequence: A sequence is monotonic if its terms are: Non‐increasing (i.e., ܽାଵܽ ∀ ݊), or Non‐decreasing (i.e., ܽାଵܽ ∀ ݊). Note that successive terms may be equal, as long as they do not turn around and head back in the direction from whence they came. Often, you can determine whether a sequence is monotonic by graphing its terms. Bounded Sequence: A sequence is bounded if it is bounded from above and below. A sequence is bounded from above if there is a number ܯ such that ܽܯ ∀ ݊. The least upper bound is called the Supremum. A sequence is bounded from below if there is a number ܰ such that ܽܰ ∀ ݊. The greatest lower bound is called the Infimum. Theorems about Sequences Consider the sequences ሼܽሽ, ሼܾሽ and ሼܿሽ. The following theorems apply: Squeeze Theorem: If ܾܽܿ ∀ ݊some ܰ and lim →ஶܽൌlim →ஶܿൌܮ, then lim →ஶܾൌܮ. Absolute Value Theorem: If lim →ஶ|ܽ| ൌ0 , then lim →ஶܽൌ0. Bounded Monotonic Sequence Theorem: If a sequence is bounded and monotonic, then it converges. Version 5.6 Page 141 of 242 April 8, 2023 Chapter 12 Sequences Limit of a Sequence ۺܑܕܑܜ: lim →ஶܽൌܮ. The limit ܮ exists if we can make ܽ as close to ܮ as we like by making ݊ sufficiently large. Convergence: If the limit of the terms ሼܽሽ exists, the sequence is said to be convergent. Divergence: If the limit of the terms ሼܽሽ does not exist, the sequence is said to be divergent. Limits are determined in the usual manner. Usual properties of limits are preserved in sequences (e. g., addition, scalar multiplication, multiplication, division of limits). Much more about limits is presented in Chapter 1. Version 5.6 Page 142 of 242 April 8, 2023 Chapter 12 Sequences Basic Recursive Sequence Theory Operators and Annihilators An operator, applied to a sequence, results in a new sequence. The operator, ܧ, shifts a sequence ݊ terms to the left. Sequences can also be added or subtracted, as well as multiplied or divided by a scalar (a number). An annihilator of a sequence is an operator that converts a sequence into the zero sequence, i.e., ሼ0ሽ. The typical example of this is the operator ሺܧെ2ሻ operating on the sequence ൛2ൟൌሼ2, 4, 8, 16, … ሽ. Examples of the algebra of operators and annihilators: Example 12.8: ܧ൛2ൟൌܧሼ2, 4, 8, 16, … ሽൌሼ4, 8, 16, 32, … ሽൌ൛2ାଵൟ Example 12.9: 2 ∙൛2ൟൌ2 ∙ሼ2, 4, 8, 16, … ሽൌሼ4, 8, 16, 32, … ሽൌ〈2ାଵ〉 Example 12.10: ሺܧെ2ሻ൛2ൟൌܧ൛2ൟെ2൛2ൟ ൌ൛2ାଵൟെ൛2ାଵൟൌሼ0ሽ Annihilators – Summary The following table summarizes some sequence forms that are annihilated by elementary operators: Form of Sequence Sample Sequence (starting with ࢇ) Annihilator ሼܣሽ ሼ3, 3, 3, 3, … ሽൌሼ3ሽ ܧെ1 ሼܣ݊ܤሽ ሼ4, 9, 14, 19, … ሽൌሼ5݊െ1ሽ ሺܧെ1ሻଶ ሼܣ݊ଶܤ݊ܥሽ ሼ6, 9, 16, 27, … ሽൌሼ2݊ଶെ3݊7ሽ ሺܧെ1ሻଷ ሼܣ∙ݎሽ ሼ6, 12, 24, 48, … ሽൌሼ3 ∙2ሽ ܧെݎ ሼሺܣ݊ܤሻ∙ݎሽ ሼ8, 28, 80, 208, … ሽൌሼሺ3݊1ሻ∙2ሽ ሺܧെݎሻଶ ሼܣ∙ݎܤ∙ݏሽ, ݎ്ݏ ሼ12, 30, 78, 210, … ሽൌሼ3 ∙22 ∙3ሽ ሺܧെݎሻሺܧെݏሻ Version 5.6 Page 143 of 242 April 8, 2023 Chapter 12 Sequences Example 12.11: Fibonacci Sequence The Fibonacci Sequence is defined by the initial conditions, ܨ ൌ0 and ܨ ଵൌ1, and the Recursive Formula: ܨ ൌܨ ିଵܨ ିଶ for ݊1. It looks like this, starting with ܨ : ሼ0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, … ሽ Let’s find the Explicit Formula (i.e., closed form) for the ݊௧ term of the Fibonacci sequence. That is, we will use annihilators to convert: ܨ ൌܨ ିଵܨ ିଶ into ܨ ൌ ఝି ట √ହ . where, ߮ൌ1ඥ5 2 , and ߰ൌ1െඥ5 2 , given the initial conditions: ܨ ൌ0 and ܨ ଵൌ1. Begin with the Fibonacci Recursive Formula: ܨ ൌܨ ିଵܨ ିଶ , or equivalently, ܨ ାଶൌܨ ାଵܨ Convert the equation to operator form in order to identify the annihilator: ܨ ାଶൌܨ ାଵܨ ܨ ାଶെܨ ାଵെܨ ൌ0 ܧଶሺܨ ሻെܧሺܨ ሻെܨ ൌ0 ሺܧଶെܧെ1ሻܨ ൌ0 The resulting operator, ሺܧଶെܧെ1ሻ, is the annihilator. Set this equal to zero, and solve it using the quadratic formula: ܧଶെܧെ1 ൌ 0 ܧൌ1േඥ5 2 . Let: ߮ൌ1ඥ5 2 , and ߰ൌ ଵି√ହ ଶ Because ߮ and ߰ are roots of the equation, the annihilator can be expressed in factored form as: ܧଶെܧെ1 ൌ ሺܧെ߮ሻሺܧെ߰ሻ ൌ 0 Using the annihilator table on the previous page, we see that: ሺܧെ߮ሻ annihilates ሼܣ∙߮ሽ ሺܧെ߰ሻ annihilates ሼܤ∙߰ሽ Version 5.6 Page 144 of 242 April 8, 2023 Chapter 12 Sequences Also from the annihilator table, ሺܧെ߮ሻሺܧെ߰ሻ annihilates sequences of the form: ሼܣ∙߮ܤ∙߰ሽ . It remains to calculate the values of the coefficients ܣ and ܤ. We do this using the initial (i.e., seed) conditions of the Fibonacci Sequence: ܨ ൌ0 and ܨ ଵൌ1, as follows: ܨ ൌ0 ൌ ܣ∙߮ܤ∙߰ ൌ ܣܤ ܨ ଵൌ1 ൌ ܣ∙߮ଵܤ∙߰ଵ ൌ ܣ∙߮ܤ∙߰ From the expression for ܨ , we get: ܤ ൌ െܣ Substituting this into the expression for ܨ ଵ, we get: 1 ൌ ܣ∙߮െܣ∙߰ ൌ ܣ∙ሺ߮െ߰ሻ ൌ ܣ∙ቆ1 √5 2 െ1 െ√5 2 ቇ ൌ ܣ∙√5 This results in the following values for the coefficients: ܣൌ1 √5 , ܤൌെ1 √5 , and, finally, the Explicit Formula for the ݊௧ term of the Fibonacci Sequence: ࡲൌ ࣐ି ࣒ √ where, ߮ൌ1ඥ5 2 , and ߰ൌ ଵି√ହ ଶ . Test Values: ܨ ൌ ఝలି టల √ହ ൌ8 ܨ ଽൌ ఝవି టవ √ହ ൌ34 Version 5.6 Page 145 of 242 April 8, 2023 Chapter 12 Sequences Process for Finding an Explicit Formula In summary, the following steps are used to convert a sequence from recursive to explicit form. Step 1: Identify the Recursive Formula (Example 12.12) ܽൌ3ܽିଵ4ܽିଶെ12ܽିଷ with seed values: ܽൌ0, ܽଵൌ1, ܽଶൌ2 So, ሼܽሽൌሼ0, 1, 2, 10, 26, 94, 266, 862, … ሽ Step 2: Shift the recursive formula so that its lowest subscript is “” ܧଷሺܽሻ ⇒ ܽାଷൌ3ܽାଶ4ܽାଵെ12ܽ Step 3: Use Algebra to get all of the “ࢇ” terms on one side of the equation ܽାଷെ3ܽାଶെ4ܽାଵ12ܽൌ0 Step 4: Express the equation in Operator Form ܧଷሺܽሻെ3ܧଶሺܽሻെ4ܧሺܽሻ12ܽൌ0 ሺܧଷെ3ܧଶെ4ܧ12ሻ ܽൌ0 Step 5: Factor the Operator Expression ሺܧെ2ሻሺܧ2ሻሺܧെ3ሻ ܽൌ0 Step 6: Use the Annihilator Table to develop a general expression for the Explicit Formula ܽൌܣ∙2ܤ∙ሺെ2ሻܥ∙3 Step 7: Use the seed values of the sequence and the formula in Step 6 to create a set of simultaneous equations for the coefficients (e.g., , , ) ݊ൌ0 ⇒ 0 ൌܣܤܥ ݊ൌ1 ⇒ 1 ൌ2ܣെ2ܤ3ܥ ݊ൌ2 ⇒ 2 ൌ4ܣ4ܤ9ܥ Step 8: Solve the simultaneous equations for the coefficients ܣ ൌ െ1 4 ൌ െ5 20 ܤ ൌ െ3 20 ܥ ൌ 2 5 ൌ 8 20 Step 9: Write the Explicit Formula for the sequence ࢇൌെ∙ െ ∙ሺെሻ ૡ∙ Step 10: Test the Explicit Formula for a couple of values in the sequence ܽସൌെ5 ∙2ସെ3 ∙ሺെ2ሻସ8 ∙3ସ 20 ൌ26 ܽହൌെ5 ∙2ହെ3 ∙ሺെ2ሻହ8 ∙3ହ 20 ൌ94 Version 5.6 Page 146 of 242 April 8, 2023 Chapter 13 Series Series Introduction A Series is an ordered summation of a sequence. If ሼ𝑎ሽ is an infinite sequence, then the associated infinite series (or simply series) is: 𝑆ൌ𝑎 ஶ ୀଵ ൌ𝑎ଵ𝑎ଶ𝑎ଷ⋯ The Partial Sum containing the first n terms of ሼ𝑎ሽ is: 𝑆ൌ𝑎 ୀଵ ൌ𝑎ଵ𝑎ଶ𝑎ଷ⋯𝑎 A sequence of partial sums can be formed as follows: ሼ𝑆ሽൌሼ𝑆ଵ, 𝑆௦, 𝑆ଷ, 𝑆ସ, … ሽ Note the following about these formulas: The symbol S is the capital Greek letter sigma, which translates into English as 𝐒, appropriate for the operation of Summation. The letter 𝒌 is used as an index variable in both formulas. The initial (minimum) value of 𝑘 is shown below the summation sign and the terminal (maximum) value of 𝑘 is shown above the summation sign. Letters other than 𝑘 may be used; 𝑖, 𝑗, and 𝑛 are common. When evaluating a series, make sure you review the initial and terminal values of the index variable. Many mistakes are made by assuming values for these instead of using the actual values in the problem. The subscript 𝒏 in 𝑺𝒏 (in the partial sum formula) indicates that the summation is performed only through term 𝑎. This is true whether the formula starts at 𝑘ൌ0, 𝑘ൌ1, or some other value of 𝑘, though alternative notations may be used if properly identified. Convergence and Divergence If the sequence of partial sums ሼ𝑆ሽ converges to 𝑺, the series converges. Not surprisingly, 𝑺 is called the sum of the series. If the sequence of partial sums ሼ𝑆ሽ diverges, the series diverges. Version 5.6 Page 147 of 242 April 8, 2023 Chapter 13 Series Key Properties of Series (these also hold for partial sums) Scalar multiplication ܿ∙ܽ ஶ ୀଵ ൌܿ∙ܽ ஶ ୀଵ Sum and difference formulas ሺܾܽሻ ஶ ୀଵ ൌܽ ஶ ୀଵ ܾ ஶ ୀଵ ሺܽെܾሻ ஶ ୀଵ ൌܽ ஶ ୀଵ െܾ ஶ ୀଵ Multiplication In order to multiply series, you must multiply every term in one series by every term in the other series. Although this may seem daunting, there are times when the products of only certain terms are of interest and we find that multiplication of series can be very useful. ‐th Term Convergence Theorems If ܽ ஶ ୀଵ converges, then lim →ஶܽൌ0. If lim →ஶ്ܽ0, then ܽ ஶ ୀଵ diverges. Power Series A Power Series is an infinite series in which each term is expressed as the product of a constant and a power of a binomial term. Generally, a power series is centered about a particular value of ݔ, which we will call ݔ in the following expression: ࢌሺ࢞ሻൌࢉሺ࢞െࢇሻ ஶ ୀ ൌࢉࢉሺ࢞െࢇሻࢉሺ࢞െࢇሻ ࢉሺ࢞െࢇሻ⋯ Examples of power series are the Taylor and Maclaurin series covered in Chapter 14. Version 5.6 Page 148 of 242 April 8, 2023 Chapter 13 Series Telescoping Series A Telescoping Series is one whose terms partially cancel, leaving only a limited number of terms in the partial sums. The general form of a telescoping series, and its sum are is: ሺܽെܽିଵሻ ஶ ୀଵ ൌܽଵെlim →ஶܽ Convergence: A telescoping series will converge if and only if the limiting term of the series, lim →ஶܽ, is a finite value. Caution: Telescoping series may be deceptive. Always take care with them and make sure you perform the appropriate convergence tests before concluding that the series sums to a particular value. Example 13.1: 1 ݇ଶ݇ ஶ ୀଵ ൌ൬1 ݇െ 1 ݇1൰ ஶ ୀଵ The Partial Sums for this example are: ܵଵൌ൬1 െ1 2൰ ܵଶൌ൬1 െ1 2൰൬1 2 െ1 3൰ൌ൬1 െ1 3൰ ܵଷൌ൬1 െ1 2൰൬1 2 െ1 3൰൬1 3 െ1 4൰ൌ൬1 െ1 4൰ . . . ܵൌ൬1 െ1 2൰൬1 2 െ1 3൰൬1 3 െ1 4൰⋯൬1 ݊െ 1 ݊1൰ൌ൬1 െ 1 ݊1൰ Then, ܵൌ1 െlim →ஶ 1 ݊1 ൌ Notice the usefulness of the telescoping approach in the case of a rational function that can be expressed as partial fractions. This approach will not work for some rational functions, but not all of them.
Version 5.6 Page 149 of 242 April 8, 2023 Chapter 13 Series Geometric Series A Geometric Series has the form: ܵൌܽݎ ஶ ୀ ൌܽܽݎܽݎଶܽݎଷ⋯ If |ݎ| ൏1, then the series converges to: ܵൌܽݎ ஶ ୀ ൌ ܽ 1 െݎ If |ݎ| 1, then the series diverges. Partial Sums Partial sums have the form: ܵൌܽ ܵଵൌܽܽݎ ܵଶൌܽܽݎܽݎଶ . . . ܵൌܽݎ ୀ ൌܽܽݎܽݎଶܽݎଷ⋯ܽݎൌܽሺ1 െݎାଵሻ ሺ1 െݎሻ Example 13.2: ܵൌ 0.9 10 ஶ ୀ ൌ 0.9 1 0.9 10 0.9 100 0.9 1000 ⋯ ൌ 0.9999 ത In this geometric series, we have ܽൌ0.9 and ݎൌ ଵ ଵ. Therefore the series converges to: ܵൌ0.9 ∙൬1 10൰ ஶ ୀ ൌ 0.9 1 െ1 10 ൌ 1 This proves, therefore, that 0.9999 തൌ1. Version 5.6 Page 150 of 242 April 8, 2023 Chapter 13 Series Estimating the Value of a Series with Positive Terms Let the following be true: ݂ሺݔሻ is a positive, decreasing, continuous function for all values of ݔ݉, ݉0. ݂ሺ݇ሻൌܽ for all integer values of ݇݉. is a convergent series with partial sums . The Remainder Term of the sum, after the ݊‐th term, is defined as: ܴൌܵെܵ. Then, න ݂ሺݔሻ ஶ ାଵ ݀ݔ ൏ ܴ ൏ න݂ሺݔሻ ஶ ݀ݔ And so, ܵන ݂ሺݔሻ ஶ ାଵ ݀ݔ ൏ ܵ ൏ ܵන݂ሺݔሻ ஶ ݀ݔ Version 5.6 Page 151 of 242 April 8, 2023 Chapter 13 Series Riemann Zeta Functions (‐Series) Definition The Riemann Zeta Function is defined by the equivalent integral and summation forms: ࣀሺ࢞ሻൌ ડሺ࢞ሻන ࢚࢞ି ࢋ࢚െ ஶ ࢊ࢚ ࣀሺ࢞ሻൌ ௫ ஶ ୀ ൌ ௫ ௫ ௫⋯ The summation form of the function is often called a ‐series (and replaces ݔ in the formula). Zeta functions are generally difficult to evaluate from basic principles. An example of how one of the world’s greatest mathematicians evaluated ࣀሺሻ in 1735 is provided later in this chapter. Positive Even Integers Values of ߞሺݔሻ for positive even integer values of ݔ in closed form (as rational expressions involving π) have been calculated by mathematicians. The formula for these is: ࣀሺ࢞ሻൌ |࢞|ሺ࣊ሻ࢞ ሺ࢞!ሻ where ܤ௫ is the ݔ‐th Bernoulli Number. The decimal approximations below were developed from up to 14 million terms of the ‐series using the Algebra App available at www.mathguy.us. Some values of ߞሺݔሻ for positive even integer values of ݔ are: ߞሺ2ሻ ൌ గమ ൌ 1.644933966 … ߞሺ8ሻ ൌ గఴ ଽସହ ൌ 1.004077356 … ߞሺ4ሻ ൌ గర ଽ ൌ 1.082323233 … ߞሺ10ሻ ൌ గభబ ଽଷହହହ ൌ 1.000994575 … ߞሺ6ሻ ൌ గల ଽସହ ൌ 1.017343061 … ߞሺ12ሻ ൌ ଽଵగభమ ଷ଼ହଵଶ଼ହൌ 1.000246086 … Positive Odd Integers Values of ߞሺݔሻ for positive odd integer values of ݔ do not have a general formula, but can be approximated. ߞሺ1ሻ diverges ߞሺ7ሻ ൌ 1.008349277 … ߞሺ3ሻ ൌ 1.202056903 … ߞሺ9ሻ ൌ 1.002008392 … ߞሺ5ሻ ൌ 1.036927755 … ߞሺ11ሻ ൌ 1.000494188 … ܤൌ1 ܤଵൌെ1 2 ܤଶൌ1 6 ܤଷൌ0 ܤସൌെ1 30 ܤହൌ0 ܤൌ1 42 ܤൌ0 ܤ଼ൌെ1 30 ܤଽൌ0 ܤଵൌ5 66 ܤଵଵൌ0 ܤଵଶൌെ691 2730 Version 5.6 Page 152 of 242 April 8, 2023 Chapter 13 Series Analytic Continuation Consider the following development: Let: ܲൌ1 െ1 1 െ1 1 െ⋯ ܲൌ 1 െ1 1 െ1 ⋯ 2ܲൌ 1 ܲൌ1 2 Next, Let: ܳൌ1 െ2 3 െ4 5 െ⋯ ܳൌ 1 െ2 3 െ4 ⋯ 2ܳൌ1 െ1 1 െ1 1 െ⋯ 2ܳൌܲൌ1 2 ܳൌ1 4 Then, Let: ܵൌ 1 2 3 4 5 6 ⋯ െܳൌെ1 2 െ3 4 െ5 6 െ⋯ ܵെܳൌ 4 8 12 ⋯ ܵെܳ ൌ 4ܵ െܳ ൌ 3ܵ ܵ ൌ െ1 3 ܳ ൌ െ1 3 ∙1 4 ൌ െ1 12 1 2 3 4 5 6 ⋯ ൌ െ1 12 And the Riemann Zeta Function value? This result is consistent with the following value of the Riemann Zeta Function: ߞሺെ1ሻൌെ1 12 How is this possible? See the column to the right for an explanation. Analytic Continuation The results in the left‐hand column are an example of a concept introduced in Complex Analysis (i.e., Calculus of Complex Variables) called Analytic Continuation. Although the results are correct for the value of the function ߞሺെ1ሻ, we cannot conclude that: 1 2 3 4 5 6 ⋯ ൌ െ1 12 Why? Because the series does not converge; therefore, it does not have a value. What does have a value is the function that overlaps the series where the series converges. For values of 1, the Zeta Function and the convergent ‐series are equal: ߞሺሻൌ݇ି ஶ ୀଵ ൌ1 ݇ ஶ ୀଵ The function also exists (i.e., continues) for values of for which the series diverges. This is Analytic Continuation. For another example, consider the following function and series: 1 1 െݔൌ1 ݔݔଶݔଷ⋯ This series converges only for െ1 ൏ݔ൏1. Yet, we can calculate the function value for ݔൌ2. 1 1 െ2 ൌെ1 This does not imply that: 1 2 2ଶ2ଷ⋯ൌെ1 Again, the function continues where the series does not. Version 5.6 Page 153 of 242 April 8, 2023 Chapter 13 Series Euler’s Development of the Value of ࣀሺሻ Definition ࣀሺሻൌ ஶ ୀ ൌ ૢ ⋯ This is also a ‐series with ൌ2. A ‐Series is defined as: ஶ ୀ Euler’s development gives us a glimpse of the extent of his genius. See if you agree. Euler’s Development 1. Begin with the Maclaurin Expansion for: sin ݔ. sin ݔൌݔെݔଷ 3! ݔହ 5! െݔ 7! ݔଽ 9! െ … 2. We know that we can fit a curve of degree ݊ through any set of ݊1 points. Euler proposed that we consider the sine function to be a polynomial of infinite degree that goes through the infinite number of points of the function. Further, he noted that the zeros of the polynomial are the zeros of the sine function, i.e., 0, േߨ, േ2ߨ, േ3ߨ, േ4ߨ… . So, the polynomial for sin ݔ is an infinite product that looks like the following, where ܿ is some constant: sin ݔൌܿ∙ݔ ሺݔଶെߨଶሻሺݔଶെ4ߨଶሻሺݔଶെ9ߨଶሻሺݔଶെ16ߨଶሻ… 3. Divide each term on the right by a factor that results in 1’s before the ݔ’s in each term. Change the lead constant to reflect this. Let’s call the new lead constant ݇. sin ݔൌ݇∙ݔ ሺݔଶെߨଶሻ െߨଶ ∙ሺݔଶെ4ߨଶሻ െ4ߨଶ ∙ሺݔଶെ9ߨଶሻ െ9ߨଶ ∙ሺݔଶെ16ߨଶሻ െ16ߨଶ … ൌ݇∙ݔ ቆ1 െݔଶ ߨଶቇቆ1 െݔଶ 4ߨଶቇቆ1 െݔଶ 9ߨଶቇቆ1 െ ݔଶ 16ߨଶቇ… ‐series converge for 1 and diverge for 1. Version 5.6 Page 154 of 242 April 8, 2023 Chapter 13 Series 4. Determine the value of ݇ by dividing each side by ݔ and evaluating the result at ݔൌ0. lim ௫ → sin ݔ ݔ ൌ݇ቆ1 െݔଶ ߨଶቇቆ1 െݔଶ 4ߨଶቇቆ1 െݔଶ 9ߨଶቇቆ1 െ ݔଶ 16ߨଶቇ… Use L’Hospital’s Rule on the left side to determine that lim ௫ → ୱ୧୬௫ ௫ ൌlim ௫ → ୡ୭ୱ௫ ଵ ൌ1. Then, 1 ൌ݇ሺ1ሻሺ1ሻሺ1ሻ… so, ݇ൌ1. 5. Rewrite the polynomial in Step 3 with ݇ൌ1. sin ݔൌ࢞ ቆ1 െ࢞ ߨଶቇቆ1 െ࢞ 4ߨଶቇቆ1 െ࢞ 9ߨଶቇቆ1 െ ࢞ 16ߨଶቇ… 6. Let’s examine the coefficient of ݔଷ in the equation in Step 5. The coefficient of the ݔଷ term in this product is obtained by multiplying ࢞ by the ࢞ part of one of the other terms and 1’s in the rest of the other terms. We sum the result of this across all of the multiplied terms to get the following ݔଷ term for the equation in Step 5: ൬െ1 ߨଶെ1 4ߨଶെ1 9ߨଶെ 1 16ߨଶെ⋯൰ݔଷ 7. The ݔଷ term in Step 1 must be equal to the ݔଷ term in Step 6, since both represent the ݔଷ term in an expansion for sin ݔ. Equating the two coefficients of ݔଷ gives: െ1 3! ൌ െ1 ߨଶെ1 4ߨଶെ1 9ߨଶെ 1 16ߨଶെ⋯ 8. Multiply both sides of the result in Step 7 by െߨଶ to get: ߨଶ 6 ൌ 1 1 1 4 1 9 1 16 ⋯ ൌߞሺ2ሻ So, ࣀሺሻൌ ஶ ୀ ൌ ࣊ Version 5.6 Page 155 of 242 April 8, 2023 Chapter 13 Series Bernoulli Numbers and the Riemann Zeta Function Summation Formulas for Powers of Positive Integers 𝑆ሺ𝑛െ1ሻൌ1 ିଵ ୀ ൌ1 1 ⋯1 ൌ𝑛 ൌ𝑛 𝑆ଵሺ𝑛െ1ሻൌ𝑘 ିଵ ୀ ൌ1 2 ⋯ሺ𝑛െ1ሻൌ𝑛ሺ𝑛െ1ሻ 2 ൌ1 2 𝑛ଶെ1 2 𝑛 𝑆ଶሺ𝑛െ1ሻൌ𝑘ଶ ିଵ ୀ ൌ1ଶ2ଶ⋯ሺ𝑛െ1ሻଶൌ𝑛ሺ𝑛െ1ሻሺ2𝑛െ1ሻ 6 ൌ1 3 𝑛ଷെ1 2 𝑛ଶ1 6 𝑛 𝑆ଷሺ𝑛െ1ሻൌ𝑘ଷ ିଵ ୀ ൌ1ଷ2ଷ⋯ሺ𝑛െ1ሻଷൌ𝑛ଶሺ𝑛െ1ሻଶ 4 ൌ1 4 𝑛ସെ1 2 𝑛ଷ1 4 𝑛ଶ 𝑆ସሺ𝑛െ1ሻൌ𝑘ସ ିଵ ୀ ൌ14 24 ⋯ሺ𝑛െ1ሻ4 ൌ 𝑛ሺ𝑛െ1ሻሺ2𝑛െ1ሻሺ3𝑛2 െ3𝑛െ1ሻ 30 ൌ 1 5 𝑛5 െ 1 2 𝑛4 1 3 𝑛3 െ 1 30 𝑛 The coefficients of the “𝑛” terms are called Bernoulli Numbers. A recursive formula for the Bernoulli numbers is: 𝐵ൌቀ𝑛 𝑘ቁ𝐵 ୀ ⇒ 𝐵ൌ𝐵ቀ𝑛 1ቁ𝐵ଵቀ𝑛 2ቁ𝐵ଶ⋯ቀ 𝑛 𝑛െ2ቁ𝐵ିଶቀ 𝑛 𝑛െ1ቁ𝐵ିଵ𝐵 0 ൌ𝐵ቀ𝑛 1ቁ𝐵ଵቀ𝑛 2ቁ𝐵ଶ⋯ቀ 𝑛 𝑛െ2ቁ𝐵ିଶ𝑛𝐵ିଵ 𝐵ିଵൌെ1 𝑛ቂ𝐵ቀ𝑛 1ቁ𝐵ଵቀ𝑛 2ቁ𝐵ଶ⋯ቀ 𝑛 𝑛െ2ቁ𝐵ିଶቃൌെ1 𝑛ቀ𝑛 𝑘ቁ𝐵 ିଶ ୀ Then, we can calculate successive Bernoulli Numbers, starting with 𝐵ൌ1 as: 𝐵ൌ1 𝐵ଵൌെ1 2 ሺ1 ∙𝐵ሻൌെ1 2 𝐵ଶൌെ1 3 ሺ1 ∙𝐵3 ∙𝐵ଵሻൌെ1 3 1 ∙ሺ1ሻ3 ∙൬െ1 2൰൨ൌ1 6 𝐵ଷൌെ1 4 ሺ1 ∙𝐵4 ∙𝐵ଵ6 ∙𝐵ଶሻൌെ1 4 1 ∙ሺ1ሻ4 ∙൬െ1 2൰6 ∙൬1 6൰൨ൌ0 𝐵ସൌെ1 5 ሺ1 ∙𝐵5 ∙𝐵ଵ10 ∙𝐵ଶ10 ∙𝐵ଷሻൌെ1 5 1 ∙ሺ1ሻ5 ∙൬െ1 2൰10 ∙൬1 6൰10 ∙0൨ൌെ1 30 Version 5.6 Page 156 of 242 April 8, 2023 Chapter 13 Series The blue numbers in the above formulas are the values from Pascal’s Triangle, excluding the last two diagonal columns: ܤൌ0 for every odd ݊1. Below are values of ܤ for even values of ݊24. Note: |ܤ2݊| ~ 4ߨ√݁ቀ గቁ ଶାଵ/ଶ as ݊ →∞ ܤଶൌ1 6 ܤ଼ൌെ1 30 ܤଵସൌ7 6 ܤଶൌെ174611 330 ܤସൌെ1 30 ܤଵൌ5 66 ܤଵൌെ3617 510 ܤଶଶൌ854513 138 ܤൌ1 42 ܤଵଶൌെ691 2730 ܤଵ଼ൌ43867 798 ܤଶସൌെ236364091 2730 Bernoulli Numbers relate to the Riemann Zeta Functions as follows: ߞሺ2݊ሻൌ1 ݇ଶ ஶ ୀଵ ൌ1 1ଶ1 2ଶ1 3ଶ⋯ൌ|ܤଶ|ሺ2ߨሻଶ 2 ሺ2݊ሻ! ߞሺ2ሻൌ1 ݇ଶ ஶ ୀଵ ൌ1 1ଶ1 2ଶ1 3ଶ⋯ൌߨଶ 6 ߞሺ6ሻൌ1 ݇ ஶ ୀଵ ൌ1 11 21 3⋯ൌߨ 945 ߞሺ4ሻൌ1 ݇ସ ஶ ୀଵ ൌ1 1ସ1 2ସ1 3ସ⋯ൌߨସ 90 ߞሺ8ሻൌ1 ଼݇ ஶ ୀଵ ൌ1 1଼1 2଼1 3଼⋯ൌ ߨ଼ 9450 The function ௫ ೣିଵ expands using Bernoulli Numbers as follows: ݔ ݁௫െ1 ൌܤ ݔ ݇!
ஶ ୀ ൌ1 െ1 2 ݔ1 6 ݔଶെ1 30 ݔସ1 42 ݔെ⋯ Version 5.6 Page 157 of 242 April 8, 2023 Chapter 13 Series Series Convergence Tests Integral Test Let ∑ܽ be a positive series, and let ݂ሺݔሻ be a continuous, positive, decreasing function on ሾ݉, ∞ሻ, ݉0, such that ݂ሺ݇ሻൌܽ for every ݇݉. Then, ܽ ஶ ୀ converges if and only if න݂ሺݔሻ݀ݔ converges ஶ . If the series converges, . That is, the sum of the series and the integral will have different values. Comparison Test Let ∑ܽ and ∑ܾ be positive series. If there is an index ݉, beyond which ܽ൏ܾ for every ݇݉, then: If ∑ܾ converges, so does ∑ܽ. If ∑ܽ diverges, so does ∑ܾ. Limit Comparison Test Let ∑ܽ and ∑ܾ be positive series such that 0 ൏lim →ஶ ൏∞. Then: ∑ܽ converges if and only if ∑ܾ converges. ∑ܽ diverges if and only if ∑ܾ diverges. Absolute and Conditional Convergence ∑ܽ is absolutely convergent if ∑|ܽ| is convergent. ∑ܽ is conditionally convergent if it is convergent but not absolutely convergent. Term Rearrangement If an infinite series is absolutely convergent, the terms can be rearranged without affecting the resulting sum. If an infinite series is conditionally convergent, a rearrangement of the terms may affect the resulting sum. Version 5.6 Page 158 of 242 April 8, 2023 Chapter 13 Series ݇ ݇!
ஶ ୀଵ Ratio Test Let ∑ܽ be a series. Then consider the ݊‐th and ሺ݊1ሻ‐th terms: If lim →ஶฬܽାଵ ܽ ฬ൏1, then: ܽ is absolutely convergent. If lim →ஶฬܽାଵ ܽ ฬ1, then: ܽ is divergent. If lim →ஶฬܽାଵ ܽ ฬൌ1, then no conclusion about convergence or divergence can be drawn. Example 13.3: Determine whether the following series converges or diverges: Ratio ൌ ተ ሺ݊1ሻାଵ ሺ݊1ሻ!
݊ ݊!
ተ ൌ ሺ݊1ሻାଵ ሺ݊1ሻ! ∙݊!
݊ ൌ ሺ݊1ሻ∙ሺ݊1ሻ ሺ݊1ሻ∙݊!
∙݊!
݊ ൌ ሺ݊1ሻ ݊ ൌ ൬݊1 ݊ ൰ ൌ ൬1 1 ݊൰ Then, lim →ஶቀ1 ଵ ቁ ൌ ݁ 1 Since ࢋ, the series diverges. Version 5.6 Page 159 of 242 April 8, 2023 Chapter 13 Series Root Test Let ∑ܽ be a series. Then consider the ݊‐th term: If lim →ஶඥ|ܽ| ൏1, then: ܽ is absolutely convergent. If lim →ஶඥ|ܽ| 1, then: ܽ is divergent. If lim →ஶඥ|ܽ| ൌ1, then no conclusion about convergence or divergence can be drawn. Example 13.4: Determine whether the following series converges or diverges: Root ൌ ඨቤ൬2݊3 3݊2൰ ቤ ൌ ඨ൬2݊3 3݊2൰ ൌ 2݊3 3݊2 ൌ 2 3 ݊ 3 2 ݊ Then, lim →ஶቆ ଶାయ ଷାమ ቇ ൌ ଶ ଷ ൏ 1 Since ൏, the series converges. ൬2݇3 3݇2൰ ஶ ୀଵ Version 5.6 Page 160 of 242 April 8, 2023 Chapter 13 Series Dirichlet’s Convergence Test Dirichlet’s Test If has bounded partial sums and ሼ𝑣𝑘ሽ is a decreasing positive sequence with lim 𝑘 →∞𝑣𝑘ൌ0, then the series converges. Example 13.5: Prove that the series converges using Dirichlet’s Convergence Test. Using the notation shown above, we will let 𝑎𝑘ൌcos 𝑘, and 𝑣𝑘ൌ 1 𝑘. We require only that be bounded since lim 𝑘 →∞ 1 𝑘 clearly decreases to 0. Start by proving the following Trigonometric identity: cos 𝑘ൌ sin ቀ𝑘1 2ቁെsin ቀ𝑘െ1 2ቁ 2 sin 1 2 ൌ ቀsin 𝑘cos 1 2 cos 𝑘sin 1 2ቁെቀsin 𝑘cos 1 2 െcos 𝑘sin 1 2ቁ 2 sin 1 2 ൌ 2 cos 𝑘sin 1 2 2 sin 1 2 ൌ cos 𝑘 Next, let’s look at the 𝑛‐th Partial Sum in light of the above identity. Note that it telescopes: cos 𝑘 𝑛 𝑘ൌ1 ൌ sin ቀ𝑘1 2ቁെsin ቀ𝑘െ1 2ቁ 2 sin 1 2 𝑛 𝑘ൌ1 ൌ sin ቀ1 1 2ቁെsin ቀ1 2ቁ 2 sin 1 2 sin ቀ2 1 2ቁെsin ቀ1 1 2ቁ 2 sin 1 2 ⋯ sin ቀ𝑛1 2ቁെsin ቀ𝑛െ1 2ቁ 2 sin 1 2 ൌ sin ቀ𝑛1 2ቁെsin ቀ1 2ቁ 2 sin ቀ1 2ቁ Note that sin ቀ𝑛 1 2ቁ is bounded in the range ሾെ1, 1ሿ. Therefore, is bounded in the range: ቈ െ1െsinቀ1 2ቁ 2 sinቀ1 2ቁ, 1െsinቀ1 2ቁ 2 sinቀ1 2ቁ, and so the original series converges. cos 𝑘 ∞ 𝑘ൌ1 cos 𝑘 𝑛 𝑘ൌ1 cos 𝑘 𝑘 ∞ 𝑘ൌ1 𝑎𝑘 ∞ 𝑘ൌ1 ሺ𝑎𝑘∙𝑣𝑘ሻ ∞ 𝑘ൌ1 Version 5.6 Page 161 of 242 April 8, 2023 Chapter 13 Series cos ቀ1 𝑘2ቁ 𝑘2 ∞ 𝑘ൌ1 Abel’s Convergence Test Abel’s Test If converges and ሼ𝑣𝑘ሽ is a monotonic bounded sequence, then the series converges. Example 13.6: Prove that the series converges using Abel’s Convergence Test. Using the notation shown above, we will let 𝑎𝑘ൌ1 𝑘2, and 𝑣𝑘ൌcos ൬1 𝑘2൰. We need to show that a) converges and b) cos ൬1 𝑘2൰ is both monotonic and bounded. First, identify as a 𝑝‐series, with 𝑝ൌ2, so it converges. Second, let’s look at some values of cos ൬1 𝑘2൰ in the table to the right. The sequence is clearly monotonic and is bounded by the value of cos 0 ൌ1. lim 𝑘 →∞cos ൬1 𝑘2൰ൌ1 We have met both requirements of Abel’s Convergence Test, and we can conclude that the given series converges. Note: the series in this example could also have been determined to be convergent (using the comparison test) by comparing it to a 𝑝‐series with 𝑝ൌ2. 𝒌 𝐜𝐨𝐬൬𝟏 𝒌𝟐൰ 1 cos ൬1 1൰ ~ 0.5403 2 cos ൬1 4൰ ~ 0.9689 3 cos ൬1 9൰ ~ 0.9938 4 cos ൬1 16൰~ 0.9980 5 cos ൬1 25൰~ 0.9992 1 𝑘2 ∞ 𝑘ൌ1 𝑎𝑘 ∞ 𝑘ൌ1 ሺ𝑎𝑘∙𝑣𝑘ሻ ∞ 𝑘ൌ1 1 𝑘2 ∞ 𝑘ൌ1 Version 5.6 Page 162 of 242 April 8, 2023 Chapter 13 Series ሺെ1ሻିଵ 6 ஶ ୀଵ Alternating Series The general form for an Alternating Series that includes an error term is: ݂ሺݔሻൌሺെ1ሻିଵ ஶ ୀଵ ܽൌܽଵെܽଶܽଷെܽସ⋯ Theorem: If the sequence, 〈ܽ〉, is positive and non‐increasing, and lim →ஶܽൌ0, Then: ∑ሺെ1ሻିଵܽ converges, and If ܴ is the nth error term, then: |ܴ| ൏ܽାଵ Error Term The maximum error in a converging alternating series after ݊ terms is term ሺെ1ሻܽାଵ. Using this, we can estimate the value of a series to a desired level of accuracy. Example 13.7: Approximate the following sum to 4 decimal places: We need to find term ݊1 to estimate the error. For the series provided, this term is ሺିଵሻ శభ. The ሺെ1ሻ term simply indicates the direction of the error. The magnitude of the error is the balance of the error term, i.e., ଵ శభ. In order to find an approximation of the series to 4 decimal places, we need an error less than 0.00005. So, we want: 1 6ାଵ ൏ 0.00005 We can solve this using logarithms or by taking successive powers of ଵ . Either way, we find: ଵ ఱ ~ 0.0001286 0.00005 and ଵ ల ~ 0.0000214 ൏0.00005, so ݊1 ൌ6, and ݊ൌ5. Using 5 terms of the alternating series, we find that the value of the sum to 4 decimal places is: ሺെ1ሻିଵ 6 ହ ୀଵ ൌ 1 6 െ1 36 1 216 െ 1 1296 1 7776 ൌ. ૡૠ The actual value of the series is ଵ ~ 0.1428571, so we can see that the desired level of accuracy has been achieved. Version 5.6 Page 163 of 242 April 8, 2023 Chapter 13 Series Absolute and Conditional Convergence ∑ܽ is absolutely convergent if ∑|ܽ| is convergent. ∑ܽ is conditionally convergent if it is convergent but not absolutely convergent. Term Rearrangement If an infinite series is absolutely convergent, the terms can be rearranged without affecting the resulting sum. If an infinite series is conditionally convergent, a rearrangement of the terms may affect the resulting sum. More Theorems about Absolutely Convergent Series The following theorems apply to absolutely convergent series (i.e., absolutely convergent alternating series and convergent series of decreasing positive terms): The commutative law applies to terms in an absolutely convergent series; i.e., terms can be rearranged without affecting the value of the series. Every sub‐series of an absolutely convergent series is absolutely convergent; i.e., terms can be omitted and the result is an absolutely convergent series. The sum, difference and product of absolutely convergent series are absolutely convergent. Furthermore, if ∑ܽ and ∑ܾ are two absolutely convergent series such that A ൌ∑ܽ and B ൌ∑ܾ, then: o ∑ܽ∑ܾൌA B. o ∑ܽെ∑ܾൌA െB. o ∑ܽ∙∑ܾൌA ∙B. Version 5.6 Page 164 of 242 April 8, 2023 Chapter 13 Series Radius and Interval of Convergence of Power Series Consider the Power Series: ࢌሺ࢞ሻൌࢉሺ࢞െࢇሻ ஶ ୀ ൌࢉࢉሺ࢞െࢇሻࢉሺ࢞െࢇሻ ࢉሺ࢞െࢇሻ⋯ Definitions: Center: The value ܽ is called the center of the power series. Many power series have a center of ܽൌ0. Coefficients: The values ܿ are called the coefficients of the power series. Radius of Convergence: The series may converge for certain values of ݔ and diverge for other values of ݔ. If the series converges for all values of ݔ within a certain distance, ܴ, from ܽ, i.e., for ݔ on the interval ሺܽെܴ, ܴܽሻ, we call ܴ the radius of convergence of the series. Interval of Convergence: The set of all values of ݔ for which the power series converges is called the interval of convergence of the series. The interval of convergence is closely related to the radius of convergence; it includes the open interval ሺܽെܴ, ܴܽሻ, and may also include one or both endpoints of that interval. Finding the Radius and Interval of Convergence The radius of convergence is found using the Ratio Test or the Root Test. To find the interval of convergence, the series defined at each endpoint of the interval must be tested separately. Example 13.8: Consider the power series: ଵ ଵା௫ ൌ1 െݔݔଶെݔଷ⋯ൌ Using the Ratio Test, we find: lim →ஶቚ శభ ቚൌlim →ஶቚ ሺିଵሻశభ௫శభ ሺିଵሻ௫ ቚൌlim →ஶ|ݔ| ൏1 … in the open interval: ݔ∈ሺെ1, 1ሻ. So, this series has a radius of convergence: ܴൌ1 about a center of ݔൌ0. To find the interval of convergence of the series, we must test the endpoints, i.e., ݔൌേ1. When ݔൌ1, we get ଵ ଵାଵ ൌ1 െ1 1ଶെ1ଷ⋯ , which diverges. When ݔൌെ1, we get ଵ ଵିଵ ൌ1 1 1ଶ1ଷ⋯ , which also diverges. The interval of convergence, then, is ሺെ1, 1ሻ. It does not include either endpoint. Version 5.6 Page 165 of 242 April 8, 2023 Chapter 13 Series Differentiating or Integrating Power Series When differentiating or integrating a Power Series, we differentiate or integrate term‐by‐term. Example 13.9: Integrate the power series: ଵ ଵା௫ ൌ1 െݔݔଶെݔଷ⋯ න 1 1 ݔ݀ݔൌනሺ1 െݔݔଶെݔଷ⋯ሻ݀ݔ ൌݔെ1 2 ݔଶ1 3 ݔଷെ1 4 ݔସ⋯ܥ The result of the integration turns out to be the power series for lnሺ1 ݔሻ, plus a constant, which we would expect to be the case because: න 1 1 ݔ݀ݔൌlnሺ1 ݔሻܥ. Theorem: Differentiation of a Power Series If a function ݂ is defined by a power series with radius of convergence ܴ, then: ݂ is differentiable on the open interval defined by ܴ. ݂ᇱሺݔሻ is found by term‐by‐term differentiation of the power series for ݂. The resulting power series for ݂′ also has radius of convergence ܴ. The interval of convergence of ݂’ may be the same as that for ݂, or it may lose either or both endpoints. Theorem: Integration of a Power Series If a function ݂ is defined by a power series with radius of convergence ܴ, then: ܨሺݔሻൌ ݂ሺݔሻ݀ݔ is found by term‐by‐term integration of the power series for ݂. The resulting power series for ܨ also has radius of convergence ܴ. The interval of convergence of ܨ may be the same as that for ݂, or it may gain either or both endpoints. Relative to ࢌሺ࢞ሻ Differentiation: ࢌ′ሺ࢞ሻ Term‐by‐term differentiation. Has same Ratio of Convergence. Interval of Convergence may lose one or both endpoints. Integration: ࡲሺ࢞ሻൌ ࢌሺ࢞ሻࢊ࢞ Term‐by‐term integration. Has same Ratio of Convergence. Interval of Convergence may gain one or both endpoints. Version 5.6 Page 166 of 242 April 8, 2023 Chapter 13 Series ܥሺെ1ሻିଵ ஶ ୀଵ ݔ ݊ Example 13.10: The Maclaurin Series for ଵ ଵା௫ is: 1 1 ݔൌ1 െݔݔଶെݔଷ⋯ As shown on a previous page, its interval of convergence is ሺെ1, 1ሻ. Integrating term‐by‐term we get: න 1 1 ݔ݀ݔൌනሺ1 െݔݔଶെݔଷ⋯ሻ݀ݔ lnሺ1 ݔሻܥൌݔെ ଵ ଶݔଶ ଵ ଷݔଷെ ଵ ସݔସ⋯ܥ ൌ For the new series, lnሺ1 ݔሻܥ, note that “ܥ” has no impact on whether the series converges or diverges at any point. Then, Using the Ratio Test, we find: lim →ஶቚ శభ ቚൌlim →ஶቚ ሺିଵሻ௫శభ/ሺାଵሻ ሺିଵሻషభ௫/ቚൌlim →ஶቚݔ∙ ାଵቚൌlim →ஶ|ݔ| ൏1 … in the open interval: ݔ∈ሺെ1, 1ሻ. So, this series also has a radius of convergence ܴൌ1 about a center of ݔൌ0. To find the interval of convergence of the series, we must test the endpoints, i.e., ݔൌേ1. When ݔൌ1, we get ln 2 ൌ1 െ ଵ ଶ ଵ ଷെ ଵ ସ⋯ , which converges by the alternating series test. When ݔൌെ1, we get ln 0 ൌെ1 െ ଵ ଶെ ଵ ଷെ ଵ ସെ⋯ , which diverges (it is the negative harmonic series, and ln 0 is undefined). The interval of convergence, then, is ሺെ1, 1ሿ. It includes the right endpoint. Conclusion: In the case of this example, the interval of convergence of the integrated series picks up the endpoint at ݔൌ1. Version 5.6 Page 167 of 242 April 8, 2023 Chapter 13 Series McCartin Table: Summary of Basic Tests for Series Test Series Form Conditions for Convergence Conditions for Divergence Comments ݊‐th term (tests for divergence only) ܽ ஶ ୀଵ lim →ஶܽൌ0 required, but not sufficient lim →ஶ്ܽ0 This test should always be performed first. Special Series Telescoping Series ሺܽെܽାଵሻ ஶ ୀଵ lim →ஶܽ is finite lim →ஶܽ not finite ܵൌܽଵെlim ݊ →∞ܽ݊ Geometric Series (്ܽ0) ܽݎିଵ ஶ ୀଵ |ݎ| ൏1 |ݎ| 1 ܵൌ ܽ 1 െݎ ‐Series 1 ݊ ஶ ୀଵ 1 1 1 ݊ ஶ ୀଵ ൌߞሺሻሺ1ሻ Alternating Series ሺ0 ൏ܽାଵܽሻ ሺെ1ሻିଵܽ ஶ ୀଵ lim →ஶܽൌ0 lim →ஶ്ܽ0 Remainder: |ܴ| ܽାଵ Integral ሺ݂ is positive, continuous, and decreasingሻ ܽ ஶ ୀଵ න݂ሺݔሻ ஶ ଵ ݀ݔ converges (2) න݂ሺݔሻ ஶ ଵ ݀ݔ diverges Remainder: 0 ൏ܴ൏න݂ሺݔሻ ஶ ݀ݔ Comparison (ܽ0, ܾ0) ܽ ஶ ୀଵ ܾܽ ሺ݊݉ሻ ܾ ஶ ୀଵ converges ܾܽ ሺ݊݉ሻ ܾ ஶ ୀଵ diverges Comparison of ܽ and ܾ need only exist for ݊ beyond some index ݉. Limit Comparison (ܽ0, ܾ0) ܽ ஶ ୀଵ 0 ൏lim →ஶ ܽ ܾ ൏∞ ܾ ஶ ୀଵ converges lim →ஶ ܽ ܾ 0 ܾ ஶ ୀଵ diverges Could use lim →ஶ instead of lim →ஶ in the conditions. Ratio ܽ ஶ ୀଵ lim →ஶቚ శభ ቚ൏1 (absolute convergence) lim →ஶฬܽାଵ ܽ ฬ1 Test inconclusive if: lim →ஶฬܽାଵ ܽ ฬൌ1. Use another test. Root ܽ ஶ ୀଵ lim →ஶඥ|ܽ| ൏1 (absolute convergence) lim →ஶඥ|ܽ| 1 Test inconclusive if: lim →ஶඥ|ܽ| ൌ1. Use another test. Notes: (1) Riemann zeta function. (2) If the series converges, ∑്ܽ ݂ሺݔሻ݀ݔ ஶ ଵ . Version 5.6 Page 168 of 242 April 8, 2023 Chapter 14 Taylor and Maclaurin Series Taylor and Maclaurin Series Taylor Series A Taylor series is an expansion of a function around a given value of ݔ. Generally, it has the following form around the point ݔൌܽ: ݂ሺݔሻൌ݂ሺሻሺܽሻ ݇!
ሺݔെܽሻ ஶ ୀ ൌ݂ሺܽሻ݂ᇱሺܽሻ 1!
ሺݔെܽሻ݂ᇱᇱሺܽሻ 2!
ሺݔെܽሻଶ ݂ᇱᇱᇱሺܽሻ 3!
ሺݔെܽሻଷ⋯ Maclaurin Series A Maclaurin series is a Taylor Series around the value ݔൌ0. Generally, it has the following form: ݂ሺݔሻൌ݂ሺሻሺ0ሻ ݇!
ஶ ୀ ݔൌ݂ሺ0ሻ݂ᇱሺ0ሻ 1! ݔ݂ᇱᇱሺ0ሻ 2! ݔଶ ݂ᇱᇱᇱሺ0ሻ 3! ݔଷ⋯ Example . : Find the Maclaurin expansion for ݂ሺݔሻൌ݁௫: ݂ሺݔሻൌ݁௫ ݂ሺ0ሻൌ݁ൌ1 ݂ᇱሺݔሻൌ݁௫ ݂ᇱሺ0ሻൌ݁ൌ1 ݂ᇱᇱሺݔሻൌ݁௫ ݂ᇱᇱሺ0ሻൌ݁ൌ1 . . . ݂ሺሻሺݔሻൌ݁௫ ݂ሺሻሺ0ሻൌ݁ൌ1 Substituting these values into the Maclaurin expansion formula (and recalling that 0! ൌ1) we get: ݁௫ ൌ 1 ݔݔଶ 2! ݔଷ 3! ݔସ 4! ⋯ ൌ ݔ ݇!
ஶ ୀ Version 5.6 Page 169 of 242 April 8, 2023 Chapter 14 Taylor and Maclaurin Series Example . : Find the Maclaurin expansion for ݂ሺݔሻൌlnሺ1 ݔሻ: ݂ሺݔሻൌlnሺ1 ݔሻ ݂ሺ0ሻൌlnሺ1 0ሻൌ0 ݂ᇱሺݔሻൌ 1 1 ݔ ݂ᇱሺ0ሻൌ 1 1 0 ൌ1 ݂ᇱᇱሺݔሻൌെ 1 ሺ1 ݔሻଶ ݂ᇱᇱሺ0ሻൌെ 1 ሺ1 0ሻଶൌെ1 ൌെ1! ݂ᇱᇱᇱሺݔሻൌ 2 ሺ1 ݔሻଷ ݂ᇱᇱᇱሺ0ሻൌ 2 ሺ1 0ሻଷൌ2 ൌ2! ݂௩ሺݔሻൌെ 6 ሺ1 ݔሻସ ݂௩ሺ0ሻൌെ 6 ሺ1 ݔሻସൌെ6 ൌെ3! . . . ݂ሺሻሺݔሻൌሺെ1ሻିଵሺ݊െ1ሻ!
ሺ1 ݔሻ ݂ሺሻሺ0ሻൌሺെ1ሻିଵሺ݊െ1ሻ! Substituting these values into the Maclaurin expansion formula, we get: lnሺ1 ݔሻ ൌ ݔെ 1 2! ݔଶ 2!
3! ݔଷെ 3!
4! ݔସ⋯ሺെ1ሻିଵሺ݊െ1ሻ!
݊! ݔ⋯ ൌ ݔെݔଶ 2 ݔଷ 3 െݔସ 4 ݔହ 5 െݔ 6 ⋯ ൌ ሺെ1ሻିଵ ݔ ݊ ஶ ୀଵ Taylor Series Convergence Theorem A Taylor Series for a function ݂ሺݔሻ that has derivatives of all orders on an open interval centered at ݔൌܽ converges if and only if: lim →ஶܴሺݔሻൌlim →ஶ ݂ሺାଵሻሺݔ∗ሻ ሺ݊1ሻ! ሺݔെܽሻାଵൌ0 The term ܴሺݔሻ is called the Lagrange Remainder; ݔ∗ is the value of ݔ that produces the greatest value of ݂ሺାଵሻሺݔሻ between ܽ and ݔ. See more on the Lagrange Remainder on the next page. Version 5.6 Page 170 of 242 April 8, 2023 Chapter 14 Taylor and Maclaurin Series LaGrange Remainder The form for a Taylor Series about ݔൌܽ that includes an error term is: ݂ሺݔሻൌ݂ሺሻሺܽሻ ݇!
ሺݔെܽሻ ୀ ܴሺݔሻ ൌ݂ሺܽሻ݂ᇱሺܽሻሺݔെܽሻ݂ᇱᇱሺܽሻ 2!
ሺݔെܽሻଶ⋯݂ሺሻሺܽሻ ሺ݊ሻ!
ሺݔെܽሻܴሺݔሻ The term ܴሺݔሻ is called the Lagrange Remainder, and has the form: ܴሺݔሻൌ݂ሺାଵሻሺݔ∗ሻ ሺ݊1ሻ! ሺݔെܽሻାଵ where ݔ∗ produces the greatest value of ݂ሺାଵሻሺݔሻ between ܽ and ݔ. This form is typically used to approximate the value of a series to a desired level of accuracy. Example 14.3: Approximate √݁ using five terms of the Maclaurin Series (i.e., the Taylor Series about ݔൌ0) for ݁௫ and estimate the maximum error in the estimate. Using five terms and letting ݔൌ ଵ ଶ, we get: ݁௫ ൌ 1 ݔݔଶ 2! ݔଷ 3! ݔସ 4! ܴସሺݔሻ ݁ ଵଶ ൗ ~ 1 1 2 ቀ1 2ቁ ଶ 2! ቀ1 2ቁ ଷ 3! ቀ1 2ቁ ସ 4! ൌ 1 1 2 1 8 1 48 1 384 ൌ . ૡૠ To find the maximum potential error in this estimate, calculate: ܴସሺݔሻൌ ሺఱሻሺ௫∗ሻ ହ!
ݔହ for ݔൌ ଵ ଶ and ݔ∗ between 0 and ଵ ଶ. Since ݂ሺݔሻൌ݁௫, the fifth derivative of ݂ is: ݂ሺହሻሺݔሻൌ݁௫. The maximum value of this between ݔൌ0 and ݔൌ ଵ ଶ occurs at ݔൌ ଵ ଶ. Then, ݂ሺହሻቀ ଵ ଶቁൌ݁ ଵଶ ൗ൏1.65 based on our estimate of 1.6484375 above (we will check this after completing our estimate of the maximum error). Combining all of this, ܴସ൬1 2൰ ൌ ݂ሺହሻቀ1 2ቁ 5!
൬1 2൰ ହ ൏ 1.65 5! ൬1 2൰ ହ ൌ. ૢૠ Note that the maximum value of √݁, then, is 1.6484375 0.0004297 ൌ1.6488672, which is less than the 1.65 used in calculating ܴସቀ ଵ ଶቁ, so our estimate is good. The actual value of √݁ is 1.6487212 … . Version 5.6 Page 171 of 242 April 8, 2023 Chapter 15 Cool Stuff e What is “e”? Euler’s number, e is the base of the natural logarithms. e is a transcendental number, meaning that it is not the root of any polynomial with integer coefficients. What Makes “e” so Special? e shows up over and over in mathematics, especially in regard to limits, derivatives, and integrals. In particular, it is noteworthy that: ݁ൌlim →ஶ൬1 1 ݊൰ ݁ൌlim →ஶ൬݊ √݊!
൰ ݀ ݀ݔሺ݁௫ሻൌ݁௫ න݀ݔ ݔ ଵ ൌ1 lim →ஶ൬1 1 ܾ݊൰ ൌ݁ ൗ Perhaps, most interestingly, the following equation, called Euler’s Equation, relates five seemingly unrelated mathematical constants to each other. ࢋ࣊ൌ Some Series Representations of e ݁ ൌ 1 ݇!
ஶ ୀ ൌ 1 1 1 2 1 6 1 24 1 120 ⋯ ݁ൌ ሺെ1ሻ ݇!
ஶ ୀ ൩ ିଵ ൌ 1 1 െ1 1 2 െ1 6 1 24 െ1 120 ⋯ Decimal Expansion ݁ൌ2.7 1828 1828 4590 4523 5360 2874 7135 2662 4977 5724 7093 6999 5957 4966 … The web site shows the decimal expansion of e to over 2 million digits. There are many more series involving e. A sampling of these is provided at: Version 5.6 Page 172 of 242 April 8, 2023 Chapter 15 Cool Stuff Derivation of Euler’s Formula by Integration Start with: ݕൌ cos ݔ݅sin ݔ [note that ሺ0, 1ሻ is a point on this function] Then: ݀ݕൌሺെsin ݔ݅cos ݔሻ ݀ݔ ݀ݕൌሺ݅cos ݔെ sin ݔሻ ݀ݔ ݀ݕൌ݅ݕ ݀ݔ ௗ௬ ௬ൌ݅ ݀ݔ Integrate: ௗ௬ ௬ൌ ݅ ݀ݔ ln ݕൌ݅ݔܥ [note that ܥൌ0 since ሺ0, 1ሻ is a point on this function] ݕൌ ݁௫ Final Result: Very Cool Sub‐Case When ࢞ൌ࣊, Euler’s equation becomes: ݁గൌ cos ߨ݅sin ߨ or, ݁గൌ െ1 Rewriting this provides an equation that relates five of the most important mathematical constants to each other: ࢋ࢞ൌ܋ܗܛ࢞ܛܑܖ࢞ ࢋ࣊ൌ Note that this will allow us to calculate logarithms of negative numbers. Version 5.6 Page 173 of 242 April 8, 2023 Chapter 15 Cool Stuff Derivation of Euler’s Formula Using Power Series A Power Series about zero is an infinite series of the form: 𝑓ሺ𝑥ሻൌ𝑎𝑥ൌ𝑎𝑎ଵ𝑥𝑎ଶ𝑥ଶ𝑎ଷ𝑥ଷ⋯ ஶ ୀ Many mathematical functions can be expressed as power series. Of particular interest in deriving Euler’s Identity are the following: sin 𝑥ൌሺെ1ሻ𝑥ଶାଵ ሺ2𝑛1ሻ!
ൌ𝑥െ𝑥ଷ 3! 𝑥ହ 5! െ𝑥 7! ⋯ ஶ ୀ cos 𝑥ൌሺെ1ሻ𝑥ଶ ሺ2𝑛ሻ!
ൌ1 െ𝑥ଶ 2! 𝑥ସ 4! െ𝑥 6! ⋯ ஶ ୀ 𝑒௫ൌ𝑥 𝑛! ൌ1 𝑥𝑥ଶ 2! 𝑥ଷ 3! 𝑥ସ 4! 𝑥ହ 5! 𝑥 6! 𝑥 7! ⋯ ஶ ୀ Then, we have: i ∙sinሺ𝑥ሻൌ𝑖∙ሺെ1ሻ𝑥ଶାଵ ሺ2𝑛1ሻ!
ൌ𝑖𝑥െ𝑖∙𝑥ଷ 3!
𝑖∙𝑥ହ 5!
െ𝑖∙𝑥 7!
⋯ ஶ ୀ cos 𝑥ൌሺെ1ሻ𝑥ଶ ሺ2𝑛ሻ!
ൌ1 െ𝑥ଶ 2! 𝑥ସ 4! െ𝑥 6! ⋯ ஶ ୀ 𝑒௫ൌሺ𝑖𝑥ሻ 𝑛!
ൌ1 𝑖𝑥െ𝑥ଶ 2! െ𝑖∙𝑥ଷ 3!
𝑥ସ 4! 𝑖∙𝑥ହ 5!
െ𝑥 6! െ𝑖∙𝑥 7!
⋯ ஶ ୀ 𝑒௫ൌቆ1 െ𝑥ଶ 2! 𝑥ସ 4! െ𝑥 6! ⋯ቇ𝑖ሺ𝑥െ𝑥ଷ 3! 𝑥ହ 5! െ𝑥 7! ⋯ሻ This implies: and, substituting 𝑥ൌ𝜋 yields: 𝒆𝒊𝒙ൌ 𝐜𝐨𝐬𝒙𝒊𝐬𝐢𝐧𝒙 𝒆𝒊𝝅 𝟏ൌ 𝟎 Version 5.6 Page 174 of 242 April 8, 2023 Chapter 15 Cool Stuff Logarithms of Negative Real Numbers and Complex Numbers Natural Logarithm of a Negative Real Number From Euler’s Formula, we have: ݁గൌ െ1 Taking the natural logarithm of both sides gives: ln ݁గൌ lnሺെ1ሻ which implies that ݅ߨൌlnሺെ1ሻ Next, let ݔ be a positive real number. Then: lnሺെݔሻൌlnሺെ1 ∙ݔሻൌlnሺെ1ሻln ݔ ܔܖሺെ࢞ሻൌ࣊ܔܖ࢞ Logarithm (Any Base) of a Negative Real Number To calculate logሺെݔሻ, use the change of base formula: logሺ݉ሻൌ ୪୭ೌ ୪୭ೌ. Let the new base be ݁ to get: logሺെݔሻൌ ୪୬ሺି௫ሻ ୪୬ ܔܗ࢈ሺെ࢞ሻൌ ࣊ ା ܔܖ࢞ ܔܖ࢈ Logarithm of a Complex Number (Principal Value) Define ݖൌݔ݅ݕ in polar form as: ݖൌݎ݁ఏ, where ݎൌඥݔଶݕଶ is the modulus (i.e., magnitude) of ݖ and ߠൌtanିଵቀ ௬ ௫ቁ is the argument (i.e., angle), in radians, of complex number ݖ. Then, ܔܖࢠൌܔܖ൫࢘ࢋࣂ൯ൌܔܖ࢘ࣂ and ܔܗ࢈ࢠൌ ܔܖ࢘ ା ࣂ ܔܖ࢈ where, ሼെߨ൏ߠߨሽ Version 5.6 Page 175 of 242 April 8, 2023 Chapter 15 Cool Stuff What Is ( to the power of ) Start with: ࢋ࣊ ൌ (Euler’s Formula – special case) Then: ݁గൌ െ1 √݁గൌ √െ1 ൫݁గ൯ ଵଶ ൗൌ ݅ ݁ గଶ ൗൌ ݅ ቀ݁ గଶ ൗቁ ൌ ݅ ݁ మగଶ ൗൌ ݅ ݁ ିగଶ ൗൌ ݅ Calculate ݁ ିగଶ ൗ to obtain: So we see that it is possible to take an imaginary number to an imaginary power and return to the realm of real numbers. ൌࢋ ି࣊ ൗ ~ . ૠૡૡ~ Version 5.6 Page 176 of 242 April 8, 2023 Chapter 15 Cool Stuff Derivative of e to a Complex Power (܍ܢ) Start with: ݖൌݔ݅ݕ ݁௬ൌ cos ݕ݅sin ݕ Then: ݁௭ൌ݁௫ା௬ൌ݁௫݁௬ൌ݁௫ሺcos ݕ݅sin ݕሻ Cauchy‐Riemann Equations A complex function, ݂ሺݖሻൌݑሺݔ, ݕሻ݅∙ݒሺݔ, ݕሻ, is differentiable at point ݖൌݖ if and only if the functions ݑ and ݒ are differentiable and: ߲ݑሺݖሻ ߲ݔ ൌ߲ݒሺݖሻ ߲ݕ and ߲ݑሺݖሻ ߲ݕ ൌെ߲ݒሺݖሻ ߲ݔ These are called the Cauchy‐Riemann Equations for the functions ݑ and ݒ: ߲ݑ ߲ݔൌ߲ݒ ߲ݕ and ߲ݑ ߲ݕൌെ߲ݒ ߲ݔ in Cartesian form ݎ߲ݑ ߲ݎൌ߲ݒ ߲ߠ and ߲ݑ ߲ߠൌെݎ߲ݒ ߲ݎ in Polar form Derivative of ࢋࢠ For a differentiable complex function, ݂ሺݖሻൌݑሺݔ, ݕሻ݅∙ݒሺݔ, ݕሻ: ݂݀ ݀ݖൌ൬߲ݑ ߲ݔ߲݅ݒ ߲ݔ൰ൌ൬߲ݒ ߲ݕെ߲݅ݑ ߲ݕ൰ Then, let ݂ሺݖሻൌ݁௭ൌ݁௫ሺcos ݕ݅sin ݕሻ: ݑൌ݁௫∙cos ݕ and ݒൌ݁௫∙sin ݕ ݀ ݀ݖሺ݁௭ሻൌ߲ ߲ݔሺ݁௫∙cos ݕሻ߲݅ ߲ݔሺ݁௫∙sin ݕሻൌ݁௫∙cos ݕ݅∙݁௫∙sin ݕൌ݁௭ So, ௗ ௗ௭ ሺ݁௭ሻൌ݁௭. Cool, huh? Version 5.6 Page 177 of 242 April 8, 2023 Chapter 15 Cool Stuff Derivatives of a Circle The general equation of a circle centered at the Origin is: ݔଶݕଶൌݎଶ, where ݎ is the radius of the circle. First Derivative ݔଶݕଶൌݎଶ Note that ݎଶ is a constant, so its derivative is zero. Using Implicit Differentiation (with respect to ݔ), we get: 2ݔ2ݕ∙݀ݕ ݀ݔൌ0 ࢊ࢟ ࢊ࢞ൌെ࢞ ࢟ Second Derivative We have a couple of options at this point. We could do implicit differentiation on 2ݔ2ݕ∙ ௗ௬ ௗ௫ ൌ0, but given the simplicity of ௗ௬ ௗ௫ൌെ ௫ ௬, let’s work from there. ݀ݕ ݀ݔൌെݔ ݕ Use the Quotient Rule, simplify and substitute in ௗ௬ ௗ௫ൌെ ௫ ௬ in the expression. ݀ଶݕ ݀ݔଶൌെ ݕ∙݀ ݀ݔሺݔሻെݔ∙݀ݕ ݀ݔ ݕଶ ൌെ ݕെݔቀെݔ ݕቁ ݕଶ ൌെ൦ ݕଶ ݕݔଶ ݕ ݕଶ ൪ൌെቈݔଶݕଶ ݕଷ Notice that the numerator is equal to the left hand side of the equation of the circle. We can simplify the expression for the second derivative by substituting ݎଶ for ݔଶݕଶ to get: ࢊ࢟ ࢊ࢞ൌെ࢘ ࢟ Version 5.6 Page 178 of 242 April 8, 2023 Chapter 15 Cool Stuff Derivatives of an Ellipse The general equation of an ellipse centered at the Origin is: ௫మ మ ௬మ మ ൌ1, where ܽ is the radius of the ellipse in the ݔ‐direction and ܾ is the radius of the ellipse in the ݕ‐direction. First Derivative ௫మ మ ௬మ మ ൌ1 which can also be written ܾଶݔଶܽଶݕଶൌܽଶܾଶ Note that ܽଶܾଶ is a constant, so its derivative is zero. Using Implicit Differentiation (with respect to ݔ), we get: 2ܾଶݔ2ܽଶݕ∙݀ݕ ݀ݔൌ0 ࢊ࢟ ࢊ࢞ൌെ࢈࢞ ࢇ࢟ Second Derivative Given the simplicity of ௗ௬ ௗ௫ൌെ మ௫ మ௬, let’s work from there to calculate ௗమ௬ ௗ௫మ. ݀ݕ ݀ݔൌെܾଶݔ ܽଶݕൌെܾଶ ܽଶ൬ݔ ݕ൰ Use the Quotient Rule, simplify and substitute in ௗ௬ ௗ௫ൌെ మ௫ మ௬ in the expression. ݀ଶݕ ݀ݔଶൌെܾଶ ܽଶ ݕ∙݀ ݀ݔሺݔሻെݔ∙݀ݕ ݀ݔ ݕଶ ൌെܾଶ ܽଶ൦ ݕെݔ൬െܾଶݔ ܽଶݕ൰ ݕଶ ൪ൌെܾଶ ܽଶ൦ ܽଶݕଶ ܽଶݕܾଶݔଶ ܽଶݕ ݕଶ ൪ ൌെܾଶ ܽଶቈܽଶݕଶܾଶݔଶ ܽଶݕଷ Notice that the numerator inside the brackets is equal to the left hand side of the equation of the ellipse. We can simplify this expression by substituting ܽଶܾଶ for ܽଶݕଶܾଶݔଶ to get: ࢊ࢟ ࢊ࢞ൌെ࢈ ࢇ࢟ Version 5.6 Page 179 of 242 April 8, 2023 Chapter 15 Cool Stuff Derivatives of a Hyperbola The general equation of a hyperbola with a vertical transverse axis, centered at the Origin is: ௬మ మെ ௫మ మ ൌ1, where ሺേܽ, 0ሻ are the vertices of the hyperbola. First Derivative ௬మ మെ ௫మ మ ൌ1 which can also be written ܾଶݕଶെܽଶݔଶൌܽଶܾଶ Note that ܽଶܾଶ is a constant, so its derivative is zero. Using Implicit Differentiation (with respect to ݔ), we get: 2ܾଶݕ∙݀ݕ ݀ݔെ2ܽଶݔൌ0 ࢊ࢟ ࢊ࢞ൌࢇ࢞ ࢈࢟ Second Derivative Given the simplicity of ௗ௬ ௗ௫ൌെ ௫ ௬, let’s work from there to calculate ௗమ௬ ௗ௫మ. ݀ݕ ݀ݔ ൌ ܽଶݔ ܾଶݕ ൌ ܽଶ ܾଶ൬ݔ ݕ൰ Use the Quotient Rule, simplify and substitute in ௗ௬ ௗ௫ൌ మ௫ మ௬ in the expression. ݀ଶݕ ݀ݔଶ ൌ ܽଶ ܾଶ ݕ∙݀ ݀ݔሺݔሻെݔ∙݀ݕ ݀ݔ ݕଶ ൌ ܽଶ ܾଶ൦ ݕെݔ൬ܽଶݔ ܾଶݕ൰ ݕଶ ൪ ൌ ܽଶ ܾଶ൦ ܾଶݕଶ ܾଶݕെܽଶݔଶ ܾଶݕ ݕଶ ൪ ൌ ܽଶ ܾଶቈܾଶݕଶെܽଶݔଶ ܾଶݕଷ Notice that the numerator inside the brackets is equal to the left hand side of the equation of the hyperbola. We can simplify this expression by substituting ܽଶܾଶ for ܾଶݕଶെܽଶݔଶ to get: ࢊ࢟ ࢊ࢞ ൌ ࢇ ࢈࢟ Version 5.6 Page 180 of 242 April 8, 2023 Chapter 15 Cool Stuff Derivative of: ሺ࢞࢟ሻൌ࢞࢟ Starting expression: ሺݔݕሻଷൌݔଷݕଷ Expand the cubic of the binomial: ݔଷ3ݔଶݕ3ݔݕଶݕଷൌݔଷݕଷ Subtract ሺݔଷݕଷሻ from both sides: 3ݔଶݕ3ݔݕଶൌ0 Divide both sides by 3: ݔଶݕݔݕଶൌ0 Investigate this expression: Factor it: ݔݕሺݔݕሻൌ0 Solutions are the three lines: ݔൌ0, ݕൌ0, ݕൌെݔ Note the slopes of these lines: undefined, 0, െ1 Obtain the derivative: Start with: ݔଶݕݔݕଶൌ0 Implicit differentiation: ቀݔଶ ∙ ௗ௬ ௗ௫2ݔݕቁቀݔ∙2ݕ ௗ௬ ௗ௫ݕଶቁൌ0 Rearrange terms: ሺݔଶ2ݔݕሻௗ௬ ௗ௫ሺ2ݔݕݕଶሻൌ0 Solve for ௗ௬ ௗ௫: ௗ௬ ௗ௫ൌ ିሺଶ௫௬ା௬మሻ ௫మାଶ௫௬ Factored form: ࢊ࢟ ࢊ࢞ൌ ି࢟ሺ࢞ା࢟ሻ ࢞ሺ࢞ା࢟ሻ Consider each solution separately: ݔൌ0: ௗ௬ ௗ௫ൌ ି௬ሺଶ∙ା௬ሻ ሺାଶ௬ሻ ൌundefined ݕൌ0: ௗ௬ ௗ௫ൌ ିሺଶ௫ାሻ ௫ሺ௫ାଶ∙ሻ ൌ0 ݕൌെݔ: ௗ௬ ௗ௫ൌ ௫ሺଶ௫ି௫ሻ ௫ሺ௫ିଶ௫ሻ ൌെ1 Conclusion: ࢊ࢟ ࢊ࢞ൌ ି࢟ሺ࢞ା࢟ሻ ࢞ሺ࢞ା࢟ሻ is an elegant way to describe the derivative of ݕ with respect to ݔ for the expression ሺݔݕሻଷൌݔଷݕଷ (which is not a function). However, it is noteworthy, that this derivative can only take on three possible values (if we allow “undefined” to count as a value) – undefined, 0 and െ1. Version 5.6 Page 181 of 242 April 8, 2023 Chapter 15 Cool Stuff Inflection Points of the PDF of the Normal Distribution The equation for the Probability Density Function (PDF) of the Normal Distribution is: ࡼሺ࢞ሻൌ ࣌√࣊ ࢋ ି ሺ࢞ିࣆሻ ࣌ where ࣆ and ࣌ are the mean and standard deviation of the distribution. ܲᇱሺݔሻൌቆ 1 ߪ√2ߨ ݁ି ሺ௫ିఓሻమ ଶఙమቇ∙݀ ݀ݔቆെ ሺݔെߤሻଶ 2ߪଶ ቇ ܲᇱሺݔሻൌܲሺݔሻ∙ቆെ 2ሺݔെߤሻ 2ߪଶ ቇ ࡼᇱሺ࢞ሻൌെ ࣌∙ሾࡼሺ࢞ሻ∙ሺ࢞െࣆሻሿ ܲᇱᇱሺݔሻൌെ1 ߪଶܲሺݔሻ∙݀ ݀ݔሺݔെߤሻሺݔെߤሻ∙݀ ݀ݔሺܲሺݔሻሻ൨ ܲᇱᇱሺݔሻൌെ1 ߪଶቈܲሺݔሻ ሺݔെߤሻ∙ቆെ1 ߪଶ∙ܲሺݔሻ∙ሺݔെߤሻቇ ࡼᇱᇱሺ࢞ሻൌെ1 ߪଶቈܲሺݔሻെܲሺݔሻ∙ሺݔെߤሻଶ ߪଶ ൌെࡼሺ࢞ሻ ࣌ቈെሺ࢞െࣆሻ ࣌ Setting ܲᇱᇱሺݔሻൌ0, and noting that ܲሺݔሻ്0 for all values of ݔ, we get: 1 െ ሺ௫ିఓሻమ ఙమ ൌ0 So that: ࢞ൌࣆേ࣌. Further, noting that the value of the second derivative changes signs at each of these values, we conclude that inflection points exist at ࢞ൌࣆേ࣌. In English, the inflection points of the Probability Density Function of the Normal Distribution exist at points one standard deviation above or below the mean. Version 5.6 Page 182 of 242 April 8, 2023 𝑭ሺ𝒙ሻൌන𝒇ሺ𝒙ሻ 𝒅𝒙. Appendix A Key Definitions in Calculus Absolute Maximum See entry on Global Maximum. May also simply be called the “maximum.” Absolute Minimum See entry on Global Minimum. May also simply be called the “minimum.” Antiderivative Also called the indefinite integral of a function, 𝑓ሺ𝑥ሻ, an antiderivative of 𝑓ሺ𝑥ሻ is a function 𝐹ሺ𝑥ሻ, such that 𝐹ᇱሺ𝑥ሻൌ𝑓ሺ𝑥ሻ on an interval of 𝑥. The general antiderivative of 𝑓ሺ𝑥ሻ is the antiderivative expressed as a function which includes the addition of a constant 𝐶, which is called the constant of integration. Example: 𝐹ሺ𝑥ሻൌ2𝑥ଷ is an antiderivative of 𝑓ሺ𝑥ሻൌ6𝑥ଶ because 𝐹′ሺ𝑥ሻൌ𝑓ሺ𝑥ሻ. 𝐹ሺ𝑥ሻൌ2𝑥ଷ𝐶 is the general antiderivative of 𝑓ሺ𝑥ሻൌ6𝑥ଶ because 𝐹′ሺ𝑥ሻൌ 𝑓ሺ𝑥ሻ for all values of 𝐶. Notation: the antiderivative of a function, 𝑓ሺ𝑥ሻ, is expressed as: Version 5.6 Page 183 of 242 April 8, 2023 Appendix A Key Definitions An Ault table facilitates the graphing of a function like the one above: 𝑠ሺ𝑡ሻൌ2𝑡ଷ – 9𝑡ଶ 12𝑡– 4 Ault Table Named for A’Laina Ault, the Math Department Chair at Damonte Ranch High School in Reno, Nevada, an Ault Table is a chart that shows the signs and the behavior of a function and its derivatives over key intervals of the independent variable (usually 𝑥 or 𝑡). It is very useful in curve sketching because it makes the process of finding extrema and inflection points relatively easy. The steps to building an Ault Table are: 1. Calculate the first and second derivatives of the function being considered. Additional derivatives may be taken if needed. 2. Find the zeros of each derivative; these form the interval endpoints for the table. Note that the zeros of the first derivative are critical values, representing potential maxima and minima, and the zeros of the second derivative are potential inflection points. 3. Arrange the zeros of the first two derivatives in numerical order, and create mutually exclusive open intervals with the zeros as endpoints. If appropriate, include intervals extending to െ∞ and/or ∞. 4. Create a set of rows as shown in the table on the next page. At this point the boxes in the table will be empty. 5. Determine the sign of each derivative in each interval and record that information in the appropriate box using a “” or a “െ“. 6. Use the signs determined in Step 5 to identify for each interval a) whether the function is increasing or decreasing (green lines in the table), b) whether the first derivative is increasing or decreasing (red lines in the table), c) whether the function is concave up or down (bottom red line in the table), and d) the shape of the curve on the interval. From the information in the table, you can determine the location of all extrema and inflection points of the curve. You can also determine where the speed is positive; the signs of both the first and second derivatives are the same. An example is provided on the next page: Version 5.6 Page 184 of 242 April 8, 2023 Appendix A Key Definitions Example: develop an Ault Table for the function: s(t) = 2t3 – 9t2 + 12t – 4 First find the key functions: 𝑠ሺ𝑡ሻൌ2𝑡ଷ – 9𝑡ଶ 12𝑡 – 4 Position function 𝑠′ሺ𝑡ሻൌ6𝑡ଶെ18𝑡12 Velocity function |𝑠′ሺ𝑡ሻ| ൌ|6𝑡ଶെ18𝑡12| Speed function 𝑠ᇱᇱሺ𝑡ሻൌ12𝑡െ18 Acceleration function Next, find the function’s critical values, inflection points, and maybe a couple more points. 𝑠ሺ𝑡ሻൌ2𝑡ଷ – 9𝑡ଶ 12𝑡 – 4 𝑠ሺ0ሻൌെ4 𝑠′ሺ𝑡ሻൌ6ሺ𝑡െ1ሻሺ𝑡െ2ሻ 𝑠ᇱሺ𝑡ሻൌ0 ⇒ Critical Values of 𝑡 are: 𝑡ൌሼ1, 2ሽ Critical Points are: ሼሺ1, 1ሻ, ሺ2, 0ሻሽ 𝑠ᇱᇱሺ𝑡ሻൌ6ሺ2𝑡െ3ሻ 𝑠ᇱᇱሺ𝑡ሻൌ0 ⇒ Inflection Point at: 𝑡ൌ1.5 𝑠ሺ𝑡ሻൌ2𝑡ଷ – 9𝑡ଶ 12𝑡 – 4 𝑠ሺ3ሻൌ5, just to get another point to plot Then, build an Ault Table with intervals separated by the key values: Key values of 𝑡 that define the intervals in the table are 𝑡ൌሼ1, 1.5, 2ሽ Note: Identify the signs (i.e., “, “െ“) first. The word descriptors are based on the signs. 𝒔ሺ𝒕ሻൌ𝟐𝒕𝟑– 𝟗𝒕𝟐𝟏𝟐𝒕– 𝟒 ሺ𝟎, 𝟏ሻ ሺ𝟏, 𝟏. 𝟓ሻ ሺ𝟏. 𝟓, 𝟐ሻ ሺ𝟐, ∞ሻ 𝒔ሺ𝒕ሻ increasing decreasing decreasing increasing 𝒔ᇱሺ𝒕ሻ െ െ and is: decreasing decreasing increasing increasing 𝒔ᇱᇱሺ𝒕ሻ െ െ 𝒔𝒐 𝒔ሺ𝒕ሻ 𝒊𝒔: concave down concave down concave up concave up Curve Shape Results. This function has: A maximum at 𝑡ൌ1. A minimum at 𝑡ൌ2. An inflection point at 𝑡ൌ1.5. Version 5.6 Page 185 of 242 April 8, 2023 Appendix A Key Definitions Concavity A function, 𝑓, is concave upward on an interval if 𝑓’ሺ𝑥ሻ is increasing on the interval, i.e., if 𝑓ᇱᇱሺ𝑥ሻ0. A function, 𝑓, is concave downward on an interval if 𝑓’ሺ𝑐ሻ is decreasing on the interval, i.e., if 𝑓ᇱᇱሺ𝑥ሻ൏0. Concavity changes at inflection points, from upward to downward or from downward to upward. Continuity A function, 𝑓, is continuous at 𝑥ൌ𝑐 iff: a. 𝑓ሺ𝑐ሻ is defined, b. lim ௫ →𝑓ሺ𝑥ሻ exists, and c. lim ௫ →𝑓ሺ𝑥ሻൌ𝑓ሺ𝑐ሻ d. If 𝑥ൌ𝑎 is an endpoint, then the limit need only exist from the left or the right. Basically, the function value and limit at a point must both exist and be equal to each other. Critical Numbers or Critical Values (and Critical Points) If a function, 𝑓, is defined at c, then the critical numbers (also called critical values) of 𝑓 are 𝑥‐ values where 𝑓’ሺ𝑐ሻൌ0 and where 𝑓’ሺ𝑐ሻ does not exist (i.e., 𝑓 is not differentiable at 𝑐). This includes 𝑥‐values where the slope of the curve is horizontal, and where cusps and discontinuities exist in an interval. The points where the critical numbers exist are called critical points. Note: endpoints are excluded from this definition, but must also be tested in cases where the student seeks an absolute (i.e., global) maximum or minimum of an interval. The curve shown is continuous everywhere except at the holes and the vertical asymptote. Version 5.6 Page 186 of 242 April 8, 2023 Appendix A Key Definitions Decreasing Function A function, 𝑓, is decreasing on an interval if for any two values in the interval, 𝑎 and 𝑏, with 𝑎൏ 𝑏, it is true that 𝑓ሺ𝑎ሻ𝑓ሺ𝑏ሻ. Degree of a Differential Equation The degree of a differential equation is the power of the highest derivative term in the equation. Contrast this with the order of a differential equation. Examples: 𝒅𝟐𝒚 𝒅𝒙𝟐𝟒ቀ 𝒅𝒚 𝒅𝒙ቁ𝟒𝒚ൌ𝟎 Degree ൌ1 ቀ 𝒅𝟐𝒚 𝒅𝒙𝟐ቁ 𝟐 ቀ 𝒅𝒚 𝒅𝒙ቁ 𝟑 ൌ 𝒚𝒂𝒙ሺ𝒙𝟐𝒚𝟐ሻ Degree ൌ2 𝟑 ቀ 𝒅𝟒𝒚 𝒅𝒙𝟒ቁ 𝟓 ቀ 𝒅𝟑𝒚 𝒅𝒙𝟑ቁ 𝟐 െ𝟐𝒚ൌ𝟐𝐜𝐨𝐬𝒙 Degree ൌ5 Version 5.6 Page 187 of 242 April 8, 2023 Appendix A Key Definitions Derivative The measure of the slope of a curve at each point along the curve. The derivative of a function 𝑓ሺ𝑥ሻ is itself a function, generally denoted 𝑓ᇱሺ𝑥ሻ or ௗ௬ ௗ௫. The derivative provides the instantaneous rate of change of a function at the point at which it is measured. The derivative function is given by either of the two following limits, which are equivalent: lim → 𝑓ሺ𝑥ℎሻെ𝑓ሺ𝑥ሻ ℎ or lim ௫ → 𝑓ሺ𝑥ሻെ𝑓ሺ𝑐ሻ 𝑥െ𝑐 In the figure below, the derivative of the curve 𝑓ሺ𝑥ሻൌ√25 െ𝑥ଶ at ሺെ3, 4ሻ is the slope of the tangent line at ሺെ3, 4ሻ, which is ଷ ସ. Differentiable A function is differentiable at a point, if a derivative can be taken at that point. A function is not differentiable at any 𝑥‐value that is not in its domain, at discontinuities, at sharp turns and where the curve is vertical. To find where a function is not differentiable by inspection, look for points of discontinuity, sharp turns, and vertical slopes in the curve. In the curve shown at right, the curve is not differentiable at the points of discontinuity (𝑥ൌ5ሻ nor at the cusp (𝑥ൌ2). Version 5.6 Page 188 of 242 April 8, 2023 Appendix A Key Definitions Differential Consider a function 𝑓ሺ𝑥ሻ, that is differentiable on an open interval around 𝑥. ∆𝑥 and ∆𝑦 represent small changes in the variables 𝑥 and 𝑦 around 𝑥 on 𝑓. Then, The differential of 𝑥 is denoted as 𝑑𝑥, and 𝑑𝑥ൌ∆𝑥. The differential of 𝑦 is denoted as 𝑑𝑦, and 𝑑𝑦ൌ𝑓ᇱሺ𝑥ሻ∙𝑑𝑥 ∆𝑦 is the actual change is 𝑦 resulting from a change in 𝑥 of ∆𝑥. 𝑑𝑦 is an approximation of ∆𝑦. Differential Equation An equation which includes variables and one or more of their derivatives. An ordinary differential equation (ODE) is a differential equation that includes an independent variable (e.g., 𝑥), a dependent variable (e. g., 𝑦), and one or more derivatives of the dependent varaiable, (e.g., ௗ௬ ௗ௫, ௗమ௬ ௗ௫మ, ௗయ௬ ௗ௫య, etc.). If the differential equation includes partial derivatives, it is a partial differential equation (PDE), and not an ordinary differential equation. See Chapter 10 for more definitions. Examples: 𝒅𝒚 𝒅𝒙ൌ𝒆𝒙 𝟏 𝟒ቀ 𝒅𝟐𝒚 𝒅𝒙𝟐ቁ𝒙 𝒅𝒚 𝒅𝒙𝟏ൌ𝟎 𝒅𝒚 𝒅𝒙ൌ 𝒚𝒂𝒙ሺ𝒙𝟐𝒚𝟐ሻ 𝟑 ቀ 𝒅𝟐𝒚 𝒅𝒙𝟐ቁ 𝒅𝒚 𝒅𝒙െ𝟐𝒚ൌ𝟐𝐜𝐨𝐬𝒙 𝒅𝒚 ∆𝒚 ∆𝒙ൌ𝒅𝒙 Version 5.6 Page 189 of 242 April 8, 2023 Appendix A Key Definitions Displacement Displacement is a measure of the shortest path between two points. So if you start at Point A and end at Point B, the length of the line segment connecting them is the displacement. To get displacement from velocity: Integrate velocity over the entire interval, without any breaks. Distance Distance is a measure of the length of the path taken to get from one point to another. So, traveling backward adds to distance and reduces displacement. To get distance from velocity, over an interval ሾ𝑎, 𝑏ሿ: Integrate velocity over the ሾ𝑎, 𝑏ሿ in pieces, breaking it up at each point where velocity changes sign from " " to "– " or from "– " to " ". Take the absolute value of each separate definite integral to get the distance for that interval. Add the distances over each interval to get the total distance. Version 5.6 Page 190 of 242 April 8, 2023 Appendix A Key Definitions 𝒆 𝑒 is the base of the natural logarithms. It is a transcendental number, meaning that it is not the root of any polynomial with integer coefficients. 𝑒ൌlim →ஶ൬1 1 𝑛൰ 𝑒ൌlim →ஶ൬𝑛 √𝑛!
൰ 𝑑 𝑑𝑥ሺ𝑒௫ሻൌ𝑒௫ න1 𝑥 ଵ 𝑑𝑥ൌ1 𝑒 ൌ 1 𝑘!
ஶ ୀ ൌ 1 1 1 2 1 6 1 24 1 120 ⋯ 𝑒ൌ ሺെ1ሻ 𝑘!
ஶ ୀ ൩ ିଵ ൌ 1 1 െ1 1 2 െ1 6 1 24 െ1 120 ⋯ Euler’s Equation: 𝑒గ1 ൌ0 shows the interconnection of five seemingly unrelated mathematical constants. Decimal Expansion of 𝒆: 𝑒ൌ2.7 1828 1828 4590 4523 5360 2874 7135 2662 4977 5724 7093 6999 5957 4966 … The web site shows the decimal expansion of e to over 2 million digits. Version 5.6 Page 191 of 242 April 8, 2023 Appendix A Key Definitions Global Maximum A global maximum is the function value at point 𝑐 on an interval if 𝑓ሺ𝑥ሻ൏𝑓ሺ𝑐ሻ for all 𝑥 in the interval. That is, 𝑓ሺ𝑐ሻ is a global maximum if there is an interval containing 𝑐 where 𝑓ሺ𝑐ሻ is the greatest value in the interval. Note that the interval may contain multiple relative maxima but only one global maximum. Global Minimum A global minimum is the function value at point 𝑐 on an interval if 𝑓ሺ𝑥ሻ𝑓ሺ𝑐ሻ for all 𝑥 in the interval. That is, 𝑓ሺ𝑐ሻ is a global minimum if there is an interval containing 𝑐 where 𝑓ሺ𝑐ሻ is the least value in the interval. Note that the interval may contain multiple relative minima but only one global minimum. Horizontal Asymptote If: lim ௫ →ஶ𝑓ሺ𝑥ሻൌ𝐿, or lim ௫ →ିஶ𝑓ሺ𝑥ሻൌ𝐿, then the line 𝑦ൌ𝐿 is a horizontal asymptote of 𝑓. Version 5.6 Page 192 of 242 April 8, 2023 Appendix A Key Definitions Hyperbolic Functions The set of hyperbolic functions relate to the unit hyperbola in much the same way that trigonometric functions relate to the unit circle. Hyperbolic functions have the same shorthand names as their corresponding trigonometric functions, but with an “h” at the end of the name to indicate that the function is hyperbolic. The names are read “hyperbolic sine,” “hyperbolic cosine,” etc. 𝐬𝐢𝐧𝐡𝒙ൌ𝒆𝒙െ𝒆ି𝒙 𝟐 𝐭𝐚𝐧𝐡𝒙ൌ𝒆𝒙െ𝒆ି𝒙 𝒆𝒙𝒆ି𝒙 𝐬𝐞𝐜𝐡𝒙ൌ 𝟐 𝒆𝒙𝒆ି𝒙 𝐜𝐨𝐬𝐡𝒙ൌ𝒆𝒙𝒆ି𝒙 𝟐 𝐜𝐨𝐭𝐡𝒙ൌ𝒆𝒙𝒆ି𝒙 𝒆𝒙െ𝒆ି𝒙 𝐜𝐬𝐜𝐡𝒙ൌ 𝟐 𝒆𝒙െ𝒆ି𝒙 Graphs of Hyperbolic Functions Increasing Function A function, 𝑓, is increasing on an interval if for any two values in the interval, 𝑎 and 𝑏, with 𝑎൏𝑏, it is true that 𝑓ሺ𝑎ሻ൏𝑓ሺ𝑏ሻ. Version 5.6 Page 193 of 242 April 8, 2023 Appendix A Key Definitions Inflection Point An inflection point is a location on a curve where concavity changes from upward to downward or from downward to upward. At an inflection point, the curve has a tangent line and 𝑓′ᇱሺ𝑥ሻൌ0 or 𝑓′ᇱሺ𝑥ሻ does not exist. However, it is not necessarily true that if 𝑓′ᇱሺ𝑥ሻൌ0, then there is an inflection point at 𝑥ൌ𝑐. Inverse Function Two functions 𝑓ሺ𝑥ሻ and 𝑔ሺ𝑥ሻ are inverses if and only if: 𝑓ሺ𝑔ሺ𝑥ሻሻൌ𝑥 for every 𝑥 in the domain of 𝑔, and 𝑔ሺ𝑓ሺ𝑥ሻሻൌ𝑥 for every 𝑥 in the domain of 𝑓. Important points about inverse functions: Each function is a reflection of the other over the line 𝑦ൌ𝑥. The domain of each function is the range of the other. Sometimes a domain restriction is needed to make this happen. If 𝑓ሺ𝑎ሻൌ𝑏, then 𝑓ିଵሺ𝑏ሻൌ𝑎. The slopes of inverse functions at a given value of 𝑥 are reciprocals. Local Maximum See entry on Relative Maximum. Local Minimum See entry on Relative Minimum. Version 5.6 Page 194 of 242 April 8, 2023 Appendix A Key Definitions Monotonic Function A function 𝑓 is monotonic if it is either entirely non‐increasing or entirely non‐decreasing. The derivative of a monotonic function never changes sign. A strictly monotonic function is either entirely increasing or entirely decreasing. The derivative of a strictly monotonic function is either always positive or always negative. Strictly monotonic functions are also one‐to‐one. Natural Exponential Function The natural exponential function is defined as: 𝑓ሺ𝑥ሻൌ𝑒௫. It is the inverse of the natural logarithmic function. Natural Logarithmic Function The natural logarithmic function is defined as: ln 𝑥ൌන1 𝑡 ௫ ଵ 𝑑𝑡, 𝑥0. The base of the natural logarithm is 𝑒. So, ln 𝑥ൌlog𝑥 ln 4 ൌන1 𝑡 4 1 𝑑𝑡 ~ 1.38629 Version 5.6 Page 195 of 242 April 8, 2023 Appendix A Key Definitions One‐to‐One Function A function 𝑓 is one‐to‐one if: for every 𝑥 in the domain of 𝑓, there is exactly one 𝑦 such that 𝑓ሺ𝑥ሻൌ𝑦, and for every 𝑦 in the range of 𝑓, there is exactly one 𝑥 such that 𝑓ሺ𝑥ሻൌ𝑦. A function has an inverse if and only if it is one‐to‐one. One‐to‐one functions are also monotonic. Monotonic functions are not necessarily one‐to‐one, but strictly monotonic functions are necessarily one‐to‐one. Order of a Differential Equation The order of a differential equation is the highest derivative that occurs in the equation. Contrast this with the degree of a differential equation. Examples: 𝒅𝟒𝒚 𝒅𝒙𝟒𝟒ቀ 𝒅𝟐𝒚 𝒅𝒙𝟐ቁ𝟒𝒚ൌ𝟎 Order ൌ4 𝒅𝒚 𝒅𝒙ൌ 𝒚𝒂𝒙ሺ𝒙𝟐𝒚𝟐ሻ Order ൌ1 𝟑൬𝒅𝟐𝒚 𝒅𝒙𝟐൰𝒅𝒚 𝒅𝒙െ𝟐𝒚ൌ𝟐𝐜𝐨𝐬𝒙 Order ൌ2 Ordinary Differential Equation (ODE) An ordinary differential equation is one that involves a single independent variable. Examples of ODEs: 𝒅𝟒𝒚 𝒅𝒙𝟒𝟒ቀ 𝒅𝟐𝒚 𝒅𝒙𝟐ቁ𝟒𝒚ൌ𝟎 𝒅𝒚 𝒅𝒙ൌ 𝒚𝒂𝒙ሺ𝒙𝟐𝒚𝟐ሻ 𝟑൬𝒅𝟐𝒚 𝒅𝒙𝟐൰𝒅𝒚 𝒅𝒙െ𝟐𝒚ൌ𝟐𝐜𝐨𝐬𝒙 Not ODEs (Partial Differential Equations): 𝝏𝒛 𝝏𝒕ൌ 𝝏𝟐𝒛 𝝏𝒙𝟐 𝝏𝟐𝒛 𝝏𝒚𝟐െ𝒛 𝝏𝒖 𝝏𝒙ൌ 𝝏𝒗 𝝏𝒚 and 𝝏𝒗 𝝏𝒙ൌെ 𝝏𝒖 𝝏𝒚 𝝏𝒖 𝝏𝒕ൌ 𝝏𝟐𝒖 𝝏𝒙𝟐 Version 5.6 Page 196 of 242 April 8, 2023 Appendix A Key Definitions Partial Differential Equation (PDE) A partial differential equation is one that involves more than one independent variable. Examples of PDEs: 𝝏𝒛 𝝏𝒕ൌ 𝝏𝟐𝒛 𝝏𝒙𝟐 𝝏𝟐𝒛 𝝏𝒚𝟐െ𝒛 𝝏𝒖 𝝏𝒙ൌ 𝝏𝒗 𝝏𝒚 and 𝝏𝒗 𝝏𝒙ൌെ 𝝏𝒖 𝝏𝒚 𝝏𝒖 𝝏𝒕ൌ 𝝏𝟐𝒖 𝝏𝒙𝟐 Position Function A position function is a function that provides the location (i.e., position) of a point moving in a straight line, in a plane or in space. The position function is often denoted 𝑠ሺ𝑡ሻ, where 𝑡 is time, the independent variable. When position is identified along a straight line, we have: 𝑠ሺ𝑡ሻ Position function 𝑠′ሺ𝑡ሻ Velocity function (rate of change in position; may be positive, negative, or zero) |𝑠′ሺ𝑡ሻ| Speed function (absolute value of velocity; it is zero or positive by definition) 𝑠ᇱᇱሺ𝑡ሻ Acceleration function (rate of change in velocity) 𝑠ᇱᇱᇱሺ𝑡ሻ Jerk function (rate of change in acceleration) Note that the inverse relationships hold for the functions as well. For example, consider the position function 𝑠ሺ𝑡ሻ and the velocity funtion 𝑣ሺ𝑡ሻ: 𝑣ሺ𝑡ሻൌ𝑠ᇱሺ𝑡ሻ and 𝑠ሺ𝑡ሻൌන𝑣ሺ𝑡ሻ𝑑𝑡 General Case of Integrating the Position Function in Problems Involving Gravity Given intial position 𝑠ሺ0ሻ, and intial velocity 𝑣ሺ0ሻ, the position function is given as: 𝑠ሺ𝑡ሻൌെ16𝑡ଶ𝑣ሺ0ሻ𝑡𝑠ሺ0ሻ where all functions involve the units feet and seconds. Note: The force of gravity is െ32 𝑓𝑡/𝑠𝑒𝑐ଶ or െ9.8 𝑚/𝑠𝑒𝑐ଶ. Version 5.6 Page 197 of 242 April 8, 2023 Appendix A Key Definitions Relative Maximum A relative maximum is the function value at point 𝑐 in an open interval if 𝑓ሺ𝑐െ𝛿ሻ൏ 𝑓ሺ𝑐ሻ and 𝑓ሺ𝑐𝛿ሻ൏𝑓ሺ𝑐ሻ for arbitrarily small 𝛿. That is, 𝑓ሺ𝑐ሻ is a relative maximum if there is an open interval containing 𝑐 where 𝑓ሺ𝑐ሻ is the greatest value in the interval. Relative Minimum A relative minimum is the function value at point 𝑐 in an open interval if 𝑓ሺ𝑐െ𝛿ሻ𝑓ሺ𝑐ሻ and 𝑓ሺ𝑐𝛿ሻ𝑓ሺ𝑐ሻ for arbitrarily small 𝛿. That is, 𝑓ሺ𝑐ሻ is a relative minimum if there is an open interval containing 𝑐 where 𝑓ሺ𝑐ሻ is the least value in the interval. Riemann Integral If ∑ 𝑓ሺ𝑥 ∗ሻ∙ ୀଵ ∆𝑥 is a Riemann Sum (see the entry on “Riemann Sum” below), then the corresponding definite integral, 𝑓ሺ𝑥ሻ𝑑𝑥 is called the Riemann Integral of 𝑓ሺ𝑥ሻ on the interval ሾ𝑎, 𝑏ሿ. Riemann Integrals in one, two and three dimensions are: න𝑓ሺ𝑥ሻ 𝑑𝑥 ൌ lim ୫ୟ୶ ∆௫ →𝑓ሺ𝑥 ∗ሻ∙∆𝑥 ୀଵ ඵ𝑓ሺ𝑥, 𝑦ሻ 𝑑𝐴ൌ lim ୫ୟ୶ ∆ →𝑓ሺ𝑥 ∗, 𝑦 ∗ሻ∙∆𝐴 ୀଵ ම𝑓ሺ𝑥, 𝑦, 𝑧ሻ 𝑑𝑉ൌ lim ୫ୟ୶ ∆ →𝑓ሺ𝑥 ∗, 𝑦 ∗, 𝑧 ∗ሻ∙∆𝑉 ୀଵ Version 5.6 Page 198 of 242 April 8, 2023 Appendix A Key Definitions Riemann Sum A Riemann Sum is the sum of the areas of a set of rectangles that can be used to approximate the area under a curve over a closed interval. Consider a closed interval ሾ𝑎, 𝑏ሿ on 𝑥 that is partitioned into 𝑛 sub‐intervals of lengths ∆𝑥ଵ, ∆𝑥ଶ, ∆𝑥ଷ, … ∆𝑥. Let 𝑥 ∗ be any value of 𝑥 on the 𝑖‐th sub‐interval. Then, the Riemann Sum is given by: A graphical representation of a Riemann sum on the interval ሾ2, 5ሿ is provided at right. Note that the area under a curve from 𝑥ൌ𝑎 to 𝑥ൌ𝑏 is: The largest ∆𝑥 is called the mesh size of the partition. A typical Riemann Sum is developed with all ∆𝑥 the same (i.e., constant mesh size), but this is not required. The resulting definite integral, 𝑓ሺ𝑥ሻ𝑑𝑥 is called the Riemann Integral of 𝑓ሺ𝑥ሻ on the interval ሾ𝑎, 𝑏ሿ. Scalar Field A Scalar Field in three dimensions provides a value at each point in space. For example, we can measure the temperature at each point within an object. The temperature can be expressed as T=ϕ(x,y,z). (note: ϕ is the Greek letter phi, corresponding to the English letter “f”.) 𝑆ൌ𝑓ሺ𝑥 ∗ሻ∙ ୀଵ ∆𝑥 lim ୫ୟ୶ ∆௫ →𝑓ሺ𝑥 ∗ሻ∙∆𝑥 ୀଵ ൌන𝑓ሺ𝑥ሻ𝑑𝑥 Version 5.6 Page 199 of 242 April 8, 2023 Appendix A Key Definitions Separation of Variables Separation of Variables is a technique used to assist in the solution of differential equations. The process involves using algebra to collect all terms involving one variable on one side of an equation and all terms involving the other variable on the other side of an equation. Example: Original differential equation: ௗ௬ ௗ௫ൌ ௬ ଶ√௫ Revised form with variables separated: ௗ௬ ௬ൌ ௗ௫ ଶ√௫ Singularity A singularity is a point at which a mathematical expression or other object is not defined or fails to be well‐behaved. Typically, singularities exist at discontinuities. Example: In evaluating the following integral, we notice that 𝑒 ଵ௫ ൗ does not exist at 𝑥ൌ0. We say, then, that 𝑒 ଵ௫ ൗ has a singularity at 𝑥ൌ0. Special techniques must often be employed to solve integrals with singularities. න𝑒 ଵ௫ ൗ ଵ ିଵ 𝑑𝑥 Version 5.6 Page 200 of 242 April 8, 2023 Appendix A Key Definitions Slope Field A slope field (also called a direction field) is a graphical representation of the slopes of a curve at various points that are defined by a differential equation. Each position in the graph (i.e., each point ሺ𝑥, 𝑦ሻ) is represented by a line segment indicating the slope of the curve at that point. Examples: 𝒅𝒚 𝒅𝒙 ൌ𝒆𝐬𝐢𝐧𝒙𝐜𝐨𝐬𝒙𝐬𝐢𝐧𝒙 𝒅𝒚 𝒅𝒙 ൌ𝒙𝟐െ𝒚𝟐 If you know a point on a curve and if you have its corresponding slope field diagram, you can plot your point and then follow the slope lines to determine the curve. Slope field plotters are available online at: Vector Field A Vector Field in three dimensions provides a vector at each point in space. For example, we can measure a magnetic field (magnitude and direction of the magnetic force) at each point in space around a charged particle. The magnetic field can be expressed as 𝑀 ሬ ሬ⃑ൌ𝑉 ሬ⃑ሺ𝑥, 𝑦, 𝑧ሻ. Note that the half‐arrow over the letters 𝑀 and 𝑉 indicate that the function generates a vector field. Version 5.6 Page 201 of 242 April 8, 2023 Appendix A Key Definitions Vertical Asymptote If lim ௫ →ష𝑓ሺ𝑥ሻൌേ∞ or lim ௫ →శ𝑓ሺ𝑥ሻൌേ∞, then the line 𝑥ൌ𝑐 is a vertical asymptote of 𝑓. Version 5.6 Page 202 of 242 April 8, 2023 Appendix B Key Theorems in Calculus Functions and Limits Inverse Function Theorem A function has an inverse function if and only if it is one‐to‐one. Intermediate Value Theorem (IVT) If a function, ݂, is continuous on the closed interval ሾܽ, ܾሿ, and ݀ is a value between ݂ሺܽሻ and ݂ሺܾሻ, Then there is a value ܿ in ሾܽ, ܾሿ such that ݂ሺܿሻൌ݀. Extreme Value Theorem (EVT) If a function, ݂, is continuous on the closed interval ሾܽ, ܾሿ, Then ݂ has both an absolute maximum and an absolute minimum on ሾܽ, ܾሿ. Squeeze Theorem (Limits): If ݃ሺݔሻ݂ሺݔሻ݄ሺݔሻ, and lim ௫ →݃ሺݔሻ ൌ ܮൌ lim ௫ →݄ሺݔሻ Then lim ௫ →݂ሺݔሻൌܮ Version 5.6 Page 203 of 242 April 8, 2023 Appendix B Key Theorems Differentiation Rolle's Theorem If a function, 𝑓, is continuous on the closed interval ሾ𝑎, 𝑏ሿ, and 𝑓 is differentiable on the open interval ሺ𝑎, 𝑏ሻ, and 𝑓ሺ𝑎ሻ ൌ 𝑓ሺ𝑏ሻ, Then there is at least one value 𝑐 in ሺ𝑎, 𝑏ሻ where 𝑓 ′ሺ𝑐ሻ ൌ 0. Mean Value Theorem (MVT) If a function, 𝑓, is continuous on the closed interval ሾ𝑎, 𝑏ሿ, and 𝑓 is differentiable on the open interval ሺ𝑎, 𝑏ሻ, Then There is at least one value 𝑐 in ሺ𝑎, 𝑏ሻ where 𝑓ᇱሺ𝑐ሻൌ ሺሻିሺሻ ି Increasing and Decreasing Interval Theorem If a function, 𝑓, is continuous on the closed interval ሾ𝑎, 𝑏ሿ, and 𝑓 is differentiable on the open interval ሺ𝑎, 𝑏ሻ, Then If 𝑓ᇱሺ𝑥ሻ0 for every 𝑥∈ሺ𝑎, 𝑏ሻ, then 𝑓 is increasing on ሾ𝑎, 𝑏ሿ. If 𝑓ᇱሺ𝑥ሻ൏0 for every 𝑥∈ሺ𝑎, 𝑏ሻ, then 𝑓 is decreasing on ሾ𝑎, 𝑏ሿ. If 𝑓ᇱሺ𝑥ሻൌ0 for every 𝑥∈ሺ𝑎, 𝑏ሻ, then 𝑓 is constant on ሾ𝑎, 𝑏ሿ. Concave Interval Theorem If a function, 𝑓, is continuous on the closed interval ሾ𝑎, 𝑏ሿ, and 𝑓ᇱᇱሺ𝑥ሻ exists on the open interval ሺ𝑎, 𝑏ሻ, Then If 𝑓ᇱᇱሺ𝑥ሻ0 for every 𝑥∈ሺ𝑎, 𝑏ሻ, then 𝑓 is concave upward on ሾ𝑎, 𝑏ሿ. If 𝑓ᇱᇱሺ𝑥ሻ൏0 for every 𝑥∈ሺ𝑎, 𝑏ሻ, then 𝑓 is concave downward on ሾ𝑎, 𝑏ሿ. Version 5.6 Page 204 of 242 April 8, 2023 Appendix B Key Theorems First Derivative Test (for finding extrema) If a function, ݂, is continuous on the open interval ሺܽ, ܾሻ, and ܿ is a critical number ∈ሺܽ, ܾሻ, ݂ is differentiable on the open interval ሺܽ, ܾሻ, except possibly at c, Then If ݂ᇱሺݔሻ changes from positive to negative at ܿ, then ݂ሺܿሻ is a relative maximum. If ݂ᇱሺݔሻ changes from negative to positive at ܿ, then ݂ሺܿሻ is a relative minimum. Second Derivative Test (for finding extrema) If a function, ݂, is continuous on the open interval ሺܽ, ܾሻ, and ܿ∈ሺܽ, ܾሻ, and ݂ᇱሺܿሻൌ0 and ݂′ᇱሺܿሻ exists, Then If ݂ᇱᇱሺܿሻ൏0, then ݂ሺܿሻ is a relative maximum. If ݂ᇱᇱሺܿሻ0, then ݂ሺܿሻ is a relative minimum. Inflection Point Theorem If a function, ݂, is continuous on the open interval ሺܽ, ܾሻ, and ܿ∈ሺܽ, ܾሻ, and ݂′ᇱሺܿሻൌ0 or ݂′ᇱሺܿሻ does not exist, Then ሺܿ, ݂ሺܿሻሻ may be an inflection point of ݂. Inverse Function Continuity and Differentiability If a function, ݂, has an inverse, Then If ݂ is continuous on its domain, then so is ݂ିଵ on its domain. If ݂ is increasing on its domain, then so is ݂ିଵ on its domain. If ݂ is decreasing on its domain, then so is ݂ିଵ on its domain. If ݂ is differentiable on its domain, then so is ݂ିଵ on its domain (wherever ݂ᇱሺݔሻ്0). Note: this exception exists because the derivatives of ݂ and ݂ᇱ are inverses. Version 5.6 Page 205 of 242 April 8, 2023 Appendix B Key Theorems Derivative of an Inverse Function If a function, ݂, is differentiable at ݔൌܽ, and ݂ has an inverse function ݃, and ݂ሺܽሻൌܾ, Then ݂ᇱሺܽሻൌ ଵ ᇱሺሻ (i.e., the derivatives of inverse functions are reciprocals). Integration First Fundamental Theorem of Calculus If ݂ሺݔሻ is a continuous function on ሾܽ, ܾሿ, ܨሺݔሻ is any antiderivative of ݂ሺݔሻ, then Then Second Fundamental Theorem of Calculus If ݂ሺݔሻ is a continuous function on ሾܽ, ܾሿ, Then For every ݔ∈ሾܽ, ܾሿ, Mean Value Theorem for Integrals (MVT) If ݂ሺݔሻ is a continuous function on ሾܽ, ܾሿ, Then there is a value ܿ∈ሾܽ, ܾሿ, such that න݂ሺݔሻ ݀ݔൌሺܾെܽሻ∙݂ሺܿሻ ݀ ݀ݔ න݂ሺݐሻ ௫ ݀ݐൌ݂ሺݔሻ න݂ሺݔሻ݀ݔൌܨሺܾሻെܨሺܽሻ Version 5.6 Page 206 of 242 April 8, 2023 Appendix C Summary of Key Derivatives and Integrals Version 5.6 Page 207 of 242 April 8, 2023 Appendix C Key Derivatives and Integrals Derivatives of Special Functions Common Functions Power Rule ݀ ݀ݔሺݔሻൌ݊∙ݔିଵ ݀ ݀ݔሺݑሻൌ݊∙ݑିଵ݀ݑ ݀ݔ Exponential and Logarithmic Functions ሺܽ0, ്ܽ1ሻ ݀ ݀ݔ݁௫ൌ ݁௫ ݀ ݀ݔ݁௨ൌ݁௨∙݀ݑ ݀ݔ ݀ ݀ݔܽ௫ൌ ܽ௫ ∙ ln ܽ ݀ ݀ݔܽ௨ൌܽ௨∙ln ܽ∙݀ݑ ݀ݔ ݀ ݀ݔln ݔൌ 1 ݔ ݀ ݀ݔln ݑൌ1 ݑ∙݀ݑ ݀ݔ ݀ ݀ݔlogݔൌ 1 ݔ ln ܽ ݀ ݀ݔlogݑൌ 1 ݑln ܽ∙݀ݑ ݀ݔ Trigonometric Functions ݀ ݀ݔsin ݔൌcos ݔ ݀ ݀ݔsin ݑൌcos ݑ∙݀ݑ ݀ݔ ݀ ݀ݔcos ݔൌെsin ݔ ݀ ݀ݔcos ݑൌെsin ݑ∙݀ݑ ݀ݔ ݀ ݀ݔtan ݔൌ secଶݔ ݀ ݀ݔtan ݑൌsecଶݑ∙݀ݑ ݀ݔ ݀ ݀ݔcot ݔൌെ cscଶݔ ݀ ݀ݔcot ݑൌെcscଶݑ∙݀ݑ ݀ݔ ݀ ݀ݔsec ݔൌ sec ݔ tan ݔ ݀ ݀ݔsec ݑൌsec ݑtan ݑ∙ ݀ݑ ݀ݔ ݀ ݀ݔcsc ݔൌെcsc ݔcot ݔ ݀ ݀ݔcsc ݑൌെcsc ݑcot ݑ∙ ݀ݑ ݀ݔ Version 5.6 Page 208 of 242 April 8, 2023 Appendix C Key Derivatives and Integrals Derivatives of Special Functions Common Functions Inverse Trigonometric Functions ݀ ݀ݔsinିଵݔൌ 1 √1 െݔଶ ݀ ݀ݔsinିଵݑൌ 1 √1 െݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q IV ݀ ݀ݔcosିଵݔൌ െ1 √1 െݔଶ ݀ ݀ݔcosିଵݑൌ െ1 √1 െݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q II ݀ ݀ݔtanିଵݔൌ 1 1 ݔଶ ݀ ݀ݔtanିଵݑൌ 1 1 ݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q IV ݀ ݀ݔcotିଵݔൌ െ1 1 ݔଶ ݀ ݀ݔcotିଵݑൌ െ1 1 ݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q II ݀ ݀ݔsecିଵݔൌ 1 |ݔ| √ݔଶെ1 ݀ ݀ݔsecିଵݑൌ 1 |ݑ| √ݑଶെ1 ∙݀ݑ ݀ݔ Angle in Q I or Q II ݀ ݀ݔcscିଵݔൌ െ1 |ݔ| √ݔଶെ1 ݀ ݀ݔcscିଵݑൌ െ1 |ݑ| √ݑଶെ1 ∙݀ݑ ݀ݔ Angle in Q I or Q IV ݀ ݀ݔsinିଵቀݔ ܽቁൌ 1 √ܽଶെݔଶ ݀ ݀ݔsinିଵቀݑ ܽቁൌ 1 √ܽଶെݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q IV ݀ ݀ݔcosିଵቀݔ ܽቁൌ െ1 √ܽଶെݔଶ ݀ ݀ݔcosିଵቀݑ ܽቁൌ െ1 √ܽଶെݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q II ݀ ݀ݔtanିଵቀݔ ܽቁൌ ܽ ܽଶݔଶ ݀ ݀ݔtanିଵቀݑ ܽቁൌ ܽ ܽଶݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q IV ݀ ݀ݔcotିଵቀݔ ܽቁൌ െܽ ܽଶݔଶ ݀ ݀ݔcotିଵቀݑ ܽቁൌ െܽ ܽଶݑଶ∙݀ݑ ݀ݔ Angle in Q I or Q II ݀ ݀ݔsecିଵቀݔ ܽቁൌ ܽ |ݔ| √ݔଶെܽଶ ݀ ݀ݔsecିଵቀݑ ܽቁൌ ܽ |ݑ| √ݑଶെܽଶ∙݀ݑ ݀ݔ Angle in Q I or Q II ݀ ݀ݔcscିଵቀݔ ܽቁൌ െܽ |ݔ| √ݔଶെܽଶ ݀ ݀ݔcscିଵቀݑ ܽቁൌ െܽ |ݑ| √ݑଶെܽଶ∙݀ݑ ݀ݔ Angle in Q I or Q IV Version 5.6 Page 209 of 242 April 8, 2023 Appendix C Key Derivatives and Integrals Indefinite Integrals Note: the rules presented in this section omit the “ ܥ” term that must be added to all indefinite integrals in order to save space and avoid clutter. Please remember to add the “ ܥ” term on all work you perform with indefinite integrals. Basic Rules නܿ݀ݑ ൌܿݑ න݂ܿሺݑሻ ݀ݑ ൌܿ ݂ሺݑሻ න݂ሺݑሻ݃ሺݑሻ݀ݑ ൌන݂ሺݑሻ݀ݑ න݃ሺݑሻ݀ݑ Integration by Parts නݑ݀ݒൌݑݒെනݒ݀ݑ Power Rule නሺݑሻ ݀ݑൌ 1 ݊1 ∙ݑାଵ ሺ്݊1ሻ න1 ݑ݀ݑൌln|ݑ| Exponential and Logarithmic Functions ሺܽ0, ്ܽ1ሻ න݁௨݀ݑൌ ݁௨ නln ݑ݀ݔൌݑln ݑെݑ නܽ௨݀ݑൌ1 ln ܽ ܽ௨ න 1 ݑln ݑ݀ݑൌlnሺln ݑሻ Version 5.6 Page 210 of 242 April 8, 2023 Appendix C Key Derivatives and Integrals Indefinite Integrals of Trigonometric Functions Trigonometric Functions නsin ݑ݀ݑൌെcos ݑ නcos ݑ݀ݑൌsin ݑ නtan ݑ݀ݑൌln |sec ݑ| ൌെln |cos ݑ| නsecଶݑ݀ݑൌtan ݑ නcot ݑ݀ݑൌെln |csc ݑ| ൌln |sin ݑ| නcscଶݑ݀ݑൌെcot ݑ නsec ݑ݀ݑൌln |sec ݑtan ݑ| නsec ݑtan ݑ݀ݑൌsec ݑ නcsc ݑ݀ݑൌെ ln |csc ݑcot ݑ| නcsc ݑcot ݑ݀ݑൌെcsc ݑ Version 5.6 Page 211 of 242 April 8, 2023 Appendix C Key Derivatives and Integrals Indefinite Integrals of Inverse Trigonometric Functions Inverse Trigonometric Functions නsinିଵ𝑢𝑑𝑢ൌ𝑢 sinିଵ𝑢 ඥ1 െ𝑢ଶ නcosିଵ𝑢𝑑𝑢ൌ𝑢 cosିଵ𝑢െ ඥ1 െ𝑢ଶ නtanିଵ𝑢𝑑𝑢ൌ𝑢 tanିଵ𝑢െ1 2 lnሺ𝑢ଶ1ሻ නcotିଵ𝑢𝑑𝑢ൌ𝑢 cotିଵ𝑢1 2 lnሺ𝑢ଶ1ሻ නsecିଵ𝑢𝑑𝑢ൌ𝑢 secିଵ𝑢െ ln ቀ𝑢ඥ𝑢ଶെ1ቁ secିଵ𝑢∈ቄቀ0, 𝜋 2ቁቅ ൌ𝑢 secିଵ𝑢 ln ቀ𝑢ඥ𝑢ଶെ1ቁ secିଵ𝑢∈ቄቀ𝜋 2 , 𝜋ቁቅ නcscିଵ𝑢𝑑𝑢ൌ𝑢 cscିଵ𝑢 ln ቀ𝑢ඥ𝑢ଶെ1ቁ cscିଵ𝑢∈ቄቀ0, 𝜋 2ቁቅ ൌ𝑢 cscିଵ𝑢െ ln ቀ𝑢ඥ𝑢ଶെ1ቁ cscିଵ𝑢∈ቄቀെ𝜋 2 , 0ቁቅ Involving Inverse Trigonometric Functions න 1 √1 െ𝑢ଶ 𝑑𝑢ൌsinିଵ𝑢 න 1 √𝑎ଶെ𝑢ଶ𝑑𝑢ൌsinିଵቀ𝑢 𝑎ቁ න 1 1 𝑢ଶ 𝑑𝑢ൌtanିଵ𝑢 න 1 𝑎ଶ𝑢ଶ𝑑𝑢ൌ1 𝑎tanିଵቀ𝑢 𝑎ቁ න 1 𝑢 √𝑢ଶെ1 𝑑𝑢ൌsecିଵ|𝑢| න 1 𝑢√𝑢ଶെ𝑎ଶ𝑑𝑢ൌ1 𝑎secିଵቆ|𝑢| 𝑎ቇ Version 5.6 Page 212 of 242 April 8, 2023 Appendix C Key Derivatives and Integrals Integrals of Special Functions Selecting the Right Function for an Integral Form Function Integral න 1 √ܽଶെݑଶ ݀ݑ sinିଵݑ න 1 √ܽଶെݑଶ݀ݑൌsinିଵቀݑ ܽቁ න 1 ܽଶݑଶ ݀ݑ tanିଵݑ න 1 ܽଶݑଶ݀ݑൌ1 ܽtanିଵቀݑ ܽቁ න 1 ݑ √ݑଶെܽଶ ݀ݔ secିଵݑ න 1 ݑ√ݑଶെܽଶ݀ݔൌ1 ܽ secିଵቆ|ݑ| ܽቇ න 1 √ݑଶܽଶ ݀ݑ sinhିଵݑ න 1 √ݑଶܽଶ݀ݑൌln ቀݑඥݑଶܽଶቁ න 1 √ݑଶെܽଶ ݀ݑ coshିଵݑ න 1 √ݑଶെܽଶ݀ݑൌln ቀݑඥݑଶെܽଶቁ න 1 ܽଶെݑଶ ݀ݑቚ ܽݑ tanhିଵݑ න 1 ܽଶെݑଶ ݀ݑൌ1 2ܽln ฬܽݑ ܽെݑฬ න 1 ݑଶെܽଶ ݀ݑቚ ݑܽ cothିଵݑ න 1 ݑ √ܽଶെݑଶ ݀ݑ sechିଵݑ න 1 ݑ√ܽଶെݑଶ݀ݑൌെ1 ܽln ቆܽ√ܽଶെݑଶ |ݑ| ቇ න 1 ݑ √ܽଶݑଶ݀ݑ cschିଵݑ න 1 ݑ√ܽଶݑଶ݀ݑൌെ1 ܽln ቆܽ√ܽଶݑଶ |ݑ| ቇ This is an inverse hyperbolic function. For more information, see Chapter 6. Note that you do not need to know about inverse hyperbolic functions to use the formulas on this page. Version 5.6 Page 213 of 242 April 8, 2023 Appendix D Key Functions and Their Derivatives Version 5.6 Page 214 of 242 April 8, 2023 Appendix D Functions and Their Derivatives Functions and Their Derivatives Function ࢌሺ࢞ሻൌ࢞࢞ ࢌሺ࢞ሻൌ ࢞ ࢌሺ࢞ሻൌࢋି࢞ ࢌሺ࢞ሻൌࢋି࢞ Description The function is always concave up and the limit of f(x) as x approaches 0 is 1. The graph of the function has the ݔ‐ and ݕ‐axes as horizontal and vertical asymptotes. The function is always decreasing and has the x‐axis as an asymptote. The function has one absolute maximum and the x‐axis is an asymptote. Function Graph First Derivative Graph Second Derivative Graph Version 5.6 Page 215 of 242 April 8, 2023 Appendix D Functions and Their Derivatives Functions and Their Derivatives Function ࢌሺ࢞ሻൌ . ି.࢞ ࢌሺ࢞ሻൌห࢞െ࢞ห ࢌሺ࢞ሻൌܔܖ|࢞| ࢌሺ࢞ሻൌܛܑܖ࢞ Description The logistic curve. It is always increasing and has one point of inflection. The function has two relative minima and one relative maximum. The function is always increasing on the right and always decreasing on the left. The y‐axis as an asymptote. The function is periodic with domain Թ and range ሾെ1, 1ሿ. Function Graph First Derivative Graph Second Derivative Graph Version 5.6 Page 216 of 242 April 8, 2023 Appendix D Functions and Their Derivatives Functions and Their Derivatives Function ࢌሺ࢞ሻൌ࢞ ࢌሺ࢞ሻൌ ሺ࢞ሻሺ࢞െሻሺ࢞െሻ ࢌሺ࢞ሻൌ ሺ࢞െሻሺ࢞ሻሺ࢞െሻ ࢌሺ࢞ሻൌ ࢞ሺ࢞െሻሺ࢞ሻሺ࢞െሻ Description The function has one absolute minimum and no points of inflection. The graph has three zeros, one relative minimum, one relative maximum, and one point of inflection. The function has one relative maximum, two relative minima, and two points of inflection. The function has two relative maxima, two relative minima, and three points of inflection. Function Graph First Derivative Graph Second Derivative Graph Version 5.6 Page 217 of 242 April 8, 2023 Appendix E Geometry and Trigonometry Formulas Version 5.6 Page 218 of 242 April 8, 2023 Appendix E Geometry and Trigonometry Formulas Geometry Summary of Perimeter and Area Formulas – 2D Shapes Shape Figure Perimeter Area Kite ࡼൌ࢈ࢉ ܾ, ܿൌݏ݅݀݁ݏ ൌ ሺࢊࢊሻ ݀ଵ, ݀ଶൌ݈݀݅ܽ݃݊ܽݏ Trapezoid ࡼൌ࢈࢈ࢉࢊ ܾଵ, ܾଶൌܾܽݏ݁ݏ ܿ, ݀ൌݏ݅݀݁ݏ ۯ ൌ ሺ܊܊ሻܐ bଵ, bଶൌbases h ൌheight Parallelogram ࡼൌ࢈ࢉ ܾ, ܿൌݏ݅݀݁ݏ ۯ ൌ܊ܐ ܾൌܾܽݏ݁ ݄ൌ݄݄݁݅݃ݐ Rectangle ࡼൌ࢈ࢉ ܾ, ܿൌݏ݅݀݁ݏ ۯ ൌ܊ܐ ܾൌܾܽݏ݁ ݄ൌ݄݄݁݅݃ݐ Rhombus ࡼൌ࢙ ݏൌݏ݅݀݁ ൌ࢈ࢎൌ ሺࢊࢊሻ ݀ଵ, ݀ଶൌ݈݀݅ܽ݃݊ܽݏ Square ࡼൌ࢙ ݏൌݏ݅݀݁ ൌ࢙ൌ ሺࢊࢊሻ ݀ଵ, ݀ଶൌ݈݀݅ܽ݃݊ܽݏ Regular Polygon ࡼൌ࢙ ݊ൌ݊ݑܾ݉݁ݎ ݂ ݏ݅݀݁ݏ ݏൌݏ݅݀݁ ൌ ࢇ∙ࡼ ܽൌܽݐ݄݁݉ ܲൌ݁ݎ݅݉݁ݐ݁ݎ Circle ൌ࣊࢘ൌ࣊ࢊ ݎൌݎܽ݀݅ݑݏ ݀ൌ݀݅ܽ݉݁ݐ݁ݎ ൌ࣊࢘ ݎൌݎܽ݀݅ݑݏ Ellipse ࡼൎ࣊ට ሺ࢘࢘ሻ ݎ ଵൌ݆݉ܽݎܽݔ݅ݏݎܽ݀݅ݑݏ ݎ ଶൌ݉݅݊ݎܽݔ݅ݏݎܽ݀݅ݑݏ ൌ࣊࢘࢘ ݎ ଵൌ݆݉ܽݎ ܽݔ݅ݏݎܽ݀݅ݑݏ ݎ ଶൌ݉݅݊ݎ ܽݔ݅ݏݎܽ݀݅ݑݏ Version 5.6 Page 219 of 242 April 8, 2023 Appendix E Geometry and Trigonometry Formulas Geometry Summary of Surface Area and Volume Formulas – 3D Shapes Shape Figure Surface Area Volume Sphere ࡿൌ࣊࢘ ݎൌݎܽ݀݅ݑݏ ࢂൌ ࣊࢘ ݎൌݎܽ݀݅ݑݏ Right Cylinder ࡿൌ࣊࢘ࢎ࣊࢘ ݄ൌ݄݄݁݅݃ݐ ݎൌݎܽ݀݅ݑݏ݂ܾܽݏ݁ ࢂൌ࣊࢘ࢎ ݄ൌ݄݄݁݅݃ݐ ݎൌݎܽ݀݅ݑݏ ݂ ܾܽݏ݁ Cone ࡿൌ࣊࢘࣊࢘ ݈ൌݏ݈ܽ݊ݐ ݄݄݁݅݃ݐ ݎൌݎܽ݀݅ݑݏ ݂ ܾܽݏ݁ ࢂൌ ࣊࢘ࢎ ݄ൌ݄݄݁݅݃ݐ ݎൌݎܽ݀݅ݑݏ ݂ ܾܽݏ݁ Square Pyramid ࡿൌ࢙࢙ ݏൌܾܽݏ݁ ݏ݅݀݁ ݈݁݊݃ݐ݄ ݈ൌݏ݈ܽ݊ݐ݄݄݁݅݃ݐ ࢂൌ ࢙ࢎ ݏൌܾܽݏ݁ ݏ݅݀݁ ݈݁݊݃ݐ݄ ݄ൌ݄݄݁݅݃ݐ Rectangular Prism ࡿൌ∙ሺ࢝ࢎ࢝ࢎሻ ݈ൌ݈݁݊݃ݐ݄ ݓൌݓ݅݀ݐ݄ ݄ൌ݄݄݁݅݃ݐ ࢂൌ࢝ࢎ ݈ൌ݈݁݊݃ݐ݄ ݓൌݓ݅݀ݐ݄ ݄ൌ݄݄݁݅݃ݐ Cube ࡿൌ࢙ ݏൌݏ݈݅݀݁݁݊݃ݐ݄ሺ݈݈ܽݏ݅݀݁ݏሻ ࢂൌ࢙ ݏൌݏ݅݀݁ ݈݁݊݃ݐ݄ ሺ݈݈ܽݏ݅݀݁ݏሻ General Right Prism ࡿൌࡼࢎ ܲൌܲ݁ݎ݅݉݁ݐ݁ݎ ݂ ܤܽݏ݁ ݄ൌ݄݄݁݅݃ݐ ሺݎ ݈݁݊݃ݐ݄ሻ ܤൌܽݎ݂݁ܽܤܽݏ݁ ࢂൌࢎ ܤൌܽݎ݁ܽ ݂ ܤܽݏ݁ ݄ൌ݄݄݁݅݃ݐ Version 5.6 Page 220 of 242 April 8, 2023 Appendix E Geometry and Trigonometry Formulas Trigonometry Function Relationships sin θ ൌ 1 csc θ csc θ ൌ 1 sin θ cos θ ൌ 1 sec θ sec θ ൌ 1 cos θ tan θ ൌ 1 cot θ cot θ ൌ 1 tan θ tan θ ൌsin θ cos θ cot θ ൌcos θ sin θ Pythagorean Identities sinଶߠ cosଶߠൌ1 tanଶߠ 1 ൌ secଶߠ cotଶߠ 1 ൌ cscଶߠ Cofunction Formulas (in Quadrant I) sin ߠൌcos ቀߨ 2 െߠቁ cos ߠൌsin ቀߨ 2 െߠቁ tan ߠൌcot ቀߨ 2 െߠቁ cot ߠൌtan ቀߨ 2 െߠቁ sec ߠൌcsc ቀߨ 2 െߠቁ csc ߠൌsec ቀߨ 2 െߠቁ Angle Addition Formulas sin ሺܣܤሻൌsin ܣcos ܤ cos ܣsin ܤ sin ሺܣെܤሻൌsin ܣcos ܤ െ cos ܣsin ܤ cos ሺܣܤሻൌcos ܣcos ܤെ sin ܣsin ܤ cos ሺܣെܤሻൌcos ܣcos ܤ sin ܣsin ܤ tan ሺܣܤሻൌtan ܣ tan ܤ 1 – tan ܣ tan ܤ tan ሺܣെܤሻൌtan ܣ െ tan ܤ 1 tan ܣ tan ܤ Double Angle Formulas sin 2ߠൌ2 sin ߠcos ߠ cos 2ߠൌcosଶߠെsinଶߠ ൌ 1 െ2 sinଶߠ ൌ 2 cosଶߠെ1 tan 2ߠൌ 2 tan ߠ 1 െ tanଶߠ Triple Angle Formulas sin 3ߠൌ3 sin ߠെ4 sinଷߠ cos 3ߠൌ4 cosଷߠെ3 cos ߠ tan 3ߠൌ3 tan ߠെtanଷߠ 1 െ3 tanଶߠ Half Angle Formulas sin ߠ 2 ൌ േඨ 1 െ cos ߠ 2 cos ߠ 2 ൌ േඨ 1 cos ߠ 2 tan ߠ 2 ൌ േඨ 1 െ cos ߠ 1 cos ߠ ൌ 1 െ cos ߠ sin ߠ ൌ sin ߠ 1 cos ߠ Power Reducing Formulas sinଶߠൌ1 െcos 2ߠ 2 cosଶߠൌ1 cos 2ߠ 2 tanଶߠൌ1 െcos 2ߠ 1 cos 2ߠ Product‐to‐Sum Formulas sin ܣ∙sin ܤ ൌ 1 2 ሾ cosሺܣെܤሻെcosሺܣܤሻ ሿ cos ܣ∙cos ܤ ൌ 1 2 ሾ cosሺܣെܤሻcosሺܣܤሻ ሿ sin ܣ∙cos ܤ ൌ 1 2 ሾ sinሺܣܤሻsinሺܣെܤሻ ሿ cos ܣ∙sin ܤൌ1 2 ሾsinሺܣܤሻെsinሺܣെܤሻሿ Sum‐to‐Product Formulas sin ܣ sin ܤൌ2 ∙sin ൬ܣܤ 2 ൰∙cos ൬ܣെܤ 2 ൰ sin ܣെ sin ܤൌ2 ∙sin ൬ܣെܤ 2 ൰∙cos ൬ܣܤ 2 ൰ cos ܣ cos ܤൌ2 ∙cos ൬ܣܤ 2 ൰∙cos ൬ܣെܤ 2 ൰ cos ܣെcos ܤൌെ2 ∙sin ൬ܣܤ 2 ൰∙sin ൬ܣെܤ 2 ൰ Law of Sines ୱ୧୬ ൌ ୱ୧୬ൌ ୱ୧୬ Law of Cosines ܽଶ ൌ ܾଶ ܿଶ െ 2ܾܿ cos ܣ ܾଶ ൌ ܽଶ ܿଶ െ 2ܽܿ cos ܤ ܿଶ ൌ ܽଶ ܾଶ െ 2ܾܽ cos ܥ Law of Tangents ܽെܾ ܾܽൌ tan ቂ1 2 ሺܣെܤሻቃ tan ቂ1 2 ሺܣܤሻቃ Arc Length ܵൌݎߠ Opposite Angle Formulas sin ሺെߠሻൌെsin ሺߠሻ cos ሺെߠሻൌcos ሺߠሻ tan ሺെߠሻൌെtan ሺߠሻ cot ሺെߠሻൌെcot ሺߠሻ sec ሺെߠሻൌsec ሺߠሻ csc ሺെߠሻൌെcsc ሺߠሻ Mollweide’s Formulas ܾܽ ܿ ൌ cos ቂ1 2 ሺܣെܤሻቃ sin ቀ1 2 ܥቁ ܽെܾ ܿ ൌ sin ቂ1 2 ሺܣെܤሻቃ cos ቀ1 2 ܥቁ Euler’s Formula ݁ఏൌcos ߠ݅sin ߠൌcis ߠ DeMoivre’s Formula ሺݎ cis ߠሻൌݎ cis ሺ݊ߠሻ Polar Multiplication and Division Let: ܽൌݎ ଵcis ߠ ܾൌݎ ଶcis ߮ ܽ∙ܾൌݎ ଵݎ ଶcis ሺߠ߮ሻ ܽ ܾൌݎ ଵ ݎ ଶ cis ሺߠെ߮ሻ mathguy.us Version 5.6 Page 221 of 242 April 8, 2023 Appendix E Geometry and Trigonometry Formulas Trigonometry Trig Functions of Special Angles (Unit Circle) 𝜽 Rad 𝜽° 𝐬𝐢𝐧𝜽 𝐜𝐨𝐬𝜽 𝐭𝐚𝐧𝜽 0 0⁰ 0 1 0 𝜋6 ൗ 30⁰ 1/2 √3/2 √3/3 𝜋4 ൗ 45⁰ √2/2 √2/2 1 𝜋3 ൗ 60⁰ √3/2 1/2 √3 𝜋2 ൗ 90⁰ 1 0 undefined Rectangular/Polar Conversion Rectangular Polar ሺ𝑥, 𝑦ሻ ሺ𝑟, 𝜃ሻ 𝑥ൌ𝑟cos 𝜃 𝑦ൌ𝑟sin 𝜃 𝑟ൌඥ𝑥ଶ𝑦ଶ 𝜃ൌtanିଵቀ𝑦 𝑥ቁ 𝑎𝑏𝑖 𝑟 ሺcos 𝜃𝑖sin 𝜃ሻ or 𝑟 𝑐𝑖𝑠𝜃 𝑎ൌ𝑟cos 𝜃 𝑏ൌ𝑟sin 𝜃 𝑟ൌඥ𝑎ଶ𝑏ଶ 𝜃ൌtanିଵ൬𝑏 𝑎൰ 𝑎𝐢𝑏𝐣 ‖𝐯‖ ∠𝜃 𝑎ൌ‖𝐯‖ cos 𝜃 𝑏ൌ‖𝐯‖ sin 𝜃 ‖𝐯‖ ൌඥ𝑎ଶ𝑏ଶ 𝜃ൌtanିଵ൬𝑏 𝑎൰ 𝒚ൌ𝑨∙𝒇ሺ𝑩𝒙െ𝑪ሻ𝑫 Amplitude: |𝑨| Period: ௧ "𝒇" ௗ 𝑩 Phase Shift: 𝑪 𝑩 Vertical Shift: 𝑫 Harmonic Motion 𝑑ൌ𝑎cos 𝜔𝑡 or 𝑑ൌ𝑎sin 𝜔𝑡 𝑓ൌ ଵ ୮ୣ୰୧୭ୢൌ ఠ ଶగ 𝜔ൌ2𝜋𝑓, 𝜔0 Triangle Area 𝐴ൌ1 2 𝑏ℎ 𝐴ൌඥ𝑠ሺ𝑠െ𝑎ሻሺ𝑠െ𝑏ሻሺ𝑠െ𝑐ሻ 𝑠ൌ1 2 𝑃ൌ1 2 ሺ𝑎𝑏𝑐ሻ 𝐴 ൌ 1 2 ቆ𝑎ଶsin 𝐵sin 𝐶 sin 𝐴 ቇ 𝐴 ൌ 1 2 𝑎𝑏sin 𝐶 𝐴 ൌ 1 2 ቮ ቮ 𝑥1 𝑦1 1 𝑥2 𝑦2 1 𝑥3 𝑦3 1 ቮ ቮ 𝐴ൌ1 2 ‖𝐮‖ ‖𝐯‖ sin θ Vector Properties 0 𝐮ൌ𝐮0 ൌ𝐮 𝐮ሺെ𝐮ሻൌሺെ𝐮ሻ𝐮ൌ0 𝐮𝐯ൌ𝐯𝐮 𝐮ሺ𝐯𝐰ሻൌሺ𝐮𝐯ሻ𝐰 𝑚ሺ𝑛𝐮ሻൌሺ𝑚𝑛ሻ𝐮 𝑚ሺ𝐮𝐯ሻൌ𝑚𝐮𝑚𝐯 ሺ𝑚𝑛ሻ𝐮ൌ𝑚𝐮𝑛𝐮 1ሺ𝐯ሻൌ𝐯 ‖𝑚𝐯‖ ൌ|𝑚| ‖𝐯‖ Unit Vector: 𝐯 ‖𝐯‖ Vector Dot Product 𝐮∘𝐯ൌሺ𝑢ଵ∙𝑣ଵሻሺ𝑢ଶ∙𝑣ଶሻ 𝐮∘ሺ𝐯𝐰ሻൌሺ𝐮∘𝐯ሻሺ𝐮∘𝐰ሻ Vector Cross Product 𝐮 x 𝐯ൌቚuଵ uଶ vଵ vଶቚൌuଵvଶെuଶvଵ 𝐮 x ሺ𝐯𝐰ሻൌሺ𝐮x 𝐯ሻሺ𝐮x 𝐰ሻ Angle between Vectors cos 𝜃ൌ 𝐮 ∘ 𝐯 ‖𝐮‖ ‖𝐯‖ sin 𝜃ൌ ‖𝐮୶𝐯‖ ‖𝐮‖ ‖𝐯‖ ⊥ iff 𝐮∘ 𝐯ൌ0 ∥ iff 𝐮x 𝐯ൌ0 Period ൌ2𝜋 Period ൌ2𝜋 Period ൌ2𝜋 Period ൌ2𝜋 Period ൌ𝜋 Period ൌ𝜋 Version 5.6 Page 222 of 242 April 8, 2023 Appendix F Polar and Parametric Equations Version 5.6 Page 223 of 242 April 8, 2023 Appendix F Polar and Parametric Equations Polar Graphs Typically, Polar Graphs will be plotted on polar graph paper such as that illustrated at right. On this graph, a point ሺݎ, ߠሻ can be considered to be the intersection of the circle of radius ݎ and the terminal side of the angle ߠ (see the illustration below). Note: a free PC app that can be used to design and print your own polar graph paper is available at www.mathguy.us. Parts of the Polar Graph The illustration below shows the key parts of a polar graph, along with a point, ቀ4, గ ଷቁ. Polar Equations – Symmetry Following are the three main types of symmetry exhibited in many polar equation graphs: Symmetry about: Quadrants Containing Symmetry Symmetry Test(1) Pole Opposite (I and III or II and IV) Replace ݎ with – ݎ in the equation ࢞‐axis Left hemisphere (II and III) or right hemisphere (I and IV) Replace ߠ with – ߠ in the equation ࢟‐axis Upper hemisphere (I and II) or lower hemisphere (III and IV) Replace ሺݎ, ߠሻ with ሺെݎ, െߠሻ in the equation (1) If performing the indicated replacement results in an equivalent equation, the equation passes the symmetry test and the indicated symmetry exists. If the equation fails the symmetry test, symmetry may or may not exist. The Pole is the point ሺ0, 0ሻ (i.e., the origin). The Polar Axis is the positive ݔ‐axis. The Line: ߠൌ గ ଶ is the positive ݕ‐axis. Many equations that contain the cosine function are symmetric about the ݔ‐axis. Many equations that contain the sine function are symmetric about the ݕ‐axis. Version 5.6 Page 224 of 242 April 8, 2023 Appendix F Polar and Parametric Equations Graphs of Polar Equations Graphing Methods Method 1: Point plotting Create a two‐column chart that calculates values of ݎ for selected values of ߠ. This is akin to a two‐column chart that calculates values of ݕ for selected values of ݔ that can be used to plot a rectangular coordinates equation (e.g., ݕൌݔଶെ4ݔ3). The ߠ‐values you select for purposes of point plotting should vary depending on the equation you are working with (in particular, the coefficient of ߠ in the equation). However, a safe bet is to start with multiples of ߨ6 ൗ (including ߠൌ0). Plot each point on the polar graph and see what shape emerges. If you need more or fewer points to see what curve is emerging, adjust as you go. If you know anything about the curve (typical shape, symmetry, etc.), use it to facilitate plotting points. Connect the points with a smooth curve. Admire the result; many of these curves are aesthetically pleasing. Method 2: Calculator Using a TI‐84 Plus Calculator or its equivalent, do the following: Make sure your calculator is set to radians and polar functions. Hit the MODE key; select RADIANS in row 4 and POLAR in row 5. After you do this, hitting CLEAR will get you back to the main screen. Hit Y= and enter the equation in the form ݎൌ݂ሺߠሻ. Use the X,T,ી,n key to enter θ into the equation. If your equation is of the form ݎଶൌ݂ሺߠሻ, you may need to enter two functions, ݎൌඥ݂ሺߠሻ and ݎൌെඥ݂ሺߠሻ, and plot both. Hit GRAPH to plot the function or functions you entered in the previous step. If necessary, hit WINDOW to adjust the parameters of the plot. o If you cannot see the whole function, adjust the X‐ and Y‐ variables (or use ZOOM). o If the curve is not smooth, reduce the value of the ીstep variable. This will plot more points on the screen. Note that smaller values of ીstep require more time to plot the curve, so choose a value that plots the curve well in a reasonable amount of time. o If the entire curve is not plotted, adjust the values of the ીmin and ીmax variables until you see what appears to be the entire plot. Note: You can view the table of points used to graph the polar function by hitting 2ND – TABLE. Version 5.6 Page 225 of 242 April 8, 2023 Appendix F Polar and Parametric Equations Graph of Polar Equations Circle Equation: ݎൌܽsin ߠ Equation: ݎൌܽcos ߠ Equation: ݎൌܽ Location: Location: Location: above ݔ‐axis if ܽ0 right of ݕ‐axis if ܽ0 Centered on the Pole below ݔ‐axis if ܽ൏0 left of ݕ‐axis if ܽ൏0 Radius: ܽ/2 Radius: ܽ/2 Radius: ܽ Symmetry: ݕ‐axis Symmetry: ݔ‐axis Symmetry: Pole, ݔ‐axis, ݕ‐axis Rose Characteristics of roses: Equation: ݎൌܽsin ݊ߠ o Symmetric about the ݕ‐axis Equation: ݎൌܽcos ݊ߠ o Symmetric about the ݔ‐axis Contained within a circle of radius ݎൌܽ If ݊ is odd, the rose has ݊ petals. If ݊ is even the rose has 2݊ petals. Note that a circle is a rose with one petal (i.e, ݊ൌ1). Version 5.6 Page 226 of 242 April 8, 2023 Appendix F Polar and Parametric Equations Graphs of Polar Equations Limaçon of Pascal Equation: ݎൌܾܽsin ߠ Equation: ݎൌܾܽcos ߠ Location: bulb above ݔ‐axis if ܾ0 Location: bulb right of ݕ‐axis if ܾ0 bulb below ݔ‐axis if ܾ൏0 bulb left of ݕ‐axis if ܾ൏0 Symmetry: ݕ‐axis Symmetry: ݔ‐axis Four Limaçon Shapes ܽ൏ܾ ܽൌܾ ܾ൏ܽ൏2ܾ ܽ2ܾ Inner loop “Cardioid” Dimple No dimple Four Limaçon Orientations (using the Cardioid as an example) sine function sine function cosine function cosine function ܾ0 ܾ൏0 ܾ0 ܾ൏0 Version 5.6 Page 227 of 242 April 8, 2023 Appendix F Polar and Parametric Equations Graph of Polar Equations Lemniscate of Bernoulli Characteristics of lemniscates: Equation: ݎଶൌܽଶsin 2ߠ o Symmetric about the line ݕൌݔ Equation: ݎଶൌܽଶcos 2ߠ o Symmetric about the ݔ‐axis Contained within a circle of radius ݎൌܽ Spirals Hyperbolic Spiral: ݎൌ ఏ Characteristics of spirals: Equation: ݎൌܽߠ, ܾ0 o Distance from the Pole increases with ߠ Equation: ݎൌ ್ ఏ, ܾ0 o Hyperbolic Spiral ሺܾൌ1ሻ: asymptotic to the line ܽ units from the ݔ‐axis o Lituus ሺܾൌ2ሻ: asymptotic to the ݔ‐axis Not contained within any circle The lemniscate is the set of all points for which the product of the distances from two points (i.e., foci) which are “2ܿ” apart is ܿଶ. Archimedes’ Spiral Fermat’s Spiral ݎൌܽߠ ݎଶൌܽଶߠ Lituus: ݎଶൌ మ ఏ Version 5.6 Page 228 of 242 April 8, 2023 Appendix F Polar and Parametric Equations Graphing Polar Equations – The Rose Example F.1: ܚൌܛܑܖી This function is a rose. Consider the forms ݎൌܽsin ܾߠ and ݎൌܽcos ܾߠ. The number of petals on the rose depends on the value of ܾ. If ܾ is an even integer, the rose will have 2ܾ petals. If ܾ is an odd integer, it will have ܾ petals. Let’s create a table of values and graph the equation: The four Rose forms: ܚൌܛܑܖࣂ ࣂ ࢘ ࣂ ࢘ 0 0 ߨ/12 2 7π/12 െ2 ߨ/6 3.464 2π/3 െ3.464 ߨ/4 4 3π/4 ‐4 ߨ/3 3.464 5ߨ/6 െ3.464 5ߨ/12 2 11ߨ/12 െ2 ߨ/2 0 ߨ 0 Once symmetry is established, these values are easily determined. Because this function involves an argument of 2ߠ, we want to start by looking at values of θ in ሾ0, 2ߨሿൊ 2 ൌሾ0, ߨሿ. You could plot more points, but this interval is sufficient to establish the nature of the curve; so you can graph the rest easily. Orange points on the graph correspond to orange values in the table. Blue points on the graph correspond to blue values in the table. The values in the table generate the points in the two petals right of the ݕ‐axis. Knowing that the curve is a rose allows us to graph the other two petals without calculating more points. Version 5.6 Page 229 of 242 April 8, 2023 Appendix F Polar and Parametric Equations Graphing Polar Equations – The Cardioid Example F.2: ܚൌܛܑܖી This cardioid is also a limaçon of form ݎൌܾܽsin ߠ with ܽൌܾ. The use of the sine function indicates that the large loop will be symmetric about the ݕ‐axis. The sign indicates that the large loop will be above the ݔ‐axis. Let’s create a table of values and graph the equation: The four Cardioid forms: ܚൌܛܑܖࣂ ࣂ ࢘ ࣂ ࢘ 0 2 ߨ/6 3 7π/6 1 ߨ/3 3.732 4π/3 0.268 ߨ/2 4 3π/2 0 2ߨ/3 3.732 5ߨ/3 0.268 5ߨ/6 3 11ߨ/6 1 ߨ 2 2ߨ 2 Once symmetry is established, these values are easily determined. Generally, you want to look at values of ߠ in ሾ0, 2ߨሿ. However, some functions require larger intervals. The size of the interval depends largely on the nature of the function and the coefficient of ߠ. Orange points on the graph correspond to orange values in the table. Blue points on the graph correspond to blue values in the table. The portion of the graph above the ݔ‐axis results from ߠ in Q1 and Q2, where the sine function is positive. Similarly, the portion of the graph below the x‐axis results from ߠ in Q3 and Q4, where the sine function is negative. Version 5.6 Page 230 of 242 April 8, 2023 Appendix F Polar and Parametric Equations Converting Between Polar and Rectangular Forms of Equations Rectangular to Polar To convert an equation from Rectangular Form to Polar Form, use the following equivalences: ݔൌݎcos ߠ Substitute ݎcos ߠ for ݔ ݕൌݎsin ߠ Substitute ݎsin ߠ for ݕ ݔଶݕଶൌ ݎଶ Substitute ݎଶ for ݔଶݕଶ Example F.3: Convert 8ݔെ3ݕ10 ൌ0 to a polar equation of the form ݎൌ݂ሺߠሻ. Starting Equation: 8ݔെ3ݕ10 ൌ0 Substitute ݔൌݎcos ߠ and ݕൌݎsin ߠ: 8 ∙ݎcos ߠെ3 ∙ݎsin ߠ10 ൌ0 Factor out ݎ: ݎ ሺ8 cos ߠെ3 sin ߠሻൌെ10 Divide by ሺ8 cos ߠെ3 sin ߠሻ: ࢘ൌ ି ૡ܋ܗܛࣂ ି ܛܑܖࣂ Polar to Rectangular To convert an equation from Polar Form to Rectangular Form, use the following equivalences: cos ߠൌݔ ݎ Substitute ݔ ݎ for cos ߠ sin ߠൌݕ ݎ Substitute ݕ ݎ for sin ߠ ݎଶൌݔଶݕଶ Substitute ݔଶݕଶ for ݎଶ Example F.4: Convert r = 8 cos ߠ + 9 sin ߠ to a rectangular equation. Starting Equation: r = 8 cos ߠ + 9 sin ߠ Substitute cos ߠൌ ௫ , sin ߠൌ ௬ : ݎൌ8 ቀ ௫ ቁ9 ቀ ௬ ቁ Multiply by ݎ: ݎଶൌ8ݔ9ݕ Substitute ݎଶൌݔଶݕଶ: ݔଶݕଶൌ8ݔ9ݕ Subtract 8ݔ9ݕ: ݔଶെ8ݔ ݕଶെ9ݕ ൌ0 Complete the square: ሺݔଶെ8ݔ16ሻቀݕଶെ9ݕ ଼ଵ ସቁൌ16 ଼ଵ ସ Simplify to standard form for a circle: ሺ࢞െሻቀ࢟െ ૢ ቁ ൌ Version 5.6 Page 231 of 242 April 8, 2023 Appendix F Polar and Parametric Equations Parametric Equations One way to define a curve is by making ݔ and ݕ (or ݎ and ߠ) functions of a third variable, often ݐ (for time). The third variable is called the Parameter, and functions defined in this manner are said to be in Parametric Form. The equations that define the desired function are called Parametric Equations. In Parametric Equations, the parameter is the independent variable. Each of the other two (or more) variables is dependent on the value of the parameter. As the parameter changes, the other variables change, generating the points of the function. Example F.5: A relatively simple example is a circle, which we can define as follows: Circle: ݔൌݎcos ݐ ݕൌݎsin ݐ As the variable ݐ progresses from 0 to 2ߨ, a circle of radius ݎ is born. The circle in the illustration at right can be defined in several ways: Cartesian form: ݔଶݕଶൌ16 Polar form: ݎൌ4 Parametric form: ݔൌ4 cos ݐ ݕൌ4 sin ݐ Familiar Curves Many curves with which the student may be familiar have parametric forms. Among those are the following: Curve Cartesian Form Polar Form Parametric Form Parabola with horizontal directrix ݕൌܽሺݔെ݄ሻଶ݇ ݎൌ 1 േsin ߠ ݔൌ2ݐ ݕൌݐଶ Ellipse with horizontal major axis ሺݔെ݄ሻଶ ܽଶ ሺݕെ݇ሻଶ ܾଶ ൌ1 ݎൌ ݁ 1 േ݁∙cos ߠ ሺ0 ൏݁൏1ሻ ݔൌܽcos ݐ ݕൌܾsin ݐ Hyperbola with horizontal transverse axis ሺݔെ݄ሻଶ ܽଶ െሺݕെ݇ሻଶ ܾଶ ൌ1 ݎൌ ݁ 1 േ݁∙cos ߠ ሺ݁1ሻ ݔൌܽsec ݐ ݕൌܾtan ݐ As can be seen from this chart, sometimes the parametric form of a function is its simplest. In fact, parametric equations often allow us to graph curves that would be very difficult to graph in either Polar form or Cartesian form. Some of these are illustrated on the next page. Version 5.6 Page 232 of 242 April 8, 2023 Appendix F Polar and Parametric Equations Some Functions Defined by Parametric Equations (Star Wars fans: are these the “oids” you are looking for?) The graphs below are examples of functions defined by parametric equations. The equations and a brief description of the curve are provided for each function. Deltoid Nephroid Astroid Parametric equations: Parametric equations: Parametric equations: ݔൌ2ܽcos ݐܽcos 2ݐ ݔൌܽሺ3 cos ݐെcos 3ݐሻ ݔൌܽcosଷݐ ݕൌ2ܽsin ݐെܽsin 2ݐ ݕൌܽሺ3 sin ݐെsin 3ݐሻ ݕൌܽsinଷݐ Cycloid Parametric equations: ݔൌܽሺݐെsin ݐሻ ݕൌܽሺ1 െcos ݐሻ The deltoid is the path of a point on the circumference of a circle as it makes three complete revolutions on the inside of a larger circle. The nephroid is the path of a point on the circumference of a circle as it makes two complete revolutions on the outside of a larger circle. The astroid is the path of a point on the circumference of a circle as it makes four complete revolutions on the inside of a larger circle. The cycloid is the path of a point on the circumference of a circle as the circle rolls along a flat surface (think: the path of a point on the outside of a bicycle tire as you ride on the sidewalk). The cycloid is both a brachistochrone and a tautochrone (look these up if you are interested). Version 5.6 Page 233 of 242 April 8, 2023 Appendix G Interesting Series and Summation Formulas ሺ𝑐ሻൌ𝑛𝑐 ୀଵ 𝑐𝑐⋯𝑐ൌ𝑛𝑐 ሺ𝑘ሻൌ𝑛ሺ𝑛1ሻ 2 ୀଵ 1 2 ⋯𝑛ൌ𝑛ሺ𝑛1ሻ 2 ሺ𝑘ଶሻൌ𝑛ሺ𝑛1ሻሺ2𝑛1ሻ 6 ୀଵ 1ଶ2ଶ⋯𝑛ଶൌ𝑛ሺ𝑛1ሻሺ2𝑛1ሻ 6 ሺ𝑘ଷሻൌቆ𝑛ሺ𝑛1ሻ 2 ቇ ଶ ୀଵ 1ଷ2ଷ⋯𝑛ଷൌቆ𝑛ሺ𝑛1ሻ 2 ቇ ଶ ሺ𝑥ሻൌ 1 1 െ𝑥 ஶ ୀ 𝑓𝑜𝑟 ሼെ1 ൏𝑥൏1ሽ 1 𝑥𝑥ଶ𝑥ଷ𝑥ସ⋯ൌ 1 1 െ𝑥 ቆ𝑥 𝑘!ቇൌ𝑒௫ ஶ ୀ 1 𝑥𝑥ଶ 2! 𝑥ଷ 3! 𝑥ସ 4! ⋯ൌ𝑒௫ ቈ1 𝑘൬𝑥െ1 𝑥 ൰ ஶ ୀଵ ൌln 𝑥 𝑓𝑜𝑟 𝑥1 2 ൬𝑥െ1 𝑥 ൰1 2 ൬𝑥െ1 𝑥 ൰ ଶ 1 3 ൬𝑥െ1 𝑥 ൰ ଷ ⋯ൌln 𝑥 ቈሺെ1ሻሺାଵሻቆ𝑥 𝑘ቇൌln ሺ1 𝑥ሻ ஶ ୀଵ 𝑓𝑜𝑟 ሼെ1 ൏𝑥1ሽ 𝑥െ𝑥ଶ 2 𝑥ଷ 3 െ𝑥ସ 4 ⋯ൌln ሺ1 𝑥ሻ ቈሺെ1ሻቆ𝑥ሺଶሻ ሺ2𝑘ሻ!ቇൌcos 𝑥 ஶ ୀ 1 െ𝑥ଶ 2! 𝑥ସ 4! െ𝑥 6! ⋯ൌcos 𝑥 ቈሺെ1ሻቆ𝑥ሺଶାଵሻ ሺ2𝑘1ሻ!ቇൌsin 𝑥 ஶ ୀ 𝑥െ𝑥ଷ 3! 𝑥ହ 5! െ𝑥 7! ⋯ൌsin 𝑥 ቈሺെ1ሻቆ𝑥ሺଶାଵሻ ሺ2𝑘1ሻቇൌtanିଵ𝑥 ஶ ୀ 𝑓𝑜𝑟 ሼെ1 𝑥1ሽ 𝑥െ𝑥ଷ 3 𝑥ହ 5 െ𝑥 7 ⋯ൌtanିଵ𝑥 Version 5.6 Page 234 of 242 April 8, 2023 Page Subject 162 Abel's Convergence Test (Series) 164 Absolute Convergence of a Series 32 Absolute Extrema - see also Integration 50, 135 Acceleration 35 Alauria Diagram 163 Alternating Series 153 Analytic Continuation 56 Antiderivatives 107, 110 Arc Length 101 Area by Integration 112, 117 Area Cross Section Method - Volume of a Solid 103 Area in Polar Form 105 Area of a Limacon 111 Area of a Surface of Revolution 233 Astroid 184 Ault Table 93 Average Rate of Change 93 Average Value of a Function 156 Bernoulli Numbers 76 Beta Function 227 Cardioid 177 Cauchy-Riemann Equations 53 Center of Curvature 18 Chain Rule 226 Circle (Polar Form) 158 Comparison Test for Series Convergence 33 Concavity 164 Conditional Convergence of a Series 110 Conversion among Rectangular, Polar, Parametric Forms 142 Convergence Tests - Sequences 158 Convergence Tests - Series 10 Continuity Examples 8 Continuity Rules Calculus Handbook Index Version 5.6 Page 235 of 242 April 8, 2023 Page Subject Calculus Handbook Index 78 Cosine Integral - Ci(x) 132 Cross Product 138 Curl 53 Curvature 39 Curve Sketching 233 Cycloid 112, 115 Cylindrical Shell Methods - Volume of a Solid of Revolution 92 Definite Integration 87 Definite Integrals 92 Fundamental Theorem of Calculus 94 Properties of Definite Integrals 87 Riemann Sums 92 Rules of Definite Integration 95 Solving Definite Integrals with Directed Line Segments 98 Special Techniques 96 u -Substitution 183 Definitions - Alphabetically 136 Del Operator 233 Deltoid 177 Derivative of e to a Complex Power (ez) 181 Derivative of: (x+y)3=x3+y3 Derivatives - see Differentiation 178 Derivatives of a Circle 179 Derivatives of a Ellipse 180 Derivatives of a Hyperbola 39 DIACIDE (curve sketching) 120 Differential Equations 52 Differentials Differentiation 17 Basic Rules 18 Exponential and Trigonometric Functions 23 Generalized Product Rule 27 Implicit Differentiation 25 Inverse Function Rule 25 Inverse Function Diagram Version 5.6 Page 236 of 242 April 8, 2023 Page Subject Calculus Handbook Index 19 Inverse Trigonometric Functions 208 List of Key Derivatives 30 Logarithmic Differentiation 110 Parametric Derivatives 26 Partial Differentiation 37 What Does the Graph of f '(x) Tell Us about f(x) ?
161 Dirichlet’s Convergence Test (Series) 9 Discontinuities 112, 113 Disk Method - Volume of a Solid of Revolution 49, 135 Displacement 49 Distance 137 Divergence 131 Dot Product 172 e 9 Essential Discontinuity 173 Euler's Formula 125 Euler's Methods (Differential Equations) 18, 57 Exponential Functions 78 Exponential Integral - Ei(x) 31 Exterema 34 Exterema of Polynomials 31 First Derivative Test 8 Functions 234 Functions and Their Derivatives (Summary) 92 Fundamental Theorems of Calculus 74 Gamma Function 46 General-Specific Method (for Related Rates Problems) 219 Geometry Formulas (Area and Volume) 136 Gradient 229 Graphing Polar Equations Hyperbolic Functions 79 Definitions 84 Derivatives 83 Graphs of Hyperbolic Functions and Their Inverses 80 Identities Version 5.6 Page 237 of 242 April 8, 2023 Page Subject Calculus Handbook Index 85 Integrals 82 Inverse Hyperbolic Functions 81 Relationship to Trigonometric Functions 228 Hyperbolic Spiral 176 i i 27 Implicit Differentiation 78 Impossible Integrals 118 Improper Integrals 158 Integral Test for Series Convergence 85 Integrals 56 Indefinite Integration 14 Indeterminate Forms 9 Infinite Discontinuity 33 Inflection Points 34 Inflection Points of Polynomials Integration 76 Beta Function 57 Exponential Functions 74 Gamma Function 78 Impossible Integrals 56 Indefinite Integration (Antiderivatives) 60 Inverse Trigonometric Functions 210 List of Key Integrals 57 Logarithmic Functions 65 Partial Fractions 68 Parts 71 Parts - Tabular Method 62 Selecting the Right Function for an Intergral 57 Trigonometric Functions 72 Trigonometric Substitution 63 u -Substitution 165 Interval of Convergence 25 Inverse Function Diagram (for derivatives) Inverse Trigonometric Functions 60 About Inverse Trig Functions 19, 22 Derivatives Version 5.6 Page 238 of 242 April 8, 2023 Page Subject Calculus Handbook Index 20, 22 Development of Derivatives 21 Graphs 60, 61 Integrals 9 Jump Discontinuity 35 Key Points on f(x), f'(x) and f''(x) 49 Kinematics (Particle Motion) 135 Kinematics (Particle Motion) - Vectors 13 L'Hospital's Rule 171 Lagrange Remainder of a Taylor Series 139 Laplacian 88 Left Endpoint Method (Riemann Sum) 228 Lemniscate of Bernoulli 68 LIATE 105, 227 Limacon 12 Limit-Finding Techniques 11 Limit Rules 11 Limits 16 Limits: Failure to Exist 228 Lituus 30 Logarithmic Differentiation 18, 57 Logarithmic Functions 78 Logarithmic Integral - li(x) 175 Logarithms of Complex Numbers 175 Logarithms of Negative Real Numbers 124 Logistic Function 169 Maclaurin Series 31 Maxima and Minima 45 Mean Value Theorem 87 Mesh Size (of a Riemann Sum) 88 Midpoint Method (Riemann Sum) 38 Natalie Chart - relationship among f(x), f'(x), f''(x) 233 Nephroid 54 Newton's Method Version 5.6 Page 239 of 242 April 8, 2023 Page Subject Calculus Handbook Index 182 Normal Distribution PDF Inflection Points 126 Order of a Numerical Method (Differential Equations) 16 Oscillating Behavior of Limits 53 Osculating Circle 152 p -Series 110 Parametric Forms - Summary 26 Partial Differentiation 65 Partial Fractions 49 Particle Motion 110 Polar Forms - Summary 224 Polar Graphs 49, 135 Position Function 18 Power Rule (differentiation) 148 Power Series 17, 23 Product Rule (differentiation) 18 Quotient Rule (differentiation) 165 Radius of Convergence 53 Radius of Curvature 159 Ratio Test for Series Convergence 110 Rectangular Forms 140, 143-145 Recursive Sequences 46 Related Rates 31 Relative Extrema 9 Removable Discontinuity 111, 112 Revolution - Volume, Surface Area 87 Riemann Sums 152 Riemann Zeta Function (p -Series) 88 Right Endpoint Method (Riemann Sum) 132 Right Hand Rule 45 Rolle's Theorem 160 Root Test for Series Convergence 226 Rose 127 Runge-Kutta Method (Differential Equations) Version 5.6 Page 240 of 242 April 8, 2023 Page Subject Calculus Handbook Index 136 Scalar Field 32 Second Derivative Test 141 Sequences 142 Absolute Value Theorem 142 Bounded Monotonic Sequence Theorem 142 Bounded Sequence 143 Convergence and Divergence 141 Explicit Sequence 14 Indeterminate Forms 143 Limit of a Sequence 142 Monotonic Sequence 140, 143-145 Recursive Sequences 142 Squeeze Theorem 141 Types of Sequences 147, 234 Series 162 Abel's Convergence Test 164 Absolute Convergence 163 Alternating Series 158 Comparison Test 164 Conditional Convergence 147 Convergence and Divergence 158 Convergence Tests 147 Definition 161 Dirichlet’s Convergence Test 151 Estimating the Value of Series with Positive Terms 150 Geometric Series 158 Integral Test 148 Key Properties 169 Maclaurin Series 148 n -th Term Convergence Theorems 152 p -Series 147 Partial Sums 148 Power Series 159 Ratio Test 160 Root Test 168 Summary of Convergence/Divergence Tests 169 Taylor Series 149 Telescoping Series 164 Term Rearrangement Version 5.6 Page 241 of 242 April 8, 2023 Page Subject Calculus Handbook Index 44 Shape of a Curve 112, 115 Shell Methods - Volume of a Solid of Revolution 78 Sine Integral - Si(x) 123 Slope Fields 112 Solids of Revolution 50, 110, 135 Speed 228 Spiral 111 Surface of Revolution #REF!
Tabular Method of Integration by Parts 169 Taylor Series 203 Theorems - Summary 88 Trapezoid Method (Riemann Sum) 19, 57 Trigonometric Functions 72 Trigonometric Substitution (Integration) 221 Trigonometry Formulas 134 Triple Products of Vectors 63, 96 u -Substitution 16 Unbounded Behavior of Limits 136 Vector Field 129 Vectors 129 Components 132 Cross Product 138 Curl 137 Divergence 131 Dot Product 136 Gradient 139 Laplacian 130 Properties 129 Special Unit Vectors 134 Triple Products 49, 135 Velocity 112 Volumes of Solids 112, 113 Washer Method - Volume of a Solid of Revolution 152 Zeta Function Version 5.6 Page 242 of 242 April 8, 2023 |
6826 | https://www.geeksforgeeks.org/maths/how-to-find-the-third-side-of-a-triangle-given-two-sides/ | How to Find the Third Side of a Triangle Given Two Sides?
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23 Jul, 2025
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Answer: To find the third side of a triangle given two sides, you need to know the type of triangle
1) Right Triangle: Use the Pythagorean Theorem
2)Non-Right Triangle: Use the Law of Cosines.
This matters because right triangles use a simple formula, while other triangles need a different one. If the triangle has a 90° angle, use the Pythagorean theorem. If not, use the Law of Cosines with the angle between the two sides.
Finding the Third Side using the Pythagorean Theorem
The third side of any triangle is found only when two sides of the Right-Angled Triangle are given. We use Pythagoras' Theorem in such cases. Let's assume that a Right Angled Triangle ABC where AB is the Perpendicular, BC is the Base, and CA is the Hypotenuse.
Then according to Pythagoras Theorem, the sum of squares of two sides is equal to the square of the third side.
(Perpendicular)2 + (Base)2 = (Hypotenuse)2
Using the above equation third side can be calculated if two sides in a right-angled triangle are known. This is explained using the example added below.
Example: If two sides of a right angled triangle are, AB(Perpendicular) = 3 cm and BC (Base) = 5 cm then find the third side Hypotenuse of the triangle.
Perpendicular (P) = 3 cm
Base (B) = 4 cm
Using Pythagoras theorem
P2 + B2 = H2
(3)2 + (4)2 = H2
9 + 16 = H2
25 = H2
H = 5
Thus, the third side hypotenuse of the triangle ABC CA is 5 cm.
Finding the Third Side using the Law of Cosines
In a non-right triangle, the sides and angles do not follow the Pythagorean theorem because there is no 90° angle. To find the third side when you know two sides and the included angle, you use the Law of Cosines. This law helps relate all three sides of the triangle to one of its angles.
It’s especially useful when:
You know two sides and the angle between them (SAS), or
You know all three sides and want to find an angle (SSS).
If you know sides a and b and the angle C between them, you can find the third side c using:
c2 = a2+b2−2abcos(C)
Then take the square root to get c:
c = √ c2=a2+b2−2abcos(C)
Example: Suppose you have a triangle with:
Side a = 5 units
Side b = 7 units
Included angle C=60o
Solution:
Using the Law of Cosines:
c2 = a2 + b2−2abcos(C)
c2 = 52 + 72−2 ✕ 5 ✕ 7 cos(60o)
c2 = 25 + 49 − 70 ✕ 0.5
c2 = 25 + 49 − 70 ✕ 0.5
c2 = 39
c= √39
Articles Related to Triangles
| Topic | Triangles: Based on Sides | Triangles: Based on Angles |
---
| Triangle | Equilateral Triangle | Acute Angled Triangle |
| Area of Triangle | Isosceles Triangle | Right-Angled Triangle |
| Perimeter of Triangle | Scalene Triangle | Obtuse Angled Triangle |
Important Formulas Related to Triangles
Heron’s Formula
Area of Isosceles Triangle
Angle Sum Property of Triangle.
Solved Examples of Finding the Third side of a Triangle
Problem 1: Find the measure of the perpendicular and hypotenuse given: perpendicular = 12 cm and hypotenuse = 13 cm.
Solution:
Given,
Perpendicular = 12 cm
Hypotenuse = 13 cm
Using Pythagoras Theorem
P2 + B2 = H2
B2 = H2 - P2
B2 = 132 - 122
B2 = 169 - 144
B2 = 25 = 5 cm
Problem 2: The perimeter of the equilateral triangle is 63 cm. Find, the side of the triangle.
Solution:
Perimeter of an Equilateral Triangle = 3×side
3×side = 63
side = 63/3
side = 21 cm
Problem 3: Find the measure of the third side of a right-angled triangle if the two sides are 6 cm and 8 cm.
Solution:
Given,
Perpendicular = 6 cm
Base = 8 cm
Using Pythagoras Theorem
H2 = P2 + B2
H2 = P2 + B2
H2 = 62 + 82
H2 = 36 + 64
H2 = 100
H = 10 cm
Problem 4: Find the hypotenuse of a right angled triangle whose base is 8 cm and whose height is 15 cm?
Solution:
Using Pythagorean theorem, a2 + b2 = c2
So 82 + 152 = c2
hence c = √(64 + 225)
c = √289 = 17 cm
Problem 5: In triangle ABC, sides AB = 9 and AC = 11, with the included angle ∠BAC = 60o . Find the length of side BC.
Solution:
Using the Law of Cosines:
BC2 = a2 + b2−2abcos(C)
BC2 = 92 + 112−2 ✕ 9 ✕ 11 ✕ cos(60o)
BC2 = 81 + 121 − 198 ✕ 0.5
BC2= 202 - 99
BC2= 103
BC2 = √103
Problem 6: Triangle XYZ has XY = 7cm, XZ = 5 cm and ∠YXZ = 100o.Find the length of side YZ.
Solution:
Using the Law of Cosines:
YZ2 = a2 + b2−2abcos(C)
YZ2 = 72 + 52−2 ✕ 7 ✕ 5 ✕ cos(100o)
YZ2= 49 + 25 − 70 ✕ (-0.1736)
YZ2 = 74 + 12.152
YZ2= 86.152
YZ2 = √86.152
Unsolved Practice Questions on Finding the Third Side of a Triangle
Question 1: What formula do we use to find the third side of a triangle when given the lengths of the two sides?
Question 2: If one side of a triangle is 5 cm and the other is 12 cm, what is the maximum length that the third side can be?
Question 3: How do you determine if three lengths can form a triangle?
Question 4: In triangle RS, side RS=8 cm, side RT=6 cm, and the included angle ∠SRT=45o. Find the length of side ST.
Question 5: If two sides of a triangle measure 7 m and 24 m., what is the length of the third side if it is known to be the longest side?
Question 6: Can the lengths of two sides of a triangle ever equal the length of the third side? What would this imply?
Question 7: In triangle DEF, side DE=7 in, side DF=10 in, and the included angle ∠EDF=75o. Find the length of side EF.
Question 8: In a right triangle, if one side is 6 cm and the hypotenuse is 10 cm, how do we find the length of the third side?
Question 9: In triangle LM, side LM=30 km, side LN=40 km, and the included angle ∠MLN=100o. Determine the length of side MN.
Question 10: If two sides of a triangle are equal, how do we find the length of the third side if the triangle is isosceles?
Conclusion
To find the third side of a triangle first determine whether the triangle is a right triangle. If it is use the Pythagorean theorem. For non-right triangles, the Law of Cosines is used. With this knowledge we can handle different types of the triangles based on the given sides and angles.
How to Find the Third Side of a Triangle Given Two Sides?
J
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2012 AMC 12B Problems/Problem 9
Contents
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1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 See Also
Problem
It takes Clea 60 seconds to walk down an escalator when it is not moving, and 24 seconds when it is moving. How many seconds would it take Clea to ride the escalator down when she is not walking?
Solution 1
She walks at a rate of units per second to travel a distance . As , we find and , where is the speed of the escalator. Setting the two equations equal to each other, , which means that . Now we divide by because you add the speed of the escalator but remove the walking, leaving the final answer that it takes to ride the escalator alone as
Solution 2
We write two equations using distance = rate time. Let be the rate she is walking, and be the speed the escalator moves. WLOG, let the distance of the escalator be , as the distance is constant. Thus, our equations are and . Solving for , we get . Thus, it will take Clea seconds.
~coolmath2017 ~Extremelysupercooldude (Latex edits)
Solution 3
Clea covers of the escalator every second. Say the escalator covers of the escalator every second. Since Clea and the escalator cover the entire escalator in seconds, we can use distance rate time to get . Solving gives us , so if Clea were to just stand on the escalator, it would take her seconds to get down. ~Extremelysupercooldude
See Also
2012 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 8Followed by
Problem 10
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 12 Problems and Solutions
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6828 | https://www.statisticshowto.com/calculus-problem-solving/optimization-problems/ | Skip to content
Optimization Problems in Calculus
Contents:
General Optimization Steps
Volume of Largest Rectangular Box Inside a Pyramid
More optimization problems:
How to maximize a function.
Maximum Profit.
Largest inscribed rectangle within a circle
Maximum Volume of Cone Inscribed in a Sphere
General Optimization Steps
Optimization problems in calculus often involve the determination of the “optimal” (meaning, the best) value of a quantity. For example, we might want to know:
The biggest area that a piece of rope could be tied around.
How high a ball could go before it falls back to the ground.
At which point of a loop does a roller coaster run the slowest.
Very often, the optimization must be done with certain constraints. In the case of the rope, we’re limited by its length. These constraints are usually very helpful to solve optimization problems (for an advanced example of using constraints, see: Lagrange Multiplier). Optimal values are often either the maximum or the minimum values of a certain function.
Optimization Problems in Calculus: Steps.
Example problem: Find the maximum area of a rectangle whose perimeter is 100 meters. (Note: This is a typical optimization problem in AP calculus).
Step 1: Determine the function that you need to optimize. In the example problem, we need to optimize the area A of a rectangle, which is the product of its length L and width W. Our function in this example is:
A = LW.
Step 2: Identify the constraints to the optimization problem. In our example problem, the perimeter of the rectangle must be 100 meters. This will be useful in the next step.
Step 3: Express that function in terms of a single variable upon which it depends, using algebra. For this example, we’re going to express the function in a single variable. “L.”
A rectangle’s perimeter is the sum of its sides, that is, 100m = 2L + 2W.
Subtract 2L from both sides of this equation, 2W = 100m – 2L.
Divide each side by 2: W = 50m – L.
Substitute 50m – L for “W” in A = LW: A = L (50m – L) = 50m L – L².
Step 4: Calculate the derivative of the function with respect to a variable.
dA/dL = 50m (1) L(1 – 1) – 2 L(2-1) = 50m – 2L.
Step 5: Set the function to zero and compute the corresponding variable’s value. For our sample problem, we set:
dA/dL = 0 = 50m – 2L.
So, L = 25m.
Step 6: Use the value from Step 5 to calculate the corresponding optimal value of the function. In our sample problem, A = 50m L – L2 = 50 m (25m) – (25m)2 = 625 m2.
The Volume of the Largest Rectangular Box Inscribed in a Pyramid
If you’re asked to find the volume of the largest rectangular box in the first octant, with three faces in the coordinate planes and one vertex in a given plane, you’re being asked to find the volume of the largest rectangular box that fits in a pyramid like the one below.
Imagine that point A is embedded in the rectangular face.
Suppose the plane is x + 2y + 3z = 6. What is the volume of the largest box?
Your first step should be to define the volume. Since your box is rectangular, the formula is:
width x depth x height.
We can write this as:
V = xyz
Since x + 2y + 3z = 6, we know z = (6 – x – 2y) / 3. We can substitute that in our volume equation to create a function that tells us the volume in terms of x and y:
V = f(x, y) = xy(6 – x – 2y) / 3
We’re looking for a maximum, so we want to find the point where the derivative of volume with respect to both x and y is zero.
We can call those two derivatives fx and fy, and we can calculate them as
fx = ⅓y (6 – 2x – 2y) = 0 → y = 0, y = 3 – x
fy = ⅓x (6 – x – 4y) = 0
At y = 0, equation B becomes (1/3)x (6 – x) = 0, so x = 0 or x = 6.
At y = 3 – x, equation B becomes x(-2 + x) = 0, so the zeros are x = 0 (in which case y = 3) or x = 2 (in which case y = 1).
We’e just found the four critical points: (6, 0), (0,0) (0, 3), (2, 1).
We can draw each of these points on a 3D graph, and use our knowledge of geometry to decide which box is the maximum: it should be fairly evident that it will be (2,1), and since x = 2 and y = 1 leads to z = 2/3, the coordinate of the largest box will be (2, 1, 2/3) .
Alternative: Second Derivative Test
If you don’t want to rely on geometry, you can use the second derivative test to find out which critical point gives us that box with maximum volume:
D = D(a, b) = fxx(a,b) fyy(a, b) – [fxy(a, b)]2
at (6, 0), (0, 0), (0, 3)
D = 0 – 4 = -4
at (2, 1)
Which leads to the same conclusion.
So the volume V = xyz = 2 · 1 · 2/3 = 4/3.
Optimization Problems: References
Miech, Ron. Calculus Problems: Volume of the Largest Rectangular Box. Retrieved from on January 1, 2019.
Formula for the volume of a rectangular box.
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6829 | https://api.pageplace.de/preview/DT0400.9781292440354_A47228037/preview-9781292440354_A47228037.pdf | 5th EDITION PHYSICS for SCIENTISTS and ENGINEERS with Modern Physics DOUGLAS GIANCOLI GLOBAL EDITION z = 11.9 z = 8.8 g 5 Other Useful Data Joule equivalent (1 cal) 4.186 J Absolute zero (0 K) -273.15°C Acceleration due to gravity at Earth’s surface (avg.) 9.80 m >s2 (= g) Speed of sound in air (20°C) 343 m >s Density of air (dry) 1.29 kg >m3 Earth: Mass 5.98 1024 kg Radius (mean) 6.38 103 km Moon: Mass 7.35 1022 kg Radius (mean) 1.74 103 km Sun: Mass 1.99 1030 kg Radius (mean) 6.96 105 km Earth – Sun distance (mean) 149.60 106 km Earth – Moon distance (mean) 384 103 km The Greek Alphabet Alpha Α a Nu Ν n Beta Β b Xi Ξ j Gamma Γ g Omicron Ο o Delta ∆ d Pi Π p Epsilon Ε ϵ, e Rho Ρ r Zeta Ζ z Sigma π s Eta Η h Tau Τ t Theta ϴ u Upsilon Υ y Iota Ι i Phi Φ f, w Kappa Κ k Chi Χ x Lambda Λ l Psi Ψ c Mu Μ m Omega Ω v Values of Some Numbers p = 3.1415927 12 = 1.4142136 ln 2 = 0.6931472 log10 e = 0.4342945 e = 2.7182818 13 = 1.7320508 ln 10 = 2.3025851 1 rad = 57.2957795° Properties of Water Density (4°C) 1.000 103 kg>m3 Heat of fusion (0°C) 334 kJ>kg (79.8 kcal>kg) Heat of vaporization (100°C) 2260 kJ>kg (539.9 kcal>kg) Specific heat (15°C) 4186 J>kg⋅C° (1.00 kcal>kg⋅C°) Index of refraction 1.33 Mathematical Signs and Symbols r is proportional to … is less than or equal to = is equal to Ú is greater than or equal to L is approximately equal to π sum of ≠ is not equal to x average value of x 7 is greater than ∆x change in x W is much greater than ∆x 9 9 T 0 ∆x approaches zero 6 is less than n!
n(n - 1) (n - 2) c (1) V is much less than Fundamental Constants Quantity Symbol Approximate Value Current Best Value† Speed of light in vacuum c 3.00 108 m>s 2.99792458 108 m>s Gravitational constant G 6.67 10-11 N⋅m2>kg2 6.67430 (15) 10-11 N⋅m2>kg2 Avogadro’s number NA 6.02 1023 mol-1 6.02214076 1023 mol-1 Gas constant R 8.314 J>mol⋅K = 1.99 cal>mol⋅K = 0.0821 L ⋅atm>mol⋅K 8.314462618 J>mol⋅K Boltzmann’s constant k 1.38 10-23 J>K 1.380649 10-23 J>K Charge on electron e 1.60 10-19 C 1.602176634 10-19 C Stefan-Boltzmann constant s 5.67 10-8 W>m2⋅K4 5.670374419 10-8 W>m2⋅K4 Permittivity of free space ϵ0 8.85 10-12 C2>N⋅m2 8.8541878128(13) 10-12 C2>N⋅m2 Permeability of free space m0 1.26 10-6 T ⋅m>A 1.25663706212(19) 10-6 T ⋅m >A Planck’s constant h 6.63 10-34 J⋅s 6.62607015 10-34 J⋅s Electron rest mass me 9.11 10-31 kg = 0.000549 u = 0.511 MeV>c2 9.1093837015(28) 10-31 kg = 5.48579909065(16) 10-4 u Proton rest mass m p 1.6726 10-27 kg = 1.00728 u = 938.27 MeV>c2 1.67262192369(51) 10-27 kg = 1.007276466621(53) u Neutron rest mass m n 1.6749 10-27 kg = 1.008665 u = 939.57 MeV>c2 1.67492749804(95) 10-27 kg = 1.00866491595(49) u Atomic mass unit (1 u) 1.6605 10-27 kg = 931.49 MeV>c2 1.66053906660(50) 10-27 kg = 931.49410242(28) MeV>c2 † Numbers in parentheses indicate one-standard-deviation experimental uncertainties in final digits (2019, new SI). Values without parentheses are exact (i.e., defined quantities).
Unit Conversions (Equivalents) Time 1 day = 8.640 104 s 1 year = 365.242 days = 3.156 107 s Length 1 in. = 2.54 cm (defined) 1 cm = 0.3937 in. 1 ft = 30.48 cm 1 m = 39.37 in. = 3.281 ft 1 mi = 5280 ft = 1.609 km 1 km = 0.6214 mi 1 nautical mile = 1.151 mi = 6076 ft = 1.852 km 1 fermi = 1 femtometer (fm) = 10-15 m 1 angstrom (Å) = 10-10 m = 0.1 nm 1 light-year (ly) = 9.461 1015 m 1 parsec = 3.26 ly = 3.09 1016 m Volume 1 liter (L) = 1000 mL = 1000 cm3 = 1.0 10-3 m3 = 1.057 qt (U.S.) = 61.02 in.3 1 gal (U.S.) = 4 qt (U.S.) = 231 in.3 = 3.785 L = 0.8327 gal (British) 1 quart (U.S.) = 2 pints (U.S.) = 946 mL 1 pint (British) = 1.20 pints (U.S.) = 568 mL 1 m3 = 35.31 ft3 Speed 1 mi>h = 1.4667 ft>s = 1.6093 km>h = 0.4470 m>s 1 km>h = 0.2778 m>s = 0.6214 mi>h 1 ft>s = 0.3048 m>s = 0.6818 mi>h = 1.0973 km>h 1 m>s = 3.281 ft>s = 3.600 km>h = 2.237 mi>h 1 knot = 1 nautical mile>h = 1.151 mi>h = 1.852 km>h = 0.5144 m>s Angle 1 radian (rad) = 57.30° = 57°18′ 1° = 0.01745 rad 1 rev>min (rpm) = 0.1047 rad>s Mass 1 atomic mass unit (u) = 1.6605 10-27 kg 1 kg = 0.06852 slug 1 ton (metric) = 1000 kg [1 kg has a weight of 2.20 lb where g = 9.80 m>s2.] Force 1 lb = 4.44822 N 1 N = 105 dyne = 0.2248 lb 1 ton (U.S.) = 2000 lbs Energy and Work 1 J = 107 ergs = 0.7376 ft ⋅lb 1 ft ⋅lb = 1.356 J = 1.29 10-3 Btu = 3.24 10-4 kcal 1 kcal = 4.19 103 J = 3.97 Btu 1 eV = 1.6022 10-19 J 1 kWh = 3.600 106 J = 860 kcal 1 Btu = 1.056 103 J Power 1 W = 1 J>s = 0.7376 ft ⋅lb>s = 3.41 Btu>h 1 hp = 550 ft ⋅lb>s = 746 W Pressure 1 atm = 1.01325 bar = 1.01325 105 N>m2 = 14.7 lb>in.2 = 760 torr 1 lb>in.2 = 6.895 103 N>m2 1 Pa = 1 N>m2 = 1.450 10-4 lb>in.2 Some SI Units in Terms of Base Units In Terms of Quantity Unit Abbreviation Base Units† Force newton N kg⋅m>s2 Energy and work joule J kg⋅m2>s2 Power watt W kg⋅m2>s3 Pressure pascal Pa kg >(m⋅s2) Frequency hertz Hz s-1 Electric charge coulomb C A ⋅s Electric potential volt V kg⋅m2>(A ⋅s3) Electric resistance ohm Ω kg⋅m2>(A 2⋅s3) Capacitance farad F A 2⋅s4>(kg⋅m2) Magnetic field tesla T kg >(A ⋅s2) Magnetic flux weber Wb kg⋅m2>(A ⋅s2) Inductance henry H kg⋅m2>(A 2⋅s2) † kg = kilogram (mass), m = meter (length), s = second (time), A = ampere (electric current) .
Metric (SI) Multipliers Prefix Abbreviation Value quetta Q 1030 ronna R 1027 yotta Y 1024 zeta Z 1021 exa E 1018 peta P 1015 tera T 1012 giga G 109 mega M 106 kilo k 103 hecto h 102 deka da 101 deci d 10-1 centi c 10-2 milli m 10-3 micro m 10-6 nano n 10-9 pico p 10-12 femto f 10-15 atto a 10-18 zepto z 10-21 yocto y 10-24 ronto r 10-27 quecto q 10-30 This page is intentionally left blank DOUGLAS GIANCOLI PHYSICS for SCIENTISTS and ENGINEERS with MODERN PHYSICS 5th EDITION Global Edition Product Management: Shabnam Dohutia, K. K.
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ISBN 10 (Print): 1-292-44027-9 ISBN 13 (Print): 978-1-292-44027-9 ISBN 13 (uPDF eBook): 978-1-292-44035-4 British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library eBook formatted by B2R Technologies Pvt. Ltd. 3 Contents Applications List 12 Preface 16 To Students 20 Use of Color 21 1 Introduction, Measurement, Estimating 23 1 – 1 How Science Works 24 1 – 2 Models, Theories, and Laws 25 1 – 3 Measurement and Uncertainty; Significant Figures 25 1 – 4 Units, Standards, and the SI System 28 1 – 5 Converting Units 31 1 – 6 Order of Magnitude: Rapid Estimating 33 1 – 7 Dimensions and Dimensional Analysis 36 Questions, MisConceptions, Problems 37 – 41 2 DescriBing Motion: Kinematics in One Dimension 42 2 – 1 Reference Frames and Displacement 43 2 – 2 Average Velocity 44 2 – 3 Instantaneous Velocity 46 2 – 4 Acceleration 49 2 – 5 Motion at Constant Acceleration 52 2 – 6 Solving Problems 55 2 – 7 Freely Falling Objects 59 2 – 8 Variable Acceleration; Integral Calculus 65 Questions, MisConceptions, Problems 67 – 75 3 Kinematics in Two or Three Dimensions; Vectors 76 3 – 1 Vectors and Scalars 77 3 – 2 Addition of Vectors—Graphical Methods 77 3 – 3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar 79 3 – 4 Adding Vectors by Components 80 3 – 5 Unit Vectors 84 3 – 6 Vector Kinematics 84 3 – 7 Projectile Motion 87 3 – 8 Solving Problems Involving Projectile Motion 89 3 – 9 Relative Velocity 95 Questions, MisConceptions, Problems 98 – 106 4 Dynamics: Newton’s Laws of Motion 107 4 – 1 Force 108 4 – 2 Newton’s First Law of Motion 108 4 – 3 Mass 110 4 – 4 Newton’s Second Law of Motion 110 4 – 5 Newton’s Third Law of Motion 113 4 – 6 Weight—the Force of Gravity; and the Normal Force 116 4 – 7 Solving Problems with Newton’s Laws: Free-Body Diagrams 119 4 – 8 Problem Solving—A General Approach 126 Questions, MisConceptions, Problems 127 – 137 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces 138 5 – 1 Using Newton’s Laws with Friction 139 5 – 2 Uniform Circular Motion—Kinematics 145 5 – 3 Dynamics of Uniform Circular Motion 148 5 – 4 Highway Curves: Banked and Unbanked 152 5 – 5 Nonuniform Circular Motion 155 5 – 6 Velocity-Dependent Forces: Drag and Terminal Velocity 156 Questions, MisConceptions, Problems 158 – 166 Estimate how much this lake weighs.
4 CONTENTS Gravitation and Newton’s Synthesis 167 6 – 1 Newton’s Law of Universal Gravitation 168 6 – 2 Vector Form of Newton’s Law of Universal Gravitation 171 6 – 3 Gravity Near the Earth’s Surface 171 6 – 4 Satellites and “Weightlessness” 174 6 – 5 Planets, Kepler’s Laws, and Newton’s Synthesis 177 6 – 6 Moon Rises an Hour Later Each Day 183 6 – 7 Types of Forces in Nature 183 6 – 8 Gravitational Field 184 6 – 9 Principle of Equivalence; Curvature of Space; Black Holes 185 Questions, MisConceptions, Problems 187 – 193 7 WorK and Energy 194 7 – 1 Work Done by a Constant Force 195 7 – 2 Scalar Product of Two Vectors 198 7 – 3 Work Done by a Varying Force 199 7 – 4 Kinetic Energy and the Work-Energy Principle 203 Questions, MisConceptions, Problems 208 – 215 8 Conservation of Energy 2 16 8 – 1 Conservative and Nonconservative Forces 217 8 – 2 Potential Energy 219 8 – 3 Mechanical Energy and Its Conservation 222 8 – 4 Problem Solving Using Conservation of Mechanical Energy 223 8 – 5 The Law of Conservation of Energy 229 8 – 6 Energy Conservation with Dissipative Forces: Solving Problems 230 8 – 7 Gravitational Potential Energy and Escape Velocity 232 8 – 8 Power 235 8 – 9 Potential Energy Diagrams; Stable and Unstable Equilibrium 237 8 – 10 Gravitational Assist (Gravitational Slingshot) 238 Questions, MisConceptions, Problems 240 – 248 Linear Momentum 249 9 – 1 Momentum and Its Relation to Force 250 9 – 2 Conservation of Momentum 252 9 – 3 Collisions and Impulse 256 9 – 4 Conservation of Energy and Momentum in Collisions 257 9 – 5 Elastic Collisions in One Dimension 258 9 – 6 Inelastic Collisions 261 9 – 7 Collisions in 2 or 3 Dimensions 263 9 – 8 Center of Mass (CM) 266 9 – 9 Center of Mass and Translational Motion 270 9 – 10 Systems of Variable Mass; Rocket Propulsion 273 Questions, MisConceptions, Problems 276 – 285 10 Rotational Motion 286 10 – 1 Angular Quantities 287 10 – 2 Vector Nature of Angular Quantities 292 10 – 3 Constant Angular Acceleration 292 10 – 4 Torque 293 10 – 5 Rotational Dynamics; Torque and Rotational Inertia 296 10 – 6 Solving Problems in Rotational Dynamics 298 10 – 7 Determining Moments of Inertia 301 10 – 8 Rotational Kinetic Energy 303 10 – 9 Rotational Plus Translational Motion; Rolling 305 10 – 10 Why Does a Rolling Sphere Slow Down? 311 Questions, MisConceptions, Problems 313 – 323 11 Angular Momentum; General Rotation 324 11 – 1 Angular Momentum : Objects Rotating About a Fixed Axis 325 11 – 2 Vector Cross Product; Torque as a Vector 329 11 – 3 Angular Momentum of a Particle 331 11 – 4 Angular Momentum and Torque for a System of Particles; General Motion 332 11 – 5 Angular Momentum and Torque for a Rigid Object 334 11 – 6 Conservation of Angular Momentum 337 11 – 7 The Spinning Top and Gyroscope 339 11 – 8 Rotating Frames of Reference; Inertial Forces 340 11 – 9 The Coriolis Effect 341 Questions, MisConceptions, Problems 344 – 352 CONTENTS 5 12 Static EquiliBrium; Elasticity and Fracture 353 12 – 1 The Conditions for Equilibrium 354 12 – 2 Solving Statics Problems 356 12 – 3 Applications to Muscles and Joints 361 12 – 4 Stability and Balance 363 12 – 5 Elasticity; Stress and Strain 364 12 – 6 Fracture 367 12 – 7 Trusses and Bridges 369 12 – 8 Arches and Domes 372 Questions, MisConceptions, Problems 375 – 386 14 Oscillations 42 1 14 – 1 Oscillations of a Spring 422 14 – 2 Simple Harmonic Motion 424 14 – 3 Energy in the Simple Harmonic Oscillator 430 14 – 4 Simple Harmonic Motion Related to Uniform Circular Motion 432 14 – 5 The Simple Pendulum 433 14 – 6 The Physical Pendulum and the Torsion Pendulum 434 14 – 7 Damped Harmonic Motion 436 14 – 8 Forced Oscillations; Resonance 439 Questions, MisConceptions, Problems 442 – 449 15 Wave Motion 450 15 – 1 Characteristics of Wave Motion 451 15 – 2 Types of Waves: Transverse and Longitudinal 453 15 – 3 Energy Transported by Waves 457 15 – 4 Mathematical Representation of a Traveling Wave 459 15 – 5 The Wave Equation 462 15 – 6 The Principle of Superposition 463 15 – 7 Reflection and Transmission 465 15 – 8 Interference 466 15 – 9 Standing Waves; Resonance 468 15 – 10 Refraction 471 15 – 11 Diffraction 472 Questions, MisConceptions, Problems 474 – 481 1 Sound 482 16 – 1 Characteristics of Sound 483 16 – 2 Mathematical Representation of Longitudinal Waves 484 16 – 3 Intensity of Sound: Decibels 486 16 – 4 Sources of Sound: Vibrating Strings and Air Columns 489 16 – 5 Quality of Sound, and Noise; Superposition 494 16 – 6 Interference of Sound Waves; Beats 495 16 – 7 Doppler Effect 498 16 – 8 Shock Waves and the Sonic Boom 502 16 – 9 Applications: Sonar, Ultrasound, and Medical Imaging 503 Questions, MisConceptions, Problems 506 – 513 B2 F 5 m2 g 5 B1 F 5 m1g 5 13 Fluids 387 13 – 1 Phases of Matter 388 13 – 2 Density and Specif ic Gravity 388 13 – 3 Pressure in Fluids 389 13 – 4 Atmospheric Pressure and Gauge Pressure 393 13 – 5 Pascal’s Principle 393 13 – 6 Measurement of Pressure; Gauges and the Barometer 394 13 – 7 Buoyancy; Archimedes’ Principle 396 13 – 8 Fluids in Motion; Flow Rate and the Equation of Continuity 400 13 – 9 Bernoulli’s Equation 402 13 – 10 Applications of Bernoulli’s Principle: Torricelli, Airplanes, Baseballs, Blood Flow 404 13 – 11 Viscosity 407 13 – 12 Flow in Tubes: Poiseuille’s Equation, Blood Flow 407 13 – 13 Surface Tension and Capillarity 408 13 – 14 Pumps, and the Heart 410 Questions, MisConceptions, Problems 412 – 420 6 CONTENTS 17 Temperature, Thermal Expansion, and the Ideal Gas Law 5 14 17 – 1 Atomic Theory of Matter 515 17 – 2 Temperature and Thermometers 517 17 – 3 Thermal Equilibrium and the Zeroth Law of Thermodynamics 519 17 – 4 Thermal Expansion 519 17 – 5 Thermal Stresses 523 17 – 6 The Gas Laws and Absolute Temperature 524 17 – 7 The Ideal Gas Law 525 17 – 8 Problem Solving with the Ideal Gas Law 526 17 – 9 Ideal Gas Law in Terms of Molecules: Avogadro’s Number 528 17 – 10 Ideal Gas Temperature Scale— a Standard 529 Questions, MisConceptions, Problems 531 – 537 18 Kinetic Theory of Gases 538 18 – 1 The Ideal Gas Law and the Molecular Interpretation of Temperature 538 18 – 2 Distribution of Molecular Speeds 542 18 – 3 Real Gases and Changes of Phase 544 18 – 4 Vapor Pressure and Humidity 546 18 – 5 Temperature Decrease of Boiling Water with Altitude 548 18 – 6 Van der Waals Equation of State 549 18 – 7 Mean Free Path 550 18 – 8 Diffusion 552 Questions, MisConceptions, Problems 554 – 559 1 Heat and the First Law of Thermodynamics 560 19 – 1 Heat as Energy Transfer 561 19 – 2 Internal Energy 562 19 – 3 Specific Heat 563 19 – 4 Calorimetry—Solving Problems 564 19 – 5 Latent Heat 567 19 – 6 The First Law of Thermodynamics 571 19 – 7 Thermodynamic Processes and the First Law 573 19 – 8 Molar Specific Heats for Gases, and the Equipartition of Energy 578 19 – 9 Adiabatic Expansion of a Gas 581 19 – 10 Heat Transfer: Conduction, Convection, Radiation 582 Questions, MisConceptions, Problems 590 – 597 20 Second Law of Thermodynamics 598 20 – 1 The Second Law of Thermodynamics—Introduction 599 20 – 2 Heat Engines 600 20 – 3 The Carnot Engine; Reversible and Irreversible Processes 602 20 – 4 Refrigerators, Air Conditioners, and Heat Pumps 606 20 – 5 Entropy 609 20 – 6 Entropy and the Second Law of Thermodynamics 612 20 – 7 Order to Disorder 615 20 – 8 Unavailability of Energy; Heat Death 616 20 – 9 Statistical Interpretation of Entropy and the Second Law 617 20 – 10 Thermodynamic Temperature; Third Law of Thermodynamics 619 20 – 11 Thermal Pollution, Global Warming, and Energy Resources 620 Questions, MisConceptions, Problems 623 – 630 CONTENTS 7 24 Capacitance, Dielectrics, Electric Energy Storage 7 14 24 – 1 Capacitors 714 24 – 2 Determination of Capacitance 716 24 – 3 Capacitors in Series and Parallel 720 24 – 4 Storage of Electric Energy 722 24 – 5 Dielectrics 725 24 – 6 Molecular Description of Dielectrics 728 Questions, MisConceptions, Problems 730 – 738 21 Electric Charge and Electric Field 63 1 21 – 1 Static Electricity; Electric Charge and Its Conservation 632 21 – 2 Electric Charge in the Atom 633 21 – 3 Insulators and Conductors 633 21 – 4 Induced Charge; the Electroscope 634 21 – 5 Coulomb’s Law 635 21 – 6 The Electric Field 640 21 – 7 Electric Field Calculations for Continuous Charge Distributions 644 21 – 8 Field Lines 648 21 – 9 Electric Fields and Conductors 649 21 – 10 Motion of a Charged Particle in an Electric Field 650 21 – 11 Electric Dipoles 651 21 – 12 Electric Forces in Molecular Biology: DNA Structure and Replication 653 Questions, MisConceptions, Problems 656 – 664 22 Gauss’s Law 665 22 – 1 Electric Flux 666 22 – 2 Gauss’s Law 667 22 – 3 Applications of Gauss’s Law 669 22 – 4 Experimental Basis of Gauss’s and Coulomb’s Laws 674 Questions, MisConceptions, Problems 675 – 681 23 Electric Potential 682 23 – 1 Electric Potential Energy and Potential Difference 683 23 – 2 Relation between Electric Potential and Electric Field 686 23 – 3 Electric Potential Due to Point Charges 688 23 – 4 Potential Due to Any Charge Distribution 691 23 – 5 Equipotential Lines and Surfaces 692 23 – 6 Potential Due to Electric Dipole; Dipole Moment 693 23 – 7 E 5 Determined from V 694 23 – 8 Electrostatic Potential Energy; the Electron Volt 696 23 – 9 Digital; Binary Numbers; Signal Voltage 698 23 – 10 TV and Computer Monitors 701 23 – 11 Electrocardiogram (ECG or EKG) 704 Questions, MisConceptions, Problems 706 – 713 25 Electric Current and Resistance 739 25 – 1 The Electric Battery 740 25 – 2 Electric Current 742 25 – 3 Ohm’s Law: Resistance and Resistors 744 25 – 4 Resistivity 746 25 – 5 Electric Power 748 25 – 6 Power in Household Circuits 751 25 – 7 Alternating Current 752 25 – 8 Microscopic View of Electric Current 754 25 – 9 Superconductivity 757 25 – 10 Electrical Conduction in the Human Nervous System 758 Questions, MisConceptions, Problems 761 – 768 2 DC Circuits 769 26 – 1 EMF and Terminal Voltage 770 26 – 2 Resistors in Series and in Parallel 771 26 – 3 Kirchhoff’s Rules 776 26 – 4 EMFs in Series and in Parallel; Charging a Battery 779 26 – 5 RC Circuits—Resistor and Capacitor in Series 781 26 – 6 Electric Hazards and Safety 786 26 – 7 Ammeters and Voltmeters—Measurement Affects Quantity Measured 789 Questions, MisConceptions, Problems 793 – 803 8 CONTENTS 2 Electromagnetic Induction and Faraday’s Law 860 29 – 1 Induced EMF 861 29 – 2 Faraday’s Law of Induction; Lenz’s Law 862 29 – 3 EMF Induced in a Moving Conductor 867 29 – 4 Electric Generators 868 29 – 5 Back EMF and Counter Torque; Eddy Currents 870 29 – 6 Transformers and Transmission of Power 873 29 – 7 A Changing Magnetic Flux Produces an Electric Field 876 29 – 8 Information Storage: Magnetic and Semiconductor 878 29 – 9 Applications of Induction: Microphone, Seismograph, GFCI 880 Questions, MisConceptions, Problems 882 – 890 30 Inductance, Electromagnetic Oscillations, and AC Circuits 89 1 30 – 1 Mutual Inductance 892 30 – 2 Self-Inductance; Inductors 894 30 – 3 Energy Stored in a Magnetic Field 896 30 – 4 LR Circuits 897 30 – 5 LC Circuits and Electromagnetic Oscillations 899 30 – 6 LC Oscillations with Resistance (LRC Circuit) 902 30 – 7 AC Circuits and Reactance 903 30 – 8 LRC Series AC Circuit; Phasor Diagrams 907 30 – 9 Resonance in AC Circuits 909 30 – 10 Impedance Matching 910 30 – 11 Three-Phase AC 911 Questions, MisConceptions, Problems 912 – 919 31 Maxwell’s Equations and Electromagnetic Waves 920 31 – 1 Changing Electric Fields Produce Magnetic Fields; Displacement Current 921 31 – 2 Gauss’s Law for Magnetism 924 31 – 3 Maxwell’s Equations 925 31 – 4 Production of Electromagnetic Waves 925 31 – 5 Electromagnetic Waves, and Their Speed, Derived from Maxwell’s Equations 927 31 – 6 Light as an Electromagnetic Wave and the Electromagnetic Spectrum 931 31 – 7 Measuring the Speed of Light 934 31 – 8 Energy in EM Waves; the Poynting Vector 935 31 – 9 Radiation Pressure 937 31 – 10 Radio and Television; Wireless Communication 939 Questions, MisConceptions, Problems 943 – 947 27 Magnetism 804 27 – 1 Magnets and Magnetic Fields 804 27 – 2 Electric Currents Produce Magnetic Fields 807 27 – 3 Force on an Electric Current in a Magnetic Field; Definition of B 5 808 27 – 4 Force on an Electric Charge Moving in a Magnetic Field 810 27 – 5 Torque on a Current Loop; Magnetic Dipole Moment 815 27 – 6 Applications: Motors, Loudspeakers, Galvanometers 817 27 – 7 Discovery and Properties of the Electron 819 27 – 8 The Hall Effect 821 27 – 9 Mass Spectrometer 822 Questions, MisConceptions, Problems 824 – 832 28 Sources of Magnetic Field 833 28 – 1 Magnetic Field Due to a Straight Wire 834 28 – 2 Force between Two Parallel Wires 835 28 – 3 Definitions of the Ampere and the Coulomb 836 28 – 4 Ampère’s Law 837 28 – 5 Magnetic Field of a Solenoid and a Toroid 841 28 – 6 Biot-Savart Law 843 28 – 7 Magnetic Field Due to a Single Moving Charge 846 28 – 8 Magnetic Materials—Ferromagnetism 846 28 – 9 Electromagnets and Solenoids—Applications 848 28 – 10 Magnetic Fields in Magnetic Materials; Hysteresis 849 28 – 11 Paramagnetism and Diamagnetism 850 Questions, MisConceptions, Problems 852 – 859 CONTENTS 9 34 The Wave Nature of Light: Interference and Polarization 1017 34 – 1 Waves vs. Particles; Huygens’ Principle and Diffraction 1018 34 – 2 Huygens’ Principle and the Law of Refraction; Mirages 1019 34 – 3 Interference—Young’s Double-Slit Experiment 1020 34 – 4 Intensity in the Double-Slit Interference Pattern 1024 34 – 5 Interference in Thin Films 1026 34 – 6 Michelson Interferometer 1032 34 – 7 Polarization 1032 34 – 8 Liquid Crystal Displays (LCD) 1036 34 – 9 Scattering of Light by the Atmosphere 1037 34 – 10 Brightness: Lumens and Luminous Intensity 1038 34 – 11 Efficiency of Lightbulbs 1038 Questions, MisConceptions, Problems 1040 – 1046 32 Light: Reflection and Refraction 948 32 – 1 The Ray Model of Light 949 32 – 2 Reflection; Image Formation by a Plane Mirror 949 32 – 3 Formation of Images by Spherical Mirrors 953 32 – 4 Seeing Yourself in a Magnifying Mirror (Concave) 958 32 – 5 Convex (Rearview) Mirrors 960 32 – 6 Index of Refraction 961 32 – 7 Refraction: Snell’s Law 961 32 – 8 The Visible Spectrum and Dispersion 963 32 – 9 Total Internal Reflection; Fiber Optics 965 32 – 10 Refraction at a Spherical Surface 968 Questions, MisConceptions, Problems 971 – 979 33 Lenses and Optical Instruments 980 33 – 1 Thin Lenses; Ray Tracing and Focal Length 981 33 – 2 The Thin Lens Equation 984 33 – 3 Combinations of Lenses 988 33 – 4 Lensmaker’s Equation 990 33 – 5 Cameras: Film and Digital 992 33 – 6 The Human Eye; Corrective Lenses 997 33 – 7 Magnifying Glass 1001 33 – 8 Telescopes 1002 33 – 9 Compound Microscope 1005 33 – 10 Aberrations of Lenses and Mirrors 1006 Questions, MisConceptions, Problems 1008 – 1016 35 Diffraction 1047 35 – 1 Diffraction by a Single Slit or Disk 1048 35 – 2 Intensity in Single-Slit Diffraction Pattern 1050 35 – 3 Diffraction in the Double-Slit Experiment 1053 35 – 4 Interference vs. Diffraction 1055 35 – 5 Limits of Resolution; Circular Apertures 1055 35 – 6 Resolution of Telescopes and Microscopes; the l Limit 1057 35 – 7 Resolution of the Human Eye and Useful Magnification 1059 35 – 8 Diffraction Grating 1059 35 – 9 The Spectrometer and Spectroscopy 1062 35 – 10 Peak Widths and Resolving Power for a Diffraction Grating 1063 35 – 11 X-Rays and X-Ray Diffraction 1065 35 – 12 X-Ray Imaging and Computed Tomography (CT Scan) 1067 35 – 13 Specialty Microscopes and Contrast 1070 Questions, MisConceptions, Problems 1071 – 1076 10 CONTENTS 39 Quantum Mechanics of Atoms 1 180 39 – 1 Quantum-Mechanical View of Atoms 1181 39 – 2 Hydrogen Atom: Schrödinger Equation and Quantum Numbers 1181 39 – 3 Hydrogen Atom Wave Functions 1185 39 – 4 Multielectron Atoms; the Exclusion Principle 1188 39 – 5 Periodic Table of Elements 1189 39 – 6 X-Ray Spectra and Atomic Number 1191 39 – 7 Magnetic Dipole Moment; Electron Spin 1193 39 – 8 Total Angular Momentum J 5 1195 39 – 9 Fluorescence and Phosphorescence 1196 39 – 10 Lasers 1197 39 – 11 Holography 1200 Questions, MisConceptions, Problems 1202 – 1207 3 The Special Theory of Relativity 1077 36 – 1 Galilean – Newtonian Relativity 1078 36 – 2 The Michelson – Morley Experiment 1080 36 – 3 Postulates of the Special Theory of Relativity 1083 36 – 4 Simultaneity 1084 36 – 5 Time Dilation and the Twin Paradox 1086 36 – 6 Length Contraction 1092 36 – 7 Four-Dimensional Space – Time 1094 36 – 8 Galilean and Lorentz Transformations 1094 36 – 9 Relativistic Momentum 1099 36 – 10 The Ultimate Speed 1101 36 – 11 E = mc2; Mass and Energy 1102 36 – 12 Doppler Shift for Light 1107 36 – 13 The Impact of Special Relativity 1108 Questions, MisConceptions, Problems 1110 – 1116 37 Early Quantum Theory and Models of the Atom 1 1 17 37 – 1 Blackbody Radiation; Planck’s Quantum Hypothesis 1118 37 – 2 Photon Theory of Light and the Photoelectric Effect 1120 37 – 3 Energy, Mass, and Momentum of a Photon 1123 37 – 4 Compton Effect 1124 37 – 5 Photon Interactions; Pair Production 1126 37 – 6 Wave – Particle Duality; the Principle of Complementarity 1127 37 – 7 Wave Nature of Matter 1128 37 – 8 Electron Microscopes 1130 37 – 9 Early Models of the Atom 1132 37 – 10 Atomic Spectra: Key to the Structure of the Atom 1133 37 – 11 The Bohr Model 1135 37 – 12 de Broglie’s Hypothesis Applied to Atoms 1142 Questions, MisConceptions, Problems 1143 – 1149 38 Quantum Mechanics 1 150 38 – 1 Quantum Mechanics—A New Theory 1151 38 – 2 The Wave Function and Its Interpretation; the Double-Slit Experiment 1151 38 – 3 The Uncertainty Principle 1153 38 – 4 Philosophic Implications; Probability versus Determinism 1157 38 – 5 The Schrödinger Equation in One Dimension—Time-Independent Form 1158 38 – 6 Time-Dependent Schrödinger Equation 1161 38 – 7 Free Particles; Plane Waves and Wave Packets 1162 38 – 8 Particle in an Infinitely Deep Square Well Potential (a Rigid Box) 1164 38 – 9 Finite Potential Well 1169 38 – 10 Tunneling through a Barrier 1171 Questions, MisConceptions, Problems 1174 – 1179 t-RNA Anticodons G C A A U G New protein chain of 4 amino acids (a 5th is being added) Ribosome 1 2 3 4 5 m-RNA Codon 4 Codon 5 C G U U A C C G U U A C A C A C C T G C A A T G T G A Growing end of m-RNA DNA Codon 5 Codon 4 40 Molecules and Solids 1208 40 – 1 Bonding in Molecules 1209 40 – 2 Potential-Energy Diagrams for Molecules 1211 40 – 3 Weak (van der Waals) Bonds 1214 40 – 4 Protein Synthesis 1216 40 – 5 Molecular Spectra 1218 40 – 6 Condensed-Matter Physics; Bonding in Solids 1224 40 – 7 Free-Electron Theory of Metals; Fermi Energy 1225 40 – 8 Band Theory of Solids 1230 40 – 9 Semiconductors and Doping 1232 40 – 10 Semiconductor Diodes, Photovoltaics, LEDs, OLEDs 1234 40 – 11 Transistors: Bipolar and MOSFETs 1240 40 – 12 Integrated Circuits, Chips, 3-nm T echnology 1241 Questions, MisConceptions, Problems 1242 – 1247 CONTENTS 11 43 Elementary Particles 13 1 1 43 – 1 High-Energy Particles and Accelerators 1312 43 – 2 Beginnings of Elementary Particle Physics—Particle Exchange 1318 43 – 3 Particles and Antiparticles 1321 43 – 4 Particle Interactions and Conservation Laws 1322 43 – 5 Neutrinos 1324 43 – 6 Particle Classification 1326 43 – 7 Particle Stability and Resonances 1328 43 – 8 Strangeness? Charm? Towards a New Model 1329 43 – 9 Quarks 1330 43 – 10 The Standard Model: QCD and Electroweak Theory 1333 43 – 11 Grand Unified Theories 1336 43 – 12 Strings and Supersymmetry 1339 Questions, MisConceptions, Problems 1340 – 1343 44 Astrophysics and Cosmology 1344 44 – 1 Stars and Galaxies 1345 44 – 2 Stellar Evolution: Birth and Death of Stars, Nucleosynthesis 1348 44 – 3 Distance Measurements 1354 44 – 4 General Relativity: Gravity and the Curvature of Space 1356 44 – 5 The Expanding Universe: Redshift and Hubble’s Law 1360 44 – 6 The Big Bang and the Cosmic Microwave Background 1364 44 – 7 The Standard Cosmological Model: Early History of the Universe 1367 44 – 8 Inflation: Explaining Flatness, Uniformity, and Structure 1370 44 – 9 Dark Matter and Dark Energy 1372 44 – 10 Large-Scale Structure of the Universe 1375 44 – 11 Gravitational Waves : LIGO and Virgo 1376 44 – 12 Finally . . . 1376 Questions, MisConceptions, Problems 1378 – 1382 Appendices A Mathematical Formulas A–1 B Derivatives and Integrals A–6 C Numerical Integration A–8 D More on Dimensional Analysis A–12 E Gravitational Force Due to a Spherical Mass Distribution A–13 F Differential Form of Maxwell’s Equations A–16 G Selected Isotopes A–18 Answers to Odd-Numbered Problems A–23 Index A–51 Photo Credits A–81 41 Nuclear Physics and Radioactivity 1248 41 – 1 Structure and Properties of the Nucleus 1249 41 – 2 Binding Energy and Nuclear Forces 1252 41 – 3 Radioactivity 1255 41 – 4 Alpha Decay 1256 41 – 5 Beta Decay 1259 41 – 6 Gamma Decay 1261 41 – 7 Conservation of Nucleon Number and Other Conservation Laws 1262 41 – 8 Half-Life and Rate of Decay 1262 41 – 9 Decay Series 1267 41 – 10 Radioactive Dating 1268 41 – 11 Detection of Particles 1270 Questions, MisConceptions, Problems 1272 – 1277 42 Nuclear Energy; Effects and Uses of Radiation 1278 42 – 1 Nuclear Reactions and the Transmutation of Elements 1279 42 – 2 Cross Section 1282 42 – 3 Nuclear Fission; Nuclear Reactors 1283 42 – 4 Nuclear Fusion 1288 42 – 5 Passage of Radiation through Matter; Biological Damage 1293 42 – 6 Measurement of Radiation—Dosimetry 1294 42 – 7 Radiation Therapy 1298 42 – 8 Tracers in Research and Medicine 1299 42 – 9 Emission Tomography: PET and SPECT 1300 42 – 10 Nuclear Magnetic Resonance (NMR) 1301 42 – 11 Magnetic Resonance Imaging (MRI) 1303 Questions, MisConceptions, Problems 1305 – 1310 12 APPLICATIONS Chapter 1 Viruses attack cell 29 Heartbeats in a lifetime 34 Number of nucleons in human body 39 Lung capacity 41 Building collapse 24, 354, 368–9 The 8000-m peaks 31 Making estimates: volume of a lake 33 Page thickness 34 Building height by triangulation 34 Earth radius estimate 35, 40 Fermi estimates 35 Particulate pollution 40 Global positioning satellite 40 Computer chips 40 Chapter 2 Airport runway design 54 Car air bag inflation time 54, 276 Car braking distance 57 , 205 CD bit size, bit rate, playing time 70, 75 Baseball 71, 104, 105, 106, 194 Basketball 72, 105, 131 Golf putt, uphill or down 74 Rapid transit system 75 Chapter 3 Helicopter supply drop 76, 94, 105 Sports 76, 87 , 91, 93, 98, 99, 101, 102, 103, 104, 405, 106 Kicked football 91, 93 Truck escape lane 101, 132 Golf on the Moon 104 Extreme sports 105 Chapter 4 How we can walk 114 Whiplash 128 Force heart exerts 129 Rocket 107 , 114, 130, 255, 274, 417 Skater pushoff 113 What force accelerates a car 114 You weigh less in a falling elevator 118 Hockey 120 Elevator, discomfort 123, 130 Mechanical advantage, pulley 124, 210 Accelerometer 124 Sports 128, 129, 130, 131, 132, 134 Bear sling 128, 377 Tug of war 128, 277 Car accident “g’s” 129 Optical tweezers 130, 938, 945 Tightrope walker 131 Basketball shot 131 Mountain climbers 132, 136, 137 , 215, 392 City planning, cars on hill 134 Bicycling 134, 136 Supermarket ramp design 135 Doomsday asteroid 136, 284 Car stuck in mud 137 Chapter 5 Centrifugation 148 Skiing 138, 143, 158 Push or pull a sled? 142 Skier speed in air vs. on snow 143 Simulating gravity 148, 158, 163, 190, 193 Uranium enrichment, reactor, bomb 148 Ferris wheel 151 Avoid skidding on a curve 152–4 Banked highway curves 154 Cross-country skiing friction 158 Rotating space station 158, 163, 190 Rotor ride 159, 165 Airplane bank /turn 159, 166 Roller coaster upside down 163 Car flying up off road 163 Rock climbing friction 165 Chapter 6 Weightlessness 176–7 Astronauts in orbit 167 , 177 , 187 Gravity on tall peaks 172 Oil and mineral exploration 172, 187 , 189 Satellites, spacecraft 167 , 174–7 , 190, 191 Geostationary satellites 175 Free fall, for athletes 177 Planets 177–80, 189 Determining the Sun’s mass 180 Planets around other stars 180, 272, 284 Ocean tides 181, 187 , 192 Lagrange point 182 Moon’s orbit, periods, phases, diagram 183, 191 Eclipses 183 Curved space 185–6 Black holes 186, 189 White dwarfs 189 Comets, asteroids, moons 190, 191, 193 GPS 191 Milky Way Galaxy 193 Chapter 7 Baseball pitch 194 Car stopping distance r v 2 205 Lever 209, 356 Pulley 210 Jet catapults 211 Bicycle, sprockets (teeth) 214, 221 Climbing rope stretch 215 Chapter 8 Stair-climbing power 235 ATP 238 Hike over logs 240 Pole vault 216, 225–6, (211) Downhill ski runs 216 Roller coaster 220, 224, 221 Escape velocity from Earth or Moon 234 Power needs of car 236 Efficiency of engine 237 Gravitational assist 238–9, 246, 285 High jump 242 Bungee jump 243 Lunar module landing 244 Escape velocity from solar system 245 Ski jump 247 Long jump 247 Chapter 9 Impulse in fall: break a leg? 279 Billiard balls 249, 252, 259, 264 Tennis serve 251, 256 Rocket propulsion 255, 274, 417 Rifle recoil 255 Karate blow 257 Nuclear reactors 260 Nuclear collisions 260, 261, 263, 265 Ballistic pendulum, speed measured 262 Distant planet discovery 272, 284 Conveyor belt 275 Car crashworthiness 283 Asteroid danger 284 Force wind exerts 285 Bowling 285 Chapter 10 Acuity of bird’s eye 288 Centrifuge 293 Biceps, triceps, torque 295, 317 , 361 Situps 313 Fast mammal 313 Rotating carnival rides 286, 289, 290 Tire iron extension 294 Flywheel, energy 304, 323 Yo-yo 309 Braking forces on a car 310–11 Bicycle odometer 313 Tightrope walking 313 Total solar eclipse 315 Wrench torque 316 Hammer throw 318 CD rotation frequency 320 Bicycle gears 321 Cue stick, ball roll 322 Bicycle turn angle 323 Chapter 11 Rotating skaters /divers 324, 326, 352 Neutron star collapse 327 , 352 Strange spinning bike wheel 329, 336 Applications (Selected) to Medicine and Biology and to Engineering, Environment, Everyday Life, Etc. (Entries with a star include material new to this edition) APPLICATIONS 13 Automobile wheel balancing 336–7 Precessing top 339–40 Gyroscope 340 Hurricanes, cyclones, typhoons 343, 416 Anticyclonic weather 343 Precession of the equinoxes 349 SUV rollover 350 Baseball bat sweet spot 352 Chapter 12 Forces in muscles & joints 361, 380, (295) What can make an athlete 361 Forces on the spine and back pain 362 Human balance with loads 364 Bone fracture 368, 381, 386 Buildings, statics 353–74 Lever, mechanical advantage 356 Balancing a seesaw 357 Cantilever 358 Fracture 367–9 Tragic collapse 368–9, (354) Trusses and bridges 369–71, 385 Architecture: arches and domes 372–4 Forces in a dome 374 Chapter 13 Pressure in cells 393 Blood flow 402, 406, 408 Human circulatory system 402 Blood loss to brain, TIA 406 Air flow in animal burrow 406 Heart disease, artery clogging 408 Walking on water, insect 409 Heart as a pump 410–11 Blood pressure measurement 411 Blood transfusion 417 , 418 Water supply pressure 391 Atmospheric pressure decrease with elevation 392 Altitude where air pressure is half 392 Finger holds water in straw 393 Hydraulic lift 394 Hydraulic brakes 394 Pressure gauges 394–5 Barometer 395 Suction 396 Hydrometer 399 Continental drift, plate tectonics 400 Lake level change, rock thrown overboard 400, 412 Helium balloon lift 400 Heating duct 402 Hot-water heating system flow 404 Perfume atomizer 405 Airplane wing lift 405 Sailing upwind 405 Baseball curve 406 Why smoke goes up a chimney 406 Soaps and detergents 409 Pumps 410–11, (396) Siphon 412 Hydraulic press 415 Rocket thrust 417 Reynolds number 417 Barrel broken by thin liquid column 419 Chapter 14 Spider web oscillations 427 Human leg as pendulum 446 Shock absorbers 421, 437 Unwanted floor vibrations 428 Loudspeaker 428–9 Pendulum clock 434, 443, 446 Geology 434, 437 Measure g with pendulum 434 Earthquake dampers 437 Child on a swing, resonance 439–40 Resonance damage 440 Q-value 441, 447 , 918 Bungee jumper 444 Metronome 446 Natural stride 446 Tall building sway 448 Chapter 15 Echolocation by bats, dolphins, whales 456 Water waves 450, 457 Sound wave 453, 482 ff Geology 457 , 474, 479 Earthquake waves 457 , 489, 472, 475 Square wave 464 Cell phone signal 473 AM and FM radio wave bending 474 Fish and fisher: internal reflection 478 Seismic reflection: oil prospecting 479 Coffee spill 479 Tsunami 481 Chapter 16 Wide range of human hearing 486 Sensitivity of the ear 489, (488) Bats use Doppler 501 Doppler blood-flow meter 501, 513 Ultrasound medical imaging 504–5 Doppler ultrasound imaging 505 Stringed instruments 482, 490–91 Wind instruments 482, 491–94 Piano strings 482, 490, 491 Distance from lightning, seconds 483 Autofocusing camera 484 Loudspeaker output 487 Musical scale 490 Guitar, violin 490, 491, 506, 509 Organ pipes 493–4 Tuning with beats 497–8 Doppler in weather forecasting 502 Radar speed gun 502 Galaxy redshift 502 Sonic boom; sound barrier 503, 511 Sonar: depth in sea, Earth “soundings” 503–4, 511 Signal-to-noise ratio 508, 512, 701 Quartz oscillator clock 509 Motion sensor 511 Audio gain 512 Chapter 17 Life under ice 522–3 Molecules in one breath 529, 536 Snorkels are short 537 Hot-air balloon 514, 537 Expansion joints 517 , 520, 523 Do holes expand? 521 Opening a tight lid 521 Gas tank overflow 522 Highway buckling 523 Closed jars in fires 525 Mass (weight) of air in a room 527 Cold and hot tire pressure 528 Thermostat 531 Pyrex glass 531 Tape measure inaccuracy 532, 535 Scuba 534, 535, 536, 537 Potato chip bag puff up 535 Chapter 18 KE of molecules in cells 541 Humidity, and comfort 547–8 Chromatography 553 Diffusion in living organisms 553–4, 558 Temperature effect on chemical reactions 543 Evaporation cools 546, 570 Humidity, weather 548 Temperature decrease of boiling water with altitude 548–9 Pressure cooker 557 Chapter 19 Working off Calories 562 Measuring Calorie content 567 , 592 Evaporation and body temperature 570–1, (546) Body heat: convection by blood 585, 596 Body’s radiative heat loss 586 Room comfort: cool air, warm walls 587 Medical thermography 589 Avoid plants freezing 590 Eating snow makes you colder 593 Heat conduction to skin, blood capillaries 595 Leaf’s energy absorption 597 Metabolizing fat 597 Cold tile, warm rugs 583 Heat loss through windows 584 Thermal windows (two panes) 584 How clothing insulates 584 R-values of thermal insulation 584 Ocean currents and wind 585 Convective home heating 585 Dark vs. light clothing 586 Radiation from the Sun, seasons 588 Astronomy—size of a star 588 Goose down loft 590 Thermos bottle 590 Emergency blanket 590 Air parcels, weather, adiabatic lapse rate 595 14 APPLICATIONS Chapter 20 Biological development, evolution 616 Trees offsetting CO2 buildup 630 Steam engine 598, 600, 604, 628 Internal combustion engines 600–2, 605–6 Engine efficiency 604–5 Refrigerators, air conditioners 606–8, 625 Heat pump 608–9, 625 SEER rating 609 Thermal pollution, climate 620–2 Carbon footprint 620 Energy resources 621, 627–8 Solar, thermal, wind energy 621, 627 Diesel engine 629, (597) Stirling cycle 629 Jet engine, Brayton cycle 629 Dehumidifier 630, (559) Chapter 21 Inside a cell: kinetic theory plus electrostatic force 653 DNA structure, replication 653–4, 662 Static electricity 631, 632, 657 , 662 Photocopiers and printers 641 Electrical shielding, safety 650 Chapter 23 Electrocardiogram (ECG) 682, 704–5, 801 Dipoles in molecular biology 694 Heart beat, depolarization process 704–5 Common voltages 10-4 V to 108 V 685 Breakdown voltage 688 Lightning rods 688 Supply voltage, signal voltage 698 Digital, bits, bytes, binary numbers 698 Analog-to-digital converter (ADC) 698 Morse code 698 Bit-rate, TV transmission 698, 700–1, 704 Data compression, jpeg 699–700 Quantization error 699 Sampling rate, bit depth 699 Digital-to-analog converter (DAC) 699, 802 Bandwidth 700 Noise, bit-flips 700–1 Digital error correction, parity bit 700 Bit error rate 701 Signal-to-noise ratio (S/ N) 701 TV and computer monitors 701–4 Digital TV, pixels, subpixels 702 Flat screens, HD 702–3 Addressing pixels 702–3 Data stream 703 Active matrix, TFT, data lines 703–4 TV refresh rate 704 Oscilloscope 704 ASCII code 710 Photocell 711 Chapter 24 Capacitor shocks, burns 725 Heart defibrillator 725, 734, 786 Capacitor use as power backup, surge protector, memory 714, 717 Condenser microphone 717 Computer key 717 Camera flash energy 723 Electrostatic air cleaner 732 Tiny distance measurement 732 Coaxial cable 736, 840, 896, 933 Dynamic random access memory (DRAM) 738, 879 Chapter 25 Electrical conduction in human nervous system, neurons 758–60 Action potential 759 Battery construction, terminals 740–1 Electric cars 742, 766 Battery connections 743, 746 Loudspeaker wire thickness 747 Heating element 748–50 Resistance thermometer 748 Lightning bolt 750, (712, 738) Household circuits, shorts 751–2 Fuses, circuit breakers 751, 788 Safety—wires getting hot 751, 786–8 Extension cord danger 752 Hair dryer 754 Strain gauge 768 Chapter 26 Blood sugar phone app 769 Heart pacemaker 786 Electricity dangers to humans 786–8 Ventricular fibrillation 786 Two-speed fan 774–75 Car battery charging 779 Jump starting a car, safely 780–1 RC: sawtooth, flashers, wipers 785, 802 Hazards, electric safety 786–8 Proper grounding, plugs 787–8 Leakage current 788 Dangerous downed power line 788 Ammeters, voltmeters, ohmmeters 789–91 Meter connection, corrections 790–1, 803 Measurement affects quantity measured 791 Voltage divider 796 Solar panel 800 Potentiometer and bridge circuits 800–1 Car battery corrosion 802 Digital-to-analog converter (DAC) 802, (699) Chapter 27 Electromagnetic blood pump 824 Blood flow rate, Hall effect 829 Use of a compass 806 Magnetic declination 806 Maps and true north 806 Aurora borealis 814 Electric motors, DC and AC 817–8 Loudspeakers and headsets 818 Chapter 28 Coaxial cable 840, 896, 933 Solenoid switches: doorbell, car starter 848 Magnetic circuit breakers 848 Relay (magnetic) 852 Chapter 29 EM blood-flow measurement 867 Induction stove 864 Generators, power plants 868–9 Alternators, in cars 870 Motor overload 871 Eddy-current damping 872, 883 Airport metal detector 872 Transformers, power transmission 873–5 Cell phone charger 874 Car ignition system 874 Wireless electric power transmission 876 Inductive charger 876 Magnetic information storage 878 Semiconductor memory 879–80 RAM, DRAM 879 Bit-line & word-line 879 Writing and reading memory 879 Volatile and nonvolatile memory 880 Flash memory, MOSFET, MRAM 880 Microphone 880 Card reader, magnetic strip 880 Seismograph 881 Ground fault circuit interrupter (GFCI) 881 Shielded cable 883 Recycling solid waste 883 Chapter 30 Electric car inductive charging 891 Surge protection 899 Capacitors as filters 906, 918, 919 Loudspeaker cross-over 906 Impedance matching 910 Three-phase AC 911 Q-value 918, (441, 447) Filter circuit 918 Chapter 31 Optical tweezers 938, 945 TV from the Moon 920, 942, 946 Wireless devices, transmission 920, 939–42 Antennas 933, 941 Phone call time lag 934 Solar sail 938, 947 Radio and TV 939–41 AM and FM 940 Cell phones, remotes, cable TV, satellite TV 942 GPS 946 Solar power use 946 APPLICATIONS 15 Chapter 32 Medical endoscope, bronchoscope, colonoscope 967 How tall a mirror do you need 952 Seeing yourself in a magnifying mirror (concave) 958–9 Convex (rearview mirrors) 960 Optical illusions 962, 1020 Apparent water depth 961–62, 963 Rainbows 964, 979 Colors underwater 965 Diamonds sparkle 966 Prism binoculars 966 Fiber-optic cables 967 , 976, 978 High-frequency trading, interception 967 Solar cooker 973 Washing machine water level detector 978 Road reflectors 979 Chapter 33 Human eye 997–1000 Fovea, denser in cones 998, 1059 Near- and far-sighted 998–1000 Corrective lenses 998–9, 1009 Contact lenses 1000 Seeing underwater 1000 Light microscopes 1005–6, 1070 Where your eye can see a lens image 983 Cameras, film and digital 992–7 CCD, CMOS sensors, potential well 992–3 Bayer pixels, Foveon 993 Digital artifacts 993 Camera adjustments, f-stop 993–5 Depth of field 995 Resolution, compression, JPEG, raw 995–6 Telephoto, wide angle 997 Optical vs. digital zoom 997 Magnifying glass 1001–2 Telescopes 1002–4 Microscopes 1005–6, 1070 Lens aberrations 1006–7 Film projector 1011 Pinhole camera 1012 Chapter 34 Soap bubbles, oil films, colors 1017 , 1026–30 Highway mirages 1020 Lens coatings 1030–1 Polarizing sunglasses 1034–5 Liquid crystal displays, TV and computer screens 1036–7 Sky color 1037–8 Lightbulb efficiency, LED 1038–9 Stealth aircraft coating 1044 CD bits, pits and lands 1046 Chapter 35 Resolution of eye 1057 , 1059 Useful magnification 1059 Spectroscopy in biology 1063 X-ray diffraction in biology 1066 Medical imaging: X-rays, CT 1067–9 Interference microscope 1070 Phase-contrast microscope 1070 Hubble space telescope 1056–7 Telescope and microscope resolution 1057–9 X-rays 1065–9 Tomography 1067–9 Chapter 36 Space travel 1089–90 Global position system (GPS) 1090–91 Fantasy supertrain 1093 Radar speed gun 1114 Chapter 37 Electron microscope image: blood vessel, clot, retina, viruses 1117 , 1131, (29) Photosynthesis 1123 Measuring bone density 1125 Electron microscopes (EM), TEM, SEM 1131, 1173, (1117) Photocells 1120 Photodiodes, soundtracks 1123 Chapter 38 Scanning tunneling electron microscope 1173 Atomic force microscope 1173 Chapter 39 Fluorescence analysis 1196–7 Medical uses of lasers, surgery 1200, 1205 Neon lights 1180 Fluorescent lightbulbs 1197 Lasers 1197–201, 1238 Bar code readers 1199 DVD, CD, Blu-ray 1199–200 Holography 1200–1 Chapter 40 Cell energy—ATP 1214 Weak bonds, DNA 1214–6 Protein synthesis 1216–8 Pulse oximeter 1238 Computer processor chips 1208 Transparent objects 1232 Zener diode voltage regulator 1235–6, 1247 Rectifiers 1236 Photovoltaic cells 1236–7 LED displays, bulbs 1237–8 TV remote 1237 , 1247 Solid-state lighting 1237–8 pn diode laser 1238 OLED, AMOLED displays 1238–9 Amplifiers 1240 MOSFET switch 1240–1 Technology generation 1241 Chapter 41 Earliest life 1270 Radiation film badges 1271, 1296 Smoke detector 1259 Radioactive activity and safety 1265–6 Carbon-14 dating 1268–9 Archeological and geological dating 1268–70 Oldest Earth rocks 1270 Geiger counter 1270 Rubidium–strontium dating 1275 Tritium dating 1276 Mass excess, mass defect 1276 Chapter 42 Biological radiation damage 1293–8 Radiation dosimetry, RBE 1294–8 Radon exposure 1296, 1298 Natural radioactive background 1296 Radiation exposure, film badge 1296 Radiation sickness 1296 Whole-body dose 1297 Radiation therapy 1298–9 Proton therapy 1299 Radioactive tracers 1299–300 Gamma camera 1300 Medical imaging, PET, SPECT, MRI 1300–4 Brain PET scan using cell phone 1301 Imaging resolutions compared 1304 Radiation and thyroid 1308 Nuclear reactors, power plants 1278, 1285–7 , 1291–3 Breeder reactors 1287 Manhattan Project 1288 Nuclear fusion 1288–93 Why stars shine 1289–91 Thermonuclear devices 1291 Fusion energy reactors 1291–3 Chapter 43 Linacs and tumor irradiation 1316 Chapter 44 Stars and galaxies 1345–54 Black holes 1353, 1359–60 Big Bang storyline 1367–70 16 PREFACE Preface New Stuff! 1. MisConceptual Questions, 10 or 15 at the end of each Chapter. The multiple-choice answers include common misconceptions as well as correct responses. Pedagogically, asking students to think, to consider the options, is more effective than just telling them what is valid and what is wrong. (These are in addition to the one at the start of each Chapter). 2. Digital is all around us. Yet that word is not always used carefully. In this new edition we have 20 new pages describing the basics from the ground up. Binary numbers, bits and bytes, are introduced in Chapter 23 along with analog-to-digital conversion (ADC), and vice versa, including digital audio and how video screens work. Also information compression, sampling rate, bit depth, pixel addressing, digital transmission and, in later chapters, information storage (RAM, DRAM, flash), digital cameras and their sensors (CCD, CMOS). 3. Gravitational Assist (Slingshot) to accelerate spacecraft (Chapter 8). 4. Magnetic field of a single moving charge, rarely treated (and if it is, maybe not well), and it shows the need for relativity theory. 5. Seeing yourself in a magnifying mirror (concave), angular magnification and blurriness with a paradox. Also convex (rearview) mirrors (Chapter 32). 6. Pedagogical clarification on defining potential energy, and energy itself (Chapter 8), and on hundreds of other topics. 7. The Moon rises an hour later each day (Chapter 6), its phases, periods, and diagram. 8. Efficiency of lightbulbs (Chapter 34). 9. Idealization vs. reality emphasized—such as PV diagrams (Chapter 19) as an idealized approximation. 10. Many new Problems (' 500) plus new Questions as well as the 500 or so MisConceptual Questions (point 1 above). 11. Many new worked-out Examples. 12. More math steps included in derivations and Examples. 13. New phrases to remind students of our objective in the middle of a long discussion or derivation (it is so easy to lose track). 14. State of a system and state variables clarified (Chapter 17). 15. Contemporary physics: Gravitational waves, LIGO and Virgo, Higgs, WIMPS, OLEDS and other semiconductor physics, nuclear fusion updates, neutrino-less double beta decay. 16. New SI units (Chapters 1, 21, Tables). 17. Boiling temperature of water vs. elevation (Chapter 18). 18. Modern physics in earlier classical Chapters (sometimes in Problems): Light-years, observable universe (Chapter 1); optical tweezers (Chapter 4); uranium enrichment (Chapter 5); black holes and curved space, white dwarfs (Chapter 6); crystal structure (Chapter 7); Yukawa potential, Lennard-Jones potential (Chap-ter 8); neutrons, nuclear reactors, moderator, nuclear collisions, radioactive decay, neutron star collapse (Chapter 9); galaxy redshift (Chapter 16); gas diffusion of uranium (Chapter 18); quarks (Chapter 21); liquid-drop model of nucleus, Gei-ger counter, Van de Graaff (Chapter 23); transistors (Chapters 23, 29); isotopes, cyclotron (Chapter 27); MOSFET (Chapter 29); semiconductor (camera sensor), photon (Chapter 33); line spectra, X-ray crystallography (Chapter 35). 19. Second law of thermodynamics and heat energy reorganized (Chapter 20). 20. Symmetry emphasized throughout. 21. Uranium enrichment, % needed in reactors, bombs (Chapters 5, 42). 22. Mass excess, mass defect (Chapter 41). 23. The mole, more careful definition (Chapter 17). 24. Liquid-gas ambiguity above critical temperature (Chapter 18). 25. Measurement affects quantity measured, new emphasis.
PREFACE 17 26. New clarifications and reminders in longer discussions and derivations. Because students can lose track of what our aim is, it is mentioned again part way through (often replacing “it” or “this”). 27. More New Applications: • Ocean Tides (Chapter 6) • Anticyclonic weather (Chapter 11) • Jump starting a car safely (Chapter 26) • Lightbulb efficiency (Chapter 34) • Specialty microscopes and contrast (Chapter 35) • Forces on Muscles and Joints (Chapter 12) • Doppler ultrasound imaging (Chapter 16) • Lake level change when rock thrown from boat (Chapter 13) • Skier speed on snow vs. flying through the air (Chapter 5) • Inductive charging (Chapter 29) • Human body internal heat transfer is convection (blood) (Chapter 19) • Blood pressure measurement (Chapter 13) • Sports (lots) • Voltage divider (Chapter 26, Problems) • Flat screen TV (Chapters 23, 34, 40) • Carbon footprint and climate (Chapter 20) • Electrocardiogram (Chapter 23) • Wireless from the Moon unimaginable (Chapter 31) • Why snorkels are short (Chapter 17 Problem) • Electric cars (Chapter 25) • Digital (Chapters 23, 29, 33, 40) includes (in addition to details in point 2 above) quantization error, digital error correction, noise, bit error rate, digi-tal TV data stream, refresh rate, active matrix, thin film transistors, digital memory, bit-line, reading and writing of memory cells (MOSFET), floating gate, volatile and nonvolatile memory, Bayer, JPEG, ASCII code, and more.
• Importance of Latin, in footnote on page 1021. Other references on pages 84, 354, 698, 805, 950, 997 , 1069.
Seeing the World through Eyes that Know Physics I was motivated to write a textbook for 2 reasons. First, non-physics students in engi-neering, pre-med, biology, and architecture sometimes asked me “why do I have to take physics.” They were right a new textbook was needed that showed how physics is the basis for so much of their fields and in everyday life. More importantly, I saw that phys-ics textbooks were in the style of an instruction manual. Even as a freshman in college I saw that physics books were not telling the truth about how physics is actually prac-ticed. In this book I start each topic by appealing to the student’s intuition, which is how physics developed and is actually practiced. Instead of beginning formally and dog-matically, I begin each topic with everyday observations and experiences the students can relate to: start with specifics, the real world, and then go to the great generalizations and more formal aspects of the physics, showing why we believe what we believe.
Much effort has gone into approaches for the practical techniques of solving prob-lems: worked-out Examples, Problem Solving sections, and Problem Solving Strategies.
Chapter 1 is not a throwaway. It is fundamental to physics to realize that ev-ery measurement has an uncertainty, and how significant figures are used. Being able to make rapid estimates is a powerful tool useful for every student, and used throughout the book starting in Chapter 1 (you can estimate the Earth’s radius!). Mathematics can be an obstacle to students. I have aimed at including all steps in a derivation. Important mathematical tools, such as addition of vectors and vector product, are incorporated in the text where first needed, so they come with a context rather than in a forbidding introductory Chapter. Appendices contain a basic math re-view, derivatives and integrals, plus some more advanced topics including numerical integration, gravitational field of spherical mass distribution, Maxwell’s equations in differential form, and a Table of selected nuclear isotopes (carefully updated, as are the Periodic Table and the Fundamental Constants found inside the back and front covers).
Versions of this Book Complete version: 44 Chapters including 9 Chapters of modern physics.
Volume 1: Chapters 1–20 on mechanics, including fluids, oscillations, waves, plus heat and thermodynamics.
Volume 2: Chapters 21–35 on electricity and magnetism, plus light and optics.
Volume 3: Chapters 36–44 on modern physics: relativity, quantum theory, atomic physics, condensed matter, nuclear physics, elementary particles, cosmology and astrophysics.
18 PREFACE Edward Adelson, The Ohio State University Lorraine Allen, United States Coast Guard Academy Zaven Altounian, McGill University Leon Amstutz, Taylor University Kim Arvidsson, Schreiner University Philip S. Baringer, Kansas University Bruce Barnett, Johns Hopkins University Michael Barnett, Lawrence Berkeley Lab Anand Batra, Howard University David Branning, Trinity College Bruce Bunker, University of Notre Dame Wayne Carr, Stevens Institute of Technology Charles Chiu, University of Texas Austin Roger N. Clark, U. S. Geological Survey Russell Clark, University of Pittsburgh Robert Coakley, University of Southern Maine David Curott, University of North Alabama Biman Das, SUNY Potsdam Bob Davis, Taylor University Kaushik De, University of Texas Arlington Michael Dennin, University of California Irvine Kathryn Dimiduk, Cornell University John DiNardo, Drexel University Scott Dudley, United States Air Force Academy John Essick, Reed College Cassandra Fesen, Dartmouth College Leonard Finegold, Drexel University Alex Filippenko, University of California Berkeley Richard Firestone, Lawrence Berkeley Lab Tom Furtak, Colorado School of Mines Gill Gabelmann, Washburn University Gabriel Orebi Gann, University of California Berkeley Edward Gibson, California State University Sacramento John Hamilton, University of Hawai’i – Hilo John Hardy, Texas A&M J. Erik Hendrickson, University of Wisconsin-Eau Claire Charles Hibbard, Lowell High School Dr. Laurent Hodges, Iowa State University David Hogg, New York University Mark Hollabaugh, Normandale Community College Russell Holmes, University of Minnesota Twin Cities William Holzapfel, University of California Berkeley Bob Jacobsen, University of California Berkeley Arthur W. John, Northeastern University David Jones, Florida International University Andrew N. Jordan, University of Rochester Teruki Kamon, Texas A&M Thomas Hemmick, State University of New York Stonybrook Daryao Khatri, University of the District of Columbia Woo-Joong Kim, Seattle University John Kinard, Greenwood High School Jay Kunze, Idaho State University Jim LaBelle, Dartmouth College Andrei Linde, Stanford University M.A.K. Lodhi, Texas Tech Lisa Madewell, University of Wisconsin Ponn Maheswaranathan, Winthrop University Bruce Mason, University of Oklahoma Mark Mattson, James Madison University Linda McDonald, North Park College Raj Mohanty, Boston University Giuseppe Molesini, Isituto Nazionale di ottica Florence Lisa K. Morris, Washington State University Richard Muller, University of California Berkeley Blaine Norum, University of Virginia Lauren Movatne, Reedley College Alexandria Oakes, Eastern Michigan University Ralph Oberly, Marshall University Michael Ottinger, San Juan College Lyman Page, Princeton Laurence Palmer, University of Maryland Bruce Partridge, Haverford College R. Daryl Pedigo, University of Washington Robert Pelcovitz, Brown University Saul Perlmutter, University of California Berkeley Vahe Peroomian, UCLA Harvey Picker, Trinity College Amy Pope, Clemson University James Rabchuk, Western Illinois University Michele Rallis, Ohio State University Andrew Resnick, Cleveland State University Paul Richards, University of California Berkeley Peter Riley, University of Texas Austin Dennis Rioux, University of Wisconsin Oshkosh John Rollino, Rutgers University Larry Rowan, University of North Carolina Chapel Hill Arthur Schmidt, Northwestern University Cindy Schwarz, Vassar College Peter Sheldon, Randolph-Macon Woman’s College James Siegrist, University of California Berkeley Christopher Sirola, University of Southern Mississippi Earl Skelton, Georgetown University George Smoot, University of California Berkeley Stanley Sobolewski, Indiana University of Pennsylvania Mark Sprague, East Carolina University Michael Strauss, University of Oklahoma Leo Takahashi, Pennsylvania State University Richard Taylor, University of Oregon Oswald Tekyi-Mensah, Alabama State University Ray Turner, Clemson University Som Tyagi, Drexel University David Vakil, El Camino College Robert Webb, Texas A&M Robert Weidman, Michigan Technological University Edward A. Whittaker, Stevens Institute of Technology Lisa M. Will, San Diego City College Suzanne Willis, Northern Illinois University Michael Winokur, University of Wisconsin-Madison Stanley George Wojcicki, Stanford University Mark Worthy, Mississippi State University Edward Wright, UCLA Todd Young, Wayne State College Some instructors may find this book contains more material than can be covered in their courses. The text offers great flexibility. Sections marked with a star may be considered optional. These contain slightly more advanced physics material, or material not usually covered in typical courses, or interest-ing applications; they contain no material needed in later Chapters (except perhaps in later optional Sections). For a brief course, all optional material could be dropped as well as significant parts of Chapters 13, 16, 26, 30, and 35, and selected parts of Chapters 9, 12, 19, 20, and 33. Topics not covered in class can be a valuable resource for outside study by students. Indeed, this text can serve as a useful reference for years because of its wide range of coverage.
Thanks Many physics professors provided input or direct feedback on every aspect of this textbook. They are listed below, and I owe each a debt of gratitude.
PREFACE 19 I owe special thanks to Prof. Bob Davis for much valuable input, and especially for working out all the Problems and producing the Solutions Manual for all Problems, as well as for providing the answers to odd-numbered Problems at the back of the book. Many thanks also to J. Erik Hendrickson who collaborated with Bob Davis on the solutions, and to the team they managed (Michael Ottinger, John Kinard, David Jones, Kristi Hatch, Lisa Will).
I am especially grateful to Profs. Lorraine Allen, Kathryn Dimiduk, Michael Strauss, Cindy Schwarz, Robert Coakley, Robert Pelcovitz, Mark Hollabaugh, Charles Hibbard, and Michael Winokur, who helped root out errors and offered significant improvements and clarifications.
For Chapters 43 and 44 on Particle Physics and Cosmology and Astrophys-ics, I was fortunate to receive generous input from some of the top experts in the field, to whom I owe a debt of gratitude: Saul Perlmutter, George Smoot, Alex Filippenko, Paul Richards, Gabriel Orebi Gann, James Siegrist, and William Holzapfel (UC Berkeley), Andreí Linde (Stanford U.), Lyman Page (Princeton), Edward Wright (UCLA), Michael Strauss (University of Oklahoma), and Bob Jacobsen (UC Berkeley).
I also wish to thank many others at the University of California, Berkeley, Physics Department for helpful discussions, and for hospitality. Thanks also to Prof. Tito Arecchi at the Istituto Nazionale di Ottica, Florence, Italy.
Finally, I am grateful to the many people at Pearson Education with whom I worked on this project, especially Jeanne Zalesky and Paul Corey, and the perspicacious editors Margy Kuntz and Andrea Giancoli.
The final responsibility for all errors lies with me. I welcome comments, corrections, and suggestions as soon as possible to benefit students for the next reprint.
D.G.
email: chris.hess@pearson.com paper mail: Christopher Hess Pearson Education 50 California Street San Francisco, CA 94111 Acknowledgments for the Global Edition Pearson would like to acknowledge and thank the following for the Global Edition: Paul McKenna, Glasgow Caledonian University Abhishek Pandey, University of the Witwatersrand Iyabo Usman, University of the Witwatersrand About the Author Doug Giancoli obtained his BA in physics (summa cum laude) from UC Berkeley, his MS in physics at MIT, and his PhD in elementary particle physics back at UC Berkeley. He spent 2 years as a post-doctoral fellow at UC Berkeley’s Virus Lab developing skills in molecular biology and biophysics. His mentors include Nobel winners Emilio Segrè, Barry Barish, and Donald Glaser. He has taught a wide range of undergraduate courses, traditional as well as innovative ones, and works to improve his textbooks meticulously, seeking ways to provide a better understanding of physics for students.
Doug loves the outdoors, especially climbing peaks. He says climbing peaks is like learning physics: it takes effort and the rewards are great.
Advice for Students HOW TO STUDY 1. Read the Chapter. Learn new vocabulary and notation. Respond to questions and exercises as they occur. Follow carefully the steps of worked-out Examples and derivations. Avoid time looking at a screen. Paper is better than pixels when it comes to learning and thinking. 2. Attend all class meetings. Listen. Take notes. Ask questions (everyone wants to, but maybe you will have the courage). You will get more out of class if you read the Chapter first. 3. Read the Chapter again, paying attention to details. Follow derivations and worked-out Examples. Absorb their logic. Answer Exercises and as many of the end-of-Chapter Questions as you can, and all MisConceptual Questions. 4. Solve at least 10 to 20 end-of-Chapter Problems, especially those assigned. In doing Problems you may find out what you learned and what you didn’t. Discuss them with other students. Problem solving is one of the great learning tools. Don’ t just look for a formula : it might be the wrong one. NOTES ON THE FORMAT AND PROBLEM SOLVING 1. Sections marked with a star () may be considered optional or advanced. They can be omitted without interrupting the main flow of topics. No later material depends on them except possibly later starred Sections. They may be fun to read, though. 2. The customary conventions are used: symbols for quantities (such as m for mass) are italicized, whereas units (such as m for meter) are not italicized. Symbols for vectors are shown in boldface with a small arrow above: F 5. 3. Few equations are valid in all situations. Where practical, the range of validity of important equations are stated in square brackets next to the equation. The equations that represent the great laws of physics are displayed with a tan background, as are a few other indispensable equations. 4. At the end of each Chapter is a set of Questions you should try to answer. Attempt all the multiple-choice MisConceptual Questions, which are inten-dend to get common misconceptions “out on the table” by including them as responses (temptations) along with correct answers. Most important are Problems which are ranked as Level I, II, or III, according to estimated dif-ficulty. Level I Problems are easiest, Level II are standard Problems, and Level III are “challenge problems.” These ranked Problems are arranged by Section, but Problems for a given Section may depend on earlier material too. There follows a group of General Problems, not arranged by Section or ranked. Problems that relate to optional Sections are starred (). Answers to odd-numbered Problems are given at the end of the book. 5. Being able to solve Problems is a crucial part of learning physics, and provides a powerful means for understanding the concepts and principles. This book contains many aids to problem solving: (a) worked-out Examples, including an Approach and a Solution, which should be studied as an integral part of the text; (b) some of the worked-out Examples are Estimation Examples, which show how rough or approximate results can be obtained even if the given data are sparse (see Section 1–6); (c) Problem Solving Strategies placed throughout the text to suggest a step-by-step approach to problem solving for a particular topic : but the basics remain the same; most of these “Strategies” are followed by an Example that is solved by explicitly following the suggested steps; (d) special problem-solving Sections; (e) “Problem Solv-ing” marginal notes which refer to hints within the text for solving Problems; (f) Exercises within the text that you should work out immediately, and then check your response against the answer given at the bottom of the last page of that Chapter; (g) the Problems themselves at the end of each Chapter. 6. Conceptual Examples pose a question which hopefully starts you to think about a response. Give yourself a little time to come up with your own response before reading the Response given. 7. Math review, plus additional topics, are found in Appendices. Useful data, conversion factors, and math formulas are found inside the front and back covers.
20 PREFACE Force ( ) Torque ( ) Displacement ( ) , a v m Vectors USE OF COLOR Electricity and magnetism Electric circuit symbols Wire, with switch S Resistor Capacitor Inductor Battery Ground S or or Electric charge (+) Electric charge (–) Electric field lines Equipotential lines Magnetic field lines A general vector resultant vector (sum) is slightly thicker components of any vector are dashed Optics Light rays Object 1.0 m Measurement lines Energy level (atom, etc.) Path of a moving object Real image (dashed) Virtual image (dashed and paler) Velocity ( ) Acceleration ( ) Momentum ( or ) Angular momentum ( ) Angular velocity ( ) Electric field ( ) Magnetic field ( ) Force on second object or third object in same figure Direction of motion or current Other L F E D v r B S S S S S S S S S S S S p V T + + PREFACE 21 This page is intentionally left blank 23 Introduction, Measurement, Estimating C H A P T E R 1 CHAPTER-OPENING QUESTIONS— Guess now!
1. How many cm3 are in 1.0 m3?
(a) 10. (b) 100. (c) 1000. (d) 10,000. (e) 100,000. (f) 1,000,000.
2. Suppose you wanted to actually measure the radius of the Earth, at least roughly, rather than taking other people’s word for what it is. Which response below describes the best approach?
(a) Use an extremely long measuring tape.
(b) It is only possible by flying high enough to see the actual curvature of the Earth.
(c) Use a standard measuring tape, a stepladder, and a large smooth lake.
(d) Use a laser and a mirror on the Moon or on a satellite.
(e) Give up; it is impossible using ordinary means.
[We start each Chapter with a Question : sometimes two. Try to answer right away. Don’t worry about getting the right answer now : the idea is to get your preconceived notions out on the table. If they are misconceptions, we expect them to be cleared up as you read the Chapter. You will get another chance at the Question later in the Chapter when the appropriate material has been covered. These Chapter-Opening Questions will also help you see the power and usefulness of physics.] Image of the Earth from out in space. The sky appears black because there are so few molecules to refl ect light. (Why the sky appears blue to us on Earth has to do with scattering of light by molecules of the atmosphere, as discussed in Chapter 34.) Note the storm off the coast of Mexico.
Important physics is covered in this fi rst Chapter, including measurement uncertainty and how to make an estimate. For example, we can determine the radius of the Earth without going out in space, but just by being near a lake or bay.
CONTENTS 1–1 How Science Works 1–2 Models, Theories, and Laws 1–3 Measurement and Uncertainty; Significant Figures 1–4 Units, Standards, and the SI System 1–5 Converting Units 1–6 Order of Magnitude: Rapid Estimating 1–7 Dimensions and Dimensional Analysis 24 CHAPTER 1 Introduction, Measurement, Estimating P hysics is the most basic of the sciences. It deals with the behavior and structure of matter. The field of physics is usually divided into classical physics which includes motion, fluids, heat, sound, light, electricity and magnetism; and modern physics which includes the topics of relativity, atomic structure, condensed matter, nuclear physics, elementary particles, and cosmology and astrophysics. We will cover all these topics in this book, beginning with motion (or mechanics, as it is often called) and ending with the most recent results in our study of the cosmos.
An understanding of physics is wonderfully useful for anyone making a career in science or technology. Engineers, for example, must know how to calculate the forces within a structure to design it so that it remains standing (Fig. 1 9 1a). Indeed, in Chapter 12 we will see a worked-out Example of how a simple physics calculation : or even intuition based on understanding the physics of forces : would have saved hundreds of lives (Fig. 1 9 1b). We will see many examples in this book of how physics is useful in many fields, and in everyday life.
1–1 How Science Works There is a real physical world out there. We could just walk through it, not thinking much about it. Or, we can instead examine it carefully. That is what scientists do. The aim of science is the search for order in our observations of the physical world so as to provide a deeper picture or description of this world around us. Sometimes we just want to understand how things work.
Some people seem to think that science is a mechanical process of collecting facts and devising theories. But it is not so simple. Science is a creative activity, and in many ways resembles other creative activities of the human mind.
One important aspect of science is observation of events (which great writers and artists also do), and includes the design and carrying out of experiments. But observation and experiment require imagination, because scientists can never include everything in a description of what they observe. In other words, scientists must make judgments about what is relevant in their observations and experiments.
Consider, for example, how two great minds, Aristotle (384 9 322 b.c.) and Galileo (1564 9 1642), interpreted motion along a horizontal surface. Aristotle noted that objects given an initial push along the ground (or on a level tabletop) always slow down and stop. Consequently, Aristotle argued, the natural state of an object is to be at rest. Galileo, in his reexamination of horizontal motion in the 1600s, had the idea that friction is a kind of force like a push or a pull; and he imagined that if friction could be eliminated, an object given an initial push along a horizontal surface would continue to move indefinitely without stopping. He concluded that for an object to be in motion was just as natural as for it to be at rest. By inventing a new approach, Galileo founded our modern view of motion (Chapters 2, 3, and 4), and he did so with a leap of the imagination. Galileo made this leap conceptually, without actually eliminating friction.
Observation, with careful experimentation and measurement, is one side of the scientific process. The other side is the invention or creation of theories to explain and order the observations. Theories are never derived directly from observations. Observations may help inspire a theory, and theories are accepted or rejected based on the results of observation and experiment.
Theories are inspirations that come from the minds of humans. For example, the idea that matter is made up of atoms (the atomic theory) was not arrived at by direct observation of atoms. Rather, the idea sprang from creative minds. The theory of relativity, the electromagnetic theory of light, and Newton’s law of universal gravitation were likewise the result of human imagination. The great theories of science may be compared, as creative achievements, with great works of art or literature. But how does science differ from these other creative activities? One important difference is that science requires testing of its ideas or theories to see if their predictions are borne out by experiment.
But theories are not “proved” by testing. First of all, no measuring instrument is perfect, so exact confirmation is not possible. Furthermore, it is not possible to test a theory in every single possible circumstance. Hence a theory cannot be absolutely verified. C A U T I O N Science is not static. It changes and develops (a) (b) FIGURE 1 – 1 (a) This bridge over the River Tiber in Rome was built 2000 years ago and still stands. (b) The Hartford Civic Center collapsed in 1978, just two years after it was built.
Indeed, the history of science tells us that long-held theories can often be replaced by new ones.
1–2 Models, Theories, and Laws When scientists are trying to understand a particular aspect of the physical world, they often make use of a model. A model, in the scientist’s sense, is a kind of analogy or mental image of the phenomena in terms of something we are familiar with. One example is the wave model of light. We cannot see waves of light as we can water waves. But it is valuable to think of light as made up of waves because experiments indicate that light behaves in many respects as water waves do.
The purpose of a model is to give us an approximate mental or visual picture : something to hold on to : when we cannot see what actually is happening in the real world. Models often give us a deeper understanding: the analogy to a known system (for instance, water waves in the above example) can suggest new experiments to perform and can provide ideas about what other related phenomena might occur.
You may wonder what the difference is between a theory and a model. Usually a model is relatively simple and provides a structural similarity to the phenomena being studied. A theory is broader, more detailed, and can give quantitatively testable predictions, often with great precision. It is important not to confuse a model or a theory with the real world and the phenomena themselves. Theories are descriptions of the physical world, and they are made up by us. Theories are invented : usually by very smart people.
Scientists give the title law to certain concise but general statements about how nature behaves (that energy is conserved, for example). Sometimes the state-ment takes the form of a relationship or equation between quantities (such as Newton’s second law, F = ma).
To be called a law, a statement must be found experimentally valid over a wide range of observed phenomena. For less general statements, the term principle is often used (such as Archimedes’ principle). We use “theory” to describe a more general picture of a large group of phenomena.
Scientific laws are different from political laws, which are prescriptive: they tell us how we ought to behave. Scientific laws are descriptive: they do not say how nature should behave, but rather are meant to describe how nature does behave. As with theories, laws cannot be tested in the infinite variety of cases possible. So we cannot be sure that any law is absolutely true. We use the term “law” when its validity has been tested over a wide range of situations, and when any limitations and the range of validity are clearly understood.
Scientists normally do their research as if the accepted laws and theories were true. But they are obliged to keep an open mind in case new information should alter the validity of any given law or theory. In other words, laws of physics, or the “laws of nature”, represent our descriptions of reality and are not inalterable facts that last forever. Laws are not lying there in nature, waiting to be discov-ered. We humans, the brightest humans, invent the laws using observations and intuition as a basis. And we hope our laws provide a good description of nature, and at a minimum give us a reliable approximation of how nature really behaves.
1–3 Measurement and Uncertainty; Significant Figures In the quest to understand the world around us, scientists seek to find relationships among physical quantities that can be measured.
Uncertainty Reliable measurements are an important part of physics. But no measurement is absolutely precise. There is an uncertainty associated with every measure-ment. Among the most important sources of uncertainty, other than blunders, are the limited accuracy of every measuring instrument and the inability to read C A U T I O N Theories and laws are NOT discovered. They are invented SECTION 1–3 Measurement and Uncertainty; Significant Figures 25 26 CHAPTER 1 Introduction, Measurement, Estimating an instrument (such as a ruler) beyond some fraction of the smallest division shown. For example, if you were to use a centimeter ruler to measure the width of a board (Fig. 1 9 2), the result could be claimed to be precise to about 0.1 cm (1 mm), the smallest division on the ruler, although half of this value might be a valid claim as well. The reason is that it is difficult for the observer to estimate (or interpolate) between the smallest divisions. Furthermore, the ruler itself may not have been manufactured to an accuracy very much better than this.
When giving the result of a measurement, it is important to state the estimated uncertainty in the measurement. For example, the width of a board might be written as 8.8 { 0.1 cm. The {0.1 cm (“plus or minus 0.1 cm”) represents the estimated uncertainty in the measurement, so that the actual width most likely lies between 8.7 and 8.9 cm. The percent uncertainty is the ratio of the uncertainty to the measured value, multiplied by 100. For example, if the measurement is 8.8 and the uncertainty about 0.1 cm, the percent uncertainty is 0.1 8.8 100% L 1%, where L means “is approximately equal to.” Often the uncertainty in a measured value is not specified explicitly. In such cases, scientists follow a general rule that uncertainty in a numerical value is assumed to be one or a few units in the last digit specified.
For example, if a length is given as 5.6 cm, the uncertainty is assumed to be about 0.1 cm or 0.2 cm, or possibly 0.3 cm. It is important in this case that you do not write 5.60 cm, for this implies an uncertainty on the order of 0.01 or 0.02 cm; it assumes that the length is probably between about 5.58 cm and 5.62 cm, when actually you believe it is between about 5.4 and 5.8 cm.
Significant Figures The number of reliably known digits in a number is called the number of significant figures. Thus there are four significant figures in the number 23.21 cm and two in the number 0.062 cm (the zeros in the latter are merely place holders that show where the decimal point goes). The number of significant figures may not always be clear. Take, for example, the number 80. Are there one or two significant figures? We need words here: If we say it is roughly 80 km between two cities, there is only one significant figure (the 8) since the zero is merely a place holder. If there is no suggestion that the 80 is a rough approximation, then we can often assume (as we will in this book) that it has two significant figures: so it is 80 km within an accuracy of about 1 or 2 km. If it is precisely 80 km, to within {0.1 or {0.2 km, then we need to write 80.0 km (three significant figures).
When specifying numerical results, you should avoid the temptation to keep more digits in the final answer than is justified: see boldface statement above. For example, to calculate the area of a rectangle 11.3 cm by 6.8 cm, the result of multi plication would be 76.84 cm2. But this answer can not be accurate to the implied 0.01 cm2 uncertainty. Why? Because (using the outer limits of the assumed uncertainty for each measurement) the result could be between 11.2 cm 6.7 cm = 75.04 cm2 and 11.4 cm 6.9 cm = 78.66 cm2. At best, we can quote the answer as 77 cm2, which implies an uncertainty of about 1 or 2 cm2. The other two digits (in the number 76.84 cm2) must be dropped (rounded off) because they are not significant. As a rough general significant figures rule, the final result of a multiplication or division should have no more digits than the numerical value with the fewest significant figures.
In our example, 6.8 cm has the least number of significant figures, namely two. Thus the result 76.84 cm2 needs to be rounded off to 77 cm2.
EXERCISE A The area of a rectangle 4.5 cm by 3.25 cm is correctly given by (a) 14.625 cm2; (b) 14.63 cm2; (c) 14.6 cm2; (d) 15 cm2.
P R O B L E M S O LV I N G Significant figures rule: Number of significant figures in final result should be same as the least significant input value FIGURE 1 – 2 Measuring the width of a board with a centimeter ruler. The uncertainty is about {1 mm. †Be careful also about other digital read-outs. If a digital bathroom scale shows 85.6, do not assume the uncertainty is {0.1 or {0.2; the scale was likely manufactured with an accuracy of perhaps only 1% or so: that is, {1 or {2. For digital scientific instruments, also be careful: the instruction manual should state the accuracy.
CONCEPTUAL EXAMPLE 1 – 1 Significant figures. Using a protractor (Fig. 1 9 4), you measure an angle to be 30°. (a) How many significant figures should you quote in this measurement? (b) Use a calculator to find the cosine of the angle you measured.
RESPONSE (a) If you look at a protractor, you will see that the precision with which you can measure an angle is about one degree (certainly not 0.1°). So you can quote two significant figures, namely 30° (not 30.0°). (b) If you enter cos 30° in your calculator, you will get a number like 0.866025403. But the angle you entered is known only to two significant figures, so its cosine is correctly given by 0.87; you must round your answer to two significant figures.
NOTE Trigonometric functions, like cosine, are reviewed in Appendix A.
FIGURE 1 – 4 Example 1 9 1. A protractor used to measure an angle.
EXERCISE C Do 0.00324 and 0.00056 have the same number of significant figures?
Scientific Notation We commonly write numbers in “powers of ten,” or “scientific” notation : for instance 36,900 as 3.69 104, or 0.0021 as 2.1 10-3. One advantage of scientific notation is that it allows the number of significant figures to be clearly expressed. For example, it is not clear whether 36,900 has three, four, or five significant figures. With powers of ten notation the ambiguity can be avoided: if the number is known to three signif-icant figures, we write 3.69 104, but if it is known to four, we write 3.690 104.
When adding or subtracting numbers, the final result should contain no more decimal places than the number with the fewest decimal places. For example, the result of subtracting 0.57 from 3.6 is 3.0 (not 3.03). Similarly 36 + 8.2 = 44, not 44.2.
Be careful not to confuse significant figures with the number of decimal places. Significant figures are related to the expected uncertainty in any measured quantity.
EXERCISE B For each of the following numbers, state the number of significant figures and the number of decimal places: (a) 1.23; (b) 0.123; (c) 0.0123.
Keep in mind when you use a calculator that all the digits it produces may not be significant. When you divide 2.0 by 3.0, the proper answer is 0.67, and not 0.666666666 as calculators give (Fig. 1 9 3a). Digits should not be quoted in a result unless they are truly significant figures. However, to obtain the most accurate result, you should normally keep one or more extra significant figures throughout a calculation, and round off only in the final result. (With a calculator, you can keep all its digits in intermediate results.) Calculators can also give too few significant figures. For example, when you multiply 2.5 3.2, a calculator may give the answer as simply 8. See Fig. 1 9 3b. But the answer is accurate to two significant figures, so the proper answer is 8.0.† P R O B L E M S O LV I N G Significant figures when adding and subtracting C A U T I O N Calculators err with significant figures P R O B L E M S O LV I N G Report only the proper number of significant figures in the final result. But keep extra digits during the calculation FIGURE 1 – 3 These two calculators show the wrong number of significant figures. In (a), 2.0 was divided by 3.0. The correct final result should be stated as 0.67. In (b), 2.5 was multiplied by 3.2. The correct result is 8.0. (a) (b) SECTION 1–3 Measurement and Uncertainty; Significant Figures 27 28 CHAPTER 1 Introduction, Measurement, Estimating EXERCISE D Write each of the following in scientific notation and state the number of significant figures for each: (a) 0.0258, (b) 42,300, (c) 344.50.
Percent Uncertainty versus Significant Figures The significant figures rule is only approximate, and in some cases may under-estimate the accuracy (or uncertainty) of the answer. Suppose for example we divide 97 by 92: 97 92 = 1.05 L 1.1.
Both 97 and 92 have two significant figures, so the rule says to give the answer as 1.1. Yet the numbers 97 and 92 both imply an uncertainty of {1 if no other uncertainty is stated. Both 92 { 1 and 97 { 1 imply an uncer-tainty of about 1% (1>92 L 0.01 = 1%). But the final result to two significant figures is 1.1, with an implied uncertainty of {0.1, which is an uncertainty of 0.1>1.1 L 0.1 L 10%. In this case it is better to give the answer as 1.05 (which is three significant figures). Why? Because 1.05 implies an uncertainty of {0.01 which is 0.01>1.05 L 0.01 L 1%, just like the uncertainty in the original numbers 92 and 97.
SUGGESTION: Use the significant figures rule, but consider the % uncer-tainty too, and add an extra digit if it gives a more realistic estimate of uncertainty.
Approximations Much of physics involves approximations, often because we do not have the means to solve a problem precisely. For example, we may choose to ignore air resistance or friction in doing a Problem even though they are present in the real world, and then our calculation is only an estimate or approximation. In doing Problems, we should be aware of what approximations we are making, and be aware that the precision of our answer may not be nearly as good as the number of significant figures given in the result.
Accuracy versus Precision There is a technical difference between “precision” and “accuracy.” Precision in a strict sense refers to the repeatability of the measurement using a given instru-ment. For example, if you measure the width of a board many times, getting results like 8.81 cm, 8.85 cm, 8.78 cm, 8.82 cm (interpolating between the 0.1 cm marks as best as possible each time), you could say the measurements give a precision a bit better than 0.1 cm. Accuracy refers to how close a measurement is to the true value. For example, if the ruler shown in Fig. 1 9 2 was manufactured with a 2% error, the accuracy of its measurement of the board’s width (about 8.8 cm) would be about 2% of 8.8 cm or about {0.2 cm. Estimated uncertainty is meant to take both accuracy and precision into account.
1–4 Units, Standards, and the SI System The measurement of any quantity is made relative to a particular standard or unit, and this unit must be specified along with the numerical value of the quantity. For example, we can measure length in British units such as inches, feet, or miles, or in the metric system in centimeters, meters, or kilometers. To specify that the length of a particular object is 18.6 is insufficient. The unit must be given, because 18.6 meters is very different from 18.6 inches or 18.6 millimeters.
For any unit we use, such as the meter for distance or the second for time, we need to define a standard which defines exactly how long one meter or one second is. It is important that standards be chosen that are readily reproducible so that anyone needing to make a very accurate measurement can refer to the standard in the laboratory and communicate results with other scientists.
TABLE 1 – 1 Some T ypical Lengths or Distances (order of magnitude) Length (or Distance) Meters (approximate) Neutron or proton (diameter) 10-15 m Atom (diameter) 10-10 m Virus [see Fig. 1 9 5a] 10-7 m Sheet of paper (thickness) 10-4 m Finger width 10-2 m Football field length 102 m Height of Mt. Everest [see Fig. 1 9 5b] 104 m Earth diameter 107 m Earth to Sun 1011 m Earth to nearest star 1016 m Earth to nearest galaxy 1022 m Earth to farthest galaxy visible 1026 m †Modern measurements of the Earth’s circumference reveal that the intended length is off by about one-fiftieth of 1%. Not bad!
Length The first truly international standard was the meter (abbreviated m) established as the standard of length by the French Academy of Sciences in the 1790s. The standard meter was originally chosen to be one ten-millionth of the distance from the Earth’s equator to either pole,† and a platinum rod to represent this length was made. (One meter is, very roughly, the distance from the tip of your nose to the tip of your finger, with arm and hand stretched out horizontally.) In 1889, the meter was defined more precisely as the distance between two finely engraved marks on a particular bar of platinum 9 iridium alloy. In 1960, to provide greater precision and reproducibility, the meter was redefined as 1,650,763.73 wavelengths of a particular orange light emitted by the gas krypton-86.
In 1983 the meter was again redefined, this time in terms of the speed of light (whose best measured value in terms of the older definition of the meter was 299,792,458 m>s, with an uncertainty of 1 m>s). The new definition reads: “The meter is the length of path traveled by light in vacuum during a time interval of 1>299,792,458 of a second.” The new definition of the meter has the effect of giving the speed of light the exact value of 299,792,458 m>s. [The newer definitions provided greater precision than the 2 marks on the old platinum bar.] British units of length (inch, foot, mile) are now defined in terms of the meter. The inch (in.) is defined as exactly 2.54 centimeters (cm; 1 cm = 0.01 m). One foot is exactly 12 in., and 1 mile is 5280 ft. Other conversion factors are given in the Table on the inside of the front cover of this book. Table 1 9 1 below presents some typical lengths, from very small to very large, rounded off to the nearest power of 10. (We call this rounded off value the order of magnitude.) See also Fig. 1 9 5. (Note that the abbreviation for inches (in.) is the only one with a period, to distinguish it from the word “in”.) [The nautical mile = 6076 ft = 1852 km is used by ships on the open sea and was originally defined as 1>60 of a degree latitude on Earth’s surface. A speed of 1 knot is 1 nautical mile per hour. Weather forecasts use it too.] Time The standard unit of time is the second (s). For many years, the second was defined as 1>86,400 of a mean solar day (24 h>day 60 min>h 60 s>min = 86,400 s>day). The standard second can be defined more precisely in terms of the frequency of radiation emitted by cesium atoms when they pass between two particular states. [Specifically, one second is the time required for 9,192,631,770 periods of this radiation. This number was chosen to keep “one second” the same as in the old definition.] There are, by definition, 60 s in one minute (min) and 60 minutes in one hour (h). Table 1 9 2 presents a range of time intervals, rounded off to the nearest power of 10.
New definition of the meter FIGURE 1 – 5 Some lengths: (a) viruses (about 10-7 m long) attacking a cell; (b) Mt. Everest’s height is on the order of 104 m (8850 m, to be precise). (a) (b) TABLE 1 – 2 Some Typical Time Intervals (order of magnitude) Time Interval Seconds (approximate) Lifetime of very unstable subatomic particle 10-23 s Lifetime of radioactive elements 10-22 s to 1028 s Lifetime of muon 10-6 s Time between human heartbeats 100 s ( = 1 s) One day 105 s One year 3 107 s Human life span 2 109 s Length of recorded history 1011 s Humans on Earth 1014 s Life on Earth 1017 s Age of Universe 4 1017 s SECTION 1–4 Units, Standards, and the SI System 29 30 CHAPTER 1 †Some exceptions are for angle (radians : see Chapter 10), solid angle (steradian), and sound level (bel or decibel, Chapter 16). Some Sections of this book, such as this subsection, may be considered optional at the discretion of the instructor and they are marked with an asterisk (). See the Preface for more details.
TABLE 1 – 3 Some Masses Object Kilograms (approximate) Electron 10-30 kg Proton, neutron 10-27 kg DNA molecule 10-17 kg Bacterium 10-15 kg Mosquito 10-5 kg Plum 10-1 kg Human 102 kg Ship 108 kg Earth 6 1024 kg Sun 2 1030 kg Galaxy 1041 kg TABLE 1 – 4 Metric (SI) Prefixes Prefix Abbreviation Value quetta Q 1030 ronna R 1027 yotta Y 1024 zetta Z 1021 exa E 1018 peta P 1015 tera T 1012 giga G 109 mega M 106 kilo k 103 hecto h 102 deka da 101 deci d 10-1 centi c 10-2 milli m 10-3 micro† m 10-6 nano n 10-9 pico p 10-12 femto f 10-15 atto a 10-18 zepto z 10-21 yocto y 10-24 ronto r 10-27 quecto q 10-30 †m is the Greek letter “mu.” Mass The standard unit of mass is the kilogram (kg). The standard mass has been, since 1889, a particular platinum 9 iridium cylinder, kept at the International Bureau of Weights and Measures near Paris, France, whose mass is defined as exactly 1 kg. A range of masses is presented in Table 1 9 3. [For practical purposes, a 1 kg mass weighs about 2.2 pounds on Earth.] 1 metric ton is 1000 kg. In the British system of units, 1 ton is 2000 pounds.
When dealing with atoms and molecules, we usually use the unified atomic mass unit (u or amu). In terms of the kilogram, 1 u = 1.6605 10-27 kg.
(Precise values of this and other numbers are given inside the front cover.) The density of a uniform object is its mass divided by its volume, commonly expressed in kg>m3. Unit Prefixes In the metric system, the larger and smaller units are defined in multiples of 10 from the standard unit, and this makes calculation particularly easy. Thus 1 kilo- me ter (km) is 1000 m, 1 centimeter is 1 100 m, 1 millimeter (mm) is 1 1000 m or 1 10 cm, and so on. The prefixes “centi-,” “kilo-,” and others are listed in Table 1 9 4 and can be applied not only to units of length but to units of volume, mass, or any other unit. For example, a centiliter (cL) is 1 100 liter (L), and a kilogram (kg) is 1000 grams (g). An 8.2-megapixel camera has a detector with 8,200,000 pixels (individual “picture elements”).
In common usage, 1 mm (= 10 - 6 m) is called 1 micron.
Systems of Units When dealing with the laws and equations of physics it is very important to use a consistent set of units. Several systems of units have been in use over the years. Today the most important is the Système International (French for International System), which is abbreviated SI. In SI units, the standard of length is the meter, the standard for time is the second, and the standard for mass is the kilogram. This system used to be called the MKS (meter-kilogram-second) system.
A second metric system is the cgs system, in which the centimeter, gram, and second are the standard units of length, mass, and time, as abbreviated in the title. The British engineering system (although more used in the U.S. than Britain) has as its standards the foot for length, the pound for force, and the second for time.
We use SI units almost exclusively in this book, although we often define the cgs and British units when a new quantity is introduced. In the SI, there have traditionally been seven base quantities, each defined in terms of a standard; seven is the smallest number of base quantities consistent with a full description of the physical world. See Table 1 9 5. All other quantities† can be defined in terms of seven base quantities; see the Table inside the front cover which lists many quantities and their units in terms of base units.
A New SI As always in science, new ideas and approaches can produce better precision and closer correspondence with the real world. Even for units and standards.
International organizations on units have proposed further changes that should make standards more readily available and reproducible. To cite one example, the standard kilogram (see above) has been found to have changed slightly in mass ( contamination is one cause).
The new redefinition of SI standards follows the method already used for the meter as being related to the defined value of the speed of light, as we mentioned on page 29 under “Length” . For example, the charge on the electron, e, instead of being a measured value, becomes defined as a certain value (its current value), and the unit of electric charge (the coulomb) follows from that. All units then become based on P R O B L E M S O LV I N G Always use a consistent set of units TABLE 1 – 5 Traditional SI Base Quantities Quantity Unit Unit Abbreviation Length meter m Time second s Mass kilogram kg Electric current ampere A Temperature kelvin K Amount of substance mole mol Luminous intensity candela cd EXAMPLE 1 – 2 The 8000-m peaks. There are only 14 peaks whose summits are over 8000 m above sea level. They are the highest peaks in the world (Fig. 1 9 6 and Table 1 9 6) and are referred to as “eight-thousanders.” What is the elevation, in feet, of an elevation of 8000 m?
APPROACH We need to convert meters to feet, and we can start with the conversion factor 1 in. = 2.54 cm, which is exact. That is, 1 in. = 2.5400 cm to any number of significant figures, because it is defined to be.
SOLUTION One foot is defined to be 12 in., so we can write 1 ft = (12 in. ) ¢2.54 cm in. ≤ = 30.48 cm = 0.3048 m, which is exact. Note how the units cancel (colored slashes). We can rewrite this equation to find the number of feet in 1 meter: 1 m = 1 ft 0.3048 = 3.28084 ft.
(We could carry the result to 6 significant figures because 0.3048 is exact, 0.304800 g.) We multiply this equation by 8000.0 (to have five significant figures): 8000.0 m = (8000.0 m ) ¢3.28084 ft m ≤ = 26,247 ft.
An elevation of 8000 m is 26,247 ft above sea level.
NOTE We could have done the unit conversions all in one line: 8000.0 m = (8000.0 m ) ¢ 100 cm 1 m ≤¢ 1 in. 2.54 cm ≤¢ 1 ft 12 in. ≤ = 26,247 ft.
The key is to multiply conversion factors, each equal to one ( = 1.0000), and to make sure which units cancel.
P H Y S I C S A P P L I E D The world’s tallest peaks FIGURE 1 – 6 The world’s second highest peak, K2, whose summit is considered the most difficult of the “8000-ers.” Example 1 9 2.
defined fundamental constants like e and the speed of light. Seven is still the number of basic standards. The new definitions maintain the values of the traditional defini-tions: the “new” meter is the same length as the “old” meter. The new definitions do not change our understanding of what length, time, or mass means.
For us, using this book, the difference between the new SI and the traditional SI is highly technical and does not affect the physics we study. We include the traditional SI because there is some good physics in explaining it. [The Table of Fundamental Constants inside the front cover would look slightly different using the new SI. The value of the charge e on the electron, for example, is defined, and so would have no uncertainty attached to it; instead, our Table inside the front cover includes the traditional SI measured uncertainty (updated) of {98 10-29 C.] 1–5 Converting Units Any quantity we measure, such as a length, a speed, or an electric current, consists of a number and a unit. Often we are given a quantity in one set of units, but we want it expressed in another set of units. For example, suppose we measure that a shelf is 21.5 inches wide, and we want to express this in centimeters. We must use a conversion factor, which in this case is, by definition, exactly 1 in. = 2.54 cm or, written another way, 1 = 2.54 cm>in.
Since multiplying by the number one does not change anything, the width of our shelf, in cm, is 21.5 inches = (21.5 in. ) a2.54 cm in. b = 54.6 cm.
Note how the units (inches in this case) cancelled out (thin red lines). A Table containing many unit conversions is found inside the front cover of this book. Let’s consider some Examples.
TABLE 1 – 6 The 8000-m Peaks Peak Height (m) Mt. Everest 8850 K2 8611 Kangchenjunga 8586 Lhotse 8516 Makalu 8462 Cho Oyu 8201 Dhaulagiri 8167 Manaslu 8156 Nanga Parbat 8125 Annapurna 8091 Gasherbrum I 8068 Broad Peak 8047 Gasherbrum II 8035 Shisha Pangma 8013 SECTION 1–5 Converting Units 31 32 CHAPTER 1 Introduction, Measurement, Estimating EXERCISE E The names and elevations of the 14 eight-thousand-meter peaks in the world (see Example 1 9 2) are given in Table 1 9 6, repeated here. They are all in the Hima-laya mountain range in India, Pakistan, Tibet, and China. Determine the elevation of the world’s three highest peaks in feet.
EXAMPLE 1 – 3 Apartment area. You have seen a nice apartment whose floor area is 880 square feet (ft2). What is its area in square meters?
APPROACH We use the same conversion factor, 1 in. = 2.54 cm, but this time we have to use it twice.
SOLUTION Because 1 in. = 2.54 cm = 0.0254 m, then 1 ft2 = (12 in.) 2(0.0254 m>in.) 2 = 0.0929 m2.
So 880 ft2 = (880 ft2) (0.0929 m2>ft2) L 82 m2.
NOTE As a rule of thumb, an area given in ft2 is roughly 10 times the number of square meters (more precisely, about 10.8).
Rule of thumb: Floor area in ft 2 is about 10 area in m2: 100 m2 L 1000 ft 2 EXAMPLE 1 – 4 Speeds. Where the posted speed limit is 55 miles per hour (mi>h or mph), what is this speed (a) in meters per second (m>s) and (b) in kilometers per hour (km>h)?
APPROACH We again use the conversion factor 1 in. = 2.54 cm, and we recall that there are 5280 feet in a mile and 12 inches in a foot; also, one hour contains (60 min>h) (60 s>min) = 3600 s>h.
SOLUTION (a) We can write 1 mile as 1 mi = (5280 ft ) ¢12 in. ft ≤¢2.54 cm in. ≤¢ 1 m 100 cm ≤ = 1609 m.
We also know that 1 hour contains 3600 s, so 55 mi h = ¢55 mi h ≤¢1609 m mi ≤¢ 1 h 3600 s ≤ = 25 m s , where we rounded off to two significant figures.
(b) Now we use 1 mi = 1609 m = 1.609 km; then 55 mi h = ¢55 mi h ≤¢1.609 km mi ≤ = 88 km h .
NOTE Each conversion factor is equal to one. You can look up most conversion factors in the Table inside the front cover.
P R O B L E M S O LV I N G Conversion factors = 1 EXERCISE G Return to the first Chapter-Opening Question, page 23, and answer it again now. Try to explain why you may have answered differently the first time.
When changing units, you can avoid making an error in the use of conversion factors by checking that units cancel out properly. For example, in our conversion of 1 mi to 1609 m in Example 1 9 4(a), if we had incorrectly used the factor (100 cm 1 m ) instead of ( 1 m 100 cm), the centimeter units would not have cancelled out; we would not have ended up with meters.
TABLE 1 – 6 The 8000-m Peaks Peak Height (m) Mt. Everest 8850 K2 8611 Kangchenjunga 8586 Lhotse 8516 Makalu 8462 Cho Oyu 8201 Dhaulagiri 8167 Manaslu 8156 Nanga Parbat 8125 Annapurna 8091 Gasherbrum I 8068 Broad Peak 8047 Gasherbrum II 8035 Shisha Pangma 8013 The first two equations in Example 1 9 2 on the previous page show how to change from feet to meters, or meters to feet. For practical purposes 1 m = 3.28 ft L 3.3 ft which means that we can change any distance or height in meters to feet by multiplying by 3 and adding 10% (0.1). For example, a 3000-m-high peak in feet is 9000 ft + 900 ft L 10,000 ft.
P R O B L E M S O LV I N G Unit conversion is wrong if units do not cancel EXERCISE F One hectare is defined as 1.000 104 m2. There are 640 acres in a square mile. Both units are used for land area. (a) How many acres are in one hectare? (b) What would be an easy everyday rule-of-thumb conversion factor?
†Formulas like this for volume, area, etc., are found inside the back cover of this book.
EXAMPLE 1 – 5 ESTIMATE Volume of a lake. Estimate how much water there is in a particular lake, Fig. 1 9 7a, which is roughly circular, about 1 km across, and you guess it has an average depth of about 10 m.
APPROACH No lake is a perfect circle, nor can lakes be expected to have a perfectly flat bottom. We are only estimating here. To estimate the volume, we can use a simple model of the lake as a cylinder: we multiply the average depth of the lake times its roughly circular surface area, as if the lake were a cylinder (Fig. 1 9 7b).
SOLUTION The volume V of a cylinder is the product of its height h times the area of its base: V = hpr2, where r is the radius of the circular base.† The radius r is 1 2 km = 500 m, so the volume is approximately V = hpr2 L (10 m) (3) (5 102 m) 2 L 8 106 m3 L 107 m3, where p was rounded off to 3. So the volume is on the order of 107 m3, ten million cubic meters. Because of all the estimates that went into this calculation, the order-of-magnitude estimate (107 m3) is probably better to quote than the 8 106 m3 figure.
NOTE To express our result in U.S. gallons, we see in the Table on the inside front cover that 1 liter = 10-3 m3 L 1 4 gallon. Hence, the lake contains about (8 106 m3) (1 gallon>4 10 - 3 m3) L 2 109 gallons of water.
P H Y S I C S A P P L I E D Estimating the volume (or mass) of a lake; see also Fig. 1–7 (b) (a) 10 m r = 500 m FIGURE 1 – 7 Example 1 9 5. (a) How much water is in this lake? (Photo is one of the Rae Lakes in the Sierra Nevada of California.) (b) Model of the lake as a cylinder. [We could go one step further and estimate the mass or weight of this lake. We will see later that water has a density of 1000 kg>m3, so this lake has a mass of about (103 kg>m3) (107 m3) L 1010 kg, which is about 10 billion kg or 10 million metric tons. (A metric ton is 1000 kg, about 2200 lb, slightly larger than a British ton, 2000 lb.)] This is an exciting and powerful Section that will be useful throughout this book, and in real life. We will see how to make approximate calculations of quantities you may never have dreamed you could do. Also, we are sometimes interested only in an approximate value for a quantity, maybe because an accurate calculation would take more time than it is worth or requires data that are not available. In other cases, we may want to make a rough estimate in order to check a calculation made on a calculator, to make sure that no blunders were made when the numbers were entered.
A rough estimate can be made by rounding off all numbers to one significant figure and its power of 10, and after the calculation is made, again keeping only one significant figure. Such an estimate is called an order-of-magnitude estimate and can be accurate within a factor of 10, and often better. In fact, the phrase “order of magnitude” is sometimes used to refer simply to the power of 10.
P R O B L E M S O LV I N G How to make a rough estimate 1–6 Order of Magnitude: Rapid Estimating SECTION 1–6 Order of Magnitude: Rapid Estimating 33 34 CHAPTER 1 Introduction, Measurement, Estimating It cannot be emphasized enough how important it is to draw a diagram when solving a physics Problem, as the next Example shows.
EXAMPLE 1 – 7 ESTIMATE Height by triangulation. Estimate the height of the building shown in Fig. 1 9 9, by “triangulation,” with the help of a bus-stop pole and a friend.
APPROACH By standing your friend next to the pole, you estimate the height of the pole to be 3 m. You next step away from the pole until the top of the pole is in line with the top of the building, Fig. 1 9 9a. You are 5 ft 6 in. tall, so your eyes are about 1.5 m above the ground. Your friend is taller, and when she stretches out her arms, one hand touches you and the other touches the pole, so you estimate that distance as 2 m (Fig. 1 9 9a). You then pace off the distance from the pole to the base of the building with big, 1-m-long steps, and you get a total of 16 steps or 16 m.
SOLUTION Now you draw, to scale, the diagram shown in Fig. 1 9 9b using these measurements. You can measure, right on the diagram, the last side of the triangle to be about x L 13 or 14 m. Alternatively, you can use similar triangles to obtain the height x: 1.5 m 2 m = x 18 m , so x L 13 1 2 m.
Finally you add in your eye height of 1.5 m above the ground to get your final result: the building is about 15 m tall.
FIGURE 1 – 8 Example 1 9 6. Micrometer used for measuring small thicknesses.
FIGURE 1 – 9 Example 1 9 7. Diagrams are really useful!
16 m 18 m 2 m 1.5 m (b) x = ?
1.5 m 3 m (a) 1.5 m ?
2 m EXAMPLE 1 – 6 ESTIMATE Thickness of a sheet of paper. Estimate the thickness of a page of this book.
APPROACH At first you might think that a special measuring device, a micro meter (Fig. 1 9 8), is needed to measure the thickness of one page since an ordinary ruler can not be read so finely. But we can use a trick or, to put it in physics terms, make use of a symmetry: we can make the reasonable assumption that all the pages of this book are equal in thickness.
SOLUTION We can use a ruler to measure hundreds of pages at once. If you measure the thickness of the first 500 pages of this book (page 1 to page 500), you might get something like 1.5 cm. Note that 500 numbered pages, counted front and back, is 250 separate pieces of paper. So one sheet must have a thickness of about 1.5 cm 250 sheets L 6 10-3 cm = 6 10-2 mm, or less than a tenth of a millimeter (0.1 mm).
EXAMPLE 1 – 8 ESTIMATE T otal number of heartbeats. Estimate the total number of beats a typical human heart makes in a lifetime.
APPROACH A typical resting heart rate is 70 beats>min. But during exercise it can be a lot higher. A reasonable average might be 80 beats>min.
SOLUTION One year, in seconds, is (24 h>d) (3600 s>h) (365 d) L 3 107 s. If an average person lives 70 years = (70 yr) (3 107 s>yr) L 2 109 s, then the total number of heartbeats would be about ¢80 beats min ≤¢ 1 min 60 s ≤(2 109 s) L 3 109, or 3 billion.
EXERCISE H Return to the second Chapter-Opening Question, page 23, and answer it again now. Try to explain why you may have answered differently the first time.
EXAMPLE 1 – 9 ESTIMATE Estimating the radius of Earth. Believe it or not, you can estimate the radius of the Earth without having to go into space (see the photograph on page 23). If you have ever been on the shore of a large lake, you may have noticed that you cannot see the beaches, piers, or rocks at water level across the lake on the opposite shore. The lake seems to bulge out between you and the opposite shore : a good clue that the Earth is round. Suppose you climb a stepladder and discover that when your eyes are 10 ft (3.0 m) above the water, you can just see the rocks at water level on the opposite shore. From a map, you estimate the distance to the opposite shore as d L 6.1 km. Use Fig. 1 9 10 with h = 3.0 m to estimate the radius R of the Earth.
APPROACH We use simple geometry, including the theorem of Pythagoras, c2 = a2 + b2, where c is the length of the hypotenuse of any right triangle, and a and b are the lengths of the other two sides.
SOLUTION For the right triangle of Fig. 1 9 10, the two sides are the radius of the Earth R and the distance d = 6.1 km = 6100 m. The hypotenuse is approx-imately the length R + h, where h = 3.0 m. By the Pythagorean theorem, R2 + d2 L (R + h) 2 L R2 + 2hR + h2.
We solve algebraically for R, after cancelled R2 on both sides: R L d2 - h2 2h = (6100 m)2 - (3.0 m)2 6.0 m = 6.2 106 m = 6200 km.
NOTE Precise measurements give 6380 km. But look at your achievement! With a few simple rough measurements and simple geometry, you made a good estimate of the Earth’s radius. You did not need to go out in space, nor did you need a very long measuring tape.† Earth Center of Earth Lake R R d h FIGURE 1 – 10 Example 1 9 9, but not to scale. You can just barely see rocks at water level on the opposite shore of a lake 6.1 km wide if you stand on a stepladder.
Another type of estimate, this one made famous by Enrico Fermi (1901 9 1954, Fig. 1 9 11), was to show his students how to estimate the number of piano tuners in a city, such as Chicago or San Francisco. To get a rough order-of-magnitude estimate of the number of piano tuners today in San Francisco, a city of about 800,000 inhabitants, we can proceed by estimating the number of functioning pianos, how often each piano is tuned, and how many pianos each tuner can tune. To estimate the number of pianos in San Francisco, we note that certainly not everyone has a piano. A guess of 1 family in 3 having a piano would correspond to 1 piano per 12 persons, assuming an average family of 4 persons.
As an order of magnitude, let’s say 1 piano per 10 people. This is certainly more reasonable than 1 per 100 people, or 1 per every person, so let’s proceed with the estimate that 1 person in 10 has a piano, or about 80,000 pianos in San Francisco. Now a piano tuner needs an hour or two to tune a piano. So let’s estimate that a tuner can tune 4 or 5 pianos a day. A piano ought to be tuned every 6 months or a year : let’s say once each year. A piano tuner tuning 4 pianos a day, 5 days a week, 50 weeks a year can tune about 1000 pianos a year. So San Francisco, with its (very) roughly 80,000 pianos, needs about 80 piano tuners. This is, of course, only a rough estimate.‡ It tells us that there must be many more than 10 piano tuners, and surely not as many as 1000.
P R O B L E M S O LV I N G Estimating how many piano tuners there are in a city FIGURE 1 – 11 Enrico Fermi. Fermi contributed significantly to both theoretical and experimental physics, a feat almost unique in modern times.
†As a teenager I had a summer job washing dishes at a camp located 350 m above famous Lake Tahoe in California. Starting the drive down to Lake Tahoe, the beaches across the lake were visible. But approaching the level of Lake Tahoe, the beaches across the lake were no longer visible! I realized that Lake Tahoe was bulging up in the middle, blocking the view. (“The Earth is round.”) ‡A search on the internet (done after this calculation) reveals over 50 listings. Each of these listings may employ more than one tuner, but on the other hand, each may also do repairs as well as tuning. In any case, our estimate is reasonable.
SECTION 1–6 Order of Magnitude: Rapid Estimating 35 36 CHAPTER 1 Introduction, Measurement, Estimating 1–7 Dimensions and Dimensional Analysis When we speak of the dimensions of a quantity, we are referring to the type of base units that make it up. The dimensions of area, for example, are always length squared, abbreviated [L 2] using square brackets; the units can be square meters, square feet, cm2, and so on. Velocity, on the other hand, can be measured in units of km>h, m>s, or mi>h, but the dimensions are always a length [L] divided by a time [T ]: that is, [L>T ].
The formula for a quantity may be different in different cases, but the dimen-sions remain the same. For example, the area of a triangle of base b and height h is A = 1 2 bh, whereas the area of a circle of radius r is A = pr2. The formulas are different in the two cases, but the dimensions of area are always [L 2].
Dimensions can be used as a help in working out relationships, a procedure referred to as dimensional analysis. One useful technique is the use of dimensions to check if a relationship is incorrect. Note that we add or subtract quantities only if they have the same dimensions (we don’t add centimeters and hours); and the quantities on each side of an equals sign must have the same dimensions. (In numer-ical calculations, the units must also be the same on both sides of an equation.) For example, suppose you derived the equation v = v0 + 1 2 at2, where v is the velocity of an object after a time t, v0 is the object’s initial velocity, and the object undergoes an acceleration a. Let’s do a dimensional check to see if this equation could be correct or is surely incorrect. Note that numerical factors, like the 1 2 here, do not affect dimensional checks. We write a dimensional equation as follows, remembering that the dimensions of velocity are [L>T ] and (as we shall see in Chapter 2) the dimensions of acceleration are [L>T 2]: B L T R ≟ B L T R + B L T 2 R [T 2] ≟ B L T R + [L].
The dimensions are incorrect: on the right side, we have the sum of quantities whose dimensions are not the same. Thus we conclude that an error was made in the derivation of the original equation.
A dimensional check can only tell you when a relationship is wrong. It can not tell you if it is completely right. For example, a dimensionless numerical factor (such as 1 2 or 2p) could be missing.
Dimensional analysis can also be used as a quick check on an equation you are not sure about. For example, consider a simple pendulum of length ℓ. Suppose that you can’t remember whether the equation for the period T (the time to make one back-and-forth swing) is T = 2p1ℓ>g or T = 2p1g>ℓ, where g is the accel-eration due to gravity and, like all accelerations, has dimensions [L>T 2]. (Do not worry about these formulas : the correct one will be derived in Chapter 11; what we are concerned about here is a person’s recalling whether it contains ℓ>g or g>ℓ.) A dimensional check shows that the former (ℓ>g) is correct: [T ] = B [L] [L>T 2] = 2[T 2] = [T ], whereas the latter (g>ℓ) is not: [T ] ≠ C [L>T 2] [L] = B 1 [T 2] = 1 [T ] ⋅ The constant 2p has no dimensions and so can’t be checked using dimensions.
Further uses of dimensional analysis are found in Appendix D.
Some Sections of this book, such as this one, may be considered optional at the discretion of the instructor, and they are marked with an asterisk (). See the Preface for more details.
Questions 37 EXAMPLE 1 – 10 Planck length. The smallest meaningful measure of length is called the “Planck length,” and is defined in terms of three fundamental constants in nature: the speed of light c = 3.00 10 8 m>s, the gravitational constant G = 6.67 10-11 m3>kg # s2, and Planck’s constant h = 6.63 10-34 kg⋅m2>s. The Planck length lP (l is the Greek letter “lambda”) is given by the following combination of these three constants: lP = A Gh c3 .
Show that the dimensions of lP are length [L], and find the order of magnitude of lP.
APPROACH We rewrite the above equation in terms of dimensions. The dimensions of c are [L>T ], of G are [L 3>MT 2], and of h are [ML 2>T ].
SOLUTION The dimensions of lP are C [L 3>MT 2][ML 2>T ] [L 3>T 3] = 2[L 2] = [L] which is a length. Good. The value of the Planck length is lP = B Gh c3 = C (6.67 10-11 m3>kg # s2) (6.63 10-34 kg # m2>s) (3.00 10 8 m>s)3 L 4 10-35 m, which is on the order of 10-34 or 10-35 m.
NOTE Some recent theories (Chapters 43 and 44) suggest that the smallest particles (quarks, leptons) have sizes on the order of the Planck length, 10-35 m. These theories also suggest that the “Big Bang,” with which the Universe is believed to have begun, started from an initial size on the order of the Planck length.
Summary [The Summary that appears at the end of each Chapter in this book gives a brief overview of the main ideas of the Chapter. The Summary cannot serve to give an understanding of the material, which can be accomplished only by a detailed reading of the Chapter.] Physics, like other sciences, is a creative endeavor. It is not simply a collection of facts. Important theories are created with the idea of explaining observations. To be accepted, theories are tested by comparing their predictions with the results of actual experiments. Note that, in general, a theory cannot be “proved” in an absolute sense.
Scientists often devise models of physical phenomena. A model is a kind of picture or analogy that helps to describe the phenomena in terms of something we already know about. A theory, often developed from a model, is usually deeper and more complex than a simple model.
A scientific law is a concise statement, often expressed in the form of an equation, which quantitatively describes a wide range of phenomena.
Measurements play a crucial role in physics, but can never be perfectly precise. It is important to specify the uncertainty of a measurement either by stating it directly using the { notation, and>or by keeping only the correct number of significant figures.
Physical quantities are always specified relative to a particular standard or unit, and the unit used should always be stated. The commonly accepted set of units today is the Système International (SI), in which the standard units of length, mass, and time are the meter, kilogram, and second.
When converting units, check all conversion factors for correct cancellation of units.
Making rough, order-of-magnitude estimates is a very useful technique in science as well as in everyday life.
[The dimensions of a quantity refer to the combination of base quantities that comprise it. Velocity, for example, has dimen-sions of [length>time] or [L>T ]. Working with only the dimensions of the various quantities in a given relationship : this technique is called dimensional analysis : makes it possible to check a relation-ship for correct form.] Questions 1. What are the merits and drawbacks of using a person’s foot as a standard? Consider both (a) a particular person’s foot, and (b) any person’s foot. Keep in mind that it is advantageous that fundamental standards be accessible (easy to compare to), invariable (do not change), indestructible, and reproducible. 2. What is wrong with this road sign: Memphis 7 mi (11.263 km)? 3. Why is it incorrect to think that the more digits you include in your answer, the more accurate it is? 4. For an answer to be complete, units need to be specified. Why? 5. You measure the radius of a wheel to be 4.16 cm. If you multiply by 2 to get the diameter, should you write the result as 8 cm or as 8.32 cm? Justify your answer. 6. Express the sine of 30.0° with the correct number of signif-icant figures. 7. List assumptions useful to estimate the number of car mechanics in (a) San Francisco, (b) your hometown, and then make the estimates.
38 CHAPTER 1 Introduction, Measurement, Estimating MisConceptual Questions [List all answers that are valid.] 1. The laws of physics (a) are permanent and unalterable.
(b) are part of nature and are waiting to be discovered.
(c) can change, but only because of evidence that convinces the community of physicists.
(d) apply to physics but not necessarily to chemistry or other fields.
(e) were basically complete by 1900, and have undergone only minor revisions since.
(f) are accepted by all major world countries, and cannot be changed without international treaties. 2. How should we write the result of the following calculation, being careful about significant figures?
(3.84 s) (37 m>s) + (5.3 s) (14.1 m>s) = (a) 200 m.
(b) 210 m.
(c) 216.81 m.
(d) 217 m.
(e) 220 m. 3. Four students use different instruments to measure the length of the same pen. Which measurement implies the greatest precision? (a) 160.0 mm. (b) 16.0 cm. (c) 0.160 m. (d) 0.00016 km.
(e) Need more information. 4. The number 0.0078 has how many significant figures? (a) 1.
(b) 2.
(c) 3.
(d) 4. 5. How many significant figures does 1.362 + 25.2 have?
(a) 2.
(b) 3.
(c) 4.
(d) 5. 6. Accuracy represents (a) repeatability of a measurement, using a given instrument. (b) how close a measurement is to the true value. (c) an ideal number of measurements to make. (d) how poorly an instrument is operating. 7. Precision represents (a) repeatability of a measurement, using a given instrument. (b) how close a measurement is to the true value. (c) an ideal number of measurements to make. (d) how poorly an instrument is operating. 8. To convert from ft2 to yd2, you should (a) multiply by 3.
(b) multiply by 1>3.
(c) multiply by 9.
(d) multiply by 1>9.
(e) multiply by 6.
(f) multiply by 1>6. 9. Which is not true about an order-of-magnitude estimation?
(a) It gives you a rough idea of the answer.
(b) It can be done by keeping only one significant figure.
(c) It can be used to check if an exact calculation is reasonable.
(d) It may require making some reasonable assumptions in order to calculate the answer.
(e) It will always be accurate to at least two significant figures. 10. [L 2] represents the dimensions for which of the following?
(a) cm2.
(b) square feet. (c) m2.
(d) All of the above.
Problems [The Problems at the end of each Chapter are ranked I, II, or III according to estimated difficulty, with (I) Problems being easiest. Level III are meant as challenges for the best students. The Problems are arranged by Section, meaning that the reader should have read up to and including that Section, but not only that Section : Problems often depend on earlier material. Next is a set of “General Problems” not arranged by Section and not ranked.] 1 – 3 Measurement, Uncertainty, Significant Figures (Note: In Problems, assume a number like 6.4 is accurate to {0.1; and 950 is accurate to 2 significant figures ({10) unless 950 is said to be “precisely” or “very nearly” 950, in which case assume 950 { 1.) 1. (I) How many significant figures do each of the following numbers have: (a) 777, (b) 81.60, (c) 7.03, (d) 0.03, (e) 0.0086, (f) 6465, and (g) 8700? 2. (I) Write the following numbers in powers of 10 notation: (a) 5.859, (b) 21.8, (c) 0.0068, (d) 328.65, (e) 0.219, (f) 444. 3. (I) Write out the following numbers in full with the correct number of zeros: (a) 8.69 10 5, (b) 9.1 10 3, (c) 2.5 10-1, (d) 4.76 10 2, and (e) 3.62 10-5. 4. (II) What is the percent uncertainty in the measurement 3.25 { 0.35 m? 5. (II) Time intervals measured with a physical stopwatch typically have an uncertainty of about 0.2 s, due to human reaction time at the start and stop moments. What is the percent uncertainty of a hand-timed measurement of (a) 4.5 s, (b) 45 s, (c) 4.5 min? 6. (II) Add (9.2 10 3 s) + (6.3 10 4 s) + (0.008 10 6 s). 7. (II) Multiply 4.079 10 2 m by 0.057 10-1 m, taking into account significant figures. 8. (II) What, approximately, is the percent uncertainty for a measurement given as 1.27 m2? 9. (II) For small angles u, the numerical value of sin u is approximately the same as the numerical value of tan u. Find the largest angle for which sine and tangent agree to within two significant figures. 10. (II) A report stated that “a survey of 215 students found that 37.2% had bought a sugar-rich soft drink the day before.” (a) How many students bought a soft drink? (b) What is wrong with the original statement? 11. (II) A watch manufacturer claims that its watches gain or lose no more than 9 seconds in a year. How accurate are these watches, expressed as a percentage? 12. (III) What is the area, and its approximate uncertainty, of a circle of radius 5.1 10 4 cm? 13. (III) What, roughly, is the percent uncertainty in the volume of a spherical beach ball of radius r = 0.64 { 0.04 m?
1 – 4 and 1 – 5 Units, Standards, SI, Converting Units 14. (I) Write the following as full (decimal) numbers without prefixes on the units: (a) 286.6 mm, (b) 74 mV, (c) 430 mg, (d) 47.2 ps, (e) 22.5 nm, (f) 2.50 gigavolts.
15. (I) Express the following using the prefixes of Table 1 9 4: (a) 3 106 volts, (b) 2 10-6 meters, (c) 5 103 days, (d) 18 102 bucks, and (e) 9 10-7 seconds. 16. (I) Determine your own height in meters, and your mass in kg. 17. (II) To the correct number of significant figures, use the information inside the front cover of this book to deter-mine the ratio of (a) the surface area of Earth compared to the surface area of the Moon, (b) the volume of Earth compared to the volume of the Moon. 18. (II) Would a driver traveling at 15 m>s in a 35 mi>h zone be exceeding the speed limit? Why or why not? 19. (II) The age of the universe is thought to be about 14 billion years. Assuming two significant figures, write this in powers of 10 in (a) years, (b) seconds. 20. (II) The Sun, on average, is 93 million miles from Earth. How many meters is this? Express (a) using powers of 10, and (b) using a metric prefix (km). 21. (II) Express the following sum with the correct number of significant figures: 1.90 m + 142.5 cm + 6.27 105 mm. 22. (II) A typical atom has a diameter of about 1.0 10-10 m. (a) What is this in inches? (b) Approximately how many atoms are along a 1.0-cm line, assuming they just touch? 23. (II) Determine the conversion factor between (a) km>h and mi>h, (b) m>s and ft>s, and (c) km>h and m>s. 24. (II) What is the conversion factor between (a) ft2 and yd2, (b) m2 and ft2? 25. (II) A light-year is the distance light travels in one year (at speed = 2.998 108 m>s). (a) How many meters are there in 1.00 light-year? (b) An astronomical unit (AU) is the average distance from the Sun to Earth, 1.50 108 km. How many AU are there in 1.00 light-year? 26. (II) How much longer (percentage) is a one-mile race than a 1500-m race (“the metric mile”)? 27. (II) How many wavelengths of orange krypton-86 light (Section 1 9 4) would fit into the thickness of one page of this book? See Example 1 9 6. 28. (II) Using the French Academy of Sciences’ original defi-nition of the meter, calculate Earth’s circumference and radius in those meters. Give % error relative to today’s accepted values (inside front cover). 29. (II) A passenger jet uses about 12 liters of fuel per km of flight. What is that value expressed as miles per gallon? 30. (II) American football uses a field that is 100.0 yd long, whereas a soccer field is 100.0 m long. Which field is longer, and by how much (give yards, meters, and percent)? 31. (II) (a) How many seconds are there in 1.00 year? (b) How many nanoseconds are there in 1.00 year? (c) How many years are there in 1.00 second? 32. (II) Use Table 1 9 3 to estimate the total number of protons or neutrons in (a) a bacterium, (b) a DNA molecule, (c) the human body, (d) our Galaxy. 33. (II) The diameter of the planet Mercury is 4879 km. (a) What is the surface area of Mercury? (b) How many times larger is the surface area of the Earth? 34. (III) A standard baseball has a circumference of approx-imately 23 cm. If a baseball had the same mass per unit volume (see Tables in Section 1 9 4) as a neutron or a proton, about what would its mass be?
1 – 6 Order-of-Magnitude Estimating (Note: Remember that for rough estimates, only round numbers are needed both as input to calculations and as final results.) 35. (I) Estimate the order of magnitude (power of 10) of: (a) 3200, (b) 86.30 103, (c) 0.076, and (d) 15.0 108. 36. (II) Estimate how many books can be shelved in a college library with 6500 m2 of floor space. Assume 8 shelves high, having books on both sides, with corridors 1.5 m wide. Assume books are about the size of this one, on average. 37. (II) Estimate how many hours it would take to run (at 10 km>h) across the U.S. from New York to California. 38. (II) Estimate the number of liters of water a human drinks in a lifetime. 39. (II) Estimate the number of cells in an adult human body, given that a typical cell has a diameter of about 10 mm, and the human body has a density of about 1000 kg>m3 . 40. (II) Estimate how long it would take one person to mow a football field using an ordinary home lawn mower (Fig. 1 9 12). (State your assumptions, such as the mower moves with a 1@km>h speed, and has a 0.5-m width.) 41. (II) Estimate the number of gallons of gasoline consumed by the total of all automobile drivers in the U.S., per year. 42. (II) Estimate the number of dentists (a) in San Francisco and (b) in your town or city. 43. (II) Estimate how many kilograms of laundry soap are used in the U.S. in one year (and therefore pumped out of washing machines with the dirty water). Assume each load of laundry takes 0.1 kg of soap. 44. (II) How big is a ton (1000 kg)? That is, what is the volume of something that weighs a ton? To be specific, estimate the diameter of a 1-ton rock, but first make a wild guess: will it be 10 cm across, 1 m, or the size of a car? [Hint: Rock has mass per volume about 3 times that of water, which is 1 kg per liter (103 cm3).] 45. (II) A hiking trail is 270 km long through varying terrain. A group of hikers cover the first 49 km in two and a half days. Estimate how much time they should allow for the rest of the trip. 46. (II) Estimate how many days it would take to walk around the circumference of the Earth, assuming 12 h walking per day at 4 km>h. 47. (II) Estimate the number of jelly beans in the jar of Fig. 1 9 13.
FIGURE 1 – 13 Problem 47. Estimate the number of jelly beans in the jar.
FIGURE 1 – 12 Problem 40.
Problems 39 40 CHAPTER 1 Introduction, Measurement, Estimating 48. (II) Estimate the number of bus drivers (a) in Washington, D.C., and (b) in your town. 49. (III) You are in a hot air balloon, 300 m above the flat Texas plains. You look out toward the horizon. How far out can you see : that is, how far is your horizon? The Earth’s radius is about 6400 km. 50. (III) I agree to hire you for 30 days. You can decide between two methods of payment: either (1) $1000 a day, or (2) one penny on the first day, two pennies on the second day and continue to double your daily pay each day up through day 30. Use quick estimation to make your decision, and justify it. 51. (III) The rubber worn from tires mostly enters the atmosphere as particulate pollution. Estimate how much rubber (in kg) is put into the air in the United States every year. To get started, a good estimate for a tire tread’s depth is 1 cm when new, and rubber has a mass of about 1200 kg per m3 of volume. 52. (III) Many sailboats are docked at a marina 4.4 km away on the opposite side of a lake. You stare at one of the sailboats because, when you are lying flat at the water’s edge, you can just see its deck but none of the side of the sailboat. You then go to that sailboat on the other side of the lake and measure that the deck is 1.5 m above the level of the water. Using Fig. 1 9 14, where h = 1.5 m, estimate the radius R of the Earth. 53. (III) You are lying on a beach, your eyes 20 cm above the sand. Just as the Sun sets, fully disappearing over the horizon, you immediately jump up, your eyes now 150 cm above the sand, and you can again just see the top of the Sun. If you count the number of seconds (= t) until the Sun fully disappears again, you can estimate the Earth’s radius. But for this Prob lem, use the known radius of the Earth to calculate the time t.
1 – 7 Dimensions 54. (I) What are the dimensions of density, which is mass per volume? 55. (II) The speed v of an object is given by the equation v = At3 - Bt, where t refers to time. (a) What are the dimensions of A and B? (b) What are the SI units for the constants A and B? 56. (II) Three students derive the following equations in which x refers to distance traveled, v the speed, a the acceleration (m>s2), t the time, and the subscript zero (0) means a quantity at time t = 0. Here are their equations: (a) x = vt2 + 2at, (b) x = v0 t + 1 2 at2, and (c) x = v0 t + 2at2. Which of these could possibly be correct according to a dimensional check, and why? 57. (II) (a) Show that the following combination of the three fundamental constants of nature that we used in Example 1 9 10 (that is G, c, and h) forms a quantity with the dimensions of time: tP = A Gh c5 .
This quantity, tP, is called the Planck time and is thought to be the earliest time, after the creation of the Universe, at which the currently known laws of physics can be applied. (b) Estimate the order of magnitude of tP using values given inside the front cover (or Example 1 9 10).
General Problems 58. Global positioning satellites (GPS) can be used to determine your position with great accuracy. If one of the satellites is 20,000 km from you, and you want to know your position to {2 m, what percent uncertainty in the distance is required? How many significant figures are needed in the distance? 59. One mole of atoms consists of 6.02 1023 individual atoms. If a mole of atoms were spread uniformly over the Earth’s surface, how many atoms would there be per square meter? 60. Computer chips (Fig. 1 9 15) can be etched on circular silicon wafers of thickness 0.300 mm that are sliced from a solid cylindrical silicon crystal of length 25 cm. If each wafer can hold 750 chips, what is the maximum number of chips that can be produced from one entire cylinder? 61. If you used only a keyboard to enter data, how many years would it take to fill up a hard drive in a computer that can store 1.0 terabytes (1.0 1012 bytes) of data? Assume 40-hour work weeks, and that you can type 150 characters per minute, and that one byte is one keyboard character. 62. An average family of four uses roughly 1200 L (about 300 gallons) of water per day (1 L = 1000 cm3). How much depth would a lake lose per year if it covered an area of 60 km2 with uniform depth and supplied a local town with a population of 40,000 people? Consider only population uses, and neglect evaporation, rain, creeks and rivers. 63. A certain compact disc (CD) contains 783.216 megabytes of digital information. Each byte consists of exactly 8 bits. When played, a CD player reads the CD’s information at a constant rate of 1.4 megabits per second. How many minutes does it take the player to read the entire CD? 64. An angstrom (symbol Å) is a unit of length, defined as 10-10 m, which is on the order of the diameter of an atom. (a) How many nanometers are in 1.0 angstrom? (b) How many femtometers or fermis (the common unit of length in nuclear physics) are in 1.0 angstrom? (c) How many angstroms are in 1.0 m? (d) How many angstroms are in 1.0 light-year (see Problem 25)?
Earth Earth center Lake R R d h FIGURE 1 – 14 Problem 52. You see a sailboat across a lake (not to scale). R is the radius of the Earth. Because of the curvature of the Earth, the water “bulges out” between you and the boat. FIGURE 1 – 15 Problem 60. The wafer held by the hand is shown below, enlarged and illuminated by colored light. Visible are rows of integrated circuits (chips). 65. A typical adult human lung contains about 300 million tiny cavities called alveoli. Estimate the average diameter of a single alveolus. 66. Hold a pencil in front of your eye at a position where its blunt end just blocks out the Moon (Fig. 1 9 16). Make appro-priate measurements to estimate the diameter of the Moon, given that the Earth 9 Moon distance is 3.8 105 km. 67. A storm dumps 1.0 cm of rain on a city 5 km wide and 7 km long in a 2-h period. How many metric tons (1 metric ton = 103 kg) of water fell on the city? (1 cm3 of water has a mass of 1 g = 10-3 kg.) How many gallons of water was this? 68. Greenland’s ice sheet covers over 1.7 106 km2 and is approximately 2.5 km thick. If it were to melt completely then by how much would you expect the ocean to rise? Assume 2 3 of Earth’s surface is ocean. See Tables inside front and back covers. 69. Noah’s ark was ordered to be 300 cubits long, 50 cubits wide, and 30 cubits high. The cubit was a unit of measure equal to the length of a human forearm, elbow to the tip of the longest finger. Express the dimensions of Noah’s ark in meters, and estimate its volume (m3). 70. One liter (1000 cm3) of oil is spilled onto a smooth lake. If the oil spreads out uniformly until it makes an oil slick just one molecule thick, with adjacent molecules just touching, estimate the diameter of the oil slick. Assume the oil mole-cules have a diameter of 2 10-10 m. 71. If you walked north along one of Earth’s lines of longitude until you had changed latitude by 1 minute of arc (there are 60 minutes per degree), how far would you have walked (in miles)? This distance is a nautical mile (page 29). 72. Determine the percent uncertainty in u, and in sin u, when (a) u = 15.0° { 0.5°, (b) u = 75.0° { 0.5°. 73. Jim stands beside a wide river and wonders how wide it is. He spots a large rock on the bank directly across from him. He then walks upstream 85 strides and judges that the angle between him and the rock, which he can still see, is now at an angle of 30° downstream (Fig. 1 9 17). Jim measures his stride to be about 0.8 m long. Estimate the width of the river. 74. Make a rough estimate of the volume of your body (in m 3). 75. Estimate the number of plumbers in San Francisco. 76. Estimate the ratio (order of magnitude) of the mass of a human to the mass of a DNA molecule. [Hint: Check the Tables in this Chapter.] 77. The following formula estimates an average person’s lung capacity V (in liters, where 1 L = 10 3 cm 3): V = 4.1H - 0.018A - 2.7, where H and A are the person’s height (in meters) and age (in years), respectively. In this formula, what are the units of the numbers 4.1, 0.018, and 2.7? 78. The density of an object is defined as its mass divided by its volume. Suppose a rock’s mass and volume are measured to be 6 g and 2.8325 cm3. To the correct number of significant figures, determine the rock’s density (mass>volume). 79. Recent findings in astrophysics suggest that the observ-able universe can be modeled as a sphere of radius R = 13.7 109 light-years = 13.0 10 25 m with an average total mass density of about 1 10-26 kg>m 3. Only about 4% of total mass is due to “ordinary” matter (such as protons, neutrons, and electrons). Estimate how much ordinary matter (in kg) there is in the observable universe. (For the light-year, see Problem 25.) A N S W E R S T O E X E R C I S E S A: (d).
B: All three have three significant figures; the number of decimal places is (a) 2, (b) 3, (c) 4.
C: No: they have three and two, respectively.
D: (a) 2.58 10-2, 3; (b) 4.23 104, 3 (probably); (c) 3.4450 102, 5.
E: Mt. Everest, 29,035 ft; K2, 28,251 ft; Kangchenjunga, 28,169 ft.
F: (a) 2.47 acres in 1 hectare; (b) 2 1 2 or even just 2 acres in 1 hectare. G: (f) 1,000,000; that is, one million.
H: (c).
FIGURE 1 – 16 Problem 66. How big is the Moon?
FIGURE 1 – 17 Problem 73.
85 Strides 30° General Problems 41 42 CHAPTER-OPENING QUESTION— Guess now!
[Don’t worry about getting the right answer now : you will get another chance later in the Chapter. See also page 23 of Chapter 1 for more explanation.] Two small heavy balls have the same diameter but one weighs twice as much as the other. The balls are dropped from a second-story balcony at the exact same time. The time to reach the ground below will be: (a) twice as long for the lighter ball as for the heavier one.
(b) longer for the lighter ball, but not twice as long.
(c) twice as long for the heavier ball as for the lighter one.
(d) longer for the heavier ball, but not twice as long.
(e) nearly the same for both balls.
T he motion of objects : baseballs, automobiles, joggers, and even the Sun and Moon : is an obvious part of everyday life. It was not until the sixteenth and seventeenth centuries that our modern understanding of motion was established. Many individuals contributed to this understanding, particularly Galileo Galilei (1564 9 1642) and Isaac Newton (1642 9 1727).
The study of the motion of objects, and the related concepts of force and energy, form the fi eld called mechanics. Mechanics is customarily divided into two parts: kinematics, which is the description of how objects move, and dynamics, which deals with force and why objects move as they do. This Chapter and the next deal with kinematics.
A space shuttle has released a parachute to reduce its speed quickly. The directions of the shuttle’s velocity and acceleration are shown by the green (v 5) and gold (a 5) arrows.
Motion is described using the concepts of velocity and acceleration. In the case shown here, the velocity v 5 is to the right, in the direction of motion. The acceleration a 5 is in the opposite direction from the velocity v 5, which means the object is slowing down.
We examine in detail motion with constant acceleration, including the vertical motion of objects falling under gravity.
v a Describing Motion: Kinematics in One Dimension CONTENTS 2–1 Reference Frames and Displacement 2–2 Average Velocity 2–3 Instantaneous Velocity 2–4 Acceleration 2–5 Motion at Constant Acceleration 2–6 Solving Problems 2–7 Freely Falling Objects 2–8 Variable Acceleration; Integral Calculus C H A P T E R 2 SECTION 2–1 Reference Frames and Displacement 43 For now we only discuss objects that move without rotating (Fig. 2 9 1a). Such motion is called translational motion. In this Chapter we will be concerned with describing an object that moves along a straight-line path, which is one- dimensional translational motion. In Chapter 3 we will describe translational motion in two (or three) dimensions along paths that are not straight. (Rotation, shown in Fig. 2 9 1b, is discussed in Chapters 10 and 11.) We will often use the concept, or model, of an idealized particle which is considered to be a mathematical point with no spatial extent (no size). A point particle can undergo only translational motion. The particle model is useful in many real situations where we are interested only in translational motion and the object’s size is not signifi cant. For example, we might consider a billiard ball, or even a spacecraft traveling toward the Moon, as a particle for many purposes.
2–1 Reference Frames and Displacement Any measurement of position, distance, or speed must be made with respect to a reference frame, or frame of reference. For example, while you are on a train traveling at 80 km>h, suppose a person walks past you toward the front of the train at a speed of, say, 5 km>h (Fig. 2 9 2). This 5 km>h is the person’s speed with respect to the train as frame of reference. With respect to the ground, that person is moving at a speed of 80 km>h + 5 km>h = 85 km>h. It is always important to specify the frame of reference when stating a speed. In everyday life, we usually mean “with respect to the Earth” without even thinking about it, but the reference frame must be specifi ed whenever there might be confusion.
(a) (b) FIGURE 2 – 1 A falling pinecone undergoes (a) pure translation; (b) it is rotating as well as translating.
FIGURE 2 – 2 A person walks toward the front of a train at 5 km>h. The train is moving at 80 km>h with respect to the ground, so the walking person’s speed, relative to the ground, is 85 km>h.
When specifying the motion of an object, it is important to specify not only the speed but also the direction of motion. Often we can specify a direction by using north, east, south, and west, and by “up” and “down.” In physics, we often draw a set of coordinate axes, as shown in Fig. 2 9 3, to represent a frame of reference. We can always place the origin 0, and the directions of the x and y axes, as we like for convenience. The x and y axes are always perpendicular to each other. The origin is where x = 0, y = 0. Objects positioned to the right of the origin of coordinates (0) on the x axis have an x coordinate which we almost always choose to be positive; objects at points to the left of 0 have a negative x coordinate. The position along the y axis is usually considered positive when above 0, and negative when below 0, although the reverse convention can be used if convenient. Any point on the xy plane can be specifi ed by giving its x and y coordinates. In three dimensions, a z axis perpendicular to the x and y axes is added.
For one-dimensional motion, we often choose the x axis as the line along which the motion takes place. Then the position of an object at any moment is given by its x coordinate. If the motion is vertical, as for a dropped object, we usually use the y axis.
FIGURE 2 – 3 Standard set of xy coordinate axes, sometimes called “rectangular coordinates.” [Also called Cartesian coordinates, after René Descartes (1596 9 1650), who invented them.] - y + y + x - x 0 44 CHAPTER 2 Describing Motion: Kinematics in One Dimension We need to make a distinction between the distance an object has traveled and its displacement, which is defined as the change in position of the object. That is, displacement is how far the object is from its starting point. To see the distinction between total distance and displacement, imagine a person walking 70 m to the east and then turning around and walking back (west) a distance of 30 m (see Fig. 2 9 4). The total distance traveled is 70 m + 30 m = 100 m, but the displace-ment is only 40 m since the person is now only 40 m from the starting point.
Displacement is a quantity that has both magnitude and direction. Such quan-tities are called vectors, and are represented by arrows in diagrams. For example, in Fig. 2 9 4, the blue arrow represents the displacement whose magnitude is 40 m and whose direction is to the right (east).
We will deal with vectors more fully in Chapter 3. For now, we deal only with motion in one dimension, along a line. In this case, vectors which point in one direction will be positive (usually to the right along the x axis). Vectors that point in the opposite direction will have a negative sign in front of their magnitude.
Consider the motion of an object over a particular time interval. Suppose that at some initial time, call it t1, the object is on the x axis at the position x1 in the coordinate system shown in Fig. 2 9 5. At some later time, t2, suppose the object has moved to position x2. The displacement of our object is x2 - x1, and is represented by the arrow pointing to the right in Fig. 2 9 5. It is convenient to write ∆x = x2 - x1, where the symbol ∆ (Greek letter delta) means “change in.” Then ∆x means “the change in x,” or “change in position,” which is in fact the displacement. The change in any quantity means the final value of that quantity, minus the initial value. Suppose x1 = 10.0 m and x2 = 30.0 m, as in Fig. 2 9 5. Then ∆x = x2 - x1 = 30.0 m - 10.0 m = 20.0 m, so the displacement is 20.0 m in the positive direction, Fig. 2 9 5.
Now consider an object moving to the left as shown in Fig. 2 9 6. Here the object, a person, starts at x1 = 30.0 m and walks to the left to the point x2 = 10.0 m. In this case her displacement is ∆x = x2 - x1 = 10.0 m - 30.0 m = -20.0 m, and the blue arrow representing the vector displacement points to the left. For one-dimensional motion along the x axis, a vector pointing to the right is positive, whereas a vector pointing to the left has a negative sign.
EXERCISE A An ant starts at x = 20 cm on a piece of graph paper and walks along the x axis to x = -20 cm. It then turns around and walks back to x = -10 cm. Determine (a) the ant’s displacement and (b) the total distance traveled.
2–2 Average Velocity An important aspect of the motion of a moving object is how fast it is moving : its speed or velocity.
The term “speed” refers to how far an object travels in a given time interval, regardless of direction. If a car travels 240 kilometers (km) in 3 hours (h), we say its average speed was 80 km>h. In general, the average speed of an object is defined as the total distance traveled along its path divided by the time it takes to travel this distance: average speed = distance traveled time elapsed . (2 – 1) The terms “velocity” and “speed” are often used interchangeably in ordi-nary language. But in physics we make a distinction between the two. Speed is simply a positive number, with units. Velocity, on the other hand, is used to signify both the magnitude (numerical value) of how fast an object is moving and also the direction in which it is moving. Velocity is therefore a vector. C A U T I O N The displacement may not equal the total distance traveled x 0 70 m West East 40 m Displacement 30 m y FIGURE 2 – 4 A person walks 70 m east, then 30 m west. The total distance traveled is 100 m (path is shown dashed in black); but the displacement, shown as a solid blue arrow, is 40 m to the east.
FIGURE 2 – 5 The arrow represents the displacement x2 - x1. Distances are in meters.
x y x1 x2 10 0 20 30 40 Distance (m) FIGURE 2 – 6 For the displacement ∆x = x2 - x1 = 10.0 m - 30.0 m, the displacement vector points left.
y x x2 x1 10 0 20 30 40 Distance (m) ∆x SECTION 2–2 Average Velocity 45 There is a second difference between speed and velocity: namely, the average velocity is defined in terms of displacement, rather than total distance traveled: average velocity = displacement time elapsed = final position - initial position time elapsed .
Average speed and average velocity have the same magnitude when the motion is all in one direction. In other cases, they may differ: recall the walk we described earlier, in Fig. 2 9 4, where a person walked 70 m east and then 30 m west. The total distance traveled was 70 m + 30 m = 100 m, but the displacement was 40 m. Suppose this walk took 70 s to complete. Then the average speed was: distance time elapsed = 100 m 70 s = 1.4 m>s.
The magnitude of the average velocity, on the other hand, was: displacement time elapsed = 40 m 70 s = 0.57 m>s.
In everyday life, we are usually interested in average speed. If this second equation on average velocity seems strange, we will see its usefulness in the next Section. To discuss one-dimensional motion of an object in general, suppose that at some moment in time, call it t1, the object is on the x axis at position x1 in a coordi-nate system, and at some later time, t2, suppose it is at position x2. The elapsed time (= change in time) is ∆t = t2 - t1. During this time interval the displacement of our object is ∆x = x2 - x1. Then the average velocity, defined as the displacement divided by the elapsed time, can be written v = x2 - x1 t2 - t1 = ∆x ∆t , [average velocity] (2 – 2) where v stands for velocity and the bar ( ) over the v is a standard symbol meaning “average.” It is always important to choose (and state) the elapsed time, or time interval, t2 - t1, the time that passes during our chosen period of observation.
C A U T I O N Average speed is not necessarily equal to the magnitude of the average velocity C A U T I O N Time interval = elapsed time EXAMPLE 2 – 1 Runner’s average velocity. The position of a runner is plotted as moving along the x axis of a coordinate system. During a 3.00-s time interval, the runner’s position changes from x1 = 50.0 m to x2 = 30.5 m, as shown in Fig. 2 9 7. What is the runner’s average velocity?
APPROACH We want to find the average velocity, which is the displacement divided by the elapsed time.
SOLUTION The displacement is ∆x = x2 - x1 = 30.5 m - 50.0 m = -19.5 m.
In this case the displacement is negative.
The elapsed time, or time interval, is given as ∆t = 3.00 s. The average velocity (Eq. 2 9 2) is v = ∆x ∆t = -19.5 m 3.00 s = -6.50 m>s.
The displacement and average velocity are negative: that is, the runner is moving to the left along the x axis, as indicated by the arrow in Fig. 2 9 7. The runner’s average velocity is 6.50 m>s to the left.
FIGURE 2 – 7 Example 2 9 1. A person runs from x1 = 50.0 m to x2 = 30.5 m. The displacement is -19.5 m.
y x 10 0 20 30 40 50 60 Distance (m) Start (x1) Finish (x2) ∆x For one-dimensional motion in the usual case of the +x axis to the right, if x2 is less than x1, then the object is moving to the left, and ∆x = x2 - x1 is less than zero. The sign of the displacement, and thus of the average velocity, indicates the direction: the average velocity is positive for an object moving to the right along the x axis and negative when the object moves to the left. The direction of the average velocity is always the same as the direction of the displacement.
P R O B L E M S O LV I N G + or - sign can signify the direction for linear motion 46 CHAPTER 2 Describing Motion: Kinematics in One Dimension 2–3 Instantaneous Velocity If you drive a car along a straight road for 150 km in 2.0 h, the magnitude of your average velocity is 75 km>h. It is unlikely, though, that you were moving at precisely 75 km>h at every instant. To describe this situation we need the concept of instantaneous velocity, which is the velocity at any instant of time. (Its magnitude is the number, with units, indicated by a speedometer, Fig. 2 9 8.) More precisely, the instantaneous velocity at any moment is defined as the average velocity over an infinitesimally short time interval. That is, Eq. 2 9 2 is to be evalu ated in the limit of ∆t becoming extremely small, approaching zero. We can write the definition of instantaneous velocity, v, for one-dimensional motion as v = lim ∆tS0 ∆x ∆t . [instantaneous velocity] (2 – 3) The notation lim∆tS0 means the ratio ∆x>∆t is to be evaluated in the limit of ∆t approaching zero. But we do not simply set ∆t = 0 in this definition, for then ∆x would also be zero, and we would not be able to evaluate it. Rather, we consider the ratio ∆x>∆t, as a whole. As we let ∆t approach zero, ∆x approaches zero as well. But the ratio ∆x>∆t approaches some definite value, which is the instantaneous velocity at a given instant.
In Eq. 2 9 3, the limit as ∆t S 0 is written in calculus notation as dx>dt and is called the derivative of x with respect to t: v = lim ∆tS0 ∆x ∆t = dx dt . (2 – 4) This equation is the definition of instantaneous velocity for one-dimensional motion.
For instantaneous velocity we use the symbol v, whereas for average velocity we use v, with a bar above. In the rest of this book, when we use the term “velocity” it will refer to instantaneous velocity. When we want to speak of the average velocity, we will make this clear by including the word “average.” Note that the instantaneous speed always equals the magnitude of the instan-taneous velocity. Why? Because as the time interval becomes infinitesimally small (∆t S 0), an object has no time to change speed or direction, and so the distance traveled and the magnitude of the displacement have to be the same.
EXAMPLE 2 – 2 Distance a cyclist travels. How far can a cyclist travel in 2.5 h along a straight road if her average velocity is 18 km>h?
APPROACH We want to find the distance traveled, which in this case equals the displacement ∆x, so we solve Eq. 2 9 2 for ∆x.
SOLUTION In Eq. 2 9 2, v = ∆x>∆t, we multiply both sides by ∆t and obtain ∆x = v ∆t = (18 km>h) (2.5 h) = 45 km.
EXAMPLE 2 – 3 Car changes speed. A car travels at a constant 50 km>h for 100 km. It then speeds up to 100 km>h and is driven another 100 km. What is the car’s average speed for the 200-km trip?
APPROACH At 50 km>h, the car takes 2.0 h to travel 100 km. At 100 km>h, it takes only 1.0 h to travel 100 km. We use the definition of average velocity, Eq. 2 9 2.
SOLUTION Average velocity (Eq. 2 9 2) is v = ∆x ∆t = 100 km + 100 km 2.0 h + 1.0 h = 67 km>h.
NOTE Averaging the two speeds, (50 km>h + 100 km>h) >2 = 75 km>h, gives a wrong answer. Can you see why? You must use the definition of v, Eq. 2 9 2.
FIGURE 2 – 8 Car speedometer showing mi>h in white, and km>h in orange.
SECTION 2–3 Instantaneous Velocity 47 If an object moves at a uniform (that is, constant) velocity during a partic-ular time interval, then its instantaneous velocity at any instant is the same as its average velocity (see Fig. 2 9 9a). But in many situations this is not the case. For example, a car may start from rest, speed up to 50 km>h, remain at that velocity for a time, then slow down to 20 km>h in a traffic jam, and finally stop at its destination after traveling a total of 15 km in 30 min. This trip is plotted on the graph of Fig. 2 9 9b. Also shown on the graph is the average velocity (dashed line), which is v = ∆x>∆t = 15 km>0.50 h = 30 km>h.
To better understand instantaneous velocity, let us consider a graph of the position versus time (x vs. t) of a particle moving along the x axis, as shown in Fig. 2 9 10. (Note that this is different from showing the “path” of a particle moving in two dimensions on an x vs. y plot.) The particle is at position x1 at time t1, and at posi-tion x2 at time t2. P1 and P2 represent these two points on the graph. A straight line drawn from point P1 (x1, t1) to point P2 (x2, t2) forms the hypotenuse of a right triangle whose sides are ∆x and ∆t. The ratio ∆x>∆t is the slope of the straight line P1 P2. But ∆x>∆t is also the average velocity of the particle during the time interval ∆t = t2 - t1. Therefore, we conclude that the average velocity of a particle during any time interval ∆t = t2 - t1 is equal to the slope of the straight line (or chord) connecting the two points (x1, t1) and (x2, t2) on an x vs. t graph.
Consider now a time ti, intermediate between t1 and t2, at which time the particle is at xi (Fig. 2 9 11). The slope of the straight line P1 Pi is less than the slope of P1 P2 in this case. Thus the average velocity during the time interval ti - t1 is less than during the time interval t2 - t1.
Now let us imagine that we take the point Pi in Fig. 2 9 11 to be closer and closer to point P1. That is, we let the interval ti - t1, which we now call ∆t, become smaller and smaller. The slope of the line connecting the two points becomes closer and closer to the slope of a line tangent to the curve at point P1. The average velocity (equal to the slope of the chord) thus approaches the slope of the tangent at point P1. The definition of the instantaneous velocity (Eq. 2 9 3) is the limiting value of the average velocity as ∆t approaches zero. Thus the instan-taneous velocity equals the slope of the tangent to the x vs. t curve at that point (which we can simply call “the slope of the curve” at that point).
Because the velocity at any instant equals the slope of the tangent to the x vs. t graph at that instant, we can obtain the velocity at any instant from such a graph. For example, in Fig. 2 9 12 (which shows the same curve as in Figs. 2 9 10 and 2 9 11), the slope continually increases as our object moves from x1 to x2, so the velocity is increasing. For times after t2, however, the slope begins to decrease and in fact reaches zero (so v = 0) where x has its maximum value, at point P3 in Fig. 2 9 12. Beyond this point, the slope is negative, as for point P4. The velocity is therefore negative, which makes sense since x is now decreasing : the particle is moving to the left on a standard xy plot, toward decreasing values of x.
If an object moves with constant velocity over a particular time interval, its instantaneous velocity is equal to its average velocity. The graph of x vs. t in this case will be a straight line whose slope equals the velocity. The curve of Fig. 2 9 10 has no straight sections, so there are no time intervals when the velocity is constant.
FIGURE 2 – 10 Graph of a particle’s position x vs. time t. The slope of the straight line P1 P2 represents the average velocity of the particle during the time interval ∆t = t2 - t1.
P1 P2 ∆x = x2 - x1 ∆t = t2 - t1 t2 t1 x1 x2 0 x t FIGURE 2 – 11 Same position vs. time curve as in Fig. 2–10, but including an intermediate time ti. Note that the average velocity over the time interval ti - t1 (which is the slope of P1 Pi) is less than the average velocity over the time interval t2 - t1. The slope of the thin line tangent to the curve at point P1 equals the instantaneous velocity at time t1.
P1 P2 ti tangent at P 1 0 t2 t1 x1 x2 x t Pi xi FIGURE 2 – 12 Same x vs. t curve as in Figs. 2 9 10 and 2 9 11, but here showing the slope at four different points: At P3, the slope is zero, so v = 0. At P4 the slope is negative, so v 6 0.
P3 P4 0 P1 P2 t2 t1 x1 x2 x t t3 60 20 40 Velocity (km>h) Time (h) (a) 0.2 0 Time (h) (b) 0.5 0.1 0.3 0.4 0.2 0 0.5 0.1 0.3 0.4 Average velocity 0 60 20 40 Velocity (km>h) 0 FIGURE 2 – 9 Velocity of a car as a function of time: (a) at constant velocity; (b) with velocity varying in time.
48 CHAPTER 2 Describing Motion: Kinematics in One Dimension EXERCISE B What is your speed at the instant you turn around to move in the opposite direction? (a) Depends on how quickly you turn around; (b) always zero; (c) always negative; (d) none of the above.
The derivatives of various functions are studied in calculus courses, and you can find a review in this book in Appendix B. The derivatives of polynomial functions (which we use a lot) are: d dt (Ctn) = nCtn - 1 and dC dt = 0, where C is any constant.
EXAMPLE 2 – 4 Given x as a function of t. A jet engine moves along an experimental track (which we call the x axis) as shown in Fig. 2 9 13a. We will treat the engine as if it were a particle. Its position as a function of time is given by the equation x = At2 + B, where A = 2.10 m>s2 and B = 2.80 m, and this equation is plotted in Fig. 2 9 13b. (a) Determine the displacement of the engine during the time interval from t1 = 3.00 s to t2 = 5.00 s. (b) Determine the average velocity during this time interval. (c) Determine the magnitude of the instantaneous velocity at t = 5.00 s.
APPROACH (a) We substitute values for t1 and t2 in the given equation for x to obtain x1 and x2. (b) The average velocity can be found from Eq. 2 9 2. (c) To find the instantaneous velocity, we take the derivative of the given x equation with respect to t using the formulas given above.
SOLUTION (a) At t1 = 3.00 s, the position (point P1 in Fig. 2 9 13b) is x1 = At1 2 + B = (2.10 m>s2) (3.00 s) 2 + 2.80 m = 21.7 m.
At t2 = 5.00 s, the position (P2 in Fig. 2 9 13b) is x2 = (2.10 m>s2) (5.00 s) 2 + 2.80 m = 55.3 m.
The displacement is thus x2 - x1 = 55.3 m - 21.7 m = 33.6 m.
(b) The magnitude of the average velocity can then be calculated as v = ∆x ∆t = x2 - x1 t2 - t1 = 33.6 m 2.00 s = 16.8 m>s.
This equals the slope of the straight line joining points P1 and P2 shown in Fig. 2 9 13b.
(c) The instantaneous velocity at t = t2 = 5.00 s equals the slope of the tangent to the curve at point P2 shown in Fig. 2 9 13b. We could measure this slope off the graph to obtain v2. But we can calculate v more precisely for any time t, using the given formula x = At2 + B, which is the engine’s position x as a function of time t. We take the derivative of x with respect to time (see formulas at top of this page): v = dx dt = d dt (At2 + B) = 2At.
We are given A = 2.10 m>s2, so for t = t2 = 5.00 s, v2 = 2At = 2(2.10 m>s2) (5.00 s) = 21.0 m>s.
FIGURE 2 – 13 Example 2 9 4. (a) Engine traveling on a straight track. (b) Graph of x vs. t: x = At2 + B.
P1 P2 0 1 2 3 4 5 6 10 0 20 30 40 50 60 t (s) x (m) x1 0 10 20 30 40 50 60 x (m) (a) (b) ∆t = 2.00 s ∆x = 33.6 m Tangent at P2 whose slope is v2 = 21.0 m>s x2 Slope of straight line = 16.8 m>s v SECTION 2–4 Acceleration 49 2–4 Acceleration An object whose velocity is changing is said to be accelerating. For instance, a car whose velocity increases in magnitude from zero to 80 km>h is accelerating. Acceleration specifies how rapidly the velocity of an object is changing.
Average Acceleration Average acceleration is defined as the change in velocity divided by the time taken to make this change: average acceleration = change of velocity time elapsed .
In symbols, the average acceleration over a time interval ∆t = t2 - t1 during which the velocity changes by ∆v = v2 - v1 is defined as a = v2 - v1 t2 - t1 = ∆v ∆t . (2 – 5) Because velocity is a vector, acceleration is a vector too. But for one-dimensional motion, we need only use a plus or minus sign to indicate acceleration direction relative to a chosen coordinate axis.
EXAMPLE 2 – 5 Average acceleration. A car accelerates along a straight road from rest to 90 km>h in 5.0 s, Fig. 2 9 14. What is the magnitude of its average acceleration?
APPROACH Average acceleration is the change in velocity divided by the elapsed time, 5.0 s. The car starts from rest, so v1 = 0. The final velocity is v2 = 90 km>h = 90 103 m>3600 s = 25 m>s.
SOLUTION From Eq. 2 9 5, the average acceleration is a = v2 - v1 t2 - t1 = 25 m>s - 0 m>s 5.0 s = 5.0 m>s s .
This is read as “five meters per second per second” and means that, on average, the velocity changed by 5.0 m>s during each second. That is, assuming the acceleration was constant, during the first second the car’s velocity increased from zero to 5.0 m>s. During the next second its velocity increased by another 5.0 m>s, reaching a velocity of 10.0 m>s at t = 2.0 s, and so on. See Fig. 2 9 14.
FIGURE 2 – 14 Example 2 9 5. The car is shown at the start with v1 = 0 at t1 = 0. The car is shown three more times, at t = 1.0 s, t = 2.0 s, and at the end of our time interval, t2 = 5.0 s. The green arrows represent the velocity vectors, whose length represents the magnitude of the velocity at that moment and get longer with time. The acceleration vector is the orange arrow, whose magnitude is constant and is found to equal 5.0 m>s2. Distances are not to scale.
Acceleration [a = 5.0 m>s2] v1 = 0 t1 = 0 at t = 2.0 s v = 10.0 m>s at t = 1.0 s v = 5.0 m>s at t = t2 = 5.0 s v = v2 = 25 m>s 50 CHAPTER 2 Describing Motion: Kinematics in One Dimension CONCEPTUAL EXAMPLE 2 – 6 Velocity and acceleration. (a) If the velo ci ty of an object is zero, does it mean that the acceleration is zero? (b) If the acceleration is zero, does it mean that the velocity is zero? Think of some examples.
RESPONSE A zero velocity does not necessarily mean that the acceleration is zero, nor does a zero acceleration mean that the velocity is zero. (a) For example, when you put your foot on the gas pedal of your car which is at rest, the velocity starts from zero but the acceleration is not zero since the velocity of the car changes. (How else could your car start forward if its velocity weren’t changing : that is, accelerating?) (b) As you cruise along a straight highway at a constant velocity of 100 km>h, your acceleration is zero: a = 0, v ≠ 0.
EXERCISE C A powerful car is advertised to go from zero to 60 mi>h in 5.4 s. What does this say about the car: (a) it is fast (high speed); or (b) it accelerates well?
EXAMPLE 2 – 7 Car slowing down. An automobile is moving to the right along a straight highway, which we choose to be the positive x axis (Fig. 2 9 15). Then the driver puts on the brakes. If the initial velocity (when the driver hits the brakes) is v1 = 15.0 m>s, and it takes 5.0 s to slow down to v2 = 5.0 m>s, what was the car’s average acceleration?
APPROACH We put the given initial and final velocities, and the elapsed time, into Eq. 2 9 5 for a.
SOLUTION In Eq. 2 9 5, we call the initial time t1 = 0, and set t2 = 5.0 s. (Note that our choice of t1 = 0 doesn’t affect the calculation of a because only ∆t = t2 - t1 appears in Eq. 2 9 5.) Then a = 5.0 m>s - 15.0 m>s 5.0 s = -2.0 m>s2.
The negative sign appears because the final velocity is less than the initial velocity. In this case the direction of the acceleration is to the left (in the negative x direction) : even though the velocity is always pointing to the right. We say that the acceleration is 2.0 m>s2 to the left, and it is shown in Fig. 2 9 15 as an orange arrow.
Acceleration a = -2.0 m>s2 v1 = 15.0 m>s at t1 = 0 v2 = 5.0 m>s at t2 = 5.0 s FIGURE 2 – 15 Example 2 9 7, showing the position of the car at times t1 and t2, as well as the car’s velocity represented by the green arrows. We calculate that the acceleration vector (orange) points to the left as the car slows down while moving to the right.
“Deceleration” When an object is slowing down, we sometimes say it is decelerating. In physics, the concept of acceleration is all we need: it can be + or -. But if the word “deceleration” is used, be careful: deceleration does not mean that the acceleration is necessarily negative, as in Example 2 9 7. The velocity of an object moving to the right along the positive x axis is positive; if the object is slowing down (as in Fig. 2 9 15), the acceleration is negative. But the same car moving to the left (decreasing x), and slowing down, has positive acceleration that points to the right, as shown in Fig. 2 9 16. We have a deceleration whenever the magnitude of the velocity is decreasing; thus the velocity and acceleration point in opposite directions when there is deceleration.
We almost always write the units for acceleration as m>s2 (meters per second squared) instead of m>s>s. This is possible because: m>s s = m s ⋅s = m s2 .
According to the calculation in Example 2 9 5, the velocity changed on average by 5.0 m>s during each second, for a total change of 25 m>s over the 5.0 s; the average acceleration was 5.0 m>s2.
Note that acceleration tells us how quickly the velocity changes, whereas velocity tells us how quickly the position changes.
C A U T I O N Distinguish velocity from acceleration C A U T I O N If v or a is zero, is the other zero too?
FIGURE 2 – 16 The car of Example 2 9 7, now moving to the left and decelerating. The acceleration is a = v2 - v1 ∆t a = ( -5.0 m>s) - ( -15.0 m>s) 5.0 s = -5.0 m>s + 15.0 m>s 5.0 s = +2.0 m>s2.
v1 = -15.0 m>s v2 = -5.0 m>s a SECTION 2–4 Acceleration 51 EXERCISE D A car moves along the x axis. What is the sign of the car’s acceleration if it is moving in the positive x direction with (a) increasing speed or (b) decreasing speed? What is the sign of the acceleration if the car moves in the negative x direction with (c) increasing speed or (d) decreasing speed?
Instantaneous Acceleration The instantaneous acceleration, a, is defined as the limiting value of the average acceleration as we let ∆t approach zero: a = lim ∆tS0 ∆v ∆t = dv dt . (2 – 6) This limit, dv>dt, is the derivative of v with respect to t. We will use the term “acceleration” to refer to the instantaneous value. If we want to discuss the average acceleration, we will always include the word “average.” If we draw a graph of the velocity, v, vs. time, t, as shown in Fig. 2 9 17, then the average acceleration over a time interval ∆t = t2 - t1 is represented by the slope of the straight line connecting the two points P1 and P2 in Fig. 2 9 17. [Compare this to the position vs. time graph of Fig. 2 9 10 for which the slope of the straight line represents the average velocity.] The instantaneous acceleration at any time, say t1, is the slope of the tangent to the v vs. t curve at time t1, which is also shown in Fig. 2 9 17. In Fig. 2 9 17, as we go from time t1 to time t2 the velocity continually increases, but the acceleration (the rate at which the velocity changes) is decreasing since the slope of the curve is decreasing.
EXAMPLE 2 – 8 Acceleration given x (t). A particle is moving in a straight line so that its position is given by the relation x = (2.10 m>s2)t2 + (2.80 m), as in Example 2 9 4. Calculate (a) its average acceleration during the time interval from t1 = 3.00 s to t2 = 5.00 s, and (b) its instantaneous acceleration as a function of time.
APPROACH To determine acceleration, we first must find the velocity at t1 and t2 by differentiating x: v = dx>dt. Then we use Eq. 2 9 5 to find the average acceleration, and Eq. 2 9 6 to find the instantaneous acceleration.
SOLUTION (a) The velocity at any time t is v = dx dt = d dt [ (2.10 m>s2)t2 + 2.80 m] = (4.20 m>s2)t, as we already saw in Example 2 9 4c. Therefore, at time t1 = 3.00 s, v1 = (4.20 m>s2) (3.00 s) = 12.6 m>s and at t2 = 5.00 s, v2 = 21.0 m>s. Therefore, a = ∆v ∆t = 21.0 m>s - 12.6 m>s 5.00 s - 3.00 s = 4.20 m>s2.
(b) With v = (4.20 m>s2)t, the instantaneous acceleration at any time is a = dv dt = d dt [ (4.20 m>s2)t] = 4.20 m>s2.
The acceleration in this case is constant; it does not depend on time. Figure 2 9 18 shows graphs of (a) x vs. t (the same as Fig. 2 9 13b), (b) v vs. t, which is linearly increasing as calculated above, and (c) a vs. t, which is a horizontal straight line because a = constant.
P1 P2 Slope is average acceleration during ∆t = t2 - t1 Slope is instantaneous acceleration at t1 t2 0 t1 v1 v2 v t ∆v = v2 - v1 ∆t = t2 - t1 FIGURE 2 – 17 A graph of velocity v vs. time t. The average acceleration over a time interval ∆t = t2 - t1 is the slope of the straight line P1 P2 : a = ∆v>∆t. The instantaneous acceleration at time t1 is the slope of the v vs. t curve at that instant.
FIGURE 2 – 18 Example 2 9 8. Graphs of (a) x vs. t, (b) v vs. t, and (c) a vs. t for the motion x = At2 + B. Note that v increases linearly with t and that the acceleration a is constant. Also, v is the slope of the x vs. t curve, whereas a is the slope of the v vs. t curve. x = ( 2.
10 t2 ) m + 2.
80 m (a) (b) (c) 60 20 10 40 50 30 t (s) x (m) 4 0 2 3 5 1 6 4 2 3 5 1 6 4 2 3 5 1 6 25 5 20 15 10 t (s) 0 6 2 4 t (s) a (m>s2) a = 4.20 m>s2 0 v (m>s) v = (4.20 t) m>s Like velocity, acceleration is a rate. The velocity of an object is the rate at which its displacement changes with time; its acceleration, on the other hand, is the rate at which its velocity changes with time. In a sense, acceleration is a “rate of a rate.” 52 CHAPTER 2 Describing Motion: Kinematics in One Dimension FIGURE 2 – 19 Example 2 9 9.
CONCEPTUAL EXAMPLE 2 – 9 Analyzing with graphs. Figure 2 9 19 shows the velocity as a function of time for two cars accelerating from 0 to 100 km>h in a time of 10.0 s. Compare for the two cars: (a) the average acceleration; (b) the instantaneous acceleration; and (c) the total distance traveled.
RESPONSE (a) Average acceleration is ∆v>∆t. Both cars have the same ∆v (100 km>h) and the same ∆t (10.0 s), so the average acceleration is the same for both cars. (b) Instantaneous acceleration is the slope of the tangent to the v vs. t curve. For about the first 4 s, the top curve is steeper than the bottom curve, so car A has a greater instantaneous acceleration during this interval. The bottom curve is steeper during the last 6 s, so car B has the larger acceleration during this period. (c) Except at t = 0 and t = 10.0 s, car A is always going faster than car B. Since it is going faster, it will go farther in the same time. Notice what marvelous information we can get from a graph.
2–5 Motion at Constant Acceleration We now examine motion in a straight line when the magnitude of the acceleration is constant. In this case, the instantaneous and average acceleration are equal. We use the definitions of average velocity and acceleration to derive a set of valuable equations that relate x, v, a, and t when a is constant, allowing us to determine any one of these variables if we know the others.
Notation in physics varies from book to book; and different instructors use different notation. We are now going to change our notation, to simplify it a bit for our discussion here of motion at constant acceleration. First we choose the initial time in any discussion to be zero, and we call it t0 . That is, t1 = t0 = 0. (This is effectively starting a stopwatch at t0 .) We can then let t2 = t be the elapsed time. The initial position (x1) and the initial velocity (v1) of an object will now be represented by x0 and v0 , since they represent x and v at t = 0. At time t the position and velocity will be called x and v (rather than x2 and v2). The average velocity during the time interval t - t0 will be (Eq. 2 9 2) v = ∆x ∆t = x - x0 t - t0 = x - x0 t since we chose t0 = 0. The acceleration, assumed constant in time, is a = ∆v>∆t (Eq. 2 9 5), so a = v - v0 t .
A common problem is to determine the velocity of an object after any elapsed time t, when we are given the object’s constant acceleration. We can solve such problems by solving for v in the last equation: first we multiply both sides by t, which gives at = v - v0 , and then v = v0 + at. [constant acceleration] (2 – 7) If an object, such as a motorcycle, starts from rest (v0 = 0) and accelerates This can be expressed in equation form: since a = dv>dt and v = dx>dt, then a = dv dt = d dt ¢ dx dt ≤ = d2x dt2 .
Here d2x>dt2 is the second derivative of x with respect to time: we first take the derivative of x with respect to time (dx>dt), and then we again take the derivative with respect to time, (d>dt) (dx>dt), to get the acceleration.
EXERCISE E The position of a particle is given by the following equation: x = (2.00 m>s3)t3 + (2.50 m>s)t.
What is the acceleration of the particle at t = 2.00 s? Choose one: (a) 13.0 m>s2; (b) 22.5 m>s2; (c) 24.0 m>s2; (d) 2.00 m>s2; (e) 21.0 m>s2.
10 8 6 4 2 0 100 t (s) v (km>h) Car A Car B at 4.0 m>s2, then after an elapsed time t = 6.0 s its velocity will be v = 0 + at = (4.0 m>s2) (6.0 s) = 24 m>s.
Next, let us see how to calculate the position x of an object after a time t when it undergoes constant acceleration. The definition of average velocity (Eq. 2 9 2) is v = (x - x0) >t, which we can rewrite by multiplying both sides by t : x = x0 + vt. (2 – 8) Because the velocity increases at a uniform rate, the average velocity, v, will be midway between the initial and final velocities: v = v0 + v 2 . [constant acceleration] (2 – 9) (Careful: Equation 2 9 9 is not necessarily valid if the acceleration is not constant.) We combine the last two Equations with Eq. 2 9 7 and find, starting with Eq. 2 9 8, x = x0 + vt = x0 + ¢ v0 + v 2 ≤ t = x0 + ¢ v0 + v0 + at 2 ≤ t or x = x0 + v0 t + 1 2 at2. [constant acceleration] (2 – 10) Equations 2 9 7, 2 9 9, and 2 9 10 are three of the four most useful equations for motion at constant acceleration. We now derive the fourth equation, which is useful in situations where the time t is not known. We substitute Eq. 2 9 9 into Eq. 2 9 8: x = x0 + vt = x0 + ¢ v + v0 2 ≤ t.
Next we solve Eq. 2 9 7 for t, obtaining t = v - v0 a , and substituting this into the previous equation we have x = x0 + ¢ v + v0 2 ≤¢ v - v0 a ≤ = x0 + v2 - v0 2 2a .
We solve this for v2 and obtain v2 = v0 2 + 2a(x - x0), [constant acceleration] (2 – 11) which is the other useful equation we sought.
We now have four equations relating position, velocity, acceleration, and time, when the acceleration a is constant. We collect these kinematic equations for constant acceleration here in one place for further reference (the tan background is used to emphasize their importance): v = v0 + at [a = constant] (2 – 12a) x = x0 + v0 t + 1 2 at2 [a = constant] (2 – 12b) v2 = v0 2 + 2a(x - x0) [a = constant] (2 – 12c) v = v + v0 2 . [a = constant] (2 – 12d) These useful equations are not valid unless a is a constant. In many cases we can set x0 = 0, and this simplifies the above equations a bit. Note that x represents position (not distance), and that x - x0 is the displacement, whereas t is the elapsed time. Equations 2 9 12 are useful also when a is approximately constant, in order to obtain reasonable estimates.
C A U T I O N Average velocity, but only if a = constant Kinematic equations for constant acceleration (we’ll use them a lot) SECTION 2–5 Motion at Constant Acceleration 53 54 CHAPTER 2 Describing Motion: Kinematics in One Dimension P H Y S I C S A P P L I E D Airport design EXAMPLE 2 – 10 Runway design. You are designing an airport for small planes. One kind of airplane that might use this airfield must reach a speed before takeoff of at least 27.8 m>s (100 km>h), and can accelerate at 2.00 m>s2. (a) If the runway is 150 m long, can this airplane reach the required speed for takeoff? (b) If not, what minimum length must the runway have?
APPROACH Assuming the plane’s acceleration is constant, we use the kinematic equations for constant acceleration. In (a), we want to find v, and we are given: Known Wanted x0 = 0 v v0 = 0 x = 150 m a = 2.00 m>s2 SOLUTION (a) Of the four kinematic equations on page 53, Eq. 2 9 12c will give us v when we know v0, a, x, and x0 : v2 = v0 2 + 2a(x - x0) = 0 + 2(2.00 m>s2)(150 m) = 600 m2>s2 v = 2600 m2>s2 = 24.5 m>s.
This runway length is not sufficient, because the minimum speed is not reached.
(b) Now we want to find the minimum runway length, x - x0, for a plane to reach v = 27.8 m>s, given a = 2.00 m>s2. We again use Eq. 2 9 12c, but rewritten as (x - x0) = v2 - v0 2 2a = (27.8 m>s)2 - 0 2(2.00 m>s2) = 193 m.
A 200-m runway is more appropriate for this plane.
NOTE We did this Example as if the plane were a particle, so we round off our answer to 200 m.
P R O B L E M S O LV I N G Equations 2–12 are valid only when the acceleration is constant, which we assume in this Example FIGURE 2 – 20 Example 2 9 11. An air bag deploying on impact.
EXAMPLE 2 – 11 ESTIMATE Air bags. Suppose you want to design an air bag system that can protect the driver at a speed of 100 km>h (60 mph) if the car hits a brick wall. Estimate how fast the air bag must inflate (Fig. 2 9 20) to effectively protect the driver. How does the use of a seat belt help the driver?
APPROACH We assume the acceleration is roughly constant, so we can use Eqs. 2 9 12. Both Eqs. 2 9 12a and 2 9 12b contain t, our desired unknown. They both contain a, so we must first find a, which we can do using Eq. 2 9 12c if we know the distance x over which the car crumples. A rough estimate might be about 1 meter. We choose the time interval to start at the instant of impact with the car moving at v0 = 100 km>h, and to end when the car comes to rest (v = 0) after traveling 1 m.
SOLUTION We convert the given initial speed to SI units: 100 km>h = 100 103 m>3600 s = 28 m>s. We then find the acceleration from Eq. 2 9 12c: a = - v0 2 2x = - (28 m>s) 2 2.0 m = -390 m>s2.
This enormous acceleration takes place in a time given by (Eq. 2 9 12a): t = v - v0 a = 0 - 28 m>s -390 m>s2 = 0.07 s.
To be effective, the air bag would need to inflate faster than this.
What does the air bag do? It spreads the force over a large area of the chest (to avoid puncture of the chest by the steering wheel). The seat belt keeps the person in a stable position directly in front of the expanding air bag.
P H Y S I C S A P P L I E D Car safety : air bags SECTION 2–6 Solving Problems 55 EXERCISE F A car starts from rest and accelerates at a constant 10 m>s2 during a 1 4-mile (402 m) race. How fast is the car going at the finish line? (a) 8040 m>s; (b) 90 m>s; (c) 81 m>s; (d) 804 m>s.
2–6 Solving Problems Before doing more worked-out Examples, let us look at how to approach problem solving. First, it is important to note that physics is not a collection of equations to be memorized. Simply searching for an equation that might work can lead you to a wrong result and will not help you understand physics (Fig. 2 9 21). A better approach is to use the following (rough) procedure, which we present as a special “Problem Solving Strategy.” (Other such Problem Solving Strategies will be found throughout the book.) FIGURE 2 – 21 Read each Chapter of this book, study it by reading it again carefully, and work the Problems using your reasoning abilities.
an applicable equation that involves only known quantities and one desired unknown, solve the equation algebraically for the unknown. Some-times several sequential calculations, or a combi-nation of equations, may be needed. It is often preferable to solve algebraically for the desired unknown before putting in numerical values. 7. Carry out the calculation if it is a numerical problem. Keep one or two extra digits during the calculations, but round off the final answer(s) to the correct number of significant figures (Section 1 9 3). 8. Think carefully about the result you obtain: Is it reasonable? Does it make sense according to your own intuition and experience? A good check is to do a rough estimate using only powers of 10, as discussed in Section 1 9 6. Often it is preferable to do a rough estimate at the start of a numer-ical problem because it can help you focus your attention on finding a path toward a solution. 9. A very important aspect of doing problems is keep ing track of units. An equals sign implies the units on each side must be the same, just as the numbers must. If the units do not balance, a mistake has been made. This can serve as a check on your solution (but it only tells you if you’re wrong, not if you’re right). Always use a consistent set of units. P R O B L E M S O L V I N G 1. Read and reread the whole problem carefully before trying to solve it. 2. Decide what object (or objects) you are going to study, and for what time interval. You can often choose the initial time to be t = 0. 3. Draw a diagram or picture of the situation, with coordinate axes wherever applicable. [You can place the origin of coordinates and the axes wherever you like to make your calculations easier. You also choose which direction is positive and which is negative. Usually we choose the x axis to the right as positive.] 4. Write down what quantities are “known” or “given,” and then what you want to know (“unknowns”). Consider quantities both at the beginning and at the end of the chosen time interval. You may need to “translate” language into physical terms, such as “starts from rest” means v0 = 0. 5. Think about which principles of physics apply in this problem. Use common sense and your own experiences. Then plan an approach. 6. Consider which equations (and>or defini-tions) relate the quantities involved. Before using them, be sure their range of validity includes your problem (for example, Eqs. 2 9 12 are valid only when the acceleration is constant). If you find 56 CHAPTER 2 Describing Motion: Kinematics in One Dimension EXAMPLE 2 – 12 Acceleration of a car. How long does it take a 4.0-m-long car to cross a 26.0-m-wide intersection after the light turns green, if the car accelerates from rest at a constant 2.00 m>s2? The car has to travel 26.0 m + 4.0 m = 30.0 m to clear the intersection.
APPROACH We follow the Problem Solving Strategy on the previous page.
SOLUTION 1. Reread the problem. Be sure you understand what it asks for (here, a time interval: “how long does it take”). 2. The object under study is the car. We need to choose the time interval during which we look at the car’s motion: we choose t = 0, the initial time, to be the moment the car starts to accelerate from rest (v0 = 0); the time t is the instant the car has traveled the full 30.0 m of the intersection. 3. Draw a diagram: the situation is shown in Fig. 2 9 22 where the car is shown moving along the positive x axis. We choose x0 = 0 at the front bumper of the car before it starts to move. 4. The “knowns” and the “wanted” information are shown in the Table in the margin. Note that “starting from rest” means v = 0 at t = 0; that is, v0 = 0. The wanted time t is how long it takes the car to travel 30.0 m. 5. The physics: the car, starting from rest (at t0 = 0), increases in speed as it covers more distance. The acceleration is constant, so we can use the kine-matic equations, Eqs. 2 9 12. 6. Equations: we want to find the time, given the distance and acceleration; Eq. 2 9 12b (x = x0 + v0 t + 1 2 at2) is perfect since the only unknown quantity is t. Setting v0 = 0 and x0 = 0 in Eq. 2 9 12b, we have x = 1 2 at2. We solve for t t = A 2x a . 7. The calculation: t = A 2x a = D 2(30.0 m) 2.00 m>s2 = 5.48 s. This is our answer. Note that the units come out correctly. 8. We can check the reasonableness of the answer by doing an alternate calculation: we use our result in step 7 and check if the distance traveled turns out to be 30.0 m. First we find the final velocity (Eq. 2 9 12a), v = at = (2.00 m>s2) (5.48 s) = 10.96 m>s, and then find the distance traveled using the definition of average velocity (see Eq. 2 9 8).
x = x0 + vt = 0 + 1 2 (10.96 m>s + 0) (5.48 s) = 30.0 m, which checks with our given distance. 9. We checked the units in step 7, and they came out correctly (seconds).
NOTE In steps 6 and 7, when we took the square root, we should have written t = { 12x>a = {5.48 s. Mathematically there are two solutions. But the second solution, t = -5.48 s, is a time before our chosen time interval and makes no sense physically. We say it is “unphysical” and ignore it.
P R O B L E M S O LV I N G “Starting from rest” means v = 0 at t = 0 [i.e., v0 = 0] P R O B L E M S O LV I N G Check your answer P R O B L E M S O LV I N G “Unphysical” solutions FIGURE 2 – 22 Example 2 9 12. (Not to scale.) Known Wanted x0 = 0 t x = 30.0 m a = 2.00 m>s2 v0 = 0 We explicitly followed the steps of the Problem Solving Strategy in Example 2 9 12. In upcoming Examples, we will use our usual “Approach” and “Solution” to avoid being wordy.
a = 2.00 m>s2 0 a = 2.00 m>s2 x0 = 0 v x = 30.0 m = 0 SECTION 2–6 Solving Problems 57 EXAMPLE 2 – 13 ESTIMATE Braking distances. Estimate the minimum stopping distance for a car, which is important for traffic safety and traffic design. The problem is best dealt with in two parts, two separate time intervals. (1) The first time interval begins when the driver decides to hit the brakes, and ends when the foot touches the brake pedal. This is the “reaction time” during which the speed is constant, so a = 0. (2) The second time interval is the actual braking period when the vehicle slows down (a ≠ 0) and comes to a stop. The stopping distance depends on the reaction time of the driver, the initial speed of the car (the final speed is zero), and the deceleration of the car. For a dry road and good tires, good brakes can decelerate a car at a rate of about 5 m>s2 to 8 m>s2. Calculate the total stopping distance for an initial velocity of 50 km>h ( = 14 m>s L 31 mi>h) and assume the acceleration of the car is -6.0 m>s2 (the minus sign appears because the velocity is taken to be in the positive x direction and its magnitude is decreasing). Reaction time for typical drivers varies from perhaps 0.3 s to about 1.0 s; take it to be 0.50 s.
APPROACH During the “reaction time,” part (1), the car moves at constant speed of 14 m>s, so a = 0. Once the brakes are applied, part (2), the accelera-tion is a = -6.0 m>s2 and is constant over this time interval. For both parts a is constant, so we can use Eqs. 2 9 12.
SOLUTION Part (1). We take x0 = 0 for the first time interval, when the driver is reacting (0.50 s): the car travels at a constant speed of 14 m>s so a = 0. See Fig. 2 9 23 and the Table in the margin. To find x, the position of the car at t = 0.50 s (when the brakes are first applied), we cannot use Eq. 2 9 12c because x is multiplied by a, which is zero. But Eq. 2 9 12b works: x = v0 t + 0 = (14 m>s) (0.50 s) = 7.0 m.
Thus the car travels 7.0 m during the driver’s reaction time, until the instant the brakes are applied. We will use this result as input to part (2).
Part (2). During the second time interval, the brakes are applied and the car is brought to rest. The initial position is x0 = 7.0 m (result of part (1)), and other variables are shown in the second Table in the margin. Equation 2 9 12a doesn’t contain x ; Eq. 2 9 12b contains x but also the unknown t. Equation 2 9 12c, v2 - v0 2 = 2a(x - x0), is what we want; after setting x0 = 7.0 m, we solve for x, the final position of the car (where it stops): x = x0 + v2 - v0 2 2a = 7.0 m + 0 - (14 m>s)2 2( -6.0 m>s2) = 7.0 m + -196 m2>s2 -12 m>s2 = 7.0 m + 16 m = 23 m.
The car traveled 7.0 m while the driver was reacting and another 16 m during the braking period before coming to a stop, for a total distance traveled of 23 m. Figure 2 9 24 shows graphs of (a) v vs. t (we were given that v is constant from t = 0 until t = 0.50 s, and after t = 0.50 s it decreases linearly to zero), and (b) x vs. t.
NOTE From the equation above for x, we see that the stopping distance after the driver hit the brakes ( = x - x0) increases with the square of the initial speed, not just linearly with speed. If you are traveling twice as fast, it takes four times the distance to stop (when braking at the same rate, a).
P H Y S I C S A P P L I E D Car stopping distances Part 1: Reaction time Known Wanted t = 0.50 s x v0 = 14 m>s v = 14 m>s a = 0 x0 = 0 Part 2: Braking Known Wanted x0 = 7.0 m x v0 = 14 m>s v = 0 a = -6.0 m>s2 FIGURE 2 – 23 Example 2 9 13: stopping distance for a braking car.
Travel during reaction time Travel during braking x v decreases from 14 m>s to zero a = -6.0 m>s2 v = constant = 14 m>s t = 0.50 s a = 0 FIGURE 2 – 24 Example 2 9 13. Graphs of (a) v vs. t and (b) x vs. t.
10 5 20 15 (b) 0 2.0 0.5 1.0 1.5 2.5 10 8 6 2 4 14 12 0 2.0 0.5 1.0 1.5 2.5 x (m) v (m>s) t = 0.50 s t (s) t (s) (a) t = 0.50 s 58 CHAPTER 2 Describing Motion: Kinematics in One Dimension FIGURE 2 – 25 Example 2 9 14.
x v v (a) 0 15 s Police Speeder (b) 0 Police Speeder Speeder t t t (c) 0 Police EXAMPLE 2 – 14 ESTIMATE T wo Moving Objects: Police and Speeder. A car speeding at 150 km>h (over 90 mph) passes a still police car which immedi-ately takes off in hot pursuit. Using simple assumptions, such as that the speeder continues at constant speed, estimate how long it takes the police car to overtake the speeder. Then estimate the police car’s speed at that moment and decide if the assumptions were reasonable.
APPROACH When the police car takes off, it accelerates, and the simplest assumption is that its acceleration is constant. This may not be reasonable, but let’s see what happens. We can estimate the acceleration if we have noticed automobile ads, which claim cars can accelerate from rest to 100 km>h in 5.0 s. So the average acceleration of the police car could be approximately aP = 100 km>h 5.0 s = 20 km>h s ¢ 1000 m 1 km ≤¢ 1 h 3600 s ≤ = 5.6 m>s2.
SOLUTION We need to set up the kinematic equations to determine the unknown quantities, and since there are two moving objects, we need two separate sets of equations. We denote the speeding car’s position by xS and the police car’s position by xP. Because we are interested in solving for the time when the two vehicles arrive at the same position on the road, we use Eq. 2 9 12b for each car (x0 = 0 for both cars): xS = v0 S t + 1 2 aS t2 = (150 km>h)t + 0 = (42 m>s)t xP = v0 P t + 1 2 aP t2 = 0 + 1 2 (5.6 m>s2)t2, where we have set v0 P = 0 and aS = 0 (speeder assumed to move at constant speed). We want the time when the cars meet, so we set xS = xP and solve for t: (42 m>s)t = (2.8 m>s2)t2.
The solutions are t = 0 and t = 42 m>s 2.8 m>s2 = 15 s.
The first solution corresponds to the instant the speeder passed the police car. The second solution tells us when the police car catches up to the speeder, 15 s later. This is our answer, but is it reasonable? The police car’s speed at t = 15 s is vP = v0 P + aP t = 0 + (5.6 m>s2) (15 s) = 84 m>s or 300 km>h ( L 190 mi>h). Not reasonable, and highly dangerous.
NOTE More reasonable is to give up the assumption of constant acceleration. The police car surely cannot maintain constant acceleration at high speed. Also, the speeder, if a reasonable person, would slow down upon hearing the police siren. Figure 2 9 25 shows (a) x vs. t and (b) v vs. t graphs, based on the original assump-tion of aP = constant, whereas (c) shows v vs. t for more reasonable assumptions.
P R O B L E M S O LV I N G Guess the acceleration C A U T I O N Initial assumptions need to be checked out for reasonableness SECTION 2–7 Freely Falling Objects 59 2–7 Freely Falling Objects One of the most common examples of uniformly accelerated motion is that of an object allowed to fall freely near the Earth’s surface. That a falling object is accelerating may not be obvious at first. And beware of thinking, as was widely believed before the time of Galileo (Fig. 2 9 26), that heavier objects fall faster than lighter objects and that the speed of fall is proportional to how heavy the object is. Wrong. The speed of a falling object is not proportional to its mass.
Galileo made use of his new technique of imagining what would happen in idealized (simplified) cases. For free fall, he postulated that all objects would fall with the same constant acceleration in the absence of air or other resistance. He showed that this postulate predicts that for an object falling from rest, the distance traveled will be proportional to the square of the time (Fig. 2 9 27); that is, d r t2. We can see this from Eq. 2 9 12b for constant acceleration; but Galileo was the first to deduce this mathematical relation.
To support his claim that falling objects increase in speed as they fall, Galileo made use of a clever argument: a heavy stone dropped from a height of 2 m will drive a stake into the ground much further than if the same stone is dropped from a height of only 0.2 m. Clearly, the stone must be moving faster in the former case.
Galileo claimed that all objects, light or heavy, fall with the same acceleration, at least in the absence of air. If you hold a piece of paper flat and horizontal in one hand, and a heavier object like a baseball in the other, and release them at the same time as in Fig. 2 9 28a, the heavier object will reach the ground first. But if you repeat the experiment, this time crumpling the paper into a small wad, you will find (see Fig. 2 9 28b) that the two objects reach the floor at nearly the same time.
Galileo was sure that air acts as a resistance to very light objects that have a large surface area. But in many ordinary circumstances this air resistance is negli-gible. In a chamber from which the air has been removed, even light objects like a feather or a horizontally held piece of paper will fall with the same acceleration as any other object (see Fig. 2 9 29). Such a demonstration in vacuum was not possible in Galileo’s time, which makes Galileo’s achievement all the greater. Galileo is often called the “father of modern science,” not only for the content of his science: astronomical discoveries, inertia, free fall; but also for his new methods of doing science: idealization and simplification, mathematization of theory, theories that have testable consequences, experiments to test theoretical predictions.
FIGURE 2 – 26 Painting of Galileo demonstrating to the Grand Duke of Tuscany his argument for the action of gravity being uniform acceleration. He used a wooden inclined plane (center) to slow down the action. A ball rolling down the plane still accelerates. Tiny bells placed at equal distances along the inclined plane would ring at shorter time intervals as the ball “fell,” indicating that the speed was increasing.
FIGURE 2 – 27 Multiflash photograph of a falling apple, at equal time intervals. The apple falls farther during each successive interval, which means it is accelerating.
FIGURE 2 – 28 (a) A ball and a light piece of paper are dropped at the same time. (b) Repeated, with the paper wadded up. (a) (b) FIGURE 2 – 29 A rock and a feather are dropped simultaneously (a) in air, (b) in a vacuum.
Air-filled tube (a) Evacuated tube (b) 60 CHAPTER 2 Describing Motion: Kinematics in One Dimension EXAMPLE 2 – 15 Falling from a tower. Suppose that a ball is dropped (v0 = 0) from a tower. How far will it have fallen after a time t1 = 1.00 s, t2 = 2.00 s, and t3 = 3.00 s? Ignore air resistance.
APPROACH Let us take y as positive downward, so the acceleration is a = g = +9.80 m>s2. We set v0 = 0 and y0 = 0. We want to fi nd the position y of the ball after three different time intervals. Equation 2 9 12b, with x replaced by y, relates the given quantities (t, a, and v0) to the unknown y.
SOLUTION We set t = t1 = 1.00 s in Eq. 2 9 12b: y1 = v0 t1 + 1 2 at2 1 = 0 + 1 2 at2 1 = 1 2 (9.80 m>s2) (1.00 s) 2 = 4.90 m.
The ball has fallen a distance of 4.90 m during the time interval t = 0 to t1 = 1.00 s. Similarly, after 2.00 s ( = t2), the ball’s position is y2 = 1 2 at2 2 = 1 2 (9.80 m>s2) (2.00 s) 2 = 19.6 m.
Finally, after 3.00 s ( = t3), the ball’s position is (see Fig. 2 9 30a) y3 = 1 2 at3 2 = 1 2 (9.80 m>s2) (3.00 s) 2 = 44.1 m.
NOTE Whenever we say “dropped,” it means v0 = 0. Note also the graph of y vs. t (Fig. 2 9 30b): the curve is not straight but bends upward because y is proportional to t2.
NOTE Because of air resistance, all of the distances in Fig. 2 9 30, y1, y2, y3, would be smaller than shown (and as just calculated); but the difference will be small for a reasonably heavy (but small) object.
Galileo’s specifi c contribution to our understanding of the motion of falling objects can be summarized as follows: at a given location on the Earth and in the absence of air resistance, all objects fall with the same constant acceleration.
We call this acceleration the acceleration due to gravity at the surface of the Earth, and we give it the symbol g. Its magnitude is approximately g = 9.80 m>s2.
In British units g is about 32 ft>s2. Actually, g varies slightly according to latitude and elevation above sea level on the Earth’s surface, but these variations are so small that we will ignore them for most purposes. (Acceleration of gravity out in space beyond the Earth’s surface is treated in Chapter 6.) Air resistance acts to reduce the speed of a falling object, but this effect is often small, and we will neglect it for the most part. However, air resistance will be noticeable even on a reasonably heavy object if the velocity becomes large.† Acceleration due to gravity is a vector, as is any acceleration, and its direction is downward toward the center of the Earth.
When dealing with freely falling objects we can make use of Eqs. 2 9 12, where for a we use the value of g given above. Also, since the motion is vertical we will substitute y in place of x, and y0 in place of x0. We take y0 = 0 unless otherwise specifi ed. It is arbitrary whether we choose y to be positive in the upward direc-tion or in the downward direction; but we must be consistent about it throughout a problem’s solution.
EXERCISE G Return to the Chapter-Opening Question, page 42, and answer it again now, assuming minimal air resistance. Try to explain why you may have answered differ-ently the fi rst time.
[ acceleration due to gravity] at surface of Earth P R O B L E M S O LV I N G You can choose y to be positive either up or down FIGURE 2 – 30 Example 2 9 15. (a) An object dropped from a tower falls with progressively greater speed and covers greater distance with each successive second. (See also Fig. 2 9 27.) (b) Graph of y vs. t (y is + downward).
(a) (b) 40 30 20 10 y (m) 2 0 1 3 t (s) y = 0 y3 = 44.1 m (After 3.00 s) y2 = 19.6 m (After 2.00 s) y1 = 4.90 m (After 1.00 s) +y Acceleration due to gravity +y †The speed of an object falling in air (or other fl uid) does not increase indefi nitely. If the object falls far enough, it will reach a maximum velocity called the terminal velocity due to air resistance. (Section 5 9 6 deals directly with air resistance.) SECTION 2–7 Freely Falling Objects 61 EXAMPLE 2 – 16 Thrown down from a tower. Suppose the ball in Example 2 9 15 is thrown downward with an initial velocity of 3.00 m>s, instead of being dropped. (a) What then would be its position after 1.00 s and 2.00 s? (b) What would its speed be after 1.00 s and 2.00 s? Compare with the speeds of a dropped ball.
APPROACH Again we use Eq. 2 9 12b, but now v0 is not zero, it is v0 = 3.00 m>s.
SOLUTION (a) At t1 = 1.00 s, the position of the ball as given by Eq. 2 9 12b is y = v0 t + 1 2 at2 = (3.00 m>s) (1.00 s) + 1 2 (9.80 m>s2) (1.00 s) 2 = 7.90 m.
At t2 = 2.00 s (time interval t = 0 to t = 2.00 s), the position is y = v0 t + 1 2 at2 = (3.00 m>s) (2.00 s) + 1 2 (9.80 m>s2) (2.00 s) 2 = 25.6 m.
As expected, the ball falls farther each second than when it is dropped with v0 = 0, Example 2 9 15.
(b) The velocity is obtained from Eq. 2 9 12a: v = v0 + at = 3.00 m>s + (9.80 m>s2) (1.00 s) = 12.8 m>s [at t1 = 1.00 s] = 3.00 m>s + (9.80 m>s2) (2.00 s) = 22.6 m>s. [at t2 = 2.00 s] In Example 2 9 15, when the ball was dropped (v0 = 0), the first term (v0) in these equations was zero, so v = 0 + at = (9.80 m>s2) (1.00 s) = 9.80 m>s [at t1 = 1.00 s] = (9.80 m>s2) (2.00 s) = 19.6 m>s. [at t2 = 2.00 s] NOTE For both Examples 2 9 15 and 2 9 16 the speed increases linearly in time by 9.80 m>s during each second. But the speed of the downwardly thrown ball at any instant is always 3.00 m>s (its initial speed) greater than that of a dropped ball.
FIGURE 2 – 31 An object thrown into the air leaves the thrower’s hand at A, reaches its maximum height at B, and returns to the original position at C. Examples 2 9 17, 2 9 18, 2 9 19, 2 9 20, and 2 9 21.
A C (v = 0) B v v g g +y EXAMPLE 2 – 17 Ball thrown upward. A person throws a ball upward into the air with an initial velocity of 15.0 m>s. Calculate how high it goes. Ignore air resistance.
APPROACH We are not concerned here with the throwing action, but only with the motion of the ball after it leaves the thrower’s hand (Fig. 2 9 31) at point A. Let us choose y to be positive in the upward direction and negative in the downward direction. (This is a different convention from that used in Examples 2 9 15 and 2 9 16, and so illustrates our options.) The acceleration due to gravity is downward and so will have a negative sign, a = -g = -9.80 m>s2. As the ball rises, its speed decreases until it reaches the highest point (B in Fig. 2 9 31), where its speed is zero for an instant; then it descends, with increasing speed.
SOLUTION We consider the time interval from when the ball leaves the throw-er’s hand until the ball reaches the highest point. To determine the maximum height, we calculate the position of the ball when its velocity equals zero (v = 0 at the highest point). At t = 0 (point A in Fig. 2 9 31) we set y0 = 0, v0 = 15.0 m>s, and a = -9.80 m>s2. At time t (maximum height), v = 0, a = -9.80 m>s2, and we wish to find y. We use Eq. 2 9 12c, replacing x with y: v2 = v0 2 + 2ay. We solve this equation for y: y = v2 - v0 2 2a = 0 - (15.0 m>s) 2 2( -9.80 m>s2) = 11.5 m.
The ball reaches a height of 11.5 m above the hand.
62 CHAPTER 2 Describing Motion: Kinematics in One Dimension We did not consider the throwing action in these Examples. Why? Because during the throw, the thrower’s hand is touching the ball and accelerating the ball at a rate unknown to us : the acceleration is not g. We consider only the time when the ball is in the air and the acceleration is equal to g.
Every quadratic equation (where the variable is squared) mathematically produces two solutions. In physics, sometimes only one solution corresponds to the real situation, as we saw in Example 2 9 12, in which case we ignore the “unphysical” solution. But here in Example 2 9 18, both solutions to our equation in t2 are physically meaningful: t = 0 and t = 3.06 s.
CONCEPTUAL EXAMPLE 2 – 19 Two possible misconceptions. Give examples to show the error in these two common misconceptions: (1) that accel-eration and velocity are always in the same direction, and (2) that an object thrown upward has zero acceleration at the highest point (B in Fig. 2 9 31).
RESPONSE Both are wrong. (1) Velocity and acceleration are not necessarily in the same direction. When the ball in Fig. 2 9 31 is moving upward, its velocity is positive (upward), whereas the acceleration is negative (downward). (2) At the highest point (B in Fig. 2 9 31), the ball has zero velocity for an instant. Is the acceleration also zero at this point? No. The velocity near the top of the arc points upward, then becomes zero for an instant (zero time) at the highest point, and then points downward. Gravity does not stop acting, so a = -g = -9.80 m>s2 even there. Thinking that a = 0 at point B would lead to the conclusion that upon reaching point B, the ball would stay there: if the acceleration ( = rate of change of velocity) were zero, the velocity would stay zero at the highest point, and the ball would stay up there without falling. Remember: the acceleration of gravity always points down toward the center of the Earth, even when the object is moving up.
C A U T I O N (1) Velocity and acceleration are not always in the same direction; the acceleration (of gravity) always points down (2) a ≠ 0 even at the highest point of a trajectory A C (v = 0) B v v g g +y FIGURE 2 – 31 (Repeated.) An object thrown into the air leaves the thrower’s hand at A, reaches its maximum height at B, and returns to the original position at C. Examples 2 9 17 , 2 9 18, 2 9 19, 2 9 20, and 2 9 21.
C A U T I O N Quadratic equations have two solutions. Sometimes only one corresponds to reality, sometimes both EXAMPLE 2 – 18 Ball thrown upward, II. In Example 2 9 17, Fig. 2 9 31 (shown here again), how long is the ball in the air before it comes back to the hand?
APPROACH We need to choose a time interval to calculate how long the ball is in the air before it returns to the hand. We could do this calculation in two parts by first determining the time required for the ball to reach its highest point, and then determining the time it takes to fall back down. However, it is simpler to consider the time interval for the entire motion from A to B to C (Fig. 2 9 31) in one step and use Eq. 2 9 12b. We can do this because y is position (or displace-ment from the origin), and not the total distance traveled. Thus, at both points A and C we have y = 0.
SOLUTION We use Eq. 2 9 12b with a = -9.80 m>s2 and find y = y0 + v0 t + 1 2 at2 0 = 0 + (15.0 m>s)t + 1 2 ( -9.80 m>s2)t2.
This equation can be factored (we factor out one t): (15.0 m>s - 4.90 m>s2 t) t = 0.
There are two solutions: t = 0 and t = 15.0 m>s 4.90 m>s2 = 3.06 s.
The first solution (t = 0) corresponds to the initial point (A) in Fig. 2 9 31, when the ball was first thrown from y0 = 0. The second solution, t = 3.06 s, corresponds to point C, when the ball has returned to y = 0. Thus the ball is in the air 3.06 s.
NOTE We have ignored air resistance in these last two Examples, which could be significant, so our result is only an approximation to a real, practical situation.
SECTION 2–7 Freely Falling Objects 63 The acceleration of objects such as rockets and fast airplanes is often given as a multiple of g = 9.80 m>s2. For example, an airplane pulling out of a dive (see Fig. 2 9 32) and undergoing 3.00 g’s would have an acceleration of (3.00)(9.80 m>s2) = 29.4 m>s2.
P R O B L E M S O LV I N G Acceleration in g’s FIGURE 2 – 32 Several airplanes, in formation, are just coming out of a downward dive.
EXAMPLE 2 – 20 Ball thrown upward, III. Let us consider again the ball thrown upward of Examples 2 9 17 and 2 9 18, and make more calculations. Calcu-late (a) how much time it takes for the ball to reach the maximum height (point B in Fig. 2 9 31), and (b) the velocity of the ball when it returns to the thrower’s hand (point C).
APPROACH Again we assume the acceleration is constant, so we can use Eqs. 2 9 12. We have the maximum height of 11.5 m and initial speed of 15.0 m>s from Example 2 9 17. Again we take y as positive upward.
SOLUTION (a) We consider the time interval between the throw (t0 = 0, v0 = 15.0 m>s) and the top of the path (y = +11.5 m, v = 0), and we want to fi nd t. The acceleration is constant at a = -g = -9.80 m>s2. Both Eqs. 2 9 12a and 2 9 12b contain the time t with other quantities known. Let us use Eq. 2 9 12a with a = -9.80 m>s2, v0 = 15.0 m>s, and v = 0: v = v0 + at; setting v = 0 gives 0 = v0 + at, which we rearrange to solve for t: at = -v0 or t = - v0 a = - 15.0 m>s -9.80 m>s2 = 1.53 s.
This is just half the time it takes the ball to go up and fall back to its original position [3.06 s, calculated in Example 2 9 18]. Thus it takes the same time to reach the maximum height as to fall back to the starting point. We might have guessed this from the symmetry of the motion. But be careful. When air resistance cannot be neglected, the symmetry is no longer perfect.
(b) To fi nd the ball’s velocity when it returns to the hand (point C), we consider the time interval from the throw (t0 = 0, v0 = 15.0 m>s) until the ball’s return to the hand, which occurs at t = 3.06 s (as calculated in Example 2 9 18). We use Eq. 2 9 12a again to fi nd v when t = 3.06 s: v = v0 + at = 15.0 m>s - (9.80 m>s2)(3.06 s) = -15.0 m>s.
NOTE The ball has the same speed (magnitude of velocity) when it returns to the starting point as it did initially, but in the opposite direction (this is the meaning of the negative sign). And, as we saw in part (a), the time is the same up as down. Thus the motion is symmetrical about the maximum height.
64 CHAPTER 2 Describing Motion: Kinematics in One Dimension FIGURE 2 – 33 Graphs of (a) y vs. t, (b) v vs. t, for a ball thrown upward, Examples 2 9 17, 2 9 18, 2 9 20, and 2 9 21.
0 0 0.5 1 1.5 2 2.5 3 3.5 2 0 4 6 8 10 12 t (s) y (m) (a) t = 2.37 s t = 0.69 s t = 1.53 s y = 11.5 m 0.5 1 1.5 2 2.5 3 3.5 0 5 -5 -10 -15 -20 10 15 20 t (s) v (m>s) t = 1.53 s 0 (b) EXAMPLE 2 – 21 Ball thrown upward, IV; the quadratic formula. For the ball in Example 2 9 20, calculate at what time t the ball passes a point 8.00 m above the person’s hand. (See Fig. 2 9 31, repeated below.) APPROACH We choose the time interval from the throw (t0 = 0, v0 = 15.0 m>s) until the time t (to be determined) when the ball is at position y = 8.00 m, using Eq. 2 9 12b.
SOLUTION We want to find t, given y = 8.00 m, y0 = 0, v0 = 15.0 m>s, and a = -9.80 m>s2. We use Eq. 2 9 12b: y = y0 + v0 t + 1 2 at2 8.00 m = 0 + (15.0 m>s)t + 1 2 ( -9.80 m>s2)t2.
To solve any quadratic equation of the form at2 + bt + c = 0, where a, b, and c are constants (a is not acceleration here), we use the quadratic formula: t = -b { 2b2 - 4ac 2a .
We rewrite our y equation just above in standard form, at2 + bt + c = 0: (4.90 m>s2)t2 - (15.0 m>s)t + (8.00 m) = 0.
So the coefficient a is 4.90 m>s2, b is -15.0 m>s, and c is 8.00 m. Putting these into the quadratic formula, we obtain t = 15.0 m>s { 2( -15.0 m>s)2 - 4(4.90 m>s2)(8.00 m) 2(4.90 m>s2) , which gives us t = 0.69 s and t = 2.37 s. Are both solutions valid? Yes, because the ball passes y = 8.00 m when it goes up (t = 0.69 s) and again when it comes down (t = 2.37 s).
NOTE Figure 2 9 33 shows graphs of (a) y vs. t and (b) v vs. t for the ball thrown upward in Fig. 2 9 31, incorporating the results of Examples 2 9 17, 2 9 18, 2 9 20, and 2 9 21.
P R O B L E M S O LV I N G Quadratic formula is a very useful tool EXAMPLE 2 – 22 Ball thrown upward at edge of cliff. Suppose that the person of Examples 2 9 17, 2 9 18, 2 9 20, and 2 9 21 is standing on the edge of a cliff, so that the ball can fall to the base of the cliff 50.0 m below, as shown in Fig. 2 9 34. (a) How long does it take the ball to reach the base of the cliff? (b) What is the total distance traveled by the ball? Ignore air resistance (likely to be significant, so our result is an approximation).
APPROACH We again use Eq. 2 9 12b, but this time we set y = -50.0 m, the bottom of the cliff, which is 50.0 m below the initial position (y0 = 0).
A C (v = 0) B v v g g +y FIGURE 2 – 31 (Repeated.) An object thrown into the air leaves the thrower’s hand at A, reaches its maximum height at B, and returns to the original position at C. Examples 2 9 17 , 2 9 18, 2 9 19, 2 9 20, and 2 9 21.
SECTION 2–8 Variable Acceleration; Integral Calculus 65 SOLUTION (a) We use Eq. 2 9 12b with a = -9.80 m>s2, v0 = 15.0 m>s, y0 = 0, and y = -50.0 m: y = y0 + v0 t + 1 2 at2 -50.0 m = 0 + (15.0 m>s)t - 1 2 (9.80 m>s2)t2.
Rewriting in the standard form we have (4.90 m>s2)t2 - (15.0 m>s)t - (50.0 m) = 0.
Using the quadratic formula, we fi nd as solutions t = 5.07 s and t = -2.01 s. The fi rst solution, t = 5.07 s, is the answer we are seeking: the time it takes the ball to rise to its highest point and then fall to the base of the cliff. To rise and fall back to the top of the cliff took 3.06 s (Example 2 9 18); so it took an additional 2.01 s to fall to the base. But what is the meaning of the other solution, t = -2.01 s? This is a time before the throw, when our calculation begins, so it isn’t relevant here. It is outside our chosen time interval, and so is an unphysical solution just as in Example 2 9 12.
(b) For the total distance traveled, Example 2 9 17 told us that the ball moves up 11.5 m, falls 11.5 m back down to the top of the cliff; it then falls down another 50.0 m to the base of the cliff, for a total distance traveled of 2 (11.5 m) + 50.0 m = 73.0 m. Note that the displacement, however, was -50.0 m. Figure 2 9 34b shows the y vs. t graph for this situation.
NOTE The solution t = -2.01 s in part (a) could be meaningful in a different physical situation. Suppose that a person standing on top of a 50.0-m-high cliff sees a rock pass by her at t = 0 moving upward at 15.0 m>s; at what time did the rock leave the base of the cliff, and when did it arrive back at the base of the cliff? The equations will be precisely the same as for our original Example, and the answers t = -2.01 s and t = 5.07 s will be the correct answers. Note that we cannot put all the information for a problem into the mathematics, so we have to use common sense in interpreting results.
EXERCISE H Two balls are thrown from a cliff. One is thrown directly up, the other directly down, each with the same initial speed, and both hit the ground below the cliff. Which ball hits the ground at the greater speed: (a) the ball thrown upward, (b) the ball thrown downward, or (c) both the same? Ignore air resistance.
y y = 0 y = -50 m (a) 1 0 2 4 5 3 6 -40 -50 -30 -20 -10 0 10 t (s) y (m) Base of cliff Hand t = 5.07 s (b) FIGURE 2 – 34 Example 2 9 22. (a) A person stands on the edge of a cliff. A ball is thrown upward, and then falls back down past the thrower to the base of the cliff, 50.0 m below. (b) The y vs. t graph.
2–8 Variable Acceleration; Integral Calculus In this brief optional Section we use integral calculus to derive the kinematic equations for constant acceleration, Eqs. 2 9 12a and b. We also show how calculus can be used when the acceleration is not constant. If you have not yet studied simple integration in your calculus course, you may want to postpone reading this Section until you have. We discuss integration in more detail in Section 7 9 3, where we begin to use it in the physics.
First we derive Eq. 2 9 12a, assuming as we did in Section 2 9 5 that an object has velocity v0 at t = 0 and a constant acceleration a. We start with the defi nition of instantaneous acceleration, a = dv>dt, which we rewrite as dv = a dt.
We take the defi nite integral of both sides of this equation, using the same notation we did in Section 2 9 5, from v = v0 at t0 = 0 to some velocity v at time t: 3 v v = v0 dv = 3 t t0 = 0 a dt which gives, since a = constant, v - v0 = at.
This is Eq. 2 9 12a, v = v0 + at.
66 CHAPTER 2 Describing Motion: Kinematics in One Dimension Next we derive Eq. 2 9 12b starting with the definition of instantaneous velocity, Eq. 2 9 4, v = dx>dt. We rewrite this as dx = v dt or dx = (v0 + at)dt where we substituted in Eq. 2 9 12a.
Now we integrate from x = x0 at t0 = 0 to an arbitrary position x at time t: 3 x x = x0 dx = 3 t t0 = 0 (v0 + at)dt x - x0 = 3 t t0 = 0 v0 dt + 3 t t0 = 0 at dt x - x0 = v0 t + 1 2 at2 since v0 and a are constants. This result is just Eq. 2 9 12b, x = x0 + v0 t + 1 2 at2.
Finally let us use calculus to find velocity and displacement, given an accel-eration that is not constant but varies in time.
Some problems in kinematics can be solved using Numerical Integration, which is discussed in Appendix C.
EXAMPLE 2 – 23 Integrating a time-varying acceleration. An experimental vehicle starts from rest (v0 = 0) at t = 0 and accelerates for a few seconds at a rate given by a = (7.00 m>s3)t. What is (a) its velocity and (b) its displacement 2.00 s later?
APPROACH We cannot use Eqs. 2 9 12 because a is not constant. We integrate the acceleration a = dv>dt over time to find v as a function of time; and then integrate v = dx>dt to get the displacement.
SOLUTION From the definition of acceleration, a = dv>dt, we have dv = a dt.
We take the integral of both sides from v = 0 at t = 0 to velocity v at an arbitrary time t: 3 v 0 dv = 3 t 0 a dt v = 3 t 0 (7.00 m>s3)t dt = (7.00 m>s3) ¢t2 2 ≤2 0 t = (7.00 m>s3) ¢t2 2 - 0≤ = (3.50 m>s3)t2.
At t = 2.00 s, v = (3.50 m>s3)(2.00 s)2 = 14.0 m>s.
(b) To get the displacement, we assume x0 = 0 and start with v = dx>dt which we rewrite as dx = v dt. Then we integrate from x = 0 at t = 0 to position x at time t: 3 x 0 dx = 3 t 0 v dt x = 3 2.00 s 0 (3.50 m/s3)t2 dt = (3.50 m/s3) t3 3 2 0 2.00 s = 9.33 m.
In sum, at t = 2.00 s, v = 14.0 m>s and x = 9.33 m.
Questions 67 Summary [The Summary that appears at the end of each Chapter in this book gives a brief overview of the main ideas of the Chapter. The Summary cannot serve to give an understanding of the material, which can be accomplished only by a detailed reading of the Chapter.] Kinematics deals with the description of how objects move. The description of the motion of any object must always be given relative to some particular reference frame.
The displacement of an object is the change in position of the object.
Average speed is the distance traveled divided by the elapsed time or time interval, ∆t, the time period over which we choose to make our observations. An object’s average velocity over a particular time interval ∆t is its displacement ∆x during that time interval, divided by ∆t: v = ∆x ∆t . (2 – 2) The instantaneous velocity, whose magnitude is the same as the instantaneous speed, is defined as the average velocity taken over an infinitesimally short time interval (∆t S 0): v = lim ∆tS0 ∆x ∆t = dx dt , (2 – 4) where dx>dt is the derivative of x with respect to t.
On a graph of position vs. time, the slope is equal to the instantaneous velocity.
Acceleration is the change of velocity per unit time. An object’s average acceleration over a time interval ∆t is a = ∆v ∆t , (2 – 5) where ∆v is the change of velocity during the time interval ∆t.
Instantaneous acceleration is the average acceleration taken over an infinitesimally short time interval: a = lim ∆tS0 ∆v ∆t = dv dt . (2 – 6) On a graph of velocity vs. time, the slope is equal to the instantaneous acceleration.
If an object has position x0 and velocity v0 at time t = 0 and moves in a straight line with constant acceleration, the velocity v and position x at a later time t are related to the acceleration a, the initial position x0, and the initial velocity v0 by Eqs. 2 9 12: v = v0 + at, x = x0 + v0 t + 1 2 at2, v2 = v0 2 + 2a(x - x0), (2 – 12) v = v + v0 2 . Objects that move vertically near the surface of the Earth, either falling or having been projected vertically up or down, move with the constant downward acceleration due to gravity, whose magnitude is g = 9.80 m>s2 if air resistance can be ignored.
[Integral calculus can be used to derive the kinematic Equations 2 9 12 and to solve Problems involving varying accel-eration. Numerical integration is also a useful tool.] Questions 1. Does a car speedometer measure speed, velocity, or both? Explain. 2. Can an object have a varying speed if its velocity is constant? Can it have varying velocity if its speed is constant? If yes, give examples in each case. 3. When an object moves with constant velocity, does its average velocity during any time interval differ from its instantaneous velocity at any instant? Explain. 4. If one object has a greater speed than a second object, does the first necessarily have a greater acceleration? Explain, using examples. 5. Compare the acceleration of a motorcycle that accelerates from 80 km>h to 90 km>h with the acceleration of a bicycle that accelerates from rest to 10 km>h in the same time. 6. Can an object have a northward velocity and a southward acceleration? Explain. 7. Can the velocity of an object be negative when its accel-eration is positive? What about vice versa? If yes, give examples. 8. Give an example where both the velocity and acceleration are negative. 9. Two cars emerge side by side from a tunnel. Car A is traveling with a speed of 60 km>h and has an accelera-tion of 40 km>h>min. Car B has a speed of 40 km>h and has an acceleration of 60 km>h>min. Which car is passing the other as they come out of the tunnel? Explain your reasoning. 10. Can an object be increasing in speed as its acceleration decreases? If so, give an example. If not, explain. 11. A baseball player hits a ball straight up into the air. It leaves the bat with a speed of 120 km>h. Ignoring air resistance, how fast would the ball be traveling when the catcher catches it (at the same height it left the bat)? Explain. 12. As a freely falling object speeds up, what is happening to its acceleration : does it increase, decrease, or stay the same? (a) Ignore air resistance. (b) Consider air resistance. 13. You travel from point A to point B in a car moving at a constant speed of 70 km>h. Then you travel the same distance from point B to another point C, moving at a constant speed of 90 km>h. Is your average speed 80 km>h for the entire trip from A to C? Explain why or why not. 14. Can an object have zero velocity and nonzero acceleration at the same time? Give examples. 15. Can an object have zero acceleration and nonzero velocity at the same time? Give examples. 16. Which of these motions is not at constant acceleration: a rock falling from a cliff, an elevator moving from the second floor to the fifth floor making stops along the way, a dish resting on a table? Explain your answers. 17. Discuss two conditions given in Section 2 9 7 for being able to use a constant acceleration of magnitude g = 9.8 m>s2. Give an example in which one of these conditions would not be met and would not even be a reasonable approxima tion of motion. [Hint: Carefully read Section 2 9 7, especially page 60.] 68 CHAPTER 2 Describing Motion: Kinematics in One Dimension 18. Describe in words the motion plotted in Fig. 2 9 35 in terms of velocity, acceleration, etc. [Hint: First try to duplicate the motion plotted by walking or moving your hand.] 20 10 00 10 20 30 40 50 t (s) x (m) FIGURE 2 – 35 Question 18. 19. Describe in words the motion of the object graphed in Fig. 2 9 36. v (m>s) t (s) 40 30 20 10 00 10 20 30 40 50 60 70 80 90 100 110 120 FIGURE 2 – 36 Question 19.
MisConceptual Questions [List all answers that are valid, and ignore air resistance.] 1. In which of the following cases does a car have a negative velocity and a positive acceleration? A car that is traveling in the (a) -x direction at a constant 20 m>s.
(b) -x direction increasing in speed. (c) +x direction increasing in speed. (d) -x direction decreasing in speed.
(e) +x direction decreasing in speed. 2. At time t = 0 an object is traveling to the right along the +x axis at a speed of 10.0 m>s with constant acceleration of -2.0 m>s2. Which statement is true?
(a) The object will slow down, eventually coming to a complete stop.
(b) The object cannot have a negative acceleration and be moving to the right.
(c) The object will continue to move to the right, slowing down but never coming to a complete stop.
(d) The object will slow down, momentarily stopping, then pick up speed moving to the left. 3. You drive 4 km at 30 km>h and then another 4 km at 50 km>h. What is your average speed for the whole 8-km trip? (a) More than 40 km>h. (c) Less than 40 km>h.
(b) Equal to 40 km>h. (d) Not enough information. 4. Two cars start from rest and travel a distance d with constant acceleration. The acceleration of car B is four times that of car A. After each has traveled distance d, (a) car B is moving 16 times as fast as car A.
(b) car B is moving 8 times as fast as car A.
(c) car B is moving 4 times as fast as car A.
(d) car B is moving 2 times as fast as car A. 5. A rock is thrown straight up rising to a maximum height before falling back down.
(a) The acceleration is constant for the entire trip.
(b) The velocity is constant for the entire trip.
(c) The magnitudes of both the velocity and acceleration decrease on the way up and increase on the way down.
(d) The acceleration is constant for the entire trip except at the top where it is 0. 6. A ball is dropped from the top of a tall building. At the same instant, a second ball is thrown upward from ground level. When the two balls pass one another, one on the way up, the other on the way down, compare the magni tudes of their acceleration: (a) The acceleration of the dropped ball is greater.
(b) The acceleration of the ball thrown upward is greater. (c) The acceleration of both balls is the same.
(d) The acceleration changes during the motion, so you cannot predict the exact value when the two balls pass each other. (e) The accelerations are in opposite directions. 7. You drop a rock off a bridge. When the rock has fallen 4 m, you drop a second rock. As the two rocks continue to fall, what happens to their velocities?
(a) Both increase at the same rate.
(b) The velocity of the first rock increases faster than the velocity of the second.
(c) The velocity of the second rock increases faster than the velocity of the first.
(d) Both velocities stay constant. 8. Two objects are dropped from a bridge, an interval of 1.0 s apart. During the time that both objects continue to fall, their separation (a) decreases at first, but then stays constant.
(b) increases at first, but then stays constant.
(c) decreases.
(d) stays constant. (e) increases. 9. A ball is thrown downward at a speed of 20 m>s. Choosing the +y axis pointing up and neglecting air resistance, which equation(s) would correctly describe the motions? The acceleration due to gravity is g = 9.8 m>s2 downward.
(a) v = (20 m>s) - gt.
(b) y = y0 + ( -20 m>s)t - (1>2)gt2.
(c) v2 = (20 m>s) 2 - 2g(y - y0).
(d) (20 m>s) = (v + v0) >2.
(e) All of the above.
Problems 69 10. A car travels along the x axis with increasing speed. We are not sure if it is moving to the left or to the right. Which of the graphs in Fig. 2 9 37 could possibly represent the motion of the car? 11. Two objects start at the same place at the same time and move along the same straight line (the x axis). Figure 2 9 38 shows the position x as a function of time t for each object. At point A, what must be true about the motion of the objects? (More than one statement may be correct.) (a) Both have the same instantaneous speed.
(b) Both have the same instantaneous velocity. (c) Both are at the same position.
(d) Both have traveled the same total distance. (e) With respect to the start, both have the same average velocity. O x A t FIGURE 2 – 38 MisConceptual Question 11.
Problems [The Problems at the end of each Chapter are ranked I, II, or III according to estimated difficulty, with level I Problems being easiest. Level III are meant as challenges for the best students. The Problems are arranged by Section, meaning that the reader should have read up to and including that Section, but not only that Section : Problems often depend on earlier material. Next is a set of “General Problems” not arranged by Section and not ranked.] (Note: In Problems, assume a number like 6.4 is accurate to {0.1; and 950 is {10 unless 950 is said to be “precisely” or “very nearly” 950, in which case assume 950 { 1. See Section 1 9 3.) 2 – 1 to 2 – 3 Speed and Velocity 1. (I) If you are driving 85 km>h along a straight road and you look to the side for 2.0 s, how far do you travel during this inattentive period? 2. (I) What must your car’s average speed be in order to travel 235 km in 2.85 h? 3. (I) A particle at t1 = -2.0 s is at x1 = 5.2 cm and at t2 = 3.4 s is at x2 = 8.5 cm. What is its average velocity over this time interval? Can you calculate its average speed from these data? Why or why not? 4. (II) According to a rule-of-thumb, each five seconds between a lightning flash and the following thunder gives the distance to the flash in miles. (a) Assuming that the light from the flash arrives in essentially no time at all, estimate the speed of sound in m>s from this rule. (b) What would be the rule for kilometers? 5. (II) You are driving home from school steadily at 95 km>h for 210 km. It then begins to rain and you slow to 65 km>h. You arrive home after driving 4.5 h. (a) How far is your hometown from school? (b) What was your average speed? 6. (II) A horse trots away from its trainer in a straight line, moving 38 m away in 7.4 s. It then turns abruptly and gallops halfway back in 1.8 s. Calculate (a) its average speed and (b) its average velocity for the entire trip, using “away from the trainer” as the positive direction. 7. (II) A person jogs eight complete laps around a 400-m track in a total time of 14.5 min. Calculate (a) the average speed and (b) the average velocity, in m>s. 8. (II) Every year the Earth travels about 109 km as it orbits the Sun. What is Earth’s average speed in km>h? 9. (II) A car traveling 95 km>h is 310 m behind a truck traveling 75 km>h. How long will it take the car to reach the truck? 10. (II) Calculate (a) the average speed and (b) average velocity of a round trip: the outgoing 280 km is covered at 95 km>h, followed by a 1.0-h lunch break, and the return 280 km is covered at 55 km>h. 11. (II) Two locomotives approach each other on parallel tracks. Each has a speed of 155 km>h with respect to the ground. If they are initially 9.5 km apart, how long will it be before they reach each other? (See Fig. 2 9 39.) 9.5 km v = 155 km>h v = 155 km>h FIGURE 2 – 39 Problem 11.
x t (a) x t (b) x t (c) x t (d) x t (e) FIGURE 2 – 37 MisConceptual Question 10.
70 CHAPTER 2 Describing Motion: Kinematics in One Dimension 12. (II) Digital bits on a 12.0-cm diameter audio CD are encoded along an outward spiraling path that starts at radius R1 = 2.5 cm and finishes at radius R2 = 5.8 cm. The dis tance between the centers of neighboring spiral- windings is 1.6 mm ( = 1.6 10-6 m). (a) Determine the total length of the spiraling path. [Hint: Imagine “unwinding” the spiral into a straight path of width 1.6 mm, and note that the original spiral and the straight path both occupy the same area.] (b) The CD player adjusts the rotation of the CD so that the player’s readout laser reads along the spiral path at a constant rate of about 1.2 m>s. Estimate the maximum playing time of such a CD. 13. (II) The position of a small object is given by x = 27 + 10 t - 2 t3, where t is in seconds and x in meters. (a) Plot x as a function of t from t = 0 to t = 3.0 s. (b) Find the average velocity of the object between 0 and 3.0 s. (c) At what time between 0 and 3.0 s is the instantaneous velocity zero? 14. (II) An airplane travels 1900 km at a speed of 720 km>h, and then encounters a tailwind that boosts its speed to 990 km>h for the next 2700 km. What was the total time for the trip? What was the average speed of the plane for this trip? [Hint: Does Eq. 2 9 12d apply?] 15. (II) Two ships need to arrive at a site in the middle of the ocean at the same time. They start out at the same time from positions equally distant from the arrival site. They travel at different velocities but both go in a straight line. The first ship travels at an average velocity of 20 km>h for the first 600 km, 40 km>h for the next 800 km, and 20 km>h for the final 600 km. The second ship can only sail at constant velocity. What is the magnitude of that velocity? 16. (II) The position of an object along a straight tunnel as a function of time is plotted in Fig. 2 9 40. What is its instanta-neous velocity (a) at t = 10.0 s and (b) at t = 30.0 s? What is its average velocity (c) between t = 0 and t = 5.0 s, (d) between t = 25.0 s and t = 30.0 s, and (e) between t = 40.0 s and t = 50.0 s?
20 10 00 10 20 30 40 50 t (s) x (m) FIGURE 2 – 40 Problems 16, 17, and 18. 17. (II) In Fig. 2 9 40, (a) during what time intervals, if any, is the velocity constant? (b) At what time is the velocity greatest? (c) At what time, if any, is the velocity zero? (d) Does the object move in one direction or in both directions during the time shown? 18. (III) Sketch the v vs. t graph for the object whose displace ment as a function of time is given by Fig. 2 9 40. 19. (III) A bowling ball traveling with constant speed hits the pins at the end of a bowling lane 16.5 m long. The bowler hears the sound of the ball hitting the pins 2.75 s after the ball is released from his hands. What is the speed of the ball, assuming the speed of sound is 340 m>s? 20. (III) An automobile traveling 95 km>h overtakes a 1.50‑km- long train traveling in the same direction on a track parallel to the road. If the train’s speed is 75 km>h, how long does it take the car to pass it, and how far will the car have traveled in this time? See Fig. 2 9 41. What are the results if the car and train are traveling in opposite directions?
= 95 km>h = 75 km>h 1.50 km v v FIGURE 2 – 41 Problem 20.
2 – 4 Acceleration 21. (I) A sprinter accelerates from rest to 9.00 m>s in 1.48 s. What is her acceleration in (a) m>s2; (b) km>h2? 22. (I) A bicyclist in the Tour de France crests a mountain pass as he moves at 15 km>h. At the bottom, 4.0 km farther, his speed is 65 km>h. Estimate his average acceleration (in m>s2) while riding down the mountain. 23. (II) A sports car moving at constant velocity travels 120 m in 5.0 s. If it then brakes and comes to a stop in 3.7 s, what is the magnitude of its acceleration (assumed constant) in m>s2, and in g’s (g = 9.80 m>s2)? 24. (II) At highway speeds, a particular automobile is capable of an acceleration of about 1.8 m>s2. At this rate, how long does it take to accelerate from 65 km>h to 120 km>h? 25. (II) A car moving in a straight line starts at x = 0 when t0 = 0. It passes the point x = 25.0 m with a speed of 11.0 m>s at t = 3.00 s. It passes the point x = 385 m with a speed of 45.0 m>s at t = 20.0 s. Find (a) the average velocity, and (b) the average acceleration, between t = 3.00 s and t = 20.0 s. 26. (II) A particle moves along the x axis. Its position as a function of time is given by x = 4.8 t + 7.3 t2, where t is in seconds and x is in meters. What is the acceleration as a function of time? 27. (II) The position of an object is given by x = At + Bt2, where x is in meters and t is in seconds. (a) What are the units of A and B? (b) What is the acceleration as a func-tion of time? (c) What are the velocity and acceleration at t = 6.0 s? (d) What is the velocity as a function of time if x = At + Bt -3? 28. (II) The position of a race car, which starts from rest at t = 0 and moves in a straight line, is given as a function of time in the following Table. Estimate (a) its velocity and (b) its acceleration as a function of time. Display each in a Table and on a graph.
t (s) 0 0.25 0.50 0.75 1.00 1.50 2.00 2.50 x (m) 0 0.11 0.46 1.06 1.94 4.62 8.55 13.79 t (s) 3.00 3.50 4.00 4.50 5.00 5.50 6.00 x (m) 20.36 28.31 37.65 48.37 60.30 73.26 87.16 29. (II) A car traveling 25.0 m>s passes a second car which is at rest. When the cars are right next to each other, the first car slows down at a constant rate of 2.0 m>s2 and the second car starts to accelerate at the same constant rate. When will the two cars be next to each other again?
Problems 71 30. (II) Figure 2 9 42 shows the velocity of a train as a function of time. (a) At what time was its velocity greatest? (b) During what periods, if any, was the velocity constant? (c) During what periods, if any, was the acceleration constant? (d) When was the magnitude of the acceleration greatest? 31. (II) A sports car accelerates approximately as shown in the velocity 9 time graph of Fig. 2 9 43. (The short flat spots in the curve represent manual shifting of the gears.) Estimate the car’s average acceleration in (a) second gear and (b) fourth gear. v (m>s) t (s) 0 10 20 30 40 10 20 30 40 50 0 1st gear 2nd gear 5th gear 3rd gear 4th gear 32. (III) A fugitive tries to hop on a freight train traveling at a constant speed of 5.0 m>s. Just as an empty box car passes him, the fugitive starts from rest and accelerates at a = 1.4 m>s2 to his maximum speed of 6.0 m>s, which he then maintains. (a) How long does it take him to catch up to the empty box car? (b) What is the distance traveled to reach the box car? 2 – 5 to 2 – 6 Motion at Constant Acceleration 33. (I) A car slows down from 26 m>s to rest in a distance of 88 m. What was its acceleration, assumed constant? 34. (I) A car slows down from 28 m>s to rest in 6.3 s. What was its (constant) acceleration? 35. (I) A car accelerates from 13 m>s to 22 m>s in 6.5 s. What was its acceleration? How far did it travel in this time? Assume constant acceleration. 36. (II) A world-class sprinter can reach a top speed (of about 11.5 m>s) in the first 18.0 m of a race. What is the average acceleration of this sprinter and how long does it take her to reach that speed? 37. (II) A car slows down uniformly from a speed of 28.0 m>s to rest in 8.60 s. How far did it travel in that time? 38. (II) In coming to a stop, an old truck leaves skid marks 45 m long on the highway. Assuming a deceleration of 6.00 m>s2, estimate the speed of the truck just before braking. 39. (II) A baseball pitcher throws a baseball with a speed of 43 m>s. Estimate the average acceleration of the ball during the throwing motion. In throwing the baseball, the pitcher accelerates it through a displacement of about 3.5 m, from behind the body to the point where it is released (Fig. 2 9 44). 40. (II) A car traveling at 95 km>h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80 m. What was the magnitude of the average acceleration of the driver during the collision? Express the answer in terms of “g’s,” where 1.00 g = 9.80 m>s2. 41. (II) A car traveling 85 km>h slows down at a constant 0.50 m>s2 just by “letting up on the gas.” Calculate (a) the distance the car coasts before it stops, (b) the time it takes to stop, and (c) the distance it travels during the first and fifth seconds. 42. (II) Determine the stopping distances for an automobile going a constant initial speed of 95 km>h in the +x direction, and human reaction time of 0.40 s: (a) for an acceleration a = -2.5 m>s2; (b) for a = -5.5 m>s2. 43. (II) A driver is traveling 18.0 m>s when she sees a red light ahead. Her car is capable of decelerating at a rate of 3.65 m>s2. If it takes her 0.380 s to get the brakes on and she is 24.0 m from the intersection when she sees the light, will she be able to stop in time? How far from the beginning of the intersection will she be, and in which direction? 44. (II) Show that v = (v + v0) >2 (see Eq. 2 9 12d) is not valid when the acceleration a = A + Bt, where A and B are non-zero constants. 45. (II) An 85-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m>s when it passes a railway worker who is standing 180 m from where the front of the train started. What will be the speed of the last car as it passes the worker? (See Fig. 2 9 45.) 85 m v = 18 m>s FIGURE 2 – 45 Problem 45. 46. (II) A space vehicle accelerates uniformly from 85 m>s at t = 0 to 162 m>s at t = 10.0 s. How far did it move between t = 2.0 s and t = 7.0 s? 47. (II) A runner hopes to complete the 10,000-m run in less than 30.0 min. After running at constant speed for exactly 27.0 min, there are still 1200 m to go. The runner must then accelerate at 0.20 m>s2 for how many seconds in order to achieve the desired time? v (m>s) t (s) 40 30 20 10 00 10 20 30 40 50 60 70 80 90 100 110 120 FIGURE 2 – 42 Problem 30.
3.5 m FIGURE 2 – 44 Problem 39.
FIGURE 2 – 43 Problem 31. The velocity of a car as a function of time, starting from a dead stop. The flat spots in the curve represent gear shifts.
72 CHAPTER 2 Describing Motion: Kinematics in One Dimension 48. (III) Mary and Sally are in a foot race (Fig. 2 9 46). When Mary is 22 m from the fi nish line, she has a speed of 4.0 m>s and is 5.0 m behind Sally, who has a speed of 5.0 m>s. Sally thinks she has an easy win and so, during the remaining portion of the race, slows down at a constant rate of 0.40 m>s2 to the fi nish line. What constant acceleration does Mary now need during the remaining portion of the race, if she wishes to cross the fi nish line side-by-side with Sally? Finish 5.0 m 22 m Mary Sally 4.0 m>s 5.0 m>s FIGURE 2 – 46 Problem 48. 49. (III) An unmarked police car traveling a constant 95 km>h is passed by a speeder traveling 135 km>h. Precisely 1.00 s after the speeder passes, the police offi cer steps on the accelerator; if the police car’s acceleration is 2.60 m>s2, how much time passes before the police car overtakes the speeder (assumed moving at constant speed)? 50. (III) Assume in Problem 49 that the speeder’s speed is not known. If the police car accelerates uniformly at 2.60 m>s2 and overtakes the speeder after accelerating for 7.00 s, what was the speeder’s speed? 51. (III) A runner completes a 400-meter race in 55.0 s. The 55.0 seconds is made up of a 0.15 s reaction time from the starting sound until the runner starts moving followed by 30.0 m of constant acceleration and then 370 m at constant speed. What are the values of the acceleration and the constant speed?
2 – 7 Freely Falling Objects (neglect air resistance) 52. (I) A stone is dropped from the top of a cliff. It is seen to hit the ground below after 3.25 s. How high is the cliff? 53. (I) Estimate (a) how long it took King Kong to fall straight down from the top of the Empire State Building (380 m high), and (b) his velocity just before “landing.” 54. (I) If a car rolls gently (v0 = 0) off a vertical cliff, how long does it take it to reach 55 km>h? 55. (II) A ball player catches a ball 2.6 s after throwing it verti-cally upward. With what speed did he throw it, and what height did it reach? 56. (II) A baseball is hit almost straight up into the air with a speed of 22 m>s. Estimate (a) how high it goes, and (b) how long it is in the air. (c) What factors make this an estimate? 57. (II) A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth? 58. (II) The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fi xed point on their body) of about 120 cm. (a) What is their initial “launch” speed off the ground? (b) How long are they in the air? 59. (II) A stone is thrown vertically upward with a speed of 18.0 m>s. (a) How fast is it moving when it is at a height of 11.0 m? (b) How much time is required to reach this height? (c) Why are there two answers to (b)? 60. (II) For an object falling freely from rest, show that the distance traveled during each successive second increases in the ratio of successive odd integers (1, 3, 5, etc.). (This was fi rst shown by Galileo.) See Figs. 2 9 27 and 2 9 30. 61. (II) If there were no air resistance, how long would it take a free-falling skydiver to fall from a plane at 3800 m to an altitude of 450 m, where she will open her parachute? What would her speed be at 450 m? (In reality, air resistance will restrict her speed to perhaps 150 km>h.) 62. (II) Pelicans tuck their wings and free-fall straight down when diving for fi sh. Suppose a pelican starts its dive from a height of 16.0 m and cannot change its path once committed. If it takes a fi sh 0.20 s to perform evasive action, at what minimum height must it spot the pelican to escape? Assume the fi sh is at the surface of the water. 63. (II) A stone is thrown ver tically upward with a speed of 15.5 m>s from the edge of a cliff 75.0 m high (Fig. 2 9 47). (a) How much later does it reach the bottom of the cliff? (b) What is its speed just before hitting? (c) What total distance did it travel? 64. (II) A rocket rises vertically, from rest, with an accelera-tion of 3.2 m>s2 until it runs out of fuel at an altitude of 725 m. After this point, its acceleration is that of gravity, downward. (a) What is the velocity of the rocket when it runs out of fuel? (b) How long does it take to reach this point? (c) What maximum altitude does the rocket reach? (d) How much time (total) does it take to reach maximum altitude? (e) With what velocity does it strike the Earth? (f) How long (total) is it in the air? 65. (II) Suppose you adjust your garden hose nozzle for a fast stream of water. You point the nozzle vertically upward at a height of 1.8 m above the ground (Fig. 2 9 48). When you quickly turn off the nozzle, you hear the water striking the ground next to you for another 2.5 s. What is the water speed as it leaves the nozzle? FIGURE 2 – 47 Problem 63.
y y = 0 y = -75 m FIGURE 2 – 48 Problem 65.
1.8 m 66. (II) A helicopter is ascending vertically with a constant speed of 6.40 m>s. At a height of 105 m above the Earth, a package is dropped from the helicopter. How much time does it take for the package to reach the ground? [Hint: What is v0 for the pack age?] 67. (II) Roger sees water balloons fall past his window. He notices that each balloon strikes the sidewalk 0.83 s after passing his window, 15 m above the sidewalk. (a) How fast are the balloons traveling when they pass Roger’s window? (b) Assuming the balloons are being released from rest, from what height are they being released? 68. (II) A baseball is seen to pass upward by a window with a vertical speed of 13 m>s. If the ball was thrown by a person 18 m below on the street, (a) what was its initial speed, (b) what altitude does it reach, (c) when was it thrown, and (d) when does it reach the street again? 69. (III) A falling stone takes 0.28 s to travel past a window 2.2 m tall (Fig. 2 9 49). From what height above the top of the window did the stone fall? 70. (III) A toy rocket moving vertically upward passes by a 2.0-m-high window whose base is 8.0 m above the ground. The rocket takes 0.15 s to travel the 2.0 m height of the window. What was the launch speed of the rocket, and how high will it go? Assume the propellant is burned very quickly at blastoff. 71. (III) A ball is dropped from the top of a 55.0-m-high cliff. At the same time, a carefully aimed stone is thrown straight up from the bottom of the cliff with a speed of 24.0 m>s. The stone and ball collide part way up. How far above the base of the cliff does this happen?
2 – 8 Variable Acceleration; Calculus 72. (II) Given v(t) = 25 + 18 t, where v is in m>s and t is in s, use calculus to determine the total displacement from t1 = 1.3 s to t2 = 3.6 s. 73. (III) The acceleration of a particle is given by a = A1t where A = 3.0 m>s5>2. At t = 0, v = 7.5 m>s and x = 0. (a) What is the velocity as a function of time? (b) What is the displacement as a function of time? (c) What are the acceleration, velocity, and displacement at t = 5.0 s? 74. (III) Air resistance acting on a falling body can be taken into account by the approximate relation for the acceleration: a = dv dt = g - kv, where k is a constant. (a) Derive a formula for the velocity of the body as a function of time assuming it starts from rest (v = 0 at t = 0). [Hint: Change variables by setting u = g - kv.] (b) Determine an expression for the terminal velocity, which is the maximum value the velocity reaches.
General Problems 75. The acceleration due to gravity on the Moon is about one-sixth what it is on Earth. If an object is thrown verti-cally upward on the Moon, how many times higher will it go than it would on Earth, assuming the same initial velocity? 76. A person jumps out a fourth-story window 18.0 m above a firefighter’s safety net. The survivor stretches the net 1.0 m before coming to rest, Fig. 2 9 50. (a) What was the average deceleration experienced by the survivor when she was slowed to rest by the net? (b) What would you do to make it “safer” (that is, to generate a smaller deceleration): would you stiffen or loosen the net? Explain. 77. A person who is properly restrained by an over-the-shoulder seat belt has a good chance of surviving a car collision if the deceleration does not exceed 30 “g’s” (1.00 g = 9.80 m>s2). Assuming uniform deceleration at 30 g’s, calculate the dis tance over which the front end of the car must be designed to collapse if a crash brings the car to rest from 95 km>h. 78. The position of a ball rolling in a straight line is given by x = 2.0 - 3.6 t + 1.7t2, where x is in meters and t in seconds. (a) What do the numbers 2.0, 3.6, and 1.7 refer to? (b) What are the units of each of these numbers? (c) Determine the position of the ball at t = 1.0 s, 2.0 s, and 3.0 s. (d) What is the average velocity over the interval t = 1.0 s to t = 3.0 s? 79. In a lecture demonstration, a 3.0-m-long vertical string with ten bolts tied to it at equal intervals is dropped from the ceiling of the lecture hall. The string falls on a tin plate, and the class hears the clink of each bolt as it hits the plate. (a) The sounds will not occur at equal time inter vals. Why? (b) Will the time between clinks increase or decrease as the string falls? (c) How could the bolts be tied so that the clinks occur at equal intervals? (Assume the string is vertical with the bottom bolt touching the tin plate when the string is released.) 80. Two students are asked to find the height of a particular building using a barometer. Instead of using the barometer as an altitude-measuring device, they take it to the roof of the building and drop it off, timing its fall. One student reports a fall time of 2.0 s, and the other, 2.3 s. What % difference does the 0.3 s make for the estimates of the building’s height?
General Problems 73 FIGURE 2 – 49 Problem 69.
To travel this distance took 0.28 s 2.2 m 18.0 m 1.0 m FIGURE 2 – 50 Problem 76.
74 CHAPTER 2 Describing Motion: Kinematics in One Dimension 81. Consider the street pattern shown in Fig. 2 9 51. Each inter-section has a traffic signal, and the speed limit is 40 km>h. Suppose you are driving from the west at the speed limit. When you are 10.0 m from the first intersection, all the lights turn green. The lights are green for 13.0 s each. (a) Calculate the time needed to reach the third stoplight. Can you make it through all three lights without stopping? (b) Another car was stopped at the first light when all the lights turned green. It can accelerate at the rate of 2.00 m>s2 to the speed limit. Can the second car make it through all three lights without stopping? By how many seconds would it make it, or not make it?
East West Speed limit 40 km>h 50 m 15 m Your car 15 m 70 m 15 m 10 m FIGURE 2 – 51 Problem 81. 82. Suppose a car manufacturer tested its cars for front-end collisions by hauling them up on a crane and dropping them from a certain height. (a) Show that the speed just before a car hits the ground, after falling from rest a vertical distance H, is given by 12gH. What height corresponds to a colli-sion at (b) 35 km>h? (c) 95 km>h? 83. A stone is dropped from the roof of a high building. A second stone is dropped 1.50 s later. How far apart are the stones when the second one has reached a speed of 12.0 m>s? 84. A person jumps off a diving board 4.0 m above the water’s surface into a deep pool. The person’s downward motion stops 1.8 m below the surface of the water. Esti-mate the average deceleration of the person while under the water. 85. A police car at rest is passed by a speeder traveling at a constant 140 km>h. The police officer takes off in hot pursuit and catches up to the speeder in 850 m, maintaining a constant acceleration. (a) Qualitatively plot the position vs. time graph for both cars from the police car’s start to the catch-up point. Calculate (b) how long it took the police officer to overtake the speeder, (c) the required police car acceleration, and (d) the speed of the police car at the over-taking point. (e) This last result is unrealistic : so which assumptions do we have to reconsider? 86. Agent Bond is standing on a bridge, 15 m above the road below, and his pursuers are getting too close for comfort. He spots a flatbed truck approaching at 25 m>s, which he measures by knowing that the telephone poles the truck is passing are 25 m apart in this region. The roof of the truck is 3.5 m above the road, and Bond quickly calculates how many poles away the truck should be when he drops down from the bridge onto the truck, making his getaway. How many poles is it? 87. Two children are playing on two trampolines. The first child bounces up one-and-a-half times higher than the second child. The initial speed upwards of the second child is 4.0 m>s. (a) Find the maximum height the second child reaches. (b) What is the initial speed of the first child? (c) How long was the first child in the air? 88. In putting, the force with which a golfer strikes a ball is planned so that the ball will stop within some small distance of the cup, say 1.0 m long or short, in case the putt is missed. Accomplishing this from an uphill lie (that is, putting the ball downhill, see Fig. 2 9 52) is more difficult than from a downhill lie. To see why, assume that on a particular “green” the ball decelerates constantly at 1.8 m>s2 going downhill, and constantly at 2.6 m>s2 going uphill. Suppose we have an uphill lie 7.0 m from the cup. Calculate the allowable range of initial velocities we may impart to the ball so that it stops in the range 1.0 m short to 1.0 m long of the cup. Do the same for a downhill lie 7.0 m from the cup. What in your results suggests that the downhill putt is more difficult?
7.0 m 7.0 m Downhill lie Green Uphill lie FIGURE 2 – 52 Problem 88. 89. A person driving her car at 35 km>h approaches an intersec-tion just as the traffic light turns yellow. She knows that the yellow light lasts only 2.0 s before turning to red, and she is 28 m away from the near side of the intersection (Fig. 2 9 53). Should she try to stop, or should she speed up to cross the intersection before the light turns red? The intersection is 15 m wide. Her car’s maximum deceleration is -5.8 m>s2, whereas it can accelerate from 45 km>h to 65 km>h in 6.0 s. Ignore the length of her car and her reaction time. 15 m 28 m +x FIGURE 2 – 53 Problem 89. 90. A car is behind a truck going 18 m>s on the highway. The car’s driver looks for an opportunity to pass, guessing that his car can accelerate at 0.60 m>s2 and that he has to cover the 20-m length of the truck, plus 10 m of extra space at the rear of the truck and 10 m more at the front of it. In the oncoming lane, he sees a car approaching, probably at the speed limit, 25 m>s (55 mph). He estimates that the car is about 500 m away. Should he attempt the pass? Give details. 91. A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 4.1 s later. If the speed of sound is 340 m>s, how high is the cliff? 92. A conveyor belt is used to send burgers through a grill ing machine. If the grilling machine is 1.2 m long and the burgers require 2.8 min to cook, how fast must the conveyor belt travel? If the burgers are spaced 25 cm apart, what is the rate of burger production (in burgers>min)? 93. A rock is thrown vertically upward with a speed of 15.0 m>s. Exactly 1.00 s later, a ball is thrown up vertically along the same path with a speed of 22.0 m>s. (a) At what time will they strike each other? (b) At what height will the collision occur? (c) Answer (a) and (b) assuming that the order is reversed: the ball is thrown 1.00 s before the rock. 94. Figure 2 9 54 is a position versus time graph for the motion of an object along the x axis. Consider the time interval from A to B. (a) Is the object moving in the positive or negative x direction? (b) Is the object speeding up or slow ing down? (c) Is the acceleration of the object positive or negative? Now consider the time interval from D to E. (d) Is the object moving in the positive or negative x direc tion? (e) Is the object speed ing up or slowing down? ( f ) Is the acceleration of the object positive or negative? (g) Finally, answer these same three questions for the time interval from C to D. 95. In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between station stops. The more stops there are, the slower the train’s average speed. To get an idea of this problem, calculate the time it takes a train to make a 15.0-km trip in two situations: (a) the stations at which the trains must stop are 3.0 km apart (a total of 6 stations, including those at the ends); and (b) the stations are 5.0 km apart (4 stations total). Assume that at each station the train accelerates at a rate of 1.1 m>s2 until it reaches 95 km>h, then stays at this speed until its brakes are applied for arrival at the next station, at which time it decelerates at -2.0 m>s2. Assume it stops at each intermediate station for 22 s. 96. A race car driver must average 200.0 km>h over the course of a time trial lasting ten laps. If the first nine laps were done at an average speed of 196.0 km>h, what average speed must be maintained for the last lap? 97. A parachutist bails out of an airplane, and freely falls 75 m (ignore air friction). Then the parachute opens, and her acceleration is -1.5 m>s2 (up). The parachutist reaches the ground with a speed of 1.5 m>s. (a) From how high did she bail out of the plane? (b) How much time did her fall take? 98. You stand at the top of a cliff while your friend stands on a beach below you. You drop a ball from rest and see that she catches it 1.4 s later. Your friend then throws the ball up to you, and it comes to rest just as it reaches your hand. What is the speed with which your friend threw the ball? 99. A robot used in a pharmacy picks up a medicine bottle at t = 0. It accelerates at 0.20 m>s2 for 4.5 s , then travels without acceleration for 68 s and finally decelerates at -0.40 m>s2 for 2.5 s to reach the counter where the phar-macist will take the medicine from the robot. From how far away did the robot fetch the medicine? 100. Bill can throw a ball vertically at a speed 1.5 times faster than Joe can. How many times higher will Bill’s ball go than Joe’s? 101. On an audio compact disc (CD), digital bits of information are encoded sequentially along a spiral path. Each bit occupies about 0.28 mm. A CD player’s readout laser scans along the spiral’s sequence of bits at a constant rate of about 1.2 m>s as the CD spins. (a) Determine the number N of digital bits that a CD player reads every second. (b) The audio information is sent to each of the two loudspeakers 44,100 times per second. Each of these samplings requires 16 bits, and so you might expect the required bit rate for a CD player to be N0 = 2a 44,100 samplings s b a 16 bits samplingb = 1.4 106 bits s , where the 2 is for the 2 loudspeakers (the 2 stereo channels). Note that N0 is less than the number N of bits actually read per second by a CD player. The excess number of bits ( = N - N0) is needed for encoding and error correction. What percentage of the bits on a CD are dedicated to encoding and error correction? 102. Figure 2 9 55 shows the position vs. time graph for two bicy-cles, A and B. (a) Identify any instant at which the two bicycles have the same velocity. (b) Which bicycle has the larger acceleration? (c) At which instant(s) are the bicycles passing each other? Which bicycle is passing the other? (d) Which bicycle has the larger instantaneous velocity? (e) Which bicycle has the larger average velocity? 103. You are traveling at a constant speed vM, and there is a car in front of you traveling with a speed vA. You realize that vM 7 vA, so you start slowing down with a constant acceleration a when the distance between you and the other car is x. What relationship between a and x deter-mines whether or not you run into the car in front of you?
A N S W E R S T O E X E R C I S E S A: (a) displacement = -30 cm; (b) total distance = 50 cm. B: (b). C: (b). D: (a) +; (b) -; (c) -; (d) +.
E: (c). F: (b).
G: (e).
H: (c).
General Problems 75 15 20 25 30 10 5 0 1 0 2 3 4 5 6 A B C D E x (m) t (s) FIGURE 2 – 54 Problem 94.
FIGURE 2 – 55 Problem 102.
A 0 B x t 76 (b) (a) A B (c) (d) (e) Kinematics in Two or Three Dimensions; Vectors CHAPTER-OPENING QUESTION— Guess now!
[Don’t worry about getting the right answer now : you will get another chance later in the Chapter. See also page 23 of Chapter 1 for more explanation.] A small heavy box of emergency supplies is dropped from a moving helicopter at point A as it flies at constant speed in a horizontal direction. Which path in the drawing below best describes the path of the box (neglecting wind and air resistance) as seen by a person standing on the ground?
CONTENTS 3–1 Vectors and Scalars 3–2 Addition of Vectors : Graphical Methods 3–3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar 3–4 Adding Vectors by Components 3–5 Unit Vectors 3–6 Vector Kinematics 3–7 Projectile Motion 3–8 Solving Problems Involving Projectile Motion 3–9 Relative Velocity I n Chapter 2 we dealt with motion along a straight line. We now consider the motion of objects that move in paths in two (or three) dimensions. In particular, we discuss an important type of motion known as projectile motion: objects projected outward near the Earth’s surface, such as struck baseballs and golf balls, kicked footballs, and other projectiles. Before beginning our discussion of motion in two dimensions, we will need to discuss vectors, and how to add them.
C H A P T E R 3 A basketball flying through the air is an example of motion in two dimensions. In the absence of air resistance, the path would be a perfect parabola. The gold arrow represents the downward acceleration of gravity, g 5. Galileo analyzed the motion of objects in two dimensions under the action of gravity near the Earth’s surface (now called “projectile motion”) by separating its horizontal and vertical components.
We will discuss how to manipulate vectors and how to add them. Besides analyzing projectile motion, we will also discuss unit vectors and vector kinematics, plus see how to work with relative velocity.
g 5 SECTION 3–2 Addition of Vectors—Graphical Methods 77 3–1 Vectors and Scalars We mentioned in Chapter 2 that the term velocity refers not only to how fast an object is moving but also to its direction. A quantity such as velocity, which has direction as well as magnitude, is a vector quantity. Other quantities that are also vectors are displacement, force, and momentum. However, many quantities have no direction associated with them, such as mass, time, and temperature. They are spec-ified completely by a number and units. Such quantities are called scalar quantities.
Drawing a diagram of a particular physical situation is always helpful in physics, and this is especially true when dealing with vectors. On a diagram, each vector is represented by an arrow. The arrow is always drawn so that it points in the direction of the vector quantity it represents. The length of the arrow is drawn proportional to the magnitude of the vector quantity. For example, in Fig. 3 9 1, green arrows have been drawn representing the velocity of a car at various places as it rounds a curve. The magnitude of the velocity at each point can be read off Fig. 3 9 1 by measuring the length of the corresponding arrow and using the scale shown (1 cm = 90 km>h).
When we write the symbol for a vector, we will always use boldface type, with a tiny arrow over the symbol. Thus for velocity we write v 5. If we are concerned only with the magnitude of the vector, we will write simply v, in italics, as we do for other symbols.
3–2 Addition of Vectors—Graphical Methods Because vectors are quantities that have direction as well as magnitude, they must be added in a special way. In this Chapter, we will deal mainly with displacement vectors, for which we now use the symbol D 5, and velocity vectors, v 5. But the results apply for acceleration and other vectors we will encounter later.
We use simple arithmetic for adding scalars. Simple arithmetic can also be used for adding vectors if they are in the same direction. For example, if a person walks 8 km east one day, and 6 km east the next day, the person will be 8 km + 6 km = 14 km east of the point of origin. We say that the net or resultant displacement is 14 km to the east (Fig. 3 9 2a). If, on the other hand, the person walks 8 km east on the first day, and 6 km west (in the reverse direction) on the second day, then the person will end up 2 km from the origin (Fig. 3 9 2b), so the resultant displacement is 2 km to the east. In this case, the resultant displacement is obtained by subtraction: 8 km - 6 km = 2 km.
But simple arithmetic cannot be used if the two vectors are not along the same line. For example, suppose a person walks 10.0 km east and then walks 5.0 km north. These displacements can be represented on a graph in which the positive y axis points north and the positive x axis points east, Fig. 3 9 3. On this graph, we draw an arrow, labeled D 5 1, to represent the 10.0-km displacement to the east. Then we draw a second arrow, D 5 2, to represent the 5.0-km displacement to the north. Both vectors are drawn to scale, as shown in Fig. 3 9 3.
Scale for velocity: 1 cm = 90 km>h FIGURE 3 – 1 Car traveling on a road, slowing down to round the curve. The green arrows represent the velocity vector at each position.
FIGURE 3 – 2 Combining vectors in one dimension.
Resultant = 14 km (east) Resultant = 2 km (east) 6 km 8 km 8 km 6 km x (km) East x (km) East (a) (b) 0 0 FIGURE 3 – 3 A person walks 10.0 km east and then 5.0 km north. These two displacements are represented by the vectors D 5 1 and D 5 2, which are shown as arrows. Also shown is the resultant displacement vector, D 5 R, which is the vector sum of D 5 1 and D 5 2. Measurement on the graph with ruler and protractor shows that D 5 R has a magnitude of 11.2 km and points at an angle u = 27° north of east.
u 0 South Resultant displacement DR = D1 + D2 West 2 2 4 6 4 6 8 10 5 5 5 D1 5 D2 5 y (km) North x (km) East 78 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors After taking this walk, the person is now 10.0 km east and 5.0 km north of the point of origin. The resultant displacement is represented by the arrow labeled D 5 R in Fig. 3 9 3. (The subscript R stands for resultant.) Using a ruler and a protractor, you can measure on this diagram (Fig. 3 9 3 on previous page) that the person is 11.2 km from the origin at an angle u = 27° north of east. In other words, the resul-tant displacement vector has a magnitude of 11.2 km and makes an angle u = 27° with the positive x axis. The magnitude (length) of D 5 R can also be obtained using the theorem of Pythagoras in this case, because D1, D2, and DR form a right triangle with DR as the hypotenuse. Thus DR = 2D1 2 + D2 2 = 2(10.0 km) 2 + (5.0 km) 2 = 2125 km2 = 11.2 km.
You can use the Pythagorean theorem only when the vectors are perpendicular to each other.
The resultant displacement vector, D 5 R, is the sum of the vectors D 5 1 and D 5 2. That is, D 5 R = D 5 1 + D 5 2.
This is a vector equation. An important feature of adding two vectors that are not along the same line is that the magnitude of the resultant vector is not equal to the sum of the magnitudes of the two separate vectors, but is smaller than their sum. That is, DR … (D1 + D2), where the equals sign applies only if the two vectors point in the same direction. In our example (Fig. 3 9 3), DR = 11.2 km, whereas D1 + D2 equals 15 km, which is the total distance traveled. Note also that we cannot set D 5 R equal to 11.2 km, because we have a vector equation and 11.2 km is only a part of the resultant vector, its magnitude. We could write something like this, though: D 5 R = D 5 1 + D 5 2 = (11.2 km, 27° N of E).
Figure 3 9 3 illustrates the general rules for graphically adding two vectors together, no matter what angles they make, to get their sum. The rules are as follows: 1. On a diagram, draw one of the vectors : call it D 5 1 : to scale.
2. Next draw the second vector, D 5 2, to scale, placing its tail at the tip of the first vector and being sure its direction is correct.
3. The arrow drawn from the tail of the first vector to the tip of the second vector represents the sum, or resultant, of the two vectors.
The length of the resultant vector represents its magnitude. Note that vectors can be moved parallel to themselves on paper (maintaining the same length and angle) to accomplish these manipulations. The length of the resultant can be measured with a ruler and compared to the scale. Angles can be measured with a protractor. This method is known as the tail-to-tip method of adding vectors.
The resultant is not affected by the order in which the vectors are added. For example, a displacement of 5.0 km north, to which is added a displacement of 10.0 km east, yields a resultant of 11.2 km and angle u = 27° (see Fig. 3 9 4), the same as when they were added in reverse order (Fig. 3 9 3). Thus we can write, using V 5 to represent any type of vector, V 51 + V 52 = V 52 + V 51, [commutative property] (3 – 1a) which is known as the commutative property of vector addition.
The tail-to-tip method of adding vectors can be extended to three or more vectors. The resultant is drawn from the tail of the first vector to the tip of the last FIGURE 3 – 4 If the vectors are added in reverse order, the resultant is the same. (Compare to Fig. 3 9 3.) 0 West South 2 2 4 6 4 6 8 10 x (km) East D2 5 D1 5 u DR = D1 + D2 5 5 5 North y (km) SECTION 3–3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar 79 It is a common error to draw the sum vector as the diagonal running between the tips of the two vectors, as in Fig. 3 9 6c. This is incorrect: it does not represent the sum of the two vectors. (In fact, it represents their difference, V 52 - V 51, as we will see in the next Section.) C A U T I O N Be sure to use the correct diagonal on the parallelogram to get the resultant FIGURE 3 – 6 Vector addition by two different methods, (a) and (b). Part (c) is incorrect.
CONCEPTUAL EXAMPLE 3 – 1 Range of vector lengths. Suppose two vectors each have length 3.0 units. What is the range of possible lengths for the vector representing the sum of the two?
RESPONSE The sum can take on any value from 6.0 ( = 3.0 + 3.0) where the vectors point in the same direction, to 0 ( = 3.0 - 3.0) when the vectors are antiparallel. Magnitudes between 0 and 6.0 occur when the two vectors are at an angle other than 0° and 180°.
EXERCISE A If the two vectors of Example 3 9 1 are perpendicular to each other, what is the resultant vector length?
3–3 Subtraction of Vectors, and Multiplication of a Vector by a Scalar Given a vector V 5, we define the negative of this vector ( -V 5 ) to be a vector with the same magnitude as V 5 but opposite in direction, Fig. 3 9 7. Note, however, that no vector is ever negative in the sense of its magnitude: the magnitude of every vector is positive. Rather, a minus sign tells us about its direction.
FIGURE 3 – 7 The negative of a vector is a vector having the same length but opposite direction.
–V 5 V 5 FIGURE 3 – 5 The resultant of three vectors: V 5R = V 51 + V 52 + V 53.
+ + = V1 5 V2 5 V3 5 V1 5 VR 5 V3 5 V2 5 INCORRECT (a) (b) (c) Tail-to-tip Parallelogram Wrong + = = = 1 1 1 1 2 2 2 2 R V 5 V 5 V 5 V 5 V 5 V 5 R V 5 V 5 V 5 V 5 one added. An example is shown in Fig. 3 9 5; the three vectors could represent displacements (northeast, south, west) or perhaps three forces. Check for yourself that you get the same resultant no matter in which order you add the three vectors; that is, (V 51 + V 52) + V 53 = V 51 + (V 52 + V 53), [associative property] (3 – 1b) which is known as the associative property of vector addition.
A second way to add two vectors is the parallelogram method. It is fully equivalent to the tail-to-tip method. In this method, the two vectors are drawn starting from a common origin, and a parallelogram is constructed using these two vectors as adjacent sides as shown in Fig. 3 9 6b. The resultant is the diagonal drawn from the common origin. In Fig. 3 9 6a, the tail-to-tip method is shown, and we can see that both methods yield the same result.
80 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors We can now define the subtraction of one vector from another: the difference between two vectors V 52 - V 51 is defined as V 52 - V 51 = V 52 + ( -V 51).
That is, the difference between two vectors is equal to the sum of the first plus the negative of the second. Thus our rules for addition of vectors can be applied as shown in Fig. 3 9 8 using the tail-to-tip method.
A vector V 5 can be multiplied by a scalar c. We define their product so that cV 5 has the same direction as V 5 and has magnitude cV. That is, multiplication of a vector by a positive scalar c changes the magnitude of the vector by a factor c but doesn’t alter the direction. If c is a negative scalar, the magnitude of the product cV 5 is still 0 c0 V (where 0 c0 means the magnitude of c), but the direction is precisely opposite to that of V 5. See Fig. 3 9 9.
EXERCISE B What does the “incorrect” vector in Fig. 3 9 6c represent? (a) V 52 - V 51, (b) V 51 - V 52, (c) something else (specify).
3–4 Adding Vectors by Components Adding vectors graphically using a ruler and protractor is often not sufficiently accurate and is not useful for vectors in three dimensions. We discuss now a more powerful and precise method for adding vectors. But do not forget graphical methods : they are useful for visualizing, for checking your math, and thus for getting the correct result.
Consider first a vector V 5 that lies in a particular plane, and we have chosen an x and a y axis on this plane. Then V 5 can be expressed as the sum of two other vectors, called the components of the original vector, which are usually chosen to be along the x and y axes. The process of finding the components is known as resolving the vector into its components. An example is shown in Fig. 3 9 10; the vector V 5 could be a displacement vector that points at an angle u = 30° north of east, where we have chosen the positive x axis to be to the east and the positive y axis north. This vector V 5 is resolved into its x and y components by drawing dashed lines out from the tip (A) of the vector (lines AB and AC), making them perpendicular to the x and y axes. Then the lines 0B and 0C represent the x and y components of V 5, respectively, as shown in Fig. 3 9 10b. These vector components are written V 5x and V 5y. In this book we usually show vector components as arrows, like vectors, but dashed. The scalar components, Vx and Vy, are the magnitudes of the vector components, with units, accompanied by a positive or negative sign depending on whether they point along the positive or negative x or y axis. As can be seen in Fig. 3 9 10, V 5x + V 5y = V 5 by the parallelogram method of adding vectors.
Space is made up of three dimensions, and sometimes it is necessary to resolve a vector into components along three mutually perpendicular directions. In rectangular coordinates the components are V 5x, V 5y, and V 5z.
FIGURE 3 – 9 Multiplying a vector V 5 by a scalar c gives a vector whose magnitude is c times greater and in the same direction as V 5 (or opposite direction if c is negative).
= -2.0 = 1.5 2 3 V 5 V 5 V 5 V 5 V 5 FIGURE 3 – 8 Subtracting two vectors: V 52 - V 51.
– + = = – 1 – 1 1 2 – 1 2 2 2 V 5 V 5 V 5 V 5 V 5 V 5 V 5 V 5 FIGURE 3 – 10 Resolving a vector V 5 into its components along a chosen set of x and y axes. The components, once found, themselves represent the vector. That is, the components contain as much information as the vector itself.
C North East North East y x B A (a) y x (b) 0 0 (= 30°) u (= 30°) u x y V V 5 V V 5 5 5 SECTION 3–4 Adding Vectors by Components 81 The use of trigonometric functions for finding the components of a vector is illustrated in Fig. 3 9 11, where a vector and its two components are thought of as making up a right triangle. (See also Appendix A for other details on trigono-metric functions and identities.) We then see that the sine, cosine, and tangent are as given in Fig. 3 9 11, where u is the angle V 5 makes with the +x axis, measured positive counterclockwise. If we multiply the definition of sin u = Vy>V by V on both sides, we get Vy = V sin u. (3 – 2a) Similarly, from the definition of cos u, we obtain Vx = V cos u. (3 – 2b) Keep in mind that u is chosen (by convention) to be the angle that the vector makes with the positive x axis, measured positive counterclockwise.
The components of a given vector will be different for different choices of coordinate axes. It is therefore crucial to specify the choice of coordinate system when giving the components.
There are two ways to specify a vector in a given coordinate system: 1. We can give its components, Vx and Vy.
2. We can give its magnitude V and the angle u it makes with the positive x axis.
We can shift from one description to the other using Eqs. 3 9 2, and, for the reverse, by using the theorem of Pythagoras† and the definition of tangent: V = 2Vx 2 + Vy 2 (3 – 3a) tan u = Vy Vx (3 – 3b) as can be seen in Fig. 3 9 11.
We can now discuss how to add vectors using components. The first step is to resolve each vector into its components, Eqs. 3 9 2. Next we can see, using Fig. 3 9 12, that the addition of any two vectors V 51 and V 52 to give a resultant V 5R = V 51 + V 52, implies that V R x = V 1 x + V 2 x V R y = V 1 y + V 2 y. (3 – 4) That is, the sum of the x components equals the x component of the resultant vector, and the sum of the y components equals the y component of the resultant vector, as can be verified by a careful examination of Fig. 3 9 12. Note that we do not add x components to y components.
If the magnitude and direction of the resultant vector are desired, they can be obtained using Eqs. 3 9 3.
Vy V sin = Vx V cos = Vy Vx tan = V2 = V2 + V2 90° y x x y 0 x y V 5 V 5 V 5 u u u u FIGURE 3 – 11 Finding the components of a vector using trigonometric functions. The equations are valid only if u is the angle V 5 makes with the positive x axis.
†In three dimensions, the theorem of Pythagoras becomes V = 3Vx 2 + Vy 2 + Vz 2, where Vz is the component along the third, or z, axis.
FIGURE 3 – 12 The components of V 5R = V 51 + V 52 are VR x = V1 x + V2 x VR y = V1 y + V2 y.
y x VRy VRx V 1x V2x V1y V2y 0 VR 1 = + 1 2 V 5 V 5 V 5 5 2 V 5 82 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors The components of a given vector depend on the choice of coordinate axes. You can often reduce the work involved in adding vectors by a good choice of axes : for example, by choosing one of the axes to be in the same direction as one of the vectors. Then that vector will have only one nonzero component.
EXAMPLE 3 – 2 Mail carrier’s displacement. A rural mail carrier leaves the post office and drives 22.0 km in a northerly direction. She then drives in a direction 60.0° south of east for 47.0 km (Fig. 3 9 13a). What is her displacement from the post office?
APPROACH We choose the positive x axis to be east and the positive y axis to be north, since those are the compass directions used on most maps. The origin of the xy coordinate system is at the post office. We resolve each vector into its x and y components. We add the x components together, and then the y compo-nents together, giving us the x and y components of the resultant.
SOLUTION Resolve each displacement vector into its components, as shown in Fig. 3 9 13b. Since D 5 1 has magnitude 22.0 km and points north, it has only a y component: D1 x = 0, D1 y = 22.0 km.
D 5 2 has both x and y components: D2 x = + (47.0 km) (cos 60°) = + (47.0 km) (0.500) = +23.5 km D2 y = - (47.0 km) (sin 60°) = - (47.0 km) (0.866) = -40.7 km.
Notice that D2 y is negative because this vector component points along the negative y axis. The resultant vector, D 5 R, has components: DR x = D1 x + D2 x = 0 km + 23.5 km = +23.5 km DR y = D1 y + D2 y = 22.0 km + ( -40.7 km) = -18.7 km.
This specifies the resultant vector completely: DR x = 23.5 km, DR y = -18.7 km.
We can also specify the resultant vector by giving its magnitude and angle using Eqs. 3 9 3: DR = 2DR x 2 + DR y 2 = 2(23.5 km) 2 + ( -18.7 km) 2 = 30.0 km tan u = DR y DR x = -18.7 km 23.5 km = -0.796.
A calculator with a key labeled inv tan, or arc tan, or tan-1 gives u = tan-1 ( -0.796) = -38.5°. The negative sign means u = 38.5° below the x axis, Fig. 3 9 13c. So, the resultant displacement is 30.0 km directed at 38.5° in a southeasterly direction.
NOTE Always be attentive about the quadrant in which the resultant vector lies. An electronic calculator does not fully give this information, but a good diagram does.
y x East (a) y x (b) 0 0 0 D2x y x (c) D2y Post offce North 60° 60° 1 2 2 2 1 1 D 5 D 5 D 5 D 5 D 5 DR 5 D 5 u FIGURE 3 – 13 Example 3 9 2. (a) The two displacement vectors, D 5 1 and D 5 2. (b) D 5 2 is resolved into its components. (c) D 5 1 and D 5 2 are added graphically to obtain the resultant D 5 R. The component method of adding the vectors is explained in the Example.
As we saw in Example 3 9 2, any component that points along the negative x or y axis gets a minus sign. The signs of trigonometric functions depend on which “quadrant” the angle falls in: for example, the tangent is positive in the first and third quadrants (from 0° to 90°, and 180° to 270°), but negative in the second and fourth quadrants; see Appendix A 9 9, Fig. A 9 6. The best way to keep track of angles, and to check any vector result, is always to draw a vector diagram, like Fig. 3 9 13. A vector diagram gives you something tangible to look at when analyzing a problem, and provides a check on the results.
The following Problem Solving Strategy should not be considered a prescrip-tion. Rather it is a summary of things to do to get you thinking and involved in the problem at hand.
P R O B L E M S O LV I N G Identify the correct quadrant by drawing a careful diagram EXAMPLE 3 – 3 Three short trips. An airplane trip involves three legs, with two stopovers, as shown in Fig. 3 9 14a. The first leg is due east for 620 km; the second leg is southeast (45°) for 440 km; and the third leg is at 53° south of west, for 550 km, as shown. What is the plane’s total displacement?
APPROACH We follow the steps in the Problem Solving Strategy above.
SOLUTION 1. Draw a diagram such as Fig. 3 9 14a, where D 5 1, D 5 2, and D 5 3 represent the three legs of the trip, and D 5 R is the plane’s total displacement. 2. Choose axes: Axes are also shown in Fig. 3 9 14a: x is east, y north. 3. Resolve components: It is imperative to draw a good diagram. The compo-nents are drawn in Fig. 3 9 14b. Instead of drawing all the vectors starting from a common origin, as we did in Fig. 3 9 13b, here we draw them “tail-to-tip” style, which is just as valid and may make it easier to see. 4. Calculate the components: D 5 1 : D1 x = +D1 cos 0° = D1 = 620 km D1 y = +D1 sin 0° = 0 km D 5 2 : D2 x = +D2 cos 45° = + (440 km) (0.707) = +311 km D2 y = -D2 sin 45° = - (440 km) (0.707) = -311 km D 5 3 : D3 x = -D3 cos 53° = - (550 km) (0.602) = -331 km D3 y = -D3 sin 53° = - (550 km) (0.799) = -439 km. We have given a minus sign to each component that in Fig. 3 9 14b points in the -x or -y direction. The components are shown in the Table in the margin. 5. Add the components: We add the x components together, and we add the y components together to obtain the x and y components of the resultant: DR x = D1 x + D2 x + D3 x = 620 km + 311 km - 331 km = 600 km DR y = D1 y + D2 y + D3 y = 0 km - 311 km - 439 km = -750 km. The x and y components of the resultant are 600 km and -750 km, and point respectively to the east and south. This is one way to give the answer. 6. Magnitude and direction: We can also give the answer as DR = 2DR x 2 + DR y 2 = 2(600) 2 + ( -750) 2 km = 960 km tan u = DR y DR x = -750 km 600 km = -1.25, so u = -51°. Thus, the total displacement has magnitude 960 km and points 51° below the x axis (south of east), as was shown in our original sketch, Fig. 3 9 14a. Components Vector x (km) y (km) D 5 1 620 0 D 5 2 311 -311 D 5 3 -331 -439 D 5 R 600 -750 = ?
45° -x 0 53° +x +y -y (a) D2y D3y D3x -x 0 +x +y -y (b) 45° D2x 53° North East East North 1 D 5 2 D 5 3 D 5 R D 5 1 D 5 3 D 5 2 D 5 u FIGURE 3 – 14 Example 3 9 3.
SECTION 3–4 Adding Vectors by Components 83 Pay careful attention to signs: any component that points along the negative x or y axis gets a minus sign. 5. Add the x components together to get the x component of the resultant. Similarly for y: V R x = V 1 x + V 2 x + any others V R y = V 1 y + V 2 y + any others. This is the answer: the components of the resultant vector. Check signs to see if they fit the quadrant shown in your diagram (point 1 above). 6. If you want to know the magnitude and direction of the resultant vector, use Eqs. 3 9 3: V R = 2V R x 2 + V R y 2 , tan u = V R y V R x . The vector diagram you already drew helps to obtain the correct position (quadrant) of the angle u. P R O B L E M S O L V I N G Here is a brief summary of how to add two or more vectors using components: 1. Draw a diagram, adding the vectors graphically by either the parallelogram or tail-to-tip method. 2. Choose x and y axes. Choose them in a way, if possible, that will make your work easier. (For example, choose one axis along the direction of one of the vectors, which then will have only one component.) 3. Resolve each vector into its x and y components, showing each component along its appropriate (x or y) axis as a (dashed) arrow. 4. Calculate each component (when not given) using sines and cosines. If u1 is the angle that vector V 51 makes with the positive x axis, then: V 1 x = V 1 cos u1, V 1 y = V 1 sin u1.
Adding Vectors 84 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors 3–5 Unit Vectors Vectors can be conveniently written in terms of unit vectors. A unit vector is defined to have a magnitude exactly equal to one (1). (The word “unit” comes from the Latin, unus, meaning “one”.) It is useful to define unit vectors that point along coordinate axes, and in an x, y, z rectangular coordinate system these unit vectors are called i N, j N, and k N. They point, respectively, along the positive x, y, and z axes as shown in Fig. 3 9 15. Like other vectors, i N, j N, and k N do not have to be placed at the origin, but can be placed elsewhere as long as the direction and unit length remain unchanged. It is common to write unit vectors with a “hat”: i N, j N, k N (and we will do so in this book) as a reminder that each has magnitude of exactly one unit.
Because of the definition of multiplication of a vector by a scalar (Section 3 9 3), the components of a vector V 5 can be written V 5x = Vx i N, V 5y = Vy j N, and V 5z = Vz k N. Hence any vector V 5 can be written in terms of its components as V 5 = Vx i N + Vy j N + Vz k N. (3 – 5) Unit vectors are helpful when adding vectors analytically by components. For example, Eq. 3 9 4 can be seen to be true by using unit vector notation for each vector (which we write for the two-dimensional case, with the extension to three dimensions being straightforward): V 5R = (V R x) i N + (V R y) j N = V 51 + V 52 = (V 1 x i N + V 1 y j N) + (V 2 x i N + V 2 y j N) = (V 1 x + V 2 x) i N + (V 1 y + V 2 y) j N.
Comparing the first line to the third line, we get Eqs. 3 9 4.
j i k y x z ˆ ˆ ˆ FIGURE 3 – 15 Unit vectors i N, j N, and k N along the x, y, and z axes.
EXAMPLE 3 – 4 Using unit vectors. Write the vectors of Example 3 9 2 in unit vector notation, and perform the addition.
APPROACH We use the components we found in Example 3 9 2, D1 x = 0, D1 y = 22.0 km, and D2 x = 23.5 km, D2 y = -40.7 km, and we now write them in the form of Eq. 3 9 5.
SOLUTION We have D 5 1 = 0 i N + 22.0 km j N D 5 2 = 23.5 km i N - 40.7 km j N.
Then the resultant displacement is D 5 R = D 5 1 + D 5 2 = (0 + 23.5) km i N + (22.0 - 40.7) km j N = 23.5 km i N - 18.7 km j N.
The components of the resultant displacement, D 5 R, are Dx = 23.5 km and Dy = -18.7 km. The magnitude of D 5 R is DR = 1(23.5 km) 2 + (18.7 km) 2 = 30.0 km, just as in Example 3 9 2.
3–6 Vector Kinematics We can now extend our definitions of velocity and acceleration in a formal way to two- and three-dimensional motion. Suppose a particle follows a path in the xy plane as shown in Fig. 3 9 16. At time t1, the particle is at point P1, and at time t2, it is at point P2. The vector r 51 is the position vector of the particle at time t1 (it represents the displacement of the particle from the origin of the coordinate system). And r 52 is the position vector at time t2.
In one dimension, we defined displacement as the change in position of the particle. In the more general case of two or three dimensions, the displacement vector is defined as the vector representing change in position. We call it ∆r 5,† where ∆r 5 = r 52 - r 51.
This represents the displacement during the time interval ∆t = t2 - t1.
FIGURE 3 – 16 Path of a particle in the xy plane. At time t1 the particle is at point P1 given by the position vector r 51; at t2 the particle is at point P2 given by the position vector r 52 . The displacement vector for the time interval t2 - t1 is ∆r 5 = r 52 - r 51. The actual distance traveled along the path between P1 and P2 is D l.
†We used D 5 for the displacement vector earlier in the Chapter for illustrating vector addition. The new notation here, ∆r 5, emphasizes that it is the difference between two position vectors.
0 y P1 P2 2 1 ∆ ∆l x r 5 r 5 r 5 SECTION 3–6 Vector Kinematics 85 In unit vector notation, we can write r 51 = x1 i N + y1 j N + z1 k N, (3 – 6a) where x1, y1, and z1 are the coordinates of point P1. Similarly, r 52 = x2 i N + y2 j N + z2 k N.
Hence ∆r 5 = (x2 - x1) i N + (y2 - y1) j N + (z2 - z1) k N. (3 – 6b) If the motion is along the x axis only, then y2 - y1 = 0, z2 - z1 = 0, and the magnitude of the displacement is ∆r = x2 - x1, which is consistent with our earlier one-dimensional equation (Section 2 9 1). Even in one dimension, displace-ment is a vector, as are velocity and acceleration.
The average velocity vector over the time interval ∆t = t2 - t1 is defined as average velocity = ∆r 5 ∆t . (3 – 7) Now let us consider shorter and shorter time intervals : that is, we let ∆t approach zero so that the distance between points P2 and P1 also approaches zero, Fig. 3 9 17a. We define the instantaneous velocity vector as the limit of the average velocity as ∆t approaches zero: v 5 = lim ∆tS0 ∆r 5 ∆t = d r 5 dt . (3 – 8) The direction of v 5 at any moment is along the line tangent to the path at that moment (Fig. 3 9 17b).
Note that the magnitude of the average velocity in Fig. 3 9 16 is not equal to the average speed, which is the actual distance traveled along the path, D l, divided by ∆t. In some special cases, the average speed and average velocity are equal in magnitude (such as motion along a straight line in one direction), but in general they are not. However, in the limit ∆t S 0, ∆r always approaches D l, so the instantaneous speed always equals the magnitude of the instantaneous velocity at any time.
The instantaneous velocity (Eq. 3 9 8) is equal to the derivative of the position vector with respect to time. Equation 3 9 8 can be written in terms of components starting with Eq. 3 9 6a as: v 5 = d r 5 dt = dx dt i N + dy dt j N + dz dt k N = vx i N + vy j N + vz k N, (3 – 9) where vx = dx>dt, vy = dy>dt, vz = dz>dt are the x, y, and z components of the velocity. Note that d i N>dt = d j N>dt = d k N >dt = 0 since these unit vectors are constant in both magnitude and direction.
Acceleration in two or three dimensions is treated in a similar way. The average acceleration vector, over a time interval ∆t = t2 - t1 is defined as average acceleration = ∆v 5 ∆t = v 52 - v 51 t2 - t1 , (3 – 10) where ∆v 5 is the change in the instantaneous velocity vector during that time interval: ∆v 5 = v 52 - v 51. Note that v 52 in many cases, such as in Fig. 3 9 18a, may not be in the same direction as v 51. Hence the average acceleration vector may be in a different direction from either v 51 or v 52 (Fig. 3 9 18b). Furthermore, v 52 and v 51 may have the same magnitude but different directions, and the difference of two such vectors will not be zero. Hence acceleration can result from either a change in the magnitude of the velocity, or from a change in direction of the velocity, or from a change in both.
The instantaneous acceleration vector is defined as the limit of the average acceleration vector as the time interval ∆t is allowed to approach zero: a 5 = lim ∆tS0 ∆v 5 ∆t = dv 5 dt , (3 – 11) and is thus the derivative of v 5 with respect to t.
FIGURE 3 – 18 (a) Velocity vectors v 51 and v 52 at instants t1 and t2 for a particle at points P1 and P2, as in Fig. 3 9 16. (b) The direction of the average acceleration is in the direction of ∆v 5 = v 52 - v 51.
0 (a) (b) y P1 ∆ P2 x v1 5 v2 5 r1 5 r2 5 v1 5 v 5 v2 5 0 y P1 P2 (a) ¢ x 0 y P1 (b) x r 5 r1 5 r2 5 r1 5 v1 5 FIGURE 3 – 17 (a) As we take ∆t and ∆r 5 smaller and smaller [compare to Fig. 3 9 16] we see that the direction of ∆r 5 and of the instantaneous velocity (∆r 5>∆t, where ∆t S 0) is (b) tangent to the curve at P1.
86 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors We can write a 5 using components: a 5 = dv 5 dt = dvx dt i N + dvy dt j N + dvz dt k N = ax i N + ay j N + az k N, (3 – 12a) where ax = dvx>dt, etc. Because vx = dx>dt, then ax = dvx>dt = d2x>dt2, as we saw in Section 2 9 4. Thus we can also write the acceleration as a 5 = d2x dt2 i N + d2y dt2 j N + d2z dt2 k N. (3 – 12b) The instantaneous acceleration will be nonzero not only when the magnitude of the velocity changes, but also if its direction changes. For example, a person riding in a car traveling at constant speed around a curve, or a child riding on a merry-go-round, will both experience an acceleration because of a change in the direction of the velocity, even though the speed may be constant. (More on this in Chapter 5.) In general, we will use the terms “velocity” and “acceleration” to mean the instan-taneous values. If we want to discuss average values, we will use the word “average.” EXAMPLE 3 – 5 Position given as a function of time. The position of a particle as a function of time is given by r 5 = [ (5.0 m>s)t + (6.0 m>s2)t2] i N + [ (7.0 m) - (3.0 m>s3)t3] j N, where r is in meters and t is in seconds. (a) What is the particle’s displacement between t1 = 2.0 s and t2 = 3.0 s? (b) Determine the particle’s instantaneous velocity and acceleration as a function of time. (c) Evaluate v 5 and a 5 at t = 3.0 s.
APPROACH For (a), we find ∆r 5 = r 52 - r 51, inserting t1 = 2.0 s for finding r 51, and t2 = 3.0 s for r 52. For (b), we take derivatives (Eqs. 3 9 9 and 3 9 12), and for (c) we substitute t = 3.0 s into our results in (b).
SOLUTION (a) We insert t1 = 2.0 s into the given equation for r 5: r 51 = [ (5.0 m>s) (2.0 s) + (6.0 m>s 2) (2.0 s) 2] i N + [ (7.0 m) - (3.0 m>s 3) (2.0 s) 3] j N = (34 m) i N - (17 m) j N.
Similarly, at t2 = 3.0 s, r 52 = (15 m + 54 m) i N + (7.0 m - 81 m) j N = (69 m) i N - (74 m) j N.
Thus ∆r 5 = r 52 - r 51 = (69 m - 34 m) i N + ( -74 m + 17 m) j N = (35 m) i N - (57 m) j N.
That is, ∆x = 35 m, and ∆y = -57 m.
(b) To find velocity, we take the derivative of the given r 5 with respect to time, noting ( Appendix B 9 2) that d (t2) >dt = 2t, and d (t3) >dt = 3t2: v 5 = d r 5 dt = [5.0 m>s + (12 m>s 2)t] i N + [0 - (9.0 m>s 3)t2] j N.
The acceleration is (keeping only two significant figures): a 5 = dv 5 dt = (12 m>s 2) i N - (18 m>s 3)t j N.
Thus ax = 12 m>s 2 is constant; but ay = - (18 m>s 3)t depends linearly on time, increasing in magnitude with time in the negative y direction.
(c) We substitute t = 3.0 s into the equations we just derived for v 5 and a 5: v 5 = (5.0 m>s + 36 m>s) i N - (81 m>s) j N = (41 m>s) i N - (81 m>s) j N a 5 = (12 m>s 2) i N - (54 m>s 2) j N.
Their magnitudes at t = 3.0 s are v = 2(41 m>s) 2 + (81 m>s) 2 = 91 m>s, and a = 2(12 m>s2) 2 + (54 m>s 2) 2 = 55 m>s 2.
SECTION 3–7 Projectile Motion 87 Constant Acceleration In Chapter 2 we studied the important case of one-dimensional motion for which the acceleration is constant. In two or three dimensions, if the accel-eration vector, a 5, is constant in magnitude and direction, then ax = constant, ay = constant, az = constant. The average acceleration in this case is equal to the instantaneous acceleration at any moment. The equations we derived in Chapter 2 for one dimension, Eqs. 2 9 12a, b, and c, apply separately to each perpendicular component of two- or three-dimensional motion. In two dimensions we let v 50 = vx 0 i N + vy 0 j N be the initial velocity, and we apply Eqs. 3 9 6a, 3 9 9, and 3 9 12b for the position vector r 5, velocity v 5, and acceleration a 5. We can then write Eqs. 2 9 12a, b, and c for two dimensions as shown in Table 3 9 1.
TABLE 3 – 1 Kinematic Equations for Constant Acceleration in 2 Dimensions x component (horizontal) y component (vertical) vx = vx 0 + ax t (Eq. 2 9 12a) vy = vy 0 + ay t x = x0 + vx 0 t + 1 2 ax t2 (Eq. 2 9 12b) y = y0 + vy 0 t + 1 2 ay t2 vx 2 = vx 0 2 + 2ax (x - x0) (Eq. 2 9 12c) vy 2 = vy 0 2 + 2ay (y - y0) The first two of the equations in Table 3 9 1 can be written more formally in vector notation. v 5 = v 50 + a 5 t [a 5 = constant] (3 – 13a) r 5 = r 50 + v 50 t + 1 2 a 5 t2. [a 5 = constant] (3 – 13b) Here, r 5 is the position vector at any time, and r 50 is the position vector at t = 0. These equations are the vector equivalent of Eqs. 2 9 12a and b. In practical situ-ations, we usually use the component form given in Table 3 9 1.
3–7 Projectile Motion In Chapter 2, we studied one-dimensional motion of an object in terms of displacement, velocity, and acceleration, including purely vertical motion of a falling object undergoing acceleration due to gravity. Now we examine the more general translational motion of objects moving through the air in two dimensions near the Earth’s surface, such as a golf ball, a thrown or batted base-ball, kicked footballs, and speeding bullets. These are all examples of projectile motion (see Fig. 3 9 19), which we can describe as taking place in two dimensions if there is no wind.
Although air resistance is often important, in many cases its effect can be ignored, and we will ignore it in the following analysis. We will not be concerned now with the process by which the object is thrown or projected. We consider only its motion after it has been projected, and before it lands or is caught : that is, we analyze our projected object only when it is moving freely through the air under the action of gravity alone. Then the acceleration of the object is that due to gravity, which acts downward with magnitude g = 9.80 m>s 2, and we assume it is constant.† Galileo was the first to describe projectile motion accurately. He showed that it could be understood by analyzing the horizontal and vertical components of the motion separately. For convenience, we assume that the motion begins at time t = 0 at the origin of an xy coordinate system (so x0 = y0 = 0).
†This restricts us to objects whose distance traveled and maximum height above the Earth are small compared to the Earth’s radius (6400 km).
(b) (a) FIGURE 3 – 19 Photographs of (a) a bouncing ball and (b) a snowboarder, each showing the characteristic “parabolic” path of projectile motion.
88 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors Let us look at a (tiny) ball rolling off the end of a horizontal table with an initial velocity in the horizontal (x) direction, v 5x 0. See Fig. 3 9 20, where an object falling vertically is also shown for comparison. The velocity vector v 5 at each instant points in the direction of the ball’s motion at that instant and is thus always tangent to the path. Like Galileo, we treat the horizontal and vertical components of the velocity and acceleration separately, and we apply the kinematic equations (Eqs. 2 9 12) to the x and y components of the motion.
First we examine the vertical ( y ) component of the motion. At the instant the ball leaves the table’s top (t = 0), it has only an x component of velocity. Once the ball leaves the table (at t = 0), it experiences a vertically downward acceleration g, the acceleration due to gravity. Thus vy is initially zero (vy 0 = 0) but increases continually in the downward direction (until the ball hits the ground). Let us take y to be positive upward. Then the acceleration due to gravity is in the -y direction, so ay = -g. From Eq. 2 9 12a (using y in place of x) we can write vy = vy 0 + ay t = -gt since we set vy 0 = 0. The vertical displacement is given by Eq. 2 9 12b written in terms of y: y = y0 + vy 0 + 1 2 ay t2. Given y0 = 0, vy 0 = 0, and ay = -g, then y = - 1 2 gt2.
In the horizontal direction, the acceleration is zero (ignoring air resistance). With ax = 0, the horizontal component of velocity, vx, remains constant, equal to its initial value, vx 0, and thus has the same magnitude at each point on the path. The horizontal displacement (with ax = 0) is given by x = vx 0 t + 1 2 ax t2 = vx 0 t. The two vector components, v 5x and v 5y, can be added vectorially at any instant to obtain the velocity v 5 at that time (each point on the path), as shown in Fig. 3 9 20.
One result of this analysis, which Galileo himself predicted, is that an object projected horizontally will reach the ground in the same time as an object dropped vertically. The vertical motions are the same in both cases, as shown in Fig. 3 9 20. Figure 3 9 21 is a multiple-exposure photograph of an experiment that confirms this.
If an object is projected at an upward angle, as in Fig. 3 9 22, the analysis is similar, but now there is an initial vertical component of velocity, vy 0. Because of the downward acceleration of gravity, the upward component of velocity vy gradually decreases with time until the object reaches the highest point on its path, at which point vy = 0. The object then moves downward (Fig. 3 9 22) and vy increases in the downward direction (becoming more negative). As before, vx remains constant.
FIGURE 3 – 21 Multiple-exposure photograph showing positions of two balls at equal time intervals. One ball was dropped from rest at the same time the other was projected horizontally outward. The vertical position of each ball is seen to be the same at each instant. FIGURE 3 – 22 Path of a projectile fired with initial velocity v 50 at angle u0 to the horizontal. Path is shown dashed in black, the velocity vectors are green arrows, and velocity components are dashed. The acceleration a 5 = d v 5>dt is downward. That is, a 5 = g 5 = -g j N where j N is the unit vector in the positive y direction. Not shown is where the projectile hits the ground (at that point projectile motion ceases).
x y 0 = 0 at this point 0 u 0 y y0 x v y x y x x0 v 5 v 5 v 5 v 5 v 5 5 v 5 vy 5 v 5 v 5 v 5 v 5 v 5 v 5 = = -gj ˆ g 5 a 5 FIGURE 3 – 20 Projectile motion of a small ball projected horizontally with initial velocity v 5 = v 5x 0. The dashed black line represents the path of the object. The velocity vector v 5 is in the direction of motion at each point, and thus is tangent to the path. The velocity vectors are green arrows, and velocity components are dashed. (A vertically falling object starting from rest at the same place and time is shown at the left for comparison; vy is the same at each instant for the falling object and the projectile.) y x Vertical fall Projectile motion x y y vx0 5 v 5 x v 5 v 5 v 5 v 5 v 5 = a 5 g 5 SECTION 3–8 Solving Problems Involving Projectile Motion 89 CONCEPTUAL EXAMPLE 3 – 6 Where does the apple land? A child sits upright in a wagon that moves to the right at constant speed, Fig. 3 9 23. The child extends her hand and throws an apple straight up (from her own point of view, Fig. 3 9 23a), as the wagon moves forward at constant speed. Neglecting air resistance, will the apple land (a) behind the wagon, (b) in the wagon, or (c) in front of the wagon?
RESPONSE The child throws the apple straight up in her own reference frame with initial velocity v 5y 0 (Fig. 3 9 23a). But when viewed by someone on the ground, the apple also has an initial horizontal component of velocity equal to the velocity of the wagon, v 5x 0. Thus, to a person on the ground, the apple follows the path of a projectile as shown in Fig. 3 9 23b. The apple experiences no horizontal accelera tion, so v 5x 0 stays constant, equal to the speed of the wagon. As the apple follows its arc, the wagon will be directly under the apple at all times because they have the same horizontal velocity. When the apple comes down, it will drop right into the outstretched hand of the child. The answer is (b).
y x (b) Ground reference frame (a) Wagon reference frame vy0 5 v0 5 vy0 5 vx0 5 vx0 5 FIGURE 3 – 23 Example 3 9 6.
3–8 Solving Problems Involving Projectile Motion 5. Examine the horizontal (x) and vertical (y) motions separately. If you are given the initial velocity, you may want to resolve it into its x and y components. 6. List the known and unknown quantities, choosing ax = 0 and ay = -g or +g, where g = 9.80 m>s2, depending on choice of y positive up or down. Recall: vx never changes throughout the trajectory, and vy = 0 at the highest point of any trajectory that returns downward. The velocity just before landing is generally not zero. 7. Think for a minute before jumping into the equations. Apply the relevant equations (Table 3 9 2), combining them if necessary. You may need to combine components of a vector to get magnitude and direction (Eqs. 3 9 3). P R O B L E M S O L V I N G Our approach to solving Problems in Section 2 9 6 also applies here. Solving projectile motion Problems can require creativity, and cannot be done just by follow ing rules. You must avoid just plugging numbers into equations that seem to “work.” 1. As always, read carefully; choose the object (or objects) you are going to analyze. 2. Draw a careful diagram showing what is happening. 3. Choose an origin and an xy coordinate system. 4. Decide on the time interval, which for projec-tile motion can only include motion under the effect of gravity alone, not throwing or landing. The time interval must be the same for the x and y analyses. The x and y motions are connected by the common time, t.
Projectile Motion TABLE 3 – 2 Kinematic Equations for Projectile Motion ( y positive upward; ax = 0, ay = −g = −9.80 m,s2) Horizontal Motion (ax = 0, vx = constant) Vertical Motion† (ay = −g = constant) vx = vx 0 (Eq. 2 9 12a) vy = vy 0 - gt x = x0 + vx 0 t (Eq. 2 9 12b) y = y0 + vy 0 t - 1 2 gt2 (Eq. 2 9 12c) vy 2 = vy 0 2 - 2g (y - y0) †If y is taken positive downward, the minus ( - ) signs in front of g become plus ( + ) signs.
EXERCISE C Return to the Chapter-Opening Question, page 76, and answer it again now. Try to explain why you may have answered differently the first time. Describe the role of the helicopter in this example of projectile motion.
We can simplify Eqs. 2 9 12 (Table 3 9 1) for the case of projectile motion because we can set ax = 0. See Table 3 9 2, which assumes y is positive upward, so ay = -g = -9.80 m>s2. If u is chosen relative to the +x axis, as in Fig. 3 9 22, then vx 0 = v0 cos u0, and vy 0 = v0 sin u0.
In doing Problems involving projectile motion, we must consider a time interval for which our chosen object is in the air, influenced only by gravity. We do not consider the throwing (or projecting) process, nor the time after the object lands or is caught, because then other influences act on the object, and we can no longer set a 5 = g 5.
P R O B L E M S O LV I N G Choice of time interval 90 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors EXAMPLE 3 – 7 Driving off a cliff. A movie stunt driver on a motorcycle speeds horizontally off a 50.0-m-high cliff. How fast must the motorcycle leave the clifftop to land on level ground below, 90.0 m from the base of the cliff where the cameras are? Ignore air resistance.
APPROACH We explicitly follow the steps of the Problem Solving Strategy on the previous page.
SOLUTION 1. and 2. Read, choose the object, and draw a diagram. Our object is the motor-cycle and driver, taken as a single unit. The diagram is shown in Fig. 3 9 24. 3. Choose a coordinate system. We choose the y direction to be positive upward, with the top of the cliff as y0 = 0. The x direction is horizontal with x0 = 0 at the point where the motorcycle leaves the cliff. 4. Choose a time interval. We choose our time interval to begin (t = 0) just as the motorcycle leaves the clifftop at position x0 = 0, y0 = 0. Our time interval ends just before the motorcycle touches the ground below. 5. Examine x and y motions. In the horizontal (x) direction, the acceleration ax = 0, so the velocity is constant. The value of x when the motorcycle reaches the ground is x = +90.0 m. In the vertical direction, the acceleration is the acceleration due to gravity, ay = -g = -9.80 m>s2. The value of y when the motorcycle reaches the ground is y = -50.0 m. The initial velocity is hori-zontal and is our unknown, vx 0 ; the initial vertical velocity is zero, vy 0 = 0. 6. List knowns and unknowns. See the Table in the margin. Note that in addition to not knowing the initial horizontal velocity vx 0 (which stays constant until landing), we also do not know the time t when the motor-cycle reaches the ground. 7. Apply relevant equations. The motorcycle maintains constant vx as long as it is in the air. The time it stays in the air is determined by the y motion : when it reaches the ground. So we fi rst fi nd the time using the y motion, and then use this time value in the x equations. To fi nd out how long it takes the motorcycle to reach the ground below, we use Eq. 2 9 12b (Table 3 9 2) for the vertical (y) direction with y0 = 0 and vy 0 = 0: y = y0 + vy 0 t + 1 2 ay t2 = 0 + 0 + 1 2 ( -g)t2 or y = - 1 2 gt2. We solve for t and set y = -50.0 m: t = A 2y -g = B 2( -50.0 m) -9.80 m>s2 = 3.19 s. To calculate the needed initial velocity, vx 0, we again use Eq. 2 9 12b, but this time for the horizontal (x) direction, with ax = 0 and x0 = 0: x = x0 + vx 0 t + 1 2 ax t2 = 0 + vx 0 t + 0 or x = vx 0 t. So the motorcycle needs to leave the clifftop with a speed vx 0 = x t = 90.0 m 3.19 s = 28.2 m>s, which is about 100 km>h (roughly 60 mi>h).
NOTE In the time interval of the projectile motion, the only acceleration is g in the negative y direction. The acceleration in the x direction is zero.
y = -50.0 m 50.0 m = g 5 a 5 90.0 m + x + y FIGURE 3 – 24 Example 3 9 7.
Known Unknown x0 = y0 = 0 vx 0 x = 90.0 m t y = -50.0 m ax = 0 ay = -g = -9.80 m>s2 vy 0 = 0 SECTION 3–8 Solving Problems Involving Projectile Motion 91 EXAMPLE 3 – 8 A kicked football. A football is kicked at an angle u0 = 37.0° with a velocity of 20.0 m>s, as shown in Fig. 3 9 25. Calculate (a) the maximum height, (b) the time of travel before the football hits the ground, and (c) how far away it hits the ground. Assume the ball leaves the foot at ground level, and ignore air resistance, wind, and rotation of the ball.
APPROACH This may seem difficult at first because there are so many questions. But we can deal with them one at a time. We take the y direction as positive upward, and treat the x and y motions separately. The total time in the air is again determined by the y motion. The x motion occurs at constant velocity. The y component of velocity varies, being positive (upward) initially, decreasing to zero at the highest point, and then becoming negative as the football falls.
SOLUTION We resolve the initial velocity into its components (Fig. 3 9 25): vx 0 = v0 cos 37.0° = (20.0 m>s) (0.799) = 16.0 m>s vy 0 = v0 sin 37.0° = (20.0 m>s) (0.602) = 12.0 m>s.
(a) To find the maximum height, we consider a time interval that begins just after the football loses contact with the foot until the ball reaches its maximum height. During this time interval, the acceleration is g downward. At the maximum height, the velocity is horizontal (Fig. 3 9 25), so vy = 0; and this occurs at a time given by vy = vy 0 - gt with vy = 0 (see Eq. 2 9 12a in Table 3 9 2). Thus t = vy 0 g = (12.0 m>s) (9.80 m>s2) = 1.224 s L 1.22 s.
From Eq. 2 9 12b, with y0 = 0, we can solve for y at this time t = vy 0>g : y = vy 0 t - 1 2 gt2 = vy 0 2 g - 1 2 vy 0 2 g = vy 0 2 2g = (12.0 m>s) 2 2 (9.80 m>s 2) = 7.35 m.
The maximum height is 7.35 m. [Solving Eq. 2 9 12c for y gives the same result.] (b) To find the time it takes for the ball to return to the ground, we consider a different time interval, starting at the moment the ball leaves the foot (t = 0, y0 = 0) and ending just before the ball touches the ground (y = 0 again). We can use Eq. 2 9 12b with y0 = 0 and also set y = 0 (ground level): y = y0 + vy 0 t - 1 2 gt2 0 = 0 + vy 0 t - 1 2 gt2.
This equation can be easily factored: t (1 2 gt - vy 0) = 0.
There are two solutions, t = 0 (which corresponds to the initial point, y0), and t = 2vy 0 g = 2(12.0 m>s) (9.80 m>s2) = 2.45 s, which is the total travel time of the football.
(c) The total distance traveled in the x direction is found by applying Eq. 2 9 12b with x0 = 0, ax = 0, vx 0 = 16.0 m>s, and t = 2.45 s: x = vx 0 t = (16.0 m>s) (2.45 s) = 39.2 m.
NOTE In (b) the time needed for the whole trip, t = 2vy 0 >g = 2.45 s, is double the time to reach the highest point, calculated in (a). That is, the time to go up equals the time to come back down to the same level (ignoring air resistance).
P H Y S I C S A P P L I E D Sports FIGURE 3 – 25 Example 3 9 8.
37.0° y = 0 at this point 0 y v 5 v 5 x0 v 5 y0 v 5 v0 5 v 5 v 5 = g 5 a 5 x 92 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors EXAMPLE 3 – 10 Range of a cannon ball. Suppose one of Napoleon’s cannons had a muzzle speed, v0, of 60.0 m>s. At what angle should it have been aimed (ignore air resistance) to strike a target 320 m away?
APPROACH We use the equation just derived for the range, R = v0 2 sin 2u0>g, with R = 320 m.
SOLUTION We solve for sin 2u0 in the range formula: sin 2u0 = Rg v0 2 = (320 m) (9.80 m>s 2) (60.0 m>s) 2 = 0.871.
We want to solve for an angle u0 that is between 0° and 90°, which means 2u0 in this equation can be as large as 180°. Then sin-1(0.871) = 2u0 = 60.6° is a solution, EXERCISE D In Example 3 9 8, what is (a) the velocity vector at the maximum height, and (b) the acceleration vector at maximum height?
In Example 3 9 8 we treated the football as if it were a particle, ignoring its rotation. We also ignored air resistance. Because air resistance is significant on a football, our results are only estimates (mainly overestimates). We also ignored any wind. If there happened to be a cross-wind, we would need a third axis for such 3-dimensional motion.
EXERCISE E Two balls are thrown in the air at different angles, but each reaches the same height. Which ball remains in the air longer: the one thrown at the steeper angle or the one thrown at a shallower angle?
CONCEPTUAL EXAMPLE 3 – 9 The wrong strategy. A boy on a small hill aims his water-balloon slingshot horizontally, straight at a second boy hanging from a tree branch a distance d away, Fig. 3 9 26. At the instant the water balloon is released, the second boy lets go and falls from the tree, hoping to avoid being hit. Show that he made the wrong move. (He hadn’t studied physics yet.) Ignore air resistance.
RESPONSE Both the water balloon and the boy in the tree start falling at the same instant, and in a time t they each fall the same vertical distance y = 1 2 gt2, much like Fig. 3 9 21. In the time it takes the water balloon to travel the hori-zontal distance d, the balloon will have the same y position as the falling boy. Splat. If the boy had stayed in the tree, he would have avoided the humiliation.
0 v0 y d FIGURE 3 – 26 Example 3 9 9.
Level Horizontal Range The total distance the football traveled in Example 3 9 8 is called the horizontal range R. We now derive a formula for the range, which applies to a projectile that lands at the same level it started (= y0) : that is, y(final) = y0 (see Fig. 3 9 27a). Looking back at Example 3 9 8 part (c), we see that x = R = vx 0 t where (from part b) t = 2vy 0>g. Thus R = vx 0 t = vx 0¢ 2vy 0 g ≤ = 2vx 0 vy 0 g = 2v0 2 sin u0 cos u0 g , [y = y0] where vx 0 = v0 cos u0 and vy 0 = v0 sin u0. This can be rewritten, using the trigo nometric identity 2 sin u cos u = sin 2u (Appendix A or inside the rear cover): R = v0 2 sin 2u0 g . [only if y(final) = y0] Note that the maximum range, for a given initial velocity v0, is obtained when sin 2u takes on its maximum value of 1.0, which occurs for 2u0 = 90°; so u0 = 45° for maximum range, and Rmax = v0 2>g.
The maximum range increases by the square of v0, so doubling the initial velocity of a projectile increases its maximum range by a factor of 4.
When air resistance is important, the range is less for a given v0, and the maximum range is obtained at an angle smaller than 45°.
FIGURE 3 – 27 (a) The range R of a projectile. (b) There are generally two angles u0 that will give the same range. If one angle is u0 1, the other is u0 2 = 90° - u0 1. Also see Example 3 9 10.
y = 0 again here (where x = R) y x x0 = 0 y0 = 0 (b) 60° 30° y x (a) R 45° u0 SECTION 3–8 Solving Problems Involving Projectile Motion 93 FIGURE 3 – 28 Example 3 9 11: the football leaves the punter’s foot at y = 0, and reaches the ground where y = -1.00 m.
The level range formula applies only if takeoff and landing are at the same height (y = y0). Example 3 9 11 below considers a case where they are not equal heights (y ≠y0).
EXAMPLE 3 – 11 A punt. Suppose the football in Example 3 9 8 was punted and left the punter’s foot at a height of 1.00 m above the ground. How far did the football travel before hitting the ground? Set x0 = 0, y0 = 0.
APPROACH The x and y motions are again treated separately. But we cannot use the range formula as in Example 3 9 10 because the range formula is valid only if y(final) = y0, which is not the case here. Now we have y0 = 0, and the football hits the ground where y = -1.00 m (see Fig. 3 9 28). We choose our time interval to start when the ball leaves his foot (t = 0, y0 = 0, x0 = 0) and end just before the ball hits the ground (y = -1.00 m). We can get x from Eq. 2 9 12b, x = vx 0 t, and we saw that vx 0 = 16.0 m>s in Example 3 9 8. But first we must find t, the time at which the ball hits the ground, which we obtain from the y motion.
P H Y S I C S A P P L I E D Sports P R O B L E M S O LV I N G Do not use any formula unless you are sure its range of validity fits the problem; the range formula does not apply here because y ≠y0 Ground y x y0 = 0 y = -1.00 m so the cannon should be aimed at u0 = 30.3°. But 2u0 = 180° - 60.6° = 119.4° is also a solution (see Appendix A 9 9), so u0 can also be u0 = 59.7°. In general we have two solutions (see Fig. 3 9 27b), which in the present case are given by u0 = 30.3° or 59.7°.
Either angle gives the same range. Only when sin 2u0 = 1 (so u0 = 45°) is there a single solution (that is, both solutions are the same).
SOLUTION To find t with y = -1.00 m and vy 0 = 12.0 m>s (see Example 3 9 8), we use the equation y = y0 + vy 0 t - 1 2 gt2, and obtain -1.00 m = 0 + (12.0 m>s)t - (4.90 m>s2)t2.
We rearrange this equation into standard form (ax2 + bx + c = 0) so we can use the quadratic formula: (4.90 m>s2)t2 - (12.0 m>s)t - (1.00 m) = 0.
The quadratic formula (Appendix A 9 1) gives t = 12.0 m>s { 2( -12.0 m>s) 2 - 4(4.90 m>s 2) ( -1.00 m) 2(4.90 m>s2) = 2.53 s or -0.081 s.
The second solution would correspond to a time prior to our chosen time interval that begins at the kick (t = 0), so it doesn’t apply. With t = 2.53 s for the time at which the ball touches the ground, the horizontal distance the ball traveled is (using vx 0 = 16.0 m>s from Example 3 9 8): x = vx 0 t = (16.0 m>s) (2.53 s) = 40.5 m.
Our assumption in Example 3 9 8 that the ball leaves the foot at ground level gives a result (39.2 m) that is an underestimate of about 1.3 m in the distance our punt traveled. (But Example 3 9 8 would apply for a kickoff or field goal in American football.) 94 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors EXAMPLE 3 – 12 Rescue helicopter drops supplies. A rescue helicopter wants to drop a package of supplies to isolated mountain climbers on a rocky ridge 200 m below. If the helicopter is traveling horizontally with a speed of 70 m>s (250 km>h), (a) how far in advance of the recipients (horizontal distance) must the package be dropped (Fig. 3 9 29a)? (b) Suppose, instead, that the heli-copter releases the package a horizontal distance of 400 m in advance of the mountain climbers. What vertical velocity should the package be given (up or down) so that it arrives precisely at the climbers’ position (Fig. 3 9 29b)? (c) With what speed does the package land in the latter case?
APPROACH We choose the origin of our xy coordinate system at the initial position of the helicopter, taking +y upward, and use the kinematic equations (Table 3 9 2).
SOLUTION (a) We can fi nd the time to reach the climbers using the vertical distance of 200 m. The package is “dropped” so initially it has the velocity of the helicopter, vx 0 = 70 m>s, vy 0 = 0. Then, since y = - 1 2 gt2, we have t = B -2y g = B -2( -200 m) 9.80 m>s2 = 6.39 s.
The horizontal motion of the falling package is at constant speed of 70 m>s. So x = vx 0 t = (70 m>s) (6.39 s) = 447 m L 450 m, assuming the given numbers were good to two signifi cant fi gures.
(b) We are given x = 400 m, vx 0 = 70 m>s, y = -200 m, and we want to fi nd vy 0 (see Fig. 3 9 29b). Like most problems, this one can be approached in various ways. Instead of searching for a formula or two, let’s try to reason it out in a simple way, based on what we did in part (a). If we know t, perhaps we can get vy 0. Since the horizontal motion of the package is at constant speed (once it is released we don’t care what the helicopter does), we have x = vx 0 t, so t = x vx 0 = 400 m 70 m>s = 5.71 s.
Now let’s try to use the vertical motion to get vy 0 : y = y0 + vy 0 t - 1 2 gt2. Since y0 = 0 and y = -200 m, we can solve for vy 0 : vy 0 = y + 1 2 gt2 t = -200 m + 1 2 (9.80 m>s2) (5.71 s) 2 5.71 s = -7.0 m>s.
Thus, in order to arrive at precisely the mountain climbers’ position, the package must be thrown downward from the helicopter with a speed of 7.0 m>s.
(c) We want to know v of the package at t = 5.71 s. The components are: vx = vx 0 = 70 m>s vy = vy 0 - gt = -7.0 m>s - (9.80 m>s2) (5.71 s) = -63 m>s.
So v = 2(70 m>s) 2 + ( -63 m>s) 2 = 94 m>s. (Better not to release the package from such an altitude, or use a parachute.) P H Y S I C S A P P L I E D Reaching a target from a moving helicopter “Dropped” (vy0 = 0) 200 m 200 m (a) x y Thrown upward?
(vy0 7 0) 400 m (b) Thrown downward?
(vy0 6 0) vx0 FIGURE 3 – 29 Example 3 9 12.
SECTION 3–9 Relative Velocity 95 Projectile Motion Is Parabolic We now show that the path followed by any projectile is a parabola, if we can ignore air resistance and can assume that only gravity is acting with g 5 constant. To do so, we need to find y as a function of x by eliminating t between the two equations for horizontal and vertical motion (Eq. 2 9 12b in Table 3 9 2), and for simplicity we set x0 = y0 = 0 : x = vx 0 t y = vy 0 t - 1 2 gt2.
From the first equation, we have t = x>vx 0, and we substitute this into the second equation to obtain y = ¢ vy 0 vx 0 ≤ x - ¢ g 2vx 0 2 ≤ x2. (3 – 14) We see that y as a function of x has the form y = Ax - Bx2, where A and B are constants for any specific projectile motion. This is the well-known equation for a parabola. See Figs. 3 9 19 and 3 9 30.
The idea that projectile motion is parabolic was at the forefront of physics research in Galileo’s day. Today we discuss it in Chapter 3 of introductory physics!
3–9 Relative Velocity We now consider how observations made in different frames of reference are related to each other. For example, consider two trains approaching one another, each with a speed of 80 km>h with respect to the Earth. Observers on the Earth beside the train tracks will measure 80 km>h for the speed of each of the trains. Observers on either one of the trains (different frames of reference) will measure a speed of 160 km>h for the other train approaching them.
Similarly, when one car traveling 90 km>h passes a second car traveling in the same direction at 75 km>h, the first car has a speed relative to the second car of 90 km>h - 75 km>h = 15 km>h.
When the velocities are along the same line, simple addition or subtraction is sufficient to obtain the relative velocity. But if they are not along the same line, we must make use of vector addition. We emphasize, as mentioned in Section 2 9 1, that when specifying a velocity, it is important to specify what the reference frame is.
When determining relative velocity, it is easy to make a mistake by adding or subtracting the wrong velocities. It is important, therefore, to draw a diagram and use a careful labeling process. Each velocity is labeled by two subscripts: the first refers to the object, the second to the reference frame in which it has this velocity. For example, suppose a boat heads directly across a river, as shown in Fig. 3 9 31. We let v 5BW be the velocity of the Boat with respect to the Water. (This is also what the boat’s velocity would be relative to the shore if the water were still.) Similarly, v 5BS is the velocity of the Boat with respect to the Shore, and v 5WS is the velocity of the Water with respect to the Shore (this is the river current). Note that v 5BW is what the boat’s motor produces (against the water), whereas v 5BS is equal to v 5BW plus the effect of the current, v 5WS. Therefore, the velocity of the boat relative to the shore is (see vector diagram, Fig. 3 9 31) v 5BS = v 5BW + v 5WS. (3 – 15) By writing the subscripts using this convention, we see that the inner subscripts (the two W’s) on the right-hand side of Eq. 3 9 15 are the same; also, the outer subscripts on the right of Eq. 3 9 15 (the B and the S) are the same as the two subscripts for the sum vector on the left, v 5BS. FIGURE 3 – 30 Examples of projectile motion: (a) a boy leaping, (b) glowing lava from the volcano Stromboli.
FIGURE 3 – 31 A boat heads north directly across a river which flows west. Velocity vectors are shown as green arrows: v 5BS = velocity of Boat with respect to the Shore, v 5BW = velocity of Boat with respect to the Water, v 5WS = velocity of Water with respect to the Shore (river current).
As it crosses the river, the boat is dragged downstream by the current (v 5WS).
E N W S BS BW WS v 5 v 5 v 5 River current u (a) (b) 96 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors By following this convention (first subscript for the object, second for the reference frame), you can write down the correct equation relating velocities in different reference frames.† Figure 3 9 32 gives a derivation of Eq. 3 9 15, v 5BS = v 5BW + v 5WS, but using different subscripts.
Equation 3 9 15 is valid in general and can be extended to three or more velocities. For example, if a fisherman on a boat walks with a velocity v 5FB rela tive to the boat, his velocity relative to the shore is v 5FS = v 5FB + v 5BW + v 5WS. The equations involving relative velocity will be correct when adjacent inner subscripts are identical and when the outermost ones correspond exactly to the two on the velocity on the left of the equation. But this works only with plus signs (on the right), not minus signs.
It is often useful to remember that for any two objects or reference frames, A and B, the velocity of A relative to B has the same magnitude, but opposite direction, as the velocity of B relative to A: v 5BA = -v 5AB. (3 – 16) For example, if a train is traveling 100 km>h relative to the Earth in a certain direction, objects on the Earth (such as trees) appear to an observer on the train to be traveling 100 km>h in the opposite direction.
FIGURE 3 – 32 Derivation of relative velocity equation (Eq. 3 9 15), in this case for a person walking along the corridor in a train. We are looking down on the train and two reference frames are shown: xy on the Earth and x′y′ fixed on the train. We have: r 5PT = position vector of person (P) relative to train (T), r 5PE = position vector of person (P) relative to Earth (E), r 5TE = position vector of train’s coordinate system (T) relative to Earth (E).
From the diagram we see that r 5PE = r 5PT + r 5TE.
We take the derivative with respect to time to obtain d dt (r 5PE) = d dt (r 5PT) + d dt (r 5TE), or, since d r 5>dt = v 5, v 5PE = v 5PT + v 5TE.
This is the equivalent of Eq. 3 9 15 for the present situation (check the subscripts!).
†We thus can see, for example, that the equation v 5BW = v 5BS + v 5WS is wrong: the inner subscripts are not the same, and the outer ones on the right do not correspond to the subscripts on the left.
y x 0 y¿ x¿ 0¿ rPE 5 rPT 5 rTE 5 vTE 5 EXAMPLE 3 – 13 Heading upstream. A boat’s speed in still water is vBW = 1.85 m>s. If the boat is to travel north directly across a river whose westward current has speed vWS = 1.20 m>s, at what upstream angle must the boat head? See Fig. 3 9 33.
APPROACH If the boat heads straight across the river, the current will drag the boat downstream (westward). To overcome the river’s current, the boat must have an upstream (eastward) component of velocity as well as a cross-stream (northward) component. Figure 3 9 33 has been drawn with v 5BS, the velocity of the Boat relative to the Shore, pointing directly across the river because this is where the boat is supposed to go. Note that v 5BS = v 5BW + v 5WS.
SOLUTION Vector v 5BW points upstream at angle u as shown. From the diagram, sin u = vWS vBW = 1.20 m>s 1.85 m>s = 0.6486.
Thus u = 40.4°, so the boat must head upstream at a 40.4° angle.
E N W S BS BW WS v 5 v 5 v 5 River current u FIGURE 3 – 33 Example 3 9 13. A boat, in order to go directly across a moving river current, must head upstream.
SECTION 3–9 Relative Velocity 97 EXAMPLE 3 – 14 Heading across the river. The same boat (vBW = 1.85 m>s) now heads directly across the river whose current is still 1.20 m>s. (a) What is the velocity (magnitude and direction) of the boat relative to the shore? (b) If the river is 110 m wide, how long will it take to cross and how far downstream will the boat be then?
APPROACH The boat now heads directly across the river and is pulled down-stream by the current, as shown in Fig. 3 9 34. The boat’s velocity with respect to the shore, v 5BS, is the sum of the boat’s velocity with respect to the water, v 5BW, plus the velocity of the water with respect to the shore, v 5WS (= the river’s current). Just as before, v 5BS = v 5BW + v 5WS.
SOLUTION (a) Since v 5BW is perpendicular to v 5WS, we can get vBS using the theorem of Pythagoras: vBS = 2v BW 2 + v WS 2 = 2(1.85 m>s) 2 + (1.20 m>s) 2 = 2.21 m>s.
We can obtain the angle (note how u is defined in Fig. 3 9 34) from: tan u = vWS>vBW = (1.20 m>s) > (1.85 m>s) = 0.6486.
A calculator with a key such as inv tan or arc tan or tan−1 gives u = tan-1(0.6486) = 33.0°. Note that this angle is not equal to the angle calcu-lated in Example 3 9 13.
(b) To find the travel time for the boat to cross the river, recall the river’s width D = 110 m, and use the velocity component that points directly across the river, vBW = D>t. Solving for t, we get t = 110 m> (1.85 m>s) = 59.5 s. Also, the boat will have been carried downstream, in this time, a distance d = vWS t = (1.20 m>s) (59.5 s) = 71.4 m L 71 m.
NOTE There is no acceleration in this Example, so the motion involves only constant velocities (of the boat or of the river).
BS BW WS v 5 v 5 v 5 River current u FIGURE 3 – 34 Example 3 9 14. A boat heading directly across a river whose current moves at 1.20 m>s.
EXAMPLE 3 – 15 Car velocities at 90°. Two cars approach a street corner at right angles to each other with the same speed of 40.0 km>h ( = 11.11 m>s) relative to the ground, as shown in Fig. 3 9 35a. What is the relative velocity of car 1 as seen by car 2?
APPROACH Figure 3 9 35a shows the situation in a reference frame fixed to the Earth. But we want to view the situation from a reference frame in which car 2 is at rest, and this is shown in Fig. 3 9 35b. In this reference frame (the world as seen by the driver of car 2), the Earth moves toward car 2 with velocity v 5E 2 (speed of 40.0 km>h), which is of course equal and opposite to v 52 E, the velocity of car 2 with respect to the Earth (Eq. 3 9 16): v 52 E = -v 5E 2.
Then the velocity of car 1 as seen by car 2 is (see Eq. 3 9 15) v 51 2 = v 51 E + v 5E 2.
SOLUTION Because v 5E 2 = -v 52 E, then v 51 2 = v 51 E - v 52 E.
That is, the velocity of car 1 as seen by car 2 is the difference of their veloci-ties, v 51 E - v 52 E, both measured relative to the Earth; see Fig. 3 9 35c. Since the magnitudes of v 51 E, v 52 E, and v 5E 2 are equal (40.0 km>h = 11.11 m>s), we see (Fig. 3 9 35b) that v 51 2 points at a 45° angle toward car 2; the speed is v1 2 = 2(11.11 m>s) 2 + (11.11 m>s) 2 = 15.7 m>s ( = 56.6 km>h).
(a) 1 2 (b) 1 2 At rest (c) v2E 5 v1E 5 vE2 5 vE2 5 v12 5 v1E 5 -v2E 5 12 = 1E - v2E v1E 5 v 5 5 v 5 FIGURE 3 – 35 Example 3 9 15.
98 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors Summary A quantity such as velocity, that has both a magnitude and a direction, is called a vector. A quantity such as mass, that has only a magnitude, is called a scalar. On diagrams, vectors are represented by arrows.
Addition of vectors can be done graphically by placing the tail of each successive arrow at the tip of the previous one. The sum, or resultant vector, is the arrow drawn from the tail of the first vector to the tip of the last vector. Two vectors can also be added using the parallelogram method.
Vectors can be added more accurately by adding their components along chosen axes with the aid of trigonometric functions. A vector of magnitude V making an angle u with the +x axis has components Vx = V cos u, Vy = V sin u. (3 – 2) Given the components, we can find a vector’s magnitude and direction from V = 2V x 2 + V y 2, tan u = Vy Vx . (3 – 3) It is often helpful to express a vector in terms of its components along chosen axes using unit vectors, which are vectors of unit length along the chosen coordinate axes; for Cartesian coordinates the unit vectors along the x, y, and z axes are called i N, j N, and k N.
The general definitions for the instantaneous velocity, v 5, and acceleration, a 5, of a particle (in one, two, or three dimensions) are v 5 = d r 5 dt (3 – 8) a 5 = d v 5 dt , (3 – 11) where r 5 is the position vector of the particle. The kinematic equa-tions for motion with constant acceleration can be written for each of the x, y, and z components of the motion and have the same form as for one-dimensional motion (Eqs. 2 9 12). Or they can be written in the more general vector form: v 5 = v 50 + a 5 t r 5 = r 50 + v 50 t + 1 2 a 5 t2. (3 – 13) Projectile motion is the motion of an object in the air near the Earth’s surface under the effect of gravity alone. It can be analyzed as two separate motions if air resistance can be ignored. The horizontal component of motion is at constant velocity, whereas the vertical component is at constant acceleration, g 5, just as for an object falling vertically under the action of gravity.
The velocity of an object relative to one frame of refer-ence can be found by vector addition if its velocity relative to a second frame of reference, and the relative velocity of the two reference frames, are known.
Questions 1. One car travels due east at 40 km>h, and a second car travels north at 40 km>h. Are their velocities equal? Explain. 2. Can you conclude that a car is not accelerating if its speed ometer indicates a steady 60 km>h? Explain. 3. Give several examples of an object’s motion in which a great distance is traveled but the displacement is zero. 4. Can the displacement vector for a particle moving in two dimensions be longer than the length of path traveled by the particle over the same time interval? Can it be less? Discuss. 5. During baseball practice, a player hits a very high fly ball and then runs in a straight line and catches it. Which had the greater displacement, the player or the ball? Explain. 6. If V 5 = V 51 + V 52, is V necessarily greater than V1 and>or V2 ? Discuss. 7. Two vectors have lengths V1 = 4.5 km and V2 = 5.0 km. What are the maximum and minimum magnitudes of their vector sum? 8. Can two vectors, of unequal magnitude, add up to give the zero vector? Can three unequal vectors? Under what conditions? 9. Can the magnitude of a vector ever (a) equal, or (b) be less than, one of its components? 10. Does the odometer of a car measure a scalar or a vector quantity? What about the speedometer? 11. How could you determine the speed a slingshot imparts to a rock, using only a meter stick, a rock, and the slingshot? 12. In archery, should the arrow be aimed directly at the target? How should your angle of aim depend on the distance to the target? 13. Where in Fig. 3 9 22 is (a) v 5 = 0, (b) vy = 0, (c) vx = 0? 14. It was reported in World War I that a pilot flying at an alti-tude of 2 km caught in his bare hands a bullet fired at the plane! Using the fact that a bullet slows down considerably due to air resistance, explain how this incident occurred. 15. You are on the street trying to hit a friend in his dorm window with a water balloon. He has a similar idea and is aiming at you with his water balloon. You aim straight at each other and throw at the same instant. Do the water balloons hit each other? Explain why or why not. 16. A projectile is launched at an upward angle of 30° to the horizontal with a speed of 30 m>s. How does the horizon tal component of its velocity 1.0 s after launch compare with its horizontal component of velocity 2.0 s after launch, ignoring air resistance? Explain. 17. A projectile has the least speed at what point in its path? 18. Two cannonballs, A and B, are fired from the ground with identical initial speeds, but with uA larger than uB. (a) Which cannonball reaches a higher elevation? (b) Which stays longer in the air? (c) Which travels farther? Explain. 19. A person sitting in an enclosed train car, moving at constant velocity, throws a ball straight up into the air in her reference frame. (a) Where does the ball land? What is your answer if the car (b) accelerates, (c) decelerates, (d) rounds a curve, (e) moves with constant velocity but is open to the air? 20. If you are riding on a train that speeds past another train moving in the same direction on an adjacent track, it appears that the other train is moving backward. Why? 21. Two rowers, who can row at the same speed in still water, set off across a river at the same time. One heads straight across and is pulled downstream somewhat by the cur rent. The other one heads upstream at an angle so as to arrive at a point opposite the starting point. Which rower reaches the opposite side first? Explain. 22. If you stand motionless under an umbrella in a rainstorm where the drops fall vertically, you remain relatively dry. However, if you start running, the rain begins to hit your legs even if they remain under the umbrella. Why?
MisConceptual Questions 99 MisConceptual Questions 1. You are adding vectors of length 20 and 40 units. Which of the following choices is a possible resultant magnitude?
(a) 0. (b) 18. (c) 37. (d) 64. (e) 100. 2. The magnitude of a component of a vector must be (a) less than or equal to the magnitude of the vector.
(b) equal to the magnitude of the vector.
(c) greater than or equal to the magnitude of the vector.
(d) less than, equal to, or greater than the magnitude of the vector. 3. You are in the middle of a large fi eld. You walk in a straight line for 100 m, then turn left and walk 100 m more in a straight line before stopping. When you stop, you are 100 m from your starting point. By how many degrees did you turn?
(a) 90°. (b) 120°. (c) 30°. (d) 180°.
(e) This is impossible. You cannot walk 200 m and be only 100 m away from where you started. 4. Which of the following equations correctly expresses the relation between vectors A 5, B 5, and C 5, shown in Fig. 3 9 36?
(a) A 5 = B 5 + C 5.
(b) B 5 = A 5 + C 5.
(c) C 5 = A 5 + B 5.
(d) A 5 + B 5 + C 5 = 0.
FIGURE 3 – 36 MisConceptual Question 4.
B 5 A 5 C 5 5. A car is driven at a constant speed of 10.0 m>s around a circle of radius 20.0 m. As it goes 1 4 of the way around, (a) the magnitude of the average velocity is 0.
(b) the magnitude of the average velocity is 10 m>s.
(c) the magnitude of the average velocity is between 0 and 10 m>s.
(d) the magnitude of the average velocity is greater than 10 m>s. 6. A bullet fi red horizontally from a rifl e begins to fall (a) as soon as it leaves the barrel.
(b) after air friction reduces its speed.
(c) not at all if air resistance is ignored. 7. A baseball player hits a ball that soars high into the air. After the ball has left the bat, and while it is traveling upward (at point P in Fig. 3 9 37), what is the direction of accelera-tion? Ignore air resistance. 8. One ball is dropped vertically from a window. At the same instant, a second ball is thrown horizontally from the same window. Which ball has the greater speed at ground level?
(a) The dropped ball.
(b) The thrown ball.
(c) Neither : they both have the same speed on impact.
(d) It depends on how hard the ball was thrown. 9. Two balls having different speeds roll off the edge of a hori-zontal table at the same time. Which hits the fl oor sooner? (a) The faster one.
(b) The slower one.
(c) Both the same. 10. You are riding in an enclosed train car moving at 90 km>h. If you throw a baseball straight up, where will the baseball land?
(a) In front of you.
(b) Behind you.
(c) In your hand.
(d) Can’t decide from the given information.
11. Which of the three kicks in Fig. 3 9 38 is in the air for the longest time? They all reach the same maximum height h. Ignore air resistance.
(a), (b), (c), or (d) all the same time.
h (a) (b) (c) FIGURE 3 – 38 MisConceptual Question 11.
12. A baseball is hit high and far. Which of the following statements is true? At the highest point, (a) the magnitude of the acceleration is zero.
(b) the magnitude of the velocity is zero.
(c) the magnitude of the velocity is the slowest.
(d) more than one of the above is true.
(e) none of the above are true.
13. A hunter is aiming horizontally at a monkey who is sitting in a tree. The monkey is so terrifi ed when it sees the gun that it falls off the tree. At that very instant, the hunter pulls the trigger. What will happen?
(a) The bullet will miss the monkey because the monkey falls down while the bullet speeds straight forward.
(b) The bullet will hit the monkey because both the monkey and the bullet are falling downward at the same rate due to gravity.
(c) The bullet will miss the monkey because although both the monkey and the bullet are falling downward due to gravity, the monkey is falling faster.
(d) It depends on how far the hunter is from the monkey.
14. Which statements are not valid for a projectile? Take up as positive and ignore air resistance.
(a) The projectile has the same x velocity at any point on its path.
(b) The acceleration of the projectile is positive and decreasing when the projectile is moving upwards, zero at the top, and increasingly negative as the projectile descends.
(c) The acceleration of the projectile has a constant negative value.
(d) The y component of the velocity of the projectile is zero at the highest point of the projectile’s path.
(e) The velocity at the highest point is zero.
15. A car travels 10 m>s east. Another car travels 10 m>s north. The relative speed of the fi rst car with respect to the second is (a) less than 20 m>s. (b) exactly 20 m>s. (c) more than 20 m>s.
FIGURE 3 – 37 MisConceptual Question 7.
P (b) (a) (c) P (b) (a) (c) 100 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors Problems 3 – 2 to 3 – 5 Vector Addition; Unit Vectors 1. (I) A car is driven 245 km west and then 118 km southwest (45.0°). What is the displacement of the car from the point of origin (magnitude and direction)? Draw a diagram. 2. (I) A delivery truck travels 21 blocks north, 16 blocks east, and 26 blocks south. What is its final displacement from the origin? Assume the blocks are equal length. 3. (I) If Vx = 9.40 units and Vy = -6.80 units, determine the magnitude and direction of V 5. 4. (II) Graphically determine the resultant of the following three vector displacements: (1) 24 m, 36° north of east; (2) 18 m, 37° east of north; and (3) 26 m, 33° west of south. 5. (II) V 5 is a vector 21.8 units in magnitude and points at an angle of 23.4° above the negative x axis. (a) Sketch this vector. (b) Calculate Vx and Vy. (c) Use Vx and Vy to obtain (again) the magnitude and direction of V 5. [Note: Part (c) is a good way to check if you’ve resolved your vector correctly.] 6. (II) Vector V 51 is 6.2 units long and points along the nega tive x axis. Vector V 52 is 8.1 units long and points at +55° to the positive x axis. (a) What are the x and y components of each vector? (b) Determine the sum V 51 + V 52 (magnitude and angle). 7. (II) Figure 3 9 39 shows two vectors, A 5 and B 5, whose magni tudes are A = 6.8 units and B = 5.5 units. Determine C 5 if (a) C 5 = A 5 + B 5, (b) C 5 = A 5 - B 5, (c) C 5 = B 5 - A 5. Give the magnitude and direction for each.
x y A 5 B 5 FIGURE 3 – 39 Problem 7. 8. (II) An airplane is traveling 815 km>h in a direction 41.5° west of north (Fig. 3 9 40). (a) Find the components of the velocity vector in the northerly and westerly direc tions. (b) How far north and how far west has the plane traveled after 1.75 h? 9. (II) The summit of a mountain, 2450 m above base camp, is measured on a map to be 4580 m horizontally from the camp in a direction 38.4° west of north. What are the compo nents of the displacement vector from camp to summit? What is its magnitude? Choose the x axis east, y axis north, and z axis up. 10. (II) Three vectors are shown in Fig. 3 9 41. Their magnitudes are given in arbitrary units. Determine the sum of the three vectors. Give the resultant in terms of (a) components, (b) magnitude and angle with the +x axis. 11. (II) (a) Given the vectors A 5 and B 5 shown in Fig. 3 9 41, determine B 5 - A 5. (b) Determine A 5 - B 5 without using your answer in (a). (c) Compare your results and see if they are opposite. 12. (II) Determine the vector A 5 - C 5, given the vectors A 5 and C 5 in Fig. 3 9 41. 13. (II) For the vectors shown in Fig. 3 9 41, determine (a) B 5 - 3A 5, (b) 2A 5 - 3B 5 + 2C 5, and (c) C 5 - A 5 - B 5. 14. (II) Let V 51 = -6.0 i N + 8.0 j N and V 52 = -4.5 i N - 5.0 j N. Determine the magnitude and direction of (a) V 51, (b) V 52, (c) V 51 + V 52, and (d) V 51 - V 52. 15. (II) (a) Determine the magnitude and direction of the sum of the three vectors V 51 = 4.0 i N - 8.0 j N, V 52 = i N + j N, and V 53 = -2.0 i N + 4.0 j N. (b) Determine V 51 - V 52 + V 53. 16. (II) Suppose a vector V 5 makes an angle f with respect to the y axis. What could be the x and y components of the vector V 5? 17. (II) Two vectors, V 51 and V 52, add to a resultant V 5R = V 51 + V 52. Describe V 51 and V 52 if (a) V R = V 1 + V 2, (b) V R 2 = V 1 2 + V 2 2, (c) V 1 + V 2 = V 1 - V 2. 18. (III) You are given a vector in the xy plane that has a magnitude of 95.0 units and a y component of -60.0 units. (a) What are the two possibilities for its x component? (b) Assuming the x component is known to be positive, specify the vector which, if you add it to the original one, would give a resultant vector that is 80.0 units long and points entirely in the -x direction.
3 – 6 Vector Kinematics 19. (I) The position of a particular particle as a function of time is given by r 5 = (9.60 t i N + 6.45 j N - 1.50 t2 k N ) m. Determine the particle’s velocity and acceleration as a function of time. 20. (I) What was the average velocity of the particle in Problem 19 between t = 1.00 s and t = 3.00 s? What is the magnitude of the instantaneous velocity at t = 2.00 s? 21. (II) A car is moving with speed 16.0 m>s due south at one moment and 25.7 m>s due east 8.00 s later. Over this time interval, determine the magnitude and direction of (a) its average velocity, (b) its average acceleration. (c) What is its average speed? [Hint: Can you determine all these from the information given?] FIGURE 3 – 40 Problem 8.
E N W S (815 km>h) 41.5° v 5 (C = 31.0) C 5 (B = 29.7) (A = 42.0) B 5 A 5 x y 56.0° 28.0° FIGURE 3 – 41 Problems 10, 11, 12, and 13. Vector magnitudes are given in arbitrary units.
FIGURE 3 – 44 Problem 37.
2.5 m u0 Problems 101 22. (II) At t = 0, a particle starts from rest at x = 0, y = 0, and moves in the xy plane with an acceleration a 5 = (4.0 i N + 3.0 j N) m>s 2. Determine (a) the x and y compo- nents of velocity, (b) the speed of the particle, and (c) the position of the particle, all as a function of time. (d) Evaluate all the above at t = 2.0 s. 23. (II) (a) A skier is accelerating down a 30.0° hill at 1.80 m>s2 (Fig. 3 9 42). What is the vertical component of her acceleration? (b) How long will it take her to reach the bottom of the hill, assuming she starts from rest and accelerates uniformly, if the elevation change is 125 m? 24. (II) A hiker follows a winding trail for 5.5 hours while climbing a mountain. The distance along the trail is 11.5 km and the summit is 850 m above and 8.0 km due north of the starting point. What are the average speed and the magnitude and direction of the average velocity vector? 25. (II) An ant walks on a piece of graph paper straight along the x axis a distance of 10.0 cm in 2.40 s. It then turns left 40.0° and walks in a straight line another 10.0 cm in 1.80 s. Finally, it turns another 70.0° to the left and walks another 10.0 cm in 1.55 s. Determine (a) the x and y components of the ant’s average velocity, and (b) its magnitude and direction. 26. (II) Suppose the position of an object is given by r 5 = (3.0 t2 i N - 6.0 t3 j N) m. (a) Determine its velocity v 5 and acceleration a 5, as a function of time. (b) Determine r 5 and v 5 at time t = 3.5 s. 27. (II) A particle’s position as a function of time t is given by r 5 = (5.0 t + 6.0 t2) m i N + (7.0 - 3.0 t3) m j N. At t = 5.0 s, find the magnitude and direction of the particle’s displace-ment vector ∆r 5 relative to the point r 50 = (2.0 i N + 7.0 j N) m. 28. (II) On mountainous downhill roads, an escape lane is sometimes placed to the side of the road for trucks whose brakes might fail. Assuming a constant upward slope of 26°, calculate the horizontal and vertical components of the acceleration of a truck that slowed from 110 km>h to rest in 7.0 s. See Fig. 3 9 43. 29. (II) A light plane is headed due south with a speed relative to still air of 185 km>h. After 1.25 h, the pilot notices that they have covered only 135 km and their direction is not south but 15.0° east of south. What is the wind velocity? 30. (III) An object, which is at the origin at time t = 0, has initial velocity v 50 = ( -14.0 i N - 7.0 j N) m>s and constant accelera-tion a 5 = (6.0 i N + 3.0 j N) m>s2. Find the position r 5 where the object comes to rest (momentarily). 31. (III) A particle starts from the origin at t = 0 with an initial velocity of 5.0 m>s along the positive x axis. If the acceleration is ( -3.0 i N + 4.5 j N) m>s2, determine the velocity and position of the particle at the moment it reaches its maximum x coordinate.
3 – 7 and 3 – 8 Projectile Motion (neglect air resistance) 32. (I) A tiger leaps horizontally from a 7.5-m-high rock with a speed of 3.0 m>s. How far from the base of the rock will she land? 33. (I) A diver running 2.5 m>s dives out horizontally from the edge of a vertical cliff and 3.5 s later reaches the water below. How high was the cliff and how far from its base did the diver hit the water? 34. (II) A ball is thrown horizontally from the roof of a building 7.5 m tall and lands 9.5 m from the base. What was the ball’s initial speed? 35. (II) A ball thrown horizontally at 10.8 m>s from the roof of a building lands 21.0 m from the base of the building. How high is the building? 36. (II) A football is kicked at ground level with a speed of 18.0 m>s at an angle of 31.0° to the horizontal. How much later does it hit the ground? 37. (II) A fire hose held near the ground shoots water at a speed of 6.5 m>s. At what angle(s) should the nozzle point in order that the water land 2.5 m away (Fig. 3 9 44)? Why are there two different angles? Sketch the two trajectories. 38. (II) You buy a plastic dart gun, and being a clever physics student you decide to do a quick calculation to find its maximum horizontal range. You shoot the gun straight up, and it takes 3.4 s for the dart to land back at the barrel. What is the maximum horizontal range of your gun? 39. (II) A projectile is fired with an initial speed of 38.8 m>s at an angle of 42.2° above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered (that is, the range), and (d) the speed of the projectile 1.50 s after firing. 40. (II) A diver leaves the end of a diving board that is 2.0 m above the surface of the water. Her dive takes her 1.2 m above the board, and then into the water a horizontal distance of 2.2 m from the end of the board. At what speed and angle did she leave the board? 41. (II) The maximum range of a projectile is found to be 95 m. If the projectile strikes the ground a distance of 68 m away, what was the angle of launch?
Main road downhill Escape route FIGURE 3 – 43 Problem 28.
FIGURE 3 – 42 Problem 23.
30.0° a = 1.80 m>s2 v0 45° 2.5 m 10.0 m FIGURE 3 – 49 Problem 53.
102 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors 42. (II) A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 62.0 m>s at an angle of 35.0° with the horizontal, as shown in Fig. 3 9 45.
(a) Determine the time taken by the projectile to hit point P at ground level. (b) Determine the distance X of point P from the base of the vertical cliff. At the instant just before the projectile hits point P , fi nd (c) the horizontal and the vertical components of its velocity, (d) the magnitude of the velocity, and (e) the angle made by the velocity vector with the horizontal. (f) Find the maximum height above the cliff top reached by the projectile. 43. (II) An athlete performing a long jump leaves the ground at a 27.0° angle and lands 7.80 m away. (a) What was the takeoff speed? (b) If this speed were increased by just 5.0%, how much longer would the jump be? 44. (II) In Example 3 9 11 we chose the x axis to the right and y axis up. Redo this problem by defi ning the x axis to the left and y axis down, and show that the conclusion remains the same : the football lands on the ground 40.5 m to the right of where it departed the punter’s foot. 45. (II) A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp (Fig. 3 9 46). (a) With what minimum speed must he drive off the hori-zontal ramp? The vertical height of the ramp is 1.5 m above the car roofs and the horizontal distance he must clear is 22 m. (b) If the ramp is now tilted upward, so that “takeoff angle” is 9.0° above the horizontal, what is the new minimum speed?
22 m Must clear this point!
1.5 m FIGURE 3 – 46 Problem 45. 46. (II) A baseball is hit with a speed of 27.0 m>s at an angle of 45.0°. It lands on the fl at roof of a 12.5-m-tall nearby building. If the ball was hit when it was 1.2 m above the ground, what horizontal distance does it travel before it lands on the building? 47. (II) A rescue plane wants to drop supplies to isolated moun-tain climbers on a rocky ridge 265 m below. If the plane is traveling horizontally with a speed of 125 km>h , how far in advance of the recipients (horizontal distance) must the goods be dropped? 48. (II) Show that the time required for a projectile to reach its highest point is equal to the time for it to return to its original height if air resistance is negligible. 49. (II) Exactly 3.0 s after a projectile is fi red into the air from the ground, it is observed to have a velocity v 5 = (7.8 i N + 5.2 j N) m>s, where the x axis is horizontal and the y axis is positive upward. Determine (a) the horizontal range of the projectile, (b) its maximum height above the ground, and (c) its speed and angle of motion just before it strikes the ground. 50. (II) At what projection angle will the range of a projectile equal its maximum height? 51. (II) A ball is thrown horizontally from the top of a cliff with initial speed v0 (at t = 0). At any moment, its direction of motion makes an angle u to the horizontal (Fig. 3 9 47). Derive a formula for u as a function of time, t, as the ball follows a projectile’s path. 52. (II) Romeo is throwing pebbles gently up to Juliet’s window, and he wants the pebbles to hit the window with only a horizontal component of velocity. He is standing at the edge of a rose garden 8.0 m below her window and 8.5 m from the base of the wall (Fig. 3 9 48). How fast are the pebbles going when they hit her window? 53. (II) (a) A long jumper leaves the ground at 45° above the hori-zontal and lands 8.0 m away. What is her “takeoff” speed v0 ? (b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away hori-zontally and 2.5 m vertically below. If she jumps from the edge of the left bank at 45° with the speed calculated in (a), how long, or short, of the opposite bank will she land (Fig. 3 9 49)?
35.0° P h = 125 m X 0 = 62.0 m>s v FIGURE 3 – 45 Problem 42.
v 5 u FIGURE 3 – 47 Problem 51.
FIGURE 3 – 48 Problem 52.
8.5 m 8.0 m Problems 103 54. (II) Show that the speed with which a projectile leaves the ground is equal to its speed just before it strikes the ground at the end of its journey, assuming the firing level equals the landing level. 55. (II) If a ball is kicked from ground level at 15.0 m>s, there are two launch angles that will make the ball land 20.0 m away. (a) What are the two angles? (b) What maximum height does the ball reach in each case? (c) How long is the ball in the air for each case? Ignore air resistance. 56. (II) At serve, a tennis player aims to hit the ball horizontally. What minimum speed is required for the ball to clear the 0.90-m-high net about 15.0 m from the server if the ball is “launched” from a height of 2.30 m? Where will the ball land if it just clears the net (and will it be “good” in the sense that it lands within 7.0 m of the net)? How long will it be in the air? See Fig. 3 9 50. 57. (II) An Olympic long jumper is capable of jumping 8.0 m. Assuming his horizontal speed is 9.1 m>s as he leaves the ground, how long is he in the air and how high does he go? Assume that he lands standing upright : that is, the same way he left the ground. 58. (III) Revisit Example 3 9 9, and assume that the boy with the slingshot is below the boy in the tree (Fig. 3 9 51) and so aims upward, directly at the boy in the tree. Show that again the boy in the tree makes the wrong move by letting go at the moment the water balloon is shot. u0 v0 FIGURE 3 – 51 Problem 58. 59. (III) Suppose the kick in Example 3 9 8 is attempted 36.0 m from the goalposts, whose crossbar is 3.05 m above the ground. If the football is directed perfectly between the goalposts, will it pass over the bar and be a field goal? Show why or why not. If not, from what horizontal distance must this kick be made if it is to score? 60. (III) A person stands at the base of a hill that is a straight incline making an angle f with the horizontal (Fig. 3 9 52). For a given initial speed v0, at what angle u (to the hori-zontal) should objects be thrown so that the distance d they land up the hill is as large as possible?
f u d FIGURE 3 – 52 Problem 60. Given f and v0, determine u to make d maximum. 61. (III) Derive a formula for the horizontal range R of a projectile when it lands at a height h above its initial point. (For h 6 0, it lands a distance -h below the starting point.) Assume it is projected at an angle u0 with initial speed v0.
3 – 9 Relative Velocity 62. (I) A person going for a morning jog on the deck of a cruise ship is running toward the bow (front) of the ship at 2.5 m>s while the ship is moving ahead at 8.8 m>s. What is the velocity of the jogger relative to the water? Later, the jogger is moving toward the stern (rear) of the ship. What is the jogger’s velocity relative to the water now? 63. (I) Huck Finn walks at a speed of 0.70 m>s across his raft (that is, he walks perpendicular to the raft’s motion relative to the shore). The heavy raft is traveling down the Mississippi River at a speed of 1.50 m>s rela-tive to the river bank (Fig. 3 9 53). What is Huck’s veloc ity (speed and direction) relative to the river bank? 64. (II) Determine the speed of the boat with respect to the shore in Example 3 9 13. 65. (II) A motorboat whose speed in still water is 4.30 m>s must aim upstream at an angle of 23.5° (with respect to a line perpendicular to the shore) in order to travel directly across the stream. (a) What is the speed of the current? (b) What is the resultant speed of the boat with respect to the shore? (See Fig. 3 9 33.) 66. (II) A passenger on a boat moving at 1.70 m>s on a still lake walks up a flight of stairs at a speed of 0.60 m>s, Fig. 3 9 54. The stairs are angled at 45° pointing in the direction of motion as shown. Write the vector velocity of the passenger relative to the water.
0.60 m>s = 1.70 m>s 45° x y v FIGURE 3 – 54 Problem 66. 67. (II) An airplane is heading due south at a speed of 688 km>h. If a wind begins blowing from the southwest at a speed of 85.0 km>h (average), calculate (a) the velocity (magnitude and direction) of the plane, relative to the ground, and (b) how far from its intended position it will be after 11.0 min if the pilot takes no corrective action. [Hint: First draw a diagram.] 68. (II) In what direction should the pilot aim the plane in Problem 67 so that it will fly due south?
FIGURE 3 –53 Problem 63.
River current 0.70 m>s 15.0 m 7.0 m 2.30 m FIGURE 3 – 50 Problem 56.
v0 0 h u FIGURE 3 – 58 Problem 81.
104 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors 69. (II) Raindrops make an angle u with the vertical when viewed through a moving train window (Fig. 3 9 55). If the speed of the train is vT, what is the speed of the raindrops in the reference frame of the Earth in which they are assumed to fall vertically? 70. (II) A boat, whose speed in still water is 2.80 m>s, must cross a 265-m-wide river and arrive at a point 118 m upstream from where it starts (Fig. 3 9 56). To do so, the pilot must head the boat at a 45.0° upstream angle. What is the speed of the river’s current? 71. (II) A swimmer is capable of swimming 0.60 m>s in still water. (a) If she aims her body directly across a 55-m-wide river whose current is 0.50 m>s, how far downstream (from a point opposite her starting point) will she land? (b) How long will it take her to reach the other side?
72. (II) (a) At what upstream angle must the swimmer in Problem 71 aim, if she is to arrive at a point directly across the stream? (b) How long will it take her?
73. (II) Two cars approach a street corner at right angles to each other (Fig. 3 9 57). Car 1 travels at a speed relative to Earth v1 E = 35 km>h, and car 2 at v2 E = 55 km>h. What is the relative velocity of car 1 as seen by car 2? What is the velocity of car 2 relative to car 1? 74. (III) An airplane, whose air speed is 560 km>h, is supposed to fly in a straight path 38.0° N of E. But a steady 82 km>h wind is blowing from the north. In what direction should the plane head? [Hint: Use the law of sines, Appendix A 9 9.] 75. (III) Point B is located across the river from point A and at a 42.0° angle upstream from A. A boat travels from point A to point B, sailing at 18.0 km>h relative to the water. The current in the river flows at 5.60 km>h. The river is 1.45 km wide. (a) At what angle should the boat head? (b) What will be the boat’s speed relative to the ground? (c) How much time does the trip from point A to point B take?
General Problems 76. A plumber steps out of his truck, walks 55 m east and 38 m south, and then takes an elevator 12 m into the subbasement of a building where a bad leak is occurring. What is the displacement of the plumber relative to his truck? Give your answer in components. Also give the magnitude and angles, with respect to the x axis, in the vertical and horizontal planes. Assume x is east, y is north, and z is up. 77. Apollo astronauts took a “nine iron” to the Moon and hit a golf ball about 180 m. Assuming that the swing, launch angle, and so on, were the same as on Earth where the same astronaut could hit it only 32 m, estimate the accel eration due to gravity on the surface of the Moon. (We neglect air resistance in both cases, but on the Moon there is none.) 78. A car moving at 95 km>h passes a 1.40-km-long train traveling in the same direction on a track that is parallel to the road. If the speed of the train is 75 km>h, how long does it take the car to pass the train, and how far will the car have traveled in this time? What are the results if the car and train are instead traveling in opposite directions? 79. When Babe Ruth hit a homer over the 8.0-m-high right-field fence 98 m from home plate, roughly what was the minimum speed of the baseball when it left the bat? Assume the ball was hit 1.0 m above the ground and its path initially made a 36° angle with the ground. 80. A child runs down a 12° hill and then suddenly jumps upward at a 15° angle above horizontal and lands 1.3 m down the hill as measured along the hill. What was the child’s initial speed at the jump? 81. Here is something to try at a sporting event. Show that the maximum height h attained by an object projected into the air, such as a baseball, football, or soccer ball, is approximately given by h L 1.2 t2 m, where t is the total time of flight for the object in sec onds. Assume that the object returns to the same level as that from which it was launched, as in Fig. 3 9 58. For example, if you count seconds and find that a baseball was in the air for t = 5.0 s, the maximum height attained was h = 1.2 (5.0) 2 = 30 m. The fun of this relation is that h can be determined without knowledge of the launch speed v0 or launch angle u0. Why is that exactly? See Section 3 9 8.
FIGURE 3 – 57 Problem 73.
1 2 1E 2E v 5 v 5 current Start Finish 118 m River 265 m Path of boat 45.0° FIGURE 3 – 56 Problem 70.
u FIGURE 3 – 55 Problem 69.
135 m 195 m Landing point v0 u FIGURE 3 – 64 Problem 92.
12.0 m>s 5.0 m 35 m General Problems 105 82. The cliff divers of Acapulco push off hori-zontally from rock platforms about 35 m above the water, but they must clear rocky outcrops at water level that extend out into the water 5.0 m from the base of the cliff directly under their launch point. See Fig. 3 9 59. What minimum pushoff speed is necessary to clear the rocks? How long are they in the air? 83. A grasshopper hops along a level road. On each hop, the grasshopper launches itself at angle u0 = 45° and achieves a range R = 0.80 m. What is the average horizontal speed of the grasshopper as it hops along the road? Assume that the time spent on the ground between hops is negligible. 84. Spymaster Andrea, fl ying a constant 208 km>h horizontally in a low-fl ying helicopter, wants to drop secret documents into her contact’s open car which is traveling 135 km>h on a level highway 78.0 m below. At what angle (with the hori zontal) should the car be in her sights when the packet is released (Fig. 3 9 60)?
135 km>h 78.0 m 208 km>h u FIGURE 3 – 60 Problem 84. 85. A basketball leaves a player’s hands at a height of 2.10 m above the fl oor. The basket is 3.05 m above the fl oor. The player likes to shoot the ball at a 38.0° angle. If the shot is made from a horizontal distance of 11.00 m and must be accurate to {0.22 m (horizontally), what is the range of initial speeds allowed to make the basket? 86. Extreme-sports enthusiasts have been known to jump off the top of El Capitan, a sheer granite cliff of height 910 m in Yosemite National Park. Assume a jumper runs horizontally off the top of El Capitan with speed 4.0 m>s and enjoys a free fall until she is 150 m above the valley fl oor, at which point she opens her para-chute (Fig. 3 9 61). (a) How long is the jumper in free fall? Ignore air resistance. (b) It is important to be as far away from the cliff as possible before opening the parachute. How far from the cliff is this jumper when she opens her chute? 87. If a baseball pitch leaves the pitcher’s hand horizontally at a velocity of 130 km>h, by what % will the pull of gravity change the magnitude of the velocity when the ball reaches the batter, 18 m away? For this estimate, ignore air resis-tance and spin on the ball. 88. A projectile is launched from ground level to the top of a cliff which is 195 m away and 135 m high (see Fig. 3 9 62). If the projectile lands on top of the cliff 6.6 s after it is fi red, fi nd the initial velocity of the projectile (magni-tude and direc-tion). Neglect air resistance. 89. A diver leaves the end of a 3.5-m-high diving board and strikes the water 1.3 s later, 3.0 m beyond the end of the board. Considering the diver as a particle, determine: (a) her initial velocity, v 50; (b) the maximum height reached; and (c) the velocity v 5f with which she enters the water. 90. A basketball is shot from an initial height of 2.40 m (Fig. 3 9 63) with an initial speed v0 = 12 m>s directed at an angle u0 = 35° above the horizontal. (a) How far from the basket was the player if he made a basket? (b) At what angle to the horizontal did the ball enter the basket? 91. A hunter aims directly at a target (on the same level) 38.0 m away. (a) If the arrow leaves the bow at a speed of 21.3 m>s, by how much will it miss the target? (b) At what angle should the bow be aimed so the target will be hit? 92. A person in the passenger basket of a hot-air balloon throws a ball horizontally outward from the basket with speed 12.0 m>s (Fig. 3 9 64). What initial velocity (magnitude and direction) does the ball have relative to a person standing on the ground (a) if the hot-air balloon is rising at 3.0 m>s relative to the ground during this throw, and (b) if the hot-air balloon is descending at 3.0 m>s relative to the ground?
FIGURE 3 – 59 Problem 82.
4.0 m>s 150 m 910 m FIGURE 3 – 61 Problem 86.
35° x = ?
v0 = 12 m>s 10 ft = 3.05 m 2.40 m FIGURE 3 – 62 Problem 88.
FIGURE 3 – 63 Problem 90.
106 CHAPTER 3 Kinematics in Two or Three Dimensions; Vectors FIGURE 3 – 66 Problem 97.
22° Fielder runs to here from here 55° 85 m 93. A rock is kicked horizontally at 15 m>s from a hill with a 45° slope (Fig. 3 9 65). How long does it take for the rock to hit the ground? 94. A batter hits a fly ball which leaves the bat 0.90 m above the ground at an angle of 64° with an initial speed of 28 m>s heading toward centerfield. Ignore air resistance. (a) How far from home plate would the ball land if not caught? (b) The ball is caught by the centerfielder who, starting at a distance of 105 m from home plate just as the ball was hit, runs straight toward home plate at a constant speed and makes the catch at ground level. Find his speed. 95. A ball is shot from the top of a building with an initial velocity of 24 m>s at an angle u = 42° above the horizontal. (a) What are the horizontal and vertical components of the initial velocity? (b) If a nearby building is the same height and 65 m away, how far below the top of the building will the ball strike the nearby building? 96. The speed of a boat in still water is v. The boat is to make a round trip in a river whose current travels at speed u. Derive a formula for the time needed to make a round trip of total distance D if the boat makes the round trip by moving (a) upstream and back downstream, and (b) directly across the river and back. We must assume u 6 v ; why? 97. At t = 0 a batter hits a baseball with an initial speed of 28 m>s at a 55° angle to the horizontal. An outfielder is 85 m from the batter at t = 0 and, as seen from home plate, the line of sight to the outfielder makes a horizontal angle of 22° with the plane in which the ball moves (see Fig. 3 9 66). What speed and direction must the fielder take to catch the ball at the same height from which it was struck? Give the angle with respect to the outfielder’s line of sight to home plate. 98. A child, who is 45 m from the bank of a river, is being carried helplessly downstream by the river’s swift current of 1.0 m>s. As the child passes a lifeguard on the river’s bank, the lifeguard starts swimming in a straight line (Fig. 3 9 67) until she reaches the child at a point downstream. If the lifeguard can swim at a speed of 2.0 m>s relative to the water, how long does it take her to reach the child? How far downstream does the lifeguard intercept the child? 99. A particle has a velocity of v 5 = ( -3.0 i N + 4.5t j N) m>s. The particle starts at r 5 = (2.5 i N - 3.1 j N) m at t = 0. Give the position and acceleration as a function of time. What is the shape of the resulting path? 100. In hot pursuit, Agent Logan of the FBI must get directly across a 1300-m-wide river in minimum time. The river’s current is 0.80 m>s, he can row a boat at 1.60 m>s, and he can run 3.00 m>s. Describe the path he should take (rowing plus running along the shore) for the minimum crossing time, and determine the minimum time. 101. You are driving south on a highway at 12 m>s (approxi-mately 25 mi>h) in a snowstorm. When you last stopped, you noticed that the snow was coming down vertically, but it is passing the windows of the moving car at an angle of 9.0° to the horizontal. Estimate the speed of the vertically falling snowflakes relative to the ground. [Hint: Construct a relative velocity diagram similar to Fig. 3 9 33 or 3 9 34. Be careful about which angle is the angle given.] 102. A boat is traveling where there is a current of 0.20 m>s east (Fig. 3 9 68). To avoid some offshore rocks, the boat must clear a buoy that is NNE (22.5°) and 2.8 km away. The boat’s speed through still water is 2.1 m>s. If the boat wants to pass the right side of the buoy by 0.15 km, at what angle should the boat head?
15 m>s 45° FIGURE 3 – 65 Problem 93.
45 m 2.0 m>s 1.0 m>s FIGURE 3 – 67 Problem 98.
A N S W E R S T O E X E R C I S E S A: 3.0 12 L 4.2 units.
B: (a).
C: (d). Helicopter provides the initial velocity of the box.
D: (a) v = vx 0 = 16.0 m>s, horizontal; (b) 9.80 m>s2 down.
E: Both balls reach the same height, so are in the air for the same length of time.
Buoy 22.5° N 0.20 m>s Current FIGURE 3 – 68 Problem 102.
107 CHAPTER-OPENING QUESTIONS— Guess now!
1. A 150-kg football player collides head-on with a 75-kg running back. During the collision, the heavier player exerts a force of magnitude FA on the smaller player. If the smaller player exerts a force FB back on the heavier player, which response is most accurate?
(a) FB = FA.
(b) FB 6 FA.
(c) FB 7 FA.
(d) FB = 0.
(e) We need more information.
2. A line by the poet T. S. Eliot (from Murder in the Cathedral) has the women of Canterbury say “the earth presses up against our feet.” What force is this?
(a) Gravity.
(b) The normal force.
(c) A friction force.
(d) Centrifugal force.
(e) No force : they are being poetic.
A space shuttle is carried out into space by powerful rockets. They are accelerating, increasing in speed rapidly. To do so, a force must be exerted on them according to Newton’s second law, πF 5 = ma 5. What exerts this force? The rocket engines exert a force on the gases they push out (expel) from the rear of the rockets (labeled F 5GR). According to Newton’s third law, these ejected gases exert an equal and opposite force on the rockets in the forward direction. This “reaction” force exerted on the rockets by the gases, labeled F 5RG, is the net force on the rockets and accelerates the rockets. (Any other forces, such as gravity, are assumed small in comparison.) CONTENTS 4–1 Force 4–2 Newton’s First Law of Motion 4–3 Mass 4–4 Newton’s Second Law of Motion 4–5 Newton’s Third Law of Motion 4–6 Weight : the Force of Gravity; and the Normal Force 4–7 Solving Problems with Newton’s Laws: Free-Body Diagrams 4–8 Problem Solving : A General Approach Dynamics: Newton’s Laws of Motion RG F GR F 5 5 C H A P T E R 4 108 CHAPTER 4 Dynamics: Newton’s Laws of Motion W e have discussed how motion is described in terms of velocity and acceleration. Now we deal with the question of why objects move as they do: What makes an object at rest begin to move? What causes an object to accelerate or decelerate? What is involved when an object moves in a curved path? We can answer in each case that a force is required. In this Chapter,† we will investigate the connection between force and motion, which is the subject called dynamics.
4–1 Force Intuitively, we experience force as any kind of a push or a pull on an object. When you push a stalled car or a grocery cart (Fig. 4 9 1), you are exerting a force on it. When a motor lifts an elevator, or a hammer hits a nail, or the wind blows the leaves of a tree, a force is being exerted. We often call these contact forces because the force is exerted when one object comes in contact with another object. On the other hand, we say that an object falls because of the force of gravity (which is not a contact force).
If an object is at rest, to start it moving requires force : that is, a force is needed to accelerate an object from zero velocity to a nonzero velocity. For an object already moving, if you want to change its velocity : either in direction or in magnitude : a force is required. In other words, to accelerate an object, a force is always required. In Section 4 9 4 we discuss the precise relation between acceleration and net force, which is Newton’s second law.
One way to measure the magnitude (or strength) of a force is to use a spring scale (Fig. 4 9 2). Normally, such a spring scale is used to fi nd the weight of an object; by weight we mean the force of gravity acting on the object (Section 4 9 6). The spring scale, once calibrated, can be used to measure other kinds of forces as well, such as the pulling force shown in Fig. 4 9 2.
A force exerted in a different direction has a different effect. Force has direc-tion as well as magnitude, and is indeed a vector that follows the rules of vector addition discussed in Chapter 3. We can represent any force on a diagram by an arrow, just as we did with velocity. The direction of the arrow is the direction of the push or pull, and its length is drawn proportional to the magnitude of the force. Force vectors in this book are drawn as red in color, velocity is green.
†We treat everyday objects in motion here. When velocities are extremely high, close to the speed of light (3.0 108 m>s), we use the theory of relativity (Chapter 36), and in the submicroscopic world of atoms and molecules we use quantum theory (Chapter 37 ff).
FIGURE 4 – 1 A force exerted on a grocery cart : in this case exerted by a person.
FIGURE 4 – 2 A spring scale used to measure a force.
4–2 Newton’s First Law of Motion What is the relationship between force and motion? The ancient Greeks, including Aristotle (384 9 322 b.c.), believed that a force was required to keep an object moving along a horizontal plane. To Aristotle, the natural state of an object was at rest, and a force was thought necessary to keep an object in motion. Aristotle also argued that the greater the force on the object, the greater its speed.
Some 2000 years later, in the early 1600s, Galileo disagreed. Galileo main-tained that it is just as natural for an object to be in motion with a constant velocity as it is for it to be at rest. See Section 1 9 1, page 24.
0 1 2 3 4 5 6 7 8 910 SECTION 4–2 Newton’s First Law of Motion 109 To understand Galileo’s idea, consider the following observations involving motion along a horizontal plane. To push an object with a rough surface along a tabletop at constant speed requires a certain amount of force. To push an equally heavy object with a very smooth surface across the table at the same speed will require less force. If a layer of oil or other lubricant is placed between the surface of the object and the table, then almost no force is required to keep the object moving. Notice that in each successive step, less force is required. As the next step, we imagine there is no friction at all, that the object does not rub against the table : or there is a perfect lubricant between the object and the table : and theorize that once started, the object would move across the table at constant speed with no force applied. A steel ball bearing rolling on a hard horizontal surface approaches this situation. So does a puck on an air table, in which a thin layer of air reduces friction almost to zero.
It was Galileo’s genius to imagine such an idealized world : in this case, one where there is no friction : and to see that it could lead to a more accurate and richer understanding of the real world. This idealization led him to his remarkable conclusion that if no force is applied to a moving object, it will continue to move with constant speed in a straight line. An object slows down only if a force is exerted on it. Galileo thus interpreted friction as a force akin to ordinary pushes and pulls.
To push an object across a table at constant speed requires a force from your hand that can balance the force of friction (Fig. 4 9 3). When the object moves at constant speed, your pushing force is equal in magnitude to the friction force; but these two forces are in opposite directions, so the net force on the object (the vector sum of the two forces) is zero. This is consistent with Galileo’s viewpoint: the object moves with constant velocity when no net force is exerted on it.
Upon this foundation laid by Galileo, Isaac Newton (Fig. 4 9 4) built his great theory of motion. Newton’s analysis of motion is summarized in his famous “three laws of motion.” In his great work, the Principia (published in 1687), Newton readily acknowledged his debt to Galileo. In fact, Newton’s fi rst law of motion is close to Galileo’s conclusions. It states that Every object continues in its state of rest, or of uniform velocity in a straight line, as long as no net force acts on it.
The tendency of an object to maintain its state of rest or of uniform velocity in a straight line is called inertia. Newton’s fi rst law is thus often called the law of inertia.
F Ffr 5 5 FIGURE 4 – 3 F 5 represents the force applied by the person and F 5fr represents the force of friction.
FIGURE 4 – 4 Isaac Newton (1642 9 1727). Besides developing mechanics, including his three great laws of motion and the law of universal gravitation, he also tried to understand the nature of light.
NEWTON’S FIRST LAW OF MOTION Inertial Reference Frames Newton’s fi rst law does not hold in every reference frame. For example, if your reference frame is an accelerating car, an object such as a cup resting on the dashboard may begin to move toward you (it stayed at rest as long as the car’s velocity remained constant). The cup accelerated toward you, but neither you nor anything else exerted a force on it in that direction. Similarly, in the reference frame of the decelerating bus in Example 4 9 1, there was no force pushing the backpacks forward. In accelerating reference frames, Newton’s fi rst law does not hold. Physics is easier in reference frames in which Newton’s fi rst law does hold, and they are called inertial reference frames (the law of inertia is valid in them). For most purposes, we usually make the approximation that a reference frame fi xed on the Earth is an inertial frame. This is not precisely true, due to the Earth’s rotation, but usually it is close enough.
CONCEPTUAL EXAMPLE 4 – 1 Newton’s fi rst law. A school bus comes to a sudden stop, and all of the backpacks on the fl oor start to slide forward. What force causes them to do that?
RESPONSE It isn’t “force” that does it. By Newton’s fi rst law, the backpacks continue their state of motion, maintaining their velocity. The backpacks slow down if a force is applied, such as friction with the fl oor.
110 CHAPTER 4 Dynamics: Newton’s Laws of Motion Any reference frame that moves with constant velocity (say, a car or an airplane) relative to an inertial frame is also an inertial reference frame. Reference frames where the law of inertia does not hold, such as the accelerating reference frames discussed above, are called noninertial reference frames. How can we be sure a reference frame is inertial or not? By checking to see if Newton’s fi rst law holds. Thus Newton’s fi rst law serves as the defi nition of inertial reference frames.
4–3 Mass Newton’s second law, which we come to in the next Section, makes use of the concept of mass. Newton used the term mass as a synonym for “quantity of matter.” This intuitive notion of the mass of an object is not very precise because the concept “quantity of matter” is not very well defi ned. More precisely, we can say that mass is a measure of the inertia of an object. The more mass an object has, the greater the force needed to give it a particular acceleration. It is harder to start it moving from rest, or to stop it when it is moving, or to change its velocity sideways out of a straight-line path. A truck has much more inertia than a baseball moving at the same speed, and a much greater force is needed to change the truck’s velocity at the same rate as the ball’s. The truck therefore has much more mass.
To quantify the concept of mass, we must defi ne a standard. In SI units, the unit of mass is the kilogram (kg) as we discussed in Chapter 1, Section 1 9 4.
The terms mass and weight are often confused with one another, but it is important to distinguish between them. Mass is a property of an object itself (a measure of an object’s inertia, or its “quantity of matter”). Weight, on the other hand, is a force, the pull of gravity acting on an object. To see the differ-ence, suppose we take an object to the Moon. The object will weigh only about one-sixth as much as it did on Earth, since the force of gravity is weaker. But its mass will be the same. It will have the same amount of matter as on Earth, and will have just as much inertia. It will be just as hard to start it moving on the Moon as on Earth. (More on weight in Section 4 9 6.) 4–4 Newton’s Second Law of Motion Newton’s fi rst law states that if no net force is acting on an object at rest, the object remains at rest; or if the object is moving, it continues moving with constant speed in a straight line. But what happens if a net force is exerted on an object? Newton perceived that the object’s velocity will change (Fig. 4 9 5). A net force exerted on an object may make its velocity increase. Or, if the net force is in a direction opposite to the motion, that force will reduce the object’s velocity. If the net force acts sideways on a moving object, the direction of the object’s velocity changes. That change in the direction of the velocity is also an acceleration. So a sideways net force on an object also causes acceleration. In general, we can say that a net force causes acceleration.
What precisely is the relationship between acceleration and force? Everyday experience can suggest an answer. Consider the force required to push a cart when friction is small enough to ignore. (If there is friction, consider the net force, which is the force you exert minus the force of friction.) If you push the cart horizontally with a gentle but constant force for a certain period of time, you will make the cart accelerate from rest up to some speed, say 3 km>h. If you push with twice the force, the cart will reach 3 km>h in half the time. The accel eration will be twice as great. If you triple the force, the acceleration is tripled, and so on. Thus, the acceleration of an object is directly proportional to the net applied force. But the acceleration depends on the mass of the object as well. If you push an empty grocery cart with the same force as you push one that is fi lled with groceries, you will fi nd that the full cart has a smaller acceleration. The greater the mass, the less the acceleration for the same net force. The mathematical relation, as Newton argued, is that the C A U T I O N Distinguish mass from weight FIGURE 4 – 5 The bobsled accelerates because the team exerts a force.
SECTION 4–4 Newton’s Second Law of Motion 111 acceleration of an object is inversely proportional to its mass. These relationships are found to hold in general and can be summarized as follows: The acceleration of an object is directly proportional to the net force acting on it, and is inversely proportional to the object’s mass. The direction of the acceleration is in the direction of the net force acting on the object.
This is Newton’s second law of motion.
Newton’s second law can be written as an equation: a 5 = πF 5 m , where a 5 stands for acceleration, m for the mass, and πF 5 for the net force on the object. The symbol π (Greek “sigma”) stands for “sum of”; F 5 stands for force, so πF 5 means the vector sum of all forces acting on the object, which we define as the net force.
We rearrange this equation to obtain the familiar statement of Newton’s second law: πF 5 = ma 5. (4 – 1a) Newton’s second law relates the description of motion (acceleration) to the cause of motion (force). It is one of the most fundamental relationships in physics. From Newton’s second law we can make a more precise definition of force as an action capable of accelerating an object.
Every force F 5 is a vector, with magnitude and direction. Equation 4 9 1a is a vector equation valid in any inertial reference frame. It can be written in component form in rectangular coordinates as πFx = max, πFy = may, πFz = maz, (4 – 1b) where F 5 = Fx i N + Fy j N + Fz k N.
The component of acceleration in each direction is affected only by the compo-nent of the net force in that direction. (See Section 3 9 5 for unit vectors i N, j N, k N .) If the motion is all along a line (one-dimensional), we can leave out the subscripts and simply write πF = ma. Again, a is the acceleration of an object of mass m, and πF includes all the forces acting on that object, and only forces acting on that object. Sometimes the net force πF is written as Fnet, so Fnet = ma.
In SI units, with the mass in kilograms, the unit of force is called the newton (N). One newton is the force required to impart an acceleration of 1 m>s2 to a mass of 1 kg. Thus 1 N = 1 kg⋅m>s2.
In cgs units, the unit of mass is the gram (g), which is 1 1000 of a kilogram.† The unit of force is the dyne, which is defined as the net force needed to impart an acceleration of 1 cm>s2 to a mass of 1 g. Thus 1 dyne = 1 g⋅cm>s2. Because 1 g = 10-3 kg and 1 cm = 10-2 m, then 1 dyne = 10-5 N.
In the British system, which we rarely use, the unit of force is the pound (abbre-viated lb),‡ where 1 lb = 4.44822 N L 4.45 N. The unit of mass is the slug, which is defined as that mass which will undergo an acceleration of 1 ft>s2 when a force of 1 lb is applied to it. Thus 1 lb = 1 slug⋅ft>s2. Table 4 9 1 summarizes the units in the different systems.
It is very important that only one set of units be used in a given calculation or problem, with the SI being what we almost always use. If the force is given in, say, newtons, and the mass in grams, then before attempting to solve for the acceleration in SI units, we must change the mass to kilograms. For example, if the force is given as 2.0 N along the x axis and the mass is 500 g, we change the latter to 0.50 kg, and the acceleration will then automatically come out in m>s2 when Newton’s second law is used: ax = πFx m = 2.0 N 0.50 kg = 2.0 kg⋅m>s2 0.50 kg = 4.0 m>s2, where we set 1 N = 1 kg⋅m>s2.
NEWTON’S SECOND LAW OF MOTION NEWTON’S SECOND LAW OF MOTION TABLE 4–1 Units for Mass and Force System Mass Force SI kilogram (kg) newton (N) ( = kg⋅m>s2) cgs gram (g) dyne ( = g⋅cm>s2) British slug pound (lb) Conversion factors: 1 dyne = 10-5 N; 1 lb L 4.45 N; 1 slug L 14.6 kg.
P R O B L E M S O LV I N G Use a consistent set of units †Be careful not to confuse g for gram with g for the acceleration due to gravity. The latter is always italicized (or boldface when shown as a vector).
‡The abbreviation lb for pound comes from the Latin (language of the ancient Romans) word “libra.” 112 CHAPTER 4 Dynamics: Newton’s Laws of Motion Newton’s second law, like the first law, is valid only in inertial reference frames (Section 4 9 2). In the noninertial reference frame of a car that begins accelerating, a cup on the dashboard starts sliding : it accelerates : even though the net force on it is zero. Thus πF 5 = ma 5 does not work in such an accelerating reference frame (πF 5 = 0, but a 5 ≠0 in this noninertial frame).
EXERCISE A Suppose you watch a cup slide on the (smooth) dashboard of an acceler-ating car as we just discussed, but this time from an inertial reference frame outside the car, on the street. From your inertial frame, Newton’s laws are valid. What force pushes the cup off the dashboard?
Precise Definition of Mass As mentioned in Section 4 9 3, we can quantify the concept of mass using its defi-nition as a measure of inertia. How to do this is evident from Eq. 4 9 1a, where we see that the acceleration of an object is inversely proportional to its mass. EXAMPLE 4 – 3 Force to stop a car. What average net force is required to bring a 1500-kg car to rest from a speed of 100 km>h within a distance of 55 m?
APPROACH We use Newton’s second law, πF = ma, to determine the force, but first we need to calculate the acceleration a. We assume the acceleration is constant so that we can use the kinematic equations, Eqs. 2 9 12, to calculate it.
v = 0 v0 = 100 km>h x = 0 x = 55m x (m) FIGURE 4 – 6 Example 4 9 3.
SOLUTION We assume the motion is along the +x axis (Fig. 4 9 6). We are given the initial velocity v0 = 100 km>h = 27.8 m>s (Section 1 9 5), the final velocity v = 0, and the distance traveled x - x0 = 55 m. From Eq. 2 9 12c, we have v2 = v0 2 + 2a (x - x0) , so a = v2 - v0 2 2(x - x0) = 0 - (27.8 m>s) 2 2(55 m) = -7.0 m>s2.
The net force required is then πF = ma = (1500 kg) ( -7.0 m>s2) = -1.1 104 N, or 11,000 N. The force must be exerted in the direction opposite to the initial velocity, which is what the negative sign means.
NOTE If the acceleration is not precisely constant, then we are determining an “average” acceleration and we obtain an “average” net force.
EXAMPLE 4 – 2 ESTIMATE Force to accelerate a fast car. Estimate the net force needed to accelerate (a) a 1000-kg car at 1 2 g; (b) a 200-gram apple at the same rate.
APPROACH We use Newton’s second law to find the net force needed for each object; we are given the mass and the acceleration. This is an estimate (the 1 2 is not said to be precise) so we round off to one significant figure.
SOLUTION (a) The car’s acceleration is a = 1 2 g = 1 2 (9.8 m>s2) L 5 m>s2. We use Newton’s second law to get the net force needed to achieve this acceleration: πF = ma L (1000 kg) (5 m>s2) = 5000 N.
(If you are used to British units, to get an idea of what a 5000-N force is, you can divide by 4.45 N>lb and get a force of about 1000 lb.) (b) For the apple, m = 200 g = 0.2 kg, so πF = ma L (0.2 kg) (5 m>s2) = 1 N.
SECTION 4–5 Newton’s Third Law of Motion 113 If the same net force πF acts to accelerate each of two masses, m1 and m2, then the ratio of their masses can be defined as the inverse ratio of their accelerations: m2 m1 = a1 a2 .
If one of the masses is known (it could be the standard kilogram) and the two accelerations are precisely measured, then the unknown mass is obtained from this definition. For example, if m1 = 1.00 kg, and for a particular force a1 = 3.00 m>s2 and a2 = 2.00 m>s2, then m2 = 1.50 kg.
4–5 Newton’s Third Law of Motion Newton’s second law of motion describes quantitatively how forces affect motion. But where do forces come from? Observations suggest that a force exerted on any object is always exerted by another object. A horse pulls a wagon, a person pushes a grocery cart, a hammer pushes on a nail, a magnet attracts a paper clip. In each of these examples, a force is exerted on one object, and that force is exerted by another object. The force exerted on the nail is exerted by the hammer.
But Newton realized that things are not so one-sided. True, the hammer exerts a force on the nail (Fig. 4 9 7). But the nail evidently exerts a force back on the hammer as well, for the hammer’s speed is rapidly reduced to zero upon contact. Only a strong force could cause such a rapid deceleration of the hammer. Thus, said Newton, the two objects must be treated on an equal basis. The hammer exerts a force on the nail, and the nail exerts a force back on the hammer. This is the essence of Newton’s third law of motion: Whenever one object exerts a force on a second object, the second object exerts an equal force in the opposite direction on the first.
This law is sometimes stated as “to every action there is an equal and opposite reaction.” But to avoid confusion, it is very important to remember that the “action” force and the “reaction” force are acting on different objects.
Force exerted on hand by desk Force exerted on desk by hand FIGURE 4 – 8 If your hand pushes against the edge of a desk (the force vector is shown in red), the desk pushes back against your hand (this force vector is shown in a different color, violet, to remind us that this force acts on a different object).
As evidence for the validity of Newton’s third law, look at your hand when you push against the edge of a desk, Fig. 4 9 8. Your hand’s shape is distorted, clear evidence that a force is being exerted on it. You can see the edge of the desk pressing into your hand. You can even feel the desk exerting a force on your hand; it hurts! The harder you push against the desk, the harder the desk pushes back on your hand. (You only feel forces exerted on you; when you exert a force on another object, what you feel is that object pushing back on you.) The force the desk exerts on your hand has the same magnitude as the force your hand exerts on the desk. This is true not only if the desk is at rest but is true even if the desk is accelerating due to the force your hand exerts.
As another demonstration of Newton’s third law, consider the ice skater in Fig. 4 9 9. There is very little friction between her skates and the ice, so she will move freely if a force is exerted on her. She pushes against the wall; and then she starts moving backward. The force she exerts on the wall cannot make her start moving, because that force acts on the wall. Something had to exert a force on her to start her moving, and that force could only have been exerted by the wall. The wall pushes on her with a force, by Newton’s third law, equal and opposite to the force she exerts on the wall.
NEWTON’S THIRD LAW OF MOTION C A U T I O N Action and reaction forces act on different objects FIGURE 4 – 7 A hammer striking a nail. The hammer exerts a force on the nail and the nail exerts a force back on the hammer. The latter force decelerates the hammer and brings it to rest.
FIGURE 4 – 9 An example of Newton’s third law: when an ice skater pushes against the wall, the wall pushes back and this force causes her to accelerate away.
Force on skater Force on wall 114 CHAPTER 4 Dynamics: Newton’s Laws of Motion When a person throws a package out of a small boat (initially at rest), the boat starts moving in the opposite direction. The person exerts a force on the package. The package exerts an equal and opposite force back on the person, and this force propels the person (and the boat) backward slightly.
Rocket propulsion also is explained using Newton’s third law (Fig. 4 9 10). A common misconception is that rockets accelerate because the gases rushing out the back of the engine push against the ground or the atmosphere. Not true. What happens, instead, is that a rocket exerts a strong force on the gases, expel ling them; and the gases exert an equal and opposite force on the rocket. It is this latter force that propels the rocket forward : the force exerted on the rocket by the gases (see Chapter-Opening Photo, page 107). Thus, a space vehicle is maneuvered in empty space by firing its rockets in the direction opposite to that in which it needs to accelerate. When the rocket pushes on the gases in one direction, the gases push back on the rocket in the opposite direction. Jet aircraft too accelerate because the gases they thrust out back-wards exert a forward force on the engines (Newton’s third law).
Consider how we walk. A person begins walking by pushing with the foot backward against the ground. The ground then exerts an equal and opposite force forward on the person (Fig. 4 9 11), and it is this force, on the person, that moves the person forward. (If you doubt this, try walking normally where there is no friction, such as on very smooth slippery ice.) In a similar way, a bird flies forward by exerting a backward force on the air, but it is the air pushing forward (Newton’s third law) on the bird’s wings that propels the bird forward.
FIGURE 4 – 10 Another example of Newton’s third law: the launch of a rocket. The rocket engine pushes the gases downward, and the gases exert an equal and opposite force upward on the rocket, accelerating it upward. (A rocket does not accelerate as a result of its expelled gases pushing against the ground.) FIGURE 4 – 11 We can walk forward because, when one foot pushes backward against the ground, the ground pushes forward on that foot (Newton’s third law). The two forces shown act on different objects.
Horizontal force exerted on the ground by person’s foot Horizontal force exerted on the person’s foot by the ground FGP 5 FPG 5 We tend to associate forces with active objects such as humans, animals, engines, or a moving object like a hammer. It is often difficult to see how an inanimate object at rest, such as a wall or a desk (Fig. 4 9 8), or the wall of an ice rink (Fig. 4 9 9), can exert a force. The explanation is that every material, no matter how hard, is elastic (springy) at least to some degree. A stretched rubber band can exert a force on a wad of paper and accelerate it to fly across the room. Other materials may not stretch as readily as rubber, but they do stretch or compress when a force is applied to them. And just as a stretched rubber band exerts a force, so does a stretched (or compressed) wall, desk, or car fender.
From the examples discussed above, we can see how important it is to remember on what object a given force is exerted and by what object that force is exerted. A force influences the motion of an object only when it is applied on that object. A force exerted by an object does not influence that same object; it only influences the other object on which it is exerted. Thus, to avoid confusion, the two prepositions on and by must always be used : and used with care.
One way to keep clear which force acts on which object is to use double subscripts. For example, the force exerted on the Person by the Ground as the person walks in Fig. 4 9 11 can be labeled F 5PG. And the force exerted on the ground by the person is F 5GP. By Newton’s third law F 5GP = -F 5PG. (4 – 2) F 5GP and F 5PG have the same magnitude (Newton’s third law), and the minus sign reminds us that these two forces are in opposite directions.
C A U T I O N Distinguish on what object a force is exerted, and by what object NEWTON’S THIRD LAW OF MOTION CONCEPTUAL EXAMPLE 4 – 4 What exerts the force to move a car? What makes a car go forward?
RESPONSE A common answer is that the engine makes the car move forward. But it is not so simple. The engine makes the wheels and tires go around. But if the tires are on slick ice or wet mud, they just spin. Friction is needed. On firm ground, the tires push backward against the ground because of friction between the tires and the ground. By Newton’s third law, the ground pushes on the tires in the opposite direction, accelerating the car forward.
SECTION 4–5 Newton’s Third Law of Motion 115 CONCEPTUAL EXAMPLE 4 – 5 Third law clarification. Michelangelo’s assistant has been assigned the task of moving a block of marble using a sled (Fig. 4 9 12). He says to his boss, “When I exert a forward force on the sled, the sled exerts an equal and opposite force backward. So how can I ever start it moving? No matter how hard I pull, the backward reaction force always equals my forward force, so the net force must be zero. I’ll never be able to move this load.” Is he correct?
RESPONSE No. Although it is true that the action and reaction forces are equal in magnitude, the assistant has forgotten that they are exerted on different objects. The forward (“action”) force is exerted by the assistant on the sled (Fig. 4 9 12), whereas the backward “reaction” force is exerted by the sled on the assistant. To determine if the assistant moves or not, we must consider only the forces on the assistant and then apply πF 5 = ma 5, where πF 5 is the net force on the assistant, a 5 is the acceleration of the assistant, and m is the assistant’s mass. There are two forces on the assistant that affect his forward motion; they are shown as bright red (magenta) arrows in Figs. 4 9 12 and 4 9 13: they are (1) the hori zontal force F 5AG exerted on the assistant by the ground (the harder he pushes backward against the ground, the harder the ground pushes forward on him : Newton’s third law), and (2) the force F 5AS exerted on the assistant by the sled, pulling backward on him; see Fig. 4 9 13. If he pushes hard enough on the ground, the force on him exerted by the ground, F 5AG, will be larger than the sled pulling back, F 5AS, and the assistant accelerates forward (Newton’s second law). The sled, on the other hand, accelerates forward when the force on it exerted by the assistant is greater than the frictional force exerted backward on it by the ground (that is, when F 5SA has greater magnitude than F 5SG in Fig. 4 9 12).
P R O B L E M S O LV I N G A study of Newton’s second and third laws Using double subscripts to clarify Newton’s third law can become cumbersome, and we won’t usually use them in this way. We will usually use a single subscript referring to what exerts the force on the object being discussed. Nevertheless, if there is any confusion in your mind about a given force, go ahead and use two subscripts to identify on what object and by what object the force is exerted.
EXERCISE B A tennis ball collides head-on with a more massive baseball. (i) Which ball experiences the greater force of impact? (ii) Which experiences the greater acceleration during the impact? (iii) Which of Newton’s laws are useful to obtain the correct answers?
EXERCISE C If you push on a heavy desk, does it always push back on you? (a) No. (b) Yes. (c) Not unless someone else also pushes on it. (d) Yes, if it is out in space. (e) A desk never pushes.
Force on assistant exerted by sled Force on assistant exerted by ground FAG 5 FAS 5 FIGURE 4 – 13 Example 4 9 5. The horizontal forces on the assistant.
Force on sled exerted by assistant Force on assistant exerted by sled Friction force on sled exerted by ground Force on assistant exerted by ground Force on ground exerted by sled (= -GS (= - SG) Force on ground exerted by assistant FSA 5 FAS) 5 FAS 5 FSG 5 F 5 F 5 FGA 5 FAG 5 (= - FAG) 5 FIGURE 4 – 12 Example 4 9 5, showing only horizontal forces. Michelangelo has selected a fine block of marble for his next sculpture. Shown here is his assistant pulling it on a sled away from the quarry. Forces on the assistant are shown as red (magenta) arrows. Forces on the sled are purple arrows. Forces acting on the ground are orange arrows. Action 9 reaction forces that are equal and opposite are labeled by the same subscripts but reversed (such as F 5GA and F 5AG) and are of different colors because they act on different objects.
Note carefully that the two forces shown in Fig. 4 9 11 act on different objects : to emphasize this we used slightly different colors for the vector arrows representing these forces. These two forces would never appear together in a sum of forces in Newton’s second law, πF 5 = ma 5. Why not? Because they act on different objects: a 5 is the acceleration of one particular object, and πF 5 must include only the forces on that one object.
C A U T I O N The 2 forces in Newton’s third law act on different bodies. Only one can be included in πF = ma for an object 116 CHAPTER 4 Dynamics: Newton’s Laws of Motion EXERCISE D Return to the first Chapter-Opening Question, page 107 , and answer it again now. Try to explain why you may have answered differently the first time.
4–6 Weight—the Force of Gravity; and the Normal Force As we saw in Chapter 2, Galileo claimed that all objects dropped near the surface of the Earth would fall with the same acceleration, g 5, if air resistance was negligible. The force that causes this acceleration is called the force of gravity or gravitational force. What exerts the gravitational force on an object? It is the Earth, as we will discuss in Chapter 5, and the force acts vertically† downward, toward the center of the Earth. Let us apply Newton’s second law to an object of mass m falling freely due to gravity. For the acceleration, a 5, we use the downward acceleration due to gravity, g 5. Thus, the gravitational force on an object, F 5G, can be written as F 5G = mg 5. (4 – 3) The direction of this force is down toward the center of the Earth. The magnitude of the force of gravity on an object, mg, is commonly called the object’s weight.
In SI units, g = 9.80 m>s2 = 9.80 N>kg,‡ so the weight of a 1.00-kg mass on Earth is 1.00 kg 9.80 m>s2 = 9.80 N. We will mainly be concerned with the weight of objects on Earth, but we note that on the Moon, on other planets, or in space, the weight of a given mass will be different than it is on Earth. For example, on the Moon the acceleration due to gravity is about one-sixth what it is on Earth, and a 1.0-kg mass weighs only 1.6 N. Although we will not use British units, we note that for practical purposes on the Earth, a mass of 1.0 kg weighs about 2.2 lb. (On the Moon, 1 kg weighs only about 0.4 lb.) The force of gravity acts on an object when it is falling. When an object is at rest on the Earth, the gravitational force on it does not disappear, as we know if we weigh it on a spring scale. The same force, given by Eq. 4 9 3, continues to act. Why, then, doesn’t the object move? From Newton’s second law, the net force on an object that remains at rest is zero. There must be another force on the object to balance the gravitational force. For an object resting on a table, the table exerts this upward force; see Fig. 4 914a. The table is compressed slightly beneath the object, and due to its elasticity, it pushes up on the object as shown. The force exerted by the table is often called a contact force, since it occurs when two objects are in contact. (The force of your hand pushing on a cart is also a contact force.) When a contact force acts perpendicular to the common surface of contact, it is referred to as the normal force (“normal” means perpendicular); hence it is labeled F 5N in Fig. 4 9 14a.
The two forces shown in Fig. 4 9 14a are both acting on the statue, which remains at rest, so the vector sum of these two forces must be zero (Newton’s second law). Hence F 5G and F 5N must be of equal magnitude and in opposite directions. But they are not the equal and opposite forces spoken of in Newton’s third law. The action and reaction forces of Newton’s third law act on different objects, whereas the two forces shown in Fig. 4 9 14a act on the same object. For each of the forces shown in Fig. 4 9 14a, we can ask, “What is the reaction force?” The upward force F 5N on the statue is exerted by the table. The reaction to this force is a force exerted by the statue downward on the table. It is shown in Fig. 4 9 14b, where it is labeled F 5N = . This force, F 5N = , exerted on the table by the statue, is the reaction force to F 5N in accord with Newton’s third law. What about the other force on the statue, the force of gravity F 5G exerted by the Earth? Can you guess what the reaction is to this force? We will see in Chapter 5 that the reaction force is also a gravitational force, exerted on the Earth by the statue.
C A U T I O N Weight and normal force are not action–reaction pairs †The concept of “vertical” is tied to gravity. The best definition of vertical is that it is the direction in which objects fall. A surface that is “horizontal,” on the other hand, is a surface on which a round object won’t start rolling: gravity has no effect. Horizontal is perpendicular to vertical.
‡Since 1 N = 1 kg ⋅m>s2 (Section 4 9 4), then 1 m>s2 = 1 N>kg.
(a) F (b) 5 FG 5 FN 5 FN FG 5 FN 5 FIGURE 4 – 14 (a) The net force on an object at rest is zero according to Newton’s second law. Therefore the downward force of gravity (F 5G) on an object at rest must be balanced by an upward force (the normal force F 5N) exerted by the table in this case. (b) F 5N = is the force exerted on the table by the statue and is the reaction force to F 5N by Newton’s third law. (F 5N = is shown in a different color to remind us it acts on a different object.) The reaction force to F 5G is not shown.
SECTION 4–6 Weight—the Force of Gravity; and the Normal Force 117 EXERCISE E Return to the second Chapter-Opening Question, page 107 , and answer it again now. Try to explain why you may have answered differently the fi rst time.
EXAMPLE 4 – 6 Weight, normal force, and a box. A friend has given you a special gift, a box of mass 10.0 kg with a mystery surprise inside. The box is resting on the smooth (frictionless) horizontal surface of a table (Fig. 4 9 15a). (a) Determine the weight of the box and the normal force exerted on it by the table. (b) Now your friend pushes down on the box with a force of 40.0 N, as in Fig. 4 9 15b. Again determine the normal force exerted on the box by the table. (c) If your friend instead pulls upward on the box with a force of 40.0 N (Fig. 4 9 15c), what now is the normal force exerted on the box by the table?
APPROACH The box is at rest on the table, so the net force on the box in each case is zero (Newton’s fi rst or second law). The weight of the box has magni tude mg in all three cases.
SOLUTION (a) The weight of the box is mg = (10.0 kg) (9.80 m>s2) = 98.0 N, and this force acts downward. The only other force on the box is the normal force F 5N exerted upward on it by the table, as shown in Fig. 4 9 15a. We choose the upward direction as the positive y direction; then the net force πFy on the box is πFy = FN - mg; the minus sign means mg acts in the negative y direction (m and g are magnitudes). The box is at rest, so the net force on it must be zero (Newton’s second law, πFy = may, and ay = 0). Thus πFy = may FN - mg = 0, so we have FN = mg.
The normal force on the box, exerted by the table, is 98.0 N upward, and has magnitude equal to the box’s weight.
(b) Your friend is pushing down on the box with a force of 40.0 N. So instead of only two forces acting on the box, now there are three forces acting on the box, as shown in Fig. 4 9 15b. The weight of the box is still mg = 98.0 N. The net force is πFy = FN - mg - 40.0 N, and is equal to zero because the box remains at rest (a = 0). Newton’s second law gives πFy = FN - mg - 40.0 N = 0.
We solve this equation for the normal force: FN = mg + 40.0 N = 98.0 N + 40.0 N = 138.0 N, which is greater than in (a). The table pushes back with more force when a person pushes down on the box. The normal force is not always equal to the weight!
(c) The box’s weight is still 98.0 N and acts downward. The force exerted by your friend and the normal force both act upward (positive direction), as shown in Fig. 4 9 15c. The box doesn’t move since your friend’s upward force is less than the weight. The net force, again set to zero in Newton’s second law because a = 0, is πFy = FN - mg + 40.0 N = 0, so FN = mg - 40.0 N = 98.0 N - 40.0 N = 58.0 N.
The table does not push against the full weight of the box because of the upward force exerted by your friend.
NOTE The weight of the box (= mg) does not change as a result of your friend’s push or pull. Only the normal force is affected.
C A U T I O N The normal force FN is not always equal to the weight 40.0 N 40.0 N y y y FN 5 FN 5 FN 5 mg 5 mg 5 mg 5 (b) (a) (c) ΣFy = FN - mg - 40.0 N = 0 ΣFy = FN - mg + 40.0 N = 0 ΣFy = FN - mg = 0 FIGURE 4 – 15 Example 4 9 6. (a) A 10-kg gift box is at rest on a table. (b) A person pushes down on the box with a force of 40.0 N. (c) A person pulls upward on the box with a force of 40.0 N. The forces are all assumed to act along a line; they are shown slightly displaced in order to be distinguishable. Only forces acting on the box are shown.
118 CHAPTER 4 Dynamics: Newton’s Laws of Motion Recall that the normal force is elastic in origin (the table in Fig. 4 9 15 sags slightly under the weight of the box). The normal force in Example 4 9 6 is vertical, perpendicular to the horizontal table. The normal force is not always vertical, however. When you push against a wall, for example, the normal force with which the wall pushes back on you is horizontal (see Fig. 4 9 9). If an object is on a plane inclined at an angle to the horizontal, such as a skier or car on a hill, the normal force acts perpendicular to the plane and so is not vertical.
C A U T I O N The normal force, F 5N, is not necessarily vertical EXAMPLE 4 – 7 Accelerating the box. What happens when a person pulls upward on the box in Example 4 9 6c with a force equal to, or greater than, the box’s weight? For example, let FP = 100.0 N (Fig. 4 9 16) rather than the 40.0 N shown in Fig. 4 9 15c.
APPROACH We can start just as in Example 4 9 6, but be ready for a surprise.
SOLUTION The net force on the box is πFy = FN - mg + FP = FN - 98.0 N + 100.0 N, and if we set this equal to zero (thinking the acceleration might be zero), we would get FN = -2.0 N. This is nonsense, since the negative sign implies FN points downward, and the table surely cannot pull down on the box (unless there’s glue on the table). The least FN can be is zero, which it will be in this case, FN = 0. What really happens here is that the box leaves the table and accelerates upward (a ≠0, see gold vector in Fig. 4 9 16) because the net force is not zero. The net force (setting the normal force FN = 0) is πFy = FP - mg = 100.0 N - 98.0 N = 2.0 N upward. See Fig. 4 9 16. We apply Newton’s second law and see that the box moves upward with an acceleration ay = πFy m = 2.0 N 10.0 kg = 0.20 m>s2.
EXAMPLE 4 – 8 Apparent weight loss. A 65-kg woman descends in an elevator that briefl y accelerates at 0.20 g downward. She stands on a scale that reads in kg. (a) During this acceleration, what is her weight and what does the scale read? (b) What does the scale read when the elevator descends at a constant speed of 2.0 m>s?
APPROACH Figure 4 9 17 shows all the forces that act on the woman (and only those that act on her). The direction of the acceleration is downward, so we choose the positive direction as down (this is the opposite choice from Examples 4 9 6 and 4 9 7).
SOLUTION (a) From Newton’s second law, πF = ma mg - FN = m(0.20 g).
We solve for FN : FN = mg - 0.20 mg = 0.80 mg, and it acts upward. The normal force F 5N is the force the scale exerts on the person, and it is equal and opposite to the force she exerts on the scale: FN = = 0.80 mg downward. Her weight (force of gravity on her) is still mg = (65 kg) (9.8 m>s2) = 640 N. But the scale, needing to exert a force of only 0.80 mg, will give a reading of 0.80 m = 52 kg.
(b) Now there is no acceleration, a = 0, so by Newton’s second law, mg - FN = 0 and FN = mg. The scale reads her true mass of 65 kg.
NOTE The scale in (a) gives a reading of 52 kg (as an “apparent mass”), but her mass doesn’t change as a result of the acceleration: it stays at 65 kg.
(100.0 N) (98.0 N) FP 5 mg 5 a 5 FIGURE 4 – 16 Example 4 9 7. The box accelerates upward because FP 7 mg.
FIGURE 4 – 17 Example 4 9 8. The acceleration vector is shown in gold to distinguish it from the red force vectors.
FN 5 a 5 mg 5 +y SECTION 4–7 Solving Problems with Newton’s Laws: Free-Body Diagrams 119 4–7 Solving Problems with Newton’s Laws: Free-Body Diagrams Newton’s second law tells us that the acceleration of an object is proportional to the net force acting on the object. The net force, as mentioned earlier, is the vector sum of all forces acting on the object. Indeed, extensive experiments have shown that forces do add together as vectors precisely according to the rules we developed in Chapter 3. For example, in Fig. 4 9 18, two forces of equal magnitude (100 N each) are shown acting on an object at right angles to each other. Intuitively, or from symmetry, we can see that the object will start moving at a 45° angle and thus the net force acts at a 45° angle. This is just what the rules of vector addition give. From the theorem of Pythag-oras, the magnitude of the resultant force is FR = 1(100 N) 2 + (100 N) 2 = 141 N.
When solving problems involving Newton’s laws and force, it is very important to draw a diagram showing all the forces acting on each object involved. Such a diagram is called a free-body diagram, or force diagram: choose one object, and draw an arrow to represent each force acting on it. Include every force acting on that object. Do not show forces that the chosen object exerts on other objects. To help you identify each and every force that is exerted on your chosen object, ask yourself what other objects could exert a force on it. If your problem involves more than one object, a separate free-body diagram is needed for each object. For now, the likely forces that could be acting are gravity and contact forces (one object pushing or pulling another, normal force, friction). Later we will consider other types of force such as buoyancy, fluid pressure, and electric and magnetic forces.
P R O B L E M S O LV I N G Free-body diagram EXAMPLE 4 – 9 Adding force vectors. Calculate the sum of the two forces exerted on the boat by workers A and B in Fig. 4 9 19a.
APPROACH We add force vectors like any other vectors as described in Chapter 3. The first step is to choose an xy coordinate system (see Fig. 4 9 19a), and then resolve vectors into their components.
SOLUTION The two force vectors are shown resolved into components in Fig. 4 9 19b. We add the forces using the method of components. The components of F 5A are FA x = FA cos 45.0° = (40.0 N) (0.707) = 28.3 N, FA y = FA sin 45.0° = (40.0 N) (0.707) = 28.3 N.
The components of F 5B are FB x = +FB cos 37.0° = + (30.0 N) (0.799) = +24.0 N, FB y = -FB sin 37.0° = - (30.0 N) (0.602) = -18.1 N.
FBy is negative because it points along the negative y axis. The components of the resultant force are (see Fig. 4 9 19c) FR x = FA x + FB x = 28.3 N + 24.0 N = 52.3 N, FR y = FA y + FB y = 28.3 N - 18.1 N = 10.2 N.
To find the magnitude of the resultant force, we use the Pythagorean theorem, FR = 2FR x 2 + FR y 2 = 2(52.3) 2 + (10.2) 2 N = 53.3 N.
The only remaining question is the angle u that the net force F 5R makes with the x axis, Fig. 4 9 19c. We use: tan u = FR y FR x = 10.2 N 52.3 N = 0.195, and tan-1( 0.195) = 11.0°. The net force on the boat has magnitude 53.3 N and acts at an 11.0° angle to the x axis.
FB = 100 N FA = 100 N (a) (b) 45° R = + B A F 5 F 5 F 5 FA 5 FB 5 FIGURE 4 – 18 (a) Two horizontal forces, F 5A and F 5B, exerted by workers A and B, act on a crate (we are looking down from above). (b) The sum, or resultant, of F 5A and F 5B is F 5R.
FIGURE 4 – 19 Example 4 9 9: Two force vectors act on a boat.
(b) (c) (a) y x y x 45.0° 37.0° FA = 40.0 N FB = 30.0 N y x A B FA 5 FAy 5 FAx 5 FBy 5 FBx 5 FB 5 FRy 5 FR 5 FRx 5 u 120 CHAPTER 4 Dynamics: Newton’s Laws of Motion CONCEPTUAL EXAMPLE 4 – 10 The hockey puck. A hockey puck is sliding at constant velocity across a flat horizontal ice surface that is assumed to be frictionless. Which of the sketches in Fig. 4 9 20 is the correct free-body diagram for this puck? What would your answer be if the puck slowed down?
RESPONSE Did you choose (a)? If so, can you answer the question: what exerts the horizontal force labeled F 5 on the puck? If you say that it is the force needed to maintain the motion, ask yourself: what exerts this force? Remember that another object must exert any force : and there simply isn’t any possi bility here. Therefore, (a) is wrong. Besides, the force F 5 in Fig. 4 9 20a would give rise to an acceleration by Newton’s second law. It is (b) that is correct. No net force acts on the puck, and the puck slides at constant velocity across the ice.
In the real world, where even smooth ice exerts at least a tiny friction force, (c) is the correct answer. The tiny friction force is in the direction opposite to the motion, and the puck’s velocity decreases, even if very slowly.
object that force acts, and by what object that force is exerted. Only forces acting on a given object can be included in πF 5 = ma 5 for that object. 3. Newton’s second law involves vectors, and it is usually important to resolve vectors into compo-nents. Choose x and y axes in a way that simplifies the calculation. For example, it often saves work if you choose one coordinate axis to be in the direc-tion of the acceleration (if known). 4. For each object, apply Newton’s second law to the x and y components separately. That is, the x component of the net force on that object is related to the x component of that object’s acceleration: πFx = max, and similarly for the y direction. 5. Solve the equation or equations for the unknown(s). Put in numerical values only at the end, and keep track of units. P R O B L E M S O L V I N G 1. Draw a sketch of the situation, after carefully reading the problem at least twice. 2. Consider only one object (at a time), and draw a free-body diagram for that object, showing all the forces acting on that object. Include any unknown forces that you have to solve for. Do not show any forces that the chosen object exerts on other objects. Draw the arrow for each force vector reason-ably accurately for direction and magnitude. Label each force acting on the object, including forces you must solve for, according to its source (gravity, person, friction, and so on). If several objects are involved, draw a free-body diagram for each object separately. For each object, show all the forces acting on that object (and only forces acting on that object). For each (and every) force, you must be clear about: on what Newton’s Laws; Free-Body Diagrams This Problem Solving Strategy should not be considered a prescription. Rather it is a summary of things to do that will start you thinking and getting involved in the problem at hand.
When we are concerned only about translational motion, all the forces on a given object can be drawn as acting at the center of the object, thus treating the object as a point particle. However, for problems involving rotation or statics, the place where each force acts is also important, as we shall see in Chapters 10, 11, and 12.
In the Examples in this Section, we assume that all surfaces are very smooth so that friction can be ignored. (Friction, and Examples using it, are discussed in Chapter 5.) C A U T I O N Treating an object as a particle FIGURE 4 – 20 Example 4 9 10. Which is the correct free-body diagram for a hockey puck sliding across frictionless ice?
(b) (c) (a) Motion Motion Motion FN 5 F 5 FG 5 FN 5 FG 5 FN 5 F 5 FG 5 SECTION 4–7 121 EXERCISE F If the force F 5P (Fig. 4 9 21) exerted by the person is doubled in magnitude, what now will be the normal force?
EXERCISE G A 10.0-kg box is dragged on a horizontal frictionless surface by a hori-zontal force of 10.0 N. If the applied force is doubled, the normal force on the box will (a) increase; (b) remain the same; (c) decrease.
Tension in a Flexible Cord When a fl exible cord pulls on an object, the cord is said to be under tension, and the force it exerts on the object is the tension FT. If the cord has negligible mass, the force exerted at one end is transmitted undiminished to each adjacent piece of cord along the entire length to the other end. Why? Because πF 5 = ma 5 = 0 for the cord if the cord’s mass m is zero (or negligible) no matter what a 5 is. Hence the forces pulling on the cord at its two ends must add up to zero (FT and -FT). Note that fl exible cords and strings can only pull. They can’t push because they bend.
P R O B L E M S O LV I N G Cords can pull but can’t push; tension exists throughout a taut cord EXAMPLE 4 – 11 Pulling the mystery box. Suppose a friend asks to examine the 10.0-kg box you were given (Example 4 9 6, Fig. 4 9 15), hoping to guess what is inside; and you respond, “Sure, pull the box over to you.” She then pulls the box by the attached cord, as shown in Fig. 4 9 21a, along the smooth surface of the table. The magnitude of the force exerted by the person is FP = 40.0 N, and it is exerted at a 30.0° angle as shown. Calculate (a) the acceleration of the box, and (b) the magnitude of the upward force FN exerted by the table on the box. Assume that friction can be neglected.
APPROACH We follow the Problem Solving Strategy on the previous page.
SOLUTION 1. Draw a sketch: The situation is shown in Fig. 4 9 21a; it shows the box and the force applied by the person, FP. 2. Free-body diagram: Figure 4 9 21b shows the free-body diagram of the box. To draw it correctly, we show all the forces acting on the box and only the forces acting on the box. They are: the force of gravity mg 5; the normal force exerted by the table F 5N ; and the force exerted by the person F 5P. We are interested only in translational motion, so we can show the three forces acting at a point, Fig. 4 9 21c. 3. Choose axes and resolve vectors: We expect the motion to be horizontal, so we choose the x axis horizontal and the y axis vertical. The pull of 40.0 N has components FP x = (40.0 N) (cos 30.0°) = (40.0 N) (0.866) = 34.6 N, FP y = (40.0 N) (sin 30.0°) = (40.0 N) (0.500) = 20.0 N. In the horizontal (x) direction, F 5N and mg 5 have zero components. Thus the horizontal component of the net force is FPx. 4. (a) Apply Newton’s second law to determine the x component of the acceleration: FPx = max. 5. (a) Solve: ax = FPx>m = 34.6 N>10.0 kg = 3.46 m>s2. The acceleration of the box is 3.46 m>s2 to the right. (b) Next we want to fi nd FN. 4. ′ (b) Apply Newton’s second law to the vertical (y) direction, with upward as positive: πFy = may FN - mg + FPy = may. 5. ′ (b) Solve: We have mg = (10.0 kg)(9.80 m>s2) = 98.0 N and, from step 3 above, FPy = 20.0 N. Furthermore, since FPy 6 mg, the box does not move vertically, so ay = 0. Thus FN - 98.0 N + 20.0 N = 0, so FN = 78.0 N.
NOTE FN is less than mg: the table does not push against the full weight of the box because part of the pull exerted by the person is in the upward direction.
30.0° 30.0° FP = 40.0 N y y (b) x (c) x (a) m m FP 5 FN 5 FN 5 FPy 5 FPx 5 g 5 g 5 FP 5 FIGURE 4 – 21 (a) Pulling the box, Example 4 9 11; (b) is the free-body diagram for the box, and (c) is the free-body diagram considering all the forces to act at a point (translational motion only, which is what we have here).
122 CHAPTER 4 Dynamics: Newton’s Laws of Motion Our next Example involves two boxes connected by a cord. We can refer to this group of objects as a system. A system is any group of one or more objects we choose to consider and study.
EXAMPLE 4 – 12 T wo boxes connected by a cord. Two boxes, A and B, are connected by a lightweight cord and are resting on a smooth (frictionless) table. The boxes have masses of 12.0 kg and 10.0 kg. A horizontal force FP of 40.0 N is applied to the 10.0-kg box by a person pulling on a cord, as shown in Fig. 4 9 22a. Find (a) the acceleration of each box, and (b) the tension in the cord connecting the boxes.
APPROACH We streamline our approach by not listing each step. We have two boxes so we draw a free-body diagram for each. To draw them correctly, we must consider the forces on each box by itself, so that Newton’s second law can be applied to each. The person exerts a force FP on box A. Box A exerts a force FT on the connecting cord, and the cord exerts an opposite but equal magnitude force FT back on box A (Newton’s third law). The two horizontal forces on box A are shown in Fig. 4 9 22b, along with the force of gravity mA g 5 downward and the normal force F 5AN exerted upward by the table. The cord is light, so we neglect its mass. The tension at each end of the cord is thus the same. Hence the cord exerts a force FT on the second box. Figure 4 9 22c shows the forces on box B, which are F 5T, mB g 5, and the normal force F 5BN. There will be only horizontal motion. We take the positive x axis to the right.
SOLUTION (a) We apply πFx = max to box A: πFx = FP - FT = mA aA. [box A] For box B, the only horizontal force is FT, so πFx = FT = mB aB. [box B] The boxes are connected, and if the cord remains taut and doesn’t stretch, then the two boxes will have the same acceleration a. Thus aA = aB = a. We now have 2 unknowns, FT and a, and the two equations above to solve for them. We are given mA = 10.0 kg and mB = 12.0 kg. We can add the two equations above to eliminate an unknown (FT) and obtain (mA + mB) a = FP - FT + FT = FP or a = FP mA + mB = 40.0 N 22.0 kg = 1.82 m>s2.
This is what we sought.
(b) From the equation for box B above (FT = mB aB), the tension in the cord is FT = mB a = (12.0 kg) (1.82 m>s2) = 21.8 N.
Thus, FT 6 FP ( = 40.0 N), as we expect, since FT acts to accelerate only mB.
Alternate Solution to (a) We would have obtained the same result had we considered a single system, of mass mA + mB, acted on by a net horizontal force equal to FP. (The tension forces FT would then be considered internal to the system as a whole, and summed together would make zero contribution to the net force on the whole system.) NOTE It might be tempting to say that the force the person exerts, FP, acts not only on box A but also on box B. It doesn’t. FP acts only on box A. It affects box B via the tension in the cord, FT, which acts on box B and accelerates it. (You could look at it this way: FT 6 FP because FP accelerates both boxes whereas FT only accelerates box B.) C A U T I O N For any object, use only the forces on that object in calculating πF = ma FIGURE 4 – 22 Example 4 9 12. (a) Two boxes, A and B, are connected by a cord. A person pulls horizontally on box A with force FP = 40.0 N. (b) Free-body diagram for box A. (c) Free-body diagram for box B.
mB 40.0 N Box A (b) y x (a) Box B (c) x y mB mA FP 5 FBN 5 FT 5 g 5 g 5 FT 5 FAN 5 FP 5 y x mA mB = 12.0 kg mA = 10.0 kg SECTION 4–7 Solving Problems with Newton’s Laws: Free-Body Diagrams 123 y x (b) (c) (a) Elevator car mE = 1150 kg Counterweight mC = 1000 kg mE mC FT 5 FT 5 g 5 g 5 aE 5 aC 5 FIGURE 4 – 23 Example 4 9 13. (a) Atwood machine in the form of an elevator 9 counterweight system. (b) and (c) Free-body diagrams for the two objects.
EXAMPLE 4 – 13 Elevator and counterweight (Atwood machine). A system of two objects suspended over a pulley by a flexible cable, as shown in Fig. 4 9 23a, is sometimes referred to as an Atwood machine. Consider the real-life appli cation of an elevator (mE) and its counterweight (mC). To minimize the work done by the motor to raise and lower the elevator safely, mE and mC are made similar in mass. We leave the motor out of the system for this calculation, and assume that the cable’s mass is negligible and that the mass of the pulley, as well as any friction, is small and ignorable. These assumptions ensure that the tension FT in the cable has the same magnitude on both sides of the pulley. Let the mass of the counterweight be mC = 1000 kg. Assume the mass of the empty elevator is 850 kg, and its mass when carrying four passengers is mE = 1150 kg. For the latter case (mE = 1150 kg), calculate (a) the acceleration of the elevator and (b) the tension in the cable.
APPROACH Again we have two objects, and we will need to apply Newton’s second law to each of them separately. Each mass has two forces acting on it: gravity downward and the cable tension pulling upward, F 5T. Figures 4 9 23b and c show the free-body diagrams for the elevator (mE) and for the coun-terweight (mC). The elevator, being the heavier, will accelerate downward, whereas the counterweight will accelerate upward. The magnitudes of their accelerations will be equal (we assume the cable is massless and doesn’t stretch). For the counterweight, mC g = (1000 kg) (9.80 m>s2) = 9800 N, so FT must be greater than 9800 N (in order that mC will accelerate upward). For the elevator, mE g = (1150 kg) (9.80 m>s2) = 11,300 N, which must have greater magnitude than FT so that mE accelerates downward. Thus our calculation must give FT between 9800 N and 11,300 N.
SOLUTION (a) To find FT as well as the acceleration a, we apply Newton’s second law, πF = ma, to each object. We take upward as the positive y direc-tion for both objects. With this choice of axes, aC = a because mC accelerates upward, and aE = -a because mE accelerates downward. Thus FT - mE g = mE aE = -mE a FT - mC g = mC aC = +mC a.
We can subtract the first equation from the second to get (mE - mC)g = (mE + mC)a, where a is now the only unknown. We solve this for a: a = mE - mC mE + mC g = 1150 kg - 1000 kg 1150 kg + 1000 kg g = 0.070 g = 0.68 m>s2.
The elevator (mE) accelerates downward (and the counterweight mC upward) at a = 0.070 g = 0.68 m>s2.
(b) The tension in the cable FT can be obtained from either of the two πF = ma equations at the start of our solution, setting a = 0.070 g = 0.68 m>s2: FT = mE g - mE a = mE (g - a) = 1150 kg (9.80 m>s2 - 0.68 m>s2) = 10,500 N, or FT = mC g + mC a = mC (g + a) = 1000 kg (9.80 m>s2 + 0.68 m>s2) = 10,500 N, which are consistent. As predicted, our result lies between 9800 N and 11,300 N.
NOTE We can check our equation for the acceleration a in this Example by noting that if the masses were equal (mE = mC), then our equation above for a would give a = 0, as we should expect. Also, if one of the masses is zero (say, mC = 0), then the other mass (mE ≠0) would be predicted by our equation to accelerate at a = g, again as expected.
P H Y S I C S A P P L I E D Elevator (as Atwood machine) P R O B L E M S O LV I N G Check your result by seeing if it works in situations where the answer is easily guessed 124 CHAPTER 4 Dynamics: Newton’s Laws of Motion m FT 5 FT 5 FT 5 g 5 FIGURE 4 – 24 Example 4 9 14.
FIGURE 4 – 25 Example 4 9 15.
FT 5 mg 5 a 5 y x FT sinu FT cosu u u (b) (a) CONCEPTUAL EXAMPLE 4 – 14 The advantage of a pulley. A mover is trying to lift a piano (slowly) up to a second-story apartment (Fig. 4 9 24). He is using a rope looped over two pulleys as shown. What force must he exert on the rope to slowly lift the piano’s 1600-N weight?
RESPONSE The magnitude of the tension force FT within the rope is the same at any point along the rope if we assume we can ignore the mass of the rope and pulleys. First notice the forces acting on the lower pulley at the piano. The weight of the piano (= mg) pulls down on the pulley. The tension in the rope, looped through this pulley, pulls up twice, once on each side of the pulley. Let us apply Newton’s second law to the pulley 9 piano combination (of mass m), choosing the upward direction as positive: 2FT - mg = ma.
To move the piano with constant speed (set a = 0 in this equation) thus requires a tension in the rope, and hence a pull on the rope, of FT = mg>2. The piano mover can exert a force equal to half the piano’s weight. NOTE We say the pulley has given a mechanical advantage of 2, since without the pulley the mover would have to exert twice the force.
EXAMPLE 4 – 15 Accelerometer. A small mass m hangs from a thin string and can swing like a pendulum. You attach it above the window of your car as shown in Fig. 4 9 25a. When the car is at rest, the string hangs vertically. What angle u does the string make (a) when the car accelerates at a constant a = 1.20 m>s2, and (b) when the car moves at constant velocity, v = 90 km>h?
APPROACH The free-body diagram of Fig. 4 9 25b shows the pendulum at some angle u relative to the vertical, and the forces on it: mg 5 downward, and the tension F 5T in the cord (including its components). These forces do not add up to zero if u ≠0; and since we have an acceleration a, we expect u ≠0. SOLUTION (a) The acceleration a = 1.20 m>s2 is horizontal (= ax), and the only horizontal force is the x component of F 5T : FT sin u (Fig. 4 9 25b). Then from Newton’s second law, ma = FT sin u.
The vertical component of Newton’s second law gives, since ay = 0, 0 = FT cos u - mg.
So mg = FT cos u.
Dividing the two equations involving FT, we obtain tan u = FT sin u FT cos u = ma mg = a g or tan u = 1.20 m>s2 9.80 m>s2 = 0.122, so u = 7.0°.
(b) The velocity is constant, so a = 0 and therefore tan u = 0. Hence the pendulum hangs vertically (u = 0°).
NOTE This simple device is an accelerometer : it can be used to determine acceleration, by measuring the angle u.
P H Y S I C S A P P L I E D Accelerometer SECTION 4–7 Solving Problems with Newton’s Laws: Free-Body Diagrams 125 Inclines Now we consider what happens when an object slides down an incline, such as a hill or ramp. Such problems are interesting because gravity is the accelerating force, yet the acceleration is not vertical. Solving problems is usually easier if we choose the xy coordinate system so the x axis points along the incline (the direction of motion) and the y axis is perpendicular to the incline, as shown in Fig. 4 9 26a. Note also that the normal force is not vertical, but is perpendicular to the sloping surface of the plane, along the y axis in Fig. 4 9 26b.
P R O B L E M S O LV I N G Good choice of coordinate system simplifi es the calculation (a) m y x u (a) m (b) mg cos u mg sin u y x y x mg 5 FN 5 u u u FIGURE 4 – 26 Example 4 9 16. (a) Box sliding on inclined plane. (b) Free-body diagram of box.
EXAMPLE 4 – 16 Box slides down an incline. A box of mass m is placed on a smooth (frictionless) incline that makes an angle u with the horizontal, as shown in Fig. 4 9 26. (a) Determine the normal force on the box. (b) Determine the box’s acceleration. (c) Evaluate for a mass m = 10 kg and an incline of u = 30°.
APPROACH We expect the motion to be along the incline, so we choose the x axis along the slope, positive down the slope (the direction of motion). The y axis is perpendicular to the incline, positive upwards. The free-body diagram is shown in Fig. 4 9 26b. The forces on the box are its weight mg 5 vertically downward, which is shown resolved into its components parallel and perpendicular to the incline, and the normal force F 5N along the +y axis. The components of mg 5 along the x and y axes are found using the defi nitions of sine (“side opposite”) and cosine (“side adjacent”): (mg) x = mg sin u, and (mg) y = mg cos u. (The angle u of the plane equals the angle between mg 5 and its y component because the left sides of the two angles are perpendicular to each other and so are the right sides : see Appendix A 9 6, Fig. A 9 2.) SOLUTION (a) There is no motion in the y direction, so ay = 0. Applying Newton’s second law we have Fy = may FN - mg cos u = 0, where FN and the y component of gravity (mg cos u) are all the forces acting on the box in the y direction. Thus the normal force is given by FN = mg cos u.
Note carefully that unless u = 0°, FN has magnitude less than the weight mg.
(b) In the x direction the only force acting is the x component of mg 5, which is mg sin u. The acceleration a is in the x direction so Fx = max mg sin u = ma, and we see that the acceleration down the plane is a = g sin u.
Thus the acceleration along an incline is always less than g, except at u = 90°, for which sin u = 1 and a = g. This makes sense since u = 90° is pure vertical fall. For u = 0°, a = 0, which makes sense because u = 0° means the plane is horizontal so gravity causes no acceleration. Note too that the acceleration does not depend on the mass (m).
(c) For u = 30°, cos u = 0.866 and sin u = 0.500, so FN = 0.866 mg = 85 N, and a = 0.500 g = 4.9 m>s2.
126 CHAPTER 4 Dynamics: Newton’s Laws of Motion EXERCISE H Is the normal force always perpendicular to an inclined plane? Is it always vertical?
We will discuss more Examples of motion on an incline in the next Chapter, where friction will be included.
4–8 Problem Solving—A General Approach A basic part of a physics course is solving problems effectively. The approach discussed here, though emphasizing Newton’s laws, can be applied generally for other topics discussed throughout this book.
to know the limitations of each formula or rela-tionship : when it is valid and when not. In this book, the more general equations have been given numbers, but even these can have a limited range of validity (often stated in brackets to the right of the equation). 6. Try to solve the problem approximately, to see if it is doable (to check if enough information has been given) and reasonable. Use your intuition, and make rough calculations : see “Order of Magnitude Estimating” in Section 1 9 6. A rough calculation, or a reasonable guess about what the range of final answers might be, is very useful. And a rough calculation can be checked against the final answer to catch errors in calculation, such as in a decimal point or the powers of 10. 7. Solve the problem, which may include alge-braic manipulation of equations and>or numerical calculations. Recall the mathematical rule that you need as many independent equations as you have unknowns; if you have three unknowns, for example, then you need three independent equa-tions. It is usually best to work out the algebra symbolically before putting in the numbers. Why? Because (a) you can then solve a whole class of similar problems with different numerical values; (b) you can check your result for cases already understood (say, u = 0° or 90°); (c) there may be cancellations or other simplifications; (d) there is usually less chance for numerical error; and (e) you may gain better insight into the problem. 8. Be sure to keep track of units, for they can serve as a check (they must balance on both sides of any equation). 9. Again consider if your answer is reasonable. The use of dimensional analysis, described in Section 1 9 7, can also serve as a check for many problems. P R O B L E M S O L V I N G 1. Read and reread written problems carefully. A common error is to skip a word or two when reading, which can completely change the meaning of a problem. 2. Draw an accurate picture or diagram of the situation. (This is probably the most overlooked, yet most crucial, part of solving a problem.) Use arrows to represent vectors such as velocity or force, and label the vectors with appropriate symbols. When dealing with forces and applying Newton’s laws, make sure to include all forces on a given object, including unknown ones, and make clear what forces act on what object (otherwise you may make an error in determining the net force on a particular object). 3. A separate free-body diagram needs to be drawn for each object involved, and it must show all the forces acting on a given object (and only on that object). Do not show forces that act on other objects. 4. Choose a convenient xy coordinate system (one that makes your calculations easier, such as one axis in the direction of the acceleration). Vectors are to be resolved into components along the coordinate axes. When using Newton’s second law, apply πF 5 = ma 5 separately to x and y components, remembering that x direction forces are related to ax, and similarly for y. If more than one object is involved, you can choose different (convenient) coordinate systems for each. 5. List the knowns and the unknowns (what you are trying to determine), and decide what you need in order to find the unknowns. For prob-lems in the present Chapter, we use Newton’s laws. More generally, it may help to see if one or more relationships (or equations) relate the unknowns to the knowns. But be sure each relationship is applicable in the given case. It is very important In General Questions 127 Summary Newton’s three laws of motion are the basic classical laws describing motion.
Newton’s first law (the law of inertia) states that if the net force on an object is zero, an object originally at rest remains at rest, and an object in motion remains in motion in a straight line with constant velocity.
Newton’s second law states that the acceleration of an object is directly proportional to the net force acting on it, and inversely proportional to its mass: πF 5 = ma 5. (4 – 1a) Newton’s second law is one of the most important and funda-mental laws in classical physics.
Newton’s third law states that whenever one object exerts a force on a second object, the second object always exerts a force on the first object which is equal in magnitude but oppo-site in direction: F 5AB = -F 5BA, (4 – 2) where F 5BA is the force on object B exerted by object A. This is true even if objects are moving and accelerating, and>or have different masses.
The tendency of an object to resist a change in its motion is called inertia. Mass is a measure of the inertia of an object.
Weight refers to the gravitational force on an object, and is equal to the product of the object’s mass m and the acceleration of gravity g 5: F 5G = mg 5. (4 – 3) Force, which is a vector, can be considered as a push or pull; or, from Newton’s second law, force can be defined as an action capable of giving rise to acceleration. The net force on an object is the vector sum of all forces acting on that object.
For solving problems involving the forces on one or more objects, it is essential to draw a free-body diagram for each object, showing all the forces acting on only that object. Newton’s second law can be applied to the vector components for each object.
Questions 1. Why does a child in a wagon seem to fall backward when you give the wagon a sharp pull forward? 2. A box rests on a frictionless part of a truck bed. The truck driver starts the truck and accelerates forward. The box immediately starts to slide toward the rear of the truck bed. Discuss the motion of the box, in terms of Newton’s laws, as seen (a) by Francesca standing on the ground beside the truck, and (b) by Phil who is riding on the truck (Fig. 4 9 27). 3. If an object is moving, is it possible for the net force acting on it to be zero? Explain. 4. If the acceleration of an object is zero, are no forces acting on it? Explain. 5. Only one force acts on an object. Can the object have zero acceleration? Can it have zero velocity? Explain. 6. When a golf ball is dropped to the pavement, it bounces back up. (a) Is a force needed to make it bounce back up? (b) If so, what exerts the force? 7. If you walk on a log floating on a lake, why does the log move in the opposite direction? 8. (a) Why do you push down harder on the pedals of a bicycle when first starting out than when moving at constant speed? (b) Why do you need to pedal at all when cycling at constant speed? 9. A stone hangs by a fine thread from the ceiling, and a section of the same thread dangles from the bottom of the stone (Fig. 4 9 28). If a person gives a sharp pull on the dangling thread, where is the thread likely to break: below the stone or above it? What if the person gives a slow and steady pull? Explain your answers.
FIGURE 4 – 28 Question 9. 10. The force of gravity on a 2-kg rock is twice as great as that on a 1-kg rock. Why then doesn’t the heavier rock fall faster? 11. You pull a box with a constant force across a frictionless table using an attached rope held horizontally. If you now pull the rope with the same force at an angle to the hori-zontal (with the box remaining flat on the table), does the acceleration of the box increase, decrease, or remain the same? Explain. 12. When an object falls freely under the influence of gravity there is a net force mg exerted on it by the Earth. Yet by Newton’s third law the object exerts an equal and opposite force on the Earth. Does the Earth move? Explain. 13. Compare the effort (or force) needed to lift a 10-kg object when you are on the Moon with the force needed to lift it on Earth. Compare the force needed to throw a 2-kg object horizontally with a given speed on the Moon and on Earth.
a 5 Box FIGURE 4 – 27 Question 2.
128 CHAPTER 4 Dynamics: Newton’s Laws of Motion FIGURE 4 – 33 MisConceptual Question 4.
FIGURE 4 – 30 MisConceptual Question 1. 14. According to Newton’s third law, each team in a tug of war (Fig. 4 9 29) pulls with equal force on the other team. What, then, determines which team will win?
FIGURE 4 – 29 Question 14. A tug of war. Describe the forces on each of the teams and on the rope. 15. When you stand still on the ground, how large a force does the ground exert on you? Why doesn’t this force make you rise up into the air? 16. Whiplash sometimes results from an automobile accident when the victim’s car is struck violently from the rear. Explain why the head of the victim seems to be thrown backward in this situation. Is it really? 17. Mary exerts an upward force of 40 N to hold a bag of groceries. Describe the “reaction” force (Newton’s third law) by stating (a) its magnitude, (b) its direction, (c) on what object it is exerted, and (d) by what object it is exerted. 18. A father and his young daughter are ice skating. They face each other at rest and then push each other so they begin moving in opposite directions. Which one has the greater speed? Explain. 19. What would your bathroom scale read if you weighed your self on an inclined plane? Assume the mechanism functions properly, even at an angle. 20. Which of the following objects weighs about 1 N: (a) an apple, (b) a mosquito, (c) this book, (d) you? 21. Why might your foot hurt if you kick a heavy desk or a wall? 22. When you are running and want to stop quickly, you must decelerate quickly. (a) What is the origin of the force that causes you to stop? (b) Estimate (using your own expe-rience) the maximum rate of deceleration of a person running at top speed to come to rest. 23. Suppose that you are standing on top of a cardboard carton that just barely supports you. What would happen to it if you jumped up into the air? It would: (a) collapse; (b) be unaffected; (c) spring upward a bit; (d) move sideways. Explain your answer.
MisConceptual Questions 1. A truck is traveling horizontally to the right (Fig. 4 9 30). When the truck starts to slow down, the crate on the frictionless truck bed starts to slide. In what direction could the net force be on the crate? (a) No direction. The net force is zero. (b) Straight down (gravity). (c) Straight up (the normal force). (d) Horizontal and to the right. (e) Horizontal and to the left. 2. George, in the foreground of Fig. 4 9 31, is able to move the large truck because (a) he is stronger than the truck. (b) he is heavier in some respects than the truck. (c) the truck offers no resistance because its brakes are off. (d) the ground exerts a greater friction force on George than it does on the truck. (e) he exerts a greater force on the truck than the truck exerts back on him. 3. A bear sling, Fig. 4 9 32, is used in some national parks for placing backpackers’ food out of the reach of bears. As the backpacker raises the pack by pulling down on the rope, the force F needed: (a) decreases as the pack rises until the rope is straight across. (b) doesn’t change. (c) increases until the rope is straight. (d) increases but the rope always sags where the pack hangs.
FIGURE 4 – 32 MisConceptual Question 3.
F 5 4. What causes the boat in Fig. 4 9 33 to move forward? (a) The force the man exerts on the paddle. (b) The force the paddle exerts on the water. (c) The force the water exerts on the paddle. (d) The motion of the water itself.
FIGURE 4 – 31 MisConceptual Question 2.
Problems 129 5. You are trying to push your stalled car. Although you apply a horizontal force of 400 N on the car, it doesn’t budge, and neither do you. What other force(s) must also have a magnitude of 400 N?
(a) The force exerted by the car on you.
(b) The friction force exerted by the car on the road.
(c) The normal force exerted by the road on you.
(d) The friction force exerted by the road on you. 6. When a skier skis down a hill, the normal force exerted on the skier by the hill is (a) equal to the weight of the skier. (b) greater than the weight of the skier. (c) less than the weight of the skier. 7. A golf ball is hit with a golf club. While the ball flies through the air, which forces act on the ball? Neglect air resistance.
(a) The force of the golf club acting on the ball.
(b) The force of gravity acting on the ball.
(c) The force of the ball moving forward through the air.
(d) All of the above.
(e) Both (a) and (b). 8. Suppose an object is accelerated by a force of 100 N. Suddenly a second force of 100 N in the opposite direc-tion is exerted on the object, so that the forces cancel. The object (a) quickly stops. (b) decelerates gradually to rest. (c) continues at the velocity it had before the second force was applied. (d) is gradually brought to rest and then accelerates in the direction of the second force. 9. You are pushing a heavy box across a rough floor. When you are initially pushing the box and it is accelerating, (a) you exert a force on the box, but the box does not exert a force on you.
(b) the box is so heavy it exerts a force on you, but you do not exert a force on the box.
(c) the force you exert on the box is greater than the force of the box pushing back on you.
(d) the force you exert on the box is equal to the force of the box pushing back on you.
(e) the force that the box exerts on you is greater than the force you exert on the box. 10. Two 5-newton boxes are attached to opposite ends of a spring scale and suspended from pulleys as shown in Fig. 4 9 34. What is the reading on the scale?
(a) 0 N.
(b) Between 0 and 5 N.
(c) 5 N.
(d) Between 5 and 10 N.
(e) 10 N.
5 N 5 N FIGURE 4 – 34 MisConceptual Question 10. 11. Two tug of war opponents each pull with a force of 500 N on opposite ends of a rope. Assume the rope is massless. What is the tension in the rope?
(a) 0 N.
(b) 250 N.
(c) 500 N.
(d) 1000 N.
(e) Impossible to tell. 12. The normal force on an extreme skier descending a very steep slope (Fig. 4 9 35) can be zero if (a) his speed is great enough. (b) he leaves the slope (no longer touches the snow). (c) the slope is greater than 75°. (d) the slope is vertical (90°).
FIGURE 4 – 35 MisConceptual Question 12. 13. To pull an old stump out of the ground, you and a friend tie two ropes to the stump. You pull on it with a force of 500 N to the north while your friend pulls with a force of 450 N to the northwest. The total force exerted by the two ropes is (a) less than 950 N.
(b) exactly 950 N.
(c) more than 950 N.
Problems 4 – 4 to 4 – 6 Newton’s Laws, Gravitational Force, Normal Force [Assume no friction.] 1. (I) What net force is needed to accelerate a 45@kg sled at 1.4 m>s2 on horizontal frictionless ice? 2. (I) What is the weight of a 74-kg astronaut (a) on Earth, (b) on the Moon (g = 1.7 m>s2), (c) on Mars (g = 3.7 m>s2), (d) in outer space traveling with constant velocity? 3. (I) How much tension must a rope withstand if it is used to accelerate a 1210-kg car horizontally along a frictionless surface at 1.35 m>s2? 4. (I) A net force of 215 N accelerates a bike and rider at 2.30 m>s2. What is the mass of the bike and rider together? 5. (II) What average force is required to stop a 950-kg car in 8.0 s if the car is traveling at 95 km>h? 6. (II) According to a simplified model of a mammalian heart, at each pulse approximately 20 g of blood is accelerated from 0.25 m>s to 0.35 m>s during a period of 0.10 s. What is the magnitude of the force exerted by the heart muscle? 7. (II) Superman must stop a 120@km>h train in 150 m to keep it from hitting a stalled car on the tracks. If the train’s mass is 3.6 105 kg, how much force must he exert? Compare to the weight of the train (give as %). How much force does the train exert on Superman?
130 CHAPTER 4 Dynamics: Newton’s Laws of Motion FIGURE 4 – 36 Problem 13. 8. (II) A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 g’s. Calculate the force on a 65-kg person accelerating at this rate. What distance is traveled if brought to rest at this rate from 85 km>h? 9. (II) Estimate the average force exerted by a shot-putter on a 7.0-kg shot if the shot is moved through a distance of 2.8 m and is released with a speed of 13 m>s. 10. (II) A 0.140-kg baseball traveling 35.0 m>s strikes the catcher’s mitt, which, in bringing the ball to rest, recoils backward 11.0 cm. What was the average force applied by the ball on the glove? 11. (II) A fi sherman yanks a fi sh vertically out of the water with an acceleration of 2.5 m>s2 using very light fi shing line that has a breaking strength of 18 N ( L 4 lb). The fi sherman unfortu-nately loses the fi sh as the line snaps. What can you say about the mass of the fi sh? 12. (II) How much tension must a cable withstand if it is used to accelerate a 1400-kg car vertically upward at 0.70 m>s2? 13. (II) A 20.0-kg box rests on a table. (a) What is the weight of the box and the normal force acting on it? (b) A 10.0-kg box is placed on top of the 20.0-kg box, as shown in Fig. 4 9 36. Determine the normal force that the table exerts on the 20.0-kg box and the normal force that the 20.0-kg box exerts on the 10.0-kg box.
20.0 kg 10.0 kg 14. (II) A particular race car can cover a quarter-mile track (402 m) in 6.40 s starting from a standstill. Assuming the acceleration is constant, how many horizontal “g’s” does the driver experience? If the combined mass of the driver and race car is 535 kg, what horizontal force must the road exert on the tires? Ignore air resistance. 15. (II) A 14.0-kg bucket is lowered vertically by a rope in which there is 132 N of tension at a given instant. What is the acceleration of the bucket? Is it up or down? 16. (II) A 75-kg petty thief wants to escape from a third-story jail window. Unfortunately, a makeshift rope made of sheets tied together can support a mass of only 62 kg. How might the thief use this “rope” to escape? Give a quantitative answer. 17. (II) A woman stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefl y reads only 0.75 of her regular weight. Calculate the acceler-ation of the elevator, and fi nd the direction of acceleration. 18. (II) Can cars “stop on a dime”? Calculate the acceleration of a 1400-kg car if it can stop from 35 km>h on a dime (diameter = 1.7 cm). How many g’s is this? What is the force felt by the 68-kg occupant of the car? 19. (II) The cable supporting a 2375-kg elevator has a maximum strength of 24,950 N. What maximum upward acceleration can it give the elevator without breaking? 20. (II) Using focused laser light, optical tweezers can apply a force of about 10 pN (piconewtons) to a 1.0@mm-diameter polystyrene bead, which has a density about equal to that of water: a volume of 1.0 cm3 has a mass of about 1.0 g. Estimate the bead’s acceleration in g’s. 21. (II) A rocket with a mass of 2.45 106 kg is launched by exerting a vertical force of 3.55 107 N on the gases it expels. Determine (a) the acceleration of the rocket, (b) its velocity after 8.0 s, and (c) how long it takes to reach an altitude of 9500 m. Assume g remains constant, and ignore the mass of gas expelled (not realistic). 22. (II) (a) What is the acceleration of two falling sky divers (total mass = 148 kg including parachute) when the upward force of air resistance is equal to one-fourth of their weight? (b) After opening the parachute, the divers descend leisurely to the ground at constant speed. What now is the force of air resistance on the sky divers and their parachute? 23. (II) An elevator (mass 4850 kg, with people) is to be designed so that the maximum acceleration is 0.0640 g. What are the maximum and minimum forces the motor should exert on the supporting cable? 24. (II) An exceptional standing jump would raise a person 0.80 m off the ground. To do this, what force must a 68-kg person exert against the ground? Assume the person crouches a distance of 0.20 m prior to jumping, and thus the upward force has this distance to act over before he leaves the ground. 25. (II) High-speed elevators function under two limitations: (1) the maximum magnitude of vertical acceleration that a typical human body can experience without discomfort is about 1.2 m>s2, and (2) the typical maximum speed attainable is about 9.0 m>s. You board an elevator on a skyscraper’s ground fl oor and are transported 180 m above the ground level in three steps: acceleration of magnitude 1.2 m>s2 from rest to 9.0 m>s, followed by constant upward velocity of 9.0 m>s, then deceleration of magnitude 1.2 m>s2 from 9.0 m>s to rest. (a) Determine the elapsed time for each of these 3 stages. (b) Determine the change in the magnitude of the normal force, expressed as a % of your normal weight during each stage. (c) What fraction of the total transport time does the normal force not equal the person’s weight? 26. (III) The 100-m dash can be run by the best sprinters in 10.0 s. A 66-kg sprinter accelerates uniformly for the fi rst 45 m to reach top speed, which he maintains for the remaining 55 m. (a) What is the average horizontal component of force exerted on his feet by the ground during acceleration? (b) What is the speed of the sprinter over the last 55 m of the race (that is, his top speed)? 27. (III) A person jumps from the roof of a house 2.8 m high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 m. If the mass of his torso (excluding legs) is 42 kg, fi nd (a) his velocity just before his feet strike the ground, and (b) the average force exerted on his torso by his legs during deceleration.
Problems 131 FIGURE 4 – 38 Problem 29.
FIGURE 4 – 40 Problem 33.
FIGURE 4 – 41 Problems 34 and 35.
FIGURE 4 – 37 Problem 28.
32° 48° L Top view FA 5 FB 5 4 – 7 Newton’s Laws and Vectors [Ignore friction.] 28. (I) A box weighing 66.0 N rests on a table. A rope tied to the box runs vertically upward over a pulley and a weight is hung from the other end (Fig. 4 9 37). Determine the force that the table exerts on the box if the weight hanging on the other side of the pulley weighs (a) 30.0 N, (b) 60.0 N, and (c) 90.0 N. 29. (I) Draw the free-body diagram for a basketball player (a) just before leaving the ground on a jump, and (b) while in the air. See Fig. 4 9 38. 30. (I) A 650-N force acts in a northwesterly direction. A second 650-N force must be exerted in what direction so that the resultant of the two forces points westward? Illus-trate your answer with a vector diagram. 31. (I) Sketch the free-body diagram of a baseball (a) at the moment it is hit by the bat, and again (b) after it has left the bat and is fl ying toward the outfi eld. Ignore air resistance. 32. (II) Arlene is to walk across a “high wire” strung horizontally between two buildings 10.0 m apart. The sag in the rope when she is at the midpoint is 10.0°, as shown in Fig. 4 9 39. If her mass is 50.0 kg, what is the tension in the rope at this point?
10.0° FIGURE 4 – 39 Problem 32. 33. (II) A window washer pulls herself upward using the bucket 9 pulley apparatus shown in Fig. 4 9 40. (a) How hard must she pull downward to raise herself slowly at constant speed? (b) If she increases this force by 15%, what will her acceleration be? The mass of the person plus the bucket is 78 kg. 34. (II) One 3.2-kg paint bucket is hanging by a massless cord from another 3.2-kg paint bucket, also hanging by a massless cord, as shown in Fig. 4 9 41. (a) If the buckets are at rest, what is the tension in each cord? (b) If the two buckets are pulled upward with an acceleration of 1.45 m>s2 by the upper cord, calculate the tension in each cord. 35. (II) The cords accelerating the buckets in Problem 34b, Fig. 4 9 41, each have a weight of 2.0 N. Determine the tension in each cord at the three points of attachment. 36. (II) Two large snowcats are towing a housing unit north, as shown in Fig. 4 9 42. The sum of the forces F 5A and F 5B exerted on the unit by the hori-zontal cables is north, parallel to the line L, and FA = 4200 N. Determine FB and the magnitude of F 5A + F 5B. 37. (II) A train locomotive is pulling two cars of the same mass behind it, Fig. 4 9 43. Determine the ratio of the tension in the coupling (think of it as a cord) between the locomotive and the fi rst car (F T1) to that between the fi rst car and the second car (F T2), for any nonzero acceleration of the train.
FT2 5 FT1 5 Car 2 Car 1 FIGURE 4 – 43 Problem 37. 38. (II) The two forces F 51 and F 52 shown in Fig. 4 9 44a and b (looking down) act on an 18.5-kg object on a frictionless tabletop. If F1 = 10.4 N and F2 = 16.2 N, fi nd the net force on the object and its acceleration for (a) and (b).
90° x y 120° x y (a) (b) F1 5 F2 5 F2 5 F1 5 FIGURE 4 – 44 Problem 38. 39. (II) A skateboarder, with an initial speed of 2.0 m>s, rolls virtu-ally friction free down a straight incline of length 18 m in 3.3 s. At what angle u is the incline oriented above the horizontal?
FIGURE 4 – 42 Problem 36.
132 CHAPTER 4 Dynamics: Newton’s Laws of Motion a 5 2.20 2.15 2.10 2.05 2.00 kg 40. (II) At the instant a race began, a 65-kg sprinter exerted a force of 720 N on the starting block at a 22° angle with respect to the ground. (a) What was the horizontal acceleration of the sprinter? (b) If the force was exerted for 0.32 s, with what speed did the sprinter leave the starting block? 41. (II) A mass m is at rest on a horizontal frictionless surface at t = 0. Then a constant force F0 acts on it for a time t0. Suddenly the force doubles to 2F0 and remains constant until t = 2 t0. Determine the total distance traveled from t = 0 to t = 2 t0. 42. (II) A 3.0-kg object has the following two forces acting on it: F 51 = (16 i N + 12 j N) N F 52 = ( -10 i N + 22 j N) N If the object is initially at rest, determine its velocity ⃗ v at t = 4.0 s. 43. (II) A 27-kg chandelier hangs from a ceiling on a vertical 3.4-m-long wire. (a) What horizontal force would be neces sary to displace its position 0.15 m to one side? (b) What will be the tension in the wire? 44. (II) Redo Example 4 9 13 but (a) set up the equations so that the direction of the acceleration a 5 of each object is in the direction of motion of that object. (In Example 4 9 13, we took a 5 as positive upward for both masses.) (b) Solve the equations to obtain the same answers as in Example 4 9 13. 45. (II) The block shown in Fig. 4 9 45 has mass m = 7.0 kg and lies on a fixed smooth frictionless plane tilted at an angle u = 22.0° to the horizontal. (a) Determine the acceleration of the block as it slides down the plane. (b) If the block starts from rest 12.0 m up the plane from its base, what will be the block’s speed when it reaches the bottom of the incline?
u y x m FIGURE 4 – 45 Block on inclined plane. Problems 45 and 46. 46. (II) A block is given an initial speed of 5.2 m>s up the 22.0° plane shown in Fig. 4 9 45. (a) How far up the plane will it go? (b) How much time elapses before it returns to its starting point? Ignore friction. 47. (II) An object is hanging by a string from your rearview mirror. While you are accelerating at a constant rate from rest to 28 m>s in 5.0 s, what angle u does the string make with the vertical? See Fig. 4 9 46.
y x FT 5 a 5 u mg 5 FIGURE 4 – 46 Problem 47 . 48. (II) A 2.0-kg purse is dropped from the top of the Lean ing Tower of Pisa and falls 55 m before reaching the ground with a speed of 27 m>s. What was the average force of air resistance? 49. (II) Bob traverses a chasm by stringing a rope between a tree on one side of the chasm and a tree on the opposite side, 25 m away, Fig. 4 9 47 . Assume the rope can provide a tension force of up to 31 kN before breaking, and use a “safety factor” of 10 (that is, the rope should only be required to undergo a tension force of 3.1 kN). (a) If Bob’s mass is 72.0 kg, determine the distance x that the rope must sag at a point halfway across if it is to be within its recom-mended safety range. (b) If the rope sags by only one-fourth the distance found in (a), determine the tension force in the rope. Will the rope break?
x FIGURE 4 – 47 Problem 49. 50. (II) As shown in Fig. 4 9 48, five balls (masses 2.00, 2.05, 2.10, 2.15, 2.20 kg) hang from a crossbar. Each mass is supported by “5-lb test” fishing line which will break when its tension force exceeds 22.2 N ( = 5.00 lb). When this device is placed in an elevator, which accelerates upward, only the lines attached to the 2.05 and 2.00 kg masses do not break. Within what range is the elevator’s acceleration? 51. (II) A high-speed 14-car Italian train has a mass of 640 metric tons (640,000 kg). It can exert a maximum force of 400 kN horizontally against the tracks, whereas at maximum constant velocity (300 km>h), it exerts a force of about 150 kN. Calcu-late (a) its maximum acceleration, and (b) estimate the total force of friction and air resistance at top speed. 52. (II) A 450-kg piano is being unloaded from a truck by rolling it down a ramp inclined at 15°. There is negligible friction and the ramp is 4.0 m long. Two workers slow the rate at which the piano moves by pushing with a combined force of 1020 N parallel to the ramp. If the piano starts from rest, how fast is it moving at the bottom? 53. (II) Uphill escape ramps are sometimes provided to the side of steep downhill highways for trucks with overheated brakes. For a simple 11° upward ramp, what length would be needed to stop a runaway truck traveling 140 km>h? Note the large size of your calculated length. (If sand is used for the bed of the ramp, its length can be reduced by a factor of about 2.) 54. (II) A child on a sled reaches the bottom of a hill with a velocity of 10.0 m>s and travels 25.0 m along a horizontal straightaway to a stop. If the child and sled together have a mass of 60.0 kg, what is the average retarding force on the sled on the horizontal straightaway?
FIGURE 4 – 48 Problem 50.
Problems 133 FIGURE 4 – 54 Problem 62.
m M F 5 u FIGURE 4 – 50 Problem 58.
4.2 kg C 2.2 kg 55. (II) Figure 4 9 49 shows a block (mass mA) on a smooth hori zontal surface, connected by a thin cord that passes over a pulley to a second block (mB), which hangs vertically. (a) Draw a free-body diagram for each block, showing the force of gravity on each, the force (tension) exerted by the cord, and any normal force. (b) Apply Newton’s second law to fi nd formulas for the acceleration of the system and for the tension in the cord. Ignore friction and the masses of the pulley and cord.
mB mA FIGURE 4 – 49 Problems 55, 56, and 57 . Mass mA rests on a smooth horizontal surface; mB hangs vertically. 56. (II) (a) If mA = 14.0 kg and mB = 5.0 kg in Fig. 4 9 49, determine the acceleration of each block. (b) If initially mA is at rest 1.250 m from the edge of the table, how long does it take to reach the edge of the table if the system is allowed to move freely? (c) If mB = 1.0 kg, how large must mA be if the acceleration of the system is to be kept at 1 100 g? 57. (III) Determine a formula for the acceleration of the system shown in Fig. 4 9 49 (see Problem 55) if the cord has a non-negligible mass mC. Specify in terms of lA and lB, the lengths of cord from the respective masses to the pulley. (The total cord length is l = lA + lB.) 58. (III) Suppose the pulley in Fig. 4 9 50 is suspended by a cord C. Determine the tension in this cord after the masses are released and before one hits the ground. Ignore the mass of the pulley and cords. 59. (III) Suppose two boxes on a frictionless table are connected by a heavy cord of mass 1.0 kg. Calculate the acceleration of each box and the tension at each end of the cord, using the free-body diagrams shown in Fig. 4 9 51. Assume FP = 40.0 N, and ignore sagging of the cord. Compare your results to Example 4 9 12 and Fig. 4 9 22. 60. (III) Three blocks on a frictionless horizontal surface are in contact with each other as shown in Fig. 4 9 52. A force F 5 is applied to block A (mass mA). (a) Draw a free-body diagram for each block. Determine (b) the acceleration of the system (in terms of mA, mB, and mC), (c) the net force on each block, and (d) the force of contact that each block exerts on its neighbor. (e) If mA = mB = mC = 10.0 kg and F = 96.0 N, give numerical answers to (b), (c), and (d). Explain how your answers make sense intuitively.
mA mC mB F 5 FIGURE 4 – 52 Problem 60. 61. (III) A 2.5-kg block is placed on a frictionless table. The block is connected by massless ropes over massless pulleys to a 5.0-kg block on the right, and a 3.0-kg block on the left, as shown in Fig. 4 9 53. Find the acceleration of the block on the table.
FIGURE 4 – 53 Problem 61. 62. (III) A small block of mass m rests on the sloping side of a triangular block of mass M which itself rests on a hori-zontal table as shown in Fig. 4 9 54. Assuming all surfaces are frictionless, determine the magnitude of the force F 5 that must be applied to M so that m remains in a fi xed position relative to M (that is, m doesn’t move on the incline). [Hint: Take x and y axes horizontal and vertical.] FIGURE 4 – 51 Problem 59. Free-body diagrams for each of the objects of the system shown in Fig. 4 9 22a. Vertical forces, F 5N and F 5G, are not shown.
y x Cord mC = 1.0 kg (c) (b) (a) FBT 5 FTB 5 FTA 5 FAT 5 FP 5 mB = 12.0 kg mA = 10.0 kg 134 CHAPTER 4 Dynamics: Newton’s Laws of Motion mC mA mB FTA FTC FIGURE 4 – 55 Problem 63.
FIGURE 4 – 57 Problem 65.
3.1 kg 1.7 kg 1.8 m 4.8 m FIGURE 4 – 58 Problem 67. 63. (III) The double Atwood machine shown in Fig. 4 9 55 has frictionless, massless pulleys and cords. Deter-mine (a) the acceleration of masses mA, mB, and mC, and (b) the tensions F TA and F TC in the cords. 64. (III) Determine a formula for the magnitude of the force F 5 exerted on the large block (mC) in Fig. 4 9 56 so that the mass mA does not move relative to mC. Ignore all friction. Assume mB does not make contact with mC.
F 5 mB mC mA FIGURE 4 – 56 Problem 64. 65. (III) The two masses shown in Fig. 4 9 57 are each initially 1.8 m above the ground, and the massless frictionless pulley is 4.8 m above the ground. What maximum height does the lighter object reach after the system is released? [Hint: First determine the acceleration of the lighter mass and then its velocity at the moment the heavier one hits the ground. This is its “launch” speed. Assume the mass doesn’t hit the pulley or the ceiling. Ignore the mass of the cord.] 66. (III) A particle of mass m, initially at rest at x = 0, is accelerated by a force that increases in time as F = Ct2. Determine its velocity v and position x as a function of time.
General Problems 67. A crane’s trolley at point P in Fig. 4 9 58 moves for a few seconds to the right with constant acceleration, and the 780-kg load hangs on a light cable at a 5.0° angle to the vertical as shown. What is the acceleration of the trolley and load?
5.0° P 68. A 75.0-kg person stands on a scale in an elevator. What does the scale read (in N and in kg) when (a) the elevator is at rest, (b) the elevator is climbing at a constant speed of 2.0 m>s, (c) the elevator is descending at 2.0 m>s, (d) the elevator is accelerating upward at 2.0 m>s2, (e) the elevator is accelerating downward at 2.0 m>s2? 69. A city planner is working on the redesign of a hilly por tion of a city. An important consideration is how steep the roads can be so that even low-powered cars can get up the hills without slowing down. A particular small car, with a mass of 920 kg, can accelerate on a level road from rest to 21 m>s (75 km>h) in 14.5 s. Using these data, calculate the maximum steepness of a hill. 70. If a bicyclist of mass 65 kg (including the bicycle) can coast down a 6.5° hill at a steady speed of 6.0 km>h because of air resistance, how much force must be applied to climb the hill at the same speed (and the same air resistance)? 71. Tom’s hang glider supports his weight using the six ropes shown in Fig. 4 9 59. Each rope is designed to support an equal fraction of Tom’s weight. Tom’s mass is 78.0 kg. What is the tension in each of the support ropes?
30° 30° 30° 30° 30° c b a a b c 30° FIGURE 4 – 59 Problem 71. 72. A wet bar of soap (m = 150 g) slides freely down a ramp 3.0 m long inclined at 8.5°. How long does it take to reach the bottom? How would this change if the soap’s mass were 250 g? 73. A parent pulls his child in a wagon across a floor. The child and wagon have a combined mass of 35 kg. The wagon handle is inclined at 55° to the horizontal, and the child and wagon (whose wheels are nearly frictionless) are accel-erating at 0.42 m>s2. With what force is the parent pulling on the wagon handle?
General Problems 135 FIGURE 4 – 64 Problem 79.
FT1 FT2 FT4 F FT3 24° FT 5 a 5 mg 5 FIGURE 4 – 61 Problem 75.
mB mA uB uA FIGURE 4 – 66 Problem 82.
FT 5 a 5 mg 5 mB mA u FIGURE 4 – 62 Problems 76 and 77.
FIGURE 4 – 63 Problem 78. 74. Two equal masses in contact on a frictionless surface are acted on by the forces F1 and F2 as shown in Fig. 4 9 60. Determine the magnitude of the contact force exerted on each mass by the other when (a) F2 = F1; (b) F2 = 1 2 F1; (c) F2 = 0. Express your answer in terms of m and F1.
F1 5 F2 5 m m FIGURE 4 – 60 Problem 74. 75. Andrea dangles her watch from a thin piece of string while the jetliner she is in accelerates for takeoff, which takes about 21 s. Estimate the takeoff speed of the aircraft if the string makes an angle of 24° with respect to the vertical, Fig. 4 9 61. 76. A block (mass mA) lying on a fi xed frictionless inclined plane is connected to a mass mB by a cord passing over a pulley, as shown in Fig. 4 9 62. (a) Determine a formula for the acceler-ation of the system in terms of mA, mB, u, and g. (b) What conditions apply to masses mA and mB for the acceleration to be in one direction (say, mA down the plane), or in the oppo-site direction? Ignore the mass of the cord and pulley. 77. (a) In Fig. 4 9 62, if mA = mB = 1.00 kg and u = 38.0°, what will be the acceleration of the system? (b) If mA = 1.00 kg and the system remains at rest, what must the mass mB be? (c) Calculate the tension in the cord for (a) and (b). 78. The masses mA and mB slide on the smooth (frictionless) inclines fi xed as shown in Fig. 4 9 63. (a) Determine a formula for the acceleration of the system in terms of mA, mB, uA, uB, and g. (b) If uA = 34°, uB = 21°, and mA = 5.0 kg, what value of mB would keep the system at rest? What would be the tension in the cord (negligible mass) in this case? (c) What ratio, mA>mB, would allow the masses to move at constant speed along their ramps in either direction? 79. (a) What minimum force F is needed to lift a piano (mass M) using the pulley apparatus shown in Fig. 4 9 64? (b) Deter-mine the tension in each section of rope: F T1, F T2, F T3, and F T4. Assume pulleys are massless and frictionless, and that ropes are massless. 80. In the design of a supermarket, there are to be several ramps connecting different parts of the store. Customers will have to push grocery carts up the ramps and it is desirable that this not be too diffi cult. The engineer has done a survey and found that almost no one complains if the force required is no more than 23 N. Ignoring friction, at what maximum angle u should the ramps be built, assuming a full 25-kg cart? 81. A jet aircraft is accelerating at 3.8 m>s2 as it climbs at an angle of 18° above the horizontal (Fig. 4 9 65). What is the total force that the cockpit seat exerts on the 75-kg pilot?
18° FIGURE 4 – 65 Problem 81. 82. A 6750-kg helicopter accelerates upward at 0.80 m>s2 while lifting a 1080-kg frame at a construction site, Fig. 4 9 66. (a) What is the lift force exerted by the air on the helicopter rotors? (b) What is the tension in the cable (ignore its mass) which connects the frame to the helicopter? (c) What force does the cable exert on the helicopter? 83. An elevator in a tall building is allowed to reach a maximum speed of 3.5 m>s going down. What must the tension be in the cable to stop this elevator over a distance of 2.6 m if the elevator has a mass of 1650 kg including occupants?
136 CHAPTER 4 Dynamics: Newton’s Laws of Motion FIGURE 4 – 68 Problem 88.
29° FIGURE 4 – 67 Problem 86.
FIGURE 4 – 69 Problem 90.
7.5 kg 1.5 kg 5 F FIGURE 4 – 70 Problem 91.
65 m 45 m FIGURE 4 – 71 Problem 92.
F 5 mA mB uA = 59° uB = 32° 84. A fisherman in a boat is using a “10-lb test” fishing line. This means that the line can exert a force of 45 N without breaking (1 lb = 4.45 N). (a) How heavy a fish can the fisherman land if he pulls the fish up vertically at constant speed? (b) If he accelerates the fish upward at 2.0 m>s2, what maximum weight fish can he land? (c) Is it possible to land a 15-lb trout on 10-lb test line? Why or why not? 85. A “doomsday” asteroid with a mass of 2.0 1010 kg is hurtling through space. Unless the asteroid’s speed is changed by about 0.20 cm>s, it will collide with Earth and cause tremendous damage. Researchers suggest that a small “space tug” sent to the asteroid’s surface could exert a gentle constant force of 2.5 N. For how long must this force act? 86. Three mountain climbers who are roped together in a line are ascending an icefield inclined at 29° to the hori zontal (Fig. 4 9 67). The last climber slips, pulling the second climber off his feet. The first climber is able to hold them both. If each climber has a mass of 75 kg, calculate the tension in each of the two sections of rope between the three climbers. Ignore friction between the ice and the fallen climbers. 87. A bicyclist can coast down a 5.0° hill at a constant speed of 6.0 km>h. If the force of air resistance is proportional to the speed v so that Fair = cv, calculate (a) the value of the constant c, and (b) the average force that must be applied (by the rider) in order to descend the hill at 18.0 km>h. The mass of the cyclist plus bicycle is 72.0 kg. 88. Consider the system shown in Fig. 4 9 68 with mA = 8.2 kg and mB = 11.5 kg. The angles uA = 59° and uB = 32°. (a) In the absence of friction, what force F 5 would be required to pull the masses at a constant velocity up the fixed inclines? (b) The force F 5 is now removed. What are the magnitude and direc-tion of the acceleration of the two blocks? (c) In the absence of F 5, what is the tension in the string? 89. A car starts rolling down a 1-in-4 hill (1-in-4 means that for each 4 m traveled along the sloping road, the elevation change is 1 m). How fast is it going when it reaches the bottom after traveling 55 m? Ignore friction. 90. An 18-kg child is riding in a child-restraint chair, securely fastened to the seat of a car (Fig. 4 9 69). Assume the car has speed 45 km>h when it hits a tree and is brought to rest in 0.20 s. Assuming constant deceleration during the collision, estimate the net horizontal force F that the straps of the restraint chair exert on the child to hold her in the chair. 91. A 1.5-kg block rests on top of a 7.5-kg block (Fig. 4 9 70). The cord and pulley have negligible mass, and there is no significant friction anywhere. (a) What force F must be applied to the bottom block so the top block accelerates to the right at 2.2 m>s2? (b) What is the tension in the connecting cord? 92. You are driving home in your 860-kg car at 15 m>s. At a point 45 m from the beginning of an intersection, you see a green traffic light change to yellow, which you expect will last 4.0 s, and the distance to the far side of the intersection is 65 m (Fig. 4 9 71). (a) If you choose to accelerate, your car’s engine will furnish a forward force of 1200 N. Will you make it completely through the intersection before the light turns red? (b) If you decide to panic stop, your brakes will provide a force of 2300 N. Will you stop before entering the intersection? Assume your car is 4 m long.
General Problems 137 (a) (b) x y FP 5 FBR 5 FCR 5 FP 5 FRB 5 FRC 5 B C u u u u FIGURE 4 – 73 Problem 95. (a) Getting a car out of the mud, showing the forces on the boulder, on the car, and exerted by the person. (b) The free-body diagram: forces on a small segment of rope.
FIGURE 4 – 72 Problem 94. 93. A person jumps from a height of 5.0 m into a swimming pool entering feet first with her hands at her sides. She maintains this position upon entering the water and experi-ences a constant upward force equal to 350% of her weight due to the water itself. How far down does she go? 94. Two rock climbers, Paul and Jeanne, use safety ropes of simi lar length. Jeanne’s rope is more elastic, called a dynamic rope by climbers. Paul has a static rope, not recommended for safety reasons. (a) Jeanne (Fig. 4 9 72) falls freely about 2.0 m and then the rope stops her over a distance of 1.0 m. Estimate how large a force (assume constant) she will feel from the rope. (Express the result in multiples of her weight.) (b) In a similar fall, Paul’s rope stretches by only 30 cm. How many times his weight will the rope pull on him? Which climber is more likely to be hurt? 95. (a) Finding her car stuck in the mud, a bright graduate of a good physics course ties a strong rope to the back bumper of the car, and the other end to a boulder, as shown in Fig. 4 9 73a. She pushes at the midpoint of the rope with her maximum effort, which she estimates to be a force FP L 300 N. The car just begins to budge with the rope at an angle u, which she estimates to be 5°. With what force is the rope pulling on the car? Neglect the mass of the rope. (b) What is the “mechanical advantage” of this technique (see Example 4 9 14)? (c) At what angle u would this technique become counterproductive? [Hint: Consider the forces on a small segment of rope where F 5P acts, Fig. 4 9 73b.] A N S W E R S T O E X E R C I S E S A: No force is needed. The car accelerates out from under the cup, which tends to remain at rest as seen from the refer-ence frame of the street. Think of Newton’s first law (see Example 4 9 1).
B: (i) The same; (ii) the tennis ball; (iii) Newton’s third law for part (i), second law for part (ii).
C: (b).
D: (a).
E: (b).
F: 58.0 N.
G: (b).
H: Yes; no.
138 CHAPTER-OPENING QUESTION— Guess now!
You revolve a ball around you in a horizontal circle at constant speed on a string, as shown here from above. Which path will the ball follow if you let go of the string when the ball is at point P?
P (c) (d) (e) (b) (a) Using Newton’s Laws: Friction, Circular Motion, Drag Forces Newton’s laws are fundamental in physics. These photos show two situations of using Newton’s laws which involve some new elements in addition to those discussed in the previous Chapter. The downhill skier illustrates friction on an incline; she is also retarded by air resistance, a velocity-dependent force. The people on the rotating amusement park ride below illustrate the dynamics of circular motion. CONTENTS 5–1 Using Newton’s Laws with Friction 5–2 Uniform Circular Motion : Kinematics 5–3 Dynamics of Uniform Circular Motion 5–4 Highway Curves: Banked and Unbanked 5–5 Nonuniform Circular Motion 5–6 Velocity-Dependent Forces: Drag and Terminal Velocity T his chapter continues our study of Newton’s laws and emphasizes their fundamental importance in physics. We will see how to apply Newton’s laws to understand important situations including friction, and in circular motion, as well as with drag forces which are velocity-dependent (an optional-advanced Section).
C H A P T E R 5 SECTION 5–1 Using Newton’s Laws with Friction 139 5–1 Using Newton’s Laws with Friction Until now we have mostly ignored friction, but it must be taken into account in most practical situations. Friction exists between two solid surfaces because even the smoothest looking surface is quite rough on a microscopic scale, Fig. 5 9 1. When we try to slide an object across a surface, these microscopic bumps impede the motion. Exactly what is happening at the microscopic level is not fully understood. One possibility is that the atoms on a bump of one surface may come so close to the atoms of the other surface that the atoms form a sort of “bond” or brief tiny weld between the two surfaces. Sliding an object across a surface is often jerky, perhaps due to the making and breaking of these bonds. Even when a round object rolls across a surface, there is still some friction, called rolling friction, although it is generally much less than when an object slides across a surface. We focus our attention now on sliding friction, which is usually called kinetic friction (kinetic is from the Greek for “moving”). When an object slides along a rough surface, the force of kinetic friction acts opposite to the direction of the object’s velocity. The magnitude of the force of kinetic friction depends on the nature of the two sliding surfaces. For given surfaces, experiments show that the friction force is approximately proportional to the normal force between the two surfaces, which is the force that either object exerts on the other and is perpendicular to their common surface of contact (see Fig. 5 9 2). The force of friction in many cases depends very little on the total surface area of contact; that is, the friction force on a book is roughly the same whether it is being slid across a table on its wide face or on its spine, assuming the surfaces have similar smoothness. We consider a simple model of friction in which we make this assumption that the friction force is independent of area. Then we write the proportionality between the magnitudes of the friction force F fr and the normal force F N as an equation by inserting a constant of proportionality, mk : F fr = mk F N. [kinetic friction] This relation is not a fundamental law. It is an experimental relation between the magnitude of the friction force Ffr, which acts parallel to the two surfaces, and the magnitude of the normal force FN, which acts perpendicular to the surfaces. It is not a vector equation since the two forces have different directions, perpendic-ular to one another. The term mk is called the coeffi cient of kinetic friction, and its value depends on the roughness of the two surfaces. Measured values for a variety of surfaces are given in Table 5 9 1. These are only approximate, however, since m depends on whether the surfaces are wet or dry, on how much they have been sanded or rubbed, if any burrs remain, and other such factors. But mk (which has no units) is roughly independent of the sliding speed, as well as the area in contact.
v 5 FIGURE 5 – 1 An object moving to the right on a table or fl oor. The two surfaces in contact are rough, at least on a microscopic scale. FIGURE 5 – 2 When an object is pulled along a surface by an applied force (F 5A), the force of friction F 5fr opposes the motion. The magnitude of F 5fr is proportional to the magnitude of the normal force (FN). Ffr 5 FA 5 FN 5 mg 5 TABLE 5 – 1 Coeffi cients of Friction† Surfaces Coeffi cient of Kinetic Friction, Mk Coeffi cient of Static Friction, Ms Wood on wood 0.2 0.4 Ice on ice 0.03 f 0.1 Metal on metal (lubricated) 0.07 0.1 Steel on steel (unlubricated) 0.6 0.7 Rubber on dry concrete 0.8 0.8 Rubber on wet concrete 0.5 0.7 Rubber on other solid surfaces 1 1 9 4 Tefl on® on Tefl on in air 0.04 0.04 Tefl on on steel in air 0.04 0.04 Lubricated ball bearings 60.01 60.01 Synovial joints (in human limbs) 60.01 60.01 †Values are approximate and intended only as a guide.
140 CHAPTER 5 What we have been discussing up to now is kinetic friction, when one object slides over another. There is also static friction, which refers to a force parallel to the two surfaces that can arise even when they are not moving. Suppose an object such as a desk is resting on a horizontal fl oor. If no horizontal force is exerted on the desk, there also is no friction force. But now suppose you try to push the desk, and it doesn’t move. You are exerting a horizontal force, but the desk isn’t moving, so there must be another force on the desk, in the opposite direction, keeping it from moving (the net force is zero on an object at rest). This is the force of static friction exerted by the fl oor on the desk. If you push with a greater force without moving the desk, the force of static friction in the opposite direction also has increased. If you push hard enough, the desk will eventually start to move, and kinetic friction takes over. At this point, you have exceeded the maximum force of static friction, which is given by (F fr)max = ms F N, where ms is the coeffi cient of static friction (Table 5 9 1). Because the force of static friction can vary from zero to this maximum value, we write F fr … ms F N. [static friction] You may have noticed that it is often easier to keep a heavy object sliding than it is to start it sliding in the fi rst place. This is consistent with ms generally being greater than mk (see Table 5 9 1).
It seems that this simple relation for friction, F fr … m F N, was fi rst established by the great “Renaissance man” Leonardo da Vinci (1452 9 1519).
EXAMPLE 5 − 1 Friction: static and kinetic. A 10.0-kg box rests on a hori-zontal fl oor. The coeffi cient of static friction is ms = 0.40 and the coeffi cient of kinetic friction is mk = 0.30. Determine the force of friction, F fr, acting on the box if a horizontal external applied force FA is exerted on it of magnitude: (a) 0, (b) 10 N, (c) 20 N, (d) 38 N, and (e) 40 N.
APPROACH We don’t know, right off, if we are dealing with static friction or kinetic friction, nor if the box remains at rest or accelerates. We need to draw a free-body diagram, and then determine in each case whether or not the box will move: the box starts moving if FA is greater than the maximum static friction force. The forces on the box are shown in Fig. 5 9 2: gravity mg 5 ; the normal force exerted upward by the fl oor F 5N ; the horizontal applied force F 5A ; and the friction force F 5fr.
SOLUTION The free-body diagram of the box is shown in Fig. 5 9 2. In the vertical direction there is no motion, so Newton’s second law in the vertical direction gives πFy = may = 0, which tells us F N - mg = 0. Hence the normal force is F N = mg = (10.0 kg) (9.80 m>s2) = 98.0 N.
(a) Because FA = 0 in this fi rst case, the box doesn’t move, and F fr = 0.
(b) The force of static friction will oppose any applied force up to a maximum of ms F N = (0.40) (98.0 N) = 39 N.
When the applied force is FA = 10 N, the box will not move. Newton’s second law gives πFx = FA - F fr = 0, so F fr = 10 N.
(c) An applied force of 20 N is also not suffi cient to move the box. Thus Ffr = 20 N to balance the applied force.
(d) The applied force of 38 N is still not quite large enough to move the box. So the friction force has now increased to 38 N to keep the box at rest.
(e) A force of 40 N will start the box moving since it exceeds the maximum force of static friction, ms F N = (0.40) (98 N) = 39 N. Instead of static friction, we now have kinetic friction, and its magnitude is F fr = mk F N = (0.30) (98.0 N) = 29 N.
There is now a net (horizontal) force on the box of magnitude F = 40 N - 29 N = 11 N, so the box will accelerate at a rate ax = πF m = 11 N 10.0 kg = 1.1 m>s2 as long as the applied force is 40 N. NOTE Figure 5 9 3 shows a graph that summarizes this Example.
FIGURE 5 – 3 Example 5 9 1. Magnitude of the force of friction as a function of the external force applied to an object initially at rest. As the applied force is increased in magnitude, the force of static friction increases in proportion until the applied force equals ms FN. If the applied force increases further, the object will begin to move, and the friction force drops to a roughly constant value characteristic of kinetic friction. Ffr = msFN Applied force, F A msFN Static friction Kinetic friction Friction force, Ffr no motion sliding 10 10 20 30 40 50 60 70 20 30 40 50 0 Ffr 5 FA 5 FN 5 mg 5 FIGURE 5 – 2 Repeated for Example 5-1.
SECTION 5–1 Using Newton’s Laws with Friction 141 Friction can be a hindrance. It slows down moving objects and causes heating and binding of moving parts in machinery. Friction can be reduced by using lubricants such as oil. More effective in reducing friction between two surfaces is to maintain a layer of air or other gas between them. Devices using this concept, which is not practical for most situations, include air tracks and air tables in which the layer of air is maintained by forcing air through many tiny holes. Another technique to maintain the air layer is to suspend objects in air using magnetic fi elds (“magnetic levitation”).
On the other hand, friction can be helpful. Our ability to walk (see Fig. 4 9 11) depends on friction between the soles of our shoes (or feet) and the ground. Walking involves static friction, not kinetic friction. The movement of a car, and also its stability, depend on friction. When friction is low, such as on ice, safe walking or driving becomes diffi cult.
EXERCISE A If ms = 0.40 and mg = 20 N, what minimum force F will keep the box from falling: (a) 100 N; (b) 80 N; (c) 50 N; (d) 20 N; (e) 8 N?
EXERCISE B If mk FN were greater than FPx, what would you conclude?
EXAMPLE 5 − 3 Pulling against friction. A 10.0-kg box is being pulled along a horizontal surface by a force F P of 40.0 N applied at a 30.0° angle above hori-zontal. This is like Example 4 9 11 except now there is friction, and we assume a coeffi cient of kinetic friction of 0.30. Calculate the acceleration.
APPROACH The free-body diagram is shown in Fig. 5 9 5. It is much like that in Fig. 4 9 21, but with one more force, that of friction.
SOLUTION The calculation for the vertical (y) direction is just the same as in Example 4 9 11, mg = (10.0 kg) (9.80 m>s2) = 98.0 N and F Py = (40.0 N) (sin 30.0° ) = 20.0 N. With y positive upward and ay = 0, we have F N - mg + F Py = may F N - 98.0 N + 20.0 N = 0, so the normal force is FN = 78.0 N. Now we apply Newton’s second law for the horizontal (x) direction (positive to the right), and include the friction force: F Px - F fr = max.
The friction force is kinetic as long as Ffr = mk F N is less than F Px = (40.0 N) cos 30.0° = 34.6 N, which it is: F fr = mk F N = (0.30) (78.0 N) = 23.4 N.
Hence the box does accelerate: ax = F Px - Ffr m = 34.6 N - 23.4 N 10.0 kg = 1.1 m>s2.
In the absence of friction, as we saw in Example 4 9 11, the acceleration would be much greater than this.
NOTE Our fi nal answer has only two signifi cant fi gures because our least signifi cant input value (mk = 0.30) has two.
FIGURE 5 – 5 Example 5 9 3. 30.0° FP 5 mg 5 Ffr 5 FN 5 CONCEPTUAL EXAMPLE 5 − 2 A box against a wall. You can hold a box against a rough wall (Fig. 5 9 4) and prevent it from slipping down by pressing hard horizontally. How does the application of a horizontal force keep an object from moving vertically?
RESPONSE This won’t work well if the wall is slippery. You need friction. Even then, if you don’t press hard enough, the box will slip. The horizontal force you apply produces a normal force on the box exerted by the wall (net force hori-zontally is zero since box doesn’t move horizontally.) The force of gravity mg, acting downward on the box, can now be balanced by an upward static friction force whose maximum magnitude is proportional to the normal force. The harder you push, the greater FN is and the greater Ffr can be. If you don’t press hard enough, then mg 7 ms FN and the box begins to slide down.
mg 5 F 5 Ffr 5 FN 5 FIGURE 5 – 4 Example 5 9 2. 142 CHAPTER 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces EXAMPLE 5 − 5 T wo boxes and a pulley. In Fig. 5 9 7a, two boxes are connected by a cord running over a pulley. The coefficient of kinetic friction between box A and the table is 0.20. We ignore the mass of the cord and pulley and any friction in the pulley, which means we can assume that a force applied to one end of the cord will have the same magnitude at the other end. We wish to find the acceleration of the system, a, which will have the same magnitude for both boxes assuming the cord doesn’t stretch. As box B moves down, box A moves to the right.
APPROACH The free-body diagrams for each box are shown in Figs. 5 9 7b and c. The forces on box A are the pulling force of the cord F T, gravity mA g, the normal force exerted by the table F N, and a friction force exerted by the table Ffr. The forces on box B are gravity mB g, and the cord pulling up, F T.
SOLUTION Box A does not move vertically, so Newton’s second law tells us the normal force just balances the weight, F N = mA g = (5.0 kg) (9.8 m>s2) = 49 N.
In the horizontal direction, there are two forces on box A (Fig. 5 9 7b): FT, the tension in the cord (whose value we don’t know), and the force of friction F fr = mk F N = (0.20) (49 N) = 9.8 N.
The horizontal acceleration of box A is what we wish to find; we use Newton’s second law in the x direction, πFAx = mA ax, which becomes (taking the posi-tive direction to the right and setting aAx = a): πFA x = F T - F fr = mA a. [box A] Next consider box B. The force of gravity mB g = (2.0 kg) (9.8 m>s2) = 19.6 N pulls downward; and the cord pulls upward with a force F T. So we can write Newton’s second law for box B (taking the downward direction as positive): πF By = mB g - F T = m B a. [box B] [Notice that if a ≠0, then FT is not equal to mB g.] We have two unknowns, a and F T, and we also have two equations. We solve the box A equation for F T : F T = F fr + mA a, and substitute this into the box B equation: m B g - F fr - mA a = m B a.
Now we solve for a and put in numerical values: a = mB g - Ffr mA + mB = 19.6 N - 9.8 N 5.0 kg + 2.0 kg = 1.4 m>s2, which is the acceleration of box A to the right, and of box B down.
If we wish, we can calculate FT using the third equation up from here: FT = Ffr + mA a = 9.8 N + (5.0 kg) (1.4 m>s2) = 17 N.
NOTE Box B is not in free fall. It does not fall at a = g because an additional force, FT, is acting upward on it.
FIGURE 5 – 7 Example 5 9 5. B 2.0 kg 5.0 kg B A (a) (b) (c) A mA mB FT 5 FT 5 Ffr 5 FN 5 g 5 g 5 CONCEPTUAL EXAMPLE 5 − 4 T o push or to pull a sled? Your little sister wants a ride on her sled. If you are on flat snow, will you exert less force if you push her or pull her? See Figs. 5 9 6a and b. Assume the same angle u in each case.
RESPONSE Let us draw free-body diagrams for the sled 9 sister combination, as shown in Figs. 5 9 6c and d. They show, for the two cases: the force exerted by you, F 5 (an unknown); by the snow, F 5N and F 5fr ; and gravity mg 5. (a) If you push her, and u 7 0, there is a vertically downward component to your force. Hence the upward normal force exerted by the ground (Fig. 5 9 6c) will be larger than mg (where m is the mass of sister plus sled). (b) If you pull her, your force has a vertically upward component, so the normal force F N will be less than mg, Fig. 5 9 6d. Because the friction force is proportional to the normal force, F fr will be less if you pull her. So you can exert less force if you pull her.
(a) (b) (c) (d) F 5 N F 5 F 5 Ffr 5 Ffr 5 FN 5 F 5 F 5 mg 5 mg 5 u u FIGURE 5 – 6 Example 5 9 4. SECTION 5–1 Using Newton’s Laws with Friction 143 EXAMPLE 5 − 6 The skier. The skier in Fig. 5 9 8a descends a 30° slope. If the coefficient of kinetic friction is 0.10, what is her acceleration when she is in contact with the snow?
APPROACH We choose the x axis along the slope, positive downslope in the direction of the skier’s motion. The y axis is perpendicular to the surface. The forces acting on the skier (Fig. 5 9 8b) are gravity, F 5G = mg 5, which points vertically downward (not perpendicular to the slope), and the two forces exerted on her skis by the snow : the normal force perpendicular to the snowy slope (not vertical), and the friction force parallel to the surface. These three forces are shown acting at one point in Fig. 5 9 8b, which is our free-body diagram for the skier.
SOLUTION We have to resolve only one vector into components, the weight F 5G, and its components are shown as dashed lines in Fig. 5 9 8c. To be general, we use u rather than 30° for now. We use the definitions of sine (“side opposite”) and cosine (“side adjacent”) to obtain the components: FGx = mg sin u, FGy = -mg cos u where FGy is in the negative y direction. To calculate the skier’s acceleration down the hill, ax, we apply Newton’s second law to the x direction: πFx = max mg sin u - mk FN = max where the two forces are the x component of the gravity force (+x direction) and the friction force (-x direction). We want to find the value of ax, but we don’t yet know FN in the last equation. Let’s see if we can get FN from the y component of Newton’s second law: πFy = may FN - mg cos u = may = 0 where we set ay = 0 because there is no motion in the y direction (perpendicular to the slope) when the skier is in contact with the slope. Thus we can solve for FN : FN = mg cos u and we can substitute this into our equation above for max : mg sin u - mk (mg cos u) = max.
There is an m in each term which can be canceled out. Thus (setting u = 30° and mk = 0.10): ax = g sin 30° - mk g cos 30° = 0.50g - (0.10) (0.866)g = 0.41g.
The skier’s acceleration is 0.41 times the acceleration of gravity, which in numbers† is a = (0.41) (9.8 m>s2) = 4.0 m>s2.
NOTE The mass canceled out, so we have the useful conclusion that the acceleration doesn’t depend on the mass. That such a cancellation sometimes occurs, and thus may give a useful conclusion as well as saving calculation, is a big advantage of working with the algebraic equations and putting in the numbers only at the end.
NOTE The friction force on high-speed alpine skiers is very small : it seems the heat produced under the skis by friction melts the snow so the skis float on tiny balls of water. Fast skiers are slowed more by air resistance, which they can reduce by going into a tuck to reduce their surface area. When going airborne (as over the top of a hill) they are slowed more (can’t hold that tuck, especially when landing) than when in contact with the snow.
P H Y S I C S A P P L I E D Skiing P R O B L E M S O LV I N G It is often helpful to put in numbers only at the end P H Y S I C S A P P L I E D Skiers are faster on snow than when airborne In Chapter 4 we examined motion on ramps and inclines, and saw that it is usually an advantage to choose the x axis along the plane, in the direction of acceleration. There we ignored friction, but now we take it into account.
†We used values rounded off to 2 significant figures to obtain a = 4.0 m>s2. If we kept all the extra digits in our calculator, we would find a = 0.4134 g L 4.1 m>s2. This difference is within the expected precision (number of significant figures, Section 1 9 3).
Ffr 5 (a) + y + y + x (b) 30° (Ffr = mkFN) (Ffr = mkFN) G = Ffr 5 FN 5 F 5 (c) 90° + x FGy 5 FG 5 FN 5 FGx 5 mg 5 - u u u FIGURE 5 – 8 Example 5 9 6. A skier descending a slope; F 5G = mg 5 is the force of gravity (weight) on the skier. 144 CHAPTER 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces EXAMPLE 5 − 7 A ramp, a pulley, and two boxes. A box of mass mA = 10.0 kg rests on a surface inclined at u = 37° to the horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless pulley, to a second box of mass mB, which hangs freely as shown in Fig. 5 9 9a. (a) If the coefficient of static friction is ms = 0.40, determine what range of values for mass mB will keep the system at rest. (b) If the coefficient of kinetic friction is mk = 0.30, and mB = 10.0 kg, determine the acceleration of the system.
APPROACH Figure 5 9 9b shows two free-body diagrams for box mA because the force of friction can be either up or down the slope, depending on which direction the box slides: (i) if mB = 0 or is sufficiently small, mA would tend to slide down the incline, so F 5fr would be directed up the incline; (ii) if mB is large enough, mA will tend to be pulled up the plane, so F 5fr would point down the plane. The tension force exerted by the cord is labeled F 5T. SOLUTION (a) For both cases (i) and (ii), Newton’s second law for the y direc-tion (perpendicular to the plane) is the same: F N - mA g cos u = mA ay = 0 since there is no y motion. So F N = mA g cos u.
Now for the x motion. We consider case (i) first for which πF = ma gives mA g sin u - F T - F fr = mA ax.
We want no acceleration, ax = 0, and we solve for F T since F T is related to mB (whose value we are seeking) by F T = mB g (see Fig. 5 9 9c). Thus mA g sin u - F fr = F T = mB g.
We solve this for mB and set F fr at its maximum value ms F N = ms mA g cos u to find the minimum value that mB can have to prevent motion (ax = 0): mB = mA sin u - ms mA cos u = (10.0 kg) (sin 37° - 0.40 cos 37°) = 2.8 kg.
Thus if mB 6 2.8 kg, then box A will slide down the incline. In problems involving a slope or “inclined plane,” avoid making errors in the directions of the normal force and gravity. The normal force is not vertical: it is perpendicular to the slope or plane. And gravity is not perpen dicular to the slope—gravity acts vertically downward toward the center of the Earth.
C A U T I O N Directions of gravity and the normal force FIGURE 5 – 9 Example 5 9 7. Note choice of x and y axes. mB (a) (b) mA = 10.0 kg u = 37° (case i) mA y x FT 5 FN 5 Ffr 5 mAg 5 mB (c) FT 5 mBg 5 u (case ii) y x mA FT 5 FN 5 Ffr 5 mAg 5 u |
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Physics: Uniform Motion
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Ok so I have this homework problem. I dont want to give out the information because i want to put in the values myself, but basically I have one object moving at said speed. Said time later, another object leaves the same location as the object 1 and said speed faster. At some time later, object 2 is only said distance away from object 1. I have to find both their speeds. If this is too vague then i can re-write the question. Anybody got any tips?
algebra-precalculus
physics
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edited Sep 15, 2011 at 14:00
J. M. ain't a mathematician
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asked Sep 13, 2011 at 23:24
Ronnie.jRonnie.j
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$\begingroup$ You should move this question to physics.stackexchange.com $\endgroup$
Curiosity
– Curiosity
2011-09-13 23:27:49 +00:00
Commented Sep 13, 2011 at 23:27
$\begingroup$ ok thanks for the commentary $\endgroup$
Ronnie.j
– Ronnie.j
2011-09-13 23:33:48 +00:00
Commented Sep 13, 2011 at 23:33
1
$\begingroup$ Your post does not make absolutely clear what information is given. Let $v$ be the speed of car 1. Let $v+k$ be the speed of car 2. Let $t_2$ be the time car 2 waits before taking off, and let $t$ be time it has travelled. Let $d$ be the distance it is still "behind." When car 2 started, it was $vt_2$ behind. It has gained $kt$, so it is $vt_2-kt$ behind. So $vt_2-kt=d$. If we know $k$, $t_2$, $t$, $d$, we can find $v$. [But if distance only is given, maybe car 2 is now ahead, then $kt-vt_2=d$. Whether this should be looked at depends on wording details of the question.] $\endgroup$
André Nicolas
– André Nicolas
2011-09-13 23:51:26 +00:00
Commented Sep 13, 2011 at 23:51
1
$\begingroup$ @Ronnie: It would have been better if you had given the problem fully, with all numbers, and requested that people not solve the problem for you, only give some idea of how to proceed. $\endgroup$
André Nicolas
– André Nicolas
2011-09-14 00:11:06 +00:00
Commented Sep 14, 2011 at 0:11
1
$\begingroup$ Yeah, let's credit Ronnie for an ingenious way of making sure that he will only get hints (ok, may be a formula) as opposed to a solution. Wouldn't mind seeing more of that actually! $\endgroup$
Jyrki Lahtonen
– Jyrki Lahtonen
2011-09-14 06:55:38 +00:00
Commented Sep 14, 2011 at 6:55
| Show 9 more comments
3 Answers 3
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EDIT
A body is said to be in uniform motion when it travels equal distances in equal intervals of time (i.e. at a constant speed).
Wikipedia link.
My comment above. Distance traveled by the first object moving at velocity $v$ and departing at time $t=0$
$$s_{1}=vt.\tag{1}$$
Distance traveled by the second object moving at velocity $v+c$ and departing with a delay $t_{0}$ with respect to the first object
$$s_{2}=(v+c)(t-t_{0})\qquad t\geq t_{0}.\tag{2}$$
What is the distance $d$ between both objects at instant $t=t_{0}+t_{1}$ ($t_{1}$ after $t_{0}$)?
$$d=s_{1}-s_{2}.\tag{3}$$
Added. Using information provided by OP in the comments:
A train leaves chicago at 12. 2 [$t_{0}$] hours later another train leaves chicago at a speed 50 [$c$] miles an hour faster than train A. After an hour [$t_{1}$], train B is 10 [$d$] miles behind train A. what are both their speeds [$v$ and $v+c=v+50$]?
You have two options.
(1) Replace the given numerical values in equations $(1)$ to $(3)$, and solve for $v$.
(2) Combine equations $(1)$ to $(3)$ to obtain the equation $d=vt_{0}-ct_{1}$, use the numerical data and solve for $v$.
If I used your [André Nicolas'] formula correctly, I got 30 miles for train 1 and 80 for train 2.
Let's use option (1) with $t_{0}=2$ h (after 12h), $c=50$ mph, $t_{1}=1$ h (after 12+2=14h), $d=10$ miles.
$$s_{1}=vt \text{ miles}.$$
$$s_{2}=(v+50)(t-2)\text{ miles}\qquad t\geq 2\text{ hours}.$$
At $t=t_{0}+t_{1}=2+1=3\text{ hours}$, we have
$$s_{1}=v\times 3=3v\text{ miles},$$
$$s_{2}=(v+50)(3-2)=v+50\text{ miles}$$
and
$$d=10=s_{1}-s_{2}=3v-\left( v+50\right) =2v-50\text{ miles},$$
or
$$10=2v-50\text{ miles},$$
whose solution is $v=30$ mph. So $v+c=v+50=30+50=80$ mph.
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edited Sep 14, 2011 at 14:57
answered Sep 14, 2011 at 0:36
Américo TavaresAmérico Tavares
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You seem to be told the "said speeds" and then have to work them out. But in questions like these you could look at questions like:
How far away is the first when the second sets off?
What is the change in distance between them?
How much time does this change in distance take?
How fast does the distance between them change?
What is the difference between their speeds?
Added after comment:
You know how far apart they are after a known time
You know the difference in their speeds
So you can work out how far apart they were when the second car set off
This is how far the first car travelled in a known time
So you can work out how fast the first car travels
So you can work out how fast the second car travels
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edited Sep 14, 2011 at 0:09
answered Sep 13, 2011 at 23:33
HenryHenry
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$\begingroup$ Ok, well 1 (im numering your bullets) is not stated. 2. 10 miles. 3.1 hr. 4. isn't that the same as 3? and 4. object 2 is 50 mph faster than object 1 $\endgroup$
Ronnie.j
– Ronnie.j
2011-09-13 23:45:57 +00:00
Commented Sep 13, 2011 at 23:45
$\begingroup$ My original equation was to try 3/x + 1/x+50=10. But that is incredibly wrong and I don't know what to do from there $\endgroup$
Ronnie.j
– Ronnie.j
2011-09-13 23:50:25 +00:00
Commented Sep 13, 2011 at 23:50
$\begingroup$ If the change in distance between them is 10 miles and this takes 1 hour, then I would have thought their difference in speeds was 10 mph $\endgroup$
Henry
– Henry
2011-09-13 23:54:02 +00:00
Commented Sep 13, 2011 at 23:54
$\begingroup$ haha i wish it was that easy xD. Ok let me rewrite this. Car1 leaves point A at said time. Some time later, Car2 leaves point A at a speed faster than car1. After some time from car 2 departure, car2 is only said distance away from car1. I need to find both their speeds $\endgroup$
Ronnie.j
– Ronnie.j
2011-09-13 23:58:10 +00:00
Commented Sep 13, 2011 at 23:58
$\begingroup$ Thanks! with this and the help of others I was able to derive (i think) to the answer. much appreciated $\endgroup$
Ronnie.j
– Ronnie.j
2011-09-14 00:49:07 +00:00
Commented Sep 14, 2011 at 0:49
Add a comment |
1
$\begingroup$
The problem, in detail, is as follows.
A train leaves Chicago at 12. Two hours later, another train leaves Chicago at a speed 50 miles an hour faster than train A. After it has been travelling for 1 hour, train B is 10 miles behind train A. What are both their speeds?
We look at a slightly more general problem. Train A leaves a station, and travels at constant speed. Train B leaves the same station $h$ hours later, and travels in the same direction as Train A, at a speed of $k$ miles per hour more than Train A. After Train B has travelled for $t$ hours, Train B is $d$ miles behind Train A. Given $h$, $k$, $t$, and $d$, at what speed does Train A travel?
Solution: Let $v$ be the speed of Train A. When Train B leaves the station, Train A has been travelling for $h$ hours, at speed $v$. So when Train B leaves the station, Train A is $vh$ miles ahead of Train B.
After that, for every hour that elapses, the distance between the two trains decreases by $k$ miles. So after $t$ hours, the distance between the trains has decreased by $kt$ miles, and therefore it is$$vh-kt.$$ But we were told that after Train B has been travelling for $t$ hours, Train B is $d$ miles behind Train A. It follows that $$vh-kt=d.$$ If we know any four of $v$, $h$, $k$, $t$, and $d$, we can use the above equation to solve for the fifth. In this problem, we know $h$, $k$, $t$, and $d$. We can solve the above equation explicitly for $v$. We get $vh=kt+d$, and therefore $$v=\frac{kt+d}{h}.$$
Another way: Start the clock at the instant that Train A leaves the station. After $h+t$ hours, Train A has travelled a distance $v(h+t)$. The speed of Train B is $v+k$. So $t$ hours after leaving the station, Train B has travelled a distance $(v+k)t$. But then Train B is $d$ miles behind Train A, so $$v(h+t) -(v+k)t=d.$$ Simplify the left-hand side. We get $$(vh+vt)-(vt+kt)=d,$$ and end up with $vh-kt=d$, the same equation as before.
Comments: $1$. The solution by Américo Tavares, which combines the algebraic with the graphical, is a much better one. It links, in a very clear way, the kinematic intuition with the geometric intuition. There is an awful lot that can be learned from it, in particular about the deep connection between velocities and slopes of certain lines. A thorough understanding of that solution is useful preparation for the calculus.
$2$. When you are writing up the problem, do not use the formula that I derived. Instead, use the idea, with the concrete numbers of the problem. Then everything will make physical sense, it will not be simply "a bunch of symbols."
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answered Sep 14, 2011 at 5:50
André NicolasAndré Nicolas
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Also known as: Charles Philip Arthur George, prince of Wales and earl of Chester, duke of Cornwall, duke of Rothesay, earl of Carrick and Baron Renfrew, Lord of the Isles, and Prince and Great Steward of Scotland
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Last Updated: •Article History
Quick Facts
Formerly called:
: Prince Charles
Formerly in full:
: Charles Philip Arthur George, prince of Wales and earl of Chester, duke of Cornwall, duke of Rothesay, earl of Carrick and Baron Renfrew, Lord of the Isles, and Prince and Great Steward of Scotland
Born:
: November 14, 1948, Buckingham Palace, London, England (age 76)
Title / Office:
: king (2022-), United Kingdom
Notable Family Members:
: spouse Camilla
: spouse Diana, princess of Wales
: father Philip, Duke of Edinburgh
: mother Elizabeth II
: son Prince Harry, Duke of Sussex
: son William, prince of Wales
: brother Prince Andrew, duke of York
: brother Prince Edward, earl of Wessex
: sister Anne, the Princess Royal
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Charles III (born November 14, 1948, Buckingham Palace, London, England) is the king of the United Kingdom of Great Britain and Northern Ireland from September 8, 2022. He is the eldest child of Queen Elizabeth II and Prince Philip, duke of Edinburgh. After being the longest-serving monarch-in-waiting in British history, Charles ascended the throne at age 73. He was crowned in the first coronation in seven decades on May 6, 2023. The following year it was announced that Charles had been diagnosed with an undisclosed form of cancer.
Early life
After private schooling at Buckingham Palace and in London, Hampshire, and Scotland, Charles entered Trinity College, Cambridge, in 1967. He took a bachelor’s degree there in 1971, the first ever earned by an heir to the British crown. He also spent a term at the University College of Wales, Aberystwyth, learning Welsh in preparation for his investiture as prince of Wales on July 1, 1969, at Caernarvon Castle.
Charles then attended the Royal Air Force College (becoming an excellent flier) and the Royal Naval College, Dartmouth, and from 1971 to 1976 took a tour of duty with the Royal Navy. Later he became an outspoken critic of modern architecture. He expressed his views on the topic in A Vision of Britain (1989). In 1992 he founded the Prince of Wales’s Institute of Architecture, which later evolved into the BRE Trust, an organization involved with urban regeneration and development projects.
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Marriages: Diana Spencer and Camilla Parker Bowles
On July 29, 1981, Charles married Lady Diana Frances Spencer, daughter of the 8th Earl Spencer. The royal wedding was a global media event, broadcast live on television and watched by hundreds of millions of people; following the ceremony, she took the title princess of Wales. The couple’s first child, Prince William of Wales, became at his birth (June 21, 1982) second in line of succession to the throne. Their second child, Prince Henry Charles Albert David (known as Harry), was born on September 15, 1984.
Charles’s marriage to Diana gradually grew strained amid intense scrutiny from the tabloid press and rumours of infidelity. On December 9, 1992, it was announced that Charles and Diana had decided to separate but would continue to fulfill their public duties and to share the responsibility of raising their sons. The marital breakdown remained in the headlines, amid mutual recriminations, tell-all biographies, and admissions of infidelity on both sides. Diana told her side of the story to Andrew Morton for his controversial Diana: Her True Story (1992), and she did so again, in an unusually candid television interview with Martin Bashir, in 1995. The couple finally divorced on August 28, 1996. The couple finally divorced on August 28, 1996. A year later Diana died in an auto accident, and popular feeling for her, stronger even in death than in life, served to jeopardize the traditional form of monarchy that Charles represented. He subsequently spent much effort in modernizing his public image as the heir apparent. On April 9, 2005, he married Camilla Parker Bowles (born 1947), with whom he had a long-standing relationship; after the wedding, Parker Bowles took the title of duchess of Cornwall.
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Arguably, the issue that has remained closest to Charles’s heart is his concern for the environment, which dates to at least 1970, when he delivered a speech on the “horrifying effects” of all forms of pollution and called attention to the threat posed by “indestructible plastic containers.” Since then he has often highlighted the need for rapid action on global warming. He has also been a passionate champion of sustainability, not least through the efforts of his Prince’s Foundation, inspired by his philosophy of harmony: “that by understanding the balance, the order and the relationships between ourselves and the natural world we can create a more sustainable future.”
During the 2010s the attention of royal watchers in many ways shifted from Charles to his sons, whose high-profile “royal weddings” put them and their glamorous partners in the international spotlight. In 2011 William married Catherine Middleton, and in 2018 Harry married Meghan Markle. Tensions arose between Charles and Harry, when Harry and Meghan chose to “step back” from their royal duties and, after negotiations, ceased to be working members of the royal family. Their absence from royal affairs and later that of Prince Andrew (who gave up his military titles and royal patronages in 2021 in the wake of a scandal tied to his involvement with sex trafficker Jeffrey Epstein) only added to the burden on Charles, who increasingly stood in for the aging Elizabeth as her health became more fragile. Some observers even suggested that Charles had effectively become a regent for the queen. Following her death on September 8, 2022, Charles became king.
Coronation and health issues
Charles delivered his inaugural address on September 9, 2022. In the televised speech he paid tribute to his mother’s life and pledged “throughout the remaining time God grants me, to uphold the Constitutional principles at the heart of our nation.” In addition, he conferred the title prince of Wales upon his eldest son, William.
The new king toured the constituent units of the United Kingdom as they mourned Elizabeth’s death. He also participated with his siblings, his sons, and their families in a series of moving processions and ceremonies honouring the life and legacy of the queen. Notably, he stood vigil with his siblings by the queen’s coffin as it lay in state in Westminster Hall and attended Elizabeth’s sombre funeral ceremony in Westminster Abbey.
Charles had met with Prime Minister Liz Truss at Buckingham Palace on September 9. A little more than a month later she would resign as leader of the Conservative Party. By the end of October she was replaced as prime minister by Rishi Sunak, with whom Charles began holding a customary weekly audience. Among the foreign leaders with whom Charles met during his first months on the throne were Cyril Ramaphosa, the president of South Africa, who made a two-day state visit to the United Kingdom in November, and Volodymyr Zelensky, the president of war-torn Ukraine, who visited Buckingham Palace in February 2023.
In delivering his first Christmas message as monarch in December 2022, Charles reflected again on Elizabeth’s death and referred to his mother’s faith in people, using it as a jumping-off point from which to celebrate the selfless contributions of individuals and institutions throughout the Commonwealth. Charles’s welcome as the new king was mixed, however. While his public appearances were sometimes greeted with shouts of “God save the king” by well-wishers, on two occasions in late 2022 eggs were thrown at him by protesters.
The royal family’s strained relationship with Harry and Meghan also continued to pose challenges for the king, especially after the six-part documentary series Harry & Meghan debuted on Netflix in December 2022. Similarly problematic was the publication in early January 2023 of Harry’s memoir, Spare, in which, among other tell-all revelations, the prince describes in detail his feud with Prince William.
Charles’s coronation on May 6, presided over by Archbishop Justin Welby, featured a new coronation liturgy. Commissioned by Welby with the theme “called to serve,” it welcomed people of all faiths and included prayers and hymns in the Welsh, Scots Gaelic, and Irish languages. It was the first coronation service to feature female bishops, and, in recognition of the diverse communities of the United Kingdom, the presentation of the regalia was done by Jewish, Muslim, Hindu, and Sikh members of the House of Lords.
In January 2024 Charles underwent a procedure to deal with an enlarged prostate. The following month it was revealed that during the procedure a separate issue had been uncovered and was later determined to be cancer. While palace officials declined to specify the type, they did state that it was not prostate cancer. As he sought treatment, Charles stepped back from public-facing duties. The royal family faced additional difficulties as it was later revealed that Catherine, princess of Wales, was also undergoing treatment for cancer.
The Editors of Encyclopaedia BritannicaThis article was most recently revised and updated by Amy Tikkanen. |
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Constant Specific Heat
In subject area:Engineering
Constant specific heat refers to the specific heat of a material that does not vary significantly with temperature, commonly applicable to solids where the specific heat at constant volume and constant pressure are virtually indistinguishable. It can be measured by determining the heat input required to raise the temperature of the material by a certain amount.
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Review article
One-dimensional automotive catalyst modeling
2005, Progress in Energy and Combustion ScienceChristopher Depcik, Dennis Assanis
Making the assumption of incompressibility and constant velocity:
(43)
and including the definition of constant specific heats
(44)
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Journal2005, Progress in Energy and Combustion ScienceChristopher Depcik, Dennis Assanis
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Structure, aerodynamics, and geometry of premixed flamelets
2000, Progress in Energy and Combustion ScienceC.K Law, C.J Sung
For energy conservation across the flame, we note that as all the deficient reactant is consumed, and because the system is adiabatic, for constant specific heat cp we have
(2)
where qc is the chemical heat release per unit mass of fuel consumed. Eq. (2) simply states that all the chemical heat liberated is used to heat the incoming gas. Therefore the downstream temperature Tbo is just the adiabatic flame temperature Tad.
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Turbomachinery
2017, Theory of Aerospace Propulsion (Second Edition)Pasquale M. Sforza
7.2.1Compressor Thermodynamics
Assuming that the specific heat of the fluid is constant through the compressor, the work input required is
(7.4)
Here cp,2 is the constant specific heat associated with the compressor, stations 2–3, and may be considered an average value for the temperature range encountered. Eq. (7.4) assumes that all the work put into the fluid is involved in raising the total enthalpy which would imply an adiabatic compression. Though we have assumed that the compression process is adiabatic, we have not assumed a lossless system, and the efficiency of adiabatic compression may be defined as
(7.5)
This process is shown on an h−s plot in Fig. 7.3 where the primed quantity ht′ (and Tt′) arises from an isentropic process while the unprimed quantity ht (and Tt) is associated with the dissipative effects that lead to an increase in entropy, therefore representing less than ideal efficiency.
Note that the first and second laws of thermodynamics provide the following relation:
Then
For a constant pressure process the slope
(7.6)
Thus, along a constant pressure line the slope increases with temperature and therefore also with enthalpy. Thus the constant pressure lines diverge as illustrated in Fig. 7.3.
In an ideal, or isentropic, compression with constant γ = cp/cv the stagnation temperature and stagnation pressure are related as follows:
(7.7)
The ratio of specific heats γ2 has the subscript 2 to denote that it is the constant value of γ associated with the compressor, stations 2–3, and may be considered an average value for the temperature range encountered. The compressor entry stagnation temperature Tt,2 is approximately equal to the free stream stagnation temperature because the inlet flow is essentially adiabatic with little heat transfer to the inlet walls. At high subsonic flight Mach numbers in the stratosphere where T0 ~ 216 K the free stream stagnation temperature Tt,0 ~ Tt,2 ~ 260 K. One may show that even for very high compressor pressure ratios (pt,3/pt,2 = 50), the compressor exit temperature Tt,3 ~ 800 K and the ratio of specific heats at the exit is about 2% less than that evaluated at the entry temperature. Thus using a constant value of γ throughout the compressor is somewhat optimistic, but quite reasonable. It is a common practice to assign γ = 7/5 for the compressor in preliminary studies. Because the compressor processes are carried out at temperatures considerably less than those in the turbine, the compressor flow is considered the “cold” flow.
The efficiency of adiabatic compression may be expressed in terms of actual thermodynamic state variables as follows:
(7.8)
The work required to operate the compressor is then given by
(7.9)
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2009, Computational Fluid Dynamics in Fire Engineering
Specific Heat Capacity of Partially Charred Wood
For partially charred wood, the specific heat capacity varies with the degree of conversion from virgin wood to char during the pyrolysis. According to Chan et al. (1985), the effect of charring is treated implicitly. In their model, the specific heat capacity of partially charred wood is related linearly to the decreasing solid density, which effectively reflects the degree of conversion from wood to char as
(4.11.34)
where A = 1.339 kJ kg−1 K−1 and B = 3.147 × 10−3 kJ m3 kg−2 K−1. However, a widely adopted formulation is used to assume the volumetric specific heat capacity varies linearly between the virgin wood and final char and can be expressed by
(4.11.35)
where Cp is the specific heat capacity and ρ is the density with the subscripts s, w, and f denoting the pyrolysing solid, dry virgin wood, and final char. The mass fraction of virgin wood (i.e., unconverted wood) is expressed as
(4.11.36)
The preceding formulation has been adopted by previous workers such as Kung and Kalelkar (1973), Kung (1972), and Fredlund (1988). Similar expressions for the specific heat of partially charred wood have also been used by Alves and Figueiredo (1989), Bonnefoy et al. (1993), and Di Blasi (1994a).
Specific Heat Capacity of Dry Virgin Wood
The specific heat capacity of dry virgin wood as a function of temperature has been determined by Atreya (1983) for a temperature range from 0°C to 140°C. Fredlund (1988) has assumed that the relationship is valid even for temperatures above 140°C. The expression for the specific heat capacity of dry virgin wood as a linear function of temperature is given by
(4.11.37)
where Cpw,o = 1.4 kJ kg−1 K−1 and Cpw,m = 3.0 × 10−4 kJ kg−1 K−2. It is nonetheless noted that a constant specific heat capacity for dry virgin wood independent of temperature has also been assumed by a number of investigators such as Kanury and Blackshear (1970a), Kung (1972), Kung and Kalelkar (1973), Chan et al. (1985), Alves and Figueiredo (1989), Bonnefoy et al. (1993), and Di Blasi (1994a). Values ranging from 1.386 kJ kg−1 K−1 to 2.52 kJ kg−1 K−1 with most of them larger than 2.0 kJ kg−1 K−1 have been typically employed.
Specific Heat Capacity of Char
A similar expression for the specific heat capacity of char as a linear function of temperature as proposed by Fredlund (1988), can be expressed as
(4.11.38)
where Cpf,o = 0.7 kJ kg−1 K−1 and Cpf,m = 6.0 × 10−4 kJ kg−1 K−2. A value for the specific heat capacity of char of 0.672 kJ kg−1 K−1 has also been given by Fredlund (1988) and by a number of other investigators such as Kanury and Blackshear (1970a), Kung (1972), Kung and Kalelkar (1973), Chan et al. (1985), Alves and Figueiredo (1989), Bonnefoy (1993), and Di Blasi (1994a). These ranged from 0.672 kJ kg−1 K−1 to 2.52 kJ kg−1 K−1. In some pyrolysis models (Kanury and Blackshear, 1970a, Kung, 1972), the specific heat capacity of char was not distinguished from that of the dry virgin wood material.
Specific Heat Capacity of Water and Vapor
In the wet wood pyrolysis model of Alves and Figueiredo (1989), the specific heat capacity of moisture (i.e., liquid water) is taken to be constant at Cpm = 4.19 kJ kg−1 K−1. For the range of temperature between 25°C and 100°C where moisture in wood is found to exist in the liquid phase, the specific heat capacity ranges from 4.181 kJ kg−1 K−1 to 4.219 kJ kg−1 K−1 (Rogers and Mayhew, 1980).
Alves and Figueiredo (1989) applied a constant value of specific heat capacity of vapor Cpv = 1.88 kJ kg−1 K−1. Fredlund (1988) presented graphically the variation of specific heat capacity of vapor with increasing temperature from 0°C to 1000°C. The reported values varied approximately linearly from 1.86 kJ kg−1K−1 to 1.90 kJ kg−1K−1. The constant value of Alves and Figueiredo (1989) is equivalent to the average value between the lower and upper values used by Fredlund (1988). In most practical cases, a linear variation of the specific heat capacity of vapor may be more preferentially adopted.
Specific Heat Capacity of Volatile Gases
Constant values of specific heat capacity for the mixture of volatile gases ranging from 1.008 kJ kg−1 K−1–1.2 kJ kg−1 K−1 have been employed by Kanury and Blackshear (1970a), Kung (1972), Kung and Kalelkar (1973), Alves and Figueiredo (1989), and Di Blasi (1994a). The specific heat capacity as a function of temperature has nonetheless been assumed by Atreya (1983).
Fredlund (1988) obtained the proportions of the chemical compounds in the pyrolysis products from the experimental result. The specific heat capacity for the volatile gases Cpg was evaluated based on the proportions of pyrolysis products. It is expressed as a function of temperature and is given by
(4.11.39)
where Cpg,o = 1.0 kJ kg−1 K−1 and Cpg,m = 8.0 × 10−5 kJ kg−1 K−2.
Specific Heat Capacity of Dry Air
Dry air (i.e., mixture of gases excluding the vapor portion, which is treated separately) is present initially in wood, and contributes to the total pressure inside the wood throughout the pyrolysis. This mixture of gases and their migration in wood must be modeled. In order to account for the internal convection heat transfer due to dry air migration, its specific heat capacity is needed. The specific heat capacity of the dry air can be expressed as
(4.11.40)
where Cpi,o = 0.994 kJ kg−1 K−1 and Cpi,m = 2.0 × 10−4 kJ kg−1 K−2 for the temperature range between 0°C and 1200°C (Rogers and Mayhew, 1980).
Thermal Conductivity
Fredlund (1988) revealed that the heat transfer during pyrolysis comprises of three components: conduction through the solid phase, radiant heat transfer in the pores, and convection heat transfer in the enclosed gases. The overall thermal conductivity is obtained as the sum of the conductivities of each of these components. Radiation heat transfer in the pore system is generally found to be negligible at room temperatures, even at high porosities. At elevated temperature and for porosities below 0.7, radiation heat transfer is still only marginally significant. In his subsequent pyrolysis modeling, Fredlund (1988) simply uses a thermal conductivity, which is a function of only the temperature of the material. On the basis of the pyrolysis models of Chan et al. (1985), the terms corresponding to the variations of porosity and radiant heat transfer in the pores are included. However, the significance of both terms has not been explored.
As applied in the one-dimensional pyrolysis models of Kung (1972), Kung and Kalelkar (1973), and Alves and Figueiredo (1989), the thermal conductivity of partially charred wood ks has been taken to vary linearly between that of virgin wood and that of final char:
(4.11.41)
where kw and kf are the thermal conductivity of the virgin wood and final char; the mass fraction of virgin wood a is as defined in equation (4.11.36). In pyrolysis models by Kanury and Blackshear (1970a), Kung and Kalelkar (1973), and Kung (1972), constant values of thermal conductivity for virgin wood materials have been applied, which ranged from 0.1134 W m−1K−1 to 0.21 W m−1K−1. Constant values of thermal conductivity ranged from 0.0412 W m−1K−1 to 0.189 W m−1K−1 for final char have nevertheless been used by Kung and Kalelkar (1973) and Kung (1972).
Di Blasi (1994a) has proposed the use of different thermal conductivities of wood and char in the longitudinal and transverse (i.e., tangential and radial) directions to the grains. In three dimensions, equation (4.11.41) can be extended to give the thermal conductivity of the partially charred wood in the Cartesian directions as
(4.11.42)
(4.11.43)
(4.11.44)
where the subscripts s, w, and f denote the pyrolysing solid, dry virgin wood, and final char, respectively. According to Di Blasi (1994a), the thermal conductivities of virgin wood in the longitudinal and transverse direction of the grains have been assigned values of 0.255 W m−1K−1 and 0.105 W m−1K−1, while the thermal conductivities of char in the longitudinal and transverse direction of the grains were set to 0.1046 W m−1K−1 and 0.071 W m−1K−1. These values have been found to be within the ranges reported in Siau (1984).
Permeability
Owing to the anatomy of wood, the permeability of wood varies greatly with direction relative to the grain orientation. Skaar (1988) has indicated that the ratio of the measured longitudinal-to-tangential permeabilities for hardwoods can be as high as 80,000 to 1, and the longitudinal-to-radial ratio may be 50,000 to 1, while for the softwoods, the longitudinal-to-transverse ratio can be as high as 4 × 108 to 1. The permeabilities in the transverse directions relative to grain orientation are, however, almost the same.
When the virgin wood gradually pyrolyses to form carbonaceous char, the porosity of the partially charred wood increases. This results in an increase of the permeability until it reaches that of the final char. Fredlund (1988) described the theoretical estimation of the permeability of the partially charred wood as a function of the varying porosity as follows:
(4.11.45)
(4.11.46)
where KD1 and KD2 are constants depending on species and orientation of the wood grains with the superscripts x and y denoting the directions. The preceding expression, however, predicted only a 100-fold increase in the permeability when the porosity changes from 0.647 to 0.9. This is considered much smaller than the change expected in reality, where cracking in the carbon layer leads to considerable increase in the permeability. Di Blasi (1994a) assumed that the permeability varied linearly between the virgin wood and char in both the longitudinal and transverse direction, respectively, according to the degree of conversion from virgin wood to char. In other words,
(4.11.47)
(4.11.48)
In her model, an activation cellulose was produced as an intermediate product during pyrolysis, which was assumed to have the same thermal conductivity as the virgin wood.
In the mathematical modeling of wood pyrolysis and combustion, a good estimate of the permeability of the partially charred wood, which governs the migration of volatile gases and vapor through wood as well as their emission from the wood surface, is needed. Suitable expressions for the permeability of the partially charred (i.e., still pyrolysing) wood remain elusive. The current authors have adopted both experimental and modeling techniques for the determination of the expression for the permeability of partially charred wood and the proposed exponential expression similar to the exponential form of Fredlund (1988) is
(4.11.49)
(4.11.50)
In the experiment, the determined permeabilities show a similar trend of increase of permeability with percentage char for the partially charred wood. Although a theoretical derivation seems impossible, Shafizadeh (1968) gave an insight into an exponential formulation of permeability of the pyrolysis wood. The pyrolysis of wood comprises of competitive reaction paths of which the formation of levoglucosan (i.e., 1,6-anhydro-β-D-glucopyranose) at temperatures around 250°C is the major by-product, followed by competitive secondary reactions to form char, vapor, and gases at elevated temperatures. This greatly suggests that at the initial stage of pyrolysis the char formation is limited, thus a slower increase of permeability. An exponential expression, which gives a slow but increasing permeability, appears to be more plausible. The proposed expression consists of a second exponential term with a multiplying factor b, which controls the rate of increase of the permeability with respect to the mass fraction of virgin wood a. This constant b has been assigned different values to obtain the best-fit prediction of the average permeability to that determined experimentally. The predicted curve using KD1 = 9.22 × 10−18 m2 and KD2 = 10.88 with a value of b = 15 is seen to provide a prediction of average permeability values, which fits extremely well to the experimentally determined values as illustrated in Figure 4.23. This expression may be employed to calculate the permeability of partially charred wood in modeling of pyrolysis and combustion of wood.
Dynamic Viscosity
The dynamic viscosity of the pyrolysis gas can be calculated by the proposal of Fredlund (1988), using the proportions of pyrolysis products according to the experimental data. For the volatile gases, the dynamic viscosity can be expressed as a function of temperature in the form of
(4.11.47)
where μg,o = 8.5 × 10−6 kg m−1 s−1 and μg,m = 2.95 × 10−8 kg m−1 s−1 K−1. The dynamic viscosity of the vapor, also by Fredlund (1988), is given by
(4.11.48)
where μv,o = 8.5 × 10−6 kg m−1 s−1 and μv,m = 3.75 × 10−8 kg m−1 s−1 K−1. The dynamic viscosity of the dry air is assumed to be constant, μi = 3.178 × 10−5 kg m−1 s−1, which is the median value for the temperature ranging between 300°C and 1000°C (Rogers and Mayhew, 1980).
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Book2009, Computational Fluid Dynamics in Fire Engineering
Chapter
Further Considerations in Field Modeling
2009, Computational Fluid Dynamics in Fire Engineering
Specific Heat Capacity of Dry Virgin Wood
The specific heat capacity of dry virgin wood as a function of temperature has been determined by Atreya (1983) for a temperature range from 0°C to 140°C. Fredlund (1988) has assumed that the relationship is valid even for temperatures above 140°C. The expression for the specific heat capacity of dry virgin wood as a linear function of temperature is given by
(4.11.37)
where Cpw,o = 1.4 kJ kg−1 K−1 and Cpw,m = 3.0 × 10−4 kJ kg−1 K−2. It is nonetheless noted that a constant specific heat capacity for dry virgin wood independent of temperature has also been assumed by a number of investigators such as Kanury and Blackshear (1970a), Kung (1972), Kung and Kalelkar (1973), Chan et al. (1985), Alves and Figueiredo (1989), Bonnefoy et al. (1993), and Di Blasi (1994a). Values ranging from 1.386 kJ kg−1 K−1 to 2.52 kJ kg−1 K−1 with most of them larger than 2.0 kJ kg−1 K−1 have been typically employed.
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Book2009, Computational Fluid Dynamics in Fire Engineering
Review article
A review of thermodynamic cycles and working fluids for the conversion of low-grade heat
2010, Renewable and Sustainable Energy ReviewsHuijuan Chen, ... Elias K. Stefanakos
3.1.2Influence of latent heat, density and specific heat
Maizza and Maizza suggested that high latent heat, high density and low liquid specific heat are preferable, as a fluid with a high latent heat and density absorbs more energy from the source in the evaporator and thus reduces the required flow rate, the size of the facility, and the pump consumption. However, Yamamoto et al. suggested that low latent heat is better because the saturated vapor at the turbine inlet would provide the best operating condition. Here, we conducted a theoretical analysis by deriving the expression of the enthalpy change through the turbine expansion, in order to obtain a comprehensive conclusion.
The transition between the two phases of matter can be characterized by Clausius–Clapayron relation, which is:
(2)
where dP/dT denotes the slope of the coexistence curve on a P–T diagram, L is the latent heat, T is the absolute temperature, and ΔV is the volume change of the phase transition. When the transition is to a gas phase, the specific volume can be many times the size of the initial specific volume, so ΔV = Vgas can be approximated, which is also applied in the current situation.
Here it is assumed that the vapor follows the ideal gas law for the sake of simplification. Since high pressure vapor cannot be considered as an ideal gas, this analysis is only for a qualitative investigation and not meant to do accurate calculations. The ideal gas law is:
(3)
Combining Eqs. (2) and (3) and after integration, the pressure ratio of any two points on the coexistence line of a phase diagram is obtained as:
(4)
From Eq. (4) one can notice that the pressure ratio of a working fluid is decided by its latent heat when the saturation temperatures are defined.
The unit isentropic enthalpy drop (i.e. the work output) through a turbine is calculated from:
(5)
where γ = Cp/Cv is the ratio of the heats, is the turbine inlet temperature and Pdis and Pin denote the turbine discharge and inlet pressures, respectively. The above expression for the enthalpy drop was widely accepted for discussion although it is derived under the assumption of ideal gas with constant specific heats [72,73].
Combining Eqs. (4) and (5) one obtains:
(6)
where T1 and T2 are the saturation temperatures of two points on the coexistence line and T1 > T2, is the turbine inlet temperature (Fig. 6), and the other notations remain the same.
Eq. (6) shows that fluids with higher latent heat give higher unit work output when the temperatures and other parameters are defined. The influence of the latent heat can also be explained by observing the T–s diagram in Fig. 6. Under defined temperatures, the length of the horizontal line segment is proportional to the latent heat. Long line segment (i.e. latent heat) is expected to obtain large work output because the area formed by the process of the cycle is the work output from the turbine. This result agrees with the conclusion from Eq. (6). Meanwhile, as it has been mentioned, Eq. (6) gives the unit mass work output from the turbine, it can be inferred that fluids with higher density need smaller equipment setup for same power production. In brief, working fluids with high density, low liquid specific heat and high latent heat are expected to give high turbine work output.
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Journal2010, Renewable and Sustainable Energy ReviewsHuijuan Chen, ... Elias K. Stefanakos
Chapter
Single Effect Evaporation – Vapor Compression
2002, Fundamentals of Salt Water DesalinationHisham T. El-Dessouky, Hisham M. Ettouney
3.2.5Summary
Analysis of the system performance by the mathematical models shows consistency of predictions and industrial practice. The specific power consumption is found to vary over a similar range, 9-17 kWh/m3 at 60 °C. In addition, the predicted evaporator specific heat transfer area is close to the industrial practice, with values between 400-600 m2/(kg/s) at 60 °C. The temperature values predicted by the model are also found consistent with reported industrial data.
Problems
1.
: An MVC system is to be designed to produce 5000 m3/d of fresh water. The boiling temperature is 70 °C and the temperatures of the compressed vapor and condensate are higher by 8 °C and 3 °C, respectively. The salinity of the feed seawater is 38000 ppm and the rejected brine salinity is 70000 ppm. For preliminary design considerations neglect thermodynamic losses, assume constant specific heat for all liquid stream (4.2 kJ/kg °C), constant specific heat for the vapor streams (1.884 kJ/kg °C), and constant overall heat transfer coefficients of 2.7, 7.2, and 7.8 for the evaporator, distillate preheater, and brine preheater, respectively. Calculate the following: Mb, Mf, Tf, To, Ab, Ad, Ae, and W.
2.
: Determine the effect of dependence of the specific heat of liquid streams, boiling point rise, latent heat, pressure, enthalpy, overall heat transfer coefficients, and specific volume on temperature and concentration on the design values obtained in problem 1. Use the correlations given in the appendices to calculate the physical properties and thermodynamic losses.
3.
: An MVC system is used to desalinate seawater at 35 °C with 42000 ppm salinity. The maximum allowable brine temperature is 100 °C. The heat transfer coefficient for the evaporator and the two preheaters is constant and equal to 5.016 kW/m2 °C. The specific heat transfer area is 109.46 m2 per (kg/s) of fresh water and the heat transfer area of the distillate preheater is 200 m2. The flow rates of the hot and cold stream in the preheaters are equal. The temperatures of the distillate and rejected brine flowing from the preheaters are 45 °C and 40 °C, respectively. Calculate the specific power consumption.
4.
: An MVC system has the following design data:
: Md = 1 kg/s, Ad = 10 m2.
: Ab = 20 m2, Ae = 500 m2.
: Ue = 2.4 kW/m2 °C, Ud = 6.7 kW/m2 °C.
: Ub = 6.3 kW/m2 °C, Xf = 42000 ppm.
: Xb = 70000 ppm.
Determine Tb, Td, T0, Tf, and Ts if Tcw = 25 °C, Cp = 4.2 kJ/kg °C, and Cpv = 1.884 kJ/kg °C. Also, determine the specific power consumption of the compressor and the flow rates of the brine and feed seawater.
5.
: For the same conditions in the previous problem determine Tb, Td, To, Tf and Ts if Tcw drops to 15 °C. Also, determine the specific power consumption and the flow rates of the brine and feed seawater.
6.
: If fouling conditions arise in the system described in problem 2, where the overall heat transfer coefficient in the evaporator is drop to 1.8 kW/m2 °C. If all other conditions are kept the same, determine Tb, Td, To, Tf, and Ts at the new fouling conditions. Also, calculate the specific power consumption and the flow rates of brine and feed streams.
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Review article
Review of thermophysical property methods applied to fueled and un-fueled molten salts
2020, Annals of Nuclear EnergyJared Magnusson, ... Troy Munro
3.4Specific heat (constant pressure)
Volumetric heat capacity represents one of the most important properties in salt selection (Ambrosek, 2011). For the same change in temperature and given mass, salt with a larger volumetric heat capacity can move more heat from the reactor to the power plant (Ambrosek, 2011). This allows for less processing of spent fuel or radiated coolant as well a reduction of heat transfer equipment (Ambrosek, 2011). Due to the temperature invariance of the data, cp values are reported as single values in Table 5 with the exception of LiF and FLiNaK, which are presented in Figs. 13 and 14.
Table 5. Coolant salt specific heat capacity data reference.
| Author | Year | Molar Composition (%) | Technique employed | cp (cal/gK) | Uncertainty quoted (%) | Temperature Range (K) | No. of Data | Form of Data |
--- --- --- ---
| LiF |
| Chase | 1998 | 100 | – | 0.5914 | – | 1121–3000 | 1 | A |
| Douglas & Dever | 1954 | 100 | Ice Calorimeter | 0.5979 | 1.5 | 1121–1169 | 1 | A |
| Macleod | 1973 | 100 | Adiabatic vacuum drop calorimeter | 0.577–0.627 | – | 1121–2000 | 10 | T |
| LiF-BeF2 |
| Cantor | 1968 | 66-34 | Estimate/Calculation | 0.284–0.29 | – | 773–973 | 3 | A |
| Douglas & Payne | 1969 | Li2BeF4 (13.85% Li, 9.34% Be, 75.95% F) | Drop method- ice Bunsen calorimeter | 0.561 | 3 | 745–900 | 8 | T |
| Hoffman & Cooke (Cited by (Cantor, 1968) | 1968 | 66-34 | Unpublished measurements at ORNL | 0.577 | – | – | 1 | A |
| Rosenthal et al. | 1968 | 66-34 | Inconel capsules with 2 copper block drop calorimeters | 0.577 | 1.4 | 773–993 | 16 | D |
| LiF-NaF-KF |
| An et al. | 2015 | 46.5-11.5-42 | ASTM E 1269-11 (DSC) | 0.45 | 4.3 | 804–973 | 34 | D |
| Blalock & Powers | 1956 | 46.5-11.5-42 | ORNL-Ice calorimeterNBS-?NRL-? | See figures | – | 373–1173773–1173373–1173 | 959 | AAA |
| Khokhlov et al. | 2009 | 46.5-11.5-42 | STA 449C Jupiter synchronous thermal analyzer (DSC) | See figures | 5 | 1066–1240 | 70 | D |
| Rogers et al. | 1982 | 46.6-11.4-42 | Perkin-Elmer DSC-Model 2 calorimeter | See figures | 2 | 748–863 | – | E |
| KCl-MgCl2 |
| Yuanyuan et al. | 2017 | 68-32 | Simultaneous DSC, thermogravimetric analysis | 0.24 | – | 704–1103 | 9 | D |
| KF-ZrF4 |
| N/A |
| NaF-KF-MgF2 |
| N/A |
Italics represent isochoric specific heats (cv).
A = averages from data/equation, D = diagram, E = equation, T = tabulated experimental data.
LiF: Douglas and Dever are commonly cited for their early measurements using an ice calorimeter (Douglas and Dever, 1954). However, their measurements only spanned about 80 K past the melting point (Fig. 13). They reported 1.5% error. Macleod’s data with an adiabatic vacuum drop calorimeter covered a much broader range, but his results are not congruent with the generally acknowledged constant specific heat for liquidus molten salts. Macleod reported heat capacity with an exponential, albeit low, dependence on temperature, though this could be well within his unreported error (Macleod, 1973). The NIST-JANAF thermochemical tables compiled in 1998 reported a constant specific heat up to 2900 K (Chase, 1998).
FLiBe: Rosenthal et al. (1968) tested two samples of FLiBe in Inconel capsules using separate copper-block drop calorimeters. Sixteen enthalpy measurements yielded 1.4% error across the temperature range, and the value of 0.577 cal/g K matched the value by Hoffman (reported by Cantor (1968)). Douglas and Payne (1969) obtained a slightly lower value of 0.561 cal/g K with an ice Bunsen calorimeter but were working with Li2BeF4 compound. This composition approaches that of the FLiBe mixture but had a purity of only 98.6%, which may explain why Douglas and Payne’s report yielded considerably more error (3%).
Cantor (1968) also reported isochoric specific heats which were calculated from isobaric values, though they are not plotted here.
FLiNaK: Four studies account for the available measurement data on FLiNaK’s specific heat (Fig. 14). The first by Blalock and Powers (Powers et al., 1956) cited studies performed at ORNL, National Bureau of Standards (NBS), and Naval Research Laboratory (NRL), but only gave the measurement technique of ORNL (ice calorimeter). However, a copper-block calorimeter was vaguely attributed to another of the sources in the method section. The large change in specific heat around 454 °C (727 K) reflected the melting point of FLiNaK; data for solid mixes are omitted in Fig. 14. No error was reported for NRL or NBS, but their heat capacity dropped off linearly with temperature, and further investigation would be needed to explain this result. The second study by Khokhlov et al. (2011) in 2011 reported well-organized data at temperatures exceeding MSR requirements. Khokhlov employed a STA 449C Jupiter synchronous thermal analyzer (DSC) to produce a data sample with 5% uncertainty. Data under-predicted the trend recorded by Rogers et al. (1982) employing a Perkin-Elmer DSC-Model 2 calorimeter, but Rogers’ composition was not quite eutectic. Rogers reported only 2% error for 23 sample measurements.
KCl-MgCl2: Yuanyuan et al. (2017) remains the sole retrievable contributor to this salt’s heat capacity. Nine measurements with a simultaneous differential scanning calorimetry (DSC) and thermogravimetric analysis system yielded a constant value of 0.239 cal/g K (0.999 J/g K) from 704 to 1103 K.
FLuZirK: None exist according to Serrano-López et al. (2013).
NaF-KF-MgF2: No available specific heat data was found for this salt.
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Chapter
Introduction
2012, Fluid Mechanics (Fifth Edition)
1.9Perfect Gas
A basic result from kinetic theory and statistical mechanics for the thermal equation of state for n identical noninteracting gas molecules confined within a container having volume V is
(1.21)
where p is the average pressure on the inside surfaces of the container, is Boltzmann's constant, and T is the absolute temperature. Equation (1.21) is the molecule-based version of the perfect gas law. It is valid when attractive forces between the molecules are negligible and when V/n is much larger than the (average) volume of an individual molecule. When used with the continuum approximation, (1.21) is commonly rearranged by noting that ρ = mn/V, where m is the (average) mass of one gas molecule. Here m is calculated (in SI units) as Mw/Ao where Mw is the (average) molecular weight in kg (kg-mole)–1 of the gas molecules, and Ao is the kilogram-based version of Avogadro's number, (kg-mole)–1. With these replacements, (1.21) becomes
(1.22)
where the product kBAo = Ru = 8314 J kmol−1 K−1 is the universal gas constant, and R = Ru/Mw is the gas constant for the gas under consideration. A perfect gas is one that obeys (1.22), even if it is a mixture of several different molecular species. For example, the average molecular weight of dry air is 28.966 kg kmol–1, for which (1.22) gives R = 287 J kg−1 K−1. At ordinary temperatures and pressures most gases can be treated as perfect gases.
The gas constant for a particular gas is related to the specific heats of the gas through the relation
(1.23)
where Cp and Cv are the specific heat capacities at constant pressure and volume, respectively. In general, Cp and Cv increase with temperature. The ratio of specific heats
(1.24)
is important in compressible fluid dynamics. For air at ordinary temperatures, γ = 1.40 and Cp = 1004 J kg−1 K−1. It can be shown that (1.21) or (1.22) is equivalent to e = e(T) and h = h(T), and conversely, so that the internal energy and enthalpy of a perfect gas are only functions of temperature (Exercise 1.10).
A process is called adiabatic if it takes place without the addition of heat. A process is called isentropic if it is adiabatic and frictionless, for then the entropy of the fluid does not change. From (1.18) it can be shown (Exercise 1.11) that isentropic flow of a perfect gas with constant specific heats obeys
(1.25)
Using (1.22) and (1.25), the temperature and density changes during an isentropic process from a reference state (subscript 0) to a current state (no subscript) are
(1.26)
(see Exercise 1.8). In addition, simple expressions can be found for the speed of sound c and the thermal expansion coefficient α for a perfect gas:
(1.27, 1.28)
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Chapter
Shock Wave Interactions and Propagation
2001, Handbook of Shock WavesMARINUS E.H. VAN DONGEN
15.4.2THERMODYNAMIC PROPERTIES OF A LIQUID GAS SUSPENSION
We now consider a mixture of gas (subscript g), vapor (v), and droplets (d). Gas and vapor are considered as calorically perfect components with constant specific heats. The liquid also has ideal properties; its specific heat cl is also considered constant. Gas and vapor satisfy the ideal gas law,
(15.4.1)
(15.4.2)
with Rg and Rv the specific gas constants of gas and vapor. The densities ρg and ρv are the actual mixture densities pertaining to a unit volume of mixture. This implies that the volume occupied by the droplets is assumed to be negligible. Also, the droplets are assumed not to contribute to the mixture pressure p:
(15.4.3)
An important thermal property is the specific enthalpy h, for which we write
(15.4.4)
(15.4.5)
(15.4.6)
In (15.4.5) we have introduced the formation enthalpy of the vapor L0. It is related to the actual latent L by
(15.4.7)
This reflects that the assumption of constant specific heats is consistent with a linear dependency of the latent heat on temperature with (Cpv – Cl) as its slope. If ρυ << ρl the vapor-liquid equilibrium pressure pvs satisfies the Clausius–Clapeyron equation:
(15.4.8)
We have introduced the dimensionless latent heat ζ(T) for later use. A combination of (15.4.7) and (15.4.8) leads to the Rankine–Kirchhoff equation for the vapor pressure,
(15.4.9)
where the subscript ref refers to an arbitrary reference state. If the liquid phase consists of monodispersed spherical droplets with radius the equilibrium vapor pressure will be different because of the curvature of the interfaces and the corresponding pressure difference between droplet and vapor,
(15.4.10)
with the Kelvin number Ke related to the surface tension σ of the liquid:
(15.4.11)
Since we neglect the liquid volume, the density of the mixture is the sum of the partial densities of the three constituents:
(15.4.12)
We have introduced here the density of the continuous phase, gas, and vapor, ρc. The droplet density ρd is related to the droplet number density nd, liquid density ρl, and average volume as
(15.4.13)
We shall make use of the mass fractions fi and of the different constituents i:
(15.4.14)
(15.4.15)
It is advantageous to define the following sets of mass-weighted quantities:
(15.4.16)
(15.4.17)
Combining Dalton's law (15.4.3) with the ideal gas law (15.4.1), (15.4.2) leads to the mixture equation of state:
(15.4.18)
For the mixture enthalpy we write
(15.4.19)
which is the mass averaged sum of the different contributions.
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Book2001, Handbook of Shock WavesMARINUS E.H. VAN DONGEN
Related terms:
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Coefficient of Performance
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Isentropic Efficiency
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6833 | https://phas.ubc.ca/~kiefl/ch13_part1.pdf | Chapter 13 : Simple Harmonic Motion Hooke’s law states that the force (F) exerted by an ideal spring is proportional to its elongation l F=k l where k is the spring constant. Consider a mass m hanging on a the spring. In equilibrium the force of gravity on the mass (-mg) is equal and opposite to the force exerted by the spring (mg=k l) One can measure the k by plotting l versus mg. In SI k has units N/m 2 -9.81ms g where l mg k If we displace the mass by an amount x from equilibrium the total force on the mass is then: kx mg x l k F ) ( Thus the force is proportional but opposite to the displacement away from equilibrium.
If we release the mass at time t=0 what is the displacement x(t) at some later time t later? Newton’s Law tells us (1) Eqn. ; m or 2 2 -kx dt x d F ma Observations about the resulting motion: 1. the motion repeats itself; that is it is periodic; where the period T or frequency f=1/T is only a function of k/m.
2. At any time the force F(t) and resulting acceleration a(t) are proportional to the displacement. Clearly a(t) is time dependent (i.e. not constant).
) ( ) ( ) ( 2 2 t x m k dt t x d t a motion the of period the is T and integer any is n where ) ( ) ( nT t x t x 3. Velocity is maximum when x=0 and zero when x= maximum. This follows from energy conservation. The sum of potential energy (U) and kinetic energy (K) must stay constant.
2 ' ' ' ) ' ( ) ( 2 0 0 x k dx x k dx x F x U x x 2 2 1 ) ( mv v K constant 2 1 2 1 constant 2 2 kx mv U K Solution to Eqn. 1 We can show experimentally that the solution of Eqn 1 with the initial condition v(t)=0 and x(t)=A at t=0 is frequency the is f and 2 2 frequency angular where 2 Eqn ; cos ) ( T f t A t x If the vector A rotates at an angular frequency such that t then the projection of A on the x-axis (point P) has the following form x(t)=Acos t. Experiment shows that this function closely matches the oscillating motion of the mass on a spring. The shadow on the screen (point P) moves with velocity 3 Eqn ; sin sin 2 sin tan t A t T A t v dt dx whereas the acceleration of point P on the x axis 4 Eqn. ; cos cos cos 2 2 tan 2 2 t A t A v t a dt x d rad Inserting Eqns 4 and 2 into Eqn 1 yields m k 2 1 2 1 f or cos cos 1 Eqn ; ) ( ) ( 2 2 2 T m k t A m k t A t x m k dt t x d Units of are radians/s (or rad/s) whereas for the frequency f has units s-1, Hertz or Hz. Note the frequency (period of the motion) does not depend on the amplitude of vibration . Check if this is true.
t A t x cos ) ( t Acos A increases from 1 to 2 to 3 mass increases from 1 to 2 to 3.
Check experimentally if k m T 2 k increases from 1 to 2 to 3 Phase angle It is possible that at time t=0 the amplitude is not at its maximum and the velocity is therefore non zero. Why? This situation is described by including an initial phase angle into the solution 5 Eqn ; ) cos( ) ( t A t x Note the period and amplitude of the function are unchanged. The phase angle simply produces an offset on the time axis t=t’-it follows that the velocity and acceleration are then given by ) cos( ) ( ) sin( ) ( 2 2 2 t A dt x d t a t A dt dx t v (1) Eqn. ; m 2 2 -kx dt x d ) cos( ) ( t A t x Show that is also solution of A?
v t A dt dx t v 2 3 ) 0 ( ); 3 / sin( ) ( 2 / ) 0 ( ); 3 / cos( ) ( 2 2 2 2 A a t A dt x d t a 2 ) 0 ( ); 3 / cos( ) ( A/ x t A t x Consider for example an oscillation with phase = /3 Note any SHM is described by 3 parameters , A and Given these you can determine the position, velocity or acceleration at any point in time. Alternatively one can determine some or all of the parameters given specific information about the position, velocity or acceleration at particular time. Each piece of information provides an equation relating the 3 parameters. For example, suppose you know =(k/m)1/2 Given the initial velocity is v0 and initial position is x0 then you can calculate A and as follows: 2 2 2 2 2 0 2 0 0 0 0 0 ] cos [sin and arctan or tan cos sin cos ; sin A A x v x v A A x v A x A v o o Suppose you know the initial acceleration is a0 and initial velocity is x0 what are the amplitude and phase?
Example: Suppose a 0.5kg mass is attached to a horizontal spring. The spring is pulled on with a force of 6N and this causes a displacement of x=0.030m. Then the mass is released. (a) What are the velocity and position of the mass 10 s later?
(b) Suppose if at t=0 the velocity is 1.0m/s and position is -0.2m what are the position and velocity 10 s later Energy in SHM Return to the hanging mass on a spring. The total energy is the sum of kinetic plus potential energy.
t kA t m k mA t mA mv K 2 2 2 2 2 2 2 2 sin 2 1 sin 2 1 sin 2 1 2 1 time of t independen ; 2 1 2 1 cos 2 1 2 1 2 1 2 1 ) 2 ( 2 1 ; ) ( 2 1 2 2 2 2 2 2 2 2 2 2 kA l k K U t kA l k kx l k mgx x l x l k mg l k mgx x l k U Simple pendulum L L g mL mg m k thus L mg k L x mg mg mg F 2 f) 2 ( g measure can One m! of t independen ; or ; sin Suppose a point mass m is suspended on a string of length L. At time t=0 the mass is displaced by an amount x or the angle away from equilibrium. Then there is a restoring force What happens when is not small?
U tube Consider a column of water of length l=0.5 m and area a in an open U tube. At time t=0 the water is stationary and displaced by an amount x0=1cm. What is the level on the right hand side 10 s later?
water x0 1cm Damped Harmonic Oscillator In all real oscillators the total energy U+K does not remain constant due to some resistance to the motion which leads to dissipation of the energy. Often this damping force is proportional to but opposite in direction to the velocity. rate damping the is 1 and frequency undamped the is where Eqn 1 x or or ; ma F using or 2 2 2 2 2 2 2 m b m k dt x d dt dx t dt x d mdt bdx x m k dt x d m dt dx b kx dt dx b kx Ftotal The solution to Eqn has 3 has a different functional forms depending on how big the magnitude of the damping rate (1/ ) . 1. Underdamped motion when or b<2(km)1/2 . The motion is a similar to an undamped oscillator.
2 2 2 4 1 ' ) ' cos( ) ( t Ae t x t 2 1 Note the amplitude of the oscillations decreases exponentially. Thus the total energy of the oscillator at the peaks is just the potential energy since K=0. peak. amplitude when the times the are 2 / t ; e E 2 1 e 2 1 ) ( p / t -0 / t -2 p p nT kA t E p Also the presence of damping reduces the oscillation frequency ’ and reaches zero when 2 2 4 1 ' 2 1 At any instant 2 2 2 ) ( 2 1 2 1 bv kx ma v dt dx kx dt dv mv dt dE kx mv E Thus the instananeous rate of energy loss varies during the period 2sin2 t. However if the amount of energy lost per period is small compared to the total energy at that time then we can replace v2 by its average value over one period which is just equal to .
peak times at the energy for the expression above with the consistent is this note ; Thus cycle for that energy total average the is E where / / 0 2 t e E E m b E dt E d m E v Critical damping: In this circumstance there is no oscillation and the solution of Eqn is just a single exponential. Verify satifies Eqn . This is useful if you want the oscillator to return to equilibrium as quickly as possible. For example shock absorbers in a car provide damping of the springs so that after a bump the potential energy in the springs is dissipated as quickly as possible. 2 1 ) ( 2 / t Ae t x Overdamping: If the damping rate then the return to equilibrium is very slow and in general is characterized by a sum of two exponentials resulting in from the imaginary argument to the cosine function since when 2 1 t a t a e C e C x 2 1 2 1 number imaginary an is 2 1 ' 2 2 2 1 Imagine an oscillator in a very viscous medium such as molasses such that 1/ >>2 or b>>(km)1/2. Then show that the approximate solution to Eqn is b kt A Ae / t 2 which may also be derived by assuming the acceleration term is zero.
Note the mass of the object does not affect the motion in this case! Driven oscillator and resonance Consider a mass m on a spring with a natural frequency k/m Now suppose on exerts a time varying force to the mass at a different frequency What does the motion look like? m t F kx t ky kx t y x k F t y t y cos cos )) ( ( cos ) ( 0 0 0 1. What happens when is far away What happens when approaches 0?
To be more quantitative we can neglect any damping so that the equation of motion is Eqn ; cos 0 2 0 2 2 t m F x dt x d To approximate this one take our mass on a spring and move the point of attachment up and down with an amplitude y0 at a frequency as shown on the right. Since the force is varying at a frequency it is reasonable to expect a solution x=Acos t. Substituting this into Eqn we get .
solution a is cos Thus / 1 / / or ; / 2 2 0 0 2 0 2 0 2 2 0 0 0 2 0 2 t /m F x(t) k F m F A m F A A A Note the negative amplitude for is equivalent to a phase shift of 0 0 for 0 for ); cos( | | t A x Note the amplitude of vibration diverges when the driving frequency matches the natural frequency of vibration. This implies that the total energy in the oscillator increases as approaches 0. Where does the energy come from?
From what you saw in the demonstration do you think there might be another solution ? What happens if at t=0 the mass is stationary but displaced from equilibrium but you still have the driving force on?
Driven harmonic oscillator with damping What do you think will happen if we include damping?
t m F dt dx x dt x d cos 1 0 2 0 2 2 As before if we start the mass in the equilibrium position with no velocity and then turn on the driving force we expect an oscillation at the driving frequency 2 2 2 2 0 0 1 / F A where ; ) cos( ) ( k t A t x Note if the first term in the square root is much greater than the second (far away from resonance) then the solution is the same as for an undamped oscillator. As before the oscillation is in phase with the driving frequency for < 0 ( =0) whereas they are 180 degrees out of phase for > 0 . b=0.2(km)1/2 or As the damping rate becomes smaller the resonance is more sharply peaked and higher. The amplitude is similar to the undamped case.
Below the resonance oscillation and driving frequency are in phase. On resonance the oscillationa nd driving force are 90 degrees out of phase and well above the resonance they are 180 degree out of phase.
Suppose the driving frequency is well above the natural frequency.
What do think will happen if instead of starting the mass off in equilibrium we start if off out of equilibrium. What will the solution look like?
The Q-factor of an oscillator is defined as 2 times the ratio of energy relative to the energy lost per cycle T e E T Ee E E Q t t 2 2 2 / / This document was created with Win2PDF available at
The unregistered version of Win2PDF is for evaluation or non-commercial use only. |
6834 | https://www.engineersedge.com/physics/speed_of_sound_in_liquid__16035.htm | Speed of Sound in Liquid or Fluid Formula
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Speed of Sound in Liquid or Fluid Formula
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Speed of Sound in a Liquid or Fluid Formula Equations and Calculator
The speed of sound (acoustic velocity or sonic velocity), in a fluid is a function of its bulk modulus (or, equivalently, of its compressibility). Equation 1 gives the speed of sound through a liquid.
Preview Premium Calculator Speed of Sound in a Liquid / Gas Calculator
Eq. 1a, SI Units
a = ( E / ρ )0.5
_Eq. 1b, U.S.
a = ( E g c / ρ )0.5_
_Eq. 2a, SI Units
a = ( 1 / ( β ρ ) ) 0.5_
_Eq. 2b U.S.
a = ( g c / ( β ρ ) ) 0.5_
Equation 3 gives the speed of sound in an ideal gas. The temperature, T, must be in degrees absolute (i.e., °R or °K). For air, the ratio of specific heats is k = 1.4, and the molecular weight is 28.967. The universal gas constant is R 1545.35 ft-lbf/lbmol-°R (8314.47 J/kmol·K). a = ( E / ρ )1/2 = ( k p / ρ )1/2
a = ( E / ρ )1/2 = ( k p / ρ )1/2
Eq. 3a, SI
a = ( k R T )1/2 = ( k R T / MW )1/2
a = ( E g c / ρ )1/2 = ( k g c p / ρ )1/2
Eq. 3b, US
a = ( k g c RT )1/2 = ( k g c R T / MW )1/2
Since k and R are constant for an ideal gas, the speed of sound is a function of temperature only. Equation 4 can be used to calculate the new speed of sound when temperature is varied.
Eq. 4
a 1 / a 2 = ( T 1 / T 2 )
Where:
a = speed of sound ft/sec, (m/s)
E = bulk modulus lbf/ft 2, (Pa)
β = compressibility ft 2/lbf, (Pa-1)
ρ = density lbm/ft 3, (kg/m 3)
g c = gravitational acceleration, 32.2 ft/sec 2, (9.81 m/s 2)
MW = Molecular weight, lbm/lbmol, (kg/kmol)
R = specific gas constant ft-lbf/lbmol-°R, (J/kmol·K)
R = universal gas constant, 1545.35 (8314.47) ft-lbf/lbmol-!R J/kmol·K
k = ratio of specific heats
The symbol c is also used for the speed of sound.
Approximate Speeds of Sound (at one atmospheric pressure)
Material Speed of sound
(ft/sec)(m/s)
air 1130 at 70° F 330 at 0° C
aluminum 16,400 4990
carbon dioxide 870 at 70° F 260 at 0° C
hydrogen 3310 at 70° F 3310 at 70° C
steel 16,900 5150
water 4880 at 70° F 1490 at 20° C
(Multiply ft/sec by 0.3048 to obtain m/s.)
Reference:
Civil Engineering Reference Manual, Fifteenth Edition, Michael R. Lindeburg, PE
Related:
Speed of Sound Table Chart
Speed of Sound in Various Liquids and Solids
Bulk Modulus for Materials Metals and Liquids
Density of Common Engineering Materials
Air Density Table and Specific Weight Table, Equations and Calculator
Densities of Metals and Elements Table
Water Vapor Density
Density Equation and Review
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6835 | https://uh.edu/~chem1p/c11/c11F99.pdf | Chapter 11 Theories of Covalent Bonding Concept 11-1 Concept 11-1. The main ideas of valence bond theory, orbital overlap, opposing electron spins, and hybridization as a means of rationalizing molecular shapes orbitals.
2 Valence Bond Theory and Orbital Hybridization • VB theory bases molecular shape on orbital characteristics.
•In VB model, a covalent bond forms as orbitals from two atoms overlap and a pair of electrons occupies space between the nuclei. • •Three central ideas of VB theory : Three central ideas of VB theory : 1. Opposing spins of the electrons must pair.
The overlapping orbitals have a capacity of two electrons that must have opposite spins.
The bond strength depends on attraction of the nuclei for the shared electrons, so the greater the orbital overlap, the stronger the bond.
2. Maximum overlap of bonding orbitals.
H• + •H → H : H 3 • Bonding in molecules like HF and F 2 results from direct overlap of s and p orbitals of atoms.
3. Hybridization of atomic orbitals.
•Shapes of many polyatomic molecules/ions aren’t easily explained using s orbitals and dumb-bell-shaped, perpendicular p orbitals.
Consider methane molecule: 4 H atoms bonded to a carbon.
•A ground-state carbon atom ([He] 2s 22p 2) is not well suited to form four bonds.
•Linus Pauling proposed that the valence atomic orbitals in the molecule are different from those in the isolated atoms.
•He mathematically combined atomic orbitals in an atom to create new atomic orbitals with better characteristics for bonding.
• Orbital mixing is called hybridization, and the mixed atomic orbitals are called hybrid orbitals hybrid orbitals (HO). Two key points are: 1. The number of hybrid orbitals obtained always equals the number of atomic orbitals (AO) mixed.
2. The type of HO’s obtained varies with the types of AO’s mixed.
•The type of HO is picked after we know the molecular shape.
sp hybridization When two electron groups surround the central atom, as in AX2 molecules like BeCl 2(g), a linear linear shape is found.
• VB theory proposes that mixing one s and one p AO on the central atom gives two equivalent sp hybrid orbitals sp hybrid orbitals that lie 180° apart.
4 Concept 11-2 Concept 11-2. How orbitals mix to form hybrid orbitals with different spatial orientations Animation of sp hybrid orbital formation An accurate quantum mechanical representation of an sp HO 5 6 • The paired 2s electrons in the Be atom occupy sp HO’s. These then form the Be–Cl bonds.
•The 4 valence e– (2 from Be + 1 from each Cl) occupy the overlapped orbitals in pairs.
7 sp 2 hybridization. To explain a trigonal planar molecule, mix one s and two p orbitals of the central atom to give three sp 2 HO’s.
•VB theory shows the B atom in BF 3 molecule to be sp 2 hybridized.
•The three sp 2 orbitals point to the corners of an equilateral triangle; the third, unhybridized 2p orbital lies perpendicular to the plane.
•The six valence electrons—three from B and one from each of the three F atoms— make up three bonding pairs.
8 sp 3 hybridization. In AX4 shapes with a tetrahedral electron-group arrangement, one s and three p orbitals of the central atom are mixed.
• Forms four sp3 hybrid orbitals, which point toward the corners of a tetrahedron.
• Carbon in methane is sp 3 hybridized.
•Its 4 valence electrons, 2- 2s and 2- 2p, half fill the four sp 3 hybrids.
•These overlap the 1s orbitals on 4 H atoms and form 4 C–H bonds.
9 •In NH 3, a lone pair fills one of the four sp 3 orbitals, whereas in H 2O, lone pairs fill two of them.
10 sp 3 d hybridization. Trigonal bipyramidal molecules have central atoms from Period 3 or higher, so d orbitals can be hybridized.
•In PCl 5 molecule, VB indicates that 3s, three 3p, and one of the five 3d orbitals of the P atom mix and form five sp3d hybrid orbitals.
•Each sp3d hybrid orbital overlaps a 3p orbital of a Cl atom.
•The five valence electrons of phosphorus, together with one from each of the five Cl atoms, pair up to form five P–Cl bonds mix mix 11 sp 3 d 2 hybridization. For octahedral SF 6 , VB mixes the 3s, three 3p, and two 3d AO’s of the central S atom forming six sp 3d 2 HO’s.
• These point to the corners of an octahedron.
• Sulfur’s 6 valence electrons half fill the HO’s, each of which overlaps a half-filled 2p orbital of an F atom forming six S–F bonds.
12 Postulating the Hybrid Orbitals in a Molecule Problem: Describe how mixing AO’s on the central atoms leads to the HO’s in : (a) Methanol, CH 3OH (b) Sulfur tetrafluoride, SF 4 Plan: Find the number and arrangement of electron groups around central atoms and postulate the type of HO’s needed. Write the box diagram for central atoms before and after orbitals are hybridized.
Solution: (a) CH 3OH The C is AX 4 and the O is AX 2E 2 so central atoms are sp 3 hybridized. The C atom has 4 half-filled sp 3 orbitals: The O atom has two half-filled sp 3 HO’s and two filled with lone pairs: 13 (b) SF 4 Molecular shape is seesaw. The central S atom is surrounded by five electron groups, which requires sp sp 3 3d d hybridization.
One hybrid orbital is filled with a lone pair lone pair and four form S–F bonds. •As the central atom becomes larger, electron repulsions decrease. Hybridization is not needed to explain the bond angles.
•H 2S molecule has a 92° bond angle, near the 90° angle between p AO’s. Such angles occur in most hydrides with large central atoms large central atoms.
NOTE 14 • VSEPR model predicts different shapes for the ethane (C2H 6), ethylene (C2H 4) and acetylene (C2H 2) molecules.
•Ethane: C–C bond: C sp 3 HO - C sp 3 HO. C–H bonds: C sp 3 HO - H 1s AO. • Both have end-to-end overlap; highest electron density along the bond axis.
• •End to end overlap End to end overlap –> sigma (σ) bond Concept 11-3. The distinction between end-to-end and side-to-side overlap and the origin of sigma (σ) and pi (π) •A π bond has lobes above and below the σ-bond axis. One π bond holds 2 electrons that move through both regions.
15 •Ethylene contains a carbon-carbon double double bond. Each C atom is AX 3, so C forms three sp 2 HO’s.
• Each carbon uses the three sp 2 HO’s to form sigma bonds; unhybridized 2p orbital lies perpendicular to the sp2 plane.
•The σ-bonded C atoms hold their half-filled unhybridized 2p AO’s close together so they overlap side to side side to side and form a pi ( π) bond.
sp2 sp2 sp2 p Concept 11-4 Concept 11-4. How two modes of orbital overlap lead to single, double, and triple bonds 16 •Side-to-side overlap is not as extensive as end-to-end overlap, so a π bond is weaker than a σ bond. •As a result, the C=C bond energy of 614 kJ/mol is 80 kJ/mol less than twice that of the C–C bond (347 kJ/mol).
•There is free rotation of the atoms around a σ bond because the extent of overlap is not affected.
• However, p orbitals must be parallel to engage in side-to-side overlap, so a π bond restricts the rotation of the atoms.
• This explains the existence of separate cis and trans structures in double-bonded molecules, such as dichloroethylene.
cis-dichloroethylene trans-dichloroethylene Concept 11-5. Why π bonding restricts rotation around double bonds.
17 • The C≡C bond in acetylene consists of one σ and two π bonds. For linear shape, each C uses 2 sp HO’s for C-C and C-H bonds.
• Two 2p orbitals on each C are unhybridized forming two π bonds.
•Side-to-side overlap of the pairs of 2p AO’s gives a pair of π bonds, oriented at 90° to one another.
• Lobes of one π bond are above and below the σ bond. Lobes of the other π bond are in front of and behind the σ bond. sp sp p p 18 Describing the Bonding in Molecules with Multiple Bonds Problem: Describe the bonds and orbitals in acetone, (CH 3) 2CO.
Solution: Each C of the two CH 3 groups are AX 4 so use sp 3 HO’s and the middle C is AX 3 so uses sp 2 HO’s.
The O atom is sp 2 hybridized also. Two of its sp 2 orbitals hold lone pairs; the third forms a σ bond with the third sp 2 orbital of the middle C atom.
The unhybridized, half-filled p orbitals on C and O form a π bond.
C O H3C H3C C O H3C H3C 19 Molecular Orbital (MO) Theory and Electron Delocalization •In MO theory, a molecule is a collection of nuclei with the electrons in orbitals that are delocalized over the structure.
•Just as atoms have AO’s of a given energy and shape, molecules have molecular orbitals (MO’s) of certain energies and shapes.
•In the LCAO approximation, used to obtain energies and shapes of MOs, MOs are derived from a L Linear C Combination of AO’ AO’s s.
• •LCAO LCAO combines atomic orbitals (atomic wave functions) of nearby atoms to form the molecular orbitals (molecular wave functions).
Concept 11-6 Concept 11-6. The distinction between the localized bonding of VB theory and the delocalized bonding of MO theory.
Concept 11-7 Concept 11-7. How addition or subtraction of AOs form bonding or antibonding MO’s.
20 •H 2 molecule: two H nuclei lie near each other so their AOs overlap.
•Adding their wave functions together forms a bonding MO, which allows a region of higher electron density between the nuclei.
(less e – density between nuclei) •The number of AOs combined always equals the number of MOs formed: two H AOs form two H2 MOs.
21 • 1s AOs of H atoms have same energy, so they interact strongly.
•Charge delocalization lowers the energy of the bonding MO. When electrons occupy 1s, H2 is more stable than the separate H atoms.
Concept 11-8 Concept 11-8. Shapes of MOs formed from combinations of two s orbitals and combinations of two p orbitals 22 • Antibonding MO (σ1s) has a node between the nuclei and most of its electron density outside the internuclear region.
•This increases nuclear repulsions making the anti- bonding MO higher in energy than the isolated AOs.
•When the antibonding orbital is occupied, the molecule is less stable than when the orbital is empty.
Filling molecular orbitals with electrons. Electrons fill MOs in the same way that they fill AOs: • Orbitals are filled in order of increasing energy.
• An MO has a maximum capacity of 2 electrons with paired spins.
• Orbitals of equal energy are half filled, with spins parallel, before any is filled with two electrons.
23 MO MO energy energy diagrams diagrams show the relative energy and number of electrons in each MO and the AOs that originally held the electrons.
•The MO bond order bond order is the number of electrons in bonding MOs minus the number in antibonding MOs, divided by two: Thus, for H2, the bond order is 1 ⁄ 2(2 – 0) = 1.
• A bond order greater than zero indicates that the molecule or ion is stable relative to the separate atoms.
• The higher the bond order, the stronger is the bond.
Concept 11-9 Concept 11-9. How MO bond order predicts the stability of molecules.
24 • •MO electron configuration MO electron configuration: symbol of each occupied MO shown in parentheses with the number of its electrons written as a superscript.
• The MO electron configuration of H2 is ( 1s) 2 . An early success of MO theory was in predicting the existence of He2 +, which is composed of two He nuclei and three electrons.
He 2 + bond order is 1 ⁄ 2 (2 – 1) = 1 ⁄ 2 He 2 bond order is 1 ⁄ 2 (2 – 2) = 0 25 Predicting Species Stability Using MO Energy Diagrams Problem: Use MO diagrams to predict whether H 2 + and H 2 – exist. Determine their bond orders and electron configurations. Plan: These ions use 1s orbitals to form MOs, so the MO diagrams are like that of H 2. Determine the number of electrons in each, place them in bonding and antibonding MOs and find the bond orders.
Solution: For H 2 + the bond order is 1 ⁄ 2(1 – 0) = 1 ⁄2 , so we predict that H 2 + does exist.
26 Since H2 has two e –, H2 – has three e –.
The bond order is 1 ⁄ 2(2 – 1) = 1 ⁄2 , so we predict that H 2 – also exists.
• The s-block diatomic molecules, Li 2 and Be 2, use their valence 2s AO’s 2s AO’s to form molecular orbitals.
•Like the MOs formed from 1s orbitals, those formed from 2s atomic orbitals are MO’s.
•A 2s AO is higher in energy than a 1s, so the σ 2s and σ2s MOs are higher in energy than the σ 1s and σ1s MOs.
Concept 11-10 Concept 11-10. How MO theory explains the bonding and properties of the diatomic molecules of Period 2 27 • Dilithium has 4 electrons in bonding MOs and 2 in antibonding MOs, so its bond order is 1 ⁄ 2(4 – 2) = 1.
• Electron configuration is ( (σ σ1s 1s) ) 2 2( (σ σ 1s 1s) ) 2 2( (σ σ2s 2s) ) 2 2.
• Usually show only configuration of the MOs formed from valence electrons: (σ2s)2.
• The MO diagram for Be 2 has filled σ σ2s 2s and σ σ 2s 2s MOs.
• The bond order is 1 ⁄ 2(4 – 4) = 0.
• The compound, Be 2 has never been observed.
28 29 Molecular orbitals from atomic p–orbital combinations.
• End-to-end overlap of p AOs gives two σMOs, the σ σ2p 2p and σ σ 2p 2p .
• Side-to-side combination gives a pair of pi (π) MOs, π π2p 2p and π π 2p 2p.
•Another pair of π MO’s from the two pz AO’s; same energy as py MO’s 30 • MOs from 2p AOs are higher in energy than MOs from 2s AOs because 2p AOs are higher in energy than 2s AOs.
• Bonding MOs are lower in energy than antibonding MOs, so σ2p is lower in energy than σ2p and π2p is lower than π2p .
3. End-to-end overlap is greater than side-to-side overlap, so the σ2p MO is lower in energy than the π2p MO. • The energy order for MOs derived from 2p orbitals is: Energy level diagram Energy level diagram • The MO energy order depends on how close the s and p AOs are in energy. If they do not interact, order is: 31 • This is true for O, F and Ne atoms but not for B, C and N.
O 2, F 2, Ne 2 B 2, C 2, N 2 32 33 • O2 molecule, by experiment, is paramagnetic paramagnetic.
• O 2 has two possible Lewis structures: •The two electrons with highest energy occupy the two π π 2p 2p MOs MOs with unpaired spins, making the molecule paramagnetic paramagnetic.
• O2 bond energy = 498 kJ/mol suggests a double bond in O2.
• By MO theory the bond order of O 2 is 2: B.O. = 1 ⁄ 2(8 – 4) = 2].
34 Using MO Theory to Explain Bond Properties Problem: Consider this data for homonuclear diatomic species: Why does bond energy decrease decrease from N 2 to N 2 + but increase increase from O 2 to O 2 +?
Predicted order of bond energy: N 2 > N 2 + and O 2 + > O 2 |
6836 | https://www2.stat.duke.edu/courses/Spring16/sta532/lec/wk-05.pdf | STA 532: Theory of Statistical Inference Robert L. Wolpert Department of Statistical Science Duke University, Durham, NC, USA 5 Bayesian & Objective Bayes Estimation For excellent accounts of Bayesian statistics in more detail than we have time for, see any of Berger (1985); Gelman et al. (2014); Hoff(2009) or Press (2003).
The Bayesian approach to statistical inference uses the language and tools of probability theory to describe all uncertain features, both observable quantities like X ∼f(x | θ) and parameters θ ∈Θ.
While this approach supports the use of subjective interpretations of probability (like Wasserman’s example concerning Einstein’s beverage choice in 1948 or, more importantly, Silver’s predictions of presidential election results and sports outcomes, or the capital management success of Bayesian investment companies like BEAM, LLC), it doesn’t require them— it simply provides a coherent structure for reflecting all uncertain aspects of a scientific problem, and for building tests, estimators, and other statistical tools whose performance may be evaluated using any criteria, objective or subjective or otherwise.
Risk & Estimation Suppose we choose to model uncertainty about an observable quantity X with a parametric model {f(x | θ) : θ ∈Θ}, and that we wish to estimate the value of θ ∈Θ that generated the data.
The most common way to quantify the error of an estimator T : X →Θ of a parameter θ ∈Θ ⊂Rk is through “squared error risk” R(θ; T) := Eθ T(X) −θ 2, the expected squared Euclidean distance between T(X) and θ— or, in the one-parameter case, the expected squared error. Often the easiest way to calculate risk is through the relation R(θ; T) := Eθ T(X) −EθT(X) 2 + EθT(X) −θ 2 = Varθ T(X) + β(θ)2 = Variance + Bias2.
Typically the risk will depend on the value of θ; for example, the MLE estimator Tn(X) = ¯ Xn of the mean λ > 0 for a random sample of size n from the Po(λ) distribution has risk R(λ; Tn) = λ/n.
In a Bayesian approach with prior distribution θ ∼π(θ) it is natural to consider the average risk r(π; T) := Z Θ R(θ; T) π(θ) dθ = E T(X) −θ 2, 1 STA 532 Week 5 R L Wolpert STA 532 Week 5 R L Wolpert STA 532 Week 5 R L Wolpert where now the expectation is with respect to both X and θ. The estimator T(x) that minimizes this risk over all possible estimators is easily seen to be the posterior mean ¯ θn(X) = E[θ | X], since r(π; T) = E [T(X) −¯ θn(X)] + [¯ θn(X) −θ] 2 = E T(X) −¯ θn(X) 2 + E ¯ θn(X) −θ] 2 ≥r(π; ˆ θn).
Thus the most common Bayesian estimator of any parameter is its posterior mean.
A similar calculation using the more daring mean absolute risk R(θ; T) = Eθ T(X) −θ would have led to the posterior median ˚ θn(X). A little more generally, the point estimate that minimizes expected risk for asymmetric loss Eθ[l(T(X) −θ)] with l(z) = |z| + αz is the γ = 1+α 2 th quantile zγ of the posterior distribution for any −1 < α < 1.
Risk & Decisions More generally, the decision theory perspective on inference is to make one of a prescribed class of possible decisions in a way that minimizes some measure of badness called “loss”. The setting is: Parameter space Θ Set of possible parameter values θ labeling possible data-generating distributions Likelihood function f(x | θ) The pdf for X at x ∈X, for a specified θ ∈Θ Outcome space X Set of possible values x of random variable X Action space A Set of possible actions a Loss function L(a, θ) Function A × Θ →R quantifying badness of action a ∈A for data generated by pdf with θ ∈Θ.
Within this structure we consider various possible decision rules δ : X →A, with the interpretation that “upon observing X = x, take action a = δ(x)”. The badness of such a decision rule when the data distribution is X ∼f(x | θ) is quantified for fixed θ ∈Θ by its “risk function” R(θ, δ) = Eθ L(δ(X), θ) and, for a prior distribution with density π(θ) on Θ, by its expectation under π, the “Bayes risk” r(π, δ) = Z Θ R(θ, δ) π(θ) dθ = E L(δ(X), θ) where now the expectation is taken over the joint distribution for X and θ, with pdf π(θ) f(x | θ).
Point estimation of θ ∈Θ ⊂Rk under squared-error loss is the special case in which the action and parameter spaces coincide, A = Θ, and L(a, θ) = ∥a −θ∥2. Earlier we found that the decision rule δ that minimizes expected squared-error loss is the posterior mean, δπ(x) = E[θ | X = x]. In a homework exercise you will verify for point estimation in k = 1 dimension under absolute error loss L(a, θ) = |a −θ| that the Bayes risk is minimized by the posterior median δ(x) = M, the element of X such that P[θ ≤M(x) | X = x] and P[θ ≥M(x) | X = x] both equal or exceed 1/2. It’s not hard to show that for ǫ > 0 the minimizer of L(a, θ) = 1{|a−θ|>ǫ} is the center of the interval [a −ǫ, a + ǫ] with the greatest posterior probability of containing θ— this is the “HPD” interval of length 2ǫ and size γ = 1 −r(π, δπ). It converges as ǫ →0 to the “MAP” (maximum a posteriori) estimator, the mode of the posterior distribution.
Page 2 Page 2 Page 2 STA 532 Week 5 R L Wolpert STA 532 Week 5 R L Wolpert STA 532 Week 5 R L Wolpert For another example: If Θ0 ⊂Θ is any subset of parameter space, and l0, l1 are any positive numbers, and A = {0, 1} with loss function L(a, θ) = 0 a = 0 and θ ∈Θ0 or a = 1 and θ / ∈Θ0 l0 a = 0 and θ / ∈Θ0 l1 a = 1 and θ ∈Θ0 where l0 and l1 represent the costs for making incorrect decisions about whether or not the “hy-pothesis” θ ∈Θ0 is true, the optimal Bayes decision is δ(X) = ( 0 if P[θ / ∈Θ0|X] P[θ∈Θ0|X] ≤l0 l1 1 if P[θ / ∈Θ0|X] P[θ∈Θ0|X] > l0 l1 = 1{P[θ∈Θ0|X]≤l0/(l0+l1)}.
The Bayesian approach is to “reject” this hypothesis if its posterior probability is below the thresh-old l0/(l0 + l1). For the symmetric loss case l0 = l1 the Bayes solution is to select whichever of “[θ ∈Θ0]” or “[θ / ∈Θ0]” has higher posterior probability, but asymmetric losses will alter the threshold for rejecting Θ0.
Admissibility of Bayes Rules Let π(θ) be a proper prior density on Θ and let L(a, θ) be a loss function such that for each θ the function L(a, θ) is strictly convex in the argument “a” (estimation under squared-error loss is an example). Under some regularity conditions (see counter-example below) this will guarantee the existence of a unique Bayes decision rule δπ that minimizes r(π, δ) over all decision functions δ : X →A, namely, δπ(x) = argmin a∈A Z Θ L(a, θ) π(θ | X = x) dθ.
Any such rule must be admissible (see last week’s lecture notes for a definition of admissibility), for if S(X) were a competitor that satisfied (∀θ)R(θ, S) ≤R(θ, δπ) then upon multiplying by π(θ) and integrating it would also satisfy r(π, S) ≤r(π, δπ), so it too would be an optimal Bayes rule for this problem. By uniqueness, it would have to satisfy S(x) = δπ(x) for almost-every x ∈X under f(x | θ) for almost-every θ ∈Θ under π(θ). SO, any estimator that is not a Bayes estimator has higher risk than some Bayes estimator.
In particular, (proper) Bayes estimators for squared-error loss are always admissible, and that for absolute error loss will be admissible if the posterior distributions have a unique median. Evidently posterior distributions whose median is not unique offer an example where the Bayes rule (for absolute-error loss) isn’t unique either. For more details see Brown (1971, 1975).
Page 3 Page 3 Page 3 STA 532 Week 5 R L Wolpert STA 532 Week 5 R L Wolpert STA 532 Week 5 R L Wolpert Objective Prior Distributions This solves the question of how to choose an estimator once the likelihood and prior distribution are chosen. In problems where it is appropriate to use expert opinion or past experience and in which these are available the problem is simple1, but how do we choose π(θ) when we need to be objective?
Let’s illustrate that the tempting approach of using uniform or “flat” densities (pioneered by Bayes and Laplace) isn’t completely satisfactory.
The success probability p for iid Bernoulli trials {Xj} can be written as p = exp(ψ)/[1+exp(ψ)] for some log odds ψ = log p 1−p , with −∞< ψ < ∞; this is often done in logistic regression when we want to explore the relationship between p and some unbounded explanatory variable, by setting pi = ψ(α + βZi) for uncertain α, β. OR, the success and failure probabilities might be written as the squared sine and cosine of some angle, p = sin2(ω), (1 −p) = cos2(ω), for ω ∈[0, π/2] given by ω = arctan p p/(1 −p) = arcsin √p. We could express ignorance about the Bernoulli trials by giving a uniform prior distribution to any one of p, or ψ, or ω. Whichever one we pick, however, the others will not have flat prior distributions due to the Jacobian of the transformations from p to ψ or ω.
The problem isn’t very extreme, however— you can show (it’s a good exercise) that these choices all lead to similar symmetric Beta Be(α, β) prior distributions for p, with α = β = 1 for the uniform p ∼Un(0, 1) = Be(1, 1); with α = β = 1/2 for the arcsin distribution with ω ∼Un(0, π/2); and with the improper α = β = 0 (i.e., p ∼p−1(1 −p)−11(0,1)(p)) for the logistic ψ ∼Un(R). The posterior distributions π(p | X) = Be(α + X, β + (n −X)) and posterior means ˆ pn = (X + α)/(n + α + β) are indistinguishable so long as there are at least several successes X and failures (n −X).
Jeffreys (1961) proposed an invariant solution to the problem— that, instead of a flat prior, he argued that one should use πJ(θ) ∝I(θ)1/2, proportional to the square-root of the Fisher information. One can show (another good exercise) that this is invariant in the sense that πJ(θ) will exactly transform into πJ(φ) under the Jacobian of the transformation θ 7→φ = H(θ) (see the discussion of change-of-variables for Fisher Information in the Week 4 notes). In the Bernoulli example, the Jeffreys’ prior in the p parametrization is the Be(1/2, 1/2) or “arcsin” prior, half-way between the other two.
Example: Poisson For n observations {Xi} iid ∼Po(λ), what is an objective posterior estimate for λ?
What is an objective interval estimate for λ?
The log-likelihood for a single observation X ∼Po(λ) is ℓ(λ) = X log λ −λ −log X!, so the Fisher information and (improper) Jeffreys’ Rule prior density are I(λ) = 1/λ πJ(λ) ∝λ−1/21{λ>0}.
1well, not as simple as it sounds— experts are notoriously over-confident about their prior opinions. A great deal of research has been done on good and bad ways of “eliciting” prior distributions from interviews with experts; ask me for references if you’re interested, or Google “Kahneman and Tversky” or “eliciting prior distributions”.
Page 4 Page 4 Page 4 STA 532 Week 5 R L Wolpert STA 532 Week 5 R L Wolpert STA 532 Week 5 R L Wolpert The resulting posterior is πJ(λ | x) ∼Ga(X+ + 1/2, n) with shape parameter one-half plus X+ := Pn j=1 Xj and rate parameter n, so the squared-error Bayes estimate is EJ[λ | x] = 1/2 + X+ n = ¯ Xn + 1/2n and a one-sided credible interval of level γ would be [0, qgamma(γ, 0.5 + sum(x), length(x))] For example, with n = 12 and X+ = 0 (zero events in a year’s monthly observations), the posterior mean and one-sided 90% credible interval would be E[λ | X] = 1/24 = 0.04167 P[λ ∈[0, R] | x] = 0.90 for R = 0.112731 A number of other approaches have been made toward the goal of objective Bayesian analysis. In some problems one would like to have procedures that are invariant under some transformations— for example, changing the units from centimeters to kilometers, or temperature from Fahrenheit to Celsius. Typically these lead to shift-invariant improper uniform prior distributions dθ for location parameters and scale-invariant improper dθ/θ priors for scale factors. See (Bernardo, 1979; Berger, 1985; Berger et al., 2009) for more on invariant, Jeffreys’, and particularly “Reference” priors.
In multiparameter problems the Jeffreys’ Rule prior πJ(θ) ∝det(I(θ))1/2 is still invariant under smooth changes of variables, but leading experts in Objective Bayesian analysis (including Harold Jeffreys himself) don’t recommend it. The Reference priors πR(θ) modern experts do recommend are often quite difficult to derive, but a simple alternative that often works well is the “component-wise Jeffreys’ prior”, the product of one-dimensional Jeffreys’ priors for the components of θ.
Consistency, Efficiency, Asymptotic Normality, & such We know that the MLE ˆ θn(X) based on the first n observations {Xi} iid ∼f(x | θ) converges to θ as n →∞under mild regularity conditions, and that √n θn(X) −θ is asymptotically normally-distributed with mean zero and variance I(θ)−1, the minimum possible. Does something like this happen for Bayesian estimators?
In a word, yes. Provided that π(θ) > 0 throughout Θ and that the LHF Ln(θ | x) separates points (i.e., that Ln(θ′ | x)/Ln(θ | x) →0 as n →∞for any θ′ ̸= θ), then the posterior mean ¯ θn converges to θ almost-surely as n →∞under mild regularity conditions (principally that π(θ) is continuous at θ, that θ is not a boundary point of Θ, and that Ln(θ | x) has a unique maximum ˆ θn(x) for each x ∈X and is twice continuously differentiable there). Furthermore, the Bernstein-von Mises theorem (sometimes called the “Bayesian Central Limit Theorem” or BCLT) asserts under minor regularity that √n θ −¯ θn has approximately a No 0, I(θ⋆)−1 distribution, for any prior density π(θ) continuous and positive in a neighborhood of the value θ⋆used to generate the data.
Thus the large-sample properties of Bayes estimators are identical to those of the MLE: both are asymptotically unbiased, consistent, efficient, and normally distributed. The differences arise for small samples, where the prior distribution may play more of a role, and where “shrinkage” may lead to substantially smaller risk for some regions in Θ.
Page 5 Page 5 Page 5 STA 532 Week 5 R L Wolpert STA 532 Week 5 R L Wolpert STA 532 Week 5 R L Wolpert Bayesian Hierarchical Models Give example— say, volcanoes References Berger, J. O. (1985), Statistical Decision Theory and Bayesian Analysis, Springer-Verlag, second edition.
Berger, J. O., Bernardo, J. M., and Sun, D. (2009), “The Formal Definition of Reference Priors,” Annals of Statistics, 37, 905–938.
Bernardo, J. M. (1979), “Reference posterior distributions for Bayesian inference (with discussion),” Journal of the Royal Statistical Society, Ser. B: Statistical Methodology, 41, 113–147.
Brown, L. D. (1971), “Admissible Estimators, Recurrent Diffusions, and Insoluble Boundary Value Problems,” Annals of Mathematical Statistics, 42, 855–903, doi:10.1214/aoms/1177693318, see also corrections in (Brown, 1975).
Brown, L. D. (1975), “Correction to (Brown, 1971),” Annals of Statistics, 1, 594–596, doi: 10.1214/aos/1176342433.
Gelman, A., Carlin, J. B., Stern, H. S., Dunson, D. B., Vehtari, A., and Rubin, D. B. (2014), Bayesian Data Analysis, Chapman & Hall/CRC, 3rd edition.
Hoff, P. D. (2009), A First Course in Bayesian Statistical Methods, New York, NY: Springer-Verlag.
Jeffreys, H. (1961), Theory of Probability, Oxford University Press.
Press, S. J. (2003), Subjective and Objective Bayesian Statistics, John Wiley & Sons, 2nd edition.
Silver, N. (2012), The Signal and the Noise: Why So Many Predictions Fail— but Some Don’t, New York, NY: Penguin Press.
Wasserman, L. (2004), All of Statistics, New York, NY: Springer-Verlag.
Last edited: February 1, 2019 Page 6 Page 6 Page 6 |
6837 | https://www.khanacademy.org/math/geometry-home/transformations | Transformations | Geometry (all content) | Math | Khan Academy
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Geometry (all content)17 units · 180 skillsUnit 1 LinesUnit 2 AnglesUnit 3 ShapesUnit 4 TrianglesUnit 5 QuadrilateralsUnit 6 Coordinate planeUnit 7 Area and perimeterUnit 8 Volume and surface areaUnit 9 Pythagorean theoremUnit 10 TransformationsUnit 11 CongruenceUnit 12 SimilarityUnit 13 TrigonometryUnit 14 CirclesUnit 15 Analytic geometryUnit 16 Geometric constructionsUnit 17 Miscellaneous
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Geometry (all content)
Unit 10: Transformations
About this unit
In this topic you will learn about the most useful math concept for creating video game graphics: geometric transformations, specifically translations, rotations, reflections, and dilations.
You will learn how to perform the transformations, and how to map one figure into another using these transformations.
Introduction to rigid transformations
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Rigid transformations intro (Opens a modal)
Translations intro (Opens a modal)
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Identify transformations 4 questionsPractice
Translations
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Translating shapes (Opens a modal)
Determining translations (Opens a modal)
Determining translations (Opens a modal)
Translating shapes (Opens a modal)
Translation challenge problem (Opens a modal)
Properties of translations (Opens a modal)
Translations review (Opens a modal)
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Translate points 4 questionsPractice
Determine translations 4 questionsPractice
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Identify your areas for growth in these lessons:
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Rotations
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Determining rotations (Opens a modal)
Rotating shapes about the origin by multiples of 90° (Opens a modal)
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Rotating shapes: center ≠ (0,0) (Opens a modal)
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Rotate points 4 questionsPractice
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Reflections
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Reflecting shapes: diagonal line of reflection (Opens a modal)
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Reflect points 4 questionsPractice
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Rigid transformations overview
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Dilations
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Performing dilations (Opens a modal)
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Sequences of transformations 4 questionsPractice
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Symmetry
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Intro to reflective symmetry (Opens a modal)
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Performing reflections: line (old) (Opens a modal)
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Unit test
Test your understanding of Transformations with these 20 questions.Start test
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6838 | https://www.youtube.com/watch?v=M1FAIWESATU | Interval of convergence testing endpoints
Abigail Payne
2150 subscribers
299 likes
Description
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Posted: 12 Apr 2020
12 comments
Transcript:
thankful beef um okay so here's another example where you're asked to find the interval of convergence for a power series since it was important in that last example I'll start out this example by noting that this power series is centered at 3 and I get that from this X minus 3 expression here so I know that this power series converges at least when x equals 3 and the interval of convergence may extend further out from 3 but whenever I come up with it for an interval convergence it should be centered around 3 if it ends up a lopsided on either side of 3 then I know something's gone wrong ok and again I have that annuit centered at 3 because the general form for the power series is something like this when it's centered at a and so in this case the a is 3 ok so now where does it converge I would start this with the ratio test like I did with that last example I use ratio test most often to start out these interval of convergence problems if you're considering the route test just you're gonna end up taking the nth root of n plus 2 which could I don't know be a little bit of a pain to deal with so that's why I would just stick with ratio test in this example so I take the limit as n approaches infinity of the absolute value of the n plus 1 term and I've said in a previous video so powers of negative 1 are just either 1 or negative 1 when you take the absolute value they're always just 1 so if it's going to turn absolute value we can drop out that negative 1 to the N because I know it's just gonna be a 1 times X minus 3 so in place of in put in plus 1 over in place of the N here put in plus 1 plus 2 so all of that over so I'm dropping the negative 1 to the N because it's in the absolute value X minus 3 to the N over n plus 2 there's in turn this is a n plus one over n am okay so now we need to simplify so this is equal to the limit as n approaches infinity so this expression right here I will write as X minus 3 to the N times X minus 3 to the first I'm going to run out of room aren't I I'm gonna start over down below go ahead and work to the N times X minus 3 to the first over n plus 3 and then I'm going to multiply by the reciprocal of what's down here so in plus 2 over X minus 3 to the N okay so X minus 3 to the n cancels and that's it right but now this expression X minus 3 to the first is no longer dependent on it so again it doesn't care about in approaching infinity it's unaffected by that limit so I can pull that out in front so this is equal to it needs to stay inside the absolute value absolute value of x minus 3 times the limit as n approaches infinity so what's left just n plus 2 over n plus 3 which that right there so ratio of two polynomials with the same degree so we limit is the ratio of the leading coefficients which is 1 1 over 1 so the limit here using ratio test turns out it's equal to the absolute value of X minus 3 okay me erase this stuff erase the work that I did below so remember ratio test says that this series converges when the limit which in this case is equal to absolute value of X minus 3 is strictly less than 1 right okay so solving that inequality if you have absolute value less than a number then to wait the way to solve that let me write a rule for you off to the side here so the rule to apply in these absolute value problems if the absolute value of U is less than a number in then that expression U is between positive and negative n whenever in is a positive number so if you have the absolute value less than some positive number then you should take what's in the absolute value and put it between positive and negative whatever that number is so maybe that's hard to read but when I meet what I mean is so in this case the U is X minus 3 so absolute value back to minus 3 is less than 1 then that means that X minus 3 must be between negative 1 and positive 1 and then if you add 3 across all three parts of this inequality you have 2 is less than X is less than 4 so interval of convergence goes from 2 to 4 which is good to see cuz that's an interval that's centered around 3 where the radius of convergence so if it converges here the radius of convergence how far it extends out is 1 in this case radius of convergence is just how far out from the center does this series converge now something else the integral sorry ratio test also says that this series diverges when the absolute Gatti of X minus 3 is greater than 1 which is going to be when you're out here either less than 2 or greater than 4 then the absolute value of x minus 3 will be greater than 1 so diverges out there also ratio test is inconclusive but when x equals 2 and when x equals 4 because so if this is our limit if we plug in 4 for X then this limit is equal to 1 and ratio test is inconclusive same thing if we plug in 2 for X here you'll have absolute value of negative 1 which is 1 make sure your test is inconclusive and so this is the way it's gonna play out every time in these types of series where you have a finite interval of integration you're going to be probably be using ratio test or root test so you'll say it definitely converges inside this interval definitely diverges outside but it's inconclusive at the endpoints so you end up having to check the endpoints individually let me do that now so checking the endpoints let's start with X equals two okay so if x equals two first thing you do when you're checking in points is substitute that X into this power series expression so this will become the Sun from 0 to infinity of negative 1 to the N times if we're letting X equals 2 then this is 2 minus 3 to the N over n plus 2 okay so that's negative 1 to the N times this is also negative 1 to the N over N plus 2 and so that's negative 1 so the N plus in so 2 in over n plus 2 and so for any integer in you're gonna have negative 1 to an even power I mean that's what an even power is it's a multiple of 2 so if this is always to an even power then that numerator is just 1 so you're taking a step back for a minute when you're checking in points you substitute the X and then most of your effort is gonna be spent just simplifying the series for that particular x value so we've simplified this to the series 1 over n plus 2 and now you just decide okay so how do I want to check convergence or divergence of that series you could do the integral test you could also do limit comparison test that's probably what I'm gonna choose here so with that series I would do limit comparison test with the series 1 over the good old harmonic series which we know diverges if you take the limit as n approaches infinity of our given series here over 1 over N you'll have the limit as n approaches infinity bring this up flip it over it's gonna be in over n plus 2 when you multiply that out so ratio of the leading coefficients is 1 over 1 the limits one with limit comparison test if your if your limit is a positive finite number here then the series behave the same way so since we're comparing it to a divergent p series here or the harmonic series which we know diverges then this series must also diverge so when x equals two we've determined that the series diverges so the interval of convergence what we know so far is it's gonna go up to two but not include two on the other side it's gonna go up to four now we just need to know is it a bracket or a parentheses it for should we include four or not include four so we need to check that endpoint so I'm gonna erase this and this is the one last bit these problems are very long when you're finding interval of convergence and you end up having to test endpoints let's see so now if we check x equals four so if I substitute 4 for X in this expression for the series then this right here will be 1 to the N which will always just be 1 right this is gonna be some power of 1 so this right here wonder the end is will just be 1 and so we end up with the series negative 1 to the N over n plus 2 when you plug in x equals 4 here and simplify this just becomes 1 so you end up with just that series okay which you take a step back you decide ok how would I like to test the series that looks like this and I would say your best bet is alternating series test so you check those two conditions is the as in approaches infinity of these terms the non alternating piece so we get to just call this 1 over N plus 2 that's 0 which is what we need to see so we move on to the second condition is a n plus 1 less than is it a deep are these decreasing terms so 1 over n plus 1 plus 2 would be less than 1 over n plus 2 yes so it satisfies both of those conditions so it converges by the alternating series fest this series converges which is when x equals 4 drop the marker okay converges when x equals 4 so we do include 4 so all of that to get this answer here the interval of convergence is parenthesis 2 comma 4 with a bracket in it does not include 2 it does include 4 |
6839 | https://www.youtube.com/watch?v=m5C8tIE2hv0 | Inscribed Angles in Circles and Tangent Lines
ProfRobBob
229000 subscribers
101 likes
Description
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Posted: 22 Mar 2013
I define Inscribed Angles, Inscribed Angles involving Tangent Lines, and 3 Corollaries of Inscribed Angles. I work through five examples to help your understanding. Find free review test, useful notes and more at If you'd like to make a donation to support my efforts look for the "Tip the Teacher" button on my channel's homepage www.YouTube.com/Profrobbob
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Transcript:
BAM! Mr. Tarrou In this video I'm going to define
for you what an inscribed angle is in a circle and relate the measure of this inscribed angle to the
measure of the arc I'm going to work through a few examples and I'm going to use those examples to
kind of lead us into some corollaries dealing with inscribed angles and how they relate to each
other in the circle and such we're also going to be looking at how an Junt lines can also help
to create some inscribed angles and work through two more examples dealing with those so it'd be
five examples total in this video along with a lot of definitions and corollaries to help you of
course do your homework so we have a circle here the measure of an inscribed angle is half of the
measure of the intercepted arc well that's fine as far as the measurements go but what exactly
is an inscribed angle well an inscribed angle is simply made up of two chords in a circle and
the vertex of that angle must be on the circle itself now there's two types of angles that we
deal with in with circles quite often those are central angles if you have a central angle where
the vertex is the center of the circle then the measure of that central angle is equal to the arc
but an inscribed angle the measure of that angle is equal to half of the arc okay so our first
example here is we have a couple of inscribed angles we have angle B F C and we have angle a
D C both of those angles have the vertex on the circle so they're inscribed okay so the first
thing I'm saying here is what is the measure of Arc AC well arc AC here remember when you name a
minor arc you want to use two letters major arcs which means more than half way around the circle
three letters so two letters we're talking about a minor arc so we're talking about arc a and around
the circle to see well we have an inscribed angle here and angle ad see here see starting and ending
basically with points a and C that inscribed angle is equal to 80 degrees so the arc the measurement
arc is going to be twice that inscribed angle so two times 80 is going to be 160 degrees okay
so now we got the measure of Arc BC well we had the segment addition postulate and then we
also have that postulate about the measurement or the combining the measurements of arcs now if
arc a C is 160 degrees and let's see your arc AC well BC is a large portion of that arc AC so if
I take the measure of Arc AC which is equal to a hundred and sixty degrees and then we can subtract
C if I just take out that little bit from here to here in degrees is again 160 and if I just back
out that fifteen degrees then I'm gonna have the measure of Arc BC so it's going to be 160 degrees
minus 15 degrees and that's going to be equal to 160 150 145 okay excellent so I'm going to go
ahead and add in that 145 here so I find it a good idea to measure my not measure but put my
numbers back into my diagram as I work through these problems often it helps to solve additional
parts of it and exactly what's what's going on here I want to find the measure of angle F well
measure of angle F it's in its vertex on the circle inscribed to then inscribed angle is half
of the measure of the arc so we're going to take 145 degrees divide that by 2 and that is going
to be equal to let's see here 7 times 2 is 14 so that's going to be 7 there's two twos and fives
and a 5 so 2 and 5 divided by 2 is 4 remainder of 1 1 over 2 means we're gonna have a measure
of 72 point five degrees okay and don't forget you can use a measure arcs in terms of the number
of degrees that includes around the circle and of course there's arc length I'm obviously dealing
with degrees okay let me just hands down here make sure I've got the right answers okay on my second
diagram here I want to find the diameter of this circle and this is the diameter of the circle so
if this is the diameter of the circle a C and a diameter basically cuts a circle in half then how
much is the measure of Arc AC well the measure of Arc AC is going to be 180 degrees half a circle
and now I'm looking for the length of the diameter but to do that I have to kind of work with this
inscribed angle and it kind of looks like see how well I drew this if I put my piece of paper up
here in the diagram it kind of looks like that's a right angle and well you can't trust diagram
just because I drew it to look like a 90-degree angle doesn't mean it necessarily is so I needed
like use the information given in the diagram to actually sometimes prove what looks like is the
actual case because sometimes we draw diagrams to kind of tricky it right and make things with like
right angles when they're really not and such so this is 180 degrees and this is an inscribed angle
with that angle or with that vertex on the circle the measure of that inscribed angle is going to
be half of the measure of the arc so it's kind of important we realize that's a diameter that
that's 180 degrees and half of 180 it is equal to 90 so the measure of angle B is equal to 180
minus 2 which is the 90 that I just said and now I validated that this doesn't just look like a
right triangle it actually is will be mentioning this relationship between the semicircle and the
inscribed angle when we get to the corollaries here in just a second kind of summarized what I
put it said in words so we have a right triangle at any time we're looking for the missing side
of a right triangle well if you know two sides of a right triangle and you're just looking for one
you're going to find them with Pythagorean theorem if you only have one side of that right triangle
and you know the men one of these acute angles then you're going to be using your sohcahtoa signs
up the sine of an angle is equal to opposite over hypotenuse cosine of an angle is adjacent over
hypotenuse and tangent of an angle is opposite over adjacent to find information of the missing
information of that triangle doing some verbal review we have two sides we're missing one we're
in do Pythagorean theorem so Pythagorean theorem is going to be a squared plus B squared the two
sides that are helping to make that right angle so seven squared plus 11 squared is equal
to x squared 7 squared 7 times 7 is 49 11 squared is 121 that's equal to x squared we're
gonna combine those like terms so 121 plus 40 is 161 and it's 161 plus 9 is 170 we're going to
square root both sides of this equation to finish solving for x the length of the diameter and the
square root of 1/7 you don't think there's any perfect squares in that so it's not reducible
but it does come out to be a decimal of 13.04 70 ters fighting got that one done let's get to
the next screen we're going to do one example and start introducing some corollaries related to
these idea this an inscribed angle now my last example and this example are going to go together
to actually help explain my movies done process and then I just got done making those two examples
are going to help explain to the three corollaries I'm going to write up here on the next screen to
kind of summarize some of the properties of these of these inscribed angles that I'm doing with
these examples what is the relationship between angles a and B so I've got two inscribed angles in
the circle and they happen to be sharing sort of like common endpoints to these in screen bangles
in other words we're looking at really just a quadrilateral where we have opposite angles which
are inscribed and I can do the same thing with angles C and D but I'm just going to do and B how
are they related to each other well they must be related to each other somehow otherwise wouldn't
be asking you this question so let's find that out do I know what the measure of angle a is no
I don't and do I know what the measure of angle B is I don't know that either so let's start with
something that we do know or just start with some kind of information that is going to be that let's
pick this minor arc here the measure of this minor arc is equal to what well I don't know I just
told you I don't know what the measure these angles are so how could you know I say well all
of a sudden just measure of CD so I don't know what it is so we're just going to say that the
measure of Arc CD is equal to X no that means that the measure of I want to lay this out here
let's put it here the measure of angle B now B the inscribed angle B or C BD is what is creating
sort of this divide this division of this arc CD so the measure of angle D is going to be half
of the measure of that arc so the measure of angle b is equal to x over 2 okay now let's keep
in mind that we just identified this portion of the circle that's being created or that arc that's
being created by the inscribed angle B and kind of flip to the other side and look at this portion
of the circle the measure of Arc C V D now look I'm using three letters because I'm wrapping
more than half way around the circle well this is basically just sort of like the other part of
the circle right here's arc CD now I want to talk about the other side arc CD V so it's just the
other part of the circle well the full rotation is 360 degrees right so if the measure of arc CD
we defined as being the value of x and the measure of Arc CVD is the other side of the circle well
the two parts have to up to 360 degrees so it's 360 minus X that means that the measure of angle
a remember the question is how our angles a and B related to each other I don't really care so much
about the arcs even though I'm having to use this idea that they're just the two arcs arc CD and
arc CBD without them they have to add up to 360 degrees so if this is excuse me if this is 360
minus X then the inscribed angle angle C ad or just angle a that the measure that angle has to
be half of this arc so the measure of angle a is equal to 360 minus 360 minus x over two okay
now what's supposed you know what's about that okay let's let's find out half let's find out
what happens when we add them the measure of angle A plus the measure of angle B something
cool is going to happen we add those well the measurement will be don't really know what it is
or excuse me I made first the measure of angle a is 360 minus x over two plus the measure of angle
B which is x over 2 something weirds gonna happen right when we add or subtract fractions we need
common denominators and already have that have a combinator common denominator excuse me of two
so I should just be able to combine the numerators I'm going to be done well 360 minus X plus X
minus X plus X the negative x and the positive x so those are opposite they're just going to
cancel out so we end up with 360 over two and 360 divided by two is equal to 180 degrees what
I've basically just done here is proven I don't know if it's going to be the second or third
corollary the third corollary I'm going to put up on the next screen of when you have inscribed
angles when you have a quadrilateral inscribed in a circle the opposite angles are always going
to be supplementary very very cool let's get those Corley's up there so you can write now so
to inscribed angles that intercept the same arc are going to be congruent so angle a and angle B
is the intercept the same arc and let's just say that maybe that's about 100 degrees than both of
those angles would have to be 50 thus they will be congruent to themselves an angle inscribed in
a semi circle is a writing let's want to discuss them where my just previous examples so if this
is a diameter or if we have 180 degrees which is a semicircle then that angle that inscribed
angle is gonna be half of the inscribed arc thus it will be right angle and you'll probably get be
doing Pythagorean theorem in there somewhere for those kind of questions and finally the opposite
angles of a quadrilateral inscribed in a circle are supplementary and I just snuck a proof in
that in this video for you for that particular third corollary okay here's an example but excuse
me our last theorem before we finish our last two examples the tangent lines and inscribed angles
remember a tangent line is a line that touches a circle at only one point even others they
don't overlap there and a court is also in this definition a chord is a line segment whose
endpoints are on the circle so the measure of an angle right there formed by a tangent line and
a chord is half the measure of the intercepted arc so kind of really this is you know we have an
angle whose vertex is on the circle so really this is sort of a specialized version of an inscribed
angle so let's just say for argument's sake the measure of this arc is equal to let's say 160
degrees then the measure of this angle that is formed by the chord and a tangent line but the
key point thing is though the key fact here is the vertex is on the circle so it's a specialized
version of an inscribed angle that is going to be equal to also a Gries and sensor corollary one
here says that you know to inscribed angles that intercept the same arc are congruent means that I
could draw this inscribed angle and I could draw this inscribed angle or I could even draw this
inscribed angle and every single one of those angles are going to be congruent every single one
of those angles intercepting the same arc which is 160 degrees each of those and scribed angles is
going to be 80 degree uh degrees all right let's get to those last two examples whoa thank you for
watching I said we had five examples but there's like five problems just in this one I want to
find all the missing values in this diagram now I've got a number here I want to find this the
measure this angle this angle this arc this arc and finally that angle I didn't number these
in the order that I would necessarily be able to find them in so you may want to just pause
the video and see if you can't figure out these missing parts on your own okay so hmm what piece
of information can we find first and maybe you can think of something different but I'm looking
at this arc AC and the fact that it's equal to 100 degrees and I've got a tangent line I've got a
chord and it's vertex of this angle here angle ACP has its vertex on the circle which basically means
it's an inscribed angle it's got you know tangent line special case but still inscribe so the
measure of angle five is going to be equal to half of 100 degrees so the measure of angle five is
equal to 100 divided by two which is going to be equal to 50 degrees now I'm going to come through
here and erase these numbers as I work this out so I can use them through the rest of the diagram
what am I going to find out well this is a tangent line I've got two angles which are adjacent
forming a straight line that's called a linear pair in a linear pair are supplementary so the
measure of angle two plus the measure of angle five I guess it wasn't a good idea to erase that
name but these two angles have to add up to 180 degrees so the measure of angle 2 is equal to
the hundred and eighty that we need when these are added together but I've already at fifty
taken accounted for so 180 minus fifty and so the measure of angle 2 is 130 degrees so let's
take this out and put in 130 hmm let's see here now I know that this is 130 degrees and I have
a tangent line in a cord so I do actually know that the measure of Arc ABC that this entire arc
is going to be twice that inscribed angle but I don't know you know it's in two separate pieces
so I don't really know how much or KB and arc BC is going to be equal to proportionally but
they're going to add up to 260 what else can we see in here well we have a triangle right and
the inside angles of a triangle the interior some of those angles of a triangle have to add up to
100 I'm trying to highlight that so you'd see it better but me orange kinda looks like white the
interior angles of a triangle have to add up to 180 degrees so that means that I've got 12 and
130 accounted for so I can find the measure of angle one because those three zing three angles
like I just said have to add 180 so the measure of angle 1 is equal to 180 and I'm going to
just back subtract here 180 minus the 130 - the 12 degrees so 180 minus 30 is 130 is 50
and now we're at 40 and now we're at 38 degrees excellent so now I can take this out and put in 38
degrees now if that inscribed angle is equal to 38 degrees then I know how much the opposite arc is
or the arc is that is created by that inscribed angle so arc the measure of Arc BC or unknown
value of three the measure of Arc BC is going to be twice or inscribed angle so two times 38 is
going to be well 2 times 30 is 60 2 times 8 is 16 + 60 + 16 is equal to 76 degrees finally let's
take this out and put in our 76 degrees yep okay so now what is left the measure of Arc a B and as
I look around the circle this circle which is 360 degrees we're not talking about arc length but
measuring our arcs and degrees we've got one two three arcs that this circle is broken up into and
the measure of these three arcs needs to add up to 360 degrees because that is a full rotation in
a circle so the measure of our a B is going to be 360 minus 100 minus 76 and that is going to be 360
- 60 - 60 - 70 is 190 and 190 minus 6 is 184 okay before I check out here let's just make sure that
I've got these right a B is 184 yes BC was equal to 76 measure of angle 1 was 38 130 and 50 all
right thank you for sticking around got one more example and it's gonna be a multi-part one like
this one so hopefully I hope you get rid of solve some of your further questions in your homework Oh
alrighty then we got our last attack on a diagram here lots of missing values and we want to find
all of them ok now again you don't necessarily have to find the same piece of information that
I do first but well let's start with angle 1 now this angle we can I'm going to say you know the
center of the circle is the vertex of this angle so this is a central angle so the measure of
angle C is going to be the same as the measure of the arc that it intersects or intercepts so
um let's see here since arc the measure of Arc a B is 50 then the measure of angle C or the
measure of angle 1 is going to be equal to the same thing 50 degrees know the measure of angle
to the now see this angle has its vertex on the center of the circle and I have it just sort of
off to the side it doesn't really matter like you might notice this isn't perfectly balanced and it
doesn't need to be as long as my vertex is on the circle then the measure of that inscribed angle
is going to be half of the arc so the measure of angle 2 or the measure of angle B D a is going
to be equal to 25 degrees the measure of angle 2 is equal to 50 divided by 2 which is 25 degrees
so let's take this out and put in let's see here 50 and take this out and replace it with our 25
degrees I probably made that ideally the name in there I'm just trying to make sure I keep this
diagram as readable as possible let's see here what's left I need to find this unknown arc
measure ad the mark the measure of Arc ad or the value of 3 and 4 so these kind of look like
they possibly equal but looking like the equal is not being equal so I'm not going to assume that
what I am gonna do is say well I have a tangent line touching the circle at one point and a cord
that special type of inscribed angle again with its vertex on the circle is the measure of that
angle is going to be half of the arc the measure of the arc so the value of three or the measure
of Arc ad is going to be two times 60 which is 120 degrees and again this particular circle like
the last this entire circle is broken up with the measure of one to three different arcs and they
have to add up to 109 excuse me 360 degrees and that means I'm almost done because I want to find
the measure of Arc B D so the measure of Arc BD is equal to 360 not 630 360 minus 50 minus 120 so
we've got 360 to 60 to 40 minus 50 is equal to 190 degrees and that is the end of my last example
so I missed a true back go to your homework you |
6840 | https://brainly.com/question/12986394 | [FREE] A class of 27 students has 33% pass. How many of them pass? - brainly.com
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A class of 27 students has 33% pass. How many of them pass?
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Around 9 students
Explanation
If a class of 27 students has a 33% pass rate, around 9 students pass.
33% of 27 = 8.91 or 9%
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In a class of 27 students with a 33% pass rate, about 9 students pass. This is calculated by taking 33% of 27, which equals approximately 8.91, rounded to 9. Therefore, 9 students passed the class.
Explanation
To determine how many students passed in a class of 27 students with a 33% pass rate, follow these steps:
Understanding Percentages: The percentage tells us how much of a whole is being considered. In this case, 33% means 33 out of every 100 students are passing.
Calculate the Number of Passes: To find the number of students who passed, we convert the percentage to a fraction and then multiply it by the total number of students. So:
Number of students passing=100 33×27\
Perform the Calculation:
First, calculate 100 33: This equals 0.33.
Now multiply by the total number of students:
0.33×27=8.91\
Round to Whole Students: Since we cannot have a fraction of a student, we round 8.91 to the nearest whole number, which is 9.
Thus, you can conclude that approximately 9 students passed the class.
Examples & Evidence
For example, if a class has 40 students and 75% pass, then the number of students passing would be calculated as 100 75×40=30. Similarly, using the percentage method helps in understanding the overall performance in different classes.
This calculation utilizes basic principles of percentages and multiplication, which is a fundamental mathematical concept taught in middle school.
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6841 | https://core.ac.uk/download/pdf/210829487.pdf | E L A S T I C N E T W O R K M O D E L S O F P R O T E I N S Uncovering the Internal Mechanics of Actin and Myosin vorgelegt von Diplom-Physiker Markus Düttmann geboren in Haselünne Von der Fakultät II - Mathematik und Naturwissenschaften der Technischen Universität Berlin zur Erlangung des akademischen Grades Doktor der Naturwissenschaften – Dr.rer.nat. – genehmigte Dissertation Promotionsausschuss: Vorsitzender: Prof. Dr. Dieter Breitschwerdt Gutachter: Prof. Dr. Alexander S. Mikhailov Gutachter: Prof. Dr. Harald Engel Tag der wissenschaftlichen Aussprache: 21. November 2012 Berlin 2012 D 83 Markus Düttmann : Elastic Network Models of Proteins– Uncov-ering the Internal Mechanics of Actin and Myosin , © September 2012 A B S T R A C T Throughout the course of evolution, proteins have obtained the ability to perform a variety of tasks within the cell. Two impor-tant examples are the structural protein actin and the molecu-lar motor myosin. They are so-called ATPases, i.e. binding and subsequent hydrolysis of an ATP molecule provide energy fu-eling these molecular machines. Here, an elastic-network (EN) approximation will be employed to study ATP induced confor-mational changes of actin, internal communication mechanisms of myosin and the interaction of these two molecules. We found that they essentially behave like strain sensors, both responding by well-defined domain motions to mechanical perturbations.
To describe the actin monomer, we extended the EN model by introducing a set of breakable links. These become effec-tive only when two domains approach one another. In this framework, actin possesses a metastable state corresponding to a closed conformation and appropriate perturbations in the nucleotide-binding pocket (NBP) can induce a transition to this state. Furthermore, a coarse-grained ligand model was intro-duced. Our analysis suggests that the presence of ATP stabi-lizes a closed conformation. This may play an important role in the explanation why the polymerization process is highly accel-erated in the presence of ATP. Next, we explore the sensitivity of myosin to external forces. Conformational responses of the motor protein to the application of forces to individual residues in its principal functional regions were systematically investi-gated. In this way, a detailed sensitivity map of myosin-V could be obtained. The results suggested that the intrinsic operation of this molecular motor is regulated by its strain-sensor behav-ior. In agreement with experiments, we find that such forces invoke conformational changes that should affect filament bind-ing and nucleotide release. Finally, interactions between the fila-ment and actin or myosin have been investigated. Electrostatic interactions were seen to guide proteins toward specific bind-ing sites. Here, strong binding occurs due to directly interact-ing residues which form stable bonds if they come close to each other.
iii Z U S A M M E N FA S S U N G Im Laufe der Evolution haben Proteine zahlreiche Fähigkeiten entwickelt und regulieren so gut wie jeden Prozess in der Zelle.
Wichtige Beispiele sind das Strukturprotein Aktin und der moleku-lare Motor Myosin. Sie gehören zu der Gruppe der ATPasen, d.h. sie werden durch eine chemische Reaktion mit diesem Nuk-leotid angetrieben. In dieser Arbeit werden ATP-induzierte Kon-formationsänderungen von Aktin, interne Kommunikationsmech-anismen von Myosin und die Wechselwirkung beider Moleküle im Rahmen eines elastischen Netzwerksmodells untersucht. Wir haben herausgefunden, dass Aktin und Myosin sich wie Kraft-sensoren verhalten; beide reagieren mit wohldefinierten Domä-nenbewegungen auf mechanische Störungen.
Um den Aktin-Monomer zu beschreiben, haben wir das Net-zwerkmodell durch Einführung von neuen Bindungen erweit-ert. Diese entstehen, wenn sich Domänen nahe genug annäh-ern. Dadurch besitzt Aktin einen metastabilen, geschlossenen Zustand. Geeignete Störungen in dem Bereich in dem Ligan-den binden können einen Übergang dorthin bewirken. Weiter-hin entwickelten wir ein Liganden-Modell und konnten zeigen, dass ein so modelliertes ATP den geschlossenen Zustand stabil-isiert. Dies könnte eine Rolle bei der Erklärung spielen, warum in Anwesenheit von ATP die Aktin-Polymerisierung beschleu-nigt wird. Außerdem untersuchten wir den Einfluss externer Kräfte auf Myosin. Indem wir systematisch Kräfte auf einzelne Residuen in den wichtigen Bereichen des Moleküls wirken ließen und deren Effekte quantifizierten, konnten wir die Empfind-lichkeiten einzelner Bereiche charakterisieren. Unsere Ergebnisse deuten darauf hin, dass intrinsische Mechanismen durch äußere Kräfte reguliert werden. In Übereinstimmung mit experimentellen Daten konnten wir zeigen, dass äußere Kräfte Konformation-sänderung induzieren, die das Binden von Filamenten und Nuk-leotiden beeinflussen können. Abschließend untersuchten wir die Wechselwirkung zwischen Aktin und Myosin auf der einen und dem Filament auf der anderen Seite. Wir zeigten, dass elek-trostatische Wechselwirkungen dafür sorgen, dass die beiden Proteine sich die gewissen Bindungsstellen im Filament näh-ern. Dort gehen sie stabile Bindungen ein, die für eine starke Wechselwirkung zwischen Filament und Protein sorgen.
iv P U B L I C AT I O N S The main results of this work have appeared previously in the following publications: Düttmann M, Togashi Y, Yanagida T, Mikhailov AS (2012) Myosin-V as a Mechanical Sensor: An Elastic Network Study. Biophysi-cal Journal. 102: 541–555.
Düttmann M, Mittnenzweig M, Togashi Y, Yanagida T, Mikhailov AS (2012) Complex Intramolecular Mechanics of G-actin – an Elastic Network Study. PLoS ONE. e45859.
v A C K N O W L E D G M E N T S I am truly indebted and grateful to my advisor, Prof. Alexander Mikhailov, for the support he showed me throughout the past three years. I am sure this thesis would have not been possible without his help. Besides I would like to thank my colleagues and coworkers for fruitful discussions and a pleasant working atmosphere. I would also like to show my gratitude to Prof.
Yuichi Togashi and Prof. Toshio Yanagida who heartily hosted me in their research groups. Finally, I acknowledge financial support from the Deutsche Forschungsgemeinschaft through the Research Training Group 1558 "Nonequilibrium Collective Dynamics in Condensed Matter and Biological Systems".
vii C O N T E N T S 1 introduction 1 2 background 7 2.1 Proteins . . . . . . . . . . . . . . . . . . . . . . . . .
7 2.1.1 Actin . . . . . . . . . . . . . . . . . . . . . .
9 2.1.2 Myosin . . . . . . . . . . . . . . . . . . . . .
10 2.1.3 Actomyosin Complex . . . . . . . . . . . .
13 2.2 Experimental Methods . . . . . . . . . . . . . . . .
15 2.3 Modeling of Protein Dynamics . . . . . . . . . . .
19 2.3.1 Molecular Dynamics . . . . . . . . . . . . .
19 2.3.2 Reduced Descriptions and Coarse-Graining 21 2.3.3 Elastic-Network Models . . . . . . . . . . .
23 3 mathematical methods 27 3.1 The Elastic Network Model . . . . . . . . . . . . .
27 3.2 Linearized Equations of Motion . . . . . . . . . . .
30 3.3 Immobilization Procedure . . . . . . . . . . . . . .
31 3.4 Extensions of the Model . . . . . . . . . . . . . . .
33 3.4.1 Breakable Links . . . . . . . . . . . . . . .
33 3.4.2 Thermal Fluctuations . . . . . . . . . . . . .
35 3.5 Protein-Protein Interactions . . . . . . . . . . . . .
36 3.5.1 Electrostatic Interactions . . . . . . . . . . .
36 3.5.2 Soft sphere potentials . . . . . . . . . . . .
37 3.5.3 Protein Binding . . . . . . . . . . . . . . . .
37 4 intramolecular mechanics of g-actin 39 4.1 G-actin Model . . . . . . . . . . . . . . . . . . . . .
39 4.2 Domain Motions and Metastable States . . . . . .
41 4.3 Responses to Perturbations in the Nucleotide-Binding Pocket . . . . . . . . . . . . . . . . . . . . . . . . . .
46 4.4 Ligand-Induced Conformational Changes . . . . .
49 4.4.1 Ligand Model . . . . . . . . . . . . . . . . .
50 4.4.2 Effects of Thermal Fluctuations . . . . . . .
54 4.5 Discussion . . . . . . . . . . . . . . . . . . . . . . .
55 5 myosin-v as a mechanical sensor 59 5.1 Myosin-V Model . . . . . . . . . . . . . . . . . . .
59 5.2 Forces Acting on the Tail . . . . . . . . . . . . . . .
62 5.3 Forces in the Nucleotide-Binding Pocket . . . . . .
65 ix x Contents 5.4 Forces in the Actin-Cleft Region . . . . . . . . . . .
68 5.5 Discussion . . . . . . . . . . . . . . . . . . . . . . .
70 6 interactions with the filament 75 6.1 Actin Monomer and Filament Models . . . . . . .
75 6.2 Actin-Actin and Actin-Myosin Binding Sites . . .
78 6.3 Guided by Electrostatic Interactions . . . . . . . .
80 6.3.1 Interaction of Actin Monomer with the Fil-ament . . . . . . . . . . . . . . . . . . . . . .
81 6.3.2 Interaction of Myosin with the Filament . .
83 6.4 Ligand Binding facilitates Docking . . . . . . . . .
86 6.4.1 Toy Model . . . . . . . . . . . . . . . . . . .
86 6.4.2 Nucleotide-Dependent Dynamics . . . . . .
88 6.5 Discussion . . . . . . . . . . . . . . . . . . . . . . .
90 7 summary and outlook 93 a appendix 99 a.1 Ligand Model for Actin Monomer . . . . . . . . .
99 a.2 Myosin Directions . . . . . . . . . . . . . . . . . . . 100 a.3 Myosin Sensitivity Tables . . . . . . . . . . . . . . 101 a.3.1 Forces in the Nucleotide-Binding Region . 102 a.3.2 Forces in the Actin-Cleft Region . . . . . . 103 a.4 Comparison to the Linearized Model . . . . . . . 104 a.5 Actin-Actin Binding Sites . . . . . . . . . . . . . . 107 a.6 Tables . . . . . . . . . . . . . . . . . . . . . . . . . . 108 bibliography 111 L I S T O F F I G U R E S Figure 1 ATP hydrolysis . . . . . . . . . . . . . . . .
9 Figure 2 Treadmilling . . . . . . . . . . . . . . . . . .
10 Figure 3 Actin filament mesh . . . . . . . . . . . . .
11 Figure 4 Schematic view of sarcomere and two-headed myosin . . . . . . . . . . . . . . . . . . . . .
12 Figure 5 Schematic view of sarcomere and two-headed myosin . . . . . . . . . . . . . . . . . . . . .
13 Figure 6 Lymn-Taylor cycle . . . . . . . . . . . . . . .
14 Figure 7 Example of X-ray diffraction . . . . . . . .
16 Figure 8 Cryo-electron microscopy picture of actin filaments . . . . . . . . . . . . . . . . . . . .
17 Figure 9 Experimental set-up of a single molecule strain experiment . . . . . . . . . . . . . . .
18 Figure 10 Illustration of Harmonic Links . . . . . . .
29 Figure 11 Immobilization . . . . . . . . . . . . . . . .
32 Figure 12 Truncated Lennard-Jones potential . . . . .
34 Figure 13 Actin and its elastic network . . . . . . . .
40 Figure 14 Subdomain distances and dihedral angle .
41 Figure 15 Responses to global perturbations . . . . .
42 Figure 16 Responses to global and local perturbaions 43 Figure 17 Residues in the nucleotide-binding pocket 47 Figure 18 Simple modeling of ligands . . . . . . . . .
50 Figure 19 Snapshots of ligand-induced conformational motions . . . . . . . . . . . . . . . . . . . . .
51 Figure 20 The pattern of relaxation trajectories for the ligand-network complex.
. . . . . . . .
52 Figure 21 Statistical distributions of interdomain dis-tance L24 . . . . . . . . . . . . . . . . . . . .
55 Figure 22 Myosin-V and its elastic network . . . . . .
60 Figure 23 Residues probed and labels/order param-eters . . . . . . . . . . . . . . . . . . . . . . .
61 Figure 24 Responses to forces applied to the tail . . .
63 Figure 25 Sensitivity of residues in the nucleotide-binding region . . . . . . . . . . . . . . . . .
67 Figure 26 Opening of actin cleft if forces act on tail .
69 Figure 27 Responses induced by the application of forces to the residues in the actin binding cleft region . . . . . . . . . . . . . . . . . . .
70 xi Figure 28 G-actin vs. F-actin . . . . . . . . . . . . . . .
76 Figure 29 G-actin with open loop . . . . . . . . . . . .
77 Figure 30 Charged residues . . . . . . . . . . . . . . .
81 Figure 31 Actin subunit guided by electrostatic in-teractions . . . . . . . . . . . . . . . . . . . .
82 Figure 32 Actin subunit binding to filament . . . . .
83 Figure 33 Myosin guided by electrostatic interaction with filament . . . . . . . . . . . . . . . . .
84 Figure 34 Myosin binding to filament . . . . . . . . .
85 Figure 35 Actin trimer . . . . . . . . . . . . . . . . . .
87 Figure 36 Actin trimer – ligand induced conforma-tional changes . . . . . . . . . . . . . . . . .
88 Figure 37 Distance between actin and filament . . . .
89 Figure 38 Myosin-V modeled on actin filament . . . . 101 Figure 39 Network responses to the application of a static force to the tail . . . . . . . . . . . . . 106 Figure 40 Actin-actin binding sites . . . . . . . . . . . 107 L I S T O F TA B L E S Table 1 Sensitivity of selected residues in the nucleotide-binding pocket (NBP) region.
. . . . . . . .
48 Table 2 Links between myosin and actin monomers in the filament . . . . . . . . . . . . . . . .
79 Table 3 Maximal distance changes (Å) observed when forces are applied to different residues in the nucleotide-binding pocket . . . . . . 108 Table 4 Maximal distance changes (Å) observed when forces are applied to different residues in the actin-binding pocket . . . . . . . . . 109 A C R O N Y M S ATP adenosine triphosphate ADP adenosine diphosphate xii acronyms xiii Pi γ-phosphate EN elastic network NBP nucleotide-binding pocket ABP actin binding protein FRET fluorescence resonance energy transfer MD molecular dynamics HCM hypertrophic cardiomyopathy AFM atomic force microscopy DB DNase-I binding NMR nuclear magnetic resonance BPTI bovine pancreatic trypsin inhibitor GNM Gaussian network model ANM anisotropic network model 1 I N T R O D U C T I O N Proteins are the machinery of life and their complex interac-tions regulate virtually every process in the cell [2, 170]. They are macromolecules assembled out of only 20 standard parts, the amino acids. These molecular compounds have proven to be a formidable construction kit for nature and, although be-ing created out of only these few ingredients, proteins have achieved a remarkable functional diversity throughout the course of evolution [2, 30, 104].
An important class of proteins are the enzymes, molecules that catalyze chemical reactions [13, 170]. Two very prominent examples are the structural protein actin and the molecular mo-tor myosin. Both are molecular machines that gain energy by binding and subsequent decomposition of adenosine triphos-phate (ATP) into its products adenosine diphosphate (ADP) and a so-called γ-phosphate (Pi). This chemical reaction fuels the machine cycles of the two macromolecules. Actin, e.g., uses the gained energy in a process called treadmilling to polymer-ize into long filaments and myosin is able to perform directed motion or exert forces. Despite decades of research, important aspects of their working mechanisms remain to be uncovered [42, 168].
In the beginning of protein research, the field has been mainly observation-driven. On the basis of experimental results, first ideas about the internal working mechanisms have been de-veloped [76, 173]. First successful theoretical descriptions of protein dynamics have been in the form of kinetic rate equa-tions. An outstanding example is the so-called Lymn-Taylor cy-cle, which for the first time provided a complete model of the interaction of actin and myosin within the muscle [58, 108, 155].
In such models, the machine cycle is characterized in terms of states and corresponding transition rates. In combination with experimental data, kinetic models can qualitatively describe the behavior of certain proteins and even make predictions about local structural mechanisms . In general, however, these re-sults lack the resolution to explain structural details and, ac-cordingly, no complete model of protein dynamics is to be ex-pected.
1 2 introduction In recent years, molecular dynamics (MD) simulation meth-ods have been applied to study proteins . Such models in-corporate the interactions between all atoms and, thus, provide high-resolution results. A major drawback of this approach is the enormous computational complexity involved. Despite to-day’s growing computational power and the development of a variety of acceleration methods aiming to overcome such limita-tions, it remains currently unfeasible to follow the slow confor-mational dynamics that define the actions of many biomolecules .
A solution is offered by approximate methods of intermedi-ate complexity . By appropriate coarse-graining of the pro-tein structure and interactions between atoms, such models can reduce the high computational costs of MD approaches while avoiding the lack of resolution of kinetic models. In this thesis, we will employ an elastic network (EN) approximation to model protein dynamics. This approach has initially been developed to describe thermal motions of residues and has continually been extended and advanced [9, 63, 69, 36, 161]. Here, we ex-pand the standard EN model to account for effects of external forces, protein-protein interactions and ligand binding.
The study of two major players in the cell – actin and myosin – is a challenging subject. A broad community of researchers is dedicated to studying the underlying working principles of these macromolecules and many experimental techniques and theoretical tools have been developed accordingly. Nonetheless, a final understanding of the working mechanisms of the two proteins has not been reached. The vast amount of existing data, however, provides a useful benchmark for our modeling.
Actin is one of the most versatile and, thus, most studied proteins. In the cell, it interacts with a large number of actin-binding proteins (ABPs) [41, 42]. Many detailed studies have been performed and experiments have been developed to shed light on such interactions, which are, e.g., responsible for regu-lating the treadmilling process [133, 142, 160, 172]. Interestingly, actin is only known to be functional in its filamentous form and the minimal requirements to enable polymerization have been determined in experiments: along with a sufficient amount of actin monomers, the presence of ATP molecules is needed to provide the energy for treadmilling . Thus, we will go right to the core of this process and study the influence of ligand-binding on the macromolecule.
introduction 3 In complex with actin, the molecular motor myosin is known for its role in muscle contraction . However, even outside of muscle cells, myosins have vital functions transporting loads inside the cell. Hence, this class of biomolecules is one of the most important members of the protein family. Due to its sig-nificance, all aspects of myosin have been investigated. The fun-damental question arising is how the energy, also provided in form of ATP molecules, is converted into directed motion. In-spired by recent experiments , we investigate the role of strain and external forces in the motor mechanism to under-stand how myosin molecules regulate processivity.
The overall aim of this study is to shed light on the working principles of the two proteins actin and myosin. More specifi-cally, we want to identify internal communication and regula-tion mechanisms.
Chapter 2 provides background information of properties of proteins in general. An overview of enzymes and more specifi-cally of the myosin motor and the structural protein actin will be given. Throughout the years, many experimental and theo-retical approaches have been developed within the field of pro-tein research. Different methods investigating various aspects of biomolecules will be reviewed.
The methodology followed in this study will be introduced in Chapter 3. In order to investigate slow conformational motions, a coarse-grained method, the EN model, will be applied. De-spite their high degree of simplification, EN models have turned out to be remarkably efficient and, therefore, provide a suitable compromise retaining single residue resolution while avoiding the numerical complexity of all-atom simulations. General as-pects of this method will be discussed and, moreover, several expansions of the EN model will be presented.
Our aim in Chapter 4 is to investigate complex intramolec-ular communication within the actin protein. In order to un-derstand the internal organization of its dynamics, mechanical responses of the macromolecule are systematically probed. We find that two mobile actin domains are able to perform well-defined large-scale motions. They can approach each other so that additional interactions between the residues from different domains develop; thus the actin can get locked in a metastable closed state. Furthermore, our detailed study of mechanical sen-sitivity of the molecule to application of various perturbations in the nucleotide-binding region reveals that the characteristic 4 introduction global domain motions can also be easily induced by the appli-cation of only local perturbations to some selected residues in the NBP.
Binding of nucleotides and the hydrolysis reaction lead to lo-cal mechanical perturbations in the NBP. In this way, character-istic large-magnitude motions of mobile domains are induced.
This can result in a transition to a locked closed state. In this the-sis we want to demonstrate, in a strongly simplified way, that characteristic global responses to small local structural changes in the NBP region are feasible. This is done by placing into the actual nucleotide-binding region a fictitious ligand dimer and introducing elastic links between the ligand and the nearest residues in the pocket. We find that the binding of a model lig-and, imitating the ATP, can stabilize the closed state of actin and induce a transition to this state from the equilibrium open state of the molecule.
In Chapter 5, our investigations will focus on the myosin mo-tor protein. It is known that a single macromolecule of myosin naturally exhibits coordinated movements of its parts which would have required complicated machinery in a comparable macroscopic device. In this chapter, an elastic network will be used as a model of the myosin-V molecule. We will study how the protein responds to the application of an external force of arbitrary orientation to each single residue in important func-tional regions. The goal of this study is to identify those residues which are particularly sensitive to strains, so that their pertur-bation invokes strong conformational responses. In this way, a detailed pattern of intrinsic intramolecular communication can be obtained.
ATP binding and hydrolysis, followed by product release, are localized near the ATP-binding site in the region comprising the so-called front and back doors [71, 100]. It will be shown that the application of static forces to the myosin tail induces opening and closing of the front door and in this way provides an effective control mechanism for nucleotide release. More-over, backward strain in the tail region leads to conformational changes in the actin cleft which are suited to facilitate binding to the filament. We will see that the nucleotide-binding region is divided into two functional parts: the application of forces in the front-door region, where the adenylate ring of ADP binds, leads to a movement of the tail, whereas forces in the back-door region, where phosphate is trapped after hydrolysis, induce conformational changes in the actin-binding cleft. We monitor introduction 5 the effects of perturbations in the actin-binding region and ob-serve communication between the hypertrophic cardiomyopa-thy (HCM) loop and the tail, indicating a connection between the binding to the filament and a motion of the tail. Thus, a complex pattern of intramolecular communication, based on elastic deformations and strain-sensor behavior, is revealed in myosin-V.
Interaction of actin-binding proteins with the filament will be studied in Chapter 6. More specifically, we will study how the actin-binding protein myosin and actin monomers themselves can attach to the filament. To this end, a model filament will be constructed out of single actin monomers. Different subunits can interact via their Coulomb interaction between charged residues and, if the residues are close enough, by soft sphere potentials. Most importantly, specific binding sites between pro-teins are modeled by breakable Lennard-Jones bonds.
Firstly, electrostatic interactions which guide either actin or myosin to their specific binding sites in the filament are inves-tigated. Here, the proteins strongly interact with the filament via breakable bonds and may form stable complexes. Moreover, this chapter will show that binding of the ligand indeed in-duces conformational changes that make the actin monomer more likely to attach, i.e. it facilitates the docking to the fila-ment. This is achieved by constructing a toy model and moni-toring the ligand-induced conformational changes with respect to the filament.
The final Chapter 7 will conclude this thesis. The results of this work are briefly summarized and an outlook is provided.
2 B A C K G R O U N D This chapter serves as an introduction and gives a brief overview of the field of protein research. First, general aspects of proteins will be considered, i.e. structures and specific features of these biomolecules will be presented. Afterwards, we will discuss the two proteins which form the main focus of this thesis, actin and myosin, and their functions and modes of action will be eluci-dated. Furthermore, the reader will get a review of the most important experimental techniques and theoretical approaches used in such studies.
2.1 proteins The cell is the building block of life and is organized in a very complex manner. It may be seen as a factory comprised of many specialized machines that work together in a highly coordi-nated fashion. The machinery of this factory are the proteins.
They are the workhorses of the cell and involved in virtually each intracellular process . Through evolutionary pressure proteins have developed an impressive variety of functions. En-zymes, e.g., catalyze chemical reactions, transmembrane pro-teins act as gateways in the cell membrane and motor pro-teins are responsible for intracellular transport [2, 170]. For each of the innumerable tasks in the cell there exists a specifically adapted protein ready to complete it.
Although the functional diversity of proteins is extremely large, their construction principle is rather simple – all proteins are built out of only 20 standard amino acids . These molec-ular compounds are called α-amino acids and they are simple molecules made out of an amino group, a carboxylic group and a side-chain specific for each amino acid . The blueprint of a protein is stored in genes, which are the basis for biosynthesis of proteins or gene expression [2, 170].
All amino acids can bind to each other and, in this way, they provide a formidable construction kit for nature. Every pro-tein is initially constructed as a polymer of these monomeric 7 8 background units and obtains its structure and function in a process called folding . Minimizing its energy, the amino acid polymer reaches its characteristic equilibrium conformation, the native state [11, 39, 174]. The native conformation corresponds to the most thermodynamically stable shape of a protein . It is only in this state that the protein is able to perform work.
Proteins can fold efficiently on the time scale of milliseconds and faster , whereas a stochastic search through all possible conformations would have been unfeasible (see the Levinthal paradox [86, 101]). Therefore, folding pathways need to proceed along a definite pathway on a complex but very specific energy surface [11, 39, 86, 174]. In this energy landscape, metastable conformations with different basins of attraction exist and en-sure the existence of a robust path leading toward the equilib-rium conformation . In this view, protein folding can be de-scribed as a series of hopping events between such metastable states. Moreover, such states are responsible for the separation of time scales between slow folding kinetics and fast dynamical motions of individual atoms in the protein .
An important group of proteins are enzymes which catalyze chemical reactions [2, 170]. In vivo enzymes are involved in most biological reactions. Common examples are the extraction of energy by the cell from food or the conversion of CO2 into sugars during photosynthesis in plants. Also well known are the enzymes tubulin and actin which self-assemble into the fil-aments making up the cytoskeleton. Moreover, motor proteins are enzymes that catalyze the decomposition of ATP molecules and convert the energy gained in this process into directed mo-tion [2, 175].
The mathematical description used to model enzymatic reac-tions in a simple way is the so-called Michaelis-Menten kinetics [2, 114, 170]. The enzyme E binds to a substrate S. Within the enzyme-substrate complex ES the substrate S is converted into the product P. Schematically, this process is written as E + S k1 ⇄ k−1 ES kcat →E + P (2.1) with the phenomenological rate constants k1, k−1 and kcat. Here it was assumed that the back reaction in which the enzyme and the product recombine to form the ES complex can be ne-glected.
Prominent members of the enzyme family are ATPases. An ATPase catalyzes the decomposition of adenine triphosphate 2.1 proteins 9 P O P O P O P O P O P O P O P O O O H O H O O H O H O H N N N N N H N H O H H O P O O O O O H O H P O P O P O P O P O P O P O P O P O P O P O P O O O H O H O O H O H O H O H N N N N N H N H H + O H H O O O H O H + Figure 1: ATP hydrolysis: ATP + H2O →ADP + Pi (ATP) into its products adenine diphosphate (ADP) and a phos-phate (Pi). In this process called hydrolysation energy is re-leased that the cell can use in various ways. The chemical hy-drolysis reaction is displayed in Fig. 1. ATP is used as a source of energy in a wide range of processes within the cell. Because ATP stores and transports energy, it may be understood as an energy currency within the cell. This thesis mainly deals with the two ATPases – actin and myosin.
2.1.1 Actin Actin is one of the most abundant proteins. In eukaryotic cells between 1% and 5% of all proteins are actin molecules .
The importance of actin is further highlighted by the fact that it is involved in more protein-protein interactions than any other known protein . Close to a hundred proteins can regulate the actin dynamics. Moreover, the actin molecule has been highly conserved during evolution .
An important feature of the actin macromolecule is its ability to form long and stable filaments. Such filaments are an inte-gral part of the cytoskeleton [2, 170]. It is noteworthy that actin is functional in its filamentous form only, i.e. no function of globular actin is up-to-date known .
Filaments polymerize in the dynamical process called tread-milling (see Fig. 2). While actin proteins on one side of the fila-ment preferentially attach, they tend to dissociate on the other.
This is sometimes described as the polarity of the filament; the plus end grows faster than the minus end . The process of treadmilling is the basis of cell motility, i.e. the ability of a cell to move autonomously. Such a behavior is seen, e.g., in E. coli bacteria or the vaccinia virus [19, 116, 118].
10 background Figure 2: Treadmilling. Actin monomers tend to attach on one side of the filament and detach on the other. Treadmilling strongly depends on the nucleotide state of actin.
Polymerization strongly depends on the nucleotide state of actin [95, 131]. The binding affinity of ATP-bound actin at the growing end of the filament is much higher than the binding affinity of the ADP-actin complex. At the other end of the fila-ment, the situations is contrary (see Fig. 2).
Actin filaments are an essential basis for muscle contraction and, furthermore, form meshes which give structure and stability to the cell. An electron micrograph of such structures from Svitkina et al. is shown in Fig. 3. The actin meshes can be seen as a complex highway system through the cell. Mo-tor proteins like myosin move along actin tracks, e.g. to trans-port vesicles into or out of the cell.
2.1.2 Myosin Myosins comprise a family of actin-based motor proteins .
Its members share the same general features, but exhibit minor structural differences . All myosins, however, have in com-mon that, attached to actin filaments, they can generate forces which allows them to move processively. Such a directed mo-tion is fueled by the hydrolysis of ATP [2, 170].
Myosin was first discovered in muscle fibres and, therefore, muscle myosin is sometimes called conventional whereas all other myosins have been named unconventional. A more con-cise characterization has been done in form of classes. Class-I myosin (abbreviated myosin-I) is, e.g., responsible for vesicle transport inside the cell, whereas the muscle myosin belongs 2.1 proteins 11 Figure 3: Actin filaments form meshes (electron micrograph, modi-fied from ) to class II. Other myosins are consecutively numbered (III, IV, V, ...) according to the date of their discovery and, currently, eighteen different classes of myosin are known [57, 146] Generally, the myosin motor protein consists of two main do-mains: the motor domain and the tail. Within the motor do-main, the nucleotide-binding pocket (NBP) is located [21, 170].
Here the chemical reactions that fuel the motor mechanism, i.e.
the binding of ATP, subsequent hydrolysis and product release, take place. The motor domain also interacts with the actin fila-ment. Nucleotide or strain induced conformational changes in the actin-binding region control the binding affinity to the fila-ment . Likewise, the tail region plays an important role for force generating within the myosin machine cycle [2, 146]. This role, however, is not entirely clarified .
Myosins can exist both as single monomers (e.g. myosin-I) or as dimers (e.g. myosin-II, myosin-V). In the case of a dimeric molecule, myosin heads are connected via their tails . This region also binds cargo that is transported through the cell. In Fig. 4 the two-headed myosin-V is depicted pulling a cargo along an actin filament.
12 background Figure 4: Schematic view of a two-headed myosin motor transporting a load along the actin filament.
The complex of actin and myosin, also called actomyosin, is best known for its role in muscle contraction . The mus-cle consists of muscle fibers which, in turn, consist of sarcom-eres. These cells have the ability to contract [2, 146, 170]. The muscle contraction is realized by a large number of myosin molecules interacting with actin filaments. A schematic illus-tration of a sarcomere is shown in Fig. 5. The functional basis of the sarcomere are thick bundles of muscle myosin (myosin-II) and actin filaments connected to the walls of the sarcomere (Z-discs). Myosin heads randomly attach to the filament and, in this way, pull the Z-discs toward each other . This re-sults in contraction of the sarcomere. The binding rate of the myosin heads, i.e. the strength of contraction, depends on the ATP concentration in the muscle cell [12, 29].
A different myosin protein is myosin-V. It is involved in the transport of vesicles and organelles along actin filaments in a specific direction. A schematic drawing of the two-headed mo-tor is shown in Fig. 4. Myosin-V moves in a hand-over-hand fashion in 36nm steps along actin filaments [141, 166]. An in-teresting cousin of myosin-V is the class-VI myosin. It is also two-headed and involved in the transport of vesicles, but moves with a smaller step size into the opposite direction. The struc-tural differences between these two proteins are subtle and, thus, they have challenged how people think about the motor mechanism [80, 151].
2.1 proteins 13 Figure 5: Schematic view of a sarcomere.
2.1.3 Actomyosin Complex In 1971 the two biophysicists Richard W. Lymn and Edwin W.
Taylor have considered the mechanism of the myosin motor op-eration . Their model is commonly referred to as the Lymn-Taylor cycle [58, 155]. In Fig. 6 the mechano-chemical cycle is schematically shown for myosin-V moving along actin.
The starting point to describe the machine cycle of the two-headed dimer is the state where both heads are attached to the filament. Due to the periodicity of the actin filament, a myosin head can only bind to the filament at specific binding sites (e.g.
every 36nm for myosin-V). In the initial step, the rear head is nucleotide-free, whereas in the leading head ADP is bound.
Then, ATP attaches to the rear head and, in this way, leads to a dissociation of myosin from the filament . The now fol-lowing step in the myosin motor cycle is the so-called force-generating step. This part of the Lymn-Taylor cycle still remains unclear and the discussion in the research community fo-cuses on two alternatives.
The majority of scientists think that the force is generated by the so-called power stroke [28, 72, 75, 134, 148]. The release of a nucleotide in the leading head is assumed here to lead to a lever arm swing. After that, the head finds a possible binding site through rotational diffusion. Note that in this view, there is a unique binding site in the filament where the free myosin head eventually binds.
In the opposing explanation, force is generated by a Brow-nian search-and-catch mechanism [158, 49, 80]. In this model, the tail is assumed to be very flexible. The rear, detached head moves by thermal fluctuations. The molecule finds the filament 14 background Figure 6: Lymn-Taylor cycle. Actin monomers are shown as yellow circles, pink circles show the possible binding sites for myosin.
Myosin is shown in blue. The oval structures are the myosin heads connected via their tails.
2.2 experimental methods 15 binding site in front of the other head by some biasing mech-anism. The important difference to the swinging lever-arm hy-pothesis is that, in the latter case, myosin has the possibility to attach to two different sites on the filament. The Brownian mo-tion, however, is biased, and thus the myosin head is able to per-form directed motion. Such a mechanism is possible, because ATP acts as a source of energy, and therefore the actin-myosin system is far from thermal equilibrium.
After the force generation step, however, both models are again found in the same state. ATP is hydrolyzed in the new leading head and this leads to a strong binding to the filament.
It is to note that communication between the two myosins is essential to ensure processive motion. First, it is important in the second step of the mechano-chemical cycle that the rear head detaches from the filament while the leading head is still attached to the filament. Otherwise, the dimer would have ei-ther stepped backward or fallen of the filament. Moreover, if myosin generates force by means of a Brownian search-and-catch mechanism, there has to be a bias to direct the motion.
This can be achieved in two ways: either the two myosin heads communicate via the actin filament or via their tails, their only physical connections.
2.2 experimental methods Proteins are functional in their folded state only. Thus, the struc-ture of the protein is essential to understand protein dynamics and many experimental methods have been developed for high-resolution structure determination. A standard technique to re-solve protein structures with atomic resolution is X-ray crystal-lography [2, 44, 99]. The first crystal structure of a protein was that of the iron- and oxygen-binding protein myoglobin and it was obtained by Perutz and Kendrew. For this work they were awarded the Nobel Prize in Chemistry in 1962 [89, 90]. An ex-ample of an X-ray diffraction pattern for the actin filament is shown in Fig. 7.
The main technical hindrance in using X-ray diffraction is the necessity to grow protein crystals . While being a complex procedure in itself, crystallization of proteins may also intro-duce physical perturbations which lead to erroneous structural data .
16 background Figure 7: X-ray diffraction pattern of the actin filament. The upper left and the lower right quadrants show the experimental data, the other quadrants display the calculated values .
A powerful alternative to X-ray crystallography is nuclear magnetic resonance (NMR) spectroscopy [20, 143]. This method has been pioneered by Richard R. Ernst, who was awarded the Nobel Prize in Chemistry in 1991 for ”for his contributions to the development of the methodology of high resolution nuclear magnetic resonance (NMR) spectroscopy”2 and by Kurt Wüthrich, awarded the Nobel Prize in chemistry in 2002 for ”for the devel-opment of methods for identification and structure analyses of biolog-ical macromolecules”2 [98, 125, 176, 177].The advantage of NMR spectroscopy is its ability to determine the structure of a protein in solution with atomic resolution.
A different approach to obtain protein structure information is cryo-electron microscopy [56, 167, 182]. With this approach, biological specimens can be monitored in their biological en-vironment and, similar to the NMR method, no crystallization is needed. To use cryo-electron microscopy, the bio-material is rapidly frozen and subsequently introduced into a high-vacuum electron microscope. High resolution structures can be obtained by averaging many electron microscopy images. Fig. 8 shows cryo-EM images of actin filaments .
2 2.2 experimental methods 17 Figure 8: Cryo-electron microscopy picture of actin filaments. Scale bar, 100nm. Throughout the years, a large number of proteins has been resolved using various methods. Such experimental data has been collected and made available in an openly accessible on-line repository. Accordingly, structural data of a large number of biomolecules can be found in the Protein Data Bank3.
Apart from 3-D structural data, researchers are interested in dynamic properties of protein. In recent experiments [80, 122, 123], proteins have been probed by applying external forces.
Figure 9 schematically shows the experimental set-up used by Iwaki et al. . In this experiment, a gold bead was attached to the tail of a myosin monomer. By means of an optical tweezer, the molecule was dragged along the actin filament in two op-posite directions. The transition from strong to weak binding to the filament was seen to strongly depend on the direction of the applied force. It was found that the probability of strong binding increases when forces are applied in the direction oppo-site to that of the intrinsic processive motion. Thus, the myosin molecule operates as a strain sensor, with well-defined responses 3 18 background Figure 9: Experimental set-up of a single molecule experiment. A gold attached to a myosin tail is caught by an optical tweezer. In this way, myosin can be moved along the actin filament and forces can be applied .
in its different parts induced by specific mechanical perturba-tions.
A frequently used method allowing to monitor conforma-tional motions directly is fluorescence resonance energy trans-fer (FRET), also called Förster resonance energy transfer . De-spite its low accuracy, distance changes in the range of 1-10 nm can be resolved in FRET experiments . In this method, a pro-tein is labeled by a fluorophore, serving as a donor, and a cor-responding acceptor. Because of long-range dipole-dipole cou-pling, energy is transferred between the two marker molecules.
The transfer rate depends strongly on the donor-acceptor sep-aration and, hence, distance changes on the order of Å can be followed dynamically [62, 83]. The possibility to follow distance changes between acceptor and donor over time is the main ad-vantage of the FRET technique.
2.3 modeling of protein dynamics 19 2.3 modeling of protein dynamics Proteins have been the subject of many theoretical studies. The straightforward way to investigate protein dynamics is, of course, to include all particles and all interactions between them. With growing computational power, such molecular dynamics (MD) simulations have become a standard tool in this research field.
Such descriptions, however, involve large systems of equations and suffer from high computational costs. To overcome such limitations, a wide range of simplified and coarse-grained ap-proximations have been developed to tackle the open questions of protein research.
2.3.1 Molecular Dynamics In the standard MD simulation method, Newton’s equations of motion are numerically integrated. Here, the interatomic po-tentials are usually separated into bonded and non-bonded in-teractions. The bonded interactions are characterized by the bond length ri, the bond angles θi and the dihedral angle φi.
Non-bonded interactions can, e.g., be Coulomb interactions for charged atoms or Lennard-Jones-type potentials to model van-der-Waals bonds. In this case, the interatomic potential is Uij = ∑ bonds kbond i (ri −r0)2 + ∑ angles kangle i (θi −θ0)2 + ∑ dihedrals kdihedral i [1 + cos(niφi + δi)] +∑ i ∑ j̸=i 4ϵij σij rij !12 − σij rij !6 (2.2) +∑ i ∑ j̸=i qiqj ϵrij .
While the charges qi are generally known, the parameters kbond i , kangle i , kdihedral i , ni, δi, ϵi and σij have to be obtained from ex-perimental data. Approximate values of these parameters have been collected and are stored in force field parameter databases.
Popular for protein simulations are the CHARMM or GRO-MACS force fields maintained at Harvard and Groningen Uni-versity, respectively [15, 103].
The method of MD simulations has been first applied as early as 1957 in the field of statistical physics. The two scientists 20 background Bernie Alder and Thomas Wainwright used it to describe sim-ple liquids [3, 4]. The method was advanced by Rahman who used such simulations to describe liquid argon . About ten years later, a realistic system was investigated by Rahman and Stillinger who studied properties of liquid water . Fi-nally, MD simulations found their way into the field of protein research in the late 1970s and have since then prospered. In-deed, they have become one of the principal tools of theoretical biophysics.
The first protein investigated was bovine pancreatic trypsin inhibitor (BPTI) . Because BPTI is small and stable, it has been used as a guinea pig system to test both experimental and theoretical methods. With growing computational power MD simulations have become widespread. Initially only used to refine the experimental data obtained by X-ray diffraction or NMR [18, 88], various MD studies investigating many different aspects of proteins are available today. They can be used, for instance, to understand the role of the solvent in protein func-tions or to investigate the mechanisms of protein folding .
All-atom MD methods are also used to study mechanochem-ical conformational motions in proteins. Although such simu-lations have been successfully used to describe atomic fluctua-tions on the order of nano-seconds, tracing the slow large-scale collective motions is prevented by the excessive computational costs . Such motions are typically on the scale of millisec-onds or longer, whereas, in full MD simulations, generally only the dynamics on much shorter time scales up to a microsecond can be resolved To overcome such computational limitations, various acceler-ation methods have been proposed. A typical example of such a method is provided by steered MD simulations, which were first developed to study the unbinding of avidin and biotin [81, 107, 79]. In such simulations, external harmonical forces are applied to lower energy barriers and, in this way, speed up rare events.
Another method to accelerate MD simulations is the Replica-exchange method [21, 23, 138]. Here, several copies (replicas) of the investigated system with different temperatures are evolved independently in time using conventional MD simulations. Af-ter some time interval tswap the states with neighboring temper-atures are swapped by velocity rescaling with a certain proba-bility pswap(∆E), which depends on the energy difference be-tween the two states. This method has been applied, for exam-ple, to the myosin NBP to identify possible exit pathways of 2.3 modeling of protein dynamics 21 Pi after hydrolysis . Moreover, specialized hardware for effi-cient MD simulations is being developed [52, 120, 147]. By using it., the native-state dynamics of the small protein BPTI could be followed over a time scale of one millisecond, the longest pub-lished all-atom MD simulation of a protein up-to-date .
A large number of studies using accelerated methods have been performed for myosin and actin. For instance, conforma-tional dynamics of actin monomers have been investigated in this way and responses due to interaction with ATP and other ligands have been analyzed. Moreover, nucleotide-dependent folding of the flexible DNase-I binding (DB) loop could be ob-served [128, 185]. It was demonstrated that the nucleotides in-duce significant local deformations in the region of the ATP binding pocket, which can partially close . Slow large-scale ligand-induced motions of principal actin domains, however, could not be traced. For the myosin macromolecule, mechani-cal coupling has been theoretically studied. By means of guided MD simulations, i.e. molecular dynamics with an external re-straining potential, it was demonstrated that significant confor-mational changes in the converter and actin-binding cleft could be induced by imposing a local perturbation of the protein con-figuration in the region of the ATP-binding pocket .
2.3.2 Reduced Descriptions and Coarse-Graining As general experience in dealing with complex systems reveals (see, e.g. ), the very existence of robust ordered collective motions often implies that some reduced dynamical descrip-tions should be possible. Thermodynamic approaches, for in-stance, have been used to describe protein systems [5, 16, 111].
Even after considerable simplifications that may involve the loss of all structural information, important insights can be gained by such models.
An example of a greatly simplified approach is the Brow-nian ratchet models for motor proteins [1, 6, 111]. Brownian motion is the thermally induced motion of small particles sus-pended in liquid. This behavior is named after the the botanist Robert Brown and theoretically considered by Einstein and Smoluchowski [48, 149]. The term Brownian ratchet refers to a thought experiment by Smoluchowski, which Feynman made popular in his famous lectures at the California Institute of Technology .
22 background Today, some proteins are thought to make use of thermal fluc-tuations to perform their functions . Particularly this refers to the processive motion of motor proteins along actin filaments and microtubules. Experiments suggest that motor stepping may be based on a Brownian-search mechanism [35, 80] and corresponding kinetic models have been developed [6, 14]. The basic idea is that the motor makes use of thermal fluctuations to efficiently and robustly perform its task. The motor proteins may have developed the mechanisms to bias the fluctuations, so that the motor preferentially moves into one direction. The bias is possible because, due to binding and hydrolysis of ATP, such molecular systems are far from its thermal equilibrium.
Kinetic models by Bierbaum and Lipowsky have shown that, for the myosin-V motor, the chemical transition rates for binding of ATP and ADP have to be force dependent. While this dependence provides a qualitative understanding of a possi-ble bias mechanism, its structural nature cannot be resolved by such models. The chemical activity of proteins, however, is to a large extent regulated by microscopic changes of its confor-mation [22, 140]. To resolve these effects, it is necessary to use high-resolution methods. In order to avoid the computational complexity of all-atom MD simulations, appropriate approxima-tion techniques have to be developed.
Various coarse-grained descriptions have been proposed (see a review in Ref. ) and the G¯ o-models [10, 60, 93, 157] de-serve to be particularly mentioned. In their original formula-tion, a residue is described as a bead and neighboring backbone residues are connected via harmonic potentials. Interactions be-tween the neighbors within the native conformation of a pro-tein are usually described with a phenomenological Lennard-Jones type potentials [68, 153]. Hence the equations of motion look identical to (2.2). In the case of G¯ o-models, however, the particles are not atoms, but represent residues. Usually a point particle at the Cα position of the amino acid is taken as a repre-sentative for the whole residue. The parameters are chosen in a way that the native state of the protein is the equilibrium con-formation. G¯ o-models have been developed to study the prob-lem of protein folding and have successfully described folding pathways and corresponding energy landscapes of various pro-tein structures [27, 64, 85, 119, 157]. Moreover, G¯ o-models have recently been applied to investigate large-scale conformational motions of proteins .
2.3 modeling of protein dynamics 23 A simplification of standard G¯ o-models is the EN approxi-mation. Such models have gained an increasing popularity in the last years [9, 63, 69, 36, 161]. In the EN approach, the phe-nomenological Lennard-Jones type potentials are replaced by harmonic potentials. Instead of differentiating between native and non-native bonds, all neighboring residues, irrespective of whether they are interacting over the backbone or over their sidechains, are connected by elastic links. The EN model can thus be pictured as a network of residues where all residues, that are within a certain range, are connected by elastic links.
2.3.3 Elastic-Network Models We see that the construction of EN models is very straightfor-ward. Amino acids are replaced by identical point particles, which typically correspond to a residue, and the interactions between residues are approximated by introducing elastic links between neighboring particles. The links have different natural lengths, but, as often assumed, the same stiffness. The architec-ture of the elastic network, i.e. the pattern of connections be-tween its nodes, is constructed on the basis of experimentally known equilibrium conformation of a protein in a way that its lowest energy state coincides with the known equilibrium con-formation of a protein. The exact formulations will be given below in the next chapter.
An EN model was first used by Tirion in 1996 to describe the thermal fluctuations (B-factors) in several proteins using an all-atom description and nowadays several variants of the EN approach exist. The work of Tirion laid the foundation for the development of the so-called Gaussian network model (GNM) [9, 63]. This approach considers protein networks on the level of amino acids and it has been very useful to understand vibra-tional dynamics, i.e. mean-square fluctuations and correlations between fluctuations of pairs of residues . Inspired by such results, the anisotropic network model (ANM) was put forward in 2000 [7, 43]. In this description, the directional dependence of deformations was additionally included. The ANM will be used to model actin and myosin proteins in the course of this thesis.
Despite their high degree of simplification, EN models have turned out to be remarkably efficient. Investigations revealed that such models, apart from describing the B-factors in many 24 background proteins, can also predict ligand-induced conformational changes [36, 159, 102, 24, 180]. Much attention has been paid to the normal-mode analysis corresponding to linearized EN models.
However, full nonlinear equations of relaxational EN dynamics were also explored [46, 53, 54, 65, 129, 163, 164]. It was even shown that EN models can be extended to include the possibil-ity of partial unfolding and refolding of proteins [67, 91, 110, 117].
Nonlinear EN models have been the focus of our research group at the Fritz Haber Institute of the Max Planck Society. It has been, for example, shown that motor proteins respond by well-defined motions to external perturbations. On the basis of such results, a design principle for protein machines was sug-gested and, in an evolutionary process, artificial machines have been constructed . Furthermore, Togashi et al. have examined the validity of the assumption of linearity which un-derlies the normal-mode analysis of proteins and, thus, repre-sents the basis for a wide range of works. They have shown that the protein kinesin KIF1A behaves strongly nonlinear and conformational motions of this macromolecule follow trajecto-ries much different from the ones expected in the linearized EN approximation.
Additionally, nonlinear EN models have been used to study helicases. Entire operation cycles of the molecular motor hepati-tis C helicase were described using a simple coarse-grained lig-and model . In the same framework, operation mechanism of different helicases could be understood by modeling them as elastic networks . Moreover, effects of hydrodynamic in-teractions with the solvent were incorporated into EN models [47, 34] and thermal fluctuations could also be considered in the framework of this approach. In this way, the machine cycles of the enzyme adenylate kinase interacting with solvent were reproduced by EN simulations.
There is one property of elastic-network models which makes them particularly appealing. Since all residues are pictured as identical point particles, irrespective of their actual chemical structure, and because interactions between the particles do not depend on the chemical nature of the corresponding residues, EN models turn out to be stripped of almost all chemical de-tails. In this rough approximation, a protein is viewed as a me-chanical object, i.e. as a complex elastic network. Therefore, by studying such models, one can investigate purely mechanical 2.3 modeling of protein dynamics 25 aspects of conformational protein dynamics, distinct from the chemical aspects.
3 M AT H E M AT I C A L M E T H O D S The dynamics of actin monomers, the molecular motor myosin and the complex of the two proteins are investigated in this the-sis from a theoretical point of view. This chapter is a complete collection of the mathematical tools used in our study. Both actin and myosin are molecular machines and they perform certain tasks characterized by collective domain motions. Such collective motions can be slow, even up to 1s. Accordingly, all-atom MD simulations, the standard approach to describe con-formational dynamics of proteins numerically, are currently un-feasible due to the high computational costs involved. To over-come these limits, the so-called anisotropic network model or simply elastic-network (EN) model [7, 43] is used to approxi-matethe proteins.
Below, the general EN approach is introduced. Furthermore, we discuss linearization of the corresponding equations of mo-tion. We also show how to extend such an approximation in or-der to be able to describe protein-specific features. Throughout this thesis, figures of proteins and networks will be found. For visualization we used the VMD software, developed and ad-vanced in the Theoretical and Computational Biophysics Group at the University of Illinois .
3.1 the elastic network model To approximate proteins, a structural coarse-graining takes place.
An elastic network is formed by N identical point particles (nodes) which are connected by identical elastic links. The net-work architecture is defined by the experimentally known equi-librium positions R(0) i of all residues i in the protein. For this purpose, the positions of α-carbon atoms are taken. Note that such an approach neglects the chemical character of each amino acid – they are treated as identical objects – and, thus, strips the protein of all chemical details. Having done so, it is necessary to introduce interactions between residues in a suitable way.
27 28 mathematical methods In the EN model, only residues that are located in the neigh-borhood of each other in the folded conformational state can interact. Interaction is modeled in a simple approximation. An elastic link is introduced between two nodes i and j if the equilibrium distance d(0) ij = |R(0) i −R(0) j |, i.e. the distance be-tween the α-carbons in the experimentally obtained conforma-tion used as a reference, is smaller than some cutoff l0. The natural length of the link is chosen equal to the respective equi-librium distance d(0) ij . Hence, in the EN approximation, the ex-perimentally found equilibrium state is by definition the state with the lowest elastic energy E = 0.
The total elastic energy of the network is given by the sum of the energy stored in its elastic links Eel = 1 2κ ∑ i,j Aij dij −d(0) ij 2 (3.1) where dij = |Ri −Rj| is the distance between two particles i and j and the elements of the adjacency matrix A are Aij = 1, if d(0) ij < l0, and Aij = 0 otherwise. The elastic constant κ is the same for all nodes. If external forces Fi are applied to the network, its energy is E = Eel −∑i Fi · Ri. The different quantities used in equations (3.1) are illustrated in Fig. 10.
In our study, the cutoff length in the anisotropic EN has been determined by following the arguments by Atilgan et al. .
They constructed a sequence of EN models with gradually in-creasing values of the cutoff length and, for each cutoff length, computed the eigenvalues of the linearization matrix (see be-low). If the cutoff length was too small, the network was falling into disconnected components or free rotations inside it were possible, as evidenced by the fact that more than six zero eigen-values of the linearization matrix were found. The cutoff length of 8.5Å for G-actin and 10Å for myosin-V, used in our numeri-cal investigations, was chosen as the first cutoff length at which only six zero eigenvalues were present. Note that, in the study , a cutoff length of 9.5Å was found to hold in the anisotropic network approximation for a large group of proteins (but actin and myosin were both not considered there).
On the considered time scale of milliseconds, inertial effects are negligible and conformational dynamics is purely dissipa-tive . In our present study, hydrodynamical interactions be-tween particles will be neglected. In the overdamped limit, par-3.1 the elastic network model 29 Figure 10: Illustration of the EN model. Two residues i and j are con-nected via an elastic link. Deformation from the equilib-rium position leads to a force proportional to the distance change dij −d(0) ij ticle velocities are proportional to the forces acting on it and the equations of motion are γ ˙ Ri = Fi −∂Eel ∂Ri .
(3.2) Within the coarse-grained EN approximation, we assume that the friction coefficient γ is equal for all particles. After an ap-propriate rescaling of time, the parameters κ and γ can there-fore be removed from these equations and they take the form ˙ Ri = Fi − N ∑ j=1 Aij dij −d(0) ij Ri −Rj dij .
(3.3) Note that, in the new units employed, the force F is measured in Å. A force of F = 1Å is stretching a single elastic link by 1Å.
An important property of equations (3.3) is that they are gen-erally nonlinear in terms of the coordinates Ri = (Xi, Yi, Zi) because of the nonlinear dependence of the distances dij = q (Xi −Xj)2 + (Yi −Yj)2 + (Zi −Zj)2. Such a dependence can give rise to interesting nonlinearities .
The dynamics of the EN is followed by integrating equations (3.3) using the explicit Euler method with the time step dt = 0.1. Although this integration scheme sometimes suffers from stability problems and may produce large errors, it can be used 30 mathematical methods in the dissipative systems investigated. On the example of an actin EN, we have studied the effect of different values dt. By repeating integrations for some relaxation trajectories with a smaller time step of dt = 0.01, we have checked that this does not lead to significant changes. To test this, we have compared the two stationary states in the presence of external forces that were obtained by integrating the equations of motion (3.3) with two different time steps of 0.1 and 0.01. The time evolution of the sum of distances between all residues in the network were followed in both simulations and were found to be below 0.001Å. Thus, the decrease of the time step did not lead to an accuracy improvement and the choice of the time step as dt = 0.1 was sufficient for our analysis.
Integration of the equations of motion (3.3) has been contin-ued until a stationary state was reached. The numerical crite-rion was that the sum of the corresponding order parameters (see the following chapters for exact definition of the order pa-rameters) has ceased to change by more than 10−14Å.
3.2 linearized equations of motion The equations of motions (3.3) of the EN model are nonlinear because distances are nonlinear functions of the coordinates.
EN models are often used for the normal-mode analysis [7, 9, 24, 36, 43, 63, 102, 159, 180]. Then, only small displacements of particles from their equilibrium position are considered and, thus, a linear approximation of the equations (3.3) can suffice.
The linearized equations are ˙ ri = − N ∑ j=1 Aij R(0) i −R(0) j h d(0) ij i2 h R(0) i −R(0) j · ri −rj i . (3.4) They hold for small deviations ri = Ri −R(0) i from the equi-librium. Equations (3.4) can also be written in the matrix form ˙ ri = −Λijrj, with a 3N × 3N matrix Λ. Such dynamical equa-tions can be solved analytically in terms of the normal modes of the linearization matrix Λ. In terms of the eigenvalues λα and the eigenvectors e(α) i the general solution is ri(t) = ∑ α kα exp(−λαt)e(α) i , (3.5) 3.3 immobilization procedure 31 where coefficients kα are determined by the initial conditions.
Thus, the eigenvalues determine the relaxation rate constants of the respective normal modes. The slowest relaxation processes are controlled by the normal modes with the lowest eigenval-ues. Characteristic normal motions that occur on different time scales are separated by a gap in the eigenvalue spectrum.
Assuming that only one of the normal modes is excited, the motions of network particles are given by equations (3.4) where only one term, corresponding to a particular mode, is present.
Thus, characteristic network motions in a specific normal mode can be determined and visualized.
The equations of motion (3.3) depend only on relative dis-tance changes. Therefore, they are always invariant against rigid translations and rotations of the entire elastic network. This im-plies that the linearization matrix Λ should always have six zero eigenvalues. On the other hand, if more than six zero eigenval-ues are found, this indicates that the network breaks down into disconnected components or that free internal rotations of some residue groups are possible. This property can be used in the selection of the cutoff length, as explained above.
3.3 immobilization procedure If a force acts on a single node of an elastic network, it induces not only internal deformations in the network, but also rigid translations and rotations of the entire object. If the network is pinned, e.g. by immobilizing one of its nodes, this can introduce additional internal deformations and the results of the study would depend on the location of the pinned node. To avoid such effects in our simulations, we have employed a special immobilization procedure. Compensating forces were applied to all network nodes in such a way that they could only lead themselves to a rigid translation or rotation. The magnitudes of the additional forces were determined by the condition that prevented translational and rotational motions induced by the external force. The computation of compensating forces was performed at each next integration step, so that they were auto-matically adjusted to the conformational changes. An illustra-tion of the immobilization procedure is shown in Fig. 11.
32 mathematical methods Figure 11: Illustration of the immobilization procedure. An external force Fext is applied to a single residues of an elastic net-work. Induced rotation and translation described in terms of the forces f and fi, respectively, are shown. The corre-sponding compensating forces are indicated by arrows.
To construct compensating forces, we first note that, if the same force f is applied to each particle, it would lead to a rigid translation of the entire network. Moreover, if forces fi = ω × Ri (3.6) with an arbitrary vector ω are applied to the particles with co-ordinates Ri, they can induce only rigid rotation.
Without loss of generality, the coordinates Ri can always be chosen in the coordinate frame whose origin coincides with the center of mass of the network. Suppose that an external force Fext acts on a particle with coordinates R0. Then, it generates an external torque R0 × Fext that should be balanced by some com-pensating forces fi. Because the coordinates Ri of all network nodes are known, the compensating forces are completely de-termined by the constant angular velocity ω. It describes the rigid rotation that would be induced by external forces.
The additional compensating forces should satisfy the bal-ance equation R0 × Fext + N ∑ i=1 Ri × fi = 0 , (3.7) which can be rewritten as −Mω = R0 × Fext (3.8) 3.4 extensions of the model 33 with the matrix M= N ∑ i=1 |Ri|2 −X2 i −XiYi −XiZi −XiYi |Ri|2 −Y2 i −YiZi −XiZi −YiZi |Ri|2 −Z2 i .
(3.9) Thus, to prevent rigid rotation, additional forces fi = − h M−1 (R0 × Fext) i × Ri (3.10) should be supplied. Note that, as can be easily checked, ∑i fi = 0 and, hence, such compensating forces do not induce trans-lational motion of the network. To prevent rigid translation in-duced by the external force, the compensating force f = −Fext/N should be additionally applied to each network node.
Thus, if compensating forces f + fi are applied to all nodes at every integration step, both translation and rotation induced by the external force can be balanced out in a noninvasive manner.
3.4 extensions of the model In the EN approximation, residues within a cutoff l0 should be connected by elastic links. A problem arises if some residues that are not neighbors in the equilibrium conformation of a pro-tein come too close to each other. This, for instance, can occur in the case of the actin monomer. Following the logic of the EN model, residues within the cutoff range should interact. Thus, we allow breakable links to be established between such nodes that are known to approach each other considerably. With a rea-sonable choice of parameters, these additional bonds may lead to a metastable conformational state. If such a metastable states exists, its stability is tested with and without thermal fluctua-tions. The general EN approach is extended as follows.
3.4.1 Breakable Links We may include breakable links between residues not connected by elastic links. In contrast to other intramolecular interactions, 34 mathematical methods 4 5 6 7 8 9 10 -1.0 -0.5 0.0 0.5 1.0 Figure 12: The truncated Lennard-Jones potentials with parameters lc = 8.5Å, req = 5Åand D = 1Å2 is shown. The red dashed curve shows the approximation near the equilibrium by a harmonic potential.
breakable links are described by the truncated Lennard-Jones potentials VtLJ(dij) = ( VLJ(dij) −VLJ(lc) dij < lc 0 otherwise (3.11) with the function VLJ(dij) = BijD req dij !12 −2 req dij !6 .
(3.12) The interaction parameters are the equilibrium distance req and the strength of the potential D. The matrix elements are Bij = 1, if a breakable link between residues i and j exists, and Bij = 0, otherwise. The cutoff length of the potential lc is set to be equal to the cutoff length l0 of the EN model. This reflects the fact that links that are too far away from each other should be broken.
Additional links are effective only when the distances between the two residues are below the truncation length lc.
3.4 extensions of the model 35 When breakable bonds are added between residues that come close to each other, the equations of motion (3.3) should be mod-ified by including additional forces Fadd i = −∑ j ∂VtLJ(dij) ∂Ri , (3.13) where the summation is performed over all nodes j.
The choice of the interaction parameters for breakable links should be based on the requirements that (i) this potential be-comes flat when distances between the residues exceed the cut-off length used in the construction of the elastic network from the experimental data, (ii) the minimum of the interaction po-tential is found at the distance which lies between the mini-mum and the maximum values of the natural lengths d(0) ij of elastic links, as deduced from the experimental data, and (iii) near its minimum the potential (3.12) can be approximated by a parabolic potential with a strength of the same order of mag-nitude as that of the elastic potential of the regular links.
A reasonable parameter choice is req = 5Å and D = 1Å2.
The corresponding truncated Lennard-Jones potential with cut-off length lc = 8.5Å is shown in Fig. 12. The blue line repre-sents the potential and the dashed red line shows its harmonic approximation near the equilibrium.
3.4.2 Thermal Fluctuations To study the stability of the breakable bonds, effects of thermal fluctuations can be considered. Although we disregard hydro-dynamic interactions, thermal fluctuations can be modeled by including appropriate random forces into the dynamical equa-tions, i.e. by writing them as γ ˙ Ri = Fi −∂Eel ∂Ri + ξ(t) .
(3.14) Here, ξi(t) is a Gaussian noise with the correlations ⟨ξi(t)⟩= 0 ⟨ξi(t)ξj(t′)⟩= σ2δijδ(t −t′) .
(3.15) separately in each direction. The parameter σ specifies the noise intensity. It is related to the temperature as σ2 = 2γkBT.
36 mathematical methods 3.5 protein-protein interactions Within the framework of the EN approach, amino acids are stripped of all chemical details. Accordingly, additional infor-mation is needed to model protein-protein interactions. If the corresponding binding sites are however known, such interac-tions can be included in a suitable way. In the following, we show how to extend the EN accordingly.
3.5.1 Electrostatic Interactions Side-chains of certain amino acids are charged . Arginine and Lysine, for instance, are positively and Aspartic and Glu-tamic acid are negatively charged. The amino acid Histidine is charged in 10% of all cases. Accordingly, we assign to the cor-responding residues charges of q = 1, q = −1 and q = 0.1, respectively.
The electrostatic interaction between two charged residues is Ve.stat.(dij) = Aqiqj ϵdij exp(−dij/lD) .
(3.16) The parameter ϵ corresponds to the dielectric constant and A is a prefactor. Furthermore, in the solvent charges are screened by interaction with water and ions. These are not explicitly in-cluded in the model, but effectively introduced by a decaying exponential term. Thus, the range of the electrostatic interac-tion is controlled by the Debye-Hückel length lD . Here, the value lD = 20Å is used and the parameters are chosen as A/ϵ = 2Å3, in agreement with Ref. .
Additional electrostatic interactions (3.16) are present between the residues of different molecules only. The reason for this as-sumption is that protein EN models are constructed based on the native state. It corresponds to the equilibrium conformation and the actual shape of the protein is a result of energy mini-mization, i.e. folding of the protein. Therefore, all intramolecu-lar interactions, including electrostatic ones, are effectively con-sidered in the EN approach.
Thus, electrostatic interactions introduce additional forces Fe.stat.
i = − ∑ other proteins∑ j ∂Ve.stat(dij) ∂Ri , (3.17) 3.5 protein-protein interactions 37 the sum of Coulomb interactions with all other residues of dif-ferent proteins. To speed up simulations, only residues closer than a cutoff distance rele = 160Å are interacting via electro-static interactions in our simulations.
3.5.2 Soft sphere potentials A soft sphere potential is used if residues come close to each other. These short-ranged interactions lead to a repulsion of the proteins that prevent protrusion of different protein domains.
To describe such a behavior, we use the high temperature limit of the Lennard-Jones model Vsp(dij) = D σ dij !12 .
(3.18) For the EN model, similar types of interactions have been pre-viously considered by Takano et al. in Ref. . For the soft sphere potentials, we use similar parameters. Here, the strength of the potential D is given as D = 0.06Å2. The characteristic in-teraction distance is chosen as σ = 5Å.
Due to repulsive interactions between different residues, an additional force Fsp i = − ∑ other proteins∑ j ∂Vsp(dij) ∂Ri (3.19) is introduced. The interaction is short range and, thus, we in-troduce a numerical cutoff distance of rsp c = 10Å. Note that the soft sphere potentials (3.18) even decay much earlier.
3.5.3 Protein Binding A protein may attach to a different one and form stable bonds.
Such bonds may be hydrogen bonds, salt bridges or other non-covalent bonds. Protein-protein complex can be very strong.
Well-known examples are the actin filament, comprised out of many actin subunits, or the actin-myosin complex.
We model such interactions by Lennard-Jones type potentials Vinter(dij) = CijDe req ij dij !12 −2 req ij dij !6 (3.20) 38 mathematical methods with a depth of the potential De = 0.2 Å2. The matrix elements Cij = 1 if the nodes i and j interact and Cij = 0, otherwise.
Missing information about the equilibrium distance req ij or the interaction matrix Bij must be obtained from additional sources like experimental data or MD studies. For actin and myosin, details will be provided in Ch. 6.
Forces due to this protein-protein interaction are given as Finter i = −∑ j ∂Vinter(dij) ∂Ri .
(3.21) Covalent and hydrogen bond lengths are relevant only on scale of several Å. This is also reflected by the fast decay of the Lennard-Jones type bonds. Here, we additionally introduce a cutoff of rpb c = 20Å to speed up the computation.
4 I N T R A M O L E C U L A R M E C H A N I C S O F G - A C T I N A major open question in the field of actin research is the role of the nucleotide. How are the interactions with ATP and sub-sequent hydrolysis related to the functionality of the protein?
It remains unclear, e.g., why ATP-bound actin is attached more stable to the filament than the ADP-bound monomer . In this chapter, we will study free monomeric actin (G-actin) and investigate ligand-induced conformational changes.
Systematic numerical investigations of conformational mo-tions in single actin molecules were performed by employing a simple EN model of this protein. We found that G-actin es-sentially behaves as a strain sensor, responding by well-defined domain motions to mechanical perturbations. Several sensitive residues within the NBP could be identified, such that the per-turbation of any of them can induce characteristic flattening of actin molecules and closing of the cleft between their two mobile domains. Extending the EN model by introduction of a set of breakable links which become effective only when two domains approach one another, it was observed that G-actin can possess a metastable state corresponding to a closed con-formation and that a transition to this state can be induced by appropriate perturbations in the NBP region. The ligands were roughly modeled as a single particle (ADP) or a dimer (ATP), which were placed inside the NBP and connected by elastic links to the neighbors. Our approximate analysis suggests that, when ATP is present, it stabilizes the closed conformation of actin. This may play an important role in the explanation why, in the presence of ATP, the polymerization process is highly ac-celerated.
4.1 g-actin model As a reference state for the construction of the elastic network, the uncomplexed G-actin in the ADP-bound state (PDB ID: 1J6Z) was used . For comparison, the elastic network of the F-actin model (PDB ID: 3MFP) , obtained by fitting to cryo-39 40 intramolecular mechanics of g-actin S4 S2 S1 S3 NBP Figure 13: Actin and its elastic network: (A) G-actin in the ribbon representation, colored according to its subdomains S1 (or-ange), S2 (yellow), S3 (blue) and S4 (green). The bound ADP molecule (red) is shown. (B) The elastic network of G-actin, colored in the same way. Magenta dotted lines indicate Lennard-Jones bonds between some residues (also marked magenta) in the subdomains S2 and S4. The nucleotide-binding pocket (NBP) is schematically displayed.
electron microscopy data, was chosen. The G-actin data con-sists of 372 residues divided into two major domains, known as the outer and the inner domains. They are separated by a cleft in which the nucleotide binds. Traditionally, each of them is further divided into two subdomains . The outer domain contains subdomains 1 (S1, residues 1–32, 70–144 and 338–372) and 2 (S2, residues 33–69). Part of S2 is the DB loop, playing an important role in inter-subunit binding. The inner domain con-sists of subdomains 3 (S3, residues 145–180 and 270–337) and 4 (S4, residues 181–269). Fig. 13 displays equilibrium structures of G- actin together with the elastic network of this protein with the cutoff length l0 = 8.5Å. Furthermore, the subdomain structure of the actin monomer is shown in panel (a).
The distance L24 characterizes the scissor-like motion of the two mobile domains, while the angle θ provides a characteri-zation of their scissor-like motion (see below). The three cho-sen order parameters show large variation when experimen-tally known conformations of G- and F-actin are compared. It should be noted that this order parameters agree with the dy-namical variables employed in the coarse-grained four-domain description of the actin filament by Chu and Both .
4.2 domain motions and metastable states 41 L24 A θ B Figure 14: Subdomain distances and dihedral angle. (A) Center of masses of subdomains S1 (orange), S2 (yellow), S3 (blue) and S4 (green) are shown as beads. (B) The dihedral angle θ is defined as the angle between the two planes between the center of masses of S1, S2 and S3, and the center of masses of S2, S3 and S4, respectively.
In our simulations, positions of all residues have been deter-mined at each integration step and, therefore, complete infor-mation about conformational motions was available. This full data has been used, e.g., when conformational snapshots were constructed or videos of characteristic conformational motions were generated. For concise characterization, we have addition-ally used a set of three order parameters traced in the simula-tions. Specifically, distances L13 and L24 between the centers of mass of S1 and S3 and the centers of mass of S2 and S4, respec-tively, were chosen. As the third order parameter, the dihedral angle θ, i.e. the angle between the plane defined by the mass centers of S1, S2 and S3 and the plane defined by the mass centers of S1, S3 and S4, was taken. Subdomain distances and dihedral angle are shown in Fig. 14.
4.2 domain motions and metastable states Generally, application of a static force induces rigid trans-lations and rotations of the entire protein. To eliminate such effects, additional balancing forces were computed at each inte-42 intramolecular mechanics of g-actin -15 -10 -5 0 5 10 15 20 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 -0.02 -0.015 -0.01 -0.005 0 0.005 0.01 0.015-0.15 -0.1 -0.05 0 0.05 0.1 0.15 -15 -10 -5 0 5 10 15 20 Figure 15: Responses to global perturbations. 100 relaxation trajec-tories (red curves) start from different initial conditions, generated by application of random, globally distributed static forces. Along the axes, relative changes ∆L13/L(0) 13 and ∆L24/L(0) 24 and absolute changes of the angle ∆θ are varied.
gration step and applied to the network. They were chosen in such a way that only global translations and rotations could be caused - and thus compensated - by them, without any inter-nal deformations arising. This immobilization is explained in depth in Ch. 3 where its detailed description can also be found.
We have always used it in the presence of external forces in our current investigations.
To generate static forces, for each residue a direction was randomly chosen and the force magnitude was randomly se-lected from the interval between 0Å and 0.09Å. Such indepen-dently generated random forces were applied to all network nodes and a new stationary configuration of the network in the presence of the forces was determined by integrating for suffi-ciently long time the equations of motion (3.3), until a station-ary state in the presence of forces was reached. Subsequently, the forces were lifted and a conformational relaxation process was followed by integrating the same equations.
Figure 15 displays results of such simulations for 100 differ-ent choices of random forces and, thus, for 100 different relax-ation trajectories. The initial positions of the trajectories corre-spond to the stationary states of the network in the presence of random external forces. Hence, they characterize conforma-tional responses of the network. As we see, the distance L24 be-tween subdomains S2 and S4 can change considerably, i.e. up 4.2 domain motions and metastable states 43 -15 -10 -5 0 5 10 15 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 -15 -10 -5 0 5 10 15 -0.15 -0.1 -0.05 0 0.05 0.1 0.15 Figure 16: Responses to global perturbations (A) and to local pertuba-tions of sensitive residues in the NBP region (B) are shown in the presence of breakable bonds. 100 relaxation trajec-tories (red curves) start from random initial conditions.
Along the axes, relative changes of the distance L24 and absolute changes of the dihedral angle θ are varied.
to 15%, and the dihedral angle θ between the inner and outer domains can undergo variation up to 15 degrees. Significant changes in the distance L13 between the lower subdomains S1 and S3 were not found in our simulations. The experimentally observed difference of about 13% in the distance L13 in F-actin, as compared to the equilibrium state of G-actin, can be a con-sequence of the interactions between monomers in the actin filament.
When external forces were switched off, the network was un-dergoing relaxation back to its equilibrium conformation. The relaxation trajectories starting from different initial deformed states are displayed in Fig. 15. The end points of the trajectories (green dots) correspond to the finally reached states the far-thest points of the trajectories represent the starting positions.
There are only two such end points in Fig. 15. One of them lies in the origin of coordinates and thus corresponds to the equi-librium state of the elastic network. We have checked that the second state corresponds to a small buckling of a single residue within the flexible part of subdomain S4 and thus represents only a slight local modification of the equilibrium state. Even after relatively large deformations the network always returns to the equilibrium state or to its slight modification.
Thus, we see that the two principal mobile domains of actin are able to perform large-magnitude motions characterized by substantial changes of the distances between S2 and S4, as well 44 intramolecular mechanics of g-actin as of the dihedral angle between the two mobile domains. The displacements of residues, accompanying such motions, are large and the linearized description and the normal-mode analysis are not justified in this case (cf. the discussion in Refs. and ). Nonetheless, such approximate descriptions can be still employed for qualitative understanding and interpretation of the observed motions.
We have computed the normal modes of G-actin in the frame-work of the EN approximation used in the present study (see Supplementary Movies in Ref. ). The slowest normal mode of the elastic network corresponds to the propeller-like twist of the two mobile domains, which can be well characterized by the dihedral angle (see Movie S1 in Ref. ). The second slow-est normal mode represents the scissor-like opening or closing of the two domains, as seen in Movie S2 in Ref. . These characteristic motions have been previously identified by Tirion and ben-Avraham in the framework of a different normal-mode analysis, where all actin atoms were resolved and only angle variations of the bond angles were taken into account. It is remarkable that they are also reproduced in a much simpler EN model and, furthermore, are approximately retained on the much larger scales.
Our analysis based on full nonlinear equations of the elastic network has indicated that, for some large-amplitude motions, the cleft separating subdomains S2 and S4 almost disappeared, so that the residues belonging to opposite domains could come near one to another. In such situations, the EN model needs however to be modified, as explained below.
When an elastic network for a protein is constructed, dis-tances between all pairs of residues in the equilibrium reference state are checked and elastic links are introduced whenever the distance between a pair is shorter than the cut-off length. Sup-pose now that some residues, well separated at equilibrium, come close when a perturbation is applied. If we want to fol-low the concept of the EN approximation, additional links con-necting such residues would need to be introduced once they come close one to another. Such emergent (and breakable) links cannot be elastic, instead they should be described by a pair in-teraction potential which becomes flat as the distance between the particles increases. Hence, they would be effectively present only when the two particles are close one to another - and would disappear when the particles are far apart.
4.2 domain motions and metastable states 45 To allow such bonds to emerge, we add into the EN model a set of breakable links between those G-actin residues from op-posite domains which are connected by elastic links in the EN model of F-actin. Thus, the original EN model of G-actin is ex-tended by us through the introduction of five additional break-able links connecting residues 62–204, 63–203, 63–204, 66–203 and 67–203. The new links are described by truncated Lennard-Jones potential (3.11).
Comparing the experimentally known conformations of the globular G-actin and the filamentous F-actin [56, 126], one can notice that, under the cutoff length of l0 = 8.5Å used in the EN model, there are eleven additional links between the sub-domains S2 and S4 in the F-actin structure. Generally, break-able links may be introduced between all such eleven residue pairs. We have selected, however, only the five closest pairs of residues and introduced breakable links between them. We have chosen the same value of the equilibrium distance req = 5Å for all additional links. With the choice of D = 1Å2 the in-teraction potential (3.12) near the equilibrium distance was by a factor of 2.88 stronger than that of the regular elastic links.
By choosing D in this way, we could approximately compen-sate for the smaller number of interacting pairs in our model as compared to the experimental F-actin structure.
Taking the expanded EN model, global mechanical responses of the elastic network were examined and its relaxation trajec-tories were explored using the same procedure as described above for the original network. The results are displayed in Fig. 16A. Not surprisingly, the relaxation behavior remains es-sentially the same in the neighborhood of the equilibrium ref-erence state of G-actin, which defines the origin of coordinates.
However, an important change is observed in the region cor-responding to the closed actin conformations. Previously, such conformations could be easily visited in response to mechani-cal perturbations, but the network was always returning from them back to the equilibrium reference state (cf. Fig. 15). In contrast, the expanded EN model of G-actin possesses a new stationary closed state, stable with respect to sufficiently small perturbations. Its origin is clear: if the two mobile subdomains are brought close enough one to another, cross links connecting them are established and, thus, the closed actin conformation becomes locked.
Actually, not one, but two closed metastable states of actin can be discerned in Fig. 16A. A detailed examination of them 46 intramolecular mechanics of g-actin reveals that they differ only by local buckling in the flexible re-gion of subdomain S4, involving just a single residue, whereas the global domain configuration is essentially the same in both of them. As can be seen in Fig. 16A, the metastable state is char-acterized by a closed cleft between subdomains S2 and S4 and a smaller dihedral angle θ, i.e. by a flattening of the molecule.
Summarizing the results of our investigations, we conclude that large-amplitude propeller and scissor motions of the prin-cipal mobile domains can take place in the elastic network of G-actin. These motions are generic and the protein responds by them when random globally distributed perturbations are ap-plied. Moreover, we find that, in the expanded version of the EN model, the protein can also be found in the closed station-ary conformation which represents its new metastable state. A transition from the equilibrium reference conformation to this metastable state can be induced by applying appropriate per-turbations to the network nodes.
4.3 responses to perturbations in the nucleotide-binding pocket When nucleotides (ATP or ADP) are bound to actin, this leads to local mechanical perturbations in the NBP. This pocket is lo-cated at the bottom of the cleft separating subdomains S2 and S4 (see Fig. 13). It includes a number of residues which are iden-tified below. In the second part of our study, domain motions induced by application of static forces to individual residues in the NBP region were systematically probed.
Our attention was focused on the perturbations correspond-ing to the transition from the ADP- to the ATP-bound states of G-actin. The nucleotide-free state is less relevant in the context of actin polymerization and it was not analyzed here. As the ref-erence conformation, the ADP-bound state (PDB ID: 1J6Z) was always taken. When ATP is instead bound, this means that the phosphate Pi is additionally present in the pocket. Hence, only the residues in the neighborhood of phosphate should be di-rectly affected. They are residues 12-16 in the S-loop, residues 71-75 (with the methylated histidine at the position 73) in the H-loop, residues 155-160 in the G-loop, and residue 301 in the subdomain S3.
Our analysis was performed similar to the previous investi-gation for myosin-V . To probe the responses, static forces 4.3 responses to perturbations in the nucleotide-binding pocket 47 S-loop H-loop G-loop Pi 12 13 14 15 16 155 156 157 158 159 160 71 72 73 74 75 301 Figure 17: Residues in the neighborhood of the phosphate Pi (red cross) which belong to the three sensory loops G, H and S inside the NBP. Red beads indicate the sensitive residues.
with randomly generated orientations and a fixed magnitude were applied to an individual residue in the chosen set, equa-tions of motion (3.3) were integrated and pair distances be-tween the subdomains, as well as the dihedral angle, were de-termined in the new stationary state.
To probe mechanical sensitivity of each residue, 200 simula-tions have been performed for each probed residue. In each of these simulations, a static force with randomly generated ori-entations and the magnitude f = 4Å was generated and the subdomain distances L13, L24 and the dihedral angle θ were de-termined in the resulting stationary state. For each residue the maximum response over the ensemble of 200 realizations was taken to characterize the sensitivity of this particular residue with respect to a certain distance or the dihedral angle.
The results of the sensitivity analysis are presented in Ta-ble 1. As we see, the maximal induced changes of the distance L13 between subdomains S1 and S3 were always small and not essential. In contrast, both the distance L24 and the dihedral angle θ could change substantially when perturbations to cer-48 intramolecular mechanics of g-actin Table 1: Sensitivity of selected residues in the NBP region.
Residue ID ∆θ ∆L13/L(0) 13 ∆L24/L(0) 23 12 4.3 0.012 0.09 13 5.2 0.017 0.10 14 4.3 0.018 0.11 15 8.0 0.019 0.15 16 10.0 0.016 0.15 71 5.1 0.016 0.08 72 7.8 0.021 0.13 73 7.9 0.026 0.14 74 2.3 0.020 0.10 75 4.6 0.027 0.09 155 5.2 0.014 0.08 156 3.8 0.016 0.09 157 3.8 0.017 0.12 158 6.9 0.020 0.13 159 8.4 0.018 0.11 160 6.7 0.018 0.10 301 5.9 0.014 0.05 tain residues were applied. According to Table 1, the sensitive residues are 15, 16, 72, 73, 158 and 159. Applying forces of mag-nitude f = 4Å to such residues, dihedral angle changes of more than 6.9 degrees and relative domain distance L24 changes of more than 11% could be induced. The sensitive residues are additionally displayed in Fig. 17. Note that pairs of sensitive residues are located within each of the three important loops S, H and G.
In the above sensitivity analysis, the original EN model was employed. As we have shown in the previous section, this model can be, however, expanded by including a set of breakable links which become effective when subdomains S2 and S4 come close one to another. Domain motion responses to the application of forces to sensitive residues in the NBP region have been further analyzed in the framework of the expanded EN model.
For the detailed analysis, only one sensitive residue in each of the three loops was chosen. Similar behavior could be ex-pected if its neighbor in the same loop was instead selected.
Thus, we focused on the responses induced by application of perturbations to the group of three residues: 16 (in G-loop), 73 4.4 ligand-induced conformational changes 49 (in H-loop) and 159 (in G-loop). Static forces were applied, at the same time, to all three residues in the group. The magnitude of each force was randomly chosen between 0 and 2Å and its orientation was random. For every choice of forces, evolution equations (3.3) for the expanded EN model were integrated un-til a stationary state was reached. After that, the forces were lifted and the relaxation process was followed by integrating the same equations. The results are displayed for 100 different random perturbations in Fig. 16B.
Comparing Figs. 16A and 16B, it can be noticed that, although the forces were applied to only three NBP residues, essentially the same domain responses as for the application of globally distributed perturbations could be produced. The minor metastable states in Fig. 15, corresponding to single-residue buckling in the highly flexible region of subdomain S4 were absent because forces in that region were not applied. As we see, perturba-tions of the three sensitive residues already led to characteristic propeller- and scissor-like motions of the inner and outer do-mains. Furthermore, such local perturbations were sufficient to induce a transition to the metastable closed state of G-actin.
Thus, a small number of sensitive residues lying in the NBP re-gion and belonging to three different loops could be identified.
Applying appropriate static perturbations to a group of three such residues, each from a different group, a transition from the open to the closed state of G-actin could be reproduced.
Remarkably, local deformations in the NBP region were able to spread over the elastic network and become transformed into large-amplitude global motions of mobile domains.
4.4 ligand-induced conformational changes Since EN models are coarse-grained and entire residues are re-placed by point-like particles, the detailed atomic structure of ligands (ATP or ADP) cannot be resolved in this approach. In this section, the ligands are treated by picturing them as single par-ticles. As it turns out, even this greatly simplified phenomeno-logical description allows us to understand some important as-pects of ligand-induced conformational changes.
50 intramolecular mechanics of g-actin ADP 158 73 16 Pi ADP 336 301 302 303 304 305 156 157 181 182 183 210 213 214 Figure 18: Simple modeling of ligands. (A) The ADP is modeled as an additional node (purple bead) added to the elastic network.
It is connected to all its neighbors (grey beads) by elastic links. (B) The ATP is modeled as a dimer consisting of ADP (purple bead) and Pi (grey bead), connected by an elastic link. The ADP is elastically connected to its neighbors and the phosphate is elastically linked to the three sensitive nodes (red beads).
4.4.1 Ligand Model The structure of G-actin with ADP is experimentally known and it was already used by us to construct its elastic network. Be-low, ADP is explicitly included into the EN description. We treat it as a single particle and put this particle into the equilibrium C1′ position, connecting it by elastic links to all residues within the cutoff distance l0 (see Fig. 18A). The natural lengths of the links are chosen equal to the equilibrium distances between C1′ and the respective residues. Hence, by construction, the in-troduction of such a particle does not change the equilibrium conformation of the protein network. Because the particle is only connected to one of the subdomains (i.e. to S1), its intro-duction does not also significantly affect the dynamics of the mobile domains. The equilibrium state of the elastic-network of G-actin with the additional particle, modeling ADP, is shown in Fig. 19A.
When an ATP molecule is bound to actin, we model it as a dimer consisting of two particles (Fig. 18B). The first of them corresponds to the ADP part of ATP and the second of them imi-tates the Pi. The first particle is at the same position where ADP was located in the equilibrium conformation of G-actin. The second particle is placed in the center of mass of the residues 4.4 ligand-induced conformational changes 51 Figure 19: Snapshots of ligand-induced conformational motions. Ad-ditional bonds (magenta) are marked solid or dashed lines if they are established or broken, respectively. The ligand is colored red. (A) Equilibrium conformation of G-actin with ADP bound. (B) Model of the ATP-bound state. A tran-sition to the closed conformation is observed. (C) In the closed state, the Pi is removed. The elastic network stays in a metastable closed conformation.
16, 73 and 159, and the ADP. It is connected by elastic links to these four particles (see Ch. 3 for the detailed description).
In contrast to ADP, residing entirely on one of the mobile do-mains, the phosphate interacts with the residues from different mobile domains (cf. Fig. 17) and, thus, its arrival may induce rel-ative domain motions. Both the X-ray diffraction experiments andMD simulations reveal that, in the presence of ATP, the nucleotide binding pocket becomes contracted. To ap-proximately account for this effect, we assume that the natural lengths of the elastic links, which connect the Pi ligand particle to its neighbors, are shorter than the distances between them and the Pi ligand when it is introduced. Namely, the natural lengths of the elastic links, connecting Pi to residues 16, 73 and 159, are chosen to be equal to 20% of the distances between these residues and the Pi position (i.e., the center of mass of these three residues) in the reference state which corresponds to the equilibrium conformation of G-actin with ADP bound.
Thus, these links are initially stretched; they tend to contract the nucleotide-binding region.
Binding of the ATP, imitated in our simple phenomenologi-cal model through the introduction of an additional Pi ligand, leads to a shrinking of the NBP which translates into confor-mational motions of the inner and outer domains. The two do-52 intramolecular mechanics of g-actin -12 -10 -8 -6 -4 -2 0 2 4 -0.15 -0.1 -0.05 0 0.05 Figure 20: The pattern of relaxation trajectories for the ligand-network complex. The blue trajectory shows relaxation starting from the open equilibrium conformation of G-actin without the ligand. Other start from the perturbed conformations which were obtained by applying random static forces to the three sensitive residues in the NBP re-gion. The open conformation does not correspond to a sta-tionary state of the complex and all trajectories converge to the new equilibrium closed state indicated by the green dot.
mains approach one another, so that within the expanded EN model the additional links connecting them become effectively established and they lock the closed conformation of the pro-tein. This process is illustrated in the first part of the supple-mentary Movie S3. The final closed conformation of G-actin, stabilized by binding of ATP, is displayed in Fig. 19B.
The hydrolysis reaction and the release of phosphate are roughly imitated in our model by cutting all links which connect the Pi ligand to its three neighbors and the ADP and by removing this particle from the pocket. When this takes place, the actin is in its closed conformation shown in Fig. 19B. The removal of Pi changes the interactions within the NBP and, as we ob-serve in our numerical simulations (the second part of Movie S3), leads to a certain opening of the cleft between the two mo-bile domains. However, in absence of thermal fluctuations (see 4.4 ligand-induced conformational changes 53 below), the additional links between the two domains then do not break and, after the phosphate release, the actin is found in its metastable closed state (Fig. 19C).
Binding of the artificial ligand leads to a new, unique equi-librium position. 100 relaxation trajectories in the presence of the ligand are shown in Fig. 20. Initial deformations were pre-pared by applying static external forces with random directions and an amplitude drawn from the interval [0,2Å] to the three sensitive residues in the NBP. Starting from the equilibrium con-formation of G-actin, the equations of motions (3.3) were inte-grated in the presence of the ligand until a stationary state was reached. Additionally, Fig. 20 shows the relaxation trajectory which starts from the original equilibrium state of G-actin in the absence of a ligand. As revealed by Fig. 20, binding of the ligand makes the open conformation unstable and stabilizes the closed actin conformation.
It should be noted that the choice of the interaction parame-ters still remained to some extent arbitrary. Indeed, it was also possible to choose other values of the lengths req or to make such lengths dependent on a particular pair of residues. More-over, the interaction strength D could also be somewhat varied.
We have repeated some simulations with a smaller interaction strength of D = 0.5Å2. In such simulations, we have found that the closed conformations could still be often visited as a result of perturbations, but the stationary closed conformation did not exist. When the ligand was introduced, the open con-formation became unstable and the closed stable conformation emerged also in this case. Thus, the ligand-induced stabiliza-tion of the closed conformation of G-actin actin is a robust ef-fect, which is not sensitive to possible parameter variations for the breakable links. The exact structural details of the closed conformational state may be sensitive to the specific choice of the parameters. In absence of direct experimental data, we have not however tried to optimize this choice.
So far, effects of thermal fluctuations have been excluded from our analysis. Such effects may become, however, impor-tant if metastable states are possible. If thermal fluctuations are strong enough, they can induce transitions between stable and metastable states, so that all of them can be visited by the sys-tem.
54 intramolecular mechanics of g-actin 4.4.2 Effects of Thermal Fluctuations The effects of thermal fluctuations can be taken into account by introducing additional random forces with appropriate inten-sities into the equations of motions. Such thermal fluctuations are described by equations (3.14). The noise parameter σ = 2Å has been chosen. We have checked that, with this choice of pa-rameter, the experimentally known B-factors are reproduced by the orders of magnitude. Integrating such stochastic differen-tial equations over sufficiently long time, data was gathered and statistical distributions for various order parameters in the presence of different ligands (ADP or ATP) were constructed.
Figure 21 displays statistical distributions of the distance L24 between the centers of mass of the mobile domains S2 and S4 in the ADP- or ATP-bound states, as described by our approximate model. In the ADP-bound state (black curve), the protein prefers to stay in the open conformation, with the distance between the domains approximately equal to 31.0Å. The closed conforma-tion, representing a metastable state, is however also occasion-ally visited, as evidenced by the presence of a shoulder in the statistical distribution of the interdomain distances. Binding of ATP stabilizes the closed conformation, leading to the distance distribution shown by the red curve in Fig. 21. In the presence of ATP, spontaneous transitions to the open conformation are not possible (or very rare), as evidenced by the presence of a clear distribution maximum at the distance L24 = 27.5Å in this case. The width of the distance distributions characterizes the stiffness of the monomer. With ATP bound, the variance of the distance L24 is reduced to 0.45Å2, as compared to the variance of 1.44Å2 in the ADP-bound state. Thus, the presence of ATP in the NBP stiffens the monomer considerably.
Already the rough modeling employed in this section reveals some important effects of the nucleotides. Binding of ATP can directly lead to flattening of the protein and closing of the cleft between its inner and outer domains. While the ATP-free actin shows the tendency to switch between its two equilib-rium states, the ligand can stabilize the closed conformation of actin and, furthermore, stiffen the macromolecule. Note that the structural details of the ligand-induced closed conformational state may depend on the parameters of interactions between the ligand particle and the NBP residues. Moreover, the dimer model of ATP used in the above simulations represents only a simple approximation for the actual ATP molecule. Therefore, 4.5 discussion 55 20 25 30 35 0.0 0.2 0.4 0.6 L24 density Figure 21: Statistical distributions of interdomain distance L24 in the presence of ADP (black) or ATP (red) ligands.
the results of our numerical investigations including the ligand should be viewed as only providing a demonstration that a transition to stable closed conformation can be induced by ATP binding. This model predictions need to be further confirmed in the experiments and in special MD simulations.
4.5 discussion In our study, the attention was focused on purely mechanical aspects of actin dynamics. With this purpose, a greatly simpli-fied dynamical model of this molecule was considered where all residues, independent of their chemical differences, were treated as identical particles connected by identical elastic links.
In addition to the elastic links, the mechanical model also in-cluded a small number of breakable links, which become estab-lished when pairs of residues come sufficiently close and break down at large separations. The information about the chemical structure of G-actin was effectively encoded only in the architec-ture of the elastic network, determined by the experimentally known equilibrium conformation of the molecule.
Remarkably, already this greatly simplified model allowed us to understand many aspects of the intramolecular conforma-56 intramolecular mechanics of g-actin tional motions in actin monomers. The model shows that the mobile inner and outer domains of actin are able to perform large-amplitude propeller twist and scissor-like motions, ear-lier revealed by the normal-mode analysis for small deviations from the equilibrium state . While performing such mo-tions, two upper subdomains (S2 and S4) can come so close one to another that attractive interactions between pair of residues from the opposite domains become present. Such emergent in-teractions lock the actin molecule in its closed conformation and thus lead to the formation of a metastable state.
We have found that, similar to myosin [46, 80, 123, 122], G-actin essentially behaves as a strain sensor, responding by well-defined domain motions to mechanical perturbations. In our study of actin, we identified a number of sensitive residues, such that small perturbations of these residues were translated into large-amplitude motions. Our investigations reveal three pairs of sensitive residues, belonging to different domain loops inside the NBP. Application of small perturbations to these par-ticular residues can result in large-amplitude domain motions and in the transition to the metastable closed state. As we see, the internal mechanics of an actin macromolecule is highly or-ganized and efficient communication between the NBP region and the mobile domains is present.
To demonstrate that ligand (i.e. ATP) binding can indeed in-duce large-scale conformational changes, we have imitated the ligand by a dimer; one of the particles, corresponding to phos-phate, has attractive interactions with the sensitive residues in-side the NBP. Previously, a similar ligand description was em-ployed when cyclic operation of the molecular motor hepatitis C virus helicase was analyzed . We have found that, under an appropriate choice of the interaction parameters, binding of ATP can induce a transition to the closed conformation and sta-bilize this metastable state.
In the hierarchy of coarse-grained models proposed to de-scribe actin monomers and filaments (see, e.g., review ), the employed description is most closely resolving the struc-ture of the individual proteins. Nonetheless, because of the sim-plifications involved in the formulation of the EN model and since some of the parameters, particularly referring to the inter-actions with ligands, remained arbitrary in the present study, quantitative agreement between the predictions based on the present coarse-grained description and the experimental data or the data of all-atom MD simulations should not be expected.
4.5 discussion 57 Nonetheless, the results of our approximate analysis can be used for better understanding of intramolecular dynamics of G-actin. They may provide helpful guidelines for further exper-imental investigations and act as motivation for MD studies.
The ATP-induced transition to the closed conformation of actin may play an important role in the explanation why, in the pres-ence of ATP, the growth of actin filaments is strongly acceler-ated. The closed conformation of G-actin, stabilized under ATP binding, is not identical to that of the filamentous F-actin. How-ever, in both of these conformations the cleft separating the up-per mobile subdomains S2 and S4 is strongly reduced, so that a better fit and higher affinity to the actin filament may result.
Another effect of binding of ATP observed in our model is the in-creased stiffness compared to the ADP state (see Fig. 20). This is in agreement with the experimental data showing that the pres-ence of ATP-bound protomers leads to an increased stiffness of the filaments .
In recent experiments , metastable conformational states of single actin protomers in the filament could be already de-tected. The distribution of these states was sensitive to addition of myosin. Actin binding proteins (ABPs), including myosin, play crucial roles in the cell . In the framework of our approach, interactions with actin binding protein (ABP)s can be interpreted as mechanical perturbations and can also be analyzed in future studies. It should also be possible to per-form FRET measurements in single G-actin molecules under controlled conditions, thus elucidating conformational states in-volved in polymerization (G-F transitions), and specifically, the effects of ligands. The results of such experiments can be com-pared with the predictions based on the elastic-network mod-els.
5 M Y O S I N - V A S A M E C H A N I C A L S E N S O R Even more than 50 years after Huxley proposed the first mech-anism for muscle myosin force generation [76, 77], the details of how the molecular motor functions remain unclear. Is the chemical reaction with ATP used to perform mechanical work directly in form of a lever-arm swing or does it regulate a recti-fication of thermal noise in order to perform work?
In this chapter, we investigate the internal dynamics of a sin-gle myosin and identify ways in which myosin may control its machine cycle or bias thermal fluctuations. According to exper-iments [80, 122, 123], the molecular-motor myosin behaves like a strain sensor, exhibiting different functional responses when loads in opposite directions are applied to its tail. Within an EN model, we explore the sensitivity of the protein to the forces acting on the tail and find, in agreement with experiments, that such forces invoke conformational changes that should affect filament binding and ADP release. Furthermore, conformational responses of myosin to the application of forces to individual residues in its principal functional regions are systematically investigated and a detailed sensitivity map of myosin-V is thus obtained. The results suggest that the strain-sensor behavior is involved in the intrinsic operation of this molecular motor.
5.1 myosin-v model The study is based on the elastic-network modeling of con-formational dynamics in the presence of external forces. The EN model of myosin-V heavy chain is constructed by using as a reference state its post-rigor equilibrium conformation with Mg·ADP·BeFx, an ATP-analog, as yielded by X-ray diffraction analysis (Protein Data Bank ID: 1W7J, chain A) . Myosin in this state does not attach strongly to the filament. The myosin-V experimental data and the corresponding EN with a cutoff of 10Å are shown in Fig. 22.
The entire elastic network of myosin-V includes 752 particles.
For probing of mechanical responses, a subset of 82 particles 59 60 myosin-v as a mechanical sensor TAIL REGION NBP ACTIN-BINDING CLEFT HCM loop converter front door back door L50kDa U50kDa A B Figure 22: Myosin-V (A) and its elastic network (B). In panel (a), three principal functional regions of the protein are schemat-ically shown; the ATP-analog is also displayed (orange).
Moreover, some important structural elements, such as the front and the back doors and the HCM loop, are also in-dicated here. In panel (B), each particle corresponds to a residue, the links represent elastic interactions between them.
was selected (see Fig. 23A). These residues are found in the three important functional regions, the tail, the NBP and the actin-binding cleft. The residue 792 was located in the tail re-gion and used to apply forces to the tail. The second group of residues from the NBP region is displayed in the first column in Table 3 in the Appendix. All residues in this group are ATP neighbors, i.e. they are adjacent to the ATP position. The third group (first column in Table 4 in the Appendix) included the residues located on the surface of the actin-binding cleft.
Although integration of equations (3.3) gave positions of all particles in the state with a force applied, this information was too detailed and to quantify induced deformations some mea-sures (or “order parameters”, cf. ) were needed. In our analysis, we chose a set of 10 particles as the labels and moni-5.1 myosin-v model 61 TAIL REGION NBP ACTIN-BINDING CLEFT 789 115 343 386 517 442 219 141 297 92 A B Figure 23: (A) The set of residues probed by application of static forces and (B) the labels and the distances between them, used to monitor conformational changes.
tored pair distances between them. The labels and the selected distances are shown in Fig. 23B.
The first three labels, corresponding to residues 789, 141, and 92, were selected to specify motion of the tail. Here, the first label is in the tail and the other two are attached to stiff parts of the protein.
Further, four labels were used to measure conformational changes in the NBP region. The labels, corresponding to residue 115 in the N-terminal subdomain and residue 297 in the upper 50kDa subdomain, were chosen to characterize the opening of the front door. They are in contact with the adenylate ring, gat-ing the front door through which ATP enters and ADP leaves the reaction site . The back-door opening and closing were characterized by the distance between residues 219 and 442 in switches 1 and 2 in the upper 50kDa subdomain, respectively.
These two amino acids form a salt bridge which hinders the phosphate from leaving the NBP after hydrolysis [21, 100].
62 myosin-v as a mechanical sensor The last three residues belonged to the actin-binding region.
Distances between residue 517 in the lower 50kDa and residues 343 and 386 in the upper 50kDa subdomain were monitored.
These labels lie in the actin-binding site and characterize open-ing and closing of the cleft, as well as the motion of the HCM loop.
To examine responses of the network, static forces were ap-plied to one of the chosen particles and, after integration of equations (3.3), changes of the distances between the labels with respect to their equilibrium values were determined. In such simulations, the immobilization procedure presented in Ch. 3 has always been used and, hence, only internal dynamics of the protein has been measured.
5.2 forces acting on the tail In this section, we intend to computationally reproduce the experiments [80, 122, 123] where external forces were acting on the myosin tail. We apply forces to one residue in the tail and monitor the conformational responses in the nucleotide-binding pocket and in the actin-binding cleft, depending on the force orientation and amplitude.
Experimentally, forward and backward force directions are defined with respect to the direction of the proccessive motion of myosin along the actin filament. To determine filament ori-entation with respect to myosin in the considered model, a spe-cial analysis had to be performed. First, we have built a fila-ment by using the structural data for F-actin (PDB ID: 2ZWH) . Then, we needed to determine how the filament was posi-tioned with respect to myosin in the actin-myosin complex. To do this, the results of the recent guided MD simulations , where several sites for binding of myosin to actin were identi-fied, were used. By keeping the filament stiff, we anchored it to the myosin network by elastic links at these sites. After that, relaxation in the actomyosin system was followed by integra-tion of dynamical equations and the final equilibrium config-uration of the complex was determined. Thus, the relative fil-ament direction, used in subsequent numerical investigations, could be identified. A more detailed description is given in the Appendix.
5.2 forces acting on the tail 63 0 -1 0 forward strain backward strain -0.5 1 0.5 3 6 -3 -6 force (Å) distance change (Å) Figure 24: Responses of pair distances between the labels (black) 115 and 297, (red) 219 and 442, (blue) 343 and 517, and (green) 386 and 517, as functions of the amplitude of the force ap-plied to the tail. Small abrupt changes observed at large negative forces are due to local buckling effects.
In the simulations, external forces have been always applied to the residue 792. Since the tail is a relatively stiff structure, the exact position of the force application point was not important.
The forces were parallel to the filament direction, which has been previously identified. Their amplitudes were varied from 0 to 6.5Å and both forward and backward directions have been considered. For each force, equations (3.3) were numerically in-tegrated until a stationary protein configuration was achieved.
The responses of the elastic network are displayed in Fig. 24.
As we see, changes in the nucleotide-binding pocket are in-duced. The distance between residues 115 and 297, defining the opening of the front door, grows under forward strain and decreases if backward forces are applied. However, the change of the pair distance between residues 219 and 442, characteriz-ing the salt bridge, remains smaller than 0.1Å. Hence, we con-clude that, while the front door configuration is controlled by the force acting on the tail, the responses in the back door re-gion of the NBP are negligible.
64 myosin-v as a mechanical sensor As evidenced by Fig. 24, external forces furthermore lead to conformational changes in the actin-binding region. When backward strain is applied, the actin cleft, characterized by the distance between residues 343 and 517, closes. At the same time, the distance between residues 386 and 517 increases and, thus, the HCM loop moves away from the lower 50kDa domain. The opposite conformational changes are observed when forward strain is applied.
The significance of the observed responses becomes clear if the operation mode of myosin is taken into account. ADP and Pi, the hydrolysis products, are exiting the nucleotide pocket in different directions, through the front and the back door, re-spectively. Therefore, closing of the front door, which, as we have found, is induced by backward strains, would hinder the ADP release and thus enhance the ADP affinity. This agrees with the experiments [122, 123]. In these experiments by Oguchi et al. the dependence of the ADP affinity on the force direction has been observed. Movie S1 in the Supporting Material of Ref. displays the dynamical response of the network to an external force applied in the forward direction. The induced tail motion and the pronounced changes in the actin binding cleft region are clearly seen there. Changes in the front door, however, are weaker and less apparent.
The effects of strains on binding to the actin filament have been investigated in the experiments by Iwaki et al. . They showed that backward strain increases the probability that myosin binds strongly to the filament. The original interpretation of these results in Ref. was that such forces induce opening of the back door. This, in turn, would enhance Pi release that should precede strong binding. Thus, an indirect connection between binding affinity and strain to the tail. Our simulations, however, do not reveal any considerable sensitivity of the back door to the forces acting on the tail, so that this original inter-pretation is not supported. Nonetheless, the experimentally ob-served behavior can still be understood. For that, the responses induced in the actin binding cleft should be analyzed.
Conformational changes accompanying a transition to strong binding of myosin to the actin filament can de identified by comparing the structures in the post-rigor state with the fil-ament detached and in the nucleotide-free (or rigor) state of the same myosin (PDB ID: 1OE9) which corresponds to the conformation of myosin bound to the filament. Focusing on the changes in the actin binding cleft, one can notice that 5.3 forces in the nucleotide-binding pocket 65 the transition from the post-rigor to the rigor state is charac-terized by shortening of the distance between residues 343 and 517 and, at the same time, by an increase of the distance be-tween the HCM loop and the lower 50kDa domain (character-ized by residues 386 and 517). But these are exactly the changes which we found to be induced by the application of backward strain. Hence, the experimental data is explained by con-formational changes which are directly invoked in the actin binding cleft, rather than being mediated through hypothetical responses in the back-door region of the nucleotide pocket.
5.3 forces in the nucleotide-binding pocket The nucleotide-binding pocket is an important functional re-gion of myosin. Here, the chemical reaction takes place that fuels the myosin machine cycle. It is known that ATP enters the pocket from the front door and, after the hydrolysis, the ADP product leaves through the same opening. In contrast, the phosphate, representing the second product, leaves the NBP re-gion through the back door. The processes taking place in the NBP affect, in turn, filament binding and tail motions.
Responses to perturbations localized in the NBP have earlier been investigated theoretically by using restrained targeted MD simulations , and by means of the normal mode analysis . In these previous studies, it has been demonstrated that global conformational transitions from the nucleotide-free to the ATP-bound states could be reproduced by applying special forces to a group of residues inside the NBP, with the direc-tions chosen to correspond to the experimentally known local changes between the two conformations.
The approach we pursued is different. Our aim was not to re-produce a particular known response, but, instead, to systemat-ically test the global sensitivity of the protein to arbitrary forces applied to various single residues in the nucleotide-binding re-gion. In this manner, we aimed to construct a map of the NBP in which residues responsible for particular functional behavior could be identified.
To probe mechanical responses, 27 residues adjacent to the ATP in the considered equilibrium conformation were selected (left column Table 3 and Fig. 23A). For every chosen residue, a series of 200 simulations was performed. In all simulations, the 66 myosin-v as a mechanical sensor magnitude of the applied force was the same (f0 = 1 Å), but its orientations were randomly varied. The simulations were al-ways continued until a stationary state was found. After that, changes in the monitored pair distances between the labels in different regions were determined. The induced distance changes were analyzed and, for each pair of labels, the maxi-mum absolute distance change over 200 force orientations was evaluated. Such maximum distance change was then taken as the measure of the protein’s sensitivity to forces applied to a particular residue. An example of the relaxation process lead-ing to a new stationary state is shown in Movie S2 in Ref. .
Here, forces were applied to residue 115 in the front door and conformational changes, especially in the tail, are clearly visi-ble.
The sensitivities of all 27 residues with respect to different pair distances are summarized in Table 3 in the Appendix and the directions of communication are schematically shown in Fig. 25.
As we have found, applying forces to the residues in the front-door region, a strong effect on the tail is produced (Fig. 25).
However, the application of forces to residues near the back door or the P-loop only weakly affects the tail region. Thus, the tail responds mainly to the perturbations applied at the entrance of the front door.
Proceeding further, the sensitivity with respect to responses in the actin-binding cleft has been investigated. As we have ob-served (see Appendix), the actin-binding region is mainly affected by the application of forces in the back-door region. When the force is applied to residue 442 in the back door, this produces a strong response in the distance between residues 343 and 517, charac-terizing the cleft opening. Moreover, application of forces in the back door has a pronounced effect on the HCM loop to which residue 386 belongs. The HCM loop is known to come into con-tact with actin and may play an important role in myosin-actin interactions [32, 156, 171]. It should however be noted that some sensitivity of the actin-binding cleft to the perturbations in the front door has also been seen.
Thus, we have found that myosin behaves as a strain sensor not only with respect to the forces applied to its tail. The forces acting to some residues within the NBP are also producing well-defined conformational responses, which are functionally im-portant.
5.3 forces in the nucleotide-binding pocket 67 distance change (Å) force (Å) 0 1 2 3 4 5 6 -1 -2 -3 -4 -5 -6 0 0.4 1.0 0.2 0.6 0.8 1.2 1.4 -0.2 -0.4 -0.6 -0.8 -1.0 -1.2 -1.4 Figure 25: Sensitivity of residues in the NBP region. Directions of com-munication are indicated by colored arrows. The region of the nucleotide-binding pocket marked by a box is enlarged.
Residues most sensitive with respect to the tail motion and to conformational changes in the actin cleft are colored red.
The responses induced by the forces applied at the gray colored residues are weaker.
We have furthermore discovered that, in terms of the sensi-tivity of its residues, the NBP region is clearly divided into a front-door and a back-door domain (enlargement in Fig. 25).
On the one hand, a pronounced effect on the actin cleft was observed when forces were applied to residues 219, 220 and 442 in the back-door region. Remarkably, it is exactly the salt bridge between the two sensitive residues 219 and 442 that hin-ders phosphate release after the hydrolysis. On the other hand, we have seen that the forces applied at the front door were pri-marily affecting the tail and could induce its motions. We have found that the forces applied to the residues of the P-loop re-gion, lying in the middle between the front and the back doors, do not lead to significant conformational changes either in the tail or in the actin cleft.
68 myosin-v as a mechanical sensor Importantly, the response of the actin cleft to perturbations in the NBP can be controlled by the forces applied to the tail.
Fig. 26 shows the change of the distance between the residues 343 and 517, characterizing the width of the actin cleft, as de-pendent on the force applied to residue 442 in the back door.
The force was chosen in the direction that maximizes the abso-lute distance change and its amplitude was varied. Addition-ally, analogous dependences in the presence of forward and backward strain in the tail are shown in the figure. One can see that backward strain enhances the closing of the actin cleft, whereas forward strain tends to prevent it. Thus, while the forces applied at the tail do not directly affect phosphate re-lease through the back door, they can nonetheless modulate the effects of phosphate release on the opening or closing of the actin cleft, which controls interactions of myosin with the filament.
5.4 forces in the actin-cleft region When myosin binds to an actin filament, this should lead to perturbations localized in the actin-binding cleft region. Such perturbations should have a specific form, determined by the organization of the actin-myosin interface and details of the myosin-actin interactions. As in the previous section, our aim is not to reproduce the responses to a particular local pertur-bation. Instead, we systematically investigate in this section the sensitivity of the protein to the application of forces with ar-bitrary orientations to individual residues in the actin-binding cleft region and look how strains in the cleft can affect the tail or the front and back doors.
High-resolution experimental data of myosin bound to the actin filament is currently not available. Based on cryo-EM ex-periments, models for the actomyosin interface were proposed [71, 73, 105]. Lorenz and Holmes suggested possible elec-trostatic interactions and hydrogen bonds between myosin-II and actin. Using these results, we have identified a set of 54 residues which may come in contact with the filament (left col-umn Table 4 and Fig. 23A). To probe sensitivities in the actin cleft, 200 independent simulations were performed for each sin-gle residue in the set. Forces with strength f0 = 1 Å and ran-domly chosen orientation were applied to a given residue and 5.4 forces in the actin-cleft region 69 front door back door 115 116 442 219 220 Figure 26: Changes in the distance between residues 343 and 517, characterizing opening of the actin-binding cleft, as func-tions of the force applied to the residue 442 in the back-door region. The red curve corresponds to the forward force of 5 Å applied to the tail. The green curve is for the force with the same strength applied in the backward direc-tion. The blue curve corresponds to the absence of forces.
integration was continued until the system relaxed to a new equilibrium. After that, changes in the characteristic pair dis-tances between the labels were determined. For each residue and the pair distance, the maximum response over the series of 200 simulations was recorded. The determined sensitivities are shown in Table 4 and systematically analyzed in the Appendix.
The results are also schematically represented in Fig. 27 where arrows indicate the principal communication pathways. Strong responses of the tail were induced when forces were applied to the residues in the HCM loop (see, for example, Movie S3 in Ref. ). There is also some sensitivity of the tail to per-turbations applied in the upper 50 kDa domain (residues 340 to 350). In contrast, the tail is not significantly sensitive to the forces acting in the lower 50 kDa domain. Examining responses 70 myosin-v as a mechanical sensor HCM Loop L50kDa U50kDa Figure 27: Responses induced by the application of forces to the residues in the actin binding cleft region: (red) residues 377 to 390 in the HCM loop, residues 340 to 350 in the up-per 50 kDa subdomain, residues 540 to 544 in the lower 50 kDa subdomain, and (gray) less sensitive residues in the lower 50 kDa subdomain. Arrows schematically indicate the directions and strength of intramolecular communica-tion.
in the NBP region, we observed that the front door is strongly sensitive to the forces applied in the upper 50 kDa domain and the HCM loop. The back door is sensitive to the perturbations acting in the lower 50 kDa domain.
5.5 discussion Functional responses of myosin motors to the application of ex-ternal forces have previously been experimentally investigated.
Iwaki et al. attached a bead to the tail of a myosin-VI monomer and dragged it along the actin filament in two oppo-site directions. It was found that the transition of myosin from weak to strong binding to the filament is affected by application 5.5 discussion 71 of a dragging force and strongly depends on the force direc-tion. The probability of strong binding increases when forces are applied in the direction opposite to that of the intrinsic pro-cessive motion (i.e., the backward direction for this molecular motor). In other experiments, effects of external loads on the ADP release from myosin-V attached to the actin filament were studied [122, 123, 135, 169]. It has been found that the release is hindered when forces are applied in the backward direction, opposite to the direction of processive motion for this myosin [135, 169]. Oguchi et al. [122, 123] have measured the magni-tudes of the force applied to the tail which are needed to un-bind a myosin-V monomer from an actin filament for different force directions and ADP concentrations. These experiments in-dicated that ADP release is slowed down when a force is applied in the backward direction (see also [94, 144]).
Our study provides a theoretical explanation of these results.
We have observed a direct effect of external forces on the front door, through which ADP leaves the protein. When backward strain is applied to the tail, this tends to close the front door and, accordingly, should make the ADP release less probable. Thus, external forces can regulate the ADP affinity through conforma-tional changes in the front-door region. We have not found a direct effect of the strains on the back door, through which the phosphate leaves the protein leading to filament attachment.
However, we have observed that strain directly influences the protein conformation in the actin-binding cleft region. Back-ward strains tend to favor closing of the actin cleft and to gen-erate a movement of the HCM loop away from the lower 50 kDa domain, thus enhancing the effects of phosphate release through the back door (see Fig. 24) and making strong coupling more probable. In this way, an explanation of the experiments is also provided.
The force-dependent release of ADP is important for the op-eration of the dimeric myosin-V, i.e. for its processive motion along the actin filament [113, 181]. Recent theoretical investiga-tions using a reduced kinetic description of the cycles of the myosin-V dimer suggest that the transition rates of ATP and ADP binding to the leading head need to be load-dependent . Thus, the strain-sensor behavior of myosin-V, revealed in the experiments with the application of external forces to its tail and explained in the present study, should play a profound role in natural functioning of this molecular motor.
72 myosin-v as a mechanical sensor Moreover, we have systematically analyzed conformational responses of myosin-V to the application of forces to individual residues in the nucleotide-binding pocket and on the surface of the actin-binding cleft. Thus, a detailed intrinsic sensitivity map of myosin has been constructed. As we have seen, by perturbing specific residues in the front- or back-door regions of the NBP, motions of the tail or opening and closure of the actin cleft can be induced. On the other hand, application of forces to residues in the actin-binding cleft can also invoke tail motions and have effects in the NBP region. In our numerical study, the responses have been induced by external forces. Under natural conditions, however, mechanical perturbations in the NBP and the actin-binding cleft can easily arise as a consequence of ligand binding and reaction events.
Remarkably, both the explanation of the experiments [80, 122, 123, 135, 169] with external loads and the identification of in-trinsic communication pathways inside the protein could be already obtained in our study by using a greatly simplified elastic-network model. The model is stripped of chemical de-tails, so that a protein is treated as a purely mechanical object, i.e. a network of particles connected by elastic strings. It turns out that generic elastic responses of such a network are suffi-cient to understand intramolecular communication responsible for the coordination of processes in different functional parts of the molecular motor. Our investigations have been performed in the framework of the full nonlinear elastic model, without the linearization and transition to a normal-mode description.
We have checked (see Appendix) that nonlinear effects are in-deed essential for the considered phenomena.
In the recent publication by Zheng , full nonlinear EN equations have been used and coordination between conforma-tional changes in two coupled units in the myosin dimer, re-sulting from mechanical interactions through the common tail, have been considered. The analysis has however only been per-formed along a particular conformational pathway, connecting two experimentally known myosin states, and the responses have been considered at the level of entire protein domains, rather than for individual residues. In contrast, we have sys-tematically analyzed the responses of the myosin molecule to the application of arbitrary forces with various directions to a large number of individual residues located in three principal functional regions. We have monitored how and to what ex-5.5 discussion 73 tent such forces affected individual residues in other functional residues.
Our results suggest that conformational changes, induced in myosin-V by ATP binding, the hydrolysis reaction and the prod-ucts release, should not be very sensitive to chemical details of the respective microscopic processes. Such changes can be in-voked, as we have shown, by the application of forces to the network nodes corresponding to specific residues in the NBP region. We have also found that the tail and the NBP are sensi-tive to mechanical perturbations of single residues in the actin-binding cleft region, which may naturally arise when binding of myosin-V to the actin filament takes place. It should be noted that similar behavior, based on elastic responses of the macro-molecule to mechanical perturbations in the ATP binding pocket, has recently been reported for a different molecular motor, the hepatitis C virus helicase; within the EN model the entire operation cycle of this motor could be reproduced by applying forces to only a few residues.
Based on such observations, we conjecture that the strain-sensor behavior and the intramolecular communication relying on elastic deformations should be a general property of pro-tein motors, essentially involved in the organization of their functional activity. It would be interesting to test this sugges-tion in further, specially designed experiments. In such exper-iments, forces can be applied to single selected residues (or small groups of them) and conformational responses can be monitored by using atomic force microscopy (AFM) or FRET tech-niques. Such kinds of experiments have been, e.g., performed to study partial unfolding of single proteins [38, 50]. More-over, once specific residues, responsible for important confor-mational responses, have been identified, one can try to modify them by using protein engineering methods and thus verify their functional roles.
6 I N T E R A C T I O N S W I T H T H E F I L A M E N T Many proteins bind to the actin filament. Two prominent exam-ples are the molecular motor myosin and G-actin monomers.
Myosin uses the filaments as tracks and moves processively along such actin polymers and actin monomers form the ac-tual filament in a dynamic process called treadmilling. How is binding to the filament organized?
In this chapter, we model binding to the filament. At large distances, only long-ranged electrostatic interactions are impor-tant. If residues, however, come close to each other, short-ranged interactions become stronger and control the attachment to the filament. Apart from soft sphere potentials preventing protru-sion of residues, strong bonds between specific binding sites form and lead to a stable attachment of ABPs. Such bonds may be hydrogen bonds, salt bridges or other non covalent bonds.
Within our model, they are described by Lennard-Jones poten-tials. Secondly, we investigate the influence of the nucleotide ATP on the attachment of actin monomers at the growing end of the filament. We show that the ligand-induced conformational changes of the actin monomer, as modeled earlier in Chapter 3, indeed increase the binding affinity of actin to the filament.
Specifically, binding of ATP stabilizes a conformation suitable to facilitate docking of free actin to the filament. To this end, a toy model is constructed based on which the effects of ligand binding can be directly quantified with respect to the filament.
6.1 actin monomer and filament models Monomeric actin is constructed as an elastic network based on specific conformational states. As already mentioned in Chap-ter 4, the protein is described in terms of its four subdomains (S1, residues 1–32, 70–144 and 338–372; S2, residues 33–69; S3, residues 145–180 and 270–337; S4, residues 181–269) . In our investigation, two distinct states are used to model actin: glob-ular G-actin and filamentous F-actin.
75 76 interactions with the filament A B Figure 28: Elastic networks of two actin states colored according to its subdomains S1 (orange), S2 (yellow), S3 (blue) and S4 (green): (A) G-actin and (B) F-actin In the absence of ATP, G-actins do not polymerize. They are characterized by a large dihedral angle and an open cleft be-tween inner and outer domains. Moreover, in this structural state actin is known to have a low attachment probability at the growing end of the filament . It is modeled using structural data of uncomplexed G-actin in the ADP-bound state (PDB ID: 1J6Z), which has been obtained via X-ray diffraction techniques . The elastic network of G-actin is displayed in Fig. 28A.
Inside the filament, the actin monomer exhibits a different conformation. In this state, the monomer is called F-actin and differs considerably from the G-actin state. Most important con-formational changes are a characteristic flattening and a clos-ing of the cleft between the two mobile domains. We model this state using the experimental data of F-actin within the fil-ament (PDB ID: 3MFP) . The high-resolution data has been obtained by averaging over many cryo-electron microscopy im-ages. The elastic network corresponding to this conformation is shown in Fig. 28B.
A further striking difference between the two structures in Fig. 28 is the conformation of the subdomain S2. It is here, where the highly mobile DNase-I binding loop (DB loop) is lo-cated. This loop is very flexible and in experimental monomer structures; it is either found in a stable α-helical conformation or in a disordered state. Up to now, spectroscopic methods have not been able to resolve the structure in a satisfying way.
In the G-actin experimental data, an α-helical conformation is found. This conformation, however, is thought to be an artifact of the crystallization methods used . In F-actin, one sees 6.1 actin monomer and filament models 77 Figure 29: EN models of G-actin with F-actin conformation of DB loop colored according to its subdomains S1 (orange), S2 (yel-low), S3 (blue) and S4 (green.
an open loop conformation. Note that the data for the F-actin monomer is obtained inside the filament and, hence, neighbors may stabilize the open loop conformation. The exact conforma-tion of the DB loop, however, is critical for the ability of actin to attach to the filament. An α-helical conformation is seen to significantly weaken the interaction strength between actin sub-units .
The description of the DB loop within the EN approxima-tion is problematic. Generally, α-helical conformations, a com-mon motif in protein structures, are densely packed and, hence, residues have many neighbors. In the standard EN approxima-tion such α-helices are artificially stable, because breaking of elastic links is not allowed. To overcome such limits, hybrid models have been constructed. In G-actin, e.g., the DB loop may be exchanged by the the open loop conformation found in F-actin (see Fig. 29) [56, 121]. This approach is used by Splettstößer et al. as a basis for MD studies of the actin filament. Also, Chu and Voth have used double-well network models al-lowing switching between the two conformations which are both modeled as an elastic networks.
The conformation of the DB loop is nucleotide dependent [126, 128] and may regulate binding to the filament . Fine details like the exact conformation of the DB loop may even be the final cause for a strong attachment. Our aim, however, is to identify collective domain motions that provide the overall basis for docking. Therefore, we abstain from expanding the model to suitably describe the DB loop with high accuracy.
78 interactions with the filament Actin monomers cannot polymerize and form filaments in the absence of ATP. In Chapter 4, we suggested that ligand-binding induces a transition from the structural state of G-actin to the F-actin state. Later, we will directly monitor ligand-induced conformational changes of actin with respect to the filament and, in this way, explain the ATP-dependent attachment mecha-nism.
In this thesis, the interactions of proteins like actin and myosin with the actin filament are investigated. To study such sys-tems from a theoretical point of view, a suitable filament model needs to be obtained. Here, we construct a filament out of its monomeric units modeled as elastic networks based on struc-tures in specific conformational states. In experiments, a certain symmetry is found in the actin filaments. Subunits are trans-lated by 27.50Å and rotated by 166.66◦. Using this infor-mation as a recipe, one can build actin filaments of arbitrary lengths.
An actin subunit in the filament is interacting with four neigh-boring proteins [121, 56]. We see in the experimental data that the F-actin conformation of the monomer inside the filament is stable. Thus, within the filament, the conformation does not de-pend so much on the actual nucleotide-state, which was of such important relevance when it came to attaching to the filament.
One must thus assume that interaction with its four neighbor-ing monomers stabilizes the F-actin conformation. Therefore, we will employ a filament model where the subunits are frozen, i.e. their conformation remains fixed at all times. To save com-putational time, only two or three subunits will represent the filament, depending on the situation.
6.2 actin-actin and actin-myosin binding sites While intramolecular interactions are modeled by means of elas-tic links, it is necessary to develop a way of describing protein-protein interactions within the framework of the coarse-grained EN approximation to study the attachment of proteins to the actin filament. Since in the EN model, all chemical details are neglected, there is without further knowledge no way of decid-ing where stable bonds can be established. Hence, additional experimental data has to be used to identify protein specific binding sites.
6.2 actin-actin and actin-myosin binding sites 79 Table 2: Links between myosin and actin monomers in the filament myosin 1st actin monomer 2nd actin monomer 343 328 -386 337 -517 167 -526 -50 542 -95 In the case of actin, high-resolution filament data is available . The structure can provide clues that can be used to identify probable pairs of binding residues. Such residues can directly be extracted from the data in the following way: • an EN is constructed on the basis of the filament data with a cutoff of l0 = 8.5Å • actin subunits are identified • elastic links between different subunits are marked • pairs of residues connected by such links are assumed to be actin-actin binding sites.
The corresponding actin-actin binding sites found by this pro-cedure are listed in Sec. A.5 in the Appendix.
Identifying protein-protein binding sites for actomyosin is less straightforward. No high-resolution structures for the actin-myosin complex are currently available and we, therefore, can-not use the above method. Nonetheless, a recent MD study by Lorenz and Holmes has investigated the actin-myosin-II interface. A low-resolution electron micrograph was used as constraint for MD simulations. In this set-up, myosin was fitted to the experimental data and residues that may form stable hy-drogen bonds could be identified. Such bonds provide a mech-anism for a strong interaction between myosin and the filament.
Fortunately, the structures of myosin-II and myosin-V are very similar and, therefore, we could obtain a list of pairs of binding residues for myosin-V and the actin filament by analogy. Such pairs are given in Table 2.
Knowing probable pairs of binding residues between pro-teins, one can now expand the EN model of actin or myosin.
Such residues can form breakable bonds described by Lennard-Jones potentials. More details can be found in Sec. 3.4.1 in Chap-ter 3. Note that such potentials account for the fact that specific protein-protein interactions are only short-ranged.
80 interactions with the filament If residues come very close to each other, soft sphere poten-tials prevent protrusion of the residues (see Sec. 3.5.2). Such in-teractions are also short-ranged and become important for even smaller distances. Furthermore, a small number of amino acids carry a charge. They interact via Coulomb potentials. In a sol-vent, such interactions are screened and thus described accord-ing to the Debye-Hückel theory . Electrostatic interactions are long-ranged and, hence, are the only important interactions of the expanded EN model at large distances. More details can be found in Sec. 3.5.1.
Thus, the interaction of myosin or actin with the actin fila-ment is modeled in a coarse-grained way. For large distances, only electrostatic forces can be present. At shorter distances, stronger protein-specific interactions become more important and, thus, govern short-range behavior.
6.3 guided by electrostatic interactions In this section, we focus on proteins interacting with the actin filament, i.e. we study how actin and myosin macromolecules approach it and subsequently form stable bonds. Electrostatic interactions, soft sphere potentials and bonds between protein specific binding sites are explicitly included.
A small group of amino acids carries a charge. The charge distributions of actin and myosin are displayed Fig. 30. Between these charges residues, electrostatic interactions are present.They are long-ranged and for distances larger than 20Å, only such interactions are present in our model. If proteins come close to each other, breakable bonds are formed between certain pairs of binding residues, as explained above. Furthermore, short-ranged soft sphere interactions prevent protruding of amino acids. A detailed discussion of all protein-protein interactions used in our modeling is given in Sec. 3.5.
In the following , we show how electrostatic interactions guide actin or myosin toward the filament, where specific protein-protein interactions lead to a stable attachment of those pro-teins.
6.3 guided by electrostatic interactions 81 A B Figure 30: Charged residues of (A) actin and (B) myosin-V. Arginine and Lysine (blue) carry a positive charge, Aspartic and Glu-tamic acid carry (red) carry a negative one. Histidine (light blue) carries a positive charge in 10% of all cases.
6.3.1 Interaction of Actin Monomer with the Filament The basis of our investigation are three actin monomers A1, A2 and A3. Actin subunits A1 and A2 are in the filament confor-mation and remain frozen. In this way, they provide a minimal model for the actin filament. In this part of our study, we are mainly interested in finding out how actin approaches the fil-ament and we do not want to consider ligand-induced effects.
Thus, we model the third monomer A3 as F-actin EN, i.e. its equilibrium is a closed and flat conformational state, suitable for docking to the filament.
If we position an actin monomer more than 20Å away from the closest filament residue, interactions between filament and protein can only be electrostatic. Thus, we start with such initial conditions and monitor the relaxation process. In our computer experiment, actin is shifted about 30Å away from the filament.
The initial set-up is shown in Fig. 31A.
Due to electrostatic interactions, the filament and the monomer attract each other. Figure 31 shows the corresponding docking process in a range where only Coulomb interaction are present.
82 interactions with the filament A B Figure 31: The filament is represent by two monomers (gray beads, iceblue beads) fixed in space. An actin monomer (yellow) approaches guided by electrostatic interactions. Starting from the initial position (A), the conformation of the actin trimer is shown after (B) 300 time steps in arbitrary units.
First, the actin subunit A3 mainly rotates to align its charges.
Then, the monomer approaches the filament until it comes into a range of 20Å of the filament residues (Fig. 31B).
If the monomer has come this close to the filament, links be-tween specific pairs of binding residues are established. The effect of such bonds is visualized in Fig. 32. Just as the elec-trostatic interactions have guided the protein toward the fila-ment, this newly established bonds guide the molecule toward its equilibrium position bound to the filament (Fig. 32C). The short-ranged soft sphere potentials forces that become relevant if distances between residues are close to 5Å and prevent pro-trusion of protein domains.
Electrostatic interactions between charged residues and the specific actin-actin interactions provide a framework for a con-trolled binding of actin to the filament. Placed at a distance where only Coulomb interactions are present, the free actin pro-tein first aligns its charges and then approaches the filament.
The free monomer is guided towards its corresponding bind-ing sites on the filament. If the monomers come close enough to such sites, strong bonds form between the specific pairs of binding residues of the actin monomer and the filament. Such strong bonds are necessary for a stable binding between pro-tein subunits. Protrusion is prevented by soft sphere potentials.
This provides the framework for controlled binding of actin to the filament.
6.3 guided by electrostatic interactions 83 A B C Figure 32: The filament is represent by two monomers (gray beads, iceblue beads) fixed in space. An actin monomer (yellow) approaches guided by electrostatic interactions. As soon as the actin monomer is close enough to the filament, bonds form between the binding sites between actin subunits (pink beads). Starting from position (A), the binding actin subunit interacts with the filament and quickly relaxes to an equilibrium position. (B) shows the relaxation 1 and (C) 2 timesteps in arbitrary units.
6.3.2 Interaction of Myosin with the Filament We have seen how actin approaches the filament. In this part of the thesis, we will concentrate on the interactions between the actin filament and a myosin molecule. Again, the focus will be on how electrostatic interactions guide the proteins toward specific binding sites where strong bonds can be formed.
If a myosin protein attaches to the filament, it establishes connections to three monomers. Hence, the minimal filament model is a trimer of F-actin EN A1, A2 and A3 in the fila-ment conformation. The conformations of the three subunits are fixed. Myosin is modeled as an EN on the basis of myosin-V experimental data in the post-rigor conformation (PDB ID: 84 interactions with the filament A B C Figure 33: The filament is represent by three fixed actin monomers in their filament conformation (gray, iceblue, lime beads). A myosin monomer (yellow) approaches the filament guided by electrostatic interactions. Starting from the initial posi-tion (A), the conformation of the actin trimer is shown after (B) 80 and (C) 160 time steps in arbitrary units.
1W7J, cf. Ch. 5). It can interact with the filament via pairs of residues given in Tab. 2.
In the initial configuration, the myosin-V EN is shifted with respect to the trimer representing the filament (see Fig. 33A).
The closest residues are approximately 63Å apart. It is clear that in our model actin and myosin can only communicate via elec-trostatic interactions. Qualitatively, we see the same behavior as in the previous section. The myosin aligns its charges (Fig. 33B) and then approaches the filament (Fig. 33C) until interacting residues come closer than 20Å to one another.
If the myosin comes close enough to the filament, the actin-myosin binding sites interact strongly. Such interactions lead 6.3 guided by electrostatic interactions 85 A B Figure 34: The filament is represent by three fixed actin monomers in their filament conformation (gray, iceblue, lime beads).
A myosin monomer (yellow) attaches to the filament. Af-ter being guided by electrostatic interaction in a way that binding sites (pink beads) come close to each other (A), Lennard-Jones type bonds between such binding sites form. The equilibrium conformation is shown (B).
the protein to its final equilibrium position, which is displayed in Fig. 34. Again, soft sphere potentials prevent protrusion, if residues come too close to each other.
In Ch. 4 the HCM loop was suggested to play an important role for the force-generating mechanism. It was shown that forces on this region lead to a conformational change in the myosin tail region which may result in a so-called power stroke.
Hence, it is noteworthy how well the HCM loop fits into a cer-tain pocket in the actin filament. The role of the HCM loop, there-fore, has to be further investigated.
In a nutshell, we identified the mechanism than enables the attachment of myosin to the filament. Placed at a large distance, short-ranged interactions like actin-myosin binding sites or soft sphere interactions do not play a role. In our model, only elec-trostatic interactions were present and guided the myosin to-ward its specific binding sites on the actin filament. Here, pair interactions between specific residues lead to a strong binding to the filament. Special attention should be paid to the role of 86 interactions with the filament the HCM loop. We have seen in our simulations, that it is led to a specific pocket in the actin filament. Here, it fits almost perfectly and therefore the role of the HCM loop in the machine cycle of myosin should be further investigated.
6.4 ligand binding facilitates docking In Chapter 4 ligand binding has been modeled in the frame-work of an expanded EN description. A key result of this chap-ter was that binding of such an artificial ligand introduces con-formational changes in G-actin. These conformational changes are found to describe the structural changes of the G- to F-actin transition, which have been found experimentally, quite well.
Accordingly, it has been suggested that binding of a ligand fa-cilitates docking of actin monomers to the filament. This means that ligand binding enhances the attachment affinity and, in this way, an ATP-dependent mechanism to regulate polymeriza-tion can be identified. In the following, we directly monitor the ligand-induced conformational changes and show that after nu-cleotide binding the actin monomer fits better to the filament.
Thus, binding of ATP does indeed facilitate docking.
6.4.1 Toy Model Our aim is to understand how ATP-induced conformational changes influence the probability of docking to the filament.
The effect of nucleotide binding shall be understood indepen-dently of protein-protein interactions and, therefore, we will neglect soft sphere and electrostatic interactions at this point.
Furthermore, actin-actin interactions are only used in a very specific way as explained below. It should be made clear that the toy model employed in this part of the thesis will not fully represent the interaction of an actin monomer with the actual filament, but will only focus on a small part of this process, the ligand-induced motions with respect to the filament.
First, we construct a toy system of an actin monomer interact-ing with the filament. If an actin protein attaches to the grow-ing end of the filament, it establishes contacts with two filament subunits. These are modeled as a dimer of two F-actin elastic 6.4 ligand binding facilitates docking 87 A1 A2 A3 Figure 35: Actin trimer constructed out of F-actin. Static actin monomers A1 (gray) and A2 (iceblue) are connected to the subdomain S4 of the EN of actin subunit A3. Beads that in-teract with A1 and A2 are colored pink. The distance l be-tween subdomain S2 of A3 and subdomain S3 of A1(light green) measures the distance of the DB loop to its corre-sponding binding sites in the neighboring actin subunit.
networks A1 and A2, fixed in space. A third actin monomer A3 is modeled as G-actin EN. It interacts with the filament via certain specific binding sites.
Natural motions of actin involve collective motions of the two mobile domains with respect to each other (cf. normal-mode analysis of G-actin in Chapter 3). If we only let the monomer interact with the filament via one of these domains, ligand-induced relative motions of them are not hindered. Accord-ingly, we allow only subdomain S4 of subunit A3 to interact with the filament. All other links between actin subunits are not present in this analysis. This corresponds to a situation where one mobile domain of a free actin monomer is anchored to the filament at its appropriate binding position while the other mo-bile domain can freely move with respect to its counterpart.
The set-up is shown in Fig. 35. Pink residues in A3 represent the anchored binding sites.
It is obvious that such a set-up does not reflect real actin-actin interactions. Only ligand-induced conformational changes are monitored and compared to the filament while neglecting all other effects. The aim is to quantify the nucleotide-dependent actin dynamics with respect to the filament. We will do so by 88 interactions with the filament A B Figure 36: Actin trimer constructed out of F-actin. Static actin monomers A1 and A2 (grey beads) are connected to the subdomain S4 of the EN of actin subunit A3. Beads that interact with A1 and A2 are colored pink.
monitoring the distance l. In Fig. 35, l is displayed as the length of a light green line. It connects the centers of mass of subdo-main S2 in actin subunit A3 and the center of subdomain S3 of actin subunit A1. It can be understood as a measure of how close the DB loop can approach its respective binding sites. Here one should remember that this loop plays a fundamental role in protein-protein interaction .
6.4.2 Nucleotide-Dependent Dynamics We will follow the distance l to measure conformational changes of G-actin with respect to the filament. The starting conforma-tion of A3 is the conformation of F-actin as it would be po-sitioned in the filament (see Fig. 35). Thus, A3 is initially de-formed. Accordingly, the system tends to relax to its equilib-rium, i.e. the A3 monomer will open its cleft and unflatten.
Ligand-binding in this state yet again leads to a flattened and closed state. Hence, following l over time we can quantify the transition from F-actin to G-actin and, subsequently, the tran-sition from G-actin to ATP-G-actin. The respective equilibrium states are shown in Fig. 36.
Figure 37 displays the change of the distance l. The black line shows the initial relaxation from the distorted F-actin state to the G-actin equilibrium conformation. The distance l changes from l = 24.0Å in the F-actin conformation to l = 32.7Å in the relaxed state. This change in distance means that the DB loop of actin subunit A3 moves away from subdomain S4 in the neigh-6.4 ligand binding facilitates docking 89 0 100 200 300 400 500 600 24 26 28 30 32 time (arbitrary units) l Figure 37: Distance between actin subdomain S4 and the filament.
The distance between the center of mass of subdomain S3 of actin A1 and the subdomain S4 of actin A3 is plotted.
From the starting conformation of F-actin, the G-actin EN relaxes to its equilibrium state (black line). Ligand bind-ing leads to a decrease in the distance l (red line). Initial, maximum and final values are indicated by dashed lines.
boring subunit A1, i.e. the loop retreats from its corresponding binding position.
Next, we study the effect of ligand binding. The red line in Fig. 37 monitors the distance l after binding of ATP. The dis-tance decreases by ∆l = 5.6Å. This clearly means that the DB loop moves closer to its expected binding position. Now we see how ligand-binding facilitates docking: a closed conformation is stabilized that is more suitable for binding to the growing end of the actin filament.
Between the initial F-actin conformation and the ATP-induced conformation lies a difference of ∆l = 3.1Å. Here, the structure of the loop mentioned above becomes important. In G-actin, the DB loop is found in its α-helical conformation. A mere con-formational changes from α-helical to open loop conformation accounts for a distance change of approximately ∆l = 2.0Å, 90 interactions with the filament already, because the center of mass of subdomain S2 of A3 is shifted accordingly. Furthermore, we do not expect exact quan-titative results due to the many approximations that were car-ried out.
We built a simple toy model to visualize ligand-induced ef-fects with respect to the filament. By following the conforma-tional changes, we see that the effect of ATP is to help the DB loop to approach its corresponding binding position. In this way, docking is regulated depending on the nucleotide.
6.5 discussion In this chapter, we have investigated the interactions of actin and myosin with the actin filament. We have used EN approx-imations to model the two proteins. Furthermore, we have de-veloped a minimal filament model comprised of two or three F-actin monomers, depending on the situation. These have been positioned according to the symmetry of the filament deter-mined in experimental data. To take into account the effect of proteins in the filament stabilizing their conformation, on the one hand, and to save avoid high computational costs, on the other, filament monomers were fixed in space.
In general, we have introduced a framework to describe protein-protein interactions within the EN approximation. Short-ranged soft sphere potentials prevent unphysical protrusion of residues.
Furthermore, charged residues interact via screened Coulomb potentials. Within the framework of the EN model, all chemical details are lost and, accordingly, we have used additional ex-perimental data to identify specific binding sites where strong bonds could be established. In the case of actin, filament data was used to obtain such pairs of binding residues. Moreover, we could identify probable binding sites between actin and myosin by using the results of a recent MD study by Lorenz and Holmes .
How does a free actin monomer bind to the filament? It has been found that electrostatic interactions guide the protein to-wards its respective binding sites in the actin filament. We have placed an actin monomer at a large distance from the filament, so only Coulomb forces were present. After aligning its charges, the free monomer moved towards its respective binding site at the filament end. Here, links could be established which pulled 6.5 discussion 91 the protein into its correct position in the filament, which lead to a strong attachment.
Takano et al. have found a ratchet-like behavior of myosin attached to the actin filament. Their discussion was entirely based on electrostatic interactions between charged and van der Waals interactions between all residues. Using results of a recent MD study of the actin-myosin interface , we have ex-panded their model by allowing strong bonds to form between specific binding sites.
On the basis of our model, we have in detail investigated the approach of myosin towards the filament guided by elec-trostatic forces. Placed at a large distance from the filament, the myosin protein was attracted to the filament by Coulomb forces and moved towards its respective binding sites. Thus, pairs of binding residues formed breakable Lennard-Jones-type bonds which enable a strong binding to the filament.
We therefore have found that the monomeric actin and the myosin motor protein independently have similar means of binding to the actin filament. Long-range electrostatic interac-tions guide the proteins towards a suitable binding position.
Here, the macromolecules approach specific binding sites and establish bonds that lead to strong binding. This behavior seems to be a quite general mechanism allowing various actin-binding proteins to attach to the filament.
In a next step, we have studied the role of the ligand in the polymerization process. To this means, we have constructed a toy system. Our aim was to visualize the ligand-induced con-formational changes of an actin monomer with respect to its neighbors. The filament has been modeled by two fixed F-actin monomers and a G-actin monomer has been anchored to the filament at its appropriate binding sites. An important aspect of the toy model was that it only has been attached with one of the two large mobile domains. Accordingly, relative motions of actin’s main domains have not been hindered. Moreover, soft sphere and electrostatic interactions were neglected completely, because we were only interested in internal interactions of G-actin.
Studying the toy system, we could monitor conformational changes and relative motions of the two mobile domains of G-actin with respect to the filament. By using the coarse-grained ligand model of Ch. 4, we saw that binding of ATP induces a conformational change that lets the DB loop approach its cor-responding binding sites in the neighboring actin subunit. Be-92 interactions with the filament cause we have seen that ATP-induced changes lead to a con-formation more suitable for stable binding, we could substanti-ate our earlier hypothesis that ligand-induced conformational changes facilitate docking.
7 S U M M A RY A N D O U T L O O K The main aim of this study was to understand the intrinsic ma-chinery of two important molecules, the structural protein actin and the molecular motor myosin. An important question was, e.g., how binding of an ATP nucleotide regulates their machine cycles. Apart from intramolecular communication, we further-more investigated interactions between different protein units.
We discussed the polymerization of actin monomers into fila-ments and studied the interactions between actin and myosin.
Proteins were investigated from a theoretical point of view in the framework of an EN approximation. Such an approach has the advantage of allowing a high resolution description of biomolecules (up to one amino acid) while avoiding the high computational costs of all-atom approaches.
This thesis has been organized as follows. Chapter 2 pro-vided a background of the field of protein research. Here, gen-eral properties of proteins, e.g. their role in the cell and their construction principle, were presented. Furthermore, the bio-logical functions of actin and myosin and their corresponding machine cycles were broadly explained. Additionally, a brief overview of scientific methods used in the study of proteins and biomolecules in general was presented. Throughout the last decades, a wide range of experimental techniques have been developed. In this chapter, methods relevant for the study of actin and myosin were introduced. Structural data with atomic resolution, e.g., can be obtained by X-ray crystallography, NMR or cryo-electron microscopy. Furthermore, optical tweezer may be used for manipulating single molecules and conformational changes can be dynamically followed by FRET measurements.
Moreover, theoretical approaches to investigate such macro-molecules were reviewed in depth. The standard approach of MD simulations was presented and several ways to speed up its simulations, which suffer from high computational costs, were depicted. To overcome certain limitations of the MD simu-lation approach, coarse-grained approximations have been suc-cessfully applied. The main approximation method used in this thesis was the EN model.
93 94 summary and outlook The following Chapter 3 contains the detailed description of all mathematical tools used in our study. Slow conforma-tional changes of proteins were investigated with the EN ap-proach. Here, we presented the general model together with its linearization. If one applies external forces to elastic networks, rigid translations and rotations are induced. To study communi-cation within proteins, however, only internal dynamics should be considered. Therefore, a method was established to immobi-lize such networks without pinning single residues.
If collective conformational motions are large, residues which are not neighbors in the EN framework may come close to each other. Following the logic of the model, they should, in this case, be connected by elastic links. To account for this behavior, the standard EN description was extended. Certain residues that could come close to each other were allowed to form breakable bonds. Additionally, we showed how to model protein-protein interactions within the coarse-grained EN approach.
In Chapter 4, the intramolecular communication of actin was investigated. Here, our aim was to understand the internal orga-nization of the conformational mechanics of the protein. Within the EN approach, we found that there exist two large mobile domains which perform slow collective motions. We identified these natural motions as a propeller-like twist of the two large mobile domains with respect to each other and a scissor-like opening and closing of the cleft between them. Although our model is considerably simpler, the results of Tirion and ben Avraham could be reproduced .
Due to the large domain motions, residues inside the actin monomer, which were not connected in the EN model, were seen to approach each other. Accordingly, additional interac-tions should occur between them. We modeled this by intro-ducing breakable bonds between residues that may come close.
In this way, a metastable, closed state was generated.
Furthermore, we were interested in important residues in the NBP that are able to translate small local deformations into large-scale domain motions. To identify such residues, we stud-ied the response to external forces applied to single residues in this region. We saw that, indeed, perturbations in the ATP-binding region of only a small group of residues could induce large collective domain motions. In actual biological systems, actin is activated by ATP binding and subsequent hydrolysis.
The chemical reaction in the NBP can lead to large-magnitude conformational changes. Therefore, we introduced in Chapter 4 summary and outlook 95 a coarse-grained model of the nucleotide on the basis of the previous sensitivity analysis within the nucleotide-binding re-gion. We found that interaction with such a model ligand sta-bilized the closed state and could induce a transition from the open equilibrium conformation to closed state. Based on these results, we were able to understand the nucleotide-dependent binding affinity of the actin monomer to the filament. The actin monomer can be in an open or closed conformation. The closed conformation is seen to be more suitable for docking to the fil-ament. If ATP is not present in the NBP, the monomer switches between those two states due to thermal noise. The open state, however, is much more probable and, hence, the binding affin-ity of actin to the filament is small. In the presence of ATP, how-ever, the closed state is stabilized and, accordingly, the binding affinity of this state is much larger. Hence, the monomer tends to attach to the filament in the ATP state.
Chapter 5 dealt with the myosin motor. Inspired by recent ex-periments which used optical tweezers to exert forces to myosin , an EN model of myosin-V was used to study responses to external forces. The myosin molecule is built out of three func-tional regions: the tail, the NBP and the actin-binding cleft. Here, we studied communication between these regions by probing single residues in a specific region by external forces and mea-suring the effect in the respective other two. In this way, we constructed a map of intrinsic communication.
In our study, we found that the ATP-binding region is di-vided into two parts - the front and back doors. The front-door region interacts with the adenine ring of ATP or ADP, respec-tively. Perturbations in this region lead to a large conforma-tional change in the tail. The back-door region, on the other hand, interacts with the γ-phosphate. Perturbations here in-duce conformational changes in the actin-binding cleft. On the basis of these results, we could assume that a chemical reaction with the adenine part of ATP/ADP may lead to a power stroke-like behavior, and interaction with the phosphate controls the binding affinity to the filament. We furthermore studied per-turbations in the actin-binding site. It became clear that exter-nal forces applied to the HCM loop cause large conformational changes in the tail region. Because the HCM loop is known to strongly interact with actin, binding to the filament has in this way been identified to be a control mechanism for a possible power stroke.
96 summary and outlook Furthermore, the effect of forces in the tail region was investi-gated. We tried to reproduce the experiments by Iwaki et al. and Oguchi et al. . In these experiments, the effect of load on the filament and on the ligand binding affinities to myosin were studied. In our simulations, we saw communication of forces on the tail with the actin-binding region and the NBP.
First of all, backward strain leads to a closing of the actin cleft and forward strain has the opposite effect. Comparing with the nucleotide-free state of myosin-V, thought to resemble the con-formation of myosin bound to the filament, we concluded that backward strain directly leads to conformational changes in the actin-binding region that make it more suitable for bind-ing to the filament. In this way, we could explain the results of Ref. . Moreover, we saw that strain controls the opening of the so-called front door which explains the load-dependence of the nucleotide-binding affinity . Thus, by means of the coarse-grained EN network approach we identified to a large extent the structural basis of the myosin motor mechanism.
In Chapter 6, we studied the interactions of unbound monomeric actin and the myosin motor protein with the filament. We in-vestigated how long-ranged Coulomb interactions guide actin-binding proteins toward their specific binding sites. Both, a free actin monomer and the myosin macromolecule, approach the filament due to electrostatic interactions. Here, the proteins come close to their specific binding sites on the actin filament surface. Establishing bonds between pairs of residues enables the actin-binding proteins to strongly attach to the filament.
Furthermore, we elucidated in a toy model the conforma-tional changes of actin due to binding of the artificial ligand which was modeled earlier in a coarse-grained way as a dimer of ADP and Pi nodes. We saw that induced conformational changes led to a movement of the DB loop towards a cleft in the neigh-boring actin subunit of the filament. Therefore, supporting the conclusion of Chapter 4, binding of the ligand leads to a con-formational state that makes docking easier.
The EN approximation has been proven very useful to de-scribe internal dynamics of proteins. This approximation, how-ever, suffers from serious drawbacks mainly due to neglecting the chemical properties of single residues. Hence, chemical de-tails are only included indirectly through the protein structure.
Such details, however, are especially important during binding events, i.e. the binding of a ligand or the binding of myosin summary and outlook 97 to the actin filament. In the course of this thesis, we saw that additional information about pairs of binding residues is nec-essary and it remains difficult to model chemical interactions in a suitable way. However, using EN models as a guide for experiments or computationally costly MD simulations can be a powerful tool in the field of protein research. Two examples follow below.
A widespread method to monitor conformational changes on the scale of 1Å are FRET measurements [55, 130]. A combination of the EN modeling of proteins, approximating conformational dynamics, and such experimental methods could make it pos-sible to follow tiny conformational changes dynamically. An example is the existence of certain metastable states in the actin filament . An interesting application for EN approximation would be to study such conformations on a microscopic level.
Due to high computational costs of MD simulations, a wide range of acceleration methods have already been developed. To describe myosin-V, e.g., targeted molecular dynamics have re-cently been used . In these simulations, a restraining poten-tial was constructed based on structural data of myosin in dif-ferent conformational states. A more general approach would be to investigate proteins within the EN approach and use these results as restraining potentials to understand the internal dy-namics on an atomic level.
In this thesis, we introduced a coarse-grained framework to describe the interactions between the actin filament and actin or myosin. Our findings can be the basis for future studies of larger systems. One may think about modeling the actual myosin stepping along actin filaments, i.e. a complete actomyosin machine cycle. Such a model would be an extension of the Brownian stepping model presented by Takano et al. . Fur-thermore, the actin monomer and its interaction with the ATP ligand have been approximated in the framework of the EN ap-proach. The simple model introduced in this study should be extended to resolve the complete process of dynamic polymer-ization of actin into filaments.
A A P P E N D I X In the Appendix, additional information and details are pro-vided. In Sec. A.1, the nucleotide model used for actin monomers is described in depth and, furthermore, residues interacting via truncated Lennard-Jones potentials are given. In the cell, myosin motors walk processively along actin filaments. The following Sec. A.2 describes how the direction of motion of myosin can be obtained by fitting the myosin elastic network to the actual filament. Then, in Sec. A.4 the difference between linearized equations of motions andx full-nonlinear equations is highlighted on the example of the myosin EN with external forces.. Finally, Sec. A.5 gives the binding sites of actin-actin interactions. Additionally, the Appendix contains Tab. 3 and Tab. 4, which where to large to place in the main part of the manuscript.
a.1 ligand model for actin monomer The ADP is imitated by introducing an additional node in the network whose equilibrium coordinates are taken to be the C1′ position. This carbon connects the nucleotide’s ribose with ade-nine and, therefore, is located in the central part of the molecule.
The new node is connected to all neighboring residues within the cutoff distance lc by elastic links of a stiffness κ. Note that the same stiffness also appears in equation (3.1) and, thus, ADP is treated as a regular node in the network and the confor-mation remains in the equilibrium. The neighbors of ADP are residues 156, 157, 181–181, 301–305, and 336 in subdomain S3 and residues 210, 213, and 214 in subdomain S4. The ADP model is shown in Fig. 18A. In this model, no links between the inner and outer domain are established. Therefore, the slow natural motions, i.e. collective motions of domains with respect to each other, remain unchanged.
Furthermore, Pi is modeled as a node connected to ADP and the three identified key residues 16, 73, 159. It is placed in the center of mass of these four nodes and connected by elastic 99 100 appendix links, again with the same stiffness κ. The complete model of ATP interacting in the NBP is visualized in Fig. 18B. The presence of phosphate in this region leads to a shrinking of the NBP .
This fact is accounted for by shortening the equilibrium lengths of all links between Pi and its neighbors to 20% of their initial value. In this way, the phosphate is modeled as a particle in the NBP interacting attractively with its vicinity. It is a property of ATPases, that local changes upon ligand binding and release lead to conformation changes. Movie S3 in Ref. shows the changes induced by this model of nucleotide binding.
a.2 myosin directions Myosin-V walks along the filament in the direction of the barbed end. In the experiments, the protein was dragged along the filament. The forward strain direction corresponded to the direction of processive motion, whereas the backward direction was opposite to it. If one wants to computationally reproduce the experimental situation, forward and backward directions for the elastic network of the protein must be identified. We have done this by including F-actin into the elastic network sim-ulations and determining the equilibrium conformation of the myosin-actin complex.
Employing guided MD simulations, Lorenz and Holmes [73, 105] have recently identified several possible binding sites of myosin-II to actin. Comparing structures of myosin-II and myosin-V, analogous binding sites for myosin-V can be suggested. The myosin head binds to two distinct F-actin monomers in the fil-ament (constructed from PDB ID: 2ZWH ) at the positions given in Table 2 in Ch. 5.
To determine the equilibrium position of myosin with respect to the actin filament, the following procedure has been em-ployed: elastic links, connecting myosin to the actin filament, have been introduced with equilibrium lengths of 3.5 Å and stiffness κ = 1. After that, relaxation equations (3.3) of the myosin-actin complex were numerically integrated until a sta-tionary state has been reached. In this way, we approximated the actomyosin structure as shown in Fig. 38.
Relaxation to the equilibrium state of the complex involved rotation of the myosin-V molecule. To quantify this rotation, we have chosen four residues (165, 195, 559, 691). These residues A.3 myosin sensitivity tables 101 Figure 38: Myosin-V modeled on actin filament: the elastic network (green) is anchored to the actin at specific binding sites (red balls mark corresponding residues on myosin and actin, respectively). The direction of the processive motion is shown by the gray arrow and the force applied by the red arrow.
belong to the stiff core of the protein. With these residues, one can construct three linear independent vectors that define a co-ordinate system only in terms of the network structure. Thus, the forward direction n∥can be approximated by transforming the filament axis to the coordinate frame of the reference state (PDB ID: 1W7J) that is defined by the same residues. In this way, we found the forward direction to be approximately n∥= (−0.254, −0.888, 0.383).
a.3 myosin sensitivity tables Our aim is to examine the mechanical responses of the protein to forces with varying directions applied to individual residues in the nucleotide-binding pocket and actin-binding cleft. To probe mechanical responses, we select two sets of residues in the respective regions (see Fig. 23a). For every chosen residue, a series of 200 simulations was performed. In all simulations, the magnitude of the applied force was the same (f0 = 1 Å), but its orientations were randomly varied. The simulations were con-tinued until a stationary state was found. After that, changes in the monitored pair distances between the labels in different regions were determined. The induced distance changes were 102 appendix analyzed and, for each pair of labels, the maximum absolute distance change of 200 force orientations was evaluated. In this way, we obtained the sensitivity to forces applied to a residue with respect to a particular pair distance.
The results are shown in Tables 3 and 4 and will be com-mented below. Note that absolute sensitivities for different pairs of labels cannot be compared. As a matter of fact, distance changes for the tail are always larger than for the actin-binding cleft. To show the variations of sensitivity, a color code was employed. In each column, the maximum and the minimum entries are taken and color gradations from dark blue for the minimum to dark red for the maximum are applied.
a.3.1 Forces in the Nucleotide-Binding Region In the NBP, a set of 27 residues adjacent to the ATP in the con-sidered equilibrium conformation was selected (left column Ta-ble 3 and Fig. 23a). The sensitivities of these residues with re-spect to different pair distances are shown in Table 3. The left column lists the residues and, in each of the other columns, the sensitivities with respect to the distance between a particular pair of labels (e.g., between residues 343 and 517) are given.
The last two columns of Table 3 show the sensitivity of the tail with respect to forces in the NBP. Applying forces to the residues 115 and 116 or their neighbors, a strong effect on the tail is induced. These two residues are located in the front-door region (Fig. 25). Remarkably, applying forces to residues in the back door (219, 220 and 438–442) or the P-loop (163–169) only weakly affects the tail region. Thus, the tail responds mainly to the perturbations applied at the entrance of the front door.
Moreover, sensitivity with respect to residues in the actin-binding cleft was investigated. The strongest response with re-spect to the distance 343-517 characterizing cleft opening is seen if forces are applied to the residue 442, which belongs to the back-door region. Additionally, perturbations near the front-door residues 115 and 116 show some effect. The distance be-tween residues 386 and 517 describes cleft opening, but mostly reflects movement of the HCM loop to which residue 386 be-longs. Here, large responses were observed when the forces were applied in the back-door region with the strongest sensi-tivity to the perturbations of residues 219 and 220. Only small A.3 myosin sensitivity tables 103 changes were seen if the forces were applied in the front-door area.
In terms of the sensitivity of its residues, the NBP region is clearly divided into a front-door and a back-door domain (Fig. 25). A pronounced effect on the actin cleft was observed when forces were applied to residues 219, 220 and 442 in the back-door region. Remarkably, it is exactly the salt bridge be-tween residues 219 and 442 that hinders phosphate release after the hydrolysis. The P-loop region is seen to be relatively stiff and external forces here do not induce large conformational changes in the tail or the actin cleft. Perturbations in the front door affect the tail.
a.3.2 Forces in the Actin-Cleft Region Using the results of Lorenz and Holmes [73, 105], we identified 54 residues which may come in contact with the filament (left column Table 4 and Fig. 23a). To study communication between the actin-binding region and the NBP or the tail, we repeat the simulation procedure described above and obtain the sensitivi-ties shown in Table 4.
As can be seen from the results, application of forces to the HCM loop (residues 377 to 390) can induce strong responses of the tail. Moreover, there is some effect on the distance between residues 789 and 92 in the tail region to the forces applied at the residues from 340 to 350, which belong to the upper 50kDa subdomain, as well. Remarkably, the tail is only weakly affected by the forces applied to the residues in the lower 50kDa subdo-main.
The front door (distance between residues 115 and 297) is strongly sensitive to the forces applied at the upper 50kDa sub-domain, including the HCM loop. The back door (distance be-tween residues 442 and 291) is mostly sensitive to the forces applied to residues 540 to 544 in the lower 50kDa subdomain; it should however be noted that these residues are located near the back door and, therefore, stronger sensitivity might have been expected.
104 appendix a.4 comparison to the linearized model The relaxation equations (3.3) of the elastic network are linear in terms of the distance changes between the particles. They are, however, still nonlinear in terms of the changes of the absolute coordinates of the particles ri = Ri −R(0) i , since the distance dij is a nonlinear function of the coordinates Ri and Rj. Note that not the distance, but the particle coordinates are the dy-namical variables in these equations. Hence, to proceed further to the linear (or harmonic) approximation, equations (3.3) need to be linearized with respect to the coordinate changes ri. After linearization, they take the form ˙ ri = Fi − N ∑ j=1 Aij R(0) i −R(0) j d(0) ij 2 h R(0) i −R(0) j · ri −rj i . (A.1) This system of linear equations can further be used to obtain the eigenvalues and the eigenvectors corresponding to various nor-mal modes. It should be noted that, although the overdamped limit of relaxational dynamics is considered here, the resulting eigenvalues and eigenvectors are still the same as when the purely inertial (vibrational) dynamics is assumed.
The linearized equations (A.1) can be used as long as all co-ordinate changes ri are much smaller than the (natural) lengths of the elastic links connecting neighbor particles. Therefore, to test the possible validity of the linear approximation, the ob-served coordinate changes should be compared with the typi-cal natural lengths of the elastic links. By the construction of the EN model, natural lengths of all elastic links cannot exceed the cutoff length, which has been 10Å in the present study. The average natural length lav of a link is smaller and, for a rough estimate, the value lav = 5Å can be chosen. Linearization holds if the coordinate changes ri are much smaller than lav, which re-quires that they should not exceed, e.g., 10% of lav, that is they cannot be larger than 0.5Å.
When the effects of forward strain were considered, an exter-nal force with the magnitude f = 6Å was applied to the tail and, after a new equilibrium state was reached, conformational changes have been inspected and changes of the distances be-tween the labels were analyzed. We have also checked what were the deviations in the absolute positions of some typical residues. As it turns out, when such a force is applied to the tail, the position of the characteristic residue 384 in the HCM loop A.4 comparison to the linearized model 105 gets changed by r384 = 7.4Å. Moreover, the residue 792, which is located in the tail, moves by 12.2Å from its equilibrium po-sition. Such displacements are comparable to the cutoff length and, thus, when responses in the actin cleft or in the tail are con-sidered, the linear description cannot hold. On the other hand, the respective induced changes within the nucleotide binding pocket are much smaller. For example, residue 115 in the front door region shifts its position by only r115 = 0.6Å when the same force is applied to the tail. Such weaker changes in the nucleotide-binding region could probably have also been cor-rectly reproduced within the local linear approximation for this protein region.
The limitations of the linearized normal-mode descriptions have been previously discussed for myosin-V and kinesin .
In the present study, we have decided to stay completely within the full nonlinear elastic description, so that such difficulties cannot arise. Because the linearized equations are only an ap-proximation to the full set of nonlinear equations, considered here, their analysis, once performed within the validity region of the approximation, cannot obviously yield anything which is not already contained in the nonlinear model.
As an illustration of the difficulties encountered in the lin-earized description, Fig. 39 shows the behavior described by the linearized equations (A.1) as compared with the responses described by the full nonlinear equations (3.3). Here, a con-stant force f = (−1, 1, 1)/ √ 3 Å is applied to residue 384 in the HCM loop. The dynamical responses of the elastic network are determined by integration of equations (3.3) or (A.1), respec-tively. The absolute displacements r115 and r792 of the residues 115 and 792, located in the front-door region and the tail, re-spectively, are plotted here as functions of time for both de-scriptions. As we see, the full nonlinear equations yield the re-sponses (solid curves) which saturate as the new equilibrium state of the network, under the constant applied force, is ap-proached. In contrast to this expected behavior, integration of the linearized equations yields the displacements which indefi-nitely grow with time (dashed curves in Fig. 39).
Such unphysical behavior has been observed because the lin-earized equations have been used in the above example beyond their validity limit. Indeed, the final stationary displacements of the considered residues in the full nonlinear model are of the order of tens of in this case, strongly exceeding what is required for the validity of the linearized description. The ori-106 appendix 0 2k 4k 6k 8k 0 50 100 150 200 residue displacements (Å) time (dimensionless) Figure 39: Comparison of the network responses to the application of a static force to the tail, as yielded by the full nonlinear and the linearized models. Constant force of amplitude f = 0.5 and direction (−1, 1, 1)/ √ 3 is applied to residue 384 in the HCM loop. Time-dependent displacements r792 (green) and r115 (red) of residues 792 and 115 from their equilib-rium positions are displayed, as yielded by the integration of the full nonlinear (3) (solid curves) and the linearized equations of motion (S5) (dashed curves) can be compared.
gin of the observed unphysical divergence lies in the fact that, after linearization, the energy of an elastic network does not depend on the displacement components of particles which are orthogonal to the directions of equilibrium links between them (cf. equations (A.1)). Therefore, such displacements may indeed grow indefinitely without increasing the energy of the linearized system. In the full nonlinear model, the energy is invariant, on the other hand, only under the displacements of particles which preserve distances between all of them, i.e the lengths of all elastic links. They correspond to rigid translations and rotations of the entire network, always eliminated in our simulations via the immobilization procedure.
A.5 actin-actin binding sites 107 Figure 40: Actin-Actin-Binding Sites. Actin inside the filament (blue beads) is connected to its neighbors (gray beads). Its bind-ing sites are colored green and the neighbor binding sites red, respectively. Green and red beads that are within the cutoff distance can form a bond.
a.5 actin-actin binding sites In Chap. 6, we model the actin filament with effective protein-protein interactions approximated as phenomenological Lennard-Jones bonds (3.20) between pairs of residues in different actin molecules. Fuji et al. could resolve the F-actin filament in atomic resolution by cryo-electron microscopy . We are interested in the protein-protein interaction between single actin monomers in the filament. Thus, we construct an EN model of the actin fil-ament with a cutoff distance of l0 = 8.5Å. Links that connect different actin monomers in the filament are modeled as break-able Lennard-Jones potentials (3.20). Such pairs of residues are displayed in Fig. 40 and listed below: 39–270, 40–169, 40–171, 40–268, 41–169, 42–168, 42–169, 42–170, 42–171, 43–168, 43–169, 44–139, 44–143, 44–168, 44–169, 44–170, 45–143, 45–168, 45–346, 61–289, 62–286, 62–288, 62–289, 63–285, 63–286, 63–287, 63–288, 63–289, 64–166, 64–167, 110–194, 110– 195, 110–196, 110–197, 111–195, 111–196, 111–197, 112–195, 112– 196, 112–197, 172–268, 173–267, 173–268, 202–286, 204–286, 204– 287, 204–288, 205–286, 205–287, 205–288, 208–288, 242–287, 242– 288, 243–287, 243–288, 243–289, 243–290, 243–291, 244–283, 244– 286, 244–287, 244–288, 244–289, 244–290, 244–291, 244–322, 244– 325, 245–287, 245–290, 245–291, 245–321, 245–322, 245–323, 245– 324, 245–325, 246–321, 246–322 and 247–322 108 appendix The binding sites are shown in Fig. 40.
a.6 tables Table 3: Maximal distance changes (Å) observed when forces are ap-plied to different residues in the nucleotide-binding pocket residue 343 to 517 386 to 517 789 to 141 789 to 92 111 0.126 0.069 7.199 6.346 112 0.134 0.078 8.275 7.231 113 0.148 0.067 8.029 7.222 114 0.146 0.104 9.112 8.065 115 0.171 0.140 9.517 8.533 116 0.167 0.138 9.825 8.476 163 0.124 0.167 4.304 3.719 164 0.137 0.186 4.131 3.746 165 0.140 0.203 4.001 3.869 166 0.131 0.157 4.897 4.817 167 0.123 0.126 5.439 4.827 168 0.121 0.117 5.768 5.114 169 0.114 0.156 4.859 4.279 170 0.107 0.188 5.171 4.494 171 0.109 0.148 6.395 5.468 214 0.075 0.244 4.953 5.040 215 0.048 0.305 5.523 5.615 216 0.066 0.282 5.362 5.299 217 0.068 0.292 4.168 4.303 218 0.090 0.307 3.580 3.669 219 0.087 0.336 2.644 2.983 220 0.125 0.333 2.310 2.359 438 0.106 0.255 2.364 2.159 439 0.135 0.250 2.274 2.203 440 0.134 0.229 2.852 2.722 441 0.144 0.272 3.004 2.682 442 0.228 0.286 2.956 2.470 A.6 tables 109 Table 4: Maximal distance changes (Å) observed when forces are ap-plied to different residues in the actin-binding pocket residue 789 to 141 789 to 92 115 to 297 442 to 219 340 4.911 6.086 0.359 0.095 341 5.103 6.191 0.409 0.107 342 5.431 5.790 0.430 0.111 343 6.643 6.378 0.466 0.107 344 6.884 6.007 0.480 0.118 345 6.531 5.374 0.468 0.127 346 6.919 5.619 0.442 0.111 347 6.898 6.160 0.423 0.094 348 6.226 6.104 0.401 0.090 349 6.545 6.892 0.388 0.079 350 6.376 7.288 0.357 0.077 377 7.457 6.754 0.341 0.061 378 7.395 6.492 0.332 0.069 379 8.492 7.168 0.346 0.075 380 9.036 7.613 0.370 0.098 381 9.204 7.994 0.385 0.109 382 9.377 7.998 0.395 0.106 383 9.533 8.123 0.403 0.116 384 9.374 8.130 0.411 0.107 385 9.494 8.152 0.420 0.102 386 9.359 8.028 0.412 0.103 387 9.425 8.161 0.418 0.101 388 9.163 7.815 0.399 0.071 389 8.631 7.215 0.410 0.081 390 7.634 6.752 0.397 0.078 500 3.929 3.617 0.136 0.019 501 4.952 4.752 0.155 0.053 502 5.688 5.403 0.184 0.069 503 5.905 5.307 0.199 0.055 504 5.216 4.540 0.177 0.035 505 4.631 3.732 0.163 0.021 506 4.074 3.315 0.140 0.020 516 6.723 5.423 0.207 0.103 517 6.591 5.050 0.179 0.080 518 6.065 4.319 0.171 0.058 519 5.378 3.560 0.162 0.067 520 4.713 2.870 0.163 0.075 521 5.584 3.334 0.202 0.072 522 6.114 4.012 0.207 0.038 523 5.386 3.621 0.191 0.033 524 5.261 3.627 0.208 0.064 525 6.319 4.426 0.236 0.060 526 6.227 4.830 0.228 0.037 527 5.616 4.385 0.218 0.056 528 6.419 5.106 0.252 0.089 529 6.908 5.667 0.272 0.088 530 6.689 5.728 0.249 0.086 540 6.058 4.730 0.260 0.145 541 5.986 4.442 0.267 0.172 542 6.953 4.672 0.292 0.199 543 7.004 4.175 0.287 0.197 544 5.939 3.638 0.262 0.160 545 5.990 3.735 0.253 0.125 634 4.464 3.959 0.100 0.117 B I B L I O G R A P H Y 1. R. Ait-Haddou and W. Herzog, Brownian ratchet models of molecular motors., Cell Biochemistry and Biophysics, 38 (2003), pp. 191–214.
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E I D E S S TAT T L I C H E E R K L Ä R U N G Ich erkläre hiermit, dass ich diese Dissertation selbstständig ohne Hilfe Dritter und ohne Benutzung anderer als der angegebe-nen Quellen und Hilfsmittel verfasst habe. Alle den benutzten Quellen wörtlich oder sinngemäßentnommenen Stellen sind als solche einzeln kenntlich gemacht.
Berlin , September 2012 Markus Düttmann |
6842 | https://arxiv.org/pdf/1709.03038 | Published Time: Sun, 22 Jan 2023 19:14:20 GMT
arXiv:1709.03038v2 [math.CA] 21 Feb 2018
On functional equations characterizing derivations: methods and examples
Eszter Gselmann, Gergely Kiss and Csaba Vincze February 22, 2018
Abstract
Functional equations satisfied by additive functions have a special interest not only in the theory of functional equations, but also in the theory of (commutative) algebra because the fundamental notions such as derivations and automorphisms are additive functions satisfying some further functional equations as well. It is an important question that how these morphisms can be characterized among additive mappings in general. The paper contains some multivariate characterizations of higher order derivations. The univariate characterizations are given as consequences by the diagonalization of the multivariate formulas. This method allows us to refine the process of computing the solutions of univariate functional equations of the form n∑
k=1
xpk fk(xqk ) = 0 ,
where pk and qk (k = 1 , . . . , n ) are given nonnegative integers and the unknown functions
f1, . . . , f n : R → R are supposed to be additive on the ring R. It is illustrated by some explicit examples too. As another application of the multivariate setting we use spectral analysis and spectral syn-thesis in the space of the additive solutions to prove that it is spanned by differential operators. The results are uniformly based on the investigation of the multivariate version of the functional equations.
1 Introduction
Functional equations satisfied by additive functions have a rather extensive literature , , , , , . They appear not only in the theory of functional equations, but also in the theory of (commutative) algebra , , , , . It is an important question that how special morphisms can be characterized among additive mappings in general. This paper is devoted to the case of functional equations characterizing derivations. It is motivated by some recent results , , due to B. Ebanks. We are looking for solutions of functional equations of the form
n
∑
k=1
xpk fk(xqk ) = 0 , (1) where pk and qk (k = 1 , . . . , n ) are given nonnegative integers and the unknown functions f1, . . . ,
fn : R → R are supposed to be additive on the ring R. According to the homogeneity of additive functions [1, Lemma 2.2] it is enough to investigate equations with constant pairwise sum of the powers, i.e.
p1 + q1 = . . . = pn + qn.
1The general form of the solutions is formulated as a conjecture in [1, Conjecture 4.15]. The proof can be found in by using special substitutions of the variable and an inductive argument. We have a significantly different starting point by following the method of free variables. The paper contains some multivariate characterizations of higher order derivations in Section 2. The basic re-sults of , (univariate characterizations of higher order derivations) are given as consequences by the diagonalization of the multivariate formulas, see Section 3. The method allows us to refine the process of computing the solutions of functional equations of the form (1). The examples (Ex-amples I, Examples II) illustrate that the multivariate version of the functional equations provides a more effective and subtle way to determine the structure of the unknown functions. Especially, functional equations with missing powers (Examples II) can be investigated in this way to avoid formal (identically zero) terms in the solution. As a refinement of Ebanks’ method we follow the main steps such as (i) The formulation of the multivariate version of the functional equation. (ii) The substitution of value 1 as many times as the number of the missing powers (due to the symmetry of the variables there is no need to specify the positions for the substitution). (iii) The application of Theorem 5/Corollary 4. The second step decreases the homogeneity degree of the functional equation to keep only non-identically zero terms in the solutions. Therefore it can be easily seen how the number of the nonzero coefficients is related to the maximal order of the solution . In Section 4 we present another approach to the problem. The application of the spectral ana-lysis and the spectral synthesis is a general method to find the additive solutions of a functional equation, see . It is a new and important trend in the theory of functional equations; see e.g. and . Although the domain should be specified as a finitely generated subfield over the ratio-nals, the multivariate version of the functional equation generates a system of functional equations concerning the translates of the original solutions in the space of complex valued additive functions restricted to the multiplicative subgroup of the given subfield. Taking into account the fundamen-tal result [7, Theorem 4.3] such a closed translation invariant linear subspace of additive functions contains automorphism solutions (spectral analysis) and the space of the solutions is spanned by the compositions of automorphisms and differential operators (spectral synthesis). All functional equa-tions in the paper are also discussed by the help of the spectral analysis and the spectral synthesis. We prove that the automorphism solutions must be trivial (identity) and, consequently, the space of the solutions is spanned by differential operators. According to the results presented in Section 2 and Section 3, the investigation of the detailed form of the differential operator solutions is omitted. However, Subsection 4.4 contains an alternative way to prove Theorem 5/Corollary 4 in the special case n = 3 . The proof uses a descending process instead of the inductive argument. In what follows we summarize some basic theoretical facts, terminology and notations.
Derivations
For the general theory of derivations we can refer to Kuczma , see also Zariski–Samuel and Kharchenko .
Definition 1. Let Q be a ring and consider a subring P ⊂ Q. A function f : P → Q is called a
derivation if it is additive, i.e.
f (x + y) = f (x) + f (y) (x, y ∈ P )
2and also satisfies the so-called Leibniz rule
f (xy ) = f (x)y + xf (y) (x, y ∈ P ) .
Example 1. Let F be a field, and F[x] be the ring of polynomials with coefficients from F. For a polynomial p ∈ F[x], p(x) = ∑nk=0 akxk, define the function f : F[x] → F[x] as
f (p) = p′,
where p′(x) = ∑nk=1 ka kxk−1 is the derivative of the polynomial p. Then the function f clearly fulfills
f (p + q) = f (p) + f (q)
and
f (pq ) = pf (q) + qf (p)
for all p, q ∈ F[x]. Hence f is a derivation.
Example 2. Let F be a field, and suppose that we are given a derivation f : F → F. We define the mapping f0 : F[x] → F[x] in the following way. If p ∈ F[x] has the form
p(x) =
n
∑
k=0
akxk,
then let
f0(p) = pf (x) =
n
∑
k=0
f (ak)xk.
Then f0 : F[x] → F[x] is a derivation. The following lemma says that the above two examples have rather fundamental importance.
Lemma 1. Let (K, +, ·) be a field and let (F, +, ·) be a subfield of K. If f : F → K is a derivation, then for any a ∈ F and for arbitrary polynomial p ∈ F[x] we have
f (p(a)) = pf (a) + f (a)p′(a).
As the following theorem shows, in fields with characteristic zero, with the aid of the notion of algebraic base, we are able to construct non-identically zero derivations.
Theorem 1. Let K be a field of characteristic zero, let F be a subfield of K, let S be an algebraic base of K over F, if it exists, and let S = ∅ otherwise. Let f : F → K be a derivation. Then, for every function u : S → K, there exists a unique derivation g : K → K such that g|F = f and
g|S = u.Remark 1. If R is a commutative ring and f : R → R is a derivation, then
f (x2) = 2 xf (x) (x ∈ R)
is a direct consequence of the Leibniz rule. In general this identity does not characterize derivations among additive functions as the following argument shows. Substituting z = x + y in place of xf (z2) = 2( x + y)f (x + y) = 2 ( xf (x) + xf (y) + yf (x) + yf (y)) .
3On the other hand
f (z2) = f (x2 + 2 xy + y2) = 2 xf (x) + 2 f (xy ) + 2 yf (y).
Therefore
2f (xy ) = 2 xf (y) + 2 yf (x) (x ∈ R) ,
i.e. 2f is a derivation. Unfortunately the division by 2 is not allowed without any further assumption on the ring R. Therefore some additional restrictions on the ring appear typically in the results. The notion of derivation can be extended in several ways. We will employ the concept of higher order derivations according to Reich and Unger–Reich .
Definition 2. Let R be a ring. The identically zero map is the only derivation of order zero . For each n ∈ N, an additive mapping f : R → R is termed to be a derivation of order n, if there exists
B : R × R → R such that B is a bi-derivation of order n − 1 (that is, B is a derivation of order n − 1
in each variable) and
f (xy ) − xf (y) − f (x)y = B(x, y ) (x, y ∈ R) .
The set of derivations of order n of the ring R will be denoted by Dn(R).
Remark 2. Since D0(R) = {0}, the only bi-derivation of order zero is the identically zero function, thus f ∈ D1(R) if and only if
f (xy ) = xf (y) + f (x)y
that is, the notions of first order derivations and derivations coincide.
Remark 3. Let R be a commutative ring and d : R → R be a derivation. Let further n ∈ N be arbitrary and
d0 = id , and dn = d ◦ dn−1 (n ∈ N) .
Then the mapping dn : R → R is a derivation of order n.To see this, we will use the following formula
dk(xy ) =
k
∑
i=0
(ki
)
di(x)dk−i(y) (x, y ∈ R) ,
which is valid for any k ∈ N, and can be proved by induction on k.For n = 1 the above statement automatically holds, since the notion of first order derivations and that of derivations coincide. Let us assume that there exists an n ∈ N so that the statement is fulfilled for any k ≤ n, that is,
dk ∈ Dk(R) holds. Then
dn+1 (xy ) =
n+1
∑
i=0
(n + 1
i
)
di(x)dn+1 −i(y) (x, y ∈ R) ,
yielding that
dn+1 (xy ) − xd n+1 (y) − yd n+1 (x) =
n
∑
i=1
(n + 1
i
)
di(x)dn+1 −i(y) (x, y ∈ R) .
Thus, the only thing that has to be clarified is that the mapping B : R × R → R defined by
B(x, y ) =
n
∑
i=1
(n + 1
i
)
di(x)dn+1 −i(y) (x, y ∈ R)
4is a bi-derivation of order n. Due to the induction hypothesis, for any k ≤ n, the function dk is a derivation of order k. Further, we also have Dk−1(R) ⊂ Dk(R) for all k ∈ N and due to the fact that Dk(R) is an R-module, we obtain that the function B is a derivation of order n in each of its variables. Of course, Dn(R) \ Dn−1(R) 6 = ∅ does not hold in general. To see this, it is enough to bear in mind that Dn(Z) = {0} for any n ∈ N.At the same time, in case R is an integral domain with char( R) > n or char( R) = 0 and there is a non-identically zero derivation d : R → R, then for any n ∈ N we have
dn ∈ Dn(R) and dn /∈ Dn−1(R),
yielding immediately that Dn(R) \ Dn−1(R) 6 = ∅.This can also be proved by induction on n. For n = 1 this is automatically true, since d : R → R
is a non-identically zero derivation. Assume now that there is an n ∈ N so that the statement holds for any k ≤ n − 1 and suppose to the contrary that dn ∈ Dn−1(R).Then due to the definition of higher order derivations, the mapping B : R × R → R defined by
B(x, y ) =
n−1
∑
i=1
(ni
)
di(x)dn−i(y) (x, y ∈ R)
has to be a bi-derivation of order n − 2, i.e.,
R ∋ x 7 −→ B(x, y ∗) =
n−1
∑
i=1
(ni
)
di(x)dn−i(y∗)
has to be a derivation of order n − 2, where y∗ ∈ R is kept fixed. Clearly, di ∈ Dn−2(R) holds, if i ≤ n − 2. Since Dn−2(R) is an R-module, from this
nd (y∗) dn−1 ∈ Dn−2(R) would follow, which is a contradiction.
Multiadditive functions
Concerning multiadditive functions we follow the terminology and notations of L. Sz´ ekelyhidi , .
Definition 3. Let G, S be commutative semigroups, n ∈ N and let A : Gn → S be a function. We say that A is n-additive if it is a homomorphism of G into S in each variable. If n = 1 or n = 2
then the function A is simply termed to be additive or biadditive , respectively. The diagonalization or trace of an n-additive function A : Gn → S is defined as
A∗(x) = A (x, . . . , x ) (x ∈ G) .
As a direct consequence of the definition each n-additive function A : Gn → S satisfies
A(x1, . . . , x i−1, kx i, x i+1 , . . . , x n)= kA (x1, . . . , x i−1, x i, x i+1 , . . . , x n) (x1, . . . , x n ∈ G)
for all i = 1 , . . . , n , where k ∈ N is arbitrary. The same identity holds for any k ∈ Z provided that
G and S are groups, and for k ∈ Q, provided that G and S are linear spaces over the rationals. For the diagonalization of A we have
A∗(kx ) = knA∗(x) (x ∈ G) .
5One of the most important theoretical results concerning multiadditive functions is the so-called
Polarization formula , that briefly expresses that every n-additive symmetric function is uniquely
determined by its diagonalization under some conditions on the domain as well as on the range. Suppose that G is a commutative semigroup and S is a commutative group. The action of the
difference operator ∆ on a function f : G → S is defined by the formula
∆y f (x) = f (x + y) − f (x);
note that the addition in the argument of the function is the operation of the semigroup G and the subtraction means the inverse of the operation of the group S.
Theorem 2 (Polarization formula) . Suppose that G is a commutative semigroup, S is a commuta-tive group, n ∈ N and n ≥ 1. If A : Gn → S is a symmetric, n-additive function, then for all
x, y 1, . . . , y m ∈ G we have
∆y1,...,y m A∗(x) =
{ 0 if m > n n!A(y1, . . . , y m) if m = n.
Corollary 1. Suppose that G is a commutative semigroup, S is a commutative group, n ∈ N and
n ≥ 1. If A : Gn → S is a symmetric, n-additive function, then for all x, y ∈ G
∆ny A∗(x) = n!A∗(y).
Lemma 2. Let n ∈ N, n ≥ 1 and suppose that the multiplication by n! is surjective in the com-mutative semigroup G or injective in the commutative group S. Then for any symmetric, n-additive function A : Gn → S, A∗ ≡ 0 implies that A is identically zero, as well.
The polarization formula plays the central role in the investigations of functional equations char-acterizing higher order derivations on a ring. This is another reason (see also Remark 1) why some additional restrictions on the the ring appear in the results. They essentially correspond to the con-ditions for the multiplication by n! in the domain as well as in the range in Lemma 2.
2 Multivariate characterizations of higher order derivations
In what follows we frequently use summation with respect to the cardinality of I ⊂ { 1, . . . , n + 1 }
as I runs through the elements of the power set 2{1,...,n +1 }, where n ∈ N. As another technical notation, we introduce the hat operator ̂ to delete arguments from multivariate expressions. Let
R be a commutative ring and consider the action of a second order derivation A ∈ D2(R) on the product of three independent variables as a motivation of the forthcoming results:
A(x1x2x3) − x1A(x2x3) − A(x1)x2x3 = B(x1, x 2x3) = x2B(x1, x 3) + B(x1, x 2)x3
= x2 (A(x1x3) − x1A(x3) − x3A(x1)) + ( A(x1x2) − x1A(x2) − A(x1)x2) x3. (2) In general
n
∑
i=0
(−1) i ∑
card( I)= i
(∏
j∈I
xj
)
· A
∏
k∈{ 1,...,n +1 }\ I
xk
= 0 (x1, . . . , x n+1 ∈ R) (3) as a simple inductive argument shows. Conversely suppose that equation (3) is satisfied and let us define the (symmetric) biadditive mapping by
B(x, y ) = A(xy ) − A(x)y − xA (y).
6An easy direct computation shows that A satisfies equation (3) with n ∈ N if and only if B satisfies equation (3) for each variable with n − 1 ∈ N, where n ≥ 1. By a simple inductive argument we can formulate the following result.
Theorem 3. Let A : R → R be an additive mapping, where R is a commutative ring, n ∈ N and
n ≥ 1. A ∈ Dn(R) if and only if
n
∑
i=0
(−1) i ∑
card( I)= i
(∏
j∈I
xj
)
· A
∏
k∈{ 1,...,n +1 }\ I
xk
= 0 (x1, . . . , x n+1 ∈ R). (4) As a generalization of the previous result we admit more general coefficients in equation (3). Some additional requirements for the ring R should be also formulated.
Definition 4. A commutative unitary ring R is called linear if it is a linear space over the field of rationals.
Remark 4. It can be easily seen that a linear ring admits the multiplication of the ring elements by fractions such as 1/2, . . . , 1/n, . . . . Using the linear space properties the field of the rationals Q can be isomorphically embedded into R as a subring. Therefore each element
1, 2 = 1 + 1, . . . , n = 1 + . . . + 1︸ ︷︷ ︸
n-times
is invertible, where 1 is the unity of the ring, n ∈ N and n ≥ 1. Hence the multiplication by n!
is injective in R as the domain of the higher order derivations (cf. Lemma 2). Some typical exam-ples for linear rings: fields, rings formed by matrices over a field, polynomial rings over a field. Another natural candidates are the integral domains. Since each cancellative semigroup G can be isomorphically embedded into a group we have that each integration domain R can be isomorphi-cally embedded into a field F as a subring. In the sense of Definition 1 we can take F as the range of derivations on the subring R. Such an extension of R provides the multiplication by n! to be obviously surjective in the range F, assuming that the characteristic is zero (cf. Lemma 2). In what follows we do not use special notation for the unity of the ring. The meaning of 1
depends on the context as usual.
Theorem 4. Let A : R → R ⊂ F be an additive mapping satisfying A(1) = 0 , where R is an integral domain with unity, F is its embedding field, assuming that char( F) = 0 and n ∈ N.
A ∈ Dn(R) if and only if there exist constants a1, . . . , a n+1 ∈ F, not all zero, such that
n
∑
i=0
an+1 −i
(n+1
i
) ∑
card( I)= i
(∏
j∈I
xj
)
· A
∏
k∈{ 1,...,n +1 }\ I
xk
= 0 , (5)
where x1, . . . , x n+1 ∈ R.Especially,
n+1
∑
i=1
ia i = 0 provided that A is a not identically zero solution. Proof. If A ∈ Dn(R) then we have equation (5) with
an+1 −i = ( −1) i
(n + 1
i
)
(i = 0 , . . . , n ) ;
7none of the constants a1, . . . , a n+1 is zero and
n+1
∑
i=1
ia i = 0 . Conversely, let n ∈ N, n ≥ 1 and
A : R → R ⊂ F be an additive function such that A(1) = 0 . Suppose that equation (5) holds for all
x1, . . . , x n+1 ∈ R. Substituting
x1 = . . . = xn = 1 and xn+1 = x ∈ R
it follows that (n+1 ∑
i=1
ia i
)
· A(x) = 0 (x ∈ R) ,
i.e.
n+1
∑
i=1
ia i = 0 provided that A is a not identically zero solution of equation (5). If n = 1 , then it takes the form
2a2A(xy ) + a1 (xA (y) + yA (x)) = 0 (x, y ∈ R) .
Since 2a2 + a1 = 0 , it follows that a2 6 = 0 . Otherwise 0 = a2 = a1 which is a contradiction. Therefore
a2A(xy ) − a2 (xA (y) + yA (x)) = 0 (x, y ∈ R)
and
A(xy ) = xA (y) + yA (x) (x, y ∈ R) .
This means that A ∈ D1(R), i.e. the statement holds for n = 1 . The inductive argument can be completed as follows. Taking xn+1 = 1 equation (5) gives that
0 =
n
∑
i=1
an+1 −i
(n+1
i
) ∑
n+1 ∈I
∑
card( I)= i
∏
j∈I{ n+1 }
xj
· A
∏
k∈{ 1,...,n }\ (I{ n+1 })
xk
+
n−1
∑
i=0
an+1 −i
(n+1
i
) ∑
n+1 /∈I
∑
card( I)= i
(∏
j∈I
xj
)
· A
∏
k∈{ 1,...,n }\ I
xk
+ a1
(n+1
n
) ∏
j∈{ 1,...,n }
xj
· A(1) =
n−1
∑
i=0
(
an+1 −(i+1)
(n+1
i+1
) + an+1 −i
(n+1
i
)) ∑
card( I)= i
(∏
j∈I
xj
)
· A
∏
k∈{ 1,...,n }\ I
xk
=
n−1
∑
i=0
˜a(n−1)+1 −i
(n−1+1
i
) ∑
card( I)= i
(∏
j∈I
xj
)
· A
∏
k∈{ 1,..., (n−1)+1 }\ I
xk
(x1, . . . , x n ∈ R)
because A(1) = 0 . We have that A ∈ Dn−1(R) ⊂ Dn(R) (inductive hypothesis) or all the coeffi-cients are zero, that is,
˜a(n−1)+1 −i =
(n − 1 + 1
i
) ( an+1 −(i+1)
(n+1
i+1
) + an+1 −i
(n+1
i
))
= 0 (0 ≤ i ≤ n − 1)
and, consequently,
(n + 1 − i)an+1 −i + ( i + 1) an+1 −(i+1) = 0 (0 ≤ i ≤ n − 1) .
Therefore
an+1 −(i+1) = ( −1) i+1
(n + 1
i + 1
)
an+1 (0 ≤ i ≤ n − 1) . (6) 8Substituting into (5)
an+1 ·
n
∑
i=0
(−1) i ∑
card( I)= i
(∏
j∈I
xj
)
· A
∏
k∈{ 1,...,n +1 }\ I
xk
= 0 (x ∈ R) .
This means that A ∈ Dn(R) since an+1 can not be zero due to the recursive formula (6) and the condition to provide the existence of a nonzero element among the constants a1, . . . , a n+1 .
Remark 5. The condition for R to be an integral domain with unity embeddable into a field F with characteristic zero, is used only in the last step of the proof because of the division by an+1 . If the coefficients are supposed to be rationals, then the previous statement holds for a linear commutative ring with unity, see the next theorem.
Theorem 5. Let f1, . . . , f n+1 : R → R be additive functions such that fi(1) = 0 for all i =1, . . . , n + 1 , where R is a linear commutative ring with unity, n ∈ N and n ≥ 1. Functional equation
n
∑
i=0
1
(n+1
i
) ∑
card( I)= i
(∏
j∈I
xj
)
· fn+1 −i
∏
k∈{ 1,...,n +1 }\ I
xk
= 0 ( x1, . . . , x n+1 ∈ R) (7)
holds if and only if
fn+1 −i = ( −1) ii∑
k=0
(n + 1 − i + kk
)
Dn−i+k (i = 0 , . . . , n ) ,
where for all possible indices i, we have Di ∈ Di(R).Proof. If n = 1 then the equation takes the form
2f2(x1x2) + x1f1(x2) + x2f1(x1) = 0 .
Substituting x1 = x and x2 = 1 we have that
2f2(x) + f1(x) = 0
and, consequently,
f2(x1x2) − x1f2(x2) − x2f2(x1) = 0 .
This means that f2 = D1 ∈ D1(R) and f1 = −D0 − 2D1, where D0 ∈ D0(R) is the identically zero function. The converse of the statement is clear under the choice f2 = D1 and f1 = −D0 −2D1. The inductive argument is the same as in the proof of the previous theorem by the formal identification
fn+1 −i = an+1 −iA. We have
0 =
n−1
∑
i=0
1
(n−1+1
i
) ∑
card( I)= i
(∏
j∈I
xj
)
· ˜f(n−1)+1 −i
∏
k∈{ 1,...,n }\ I
xk
, (8) where ˜f(n−1)+1 −i = ( i + 1) fn+1 −(i+1) + ( n + 1 − i)fn+1 −i (0 ≤ i ≤ n − 1) .
Using the inductive hypothesis
(n + 1 − i)fn+1 −i + ( i + 1) fn+1 −(i+1) = ( −1) ii∑
k=0
(n − i + kk
)˜D(n−1) −i+k , (9) 9where ˜Di ∈ Di(R) for all possible indices. Especially, all the unknown functions f1, . . . , f n can be expressed in terms of fn+1 and ˜D0, . . . , ˜Dn−1 in a recursive way:
fn+1 −(i+1) = ( −1) i
{( i∑
k=0
(n − ki − k
) ˜Dn−(k+1)
k + 1
)
−
(n + 1
i + 1
)
fn+1
}
, (10) where i = 0 , . . . , n − 1. Rescaling the indices,
fn+1 −i = fn+1 −(i−1+1)
= ( −1) i−1
{( i−1∑
k=0
( n − ki − 1 − k
) ˜Dn−(k+1)
k + 1
)
−
(n + 1
i
)
fn+1
}
, (11) where i = 1 , . . . , n . Substituting into equation (7)
n
∑
i=0
(−1) i ∑
card( I)= i
(∏
j∈I
xj
)
· fn+1
∏
k∈{ 1,...,n +1 }\ I
xk
= 0 ,
because the term containing ˜Dl is of the form
n
∑
i=n−l
(−1) i−1
(n+1
i
)( l + 1
i − n + l
) ∑
card( I)= i
(∏
j∈I
xj
)
· ˜Dl
n − l
∏
k∈{ 1,...,n +1 }\ I
xk
= 1
(n+1
n−l
)
n
∑
i=n−l
(−1) i−1
( in − l
) ∑
card( I)= i
(∏
j∈I
xj
)
· ˜Dl
n − l
∏
k∈{ 1,...,n +1 }\ I
xk
= (−1) n−l−1
(n+1
n−l
)
l
∑
m=0
(−1) m
(m + n − ln − l
)
· ∑
card( I)= n−l+m
(∏
j∈I
xj
)
· ˜Dl
n − l
∏
k∈{ 1,...,n +1 }\ I
xk
= (−1) n−l−1
(n+1
n−l
) ∑
1≤j1<...<j l+1 ≤n+1
∏
k∈{ 1,...,n +1 }{ j1,...,j l+1 }
xk
·
l
∑
m=0
(−1) m ∑
card( J)= m
(∏
j∈J
xj
)
· ˜Dl
n − l
∏
k∈{ j1,...,j l+1 }\ J
xk
= 0 ,
due to Theorem 3. Finally, let
Dn = fn+1 and Dn−(k+1) = − ˜Dn−(k+1)
k + 1 (k = 0 , . . . , n − 1) .
10 Taking m = i − 1 − k in formula (11) it follows that
fn+1 −i = ( −1) i−1
{( i−1∑
m=0
(n + 1 − i + mm
) ˜Dn−i+m
i − m
)
−
(n + 1
i
)
fn+1
}
= ( −1) i
{( i−1∑
m=0
(n + 1 − i + mm
)
Dn−i+m
)
+
(n + 1
i
)
fn+1
}
= ( −1) ii∑
m=0
(n + 1 − i + mm
)
Dn−i+m
as had to be proved. The converse statement is a straightforward calculation.
Remark 6. Theorem 5 can be also stated in case of integral domains provided that the range of the functions f1, . . . , f n+1 is extended to a field of characteristic zero, containing R as a subring; see Remark 4. The proof works without any essential modification.
3 Univariate characterizations of higher order derivations
Each multivariate characterization implies an univariate characterization of the higher order deriva-tions under some mild conditions on the ring R due to Lemma 2. The ring R is supposed to be a linear commutative, unitary ring or an integral domain embeddable to a field of characteristic zero; see also Remark 4. The following results can be found in , but the proofs are essentially different. We present them as direct consequences of the multivariate characterizations by taking the diagonalization of the formulas. This provides unified arguments of specific results but we can also use the idea as a general method to solve functional equations of the form (1); see Examples I and Examples II.
Corollary 2. Let A : R → R be an additive mapping, where R is a linear commutative, unitary ring or an integral domain with unity, embeddable to a field of characteristic zero and n ∈ N and
n ≥ 1. Then A ∈ Dn(R) if and only if
n
∑
i=0
(−1) i
(n + 1
i
)
xiA (xn+1 −i) = 0 (x ∈ R) . (12)
Corollary 3. Let A : R → R ⊂ F be an additive mapping satisfying A(1) = 0 , where R is an integral domain with unity, F is its embedding field with char ( F) = 0 , n ∈ N and n ≥ 1.
A ∈ Dn(R) if and only if there exist constants a1, . . . , a n+1 ∈ F , not all zero, such that
n
∑
i=0
an+1 −ixiA (xn+1 −i) = 0 (13)
for any x ∈ R.Especially,
n+1
∑
i=1
ia i = 0 provided that A is a not identically zero solution.
As an application of Theorem 5 we are going to give the general form of the solutions of func-tional equations of the form
n
∑
k=1
xpk fk(xqk ) = 0 (x ∈ R) , (14) 11 where n ∈ N, n ≥ 1, pk, q k are given nonnegative integers for all k = 1 , . . . , n and f1, . . . , f n : R →
R are additive functions on the ring R. The problem is motivated by B. Ebanks’ paper , although there are some earlier relevant results. For example the case n = 2 is an easy consequence of [5, Theorem 6]. Using the homogeneity of additive functions it can be easily seen that equation (14) can be assumed to be of constant degree of homogeneity:
pk + qk = l (k = 1 , . . . , n );
see [1, Lemma 2.2]. In other words collecting the addends of equation (14) of the same degree of homogeneity we discuss the vanishing of the left hand side term by term. To formulate the multivariate version of the functional equation let us define the mapping
Φ( x1, . . . , x l) =
n
∑
k=1
1
( lpk
) ∑
card( I)= pk
∏
j∈{ 1,...,l }\ I
xj
· fk
(∏
i∈I
xi
)
(x1, . . . , x l ∈ R) ,
where the summation is taken for all subsets I of cardinality pk of the index set {1, 2, . . . , l }. Due to the additivity of the functions f1, . . . , f n, the mapping Φ : Rl → R is a symmetric l-additive function with vanishing trace
Φ∗(x) = Φ( x, . . . , x ) =
n
∑
k=1
xpk fk(xqk ) = 0 (x ∈ R)
and we can conclude the equivalence of the following statements by Lemma 2: (i) The functions f1, . . . , f n fulfill equation (14). (ii) For any x1, . . . , x l ∈ R
n
∑
k=1
1
( lpk
) ∑
card( I)= pk
∏
j∈{ 1,...,l }\ I
xj
· fk
(∏
i∈I
xi
)
= 0 .
In what follows we summarize all the simplifications we use to solve equation (14). (C 0) R is a linear commutative, unitary ring or an integral domain with unity embeddable to a field of characteristic zero. (C 1) Each addend has the same degree of homogeneity, i.e. pk + qk = l for all k = 1 , . . . , n .If pi = pj for some different indices i 6 = j ∈ { 1, . . . , n } then qi = qj by the constancy of the degree of homogeneity and we can write that
xpi fi(xqi ) + xpj fj (xqj ) = xpi ˜f(xqi ),
where ˜f = fi + fj . Therefore the number of the unknown functions has been reduced. If qi = 0
then, by choosing j 6 = i, we can write equation (14) into the form
∑
ν∈{ 1,...,n }{ i,j }
xpν fν (xqν ) + xpj ˜f (xqj ) = 0 ,
where ˜f (x) = fi(1) x + fj (x) and the number of the unknown functions has been reduced again. Without loss of the generality we can suppose that 12 (C 2) p1, . . . , p n are pairwise different nonnegative integers, q1, . . . , q n are positive, pairwise differ-ent integers. Especially, l ≥ n because of (C 1) . By multiplying equation (14) with x if necessary we have that l ≥ n + 1 .(C 3) Moreover, for any k = 1 , . . . , n , condition fk(1) = 0 can be assumed. Otherwise, let us introduce the functions
˜fk(x) = fk(x) − fk(1) x (x ∈ R) .
They are obviously additive and, for any x ∈ R
n
∑
k=1
xpk ˜fk(xqk ) =
n
∑
k=1
xpk [fk(xqk ) − fk(1) xqk ] =
n
∑
k=1
xpk fk(xqk ) − xln∑
k=1
fk(1) = 0
because
n
∑
k=1
fk(1) = 0 due to (14). Assuming conditions (C 0) , (C 1) , (C 2) and (C 3) it is enough to investigate functional equation
n
∑
i=0
xifn+1 −i(xn+1 −i) = 0 (x ∈ R) . (15) For those exponents that do not appear in the corresponding homogeneous term of the original equation we assign the identically zero function (that is clearly additive).
Corollary 4. Let f1, . . . , f n+1 : R → R be additive functions such that fi(1) = 0 for all i =1, . . . , n + 1 , where R is a linear commutative ring with unity, n ∈ N and n ≥ 1. Functional equation n∑
i=0
xifn+1 −i
(xn+1 −i) = 0 (x ∈ R) (16)
holds, if and only if
fn+1 −i = ( −1) ii∑
k=0
(n + 1 − i + kk
)
Dn−i+k (i = 0 , . . . , n ) ,
where for all possible indices i, we have Di ∈ Di(R).Proof. The statement is a direct consequence of Theorem 5 and Lemma 2.
Remark 7. Corollary 4 can be also stated in case of integral domains provided that the range of the functions f1, . . . , f n+1 is extended to a field having characteristic zero, containing R as a subring; see Remark 4. The proof is working without any essential modification.
Examples I: equations without missing powers
As an application of the results presented above, we will show some examples. 13 Example 3. Let f : R → R be a non-identically zero additive function and assume that
f (x3) + xf (x2) − 2x2f (x) = 0
is fulfilled for any x ∈ R. Then the function A(x) = f (x) − f (1) x satisfies the conditions of Corollary 3. Therefore there exists a derivation D2 ∈ D2(R) such that
f (x) = D2(x) + f (1) x (x ∈ R) .
Moreover, if we define a3 = 1 , a2 = 1 and a1 = −2, then 3a3 + 2 a2 + a1 6 = 0 , i.e. D2 ≡ 0 in the sense of Corollary 3. The solution is f (x) = f (1) x, where f (1) 6 = 0 and x ∈ R.
Example 4. Let f : R → R be an additive function and assume that
f (x5) − 5xf (x4) + 10 x2f (x3) − 10 x3f (x2) + 5 x4f (x) = 0 (x ∈ R) .
Let
a5−i = ( −1) i
(5
i
)
(i = 0 , 1, 2, 3, 4) .
Then we have 4∑
i=0
a5−ixif (x5−i) = 0 (x ∈ R) .
Thus there exists a fourth order derivation D4 ∈ D4(R) such that
f (x) = D4(x) + f (1) x (x ∈ R) .
Since f (1) = 0 , the solution is f (x) = D4(x), where x ∈ R.
Examples II: equations with missing powers
The following examples illustrate how to use the method of free variables in the solution of func-tional equations with missing powers instead of the direct application of Theorem 5/Corollary 4. The key step is the substitution of value 1 as many times as the number of the missing powers (due to the symmetry of the variables there is no need to specify the positions for the substitution). This provides the reduction of the homogeneity degree of the functional equation to avoid formal (identically zero) terms in the solution. It is given in a more effective and subtle way. The method obviously shows how the number of the nonzero coefficients is related to the maximal order of the solution. In it was shown that if the number of nonzero coefficients ai ∈ R is m, then the solution of n∑
i=1
aixpi f (xqi ) = 0
is a derivation of order at most m − 1.
Example 5. Assume that we are given two additive functions f, g : R → R such that f (1) = g(1) = 0 and
f (x3) + x2g(x) = 0 (x ∈ R) .
If we define the functions f1, f 2, f 3 : R → R as f3 = f , f2 = 0 and f1 = g then, by using Corollary 4 with n = 2 , it follows that
f3 = D2
f2 = −3D2 − D1
f1 = 3D2 + 2 D1 + D0,
14 where Di ∈ Di(R), if i = 0 , 1, 2. It is a direct application of Corollary 4 used by B. Ebanks in , . Another way of the solution is to formulate the multivariate version of the equation in the first step:
3f (x1x2x3) + x1x2g(x3) + x1x3g(x2) + x2x3g(x1) = 0 (x1, x 2, x 3 ∈ R) .
If x1 = x2 = x and x3 = 1 , then we get that
3f (x2) + 2 xg (x) = 0 (x ∈ R) .
Applying Corollary 4 with n = 1 , f2 = 3 f and f1 = 2 g, it follows that
f2 = D1
2D1 + D0 + f1 = 0,
where Di ∈ Di(R) for all i = 0 , 1. Therefore 3f = D1 and g = −D1, where D1 ∈ D1(R).
Example 6. Let f, g, h : R → R be additive functions so that f (1) = g(1) = h(1) = 0 . Further-more, assume that
f (x5) + xg (x4) + x4h(x) = 0
is fulfilled for all x ∈ R. To determine the functions f, g, h , we will show two ways. The first is a direct application of the results above used by B. Ebanks in , . Let us define the functions f1, f 2, f 3, f 4, f 5 : R → R as f1 = h, f2 = 0 , f3 = 0 , f4 = g and f5 = f . Then the equation takes the form 4∑
i=0
xif5−i(x5−i) = 0 (x ∈ R) .
and, by Corollary 4, for any i = 0 , 1, 2, 3, 4
f5−i = ( −1) ii∑
k=0
(5 − i + kk
)
D4−i+k
holds, that is
f5 − D4 = 05 D4 + f4 + D3 = 0
−10 D4 − 4 D3 + f3 − D2 = 010 D4 + 6 D3 + 3 D2 + f2 + D1 = 0
−5 D4 − 4 D3 − 3 D2 − 2 D1 + f1 − D0 = 0,
where for all i = 0 , 1, 2, 3, 4 we have Di ∈ Di(R). Bearing in mind the above notations, for the functions f, g and h this yields that
f − D4 = 05 D4 + g + D3 = 0
−10 D4 − 4 D3 − D2 = 010 D4 + 6 D3 + 3 D2 + D1 = 0
−5 D4 − 4 D3 − 3 D2 − 2 D1 + h = 0.
The second way gives a much more precise form of the unknown functions by formulating the multivariate version of the equation in the first step:
5f (x1x2x3x4x5)+ g(x2x3x4x5)x1 + g(x1x3x4x5)x2 + g(x2x1x4x5)x3 + g(x2x3x1x5)x4 + g(x2x3x4x1)x5
x2x3x4x5h(x1) + x1x3x4x5h(x2) + x2x1x4x5h(x3) + x2x3x1x5h(x4) + x2x3x4x1h(x5)= 0 ,
15 where x1, x 2, x 3, x 4, x 5 ∈ R. Substituting x1 = x2 = x3 = x and x4 = x5 = 1 we get that
5f (x3) + 2 g(x3) + 3 xg (x2) + 3 x2h(x) = 0 .
Taking f3 = 5 f + 2 g, f2 = 3 g and f1 = 3 h it follows that
f3(x3) + xf 2(x2) + x2f1(x) = 0 (x ∈ R)
and, by Corollary 4,
f3 = D2, f2 = −D1 − 3D2, f1 = 2 D1 + 3 D2,
where Di ∈ Di(R) for all i = 1 , 2. This means that
15 f = 2D1 + 9 D2
3g = −D1 − 3D2
3h = 2D1 + 3 D2,
where Di ∈ Di(R) for all i = 1 , 2.
4 The application of the spectral synthesis in the solution of lin-ear functional equations
In what follows we present another approach to the problem of linear functional equations char-acterizing derivations among additive mappings in the special case of a finitely generated field K
over the field Q of rationals as the domain R of the equation, i.e. Q ⊂ K = Q(x1, . . . , x m) ⊂ C,where m ∈ N and C denotes the field of complex numbers. The linearity of the functional equa-tion means that the solutions form a vector space over C. The idea of using spectral synthesis to find additive solutions of linear functional equations is natural due to the fundamental work . The key result says that spectral synthesis holds in any translation invariant closed linear subspace formed by additive mappings on a finitely generated subfield K ⊂ C. Therefore such a subspace is spanned by so-called exponential monomials which can be given in terms of automorphisms of C
and differential operators (higher order derivations), see also and .
4.1 Basic theoretical facts
Let (G, ∗) be an Abelian group. By a variety V on G we mean a translation invariant closed linear subspace of CG, where CG denotes the space of complex valued functions defined on G. The space of functions is equipped with the product topology. The translation invariance of the linear subspace provides that fg : G → C, fg(x) = f (g ∗ x) is an element of V for any f ∈ V and g ∈ G. An additive mapping is a homomorphism of G into the additive group of C. The so-called polynomials
are the elements of the algebra generated by the additive and constant functions. An exponential mapping is a nonzero (and, consequently, injective) homomorphism of G into the multiplicative group of C, i .e. m ∈ CG such that m(x ∗ y) = m(x) · m(y). An exponential monomial is the product of an exponential and a polynomial function. The finite sums of exponential monomials are called polynomial-exponentials . If a variety V is spanned by exponential monomials then we say that spectral synthesis holds in V . If spectral synthesis holds in every variety on G then spectral synthesis holds on G. Especially, spectral analysis holds on G, i.e. every nontrivial variety contains an exponential; see Lemma 2.1 in . 16 To formulate the key result of let G := K∗ (the multiplicative group of K) and consider the variety Va on K∗ consisting of the restriction of additive functions on K (as an additive group) to
K∗, i.e.
Va = {A|K∗ | A(x + y) = A(x) + A(y), where x and y ∈ K}. (17) It can be easily seen that if A is an additive function then its translate Ac(x) = A(cx ) with respect to the multiplication by c ∈ K∗ is also additive. Theorem 4.3 in states that if the transcendence degree of K over Q is finite 1 then spectral synthesis holds in every variety V contained in Va.By Theorem 3.4 in , the polynomials in V correspond to mappings of the form D(x)/x (x ∈
K∗), where D is a differential operator on K. A differential operator means the (complex) linear combination of finitely many mappings of the form d1 ◦ . . . ◦ dk, where d1, . . . , dk are derivations on K. If k = 0 , then this expression is by convention the identity function. The exponentials in V
satisfy both m(x + y) = m(x) + m(y) and m(x · y) = m(x) · m(y), where x, y ∈ K and m(0) = 0 .Extending m to an automorphism of C (see Lemma 4.1 in ), a special expression for the elements (exponential monomials) spanning a variety V contained in Va can be given: they are of the form
ϕ ◦ D, where ϕ is the extension of an exponential m ∈ V to an automorphism of C and D is a differential operator on K; see Theorem 4.2 in .
4.2 Spectral synthesis in the variety generated by the solutions of equation
(12)
Let n ∈ N and consider an additive mapping A : K → C satisfying equation
n
∑
i=0
(−1) i
(n + 1
i
)
xiA (xn+1 −i) = 0 (x ∈ K) . (18) First of all observe that the solutions of functional equation (18) form a linear subspace. To get some information about the translates of the form Ac(x) = A(cx ) of the solutions we have to set the variable x free by using a symmetrization process:
Φ( x1, . . . , x n+1 ) = A(x1 · · · xn+1 ) − ∑
1≤i≤n+1
xiA(x1 · · · ˆxi · · · xn+1 )+ ∑
i≤i<j ≤n+1
xixj A (x1 · · · ˆxi · · · ˆxj · · · xn+1 ) −· · · + ( −1) n ∑
1≤i≤n+1
x1 · · · ˆxi · · · xn+1 A(xi) (x1, . . . , x n+1 ∈ K) .
Due to equation (18), the diagonal Φ∗(x) = Φ( x, . . . , x ) is identically zero. In the sense of Lemma 2, the mapping Φ is also identically zero. By some direct computations
Φ( x, . . . , x, cx )= A(cx n+1 ) −
(n
1
)
xA (cx n) − cxA (xn) +
(n
2
)
x2A(cx n−1) +
(n
1
)
cx 2A(xn−1)−
(n
3
)
x3A(cx n−2) −
(n
2
)
cx 3A(xn−2)+ . . . + ( −1) nxnA(cx ) + ( −1) n
( nn − 1
)
cx nA(x),
1If Kis finitely generated over Qthen it is automatically satisfied.
17 Φ( x, . . . , x, c )= A(cx n) −
(n
1
)
xA (cx n−1) − cA (xn) +
(n
2
)
x2A(cx n−2) +
(n
1
)
cxA (xn−1)−
(n
3
)
x3A(cx n−3) −
(n
2
)
cx 2A(xn−2)+ . . . + ( −1) nxnA(c) + ( −1) n
( nn − 1
)
cx n−1A(x)
and, consequently, for any c ∈ K∗
Φ( x, . . . , x, cx ) − xΦ( x, . . . , x, c ) =
n+1
∑
i=0
(−1) i
(n + 1
i
)
xiA (cx n+1 −i) (x ∈ K) ,
i.e. the translation invariant linear subspace generated by the solutions of equation (18) can be described by the family of equations
n+1
∑
i=0
(−1) i
(n + 1
i
)
xiA (cx n+1 −i) = 0 (x ∈ K, c ∈ K∗) . (19) Let c ∈ K∗ and x ∈ K be given and suppose that f : K → C is the limit function 2 of the sequence Al of solutions of equation (19), that is for any ε > 0 we have that
∣∣Al(cx n+1 −i) − f (cx n+1 −i)∣∣ < ε (i = 0 , . . . , n + 1)
provided that l is large enough. Then
∣∣∣∣∣
n+1
∑
i=0
(−1) i
(n + 1
i
)
xif (cx n+1 −i)∣∣∣∣∣
=
∣∣∣∣∣
n+1
∑
i=0
(−1) i
(n + 1
i
)
xif (cx n+1 −i) −
n+1
∑
i=0
(−1) i
(n + 1
i
)
xiAl
(cx n+1 −i)∣∣∣∣∣
≤
n+1
∑
i=0
(n + 1
i
)
|x|i ∣∣f (cx n+1 −i) − Al
(cx n+1 −i)∣ ∣ = ε (1 + |x|)n+1 .
Therefore the space of solutions is a translation invariant closed linear subspace, i.e. it is a variety in Va. Using the exponential element m we have that
0 =
n+1
∑
i=0
(−1) i
(n + 1
i
)
xim (cx n+1 −i)
= m(c)
n+1
∑
i=0
(−1) i
(n + 1
i
)
ximn+1 −i(x) = m(c)( m(x) − x)n+1
and, consequently, the exponential element must be the identity on the finitely generated field over
Q. This means that the space of the solutions is spanned by differential operators.
2Since Kis countable, the limit can be taken in a pointwise sense.
18 4.3 Spectral synthesis in the variety generated by the solutions of equation
(13)
Let n ∈ N and consider an additive mapping A : K → C satisfying equation
n
∑
i=0
an+1 −ixiA (xn+1 −i) = 0 (x ∈ K) . (20) First of all observe that the solutions of functional equation (20) form a linear subspace. To get some information about the translates of the form Ac(x) = A(cx ) of the solutions we have to set the variable x free by using a symmetrization process:
Φ( x1, . . . , x n+1 ) =
n
∑
i=0
an+1 −i
1
(n+1
i
) ∑
card( I)= i
(∏
j∈I
xj
)
· A
∏
k∈{ 1,...,n +1 }\ I
xk
(x1, . . . , x n+1 ∈ K) .
Then
Φ∗(x) = Φ( x, . . . , x ) =
n
∑
i=0
an+1 −ixiA (xn+1 −i) = 0 (x ∈ K)
because of equation (20). In the sense of Lemma 2, Φ is also identically zero. By some direct computations
Φ( x, . . . , x, cx ) − xΦ( x, . . . , x, c ) = 1
n + 1
n
∑
i=0
(n + 1 − i)an+1 −i
(xiA(cx n+1 −i) − xi+1 A(cx n−i)) ,
that is the translation invariant linear subspace generated by the solutions of equation (20) can be described by the family of equations
n
∑
i=0
(n + 1 − i)an+1 −i
(xiA(cx n+1 −i) − xi+1 A(cx n−i)) = 0 (x ∈ K, c ∈ K∗) . (21) By the same way as above we can prove that the space of the solutions of equation (21) is a variety in Va. Using the exponential element m we have that
0 = m(c)( m(x) − x)
n+1
∑
j=1
ja j mj−1(x)xn−(j−1) ,
where j = n + 1 − i, i = 0 , . . . , n . If m(x) 6 = x for some x ∈ K∗ then we can write that
0 =
n+1
∑
j=1
ja j
(m(x)
x
)j−1
.
Therefore m(x)/x is the root of the polynomial ∑n+1
j=1
ja j tj−1 and it has only finitely many different values. This is obviously a contradiction because for any x ∈ K∗, the function
r ∈ Q 7 −→ m(x + r)
x + r = m(x) + r
x + r
provides infinitely many different values unless m(x) = x; note that m(x) 6 = x is equivalent to
m(x + r) 6 = x + r for any rational number r ∈ Q. Therefore we can conclude that the exponential element must be the identity on any finitely generated field over Q and the space of the solutions is spanned by differential operators. 19 4.4 The solutions of equation (16) in case of n = 3
Let n ∈ N, n ≥ 1 be arbitrary and assume that the additive functions f1, . . . , f n+1 : K → C satisfy equation n∑
i=0
xifn+1 −i
(xn+1 −i) = 0 (x ∈ K) (22) under the initial conditions fi(1) = 0 for all i = 1 , . . . , n + 1 . In the first step we set the variable x
free by using a symmetrization process:
Φ( x1, . . . , x n+1 ) =
n
∑
i=0
1
(n+1
i
) ∑
card( I)= i
(∏
j∈I
xj
)
·fn+1 −i
∏
k∈{ 1,...,n +1 }\ I
xk
(x1, . . . , x n+1 ∈ K) .
Then
Φ∗(x) = Φ( x, . . . , x ) =
n
∑
i=0
xifn+1 −i
(xn+1 −i) = 0 (x ∈ K)
because of equation (22). In the sense of Lemma 2, Φ is also identically zero. We are going to investigate the explicit case of n = 3 , that is, equation
3
∑
i=0
xif4−i
(x4−i) = 0 (x ∈ K) . (23) By some direct computations
Φ( x, x, x, cx ) = f4(cx 4) + 3
4 xf 3(cx 3) + 1
4 cxf 3(x3) + 1
2 x2f2(cx 2)+ 1
2cx 2f2(x2) + 1
4 x3f1(cx ) + 3
4 cx 3f1(x),
Φ( x, x, x, c ) = f4(cx 3) + 3
4 xf 3(cx 2) + 1
4 cf 3(x3) + 1
2 x2f2(cx )+ 1
2 cxf 2(x2) + 1
4 x3f1(c) + 3
4 cx 2f1(x)
and, consequently, for any c ∈ K∗
Φ( x, x, x, cx ) − xΦ( x, x, x, c ) = f4(cx 4) + x
(3
4 f3 − f4
)
(cx 3)+ x2
(1
2 f2 − 3
4 f3
)
(cx 2) + x3
(1
4 f1 − 1
2 f2
)
(cx ) − 1
4 x4f1(c),
where x ∈ K, i.e. we can formulate the family of equations
g4(cx 4) + xg 3(cx 3) + x2g2(cx 2) + x3g1(cx ) = x4(g1 + g2 + g3 + g4)( c) (x ∈ K) , (24) where c runs through the elements of K∗ and
g1
g2
g3
g4
=
1/4 −1/2 0 00 1/2 −3/4 00 0 3/4 −10 0 0 1
f1
f2
f3
f4
.
20 The inverse formulas are
f4 = g4, f 3 = 4
3 (g3 + g4) , f 2 = 2 ( g2 + g3 + g4) , f 1 = 4 ( g1 + g2 + g3 + g4) .
Taking c = 1 we have that
g4(x4) + xg 3(x3) + x2g2(x2) + x3g1(x) = 0 (x ∈ K) (25) because of the initial conditions f1(1) = f2(1) = f3(1) = f4(1) = 0 . Therefore the space of the solutions is invariant under the action of the linear transformation represented by the matrix
M :=
1/4 −1/2 0 00 1/2 −3/4 00 0 3/4 −10 0 0 1
= 1
4
1 −2 0 00 2 −3 00 0 3 −40 0 0 4
.
As a MAPLE computation shows
with(LinearAlgebra); M:=(1/4)Matrix(); MatrixPower(M,n);
4−n −21−n + 2 4 −n 31+ n4−n − 6 2 −n + 3 4 −n −4 + 12 3 n4−n − 12 2 −n + 2 2−2 n
0 2−n −31+ n4−n + 3 2 −n 6 − 12 3 n4−n + 6 2 −n
0 0 3n4−n −4 + 2 2−2 n3n
0 0 0 1
and, consequently,
lim
n→∞
Mn =
0 0 0 −40 0 0 60 0 0 −40 0 0 1
.
Therefore
f4(x4) − 4xf 4(x3) + 6 x2f4(x2) − 4x3f4(x) = 0 (x ∈ K) (26) and, by Theorem 3 (Corollary 2), we can conclude that f4 is a differential operator on any finitely generated field K. By taking the difference of equations (23) and (26) the number of the unknown functions can be reduced:
˜f3(x3) + x ˜f2(x2) + x2 ˜f1(x) = 0 (x ∈ K) , (27) where ˜f3 = f3 + 4 f4, ˜f2 = f2 − 6f4, ˜f1 = f1 + 4 f4.
Repeating the process above we can conclude that ˜f3 is a differential operator on any finitely gener-ated field K and so on. Note that it is an alternative way to prove Theorem 5/Corollary 4 by using a descending process instead of the inductive argument.
Acknowledgement. This paper is dedicated to the 65 th birthday of Professor L´ aszl´ o Sz´ ekelyhidi. 21 The research of the first author has been supported by the Hungarian Scientific Research Fund (OTKA) Grant K 111651 and by the ´ UNKP-4 New National Excellence Program of the Ministry of Human Capacities. The work of the first and the third author is also supported by the EFOP-3.6.1-16-2016-00022 project. The project is co-financed by the European Union and the European Social Fund. The second author was supported by the internal research project R-AGR-0500 of the University of Luxembourg and by the Hungarian Scientific Research Fund (OTKA) K 104178.
References
Bruce Ebanks. Characterizing ring derivations of all orders via functional equations: results and open problems. Aequationes Math. , 89(3):685–718, 2015. Bruce Ebanks. Polynomially linked additive functions. Aequationes Math. , 91(2):317–330, 2017. Bruce Ebanks. Polynomially linked additive functions II. Aequationes Math. , 2018. to appear. Bruce Ebanks, Thomas Riedel, and Prasanna K. Sahoo. On the order of a derivation. Aequa-tiones Math. , 90(2):335–340, 2016. Eszter Gselmann. Derivations and linear functions along rational functions. Monatsh. Math. ,169(3-4):355–370, 2013. Vladislav K. Kharchenko. Automorphisms and derivations of associative rings , volume 69 of Mathematics and its Applications (Soviet Series) . Kluwer Academic Publishers Group, Dordrecht, 1991. Translated from the Russian by L. Yuzina. Gergely Kiss and Mikl´ os Laczkovich. Linear functional equations, differential operators and spectral synthesis. Aequationes Math. , 89(2):301–328, 2015. Gergely Kiss, Adrienn Varga, and Csaba Vincze. Algebraic methods for the solution of linear functional equations. Acta Math. Hungar. , 146(1):128–141, 2015. Gergely Kiss and Csaba Vincze. On spectral analysis in varieties containing the solutions of inhomogeneous linear functional equations. Aequationes Math. , 2017. Gergely Kiss and Csaba Vincze. On spectral synthesis in varieties containing the solutions of inhomogeneous linear functional equations. Aequationes Math. , 2017. Marek Kuczma. An introduction to the theory of functional equations and inequalities .Birkh¨ auser Verlag, Basel, second edition, 2009. Cauchy’s equation and Jensen’s inequality, Edited and with a preface by Attila Gil´ anyi. Ludwig Reich. Derivationen zweiter Ordnung als L¨ osungen von Funktionalgleichungen—ein ¨uberblick. In Gyula Maurer zum 70. Geburtstag , volume 337 of Grazer Math. Ber. , pages 45–65. Karl-Franzens-Univ. Graz, Graz, 1998. L´ aszl´ o Sz´ ekelyhidi. Convolution type functional equations on topological abelian groups .World Scientific Publishing Co., Inc., Teaneck, NJ, 1991. L´ aszl´ o Sz´ ekelyhidi. Discrete spectral synthesis and its applications . Springer Monographs in Mathematics. Springer, Dordrecht, 2006. 22 Josef Unger and Ludwig Reich. Derivationen h¨ oherer Ordnung als L¨ osungen von Funktion-algleichungen , volume 336 of Grazer Mathematische Berichte [Graz Mathematical Reports] .Karl-Franzens-Universit¨ at Graz, Graz, 1998. Adrienn Varga. On additive solutions of a linear equation. Acta Math. Hungar. , 128(1-2):15– 25, 2010. Adrienn Varga and Csaba Vincze. On Dar´ oczy’s problem for additive functions. Publ. Math. Debrecen , 75(1-2):299–310, 2009. Oscar Zariski and Pierre Samuel. Commutative algebra. Vol. 1 . Springer-Verlag, New York-Heidelberg-Berlin, 1975. With the cooperation of I. S. Cohen, Corrected reprinting of the 1958 edition, Graduate Texts in Mathematics, No. 28. 23 |
6843 | https://www.ck12.org/book/algebra-i/r2/section/5.2/ | Linear Equations in Point-Slope Form | CK-12 Foundation
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HomeMathematicsFlexBooksCK-12 Algebra ICh52. Linear Equations in Point Slope Form
5.2 Linear Equations in Point-Slope Form
Difficulty Level: Basic | Created by: CK-12
Last Modified: Feb 23, 2012
Read Resources Details Attributions
Learning Objectives
Write an equation in point-slope form.
Graph an equation in point-slope form.
Write a linear function in point-slope form.
Solve real-world problems using linear models in point-slope form.
Introduction
In the last lesson, we saw how to write the equation of a straight line in slope-intercept form. We can rewrite this equation in another way that sometimes makes solving the problem easier. The equation of a straight line that we are going to talk about is called point-slope form.
y−y 0=m(x−x 0)
Here m is the slope and (x 0,y 0) is a point on the line. Let’s see how we can use this form of the equation in the three cases that we talked about in the last section.
Case 1: You know the slope of the line and the y−intercept.
Case 2: You know the slope of the line and a point on the line.
Case 3: You know two points on the line.
Write an Equation in Point-Slope Form
Case 1 You know the slope and the y−intercept.
Start with the equation in point-slope form y−y 0=m(x−x 0).
Plug in the value of the slope.
Plug in 0 for x 0 and b for y 0.
Example 1
Write the equation of the line in point-slope form, given that the slope=−5 and the y−intercept =4.
Solution:
Start with the equation in point-slope form. y−y 0=m(x−x 0)
Plug in the value of the slope. y−y 0=−5(x−x 0)
Plug in 0 for x 0 and 4 for y 0. y−(−4)=−5(x−(0))
Therefore, the equation is y+4=−5 x
Case 2 You know the slope and a point on the line.
Start with the equation in point-slope form y−y 0=m(x−x 0).
Plug in the value of the slope.
Plug in the x and y values in place of x 0 and y 0.
Example 2
Write the equation of the line in point-slope form, given that the slope=3 5 and the point (2,6) is on the line.
Solution:
Start with the equation in point-slope form. y−y 0=m(x−x 0)
Plug in the value of the slope. y−y 0=3 5(x−x 0)
Plug in 2 for x 0 and 6 for y 0. y−(6)=3 5(x−(2))
The equation is y−6=3 5(x−2)
Notice that the equation in point-slope form is not solved for y.
Case 3 You know two points on the line.
Start with the equation in point-slope form y−y 0=m(x−x 0).
Find the slope using the slope formula. m=y 2−y 1 x 2−x 1
Plug in the value of the slope.
Plug in the x and y values of one of the given points in place of x 0 and y 0
Example 3
Write the equation of the line in point-slope form, given that the line contains points (-4, -2) and (8, 12).
Solution
Start with the equation in point-slope form. y−y 0=m(x−x 0)
Find the slope using the slope formula. m=12−(−2)8−(−4)=14 12=7 6
Plug in the value of the slope. y−y 0=7 6(x−x 0)
Plug in -4 for x 0 and -2 for y 0. y−(−2)=7 6(x−(−4))
Therefore, the equation is y+2=7 6(x+4)Answer 1
In the last example, you were told that for the last step you could choose either of the points you were given to plug in for the point (x 0,y 0) but it might not seem like you would get the same answer if you plug the second point in instead of the first. Let’s redo Step 4.
Plug in 8 for x 0 and 12 for y 0 y−12=7 6(x−8)Answer 2
This certainly does not see like the same answer as we got by plugging in the first point. What is going on?
Notice that the equation in point-slope form is not solved for y. Let’s change both answers into slope-intercept form by solving for y.
Answer 1 y+2=7 6(x+4)y+2=7 6 x+28 6 y=7 6 x+14 3−2 y=7 6 x+8 3 Answer 2 y−12=7 6(x−8)y−12=7 6 x−56 6 y=7 6 x−28 3+12 y=7 6 x+8 3
Now that the two answers are solved for y, you can see that they simplify to the same thing. In point-slope form you can get an infinite number of right answers, because there are an infinite number of points on a line. The slope of the line will always be the same but the answer will look different because you can substitute any point on the line for (x 0,y 0). However, regardless of the point you pick, the point-slope form should always simplify to the same slope-intercept equation for points that are on the same line.
In the last example you saw that sometimes we need to change between different forms of the equation. To change from point-slope form to slope-intercept form, we just solve for y.
Example 4
Re-write the following equations in slope-intercept form.
a) y−5=3(x−2)
b) y+7=−(x+4)
Solution
a) To re-write in slope-intercept form, solve for y.
y−5−5 y=3(x−2)=3 x−6=3 x−1
b) To re-write in slope-intercept form, solve for y.
y+7 y+7 y=−(x+4)=−x−4=−x−11
Graph an Equation in Point-Slope Form
If you are given an equation in point-slope form, it is not necessary to re-write it in slope-intercept form in order to graph it. The point-slope form of the equation gives you enough information so you can graph the line y−y 0=m(x−x 0). From this equation, we know a point on the line (x 0,y 0) and the slope of the line.
To graph the line, you first plot the point (x 0,y 0). Then the slope tells you how many units you should go up or down and how many units you should go to the right to get to the next point on the line. Let’s demonstrate this method with an example.
Example 5
Make a graph of the line given by the equation y−2=2 3(x+2)
Solution
Let’s rewrite the equation y−(2)=2 3(x+2).
Now we see that point (-2, 2) is on the line and that the slope=2 3.
First plot point (-2, 2) on the graph.
[Figure 1]
A slope of 2 3 tells you that from your point you should move. 2 units up and 3 units to the right and draw another point.
[Figure 2]
Now draw a line through the two points and extend the line in both directions.
[Figure 3]
Write a Linear Function in Point-Slope Form
The functional notation for the point-slope form of the equation of a line is:
f(x)−f(x 0)=m(x−x 0)
Note that we replaced the each y with the f(x)
y=f(x) and y 0=f(x 0)
That tells us more clearly that we find values of y by plugging in values of x into the function defined by the equation of the line. Let’s use the functional notation to solve some examples.
Example 6
Write the equation of the following linear functions in point-slope form.
a) m=25 and f(0)=250
b) m=9.8 and f(5.5)=12.5
c) f(32)=0 and f(77)=25
a) Here we are given the slope=25 and the point on the line gives x 0=0,f(x 0)=250
Start with the equation in point-slope form. f(x)−f(x 0)=m(x−x 0)
Plug in the value of the slope. f(x)−f(x 0)=25(x−x 0)
Plug in 0 for x 0 and 250 for f(x 0). f(x)−250=25(x−0)
Solution
The linear function is f(x)−250=25 x.
b) Here we are given that slope=9.8 and the point on the line gives x 0=5.5,f(x 0)=12.5
Start with the equation in point-slope form. f(x)−f(x 0)=m(x−x 0)
Plug in the value of the slope. f(x)−f(x 0)=9.8(x−x 0)
Plug in 5.5 for x 0 and 12.5 for f(x 0). f(x)−12.5=9.8(x−5.5)
Solution
The linear function is f(x)−12.5=9.8(x−5.5).
c) Here we are given two points (32, 0) and (77, 25).
Start with the equation in point-slope form. f(x)−f(x 0)=m(x−x 0)
Find the value of the slope. m=25−0 77−32=25 45=5 9
Plug in the value of the slope. f(x)−f(x 0)=5 9(x−x 0)
Plug in 32 for and 0 for f(x 0). f(x)−0=5 9(x−32)
Solution
The linear function is f(x)−0=5 9(x−32).
Solve Real-World Problems Using Linear Models in Point-Slope Form
Let’s solve some word problems where we need to write the equation of a straight line in point-slope form.
[Figure 4]
Example 7
Marciel rented a moving truck for the day. Marciel only remembers that the rental truck company charges $40 per day and some amount of cents per mile. Marciel drives 46 miles and the final amount of the bill (before tax) is $63. What is the amount per mile the truck rental company charges per day? Write an equation in point-slope form that describes this situation. How much would it cost to rent this truck if Marciel drove 220 miles?
Let’s define our variables:
x= distance in miles
y= cost of the rental truck in dollars
We see that we are given the y−intercept and the point (46, 63).
Peter pays a flat fee of $40 for the day. This is the y−intercept.
He pays $63 for 46 miles – this is the coordinate point (46, 63).
Start with the point-slope form of the line.Plug in the coordinate point.Plug in point(0,40).Solve for the slope.The slope is:So, the truck company charges 50 cents per mile.(y−y 0)63−y 0 63−40 23($0.5=m(x−x 0)=m(46−x 0)=m(46−0)=m(46)→m=23 46=0.5 0.5 dollars per mile=50 cents)Equation of line is:y=0.5 x+40
To answer the question of 220 miles we plug in x=220.
Solution
y−40=0.5(220)⇒y=$150
[Figure 5]
Example 8
Anne got a job selling window shades. She receives a monthly base salary and a $6 commission for each window shade she sells. At the end of the month, she adds up her sales and she figures out that she sold 200 window shades and made $2500. Write an equation in point-slope form that describes this situation. How much is Anne’s monthly base salary?
Let’s define our variables
x= number of window shades sold
y= Anne’s monthly salary in dollars
We see that we are given the slope and a point on the line:
Anne gets $6 for each shade, so the slope=6 dollars/shade.
She sold 200 shades and made $2500, so the point is (200, 2500).
Start with the point-slope form of the line.Plug in the coordinate point.Plug in point(200,2500).y−y 0 y−y 0 y−2500=m(x−x 0)=6(x−x 0)=6(x−200)
Anne’s base salary is found by plugging in x=0. We obtain y−2500=−1200⇒y=$1300
Solution
Anne’s monthly base salary is $1300.
Lesson Summary
The point-slope form of an equation for a line is: y−y 0=m(x−x 0).
If you are given the slope and a point on the line:
Simply plug the point and the slope into the equation.
If you are given the slope andy−intercept of a line:
Plug the value of m into the equation
Plug the y−intercept point into the equation y 0=y−intercept and x 0=0.
If you are given two points on the line:
Use the two points to find the slope using the slope formula m=y 2−y 1 x 2−x 1.
Plug the value of m into the equation.
Plug either of the points into the equation as (x 0,y 0).
The functional notation of point-slope form is f(x)−f(x 0)=m(x−x 0).
Review Questions
Write the equation of the line in point-slope form.
The line has slope −1 10 and goes through point (10, 2).
The line has slope -75 and goes through point (0, 125).
The line has slope 10 and goes through point (8, -2).
The line goes through the points (-2, 3) and (-1, -2).
The line contains points (10, 12) and (5, 25).
The line goes through points (2, 3) and (0, 3).
The line has a slope 3 5 and a y−intercept -3.
The line has a slope -6 and a y−intercept 0.5.
Write the equation of the linear function in point-slope form.
m=−1 5 and f(0)=7
m=−12 and f(−2)=5
f(−7)=5 and f(3)=−4
f(6)=0 and f(0)=6
m=3 and f(2)=−9
m=−9 5 and f(0)=32
Nadia is placing different weights on a spring and measuring the length of the stretched spring. She finds that for a 100 gram weight the length of the stretched spring is 20 cm and for a 300 gram weight the length of the stretched spring is 25 cm. Write an equation in point-slope form that describes this situation. What is the unstretched length of the spring?
Andrew is a submarine commander. He decides to surface his submarine to periscope depth. It takes him 20 minutes to get from a depth of 400 feet to a depth of 50 feet. Write an equation in point-slope form that describes this situation. What was the submarine’s depth five minutes after it started surfacing?
Review Answers
y−2=−1 10(x−10)
y−125=−75 x
y+2=10(x−8)
y+2=−5(x+1) or y−3=−5(x+2)
y−25=−13 5(x−5) or y−12=−13 5(x−10)
y−3=0
y+3=3 5 x
y−0.5=−6 x
f(x)−7=−1 5 x
f(x)−5=−12(x+2)
f(x)−5=−9 10(x+7) or f(x)+4=−9 10(x−3)
f(x)=−x(x−6) or f(x)−6=−x
f(x)+9=3(x−2)
f(x)−32=9 5 x
y−20=1 40(x−100) unstretched length =17.5 c m
y−50=−17.5(x−20) or y−400=−17.5 x depth =312.5 f e e t
Notes/Highlights
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Description
CK-12's Algebra FlexBook® textbook is an introduction to algebraic concepts for the high school student. Topics include: Equations & Functions, Real Numbers, Equations of Lines, Solving Systems of Equations & Quadratic Equations.
Difficulty Level
Basic
Tags
Problem Solving,slope,circumference,Parallel Lines,Survey,slope-intercept form,point-slope form,linear equations,linear functions,standard form,diameter,extrapolate,interpolate,line of best fit,measurement error,outliers,negative reciprocal,linear regression,scatter plot,family of lines,vertical shift,reciprocal,slope formula,dependent variable,x-intercept,y-intercept,perpendicular lines,point slope form,finding equations for lines,CK.MAT.ENG.SE.1.Algebra-I.5, (27 more)
Subjects
mathematics
Grades
9,10
Standards Correlations
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License
CC BY NC
Language
English
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Date Created
Feb 23, 2012
Last Modified
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6844 | https://www.jmp.com/content/dam/jmp/documents/en/support/jmp161/quality-and-process-methods.pdf | Version 16 16.1 Quality and Process Methods JMP, A Business Unit of SAS SAS Campus Drive Cary, NC 27513 “The real voyage of discovery consists not in seeking new landscapes, but in having new eyes.” Marcel Proust The correct bibliographic citation for this manual is as follows: SAS Institute Inc. 2020–2021. JMP® 16 Quality and Process Methods. Cary, NC: SAS Institute Inc.
JMP® 16 Quality and Process Methods Copyright © 2020–2021, SAS Institute Inc., Cary, NC, USA All rights reserved. Produced in the United States of America.
U.S. Government License Rights; Restricted Rights: The Software and its documentation is commercial computer software developed at private expense and is provided with RESTRICTED RIGHTS to the United States Government. Use, duplication or disclosure of the Software by the United States Government is subject to the license terms of this Agreement pursuant to, as applicable, FAR 12.212, DFAR 227.7202-1(a), DFAR 227.7202-3(a) and DFAR 227.7202-4 and, to the extent required under U.S. federal law, the minimum restricted rights as set out in FAR 52.227-19 (DEC 2007). If FAR 52.227-19 is applicable, this provision serves as notice under clause (c) thereof and no other notice is required to be affixed to the Software or documentation. The Government’s rights in Software and documentation shall be only those set forth in this Agreement.
SAS Institute Inc., SAS Campus Drive, Cary, North Carolina 27513-2414.
March 2021 August 2021 SAS® and all other SAS Institute Inc. product or service names are registered trademarks or trademarks of SAS Institute Inc. in the USA and other countries. ® indicates USA registration.
Other brand and product names are trademarks of their respective companies.
SAS software may be provided with certain third-party software, including but not limited to open-source software, which is licensed under its applicable third-party software license agreement. For license information about third-party software distributed with SAS software, refer to
Get the Most from JMP Whether you are a first-time or a long-time user, there is always something to learn about JMP.
Visit JMP.com to find the following: • live and recorded webcasts about how to get started with JMP • video demos and webcasts of new features and advanced techniques • details on registering for JMP training • schedules for seminars being held in your area • success stories showing how others use JMP • the JMP user community, resources for users including examples of add-ins and scripts, a forum, blogs, conference information, and so on Contents Quality and Process Methods 1 Learn about JMP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13 Documentation and Additional Resources Formatting Conventions in JMP Documentation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 JMP Help . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 JMP Documentation Library . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Additional Resources for Learning JMP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 JMP Tutorials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Sample Data Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Learn about Statistical and JSL Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Learn JMP Tips and Tricks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 JMP Tooltips . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 JMP User Community . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Free Online Statistical Thinking Course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 JMP New User Welcome Kit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Statistics Knowledge Portal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 JMP Training . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 JMP Books by Users . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 The JMP Starter Window . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 JMP Technical Support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2 Introduction to Quality and Process Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .27 Tools for Process and Product Improvement 3 Control Chart Builder. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .29 Create Control Charts Interactively Overview of Control Chart Builder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Example of Control Chart Builder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Control Chart Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Shewhart Control Charts for Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Shewhart Control Charts for Attributes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Rare Event Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 Control Chart Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Launch Control Chart Builder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 6 Contents Quality and Process Methods Control Chart Builder Interactive Workspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 Launch Windows for Specific Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 Control Chart Builder Window . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 Control Chart Builder Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Control Chart Builder Red Triangle Menu Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 Options Panel and Right-Click Chart Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Control Chart Builder Right-Click Axis Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Work with Control Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 Example of Control Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 Excluded and Hidden Samples in Control Chart Builder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Additional Examples of Control Chart Builder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 Individual Measurement and Moving Range Charts Example . . . . . . . . . . . . . . . . . . . . . . 72 XBar and R Chart Phase Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 XBar and S Charts with Varying Subgroup Sizes Example . . . . . . . . . . . . . . . . . . . . . . . . . . 76 Run Chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 P chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 NP chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 C chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 U chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 G chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 T chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Three Way Control Chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Example of Multiple Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Statistical Details for Control Chart Builder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Control Limits for XBar and R Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 Control Limits for XBar and S Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 Control Limits for Individual Measurement and Moving Range Charts . . . . . . . . . . . . . . 92 Control Limits for P and NP Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Control Limits for U Charts and C Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 Levey-Jennings Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Control Limits for G Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Control Limits for T Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 Sigma Calculations for Three Way Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 4 Measurement Systems Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Evaluate a Continuous Measurement Process Using the EMP Method Overview of Measurement Systems Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 Example of Measurement Systems Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 Launch the Measurement Systems Analysis Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Data Format . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 Quality and Process Methods Contents 7 Measurement Systems Analysis Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 Average Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 Range Chart or Standard Deviation Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 EMP Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 Effective Resolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 Shift Detection Profiler . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 Bias Comparison . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 Test-Retest Error Comparison . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 Additional Example of Measurement Systems Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 Statistical Details for Measurement Systems Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 Computation of Intraclass Correlation and Probable Error . . . . . . . . . . . . . . . . . . . . . . . . . . 124 5 Variability Gauge Charts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 Evaluate a Continuous Measurement Process Using Gauge R&R Overview of Variability Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 Example of a Variability Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 Launch the Variability/Attribute Gauge Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 Data Format . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 The Variability Gauge Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 Variability Gauge Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 Heterogeneity of Variance Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 Variance Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 About the Gauge R&R Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Gauge R&R Option . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 Discrimination Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 Misclassification Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 Bias Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 Linearity Study . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 Additional Examples of Variability Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 Example of the Heterogeneity of Variance Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 Example of the Bias Report Option . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148 Statistical Details for Variability Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Statistical Details for Variance Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Statistical Details for the Discrimination Ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 Statistical Details for the Misclassification Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 6 Attribute Gauge Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 Evaluate a Categorical Measurement Process Using Agreement Measures Overview of Attribute Gauge Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 Example of an Attribute Gauge Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 Launch the Variability/Attribute Gauge Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 8 Contents Quality and Process Methods The Attribute Gauge Chart and Reports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 Agreement Reports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 Effectiveness Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 Attribute Gauge Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 Statistical Details for Attribute Gauge Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 Statistical Details for the Agreement Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 7 Process Capability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 Measure the Variability of a Process over Time Overview of the Process Capability Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 Example of the Process Capability Platform with Normal Variables . . . . . . . . . . . . . . . . . . . 173 Example of the Process Capability Platform with Nonnormal Variables . . . . . . . . . . . . . . . . 175 Launch the Process Capability Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 Process Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 Process Subgrouping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 Moving Range Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 Historical Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 Distribution Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 Other Specifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 Entering Specification Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 Spec Limits Window . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 Limits Data Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 Spec Limits Column Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 The Process Capability Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 Goal Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 Capability Box Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 Capability Index Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 Process Capability Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 Individual Detail Reports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 Normalized Box Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 Process Performance Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 Summary Reports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 Make Goal Plot Summary Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 Additional Examples of the Process Capability Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 Process Capability for a Stable Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 Process Capability for an Unstable Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 Simulation of Confidence Limits for a Nonnormal Process Ppk . . . . . . . . . . . . . . . . . . . . 220 Statistical Details for the Process Capability Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 Variation Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 Notation for Goal Plots and Capability Box Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 Quality and Process Methods Contents 9 Goal Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 Capability Box Plots for Processes with Missing Targets . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 Capability Indices for Normal Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 Capability Indices for Nonnormal Distributions: Percentile and Z-Score Methods . . . 239 Parameterizations for Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 8 CUSUM Control Charts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 Create Tabular CUSUM Control Charts with Decision Limits Overview of the CUSUM Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 Example of CUSUM Control Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 Launch the CUSUM Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 The CUSUM Control Chart Platform Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 Control Panel for CUSUM Control Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 CUSUM Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 CUSUM Control Chart Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 Average Run Length (ARL) Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Additional Examples of CUSUM Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Example of the Data Units Option in a CUSUM Control Chart . . . . . . . . . . . . . . . . . . . . . 256 Example of CUSUM Chart with Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 Statistical Details for the CUSUM Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 Statistical Details for CUSUM Control Chart Construction . . . . . . . . . . . . . . . . . . . . . . . . . 258 Statistical Details for Shift Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 Statistical Details for Average Run Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 9 EWMA Control Charts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 Create Exponentially Weighted Control Charts Overview of the EWMA Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 Example of the EWMA Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 Launch the EWMA Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 The EWMA Control Chart Platform Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 Control Panel for EWMA Control Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 EWMA Chart Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 The EWMA Control Chart Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 Average Run Length (ARL) Report for EWMA Control Charts . . . . . . . . . . . . . . . . . . . . . 272 Additional Example of the EWMA Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 Statistical Details for the EWMA Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 10 Multivariate Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 Monitor Multiple Process Characteristics Simultaneously Overview of Multivariate Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 Example of a Multivariate Control Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 10 Contents Quality and Process Methods Launch the Multivariate Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 The Multivariate Control Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 Multivariate Control Chart Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286 T Square Partitioned . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 Change Point Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288 Principal Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 Additional Examples of Multivariate Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 Example of Monitoring a Process Using Sub-Grouped Data . . . . . . . . . . . . . . . . . . . . . . . 289 Example of T Square Partitioned . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 Example of Change Point Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294 Statistical Details for Multivariate Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 Statistical Details for Individual Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296 Statistical Details for Observations in Rational Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . 297 Statistical Details for Change Point Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 11 Model Driven Multivariate Control Charts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 Monitor and Diagnosis a Complex Process Overview of Model Driven Multivariate Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 Example of Model Driven Multivariate Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 Launch the Model Driven Multivariate Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . 310 Data Format . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 The Model Driven Multivariate Control Chart Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312 Model Driven Multivariate Control Chart Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . 312 Plot Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 Score Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 Additional Examples of the Model Driven Multivariate Control Chart Platform . . . . . . . . 315 Example of an MDMVCC with Historical Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 Example of an MDMVCC with a PLS Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 Statistical Details for the Model Driven Multivariate Control Chart Platform . . . . . . . . . . . 318 Monitoring Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 Contributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 Score Plot Group Comparisons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 Score Plot Loadings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 12 Legacy Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 Create Variable and Attribute Control Charts Example of a Legacy Control Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 Legacy Control Chart Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 Control Charts for Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 Control Charts for Attributes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 Quality and Process Methods Contents 11 Launch a Legacy Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334 Process Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 Limits Specifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 Specified Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 Legacy Control Chart Reports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340 V-Mask CUSUM Chart Reports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 Interpret a Two-Sided V-Mask CUSUM Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 Interpret a One-Sided CUSUM Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343 Legacy Control Chart Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344 Window Options for Legacy Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344 Chart Options for Legacy Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 Chart Options for V-Mask CUSUM Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349 Saving and Retrieving Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349 Excluded, Hidden, and Deleted Samples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 Additional Examples of the Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354 Presummarize Chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354 V-Mask CUSUM Chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356 One-Sided CUSUM Chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358 UWMA Chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 Statistical Details for the Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 Control Limits for Median Moving Range Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 Statistical Details for Capability Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 Statistical Details for V-Mask CUSUM Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367 Statistical Details for Weighted Moving Average Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . 370 13 Pareto Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 Focus Improvement Efforts on the Vital Few Overview of the Pareto Plot Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375 Example of the Pareto Plot Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375 Launch the Pareto Plot Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378 The Pareto Plot Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379 Pareto Plot Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 Causes Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382 Additional Examples of the Pareto Plot Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382 Threshold of Combined Causes Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 Using a Constant Size across Groups Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 Using a Non-Constant Sample Size across Groups Example . . . . . . . . . . . . . . . . . . . . . . . 386 One-Way Comparative Pareto Plot Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387 Two-Way Comparative Pareto Plot Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389 Statistical Details for the Pareto Plot Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390 12 Contents Quality and Process Methods Likelihood Ratio Chi-Square Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390 14 Cause-and-Effect Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393 Identify Root Causes Overview of Cause-and-Effect Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395 Example of a Cause-and-Effect Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395 Prepare the Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396 Launch the Diagram Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396 The Cause-and-Effect Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397 Right-Click Menus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397 Cause and Effect Diagram Menu Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400 Save the Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 Save the Diagram as a Data Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 Save the Diagram as a Journal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 Save the Diagram as a Script . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402 15 Quality Utilities. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403 Manage Specification Limits and Create OC Curves Manage Spec Limits Utility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405 Example of the Manage Spec Limits Utility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405 Manage Spec Limits Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408 Operating Characteristic Curves Utility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 Launch the OC Curves Utility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 OC Curves for Control Chart Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 OC Curves for Acceptance Sampling Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 A References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413 B Technology License Notices. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417 Chapter 1 Learn about JMP Documentation and Additional Resources Learn about JMP documentation, such as book conventions, descriptions of each JMP document, the Help system, and where to find additional support.
14 Learn about JMP Chapter 1 Quality and Process Methods Contents Formatting Conventions in JMP Documentation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15 JMP Help . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 JMP Documentation Library . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Additional Resources for Learning JMP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 JMP Tutorials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Sample Data Tables. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Learn about Statistical and JSL Terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 Learn JMP Tips and Tricks. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 JMP Tooltips. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 JMP User Community . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Free Online Statistical Thinking Course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 JMP New User Welcome Kit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 Statistics Knowledge Portal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 JMP Training . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 JMP Books by Users . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 The JMP Starter Window . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 JMP Technical Support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Chapter 1 Learn about JMP 15 Quality and Process Methods Formatting Conventions in JMP Documentation Formatting Conventions in JMP Documentation These conventions help you relate written material to information that you see on your screen: • Sample data table names, column names, pathnames, filenames, file extensions, and folders appear in Helvetica (or sans-serif online) font.
• Code appears in Lucida Sans Typewriter (or monospace online) font.
• Code output appears in Lucida Sans Typewriter italic (or monospace italic online) font and is indented farther than the preceding code.
• Helvetica bold formatting (or bold sans-serif online) indicates items that you select to complete a task: – buttons – check boxes – commands – list names that are selectable – menus – options – tab names – text boxes • The following items appear in italics: – words or phrases that are important or have definitions specific to JMP – book titles – variables • Features that are for JMP Pro only are noted with the JMP Pro icon . For an overview of JMP Pro features, visit
Note: Special information and limitations appear within a Note.
Tip: Helpful information appears within a Tip.
16 Learn about JMP Chapter 1 JMP Help Quality and Process Methods JMP Help JMP Help in the Help menu enables you to search for information about JMP features, statistical methods, and the JMP Scripting Language (or JSL). You can open JMP Help in several ways: • Search and view JMP Help on Windows by selecting Help > JMP Help.
• On Windows, press the F1 key to open the Help system in the default browser.
• Get help on a specific part of a data table or report window. Select the Help tool from the Tools menu and then click anywhere in a data table or report window to see the Help for that area.
• Within a JMP window, click the Help button.
Note: The JMP Help is available for users with Internet connections. Users without an Internet connection can search all books in a PDF file by selecting Help > JMP Documentation Library. See “JMP Documentation Library” for more information.
JMP Documentation Library The Help system content is also available in one PDF file called JMP Documentation Library. Select Help > JMP Documentation Library to open the file. You can also download the Documentation PDF Files add-in if you prefer searching individual PDF files of each document in the JMP library. Download the available add-ins from
The following table describes the purpose and content of each document in the JMP library.
Document Title Document Purpose Document Content Discovering JMP If you are not familiar with JMP, start here.
Introduces you to JMP and gets you started creating and analyzing data. Also learn how to share your results.
Using JMP Learn about JMP data tables and how to perform basic operations.
Covers general JMP concepts and features that span across all of JMP, including importing data, modifying columns properties, sorting data, and connecting to SAS.
Chapter 1 Learn about JMP 17 Quality and Process Methods JMP Documentation Library Basic Analysis Perform basic analysis using this document.
Describes these Analyze menu platforms: • Distribution • Fit Y by X • Tabulate • Text Explorer Covers how to perform bivariate, one-way ANOVA, and contingency analyses through Analyze > Fit Y by X. How to approximate sampling distributions using bootstrapping and how to perform parametric resampling with the Simulate platform are also included.
Essential Graphing Find the ideal graph for your data.
Describes these Graph menu platforms: • Graph Builder • Scatterplot 3D • Contour Plot • Bubble Plot • Parallel Plot • Cell Plot • Scatterplot Matrix • Ternary Plot • Treemap • Chart • Overlay Plot The book also covers how to create background and custom maps.
Profilers Learn how to use interactive profiling tools, which enable you to view cross-sections of any response surface.
Covers all profilers listed in the Graph menu. Analyzing noise factors is included along with running simulations using random inputs.
Document Title Document Purpose Document Content 18 Learn about JMP Chapter 1 JMP Documentation Library Quality and Process Methods Design of Experiments Guide Learn how to design experiments and determine appropriate sample sizes.
Covers all topics in the DOE menu.
Fitting Linear Models Learn about Fit Model platform and many of its personalities.
Describes these personalities, all available within the Analyze menu Fit Model platform: • Standard Least Squares • Stepwise • Generalized Regression • Mixed Model • MANOVA • Loglinear Variance • Nominal Logistic • Ordinal Logistic • Generalized Linear Model Document Title Document Purpose Document Content Chapter 1 Learn about JMP 19 Quality and Process Methods JMP Documentation Library Predictive and Specialized Modeling Learn about additional modeling techniques.
Describes these Analyze > Predictive Modeling menu platforms: • Neural • Partition • Bootstrap Forest • Boosted Tree • K Nearest Neighbors • Naive Bayes • Support Vector Machines • Model Comparison • Model Screening • Make Validation Column • Formula Depot Describes these Analyze > Specialized Modeling menu platforms: • Fit Curve • Nonlinear • Functional Data Explorer • Gaussian Process • Time Series • Matched Pairs Describes these Analyze > Screening menu platforms: • Modeling Utilities • Response Screening • Process Screening • Predictor Screening • Association Analysis • Process History Explorer Document Title Document Purpose Document Content 20 Learn about JMP Chapter 1 JMP Documentation Library Quality and Process Methods Multivariate Methods Read about techniques for analyzing several variables simultaneously.
Describes these Analyze > Multivariate Methods menu platforms: • Multivariate • Principal Components • Discriminant • Partial Least Squares • Multiple Correspondence Analysis • Structural Equation Models • Factor Analysis • Multidimensional Scaling • Item Analysis Describes these Analyze > Clustering menu platforms: • Hierarchical Cluster • K Means Cluster • Normal Mixtures • Latent Class Analysis • Cluster Variables Quality and Process Methods Read about tools for evaluating and improving processes.
Describes these Analyze > Quality and Process menu platforms: • Control Chart Builder and individual control charts • Measurement Systems Analysis • Variability / Attribute Gauge Charts • Process Capability • Model Driven Multivariate Control Chart • Legacy Control Charts • Pareto Plot • Diagram • Manage Spec Limits • OC Curves Document Title Document Purpose Document Content Chapter 1 Learn about JMP 21 Quality and Process Methods JMP Documentation Library Reliability and Survival Methods Learn to evaluate and improve reliability in a product or system and analyze survival data for people and products.
Describes these Analyze > Reliability and Survival menu platforms: • Life Distribution • Fit Life by X • Cumulative Damage • Recurrence Analysis • Degradation • Destructive Degradation • Reliability Forecast • Reliability Growth • Reliability Block Diagram • Repairable Systems Simulation • Survival • Fit Parametric Survival • Fit Proportional Hazards Consumer Research Learn about methods for studying consumer preferences and using that insight to create better products and services.
Describes these Analyze > Consumer Research menu platforms: • Categorical • Choice • MaxDiff • Uplift • Multiple Factor Analysis Scripting Guide Learn about taking advantage of the powerful JMP Scripting Language (JSL).
Covers a variety of topics, such as writing and debugging scripts, manipulating data tables, constructing display boxes, and creating JMP applications.
JSL Syntax Reference Read about many JSL functions on functions and their arguments, and messages that you send to objects and display boxes.
Includes syntax, examples, and notes for JSL commands.
Document Title Document Purpose Document Content 22 Learn about JMP Chapter 1 Additional Resources for Learning JMP Quality and Process Methods Additional Resources for Learning JMP In addition to reading JMP help, you can also learn about JMP using the following resources: • “JMP Tutorials” • “Sample Data Tables” • “Learn about Statistical and JSL Terms” • “Learn JMP Tips and Tricks” • “JMP Tooltips” • “JMP User Community” • “Free Online Statistical Thinking Course” • “JMP New User Welcome Kit” • “Statistics Knowledge Portal” • “JMP Training” • “JMP Books by Users” • “The JMP Starter Window” JMP Tutorials You can access JMP tutorials by selecting Help > Tutorials. The first item on the Tutorials menu is Tutorials Directory. This opens a new window with all the tutorials grouped by category.
If you are not familiar with JMP, start with the Beginners Tutorial. It steps you through the JMP interface and explains the basics of using JMP.
The rest of the tutorials help you with specific aspects of JMP, such as designing an experiment and comparing a sample mean to a constant.
Sample Data Tables All of the examples in the JMP documentation suite use sample data. Select Help > Sample Data Library to open the sample data directory.
To view an alphabetized list of sample data tables or view sample data within categories, select Help > Sample Data.
Sample data tables are installed in the following directory: On Windows: C:\Program Files\SAS\JMP\16\Samples\Data On macOS: \Library\Application Support\JMP\16\Samples\Data Chapter 1 Learn about JMP 23 Quality and Process Methods Additional Resources for Learning JMP In JMP Pro, sample data is installed in the JMPPRO (rather than JMP) directory.
To view examples using sample data, select Help > Sample Data and navigate to the Teaching Resources section. To learn more about the teaching resources, visit
Learn about Statistical and JSL Terms For help with statistical terms, select Help > Statistics Index. For help with JSL scripting and examples, select Help > Scripting Index.
Statistics Index Provides definitions of statistical terms.
Scripting Index Lets you search for information about JSL functions, objects, and display boxes. You can also edit and run sample scripts from the Scripting Index and get help on the commands.
Learn JMP Tips and Tricks When you first start JMP, you see the Tip of the Day window. This window provides tips for using JMP.
To turn off the Tip of the Day, clear the Show tips at startup check box. To view it again, select Help > Tip of the Day. Or, you can turn it off using the Preferences window.
JMP Tooltips JMP provides descriptive tooltips (or hover labels) when you hover over items, such as the following: • Menu or toolbar options • Labels in graphs • Text results in the report window (move your cursor in a circle to reveal) • Files or windows in the Home Window • Code in the Script Editor Tip: On Windows, you can hide tooltips in the JMP Preferences. Select File > Preferences > General and then deselect Show menu tips. This option is not available on macOS.
24 Learn about JMP Chapter 1 Additional Resources for Learning JMP Quality and Process Methods JMP User Community The JMP User Community provides a range of options to help you learn more about JMP and connect with other JMP users. The learning library of one-page guides, tutorials, and demos is a good place to start. And you can continue your education by registering for a variety of JMP training courses.
Other resources include a discussion forum, sample data and script file exchange, webcasts, and social networking groups.
To access JMP resources on the website, select Help > JMP User Community or visit
Free Online Statistical Thinking Course Learn practical statistical skills in this free online course on topics such as exploratory data analysis, quality methods, and correlation and regression. The course consists of short videos, demonstrations, exercises, and more. Visit
JMP New User Welcome Kit The JMP New User Welcome Kit is designed to help you quickly get comfortable with the basics of JMP. You’ll complete its thirty short demo videos and activities, build your confidence in using the software, and connect with the largest online community of JMP users in the world. Visit
Statistics Knowledge Portal The Statistics Knowledge Portal combines concise statistical explanations with illuminating examples and graphics to help visitors establish a firm foundation upon which to build statistical skills. Visit
JMP Training SAS offers training on a variety of topics led by a seasoned team of JMP experts. Public courses, live web courses, and on-site courses are available. You might also choose the online e-learning subscription to learn at your convenience. Visit
Chapter 1 Learn about JMP 25 Quality and Process Methods JMP Technical Support JMP Books by Users Additional books about using JMP that are written by JMP users are available on the JMP website. Visit
The JMP Starter Window The JMP Starter window is a good place to begin if you are not familiar with JMP or data analysis. Options are categorized and described, and you launch them by clicking a button. The JMP Starter window covers many of the options found in the Analyze, Graph, Tables, and File menus. The window also lists JMP Pro features and platforms.
• To open the JMP Starter window, select View (Window on macOS) > JMP Starter.
• To display the JMP Starter automatically when you open JMP on Windows, select File > Preferences > General, and then select JMP Starter from the Initial JMP Window list. On macOS, select JMP > Preferences > Initial JMP Starter Window.
JMP Technical Support JMP technical support is provided by statisticians and engineers educated in SAS and JMP, many of whom have graduate degrees in statistics or other technical disciplines.
Many technical support options are provided at including the technical support phone number.
26 Learn about JMP Chapter 1 JMP Technical Support Quality and Process Methods Chapter 2 Introduction to Quality and Process Methods Tools for Process and Product Improvement Quality and Process Methods describes a number of methods and tools that are available in JMP to help you evaluate and improve quality and process performance: • Control charts provide feedback on key variables and show when a process is in, or out of, statistical control. “Control Chart Builder” describes the JMP approach to creating control charts using an interactive control chart platform called Control Chart Builder.
• The Measurement Systems Analysis platform assesses the precision, consistency, and bias of a system. Before you can study a process, you need to make sure that you can accurately and precisely measure the process. If variation comes from the measurement itself, then you are not reliably learning about the process. Use this analysis to find out how your system is performing. See “Measurement Systems Analysis”.
• The Variability/Attribute Gauge Chart platform creates variability or attribute gauge charts. Variability charts analyze continuous measurements and reveal how your system is performing. Attribute charts analyze categorical measurements and show you measures of agreement across responses. You can also perform a gauge study to see measures of variation in your data. See “Variability Gauge Charts” and “Attribute Gauge Charts”.
• The Process Capability platform measures the ability of a process to meet specification limits. You can compare process performance, summarized by process centering and variability, to specification limits. The platform calculates capability indices based on both long-term and short-term variation. The analysis helps identify the variation relative to the specifications; this enables you to achieve increasingly higher conformance values. See “Process Capability”.
• CUSUM charts enable you to make decisions based on the cumulative sum. These charts can detect small shifts in a process. See “CUSUM Control Charts”.
• Exponentially weighted moving average (EWMA) charts can also be used to detect small shifts in a process. See “EWMA Control Charts”.
• When you need to monitor multiple process characteristics simultaneously, see “Multivariate Control Charts”.
• The Model Driven Multivariate Control Chart (MDMVCC) platform enables you to build a control chart based on principal components or partial least squares models. See “Model Driven Multivariate Control Charts”.
• “Legacy Control Charts” describes the older control chart platforms in JMP. Instead of using these platforms, you are encouraged to use the Control Chart Builder platform, as well as the new CUSUM and EWMA Control Chart platforms.
28 Introduction to Quality and Process Methods Chapter 2 Quality and Process Methods • The Pareto Plot platform shows the frequency of problems in a quality related process or operation. Pareto plots help you decide which problems to solve first by highlighting the frequency and severity of problems. See “Pareto Plots”.
• The Diagram platform constructs cause-and-effect diagrams, which organize the sources of a problem for brainstorming or as a preliminary analysis to identify variables for further experimentation. Once complete, further analysis can be done to identify the root cause of the problem. See “Cause-and-Effect Diagrams”.
• The Manage Spec Limits utility enables you to quickly add or edit many specification limits for several columns at once. See “Manage Spec Limits Utility”.
• The Operating Characteristic (OC) Curves utility enables you to construct OC curves for control charts and attribute acceptance sampling plans. See “Operating Characteristic Curves Utility”.
Chapter 3 Control Chart Builder Create Control Charts Interactively A control chart is a graphical and analytic tool for monitoring process variation. The natural variation in a process can be quantified using a set of control limits. Control limits help distinguish common-cause variation from special-cause variation. Typically, action is taken to identify and eliminate special-cause variation. It is also important to quantify the common-cause variation in a process, as this determines the capability of a process.
Use Control Chart Builder to create control charts of your process data. Control Chart Builder can be launched as an interactive workspace or from specific control chart menu options. In the interactive workspace, you select the variables that you want to chart and drag them into zones. JMP automatically chooses an appropriate chart type based on the data. You can quickly create another type of chart, or change the current settings for an existing chart.
Figure 3.1 Control Chart Builder Example 30 Control Chart Builder Chapter 3 Quality and Process Methods Contents Overview of Control Chart Builder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 Example of Control Chart Builder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Control Chart Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Shewhart Control Charts for Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Shewhart Control Charts for Attributes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Rare Event Control Charts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 Control Chart Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Launch Control Chart Builder. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 Control Chart Builder Interactive Workspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 Launch Windows for Specific Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 Control Chart Builder Window. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 Control Chart Builder Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Control Chart Builder Red Triangle Menu Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 Options Panel and Right-Click Chart Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 Control Chart Builder Right-Click Axis Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Work with Control Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 Example of Control Limits. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 Excluded and Hidden Samples in Control Chart Builder. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Additional Examples of Control Chart Builder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71 Individual Measurement and Moving Range Charts Example . . . . . . . . . . . . . . . . . . . . . . . 72 XBar and R Chart Phase Example. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 XBar and S Charts with Varying Subgroup Sizes Example . . . . . . . . . . . . . . . . . . . . . . . . . . 76 Run Chart Example. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78 P chart Example. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 NP chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 C chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 U chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83 G chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 T chart Example. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Three Way Control Chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Example of Multiple Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Statistical Details for Control Chart Builder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Control Limits for XBar and R Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90 Control Limits for XBar and S Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 Control Limits for Individual Measurement and Moving Range Charts. . . . . . . . . . . . . . . 92 Control Limits for P and NP Charts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 Control Limits for U Charts and C Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 Levey-Jennings Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Control Limits for G Charts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Chapter 3 Control Chart Builder 31 Quality and Process Methods Control Limits for T Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96 Sigma Calculations for Three Way Control Charts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 32 Control Chart Builder Chapter 3 Overview of Control Chart Builder Quality and Process Methods Overview of Control Chart Builder A control chart is a graphical and analytic tool for monitoring process variation and identifying special-cause variation in a process. Establishing control limits to filter out routine variation helps determine whether a process is stable and predictable. If the variation in a process is more than desired, the process can be adjusted to create higher quality output with potential cost savings.
All processes exhibit measurement variation as the process is monitored over time. There are two types of variation in process measurements: • Routine or common-cause variation. Measurements from a stable process still exhibit random variation. When process measurements exhibit only common-cause variation, the measurements stay within expected limits.
• Abnormal or special-cause variation. Special-cause variation is indicated by patterns observed on the control chart. Examples include a shift in the process mean, points above or below the control limits, or measurements that trend up or down. These shifts in the process measurements can be caused by factors such as a broken tool or machine, equipment degradation, or changes to raw materials. A change or defect in the process is often identifiable by abnormal variation in the process measurements.
Control Chart Builder enables you to create several types of control charts including Shewhart and Rare Event control charts. Shewhart control charts are broadly classified into control charts for variables and control charts for attributes. Rare event charts are designed for events that occur infrequently. JMP provides a flexible, user-defined approach to building control charts. You can construct control charts in the following ways: • Use the interactive Control Chart Builder workspace. When you drag a data column to the workspace, Control Chart Builder creates an appropriate chart based on the data type and sample size.
• Use the control chart menu options to build a specific control chart using a launch window.
Once an initial chart is created through either method above, use the menus and other options to change the type of chart, change the statistic on the chart, reformat the chart, or add additional charts.
Chapter 3 Control Chart Builder 33 Quality and Process Methods Example of Control Chart Builder Example of Control Chart Builder The Socket Thickness.jmp sample data table contains measurements for the thickness of sockets. There has been an increase in the number of defects during production and you want to investigate why this is occurring. This example illustrates how to perform this investigation in Control Chart Builder using either the interactive approach or the launch window approach. The second approach is convenient if you know which type of control chart you want to build.
Control Chart Builder Interactive Method Use the interactive Control Chart Builder workspace to investigate the variability in the process data.
1.
Select Help > Sample Data Library and open Quality Control/Socket Thickness.jmp.
2.
Select Analyze > Quality and Process > Control Chart Builder.
3.
Drag Thickness to the Y zone.
4.
Drag Hour to the Subgroup zone (at bottom).
Figure 3.2 Control Charts for Socket Thickness Looking at the Average chart, you can see that there are several points below the lower control limit of 7.788772. You want to see whether another variable might be contributing to the problem.
5.
Drag Cavity into the Phase zone.
6.
Click Done.
34 Control Chart Builder Chapter 3 Example of Control Chart Builder Quality and Process Methods Figure 3.3 Control Charts for Each Cavity From the Average chart, you can conclude the following: • There are differences between the cavities, indicating the need for separate control limits for each cavity.
• Cavity 1 is producing sockets with an average thickness above that of the other cavities. This indicates that further investigation of the differences between cavities is warranted.
• All of the cavities have points that are outside the control limits. Therefore, you should investigate the lack of control in the process for each cavity.
The Range chart for each cavity shows that the within-subgroup measurements are in control and are similar across cavities.
Control Chart Builder Launch Window Method Use the XBar Control Chart launch window to obtain the same chart as Figure 3.3.
1.
Select Help > Sample Data Library and open Quality Control/Socket Thickness.jmp.
2.
Select Analyze > Quality and Process > Control Chart > XBar Control Chart.
3.
Select Thickness and click Y.
4.
Select Hour and click Subgroup.
5.
Select Cavity and click Phase.
6.
Click OK.
You should see the same control chart that appears in Figure 3.3.
Chapter 3 Control Chart Builder 35 Quality and Process Methods Control Chart Types Control Chart Types Control Chart Builder enables you to create several types of control charts, including Shewhart Variable, Shewhart Attribute, and Rare Event charts.
• “Shewhart Control Charts for Variables” • “Shewhart Control Charts for Attributes” • “Rare Event Control Charts” • “Control Chart Types” Shewhart Control Charts for Variables Control charts for variables are classified according to the subgroup summary statistic plotted on the chart.
• XBar charts are a type of location chart that display subgroup means (averages).
• R charts are a type of dispersion chart that display subgroup ranges (maximum – minimum).
• S charts are a type of dispersion chart that display subgroup standard deviations.
• Individual Measurement charts are a type of location chart that display individual measurements.
• Run charts are a type of location chart that display individual measurements as a connected series of points.
• Presummarize charts display both subgroup means and standard deviations.
• Moving Range charts are a type of dispersion chart that display moving ranges of two successive measurements.
Note: If you remove a dispersion chart or turn off the preference Show Two Shewhart Charts in File > Preferences > Platforms > Control Chart Builder, you will see only the location chart. Any associated scripts will contain the JSL option Show Two Shewhart Charts set to off (0).
XBar, R, and S Charts For quality characteristics measured on a continuous scale, a typical analysis shows both the process mean and its variability with a mean chart aligned above its corresponding R or S chart.
36 Control Chart Builder Chapter 3 Control Chart Types Quality and Process Methods Individual Measurement Charts Individual Measurement charts displays individual measurements. Individual Measurement charts are appropriate when only one measurement is available for each sampling time point. If you are charting individual measurements, the individual measurement chart shows above its corresponding moving range chart. Moving Range charts displays moving ranges of two successive measurements.
Presummarize Charts If your data consist of repeated measurements of the same process unit, you can combine these into one measurement for the unit. Pre-summarizing is not recommended unless the data contain repeated measurements on each process or measurement unit.
Presummarize summarizes the process column into sample means and/or standard deviations, based either on the sample size or sample label chosen. Then it charts the summarized data based on the options chosen in the window.
Levey-Jennings Charts Levey-Jennings charts show a process mean with control limits based on a long-term sigma. The control limits are placed at 3s distance from the center line. The standard deviation, s, for the Levey-Jennings chart is calculated the same way standard deviation is in the Distribution platform.
Shewhart Control Charts for Attributes Attributes charts are applicable for count data. Attribute charts are based on binomial and Poisson models. Because the counts are measured per subgroup, it is important when comparing multiple charts to determine whether you have a similar number of items in the subgroups between the charts. Attribute charts, like variables charts, are classified according to the subgroup sample statistic plotted on the chart.
Table 3.1 Attribute Chart Determination Distribution Used to Calculate Sigma Statistic Type: Proportion Statistic Type: Count Binomial P chart NP chart Poisson U chart C chart Chapter 3 Control Chart Builder 37 Quality and Process Methods Control Chart Types Control Chart Builder makes some decisions for you based on the variable selected. Once the basic chart is created, you can use the menus and other options to change the type, the statistic, and the format of the chart.
• P charts display the proportion of nonconforming (defective) items in subgroup samples, which can vary in size. Because each subgroup for a P chart consists of Ni items, and an item is judged as either conforming or nonconforming, the maximum number of nonconforming items in a subgroup is Ni.
• NP charts display the number of nonconforming (defective) items in subgroup samples. Because each subgroup for an NP chart consists of Ni items, and an item is judged as either conforming or nonconforming, the maximum number of nonconforming items in subgroup i is Ni.
• C charts display the number of nonconformities (defects) in a subgroup sample that usually, but does not necessarily, consists of one inspection unit.
• U charts display the number of nonconformities (defects) per unit in subgroup samples that can have a varying number of inspection units.
Rare Event Control Charts A Rare Event chart is a control chart that provides information about a process where the data comes from rarely occurring events. Tracking processes that occur infrequently on a traditional control chart tend to be ineffective. Rare event charts were developed in response to the limitations of control charts in rare event scenarios. Control Chart Builder provides two types of rare event charts (G charts and T charts). The difference between a G chart and a T chart is the quantity used to measure distance between rare events. The G chart measures counts of events between incidents, whereas the T chart measures time intervals between incidents.
Table 3.2 Rare Event Chart Determination Based on Sigma Distribution Used to Calculate Sigma Chart Type Negative Binomial G chart Weibull T chart 38 Control Chart Builder Chapter 3 Control Chart Types Quality and Process Methods G charts A G chart measures the number of events between rarely occurring errors or nonconforming incidents, and creates a chart of a process over time. Each point on the chart represents the number of units between occurrences of a relatively rare event. For example, in a production setting, where an item is produced daily, an unexpected line shutdown can occur. You can use a G chart to look at the number of units produced between line shutdowns.
When reading a G chart, the points above the upper control limit indicate that the number of events between errors has increased. If the number of events between rarely occurring errors or nonconforming incidents has increased, that is good. Therefore, a point flagged as out of control above the limits is generally considered a desirable effect when working with G charts.
T charts A T chart measures the time intervals elapsed since the last event. Each point on the chart represents a number of time intervals that have passed since a prior occurrence of a rare event. A T chart can be used for numeric, nonnegative data, date/time data, and time-between data. Since a traditional plot of these data might contain many points at zero and an occasional point at one, using a T chart avoids flagging numerous points as out of control. The data points for a T chart in Control Chart Builder are restricted to integer values.
When reading a T chart, the points above the upper control limit indicate that the amount of time between events has increased. This means that the rate of adverse events has decreased. Therefore, a point flagged as out of control above the limits is generally considered a desirable effect when working with T charts.
Control Chart Types The most common control charts are available in Control Chart Builder and in the platforms in the Analyze > Quality and Process > Control Chart menu. Use Control Chart Builder as your first choice to easily and quickly generate charts. JMP automatically chooses the appropriate chart type based on the data. Table 3.3 through Table 3.7 summarize the different control chart types.
Table 3.3 Variable Charts Without Grouping (X) Variable or Nonsummarized Data Chart Types Control Chart Builder Options Points > Statistic Limits > Sigma Individual Individual Moving Range Moving Range on Individual Moving Range Moving Range Chapter 3 Control Chart Builder 39 Quality and Process Methods Control Chart Types Individual (limits computed on median moving range) Individual Median Moving Range Median Moving Range on Individual Moving Range Median Moving Range Levey Jennings Individual Levey Jennings Table 3.4 Variable Charts with Grouping (X) Variables or Summarized Data Chart Types Control Chart Builder Options Points > Statistic Limits > Sigma XBar (limits computed on range) Average Range XBar (limits computed on standard deviation) Average Standard Deviation R Range Range S Standard Deviation Standard Deviation Levey Jennings Average Levey Jennings or overall Standard Deviation Table 3.5 Presummarize Charts Chart Types Control Chart Builder Options Points > Statistic Limits > Sigma Individual on Group Means Average Moving Range Individual on Group Means (limits computed on median moving range) Average Median Moving Range Individual on Group Std Devs Standard Deviation Moving Range Table 3.3 Variable Charts Without Grouping (X) Variable or Nonsummarized Chart Types Control Chart Builder Options Points > Statistic Limits > Sigma 40 Control Chart Builder Chapter 3 Control Chart Types Quality and Process Methods Individual on Group Std Devs (limits computed on median moving range) Standard Deviation Median Moving Range Moving Range on Group Means Moving Range on Means Moving Range Median Moving Range on Group Means Moving Range on Mean Median Moving Range Moving Range on Group Std Devs Moving Range on Std Dev Moving Range Median Moving Range on Group Std Devs Moving Range on Std Dev Median Moving Range Table 3.6 Attribute Charts Chart Types Control Chart Builder Options Points > Statistic Limits > Sigma P chart Proportion Binomial NP chart Count Binomial C chart Count Poisson U chart Proportion Poisson Table 3.7 Rare Event Charts Chart Types Control Chart Builder Options Points > Statistic Limits > Sigma G chart Count Negative Binomial T chart Count Weibull Table 3.5 Presummarize Charts (Continued) Chart Types Control Chart Builder Options Points > Statistic Limits > Sigma Chapter 3 Control Chart Builder 41 Quality and Process Methods Launch Control Chart Builder Launch Control Chart Builder You can launch Control Chart Builder in the following two ways: • If you are not sure what type of control chart is appropriate for your data, select Analyze > Quality and Process > Control Chart Builder. This method enables you to drag data columns to the workspace and Control Chart Builder creates an appropriate chart based on the data type and sample size. See “Control Chart Builder Interactive Workspace”.
• If you know which type of control chart is appropriate for your data, select the appropriate chart from the Analyze > Quality and Process > Control Chart submenu. This displays a control chart launch window. See “Launch Windows for Specific Control Charts”. There are control chart builder launch windows for the following control charts: – I/MR Control Chart – XBar Control Chart – Run Chart – P Control Chart – NP Control Chart – C Control Chart – U Control Chart – Levey Jennings Control Chart – I/MR on Means Control Chart – Three Way Control Chart Note: The CUSUM Control Chart, EWMA, and Multivariate Control Charts launch in their own platforms instead of launching in Control Chart Builder, and are documented separately. See “CUSUM Control Charts”, “Multivariate Control Charts”, and “EWMA Control Charts”.
Once you click OK in a launch window, the Control Chart Builder window appears with the Control Panel hidden by default. All other options and features are the same.
42 Control Chart Builder Chapter 3 Launch Control Chart Builder Quality and Process Methods Control Chart Builder Interactive Workspace Figure 3.4 Interactive Control Chart Builder Window For more information about the options in the Select Columns red triangle menu, see Using JMP.
To begin creating a control chart, drag variables from the Select Columns box into the zones. If you drop variables in the center, JMP guesses where to put them based on whether the variables are continuous or categorical. The Control Chart Builder workspace contains the following zones: Y Assigns the process variable.
Subgroup Assigns subgroup variables. To define subgroup levels as a combination of multiple columns, add multiple variables to the Subgroup zone. When a subgroup variable is assigned, each point on the control chart corresponds to a summary statistic for all of the points in the subgroup.
Phase Assigns phase variables. When a Phase variable is assigned, separate control limits are computed for each phase. See also “XBar and R Chart Phase Example”.
The initial Control Chart Builder window contains the following buttons: Undo Reverses the last change made to the window.
Chapter 3 Control Chart Builder 43 Quality and Process Methods Launch Control Chart Builder Start Over Returns the window to the default condition, removing all data, and clearing all zones.
Done Hides the buttons and the Select Columns box and removes all drop zone outlines. In this presentation-friendly format, you can copy the graph to other programs. To restore the window to the interactive mode, click the Control Chart Builder red triangle and select Show Control Panel.
By Identifies the variable and produces a separate analysis for each value that appears in the column.
Shewhart Variables/Shewhart Attribute/Rare Event Enables you to select Shewhart Variables, Shewhart Attribute, or Rare Event control chart types. If you select an Attribute chart type, an n Trials box and zone appear on the chart.
n Trials (Available for Attribute charts.) Assigns a lot size for an attribute control chart.
New Y Chart Produces a copy of the current chart for every column selected in the Select Columns box. The new charts use the selected columns in the Y role.
Once you drag variables to the chart, other buttons and options appear at left that enable you to show, hide or switch items on the chart (Figure 3.7). Many of these functions (Points, Limits, Warnings, etc.) are the same as the functions available when you right-click the chart. See “Options Panel and Right-Click Chart Options”. For information about warnings and rules, see “Tests” and “Westgard Rules”.
Three Way Control Chart Enables you to produce a three way control chart for variable chart types. The subgroup size must be greater than one. The plotting statistic is based on subgroup averages, within-subgroup variation, or between-subgroup variation. The default set of three includes a presummarized chart of the averages using Moving Range limits, a Moving Range chart and a Range chart.
Event Chooser Allows the chart to respond in real time to selection changes. There are several standard groups of responses that are recognized and pre-scored (for example, pass/fail, yes/no, Likert Scales, conforming/non-conforming, and defective/non-defective). If you are analyzing results from a survey and want to focus solely on a specific sector of the results for one or more questions, you can make the selection on the screen. When you make the selection, the chart is scored again and replotted immediately. The levels selected in the Event Chooser are counted as events, and all other levels are counted as non-events.
The Event Chooser is available for attribute charts with response columns that have a modeling type of nominal or ordinal. If you want the Event Chooser to work on a numeric integer-valued nominal or ordinal response column, you must select the Use Event Chooser option from the Control Chart Builder red triangle menu. The Event Chooser does not appear for response columns with a modeling type of continuous.
44 Control Chart Builder Chapter 3 Launch Control Chart Builder Quality and Process Methods Launch Windows for Specific Control Charts The options that you see in the launch windows vary depending on whether you are launching variable control charts or attribute control charts.
Launch Windows for Variable Control Charts This section contains information about the launch windows for I/MR, XBar, Run, Levey Jennings, I/MR on Means, and Three Way Control Charts.
Figure 3.5 Launch Window for Variable Control Charts For more information about the options in the Select Columns red triangle menu, see Using JMP.
Y Assigns the process variables.
Subgroup (Available only for XBar, Levey Jennings, I/MR on Means, and Three Way Control Charts.) Assigns the subgroup variables. When a subgroup variable is assigned, each point on the control chart corresponds to a summary statistic for all of the points in the subgroup.
Sample Label (Available only for I/MR Charts.) Assigns a sample label variable to uniquely label individual measurements.
Phase (Not available for Run Charts.) Assigns the phase variable. When a Phase variable is assigned, separate control limits are computed for each phase.
By Identifies a variable to produce a separate analysis for each value that appears in the column.
Sort by Sample Label Sorts all variables by the sample label.
Chapter 3 Control Chart Builder 45 Quality and Process Methods Launch Control Chart Builder Constant Subgroup Size (Available only for XBar charts.) Specifies a number of observations that make up each subgroup. If the number of observations per subgroup is not constant, you should specify a column in the Subgroup role.
Note: If you specify a Subgroup variable, the platform ignores the setting for the Constant Subgroup Size option.
Dispersion Chart (Available only for XBar charts.) Specifies the type of dispersion chart that appears below the XBar chart. You can choose between no dispersion chart, a range chart (XBar & R), or a standard deviation chart (XBar & S).
Launch Windows for Attribute Control Charts This section contains information about the launch windows for NP, P, C, and U Control Charts.
Figure 3.6 Launch Window for Attribute Control Charts For more information about the options in the Select Columns red triangle menu, see Using JMP.
Y Assigns the process variables.
Subgroup Assigns the subgroup variables. When a subgroup variable is assigned, each point on the control chart corresponds to a summary statistic for all of the points in the subgroup.
n Trials Assigns the subgroup sample size.
Phase Assigns the phase variable. When a Phase variable is assigned, separate control limits are computed for each phase.
By Identifies a variable to produce a separate analysis for each value that appears in the column.
46 Control Chart Builder Chapter 3 Control Chart Builder Window Quality and Process Methods Control Chart Builder Window Use Control Chart Builder to construct control charts for process data. The analysis produces a chart that can be used to evaluate whether a process is in a state of statistical control. The report varies depending on which type of chart you select. Control charts update dynamically as data is added or changed in the data table. Figure 3.7 displays the Control Chart Builder window for the Bottle Tops.jmp sample data table.
To create the chart: 1.
Select Help > Sample Data Library and open Quality Control/Bottle Tops.jmp.
2.
Select Analyze > Quality and Process > Control Chart Builder.
3.
Drag Status to the Y zone.
4.
Drag Sample to the Subgroup zone (at bottom).
Figure 3.7 Control Chart Builder Window You can drag other variables into the various zones to augment the analysis and use the “Control Chart Builder Options” to further examine the data. Some of the right-click chart options (for example, show or hide points, limits, warnings, and zones; select statistic and sigma options) also appear on the left hand side of the chart for easy access.
Control charts have the following characteristics: • Each point plotted on the chart represents an individual process measurement or summary statistic. Subgroups should be chosen rationally, that is, they should be chosen to maximize the probability of seeing a true process signal between subgroups. Often, this center line UCL and LCL Chapter 3 Control Chart Builder 47 Quality and Process Methods Control Chart Builder Options requires knowledge of the process to determine the most effective grouping strategy. See Wheeler (2004); Woodall and Adams (1998).
• The vertical axis of a control chart is scaled in the same units as the summary statistic.
• The horizontal axis of a control chart identifies the subgroup samples and is time ordered. Observing the process over time is important in assessing if the process is changing.
The green line is the center line, or the average of the data. The center line indicates the average (expected) value of the summary statistic when the process is in statistical control. Measurements should appear equally on both sides of the center line. If not, this is possible evidence that the process average is changing.
• The two red lines are the upper and lower control limits, labeled UCL and LCL. These limits give the range of variation to be expected in the summary statistic when the process is in statistical control. If the process is exhibiting only routine variation, then all the points should fall randomly in that range.
Note: To hide the lower control limits on dispersion, attribute, and rare event charts, deselect the Show Lower Limit option in the options panel. To change the default to always hide the lower control limits, deselect the Show Lower Control Limit preference in File > Preferences > Platforms > Control Chart Builder.
• A point outside the control limits signals the presence of a special cause of variation.
Options in the Control Chart Builder window create control charts that can be updated dynamically as samples are received and recorded or added to the data table. When a control chart signals abnormal variation, action should be taken to return the process to a state of statistical control if the process degraded. If the abnormal variation indicates an improvement in the process, the causes of the variation should be studied and implemented.
When you double-click the axes, the appropriate Axis Specification window appears for you to specify the format, axis values, number of ticks, gridline, reference lines, and other options to display.
Control Chart Builder Options Control Chart Builder options appear in the red triangle menu or by right-clicking on a chart or axis. Some of the right-click chart options also appear on the bottom left hand side of the chart for easy access. You can also set preferences for many of the options in Control Chart Builder at File > Preferences > Platforms > Control Chart Builder.
• “Control Chart Builder Red Triangle Menu Options” • “Options Panel and Right-Click Chart Options” • “Control Chart Builder Right-Click Axis Options” 48 Control Chart Builder Chapter 3 Control Chart Builder Options Quality and Process Methods Control Chart Builder Red Triangle Menu Options The Control Chart Builder red triangle menu contains the following options: Show Control Panel Shows or hides the following elements: – buttons – the Select Columns box – the drop zone borders – check boxes and drop-down menus Show Limit Summaries Shows or hides the Limit Summaries report. This report shows the control limits (LCL and UCL), the center line (Avg), the Points and Limits plotted, and the Sample Size for the chart. Sample size is not shown for rare event charts.
Show Capability (Available only for Shewhart Variables charts that have specification limits.) Shows or hides the Process Capability Analysis report. Since the report is part of the Limit Summaries report, the Process Capability report appears only when the Show Limit Summaries option is selected. For more information, see “The Process Capability Report”. You can set preferences for many of the options in the Process Capability report in Control Chart Builder at File > Preferences > Platforms > Process Capability.
Note: Show Capability is not available if the response variable has no variation.
Show Alarm Report Shows or hides a report that contains information about out of control samples. The report reflects failures for currently enabled tests in each chart and updates automatically as different tests are enabled and disabled and as data and row states change. A second table lists the currently enabled tests for each chart. The first table contains the following columns: Position Indicates the numerical position of the chart, starting from the top of the report.
Total Samples Out of Control Counts the number of samples that failed at least one of the selected tests.
Alarm Rate The total number of samples out of control divided by the total number of nonmissing samples. This is also known as the Proportion Out of Control.
Note: The counts that contribute to the calculation of the alarm rate include excluded samples only if the Test Excluded Subgroups and the Show Excluded Region options are both selected.
Show Limit Labels Shows or hides labels for the limits in each chart. The limits are shown inside the right frame of the chart.
Chapter 3 Control Chart Builder 49 Quality and Process Methods Control Chart Builder Options Show Sigma Report (Available only for Shewhart Variables charts.) Shows or hides the Process Sigma Report, which is a table of sigma values. The Process Sigma Report contains the overall sample size, number of subgroups, sample mean, overall sigma, within sigma, and stability index. For three way control charts, the between-sigma and between-and-within sigma values are also shown. If a phase variable is specified, a set of values is given for each phase.
Note: The Process Sigma Report appears only if the Limit Summaries report is turned on.
Get Limits Retrieves the control limits that are stored in an open or saved data table.
Show Excluded Region Shows (on) or removes (off) the regions of the chart where samples have been excluded. When entirely excluded subgroups are shown on the location chart, they appear as dimmed points to indicate that they are excluded.
Caution: The Show Excluded Region option impacts the chart. Excluded samples are removed from the calculation of control limits, whether this option is on or off. Excluded samples are included in alarm rate calculations only if the Test Excluded Subgroups and the Show Excluded Region options are both selected.
Set Subgroup Size (Not available if a subgroup variable is specified.) Sets a subgroup size. Missing values are taken into account when computing limits and sigma.
Note: If the Set Subgroup Size option is used, the Show Excluded Region option is turned on automatically.
Save Limits Saves the control limits in one of the following ways: in Column Saves control limits as a column property in the existing data table for the response variable. If the limits are constant, LCL, Avg, and UCL values for each chart type in the report are saved. This option is not available with phase charts. In addition, the option has no effect if the sample sizes are not constant for each chart.
in New Table Saves the standard deviation and mean for each chart into a new data table. If the limits are constant, the LCL, Avg, UCL, and Sample Size for each chart are saved as well. If there are phases, a new set of values is saved for each phase. There is a row for each statistic and a column for each Y variable.
in New Tall Table (Not available for Rare Event, Attribute, or Phase charts.) Saves the standard deviation, mean, and Sigma for each chart into a new data table. If the limits are constant, the LCL, Avg, UCL, and Sample Size for each chart are saved as well. There is a row for each Y variable and a column for each statistic. A column for Sigma that can be used in the Process Screening platform is also saved.
50 Control Chart Builder Chapter 3 Control Chart Builder Options Quality and Process Methods Save Summaries Creates a new data table containing such information as the sample label, sample sizes, statistic being plotted, center line, control limits, and any tests, warnings and failures. The specific statistics included in the table depend on the type of chart.
Graph Spacing Sets the amount of space between the graphs.
Include Missing Categories Enables the graph to collect rows with missing values in a categorical column, and displays the missing values on the graph as a separate category. If this option is disabled, all rows with a missing X value are removed from the calculations, in addition to being hidden from the graph.
This option is not available for continuous X variables or categorical Y variables because there is no compelling way to display the collected missing values on the relevant axes. By default, this option is enabled.
Note: If Include Missing Categories is enabled, capability analysis results in Control Chart Builder do not match those in the Process Capability platform if a categorical X variable has missing values.
Use Event Chooser (Available only for attribute charts with numeric non-continuous Y variables.) Categorizes ordinal numeric data and offers individual numeric-level modeling selections.
Alarm Script Enables you to write and run a script that indicates when the data fail special causes tests. See “Tests”. Results can be written to a file, written to the log or sent in an email. There is an option to include an explanation of why the test failed.
As an Alarm Script is invoked, the following variables are available, both in the issued script and in subsequent JSL scripts: qc_col is the name of the column qc_test is the test that failed qc_sample is the sample number qc_phase is the label of the phase during which the failure occurred See the Scripting Guide for more information about writing custom Alarm Scripts.
Note: Alarm scripts are not available in reports that use the Local Data Filter.
Sort by Row Order Sorts all subgroup and phase variables in the order in which the levels appear in the data table. This applies to all combinations of nested subgroup and phase variables.
Test Excluded Subgroups (Available only if the Show Excluded Region option is selected.) Includes (on) or excludes (off) entirely excluded subgroups in the computation of tests. Chapter 3 Control Chart Builder 51 Quality and Process Methods Control Chart Builder Options When excluded subgroups are shown and the Text Excluded Subgroups option is not selected, the excluded subgroups are treated as missing values.
Note: For any test that relies on consecutive points (runs tests), an entirely excluded subgroup is treated as missing and counts of consecutive points are restarted.
Control Chart Dialog (Available only if the control chart is launched through a Control Chart launch window.) Opens the Control Chart launch window with the original settings that were used to create the control chart.
See Using JMP for more information about the following options: Local Data Filter Shows or hides the local data filter that enables you to filter the data used in a specific report.
Redo Contains options that enable you to repeat or relaunch the analysis. In platforms that support the feature, the Automatic Recalc option immediately reflects the changes that you make to the data table in the corresponding report window.
Save Script Contains options that enable you to save a script that reproduces the report to several destinations.
• Column Switcher is available only for a single Y variable having two or fewer associated charts. Based on the selected chart type, only columns that are appropriate for the Y role are included in the Column Switcher column list.
• In Control Chart Builder, the Automatic Recalc option is turned on by default and cannot be turned off.
• When using the local data filter, you can deselect the Show Excluded Region option for more focused exploration.
Options Panel and Right-Click Chart Options The following options appear on the left hand side of the Control Chart Builder report or when you right-click a chart. Some options appear in both locations.
Event Chooser (Available only for attribute charts with response columns that have a modeling type of nominal or ordinal.) Allows the chart to respond in real time to selection changes. See “Event Chooser”.
Points Provides the following options: Statistic Changes the statistic plotted on the chart. See “Statistic”.
52 Control Chart Builder Chapter 3 Control Chart Builder Options Quality and Process Methods Individual Points Show or hides individual observations in a subgroup. Available only with a subgroup variable or Set Sample Size. This option is not available for Attribute chart types or Rare Event charts.
Box Plots (Available only for Shewhart Variables charts.) Shows or hides box plots on the chart.
Show Connect Line Shows connecting lines between the points.
Show Points Shows or hides the points on the chart.
Limits Provides the following options: Sigma Specifies the method of computing sigma. See “Sigma”.
Zones (Available only for Variables and Attribute chart types.) Shows or hides the zones on the chart. There are three equal-width zones on either side of the mean. Zones are not drawn below the LCL or above the UCL. If the limits for a Variables chart are not centered around the mean, min(Avg-LCL,UCL-Avg)/3 is used as the width of each zone. The zones for an Attributes chart use a width of (UCL-Avg)/3.
Shade Zones Shows or hides shading zones by ranges. Zone C is shaded green, zones A and B are shaded yellow, and beyond zone A is shaded red.
Spec Limits (Available only if the data table has a Spec Limits column property or if you specify Spec Limits using the Add Spec Limits option.) Shows or hides the specification limits on the chart. By default, the spec limits are shown if the Spec Limits column property has the Show as Graph Reference Lines option selected. See Using JMP for information about adding a Spec Limits column property.
Add Spec Limits (Available in the right-click menu.) Enables you to enter specification limits.
Set Control Limits (Available in the right-click menu.) Enables you to enter control limits for tests. After you click OK in the Set Control Limits window, the specified control limits are set uniformly across groups. Select this option again to remove the specified control limits.
Add Limits (Available in the right-click menu.) Specifies additional control limits to be plotted on the chart. These limits are not used in tests.
Show Upper Limit Shows or hides the upper control limit on the chart. If you hide the upper control limit on a chart, the Test Beyond Limits and Test 1 options do not flag points associated with the hidden upper control limit.
Show Lower Limit Shows or hides the lower control limit on the chart. If you hide the lower control limit on a chart, the Test Beyond Limits and Test 1 options do not flag points associated with the hidden lower control limit.
Chapter 3 Control Chart Builder 53 Quality and Process Methods Control Chart Builder Options Show Center Line Shows or hides the center line on the chart.
Add Dispersion Chart (Available in the right-click menu.) Adds a dispersion chart to the chart area. Change the chart type with the Points options. A dispersion chart illustrates the variation in the data by plotting one of many forms of dispersion, including the range, standard deviation, or moving range. Available only for Variables chart types.
Note: You can customize the default dispersion chart type using the Dispersion Chart and Summarized Dispersion Chart preferences in File > Preferences > Platforms > Control Chart Builder.
Set Subgroup Size Sets a subgroup size. Missing values are taken into account when computing limits and sigma.
Warnings Provides the following options: Customize Tests (Available in the right-click menu.) Enables you to design custom tests and select or deselect multiple tests at once. After the option is selected, the Customize Tests window appears for designing the tests. Select a test description, and enter the desired number (n) and label. You can save the settings to preferences and also restore the default settings. Available only for Variables and Attribute chart types.
Tests Enables you to select which statistical control tests to enable. For more information about tests, see “Tests”. Available only for Variables and Attribute chart types.
Note: Hover over a flagged point on the chart to see a description of the test that failed.
Westgard Rules Specifies the set of Westgard statistical control tests that are enabled. Because Westgard rules are based on sigma and not the zones, they can be computed without regard to constant sample size. For more information about tests, see “Westgard Rules”. Available only for Variables and Attribute chart types.
Test Beyond Limits (Called Test 15 in JMP) Enables the test for any points beyond the control limits. These points are identified on the chart. This test works on all charts with limits, regardless of the sample size being equal.
Note: If you hide the upper or lower control limits, the Test Beyond Limits option does not flag points that are beyond limits that are not shown on the control chart.
Remove Graph (Available in the right-click menu.) Removes the control chart.
Remove Location Chart (Available only if you right-click a location chart.) Removes the location chart from the report.
Remove Dispersion Chart (Available only if you right-click a dispersion chart.) Removes the dispersion chart from the report.
54 Control Chart Builder Chapter 3 Control Chart Builder Options Quality and Process Methods Note: For a description of the Rows, Graph, Customize, and Edit menus, see Using JMP.
Three Way Control Chart Enables you to produce a three way control chart for variable chart types. See “Three Way Control Chart”.
Statistic You can change the statistic represented by the points on the chart. The options available depend on the chart type selected.
For Variables chart types, you can change the statistic represented by the points on the chart using the following options: Individual Creates a chart where each point represents an individual value in the data table.
Average Creates a chart where each point represents the average of the values in a subgroup.
Range Creates a chart where each point represents the range of the values in a subgroup.
Standard Deviation Creates a chart where each point represents the standard deviation of the values in a subgroup.
Moving Range on Means Computes the difference in the range between two consecutive subgroup means.
Moving Range on Std Dev Computes the difference in the range between two consecutive subgroup standard deviations.
Moving Range Creates a chart where each point is the difference between two consecutive observations.
Note: The Average, Range, Standard Deviation, Moving Range on Means, and Moving Range on Std Dev methods appear only if a subgroup variable with a sample size greater than one is specified or a sample size is set.
For Attribute chart types, you can change the statistic represented by the points on the chart using the following options: Proportion Creates a chart where each point represents the proportion of items in subgroup samples.
Count Creates a chart where each point represents the number of items in subgroup samples.
For Rare Event chart types, the statistic represented by the points on the chart uses the Count option.
Chapter 3 Control Chart Builder 55 Quality and Process Methods Control Chart Builder Options Sigma You can change the method for computing sigma for the chart. The options available depend on the chart type selected.
For Variables chart types, you can use the following options: Range Uses the range of the data in a subgroup to estimate sigma.
Standard Deviation Uses the standard deviation of the data in a subgroup to estimate sigma.
Moving Range Uses the moving ranges to estimate sigma. The moving range is the difference between two consecutive points.
Median Moving Range Uses the median moving range to estimate sigma, rather than the average moving range.
Levey-Jennings Uses the standard deviation of all the observations to estimate sigma. If your chart has phases, sigma is calculated for each phase separately.
For Attribute chart types, you can use the following options: Binomial Uses the binomial distribution model to estimate sigma. The model indicates the number of successes in a sequence of experiments, where each experiment yields success with some probability. Selecting Binomial yields either a P or NP chart.
Poisson Uses the Poisson distribution model to estimate sigma. The model indicates the number of events and the time at which these events occur in a given time interval. Selecting Poisson yields either a C or U chart.
For Rare Event chart types, you can use the following options: Negative Binomial Uses the negative binomial distribution model to estimate sigma. The model indicates the number of successes in a sequence of trials before a specified number of failures occur. Selecting Negative Binomial yields a G chart.
Weibull Uses the Weibull distribution model to estimate sigma. The model indicates the mean time between failures. Selecting Weibull yields a T chart.
Tests The Warnings option in the pop-up menu or on the left hand side of the window displays a submenu for Tests selection. You can select one or more tests for special causes (Western Electric rules) from the menu. Nelson (1984) developed the numbering notation used to identify special tests on control charts. The tests work with both equal and unequal sample sizes.
56 Control Chart Builder Chapter 3 Control Chart Builder Options Quality and Process Methods If a selected test is positive for a particular sample, that point is labeled with the test number. When you select several tests for display and more than one test signals at a particular point, the label of the numerically lowest test specified appears beside the point. You can hover over a flagged point on the chart to see a description of the test that failed.
Tip: To add or remove several tests at once, select or deselect the tests in the Control Panel under Warnings > Tests.
Table 3.8 lists and interprets the eight tests, and Figure 3.9 illustrates the tests. The following rules apply to each test: • The area between the upper and lower limits is divided into six zones, each with a width of one standard deviation.
• The zones are labeled A, B, C, C, B, A with zones C nearest the center line.
• A point lies in Zone B or beyond if it lies beyond the line separating zones C and B. That is, if it is more than one standard deviation from the center line.
• Any point lying on a line separating two zones lines is considered belonging to the innermost zone. So, if a point lies on the line between Zone A and Zone B, the point is considered to be in Zone B.
• When a Phase variable is specified, the counts for each test are reset at the start of each phase.
Notes: • Tests 1 through 8 apply to all Shewhart Variables chart types.
• Tests 1, 2, 5, and 6 apply to the upper and lower halves of the chart separately.
• Tests 3, 4, 7, and 8 apply to the whole chart.
• Once a runs test (one that is based on consecutive observations) is triggered, the counts do not reset to 0 when moving to the next sample.
• Runs tests handle excluded rows based on the setting of the Show Excluded Region and Test Excluded Subgroups options.
– By default, both options are selected, and the excluded rows are included in the runs tests calculations.
– If the Show Excluded Region option is selected and the Test Excluded Subgroups option is not selected, the excluded rows are treated as missing and the counts for the runs tests reset to 0 when moving to the next sample.
– If the Show Excluded Region option is not selected, the excluded rows are treated as if they are deleted.
• Tests 5 through 8 are not available for attribute charts.
See Nelson (1984, 1985) for further recommendations on how to use these tests.
Chapter 3 Control Chart Builder 57 Quality and Process Methods Control Chart Builder Options Figure 3.8 Zones for Western Electric Rules Table 3.8 Description and Interpretation of Tests for Special Causesa Test 1 One point beyond Zone A (upper or lower) Detects a shift in the mean, an increase in the standard deviation, or a single aberration in the process. For interpreting Test 1, any dispersion chart (R, S, or MR) can be used to rule out increases in variation.
Note that if you hide the upper or lower control limits, the Test 1 option does not flag points that are associated with limits that are not shown on the control chart.
Test 2 Nine points in a row in a single (upper or lower) side of Zone C or beyond Detects a shift in the process mean.
Test 3 Six points in a row steadily increasing or decreasing (anywhere on the chart) Detects a trend or drift in the process mean.
Test 4 Fourteen points in a row alternating up and down (anywhere on the chart) Detects systematic effects such as two alternately used machines, vendors, or operators.
3 limits center line zones 58 Control Chart Builder Chapter 3 Control Chart Builder Options Quality and Process Methods Test 5 Two out of three points in a row in or beyond Zone A and the point itself is in or beyond Zone A; the two points must be on the same side (upper or lower) Detects a shift in the process average or increase in the standard deviation. Any two out of three points provide a positive test.
Test 6 Four out of five points in a row in or beyond Zone B and the point itself is in or beyond Zone B; the four points must be on the same side (upper or lower) Detects a shift in the process mean. Any four out of five points provide a positive test.
Test 7 Fifteen points in a row in Zone C, above and below the center line Detects stratification of subgroups when the observations in a single subgroup come from various sources with different means. Also detects a reduction in variation.
Test 8 Eight points in a row on both sides of the center line with none in Zones C Detects stratification of subgroups when the observations in one subgroup come from a single source, but subgroups come from different sources with different means.
a. Nelson (1984, 1985) Table 3.8 Description and Interpretation of Tests for Special Causesa (Continued) Chapter 3 Control Chart Builder 59 Quality and Process Methods Control Chart Builder Options Figure 3.9 Illustration of Special Causes Tests1 Westgard Rules Westgard rules are implemented under the Westgard Rules submenu of the Warnings option when you right-click a chart or on the left hand side of the window. The different tests are abbreviated with the decision rule for the particular test. For example, 1 2s refers to a test where one point is two standard deviations away from the mean.
1. Nelson 1984, 1985) UCL Avg LCL A B C C B A UCL Avg LCL A B C C B A UCL Avg LCL A B C C B A UCL Avg LCL A B C C B A UCL Avg LCL A B C C B A UCL Avg LCL A B C C B A UCL Avg LCL A B C C B A UCL Avg LCL A B C C B A 1 2 3 4 5 5 5 6 7 8 Test 1: One point beyond Zone A Test 2: Nine points in a row in a single (upper or lower) side of Zone C or beyond Test 3: Six points in a row steadily increasing or decreasing Test 4: Fourteen points in a row alternating up and down Test 5: Two out of three points in a row in Zone A or beyond Test 6: Four out of five points in a row in Zone B or beyond Test 7: Fifteen points in a row in Zone C (above and below the center line) Test 8: Eight points in a row on both sides of the center line with none in Zone C 60 Control Chart Builder Chapter 3 Control Chart Builder Options Quality and Process Methods Notes: • Once a runs test (one that is based on consecutive observations) is triggered, the counts do not reset to 0 when moving to the next sample.
• Runs tests handle excluded rows based on the setting of the Show Excluded Region and Test Excluded Subgroups options.
– By default, both options are selected, and the excluded rows are included in the runs tests calculations.
– If the Show Excluded Region option is selected and the Test Excluded Subgroups option is not selected, the excluded rows are treated as missing and the counts for the runs tests reset to 0 when moving to the next sample.
– If the Show Excluded Region option is not selected, the excluded rows are treated as if they are deleted.
Rule 1 2S (called Test 9 in JMP) is commonly used with Levey-Jennings charts, where control limits are set 2 standard deviations away from the mean. The rule is triggered when any one point goes beyond these limits.
Rule 1 3S (called Test 10 in JMP) refers to a rule common to Levey-Jennings charts where the control limits are set 3 standard deviations away from the mean. The rule is triggered when any one point goes beyond these limits.
Rule 2 2S (called Test 11 in JMP) is triggered when two consecutive control measurements are farther than two standard deviations from the mean.
UCL Avg LCL +1s +2s +3s -1s -2s -3s UCL Avg LCL +1s +2s +3s -1s -2s -3s UCL Avg LCL +1s +2s +3s -1s -2s -3s Chapter 3 Control Chart Builder 61 Quality and Process Methods Control Chart Builder Options Rule R 4S (called Test 12 in JMP) is triggered when one measurement is greater than two standard deviations from the mean and the previous measurement is greater than two standard deviations from the mean in the opposite direction such that the difference is greater than 4 standard deviations.
Rule 4 1S (called Test 13 in JMP) is triggered when four consecutive measurements are more than one standard deviation from the mean.
Rule 10 X (called Test 14 in JMP) is triggered when ten consecutive points are on one side of the mean.
Control Chart Builder Right-Click Axis Options Remove Graph Removes the entire graph.
Remove Removes a variable.
Note: If there is more than one chart type on the graph, a submenu listing the different charts is displayed. You can select which chart to remove.
For more information about the Axis Settings, Revert Axis, Add or Remove Axis Label, Save to Column Property, and Edit options, see Using JMP.
UCL Avg LCL +1s +2s +3s -1s -2s -3s UCL Avg LCL +1s +2s +3s -1s -2s -3s UCL Avg LCL +1s +2s +3s -1s -2s -3s 62 Control Chart Builder Chapter 3 Work with Control Limits Quality and Process Methods Work with Control Limits Control limits are based on the performance of your process and tell you about the variability in your process. Upper control limits (UCLs), center lines, and lower control limits (LCLs) are calculated from the data when a control chart is created. You can use these calculated control limits to indicate when your process has changed.
It is important to note that control limits are different from specification limits, which are often used in capability analysis.
Example of Control Limits In this example, consider a company’s printing process. Variations can cause distortion in the line, including skew, thickness, and length problems. In this example, we will consider the length of the line. A line is considered good if it has a printed length of 16 cm +/- 0.2 cm. Any longer and the sentence might run off of the page. Any shorter and there would be a lot of wasted space on the page. For every print run, the first and last books are taken for measurement. The line lengths are measured on a specified page in the middle of each book.
You want to know: Is this process in control (stable)? Are we getting consistent print quality? What happens when we make improvements to the printing process? Does quality improve? To answer these questions, we need to create control charts and use control limits.
This example is in three parts. In most cases, you would start with “Create the Baseline Control Chart”, where you let JMP calculate the control limits for you. Then, to apply these control limits to new data, you would either “Specify Control Limits” or “Specify Multiple Sets of Control Limits” (for phase data).
Create the Baseline Control Chart First, examine whether the existing process is in control. If it is, we can use the control limits created by JMP as our baseline or historical limits.
1.
Select Help > Sample Data Library and open Quality Control/Line Length.jmp.
Table 3.9 Control Limits versus Specification Limits Control Limits Specification Limits Calculated from data Defined by the customer or design Based on variability Based on system requirements The voice of the process The voice of the customer Chapter 3 Control Chart Builder 63 Quality and Process Methods Work with Control Limits 2.
Select Analyze > Quality and Process > Control Chart Builder.
3.
Drag Length to the Y zone.
An Individual and Moving Range chart of Length appears. This chart is appropriate if you have no natural subgrouping in your data. However, in this example, there is a natural subgrouping, which is each print run.
4.
Drag Run to the Subgroup zone (at bottom).
Figure 3.10 XBar and R Chart of Line Length by Print Run Three lines are drawn horizontally across the XBar and R charts. These are the calculated LCL (lower control limit), Avg (average) and UCL (upper control limit).
Ideally, we would like for all of our points to fall within the control limits, and we would like for the points to fall randomly within these limits. Looking at the graph, we see that no points fall outside of the control limits, and there does not appear to be a pattern to the points. To investigate further, perform Western Electric tests to check for patterns and trends that would cause these tests to fail. (The Western Electric tests are also referred to as Nelson tests.) 5.
In the XBar chart, right-click and select Warnings > Tests > All Tests.
Notice that no points were circled or flagged. This means that our process is in control or stable.
If we had determined that our process was not in control, we would investigate out of control points or work to alter our process so that it is in control. For this example, since the process is already in control or stable, we can skip that step. Now, you can use these 64 Control Chart Builder Chapter 3 Work with Control Limits Quality and Process Methods control limits with new data. Proceed to “Specify Control Limits”, or “Specify Multiple Sets of Control Limits” (to see an example with phase data).
Specify Control Limits Since we established that the process is in control, we can use these historical limits with new data to see how the new data compares to the existing process. To use historical limits, we need to specify control limits instead of having JMP calculate them.
There are several ways to specify control limits in JMP: • “Set Control Limits Option” • “Add a Column Property” • “Use the Get Limits Option” • “Exclude Rows” Set Control Limits Option One simple way to specify control limits is to use the Set Control Limits option in Control Chart Builder.
1.
Select Help > Sample Data Library and open Quality Control/New Length Data.jmp.
This is the table that contains your new data.
2.
Select Analyze > Quality and Process > Control Chart Builder.
3.
Drag Length to the Y zone.
4.
Drag Run to the Subgroup zone (at bottom).
5.
Right-click in the Average (XBar) chart and select Limits > Set Control Limits.
6.
Enter these limits: – LCL - 15.90519 – Avg - 15.99825 – UCL - 16.09131 These are the historical limits from the Average (XBar) chart in Figure 3.10.
7.
Click OK.
8.
Right-click in the Range (R) chart and select Limits > Set Control Limits.
9.
Enter these limits: – LCL - 0 – Avg - 0.0495 – UCL - 0.161693 These are the historical limits from the Range (R) chart in Figure 3.10.
Chapter 3 Control Chart Builder 65 Quality and Process Methods Work with Control Limits 10. Click OK.
Figure 3.11 XBar and R Chart of Line Length with Historical Limits Rather than calculating limits from the data, JMP used the historical control limits that you defined. In the Length Limit Summaries table, notice that the Limits Sigma now says User Defined. Many points now fall outside of the limits. Also, the averages are higher than those of the baseline process. This process appears different from the original process that we used to calculate the baseline control limits.
Add a Column Property Another way to specify control limits is to add the Control Limits column property to a column in your new data table.
1.
Select Help > Sample Data Library and open Quality Control/New Length Data.jmp.
This is the table that contains your new data.
2.
Select the Length column and click Cols > Column Info.
3.
Click Column Properties > Control Limits.
4.
XBar is selected, so enter these fixed limits for the Average (XBar) chart: – Avg - 15.99825 – LCL - 15.90519 – UCL - 16.09131 These are the historical limits from the Average (XBar) chart in Figure 3.10.
66 Control Chart Builder Chapter 3 Work with Control Limits Quality and Process Methods Leave the value for Subgroup Size as missing. This value is not used in the Control Chart Builder platform.
5.
Click XBar > R. Enter these fixed limits for the Range (R) chart: – Avg - 0.0495 – LCL - 0 – UCL - 0.161693 These are the historical limits from the Range (R) chart in Figure 3.10.
Leave the value for Subgroup Size as missing. This value is not used in the Control Chart Builder platform.
6.
Click OK.
You have entered control limits for XBar and R charts for the Length column. Now you can create a control chart.
7.
Select Analyze > Quality and Process > Control Chart Builder.
8.
Drag Length to the Y zone.
9.
Drag Run to the Subgroup zone (at bottom).
The control chart is identical to Figure 3.11.
Use the Get Limits Option The Get Limits method of specifying control limits is the most flexible. You should use this method in the following cases: • If you have control limits for many different processes • If you have different control limits for each phase (see “Specify Multiple Sets of Control Limits”) To use the Get Limits method, you need a data table that defines your historical limits. For more information about how to create a limits table, see “Saving and Retrieving Limits”.
Note: When no subgroup variable is specified, the Get Limits option uses the subgroup size (_Sample Size) from the limits table. Also, when the limits are missing in the file, JMP also looks for the sigma (_Std Dev). When no LCL or UCL are specified in the limits file (if both the average and sigma are found, and the subgroup size is constant), the option sets the limits based on the average, subgroup size, and sigma.
In this example, a limits data table has already been created.
1.
Select Help > Sample Data Library and open Quality Control/New Length Data.jmp.
This is the table that contains your new data.
2.
Select Analyze > Quality and Process > Control Chart Builder.
Chapter 3 Control Chart Builder 67 Quality and Process Methods Work with Control Limits 3.
Drag Length to the Y zone.
4.
Drag Run to the Subgroup zone (at bottom).
5.
Click the Control Chart Builder red triangle and select Get Limits.
6.
Select Other and click OK.
7.
Navigate and open the limits data table for this example, called Length Limits.jmp. By default, the file is located here: – On Windows: C:\Program Files\SAS\JMP\16\Samples\Data\Quality Control – On macOS: \Library\Application Support\JMP\16\Samples\Data\Quality Control The control chart is identical to Figure 3.11.
Exclude Rows Another way to specify control limits is to exclude rows in a data table. One advantage to this method is that you can see both the historical data and new data in the same graph. This can help to visualize and investigate differences when they occur between the data collection periods.
To use this method, you must meet the following criteria: • New and old data must reside in the same data table.
• Historical data and new data must all have equal subgroup sizes.
• All new data must be excluded in the data table (using Rows > Exclude/Unexclude).
In this example, new data have already been excluded.
1.
Select Help > Sample Data Library and open Quality Control/Combined.jmp.
This table contains old and new data, and the rows corresponding to the new data are excluded.
2.
Select Analyze > Quality and Process > Control Chart Builder.
3.
Drag Length to the Y zone.
4.
Drag Run to the Subgroup zone (at bottom).
68 Control Chart Builder Chapter 3 Work with Control Limits Quality and Process Methods Figure 3.12 XBar and R Chart of Line Length with Excluded Data JMP uses only the unexcluded rows (historical data) to create the control limits. The new data (excluded data) are plotted on the graph (dimmed), but these data were not used in any of the calculations.
Specify Multiple Sets of Control Limits In this example, you want to set different control limits for different phases of a process. The column property, set control limits, and excluded row state methods will not work in this situation because these methods are limited to only one set of control limits for the entire chart. For a control chart with phases, you need to use the get limits method.
In the printing company, the goal is to reduce the variability of the force needed to break the bond between paper and the book spine for three different sites. Each site has different machines, different operators, and is also located in different countries; therefore, each site has a unique set of historical limits. For all three sites, the company does the following: 1.
Creates a baseline control chart based on the existing process data.
2.
Changes the process, based on a designed experiment.
3.
Gathers data from the new process.
4.
Creates a new control chart based on the new process data.
The goal is to plot the new data on a control chart using historical limits from the old process. In this way, the printing company can compare the new process to the old process limits.
Chapter 3 Control Chart Builder 69 Quality and Process Methods Work with Control Limits Create a Control Chart Based on Existing Process 1.
Select Help > Sample Data Library and open Quality Control/Phase Historical Data.jmp.
This table contains the existing process data for all three sites.
2.
Select Analyze > Quality and Process > Control Chart Builder.
3.
Drag Force to the Y zone.
4.
Drag Run to the Subgroup zone (at bottom).
5.
Drag Site to the Phase zone.
Figure 3.13 Baseline Control Chart for Existing Data Create a Control Chart Based on Updated Process 1.
From the report in Figure 3.13, click the Control Chart Builder red triangle and select Save Limits > in New Table.
This creates a limits table.
2.
Save this new limits table to any location, so you can access it later.
3.
Select Help > Sample Data Library and open Quality Control/Phase New Data.jmp.
This data was collected from all three sites after the change was made to the process.
4.
Select Analyze > Quality and Process > Control Chart Builder.
5.
Drag Force to the Y zone.
6.
Drag Run to the Subgroup zone (at bottom).
7.
Drag Site to the Phase zone.
70 Control Chart Builder Chapter 3 Excluded and Hidden Samples in Control Chart Builder Quality and Process Methods 8.
Click the Control Chart Builder red triangle and select Get Limits. Open the limits table that you saved in step 2.
This applies the historical limits to the new data in the Control Chart Builder report.
Figure 3.14 Control Chart for New Data Based on Historical Limits Now you can see how the new data (after the process change) compare with the historical process limits (before the process change). None of the points fall outside of the control limits for either the location or dispersion chart. The goal was to reduce variability. Looking at the moving range chart, you can see that most points fall below the average line. For sites 1 and 2, it is clear that the variability of force needed to break the bond between pages and the book spine has been decreased. The decrease at Site 3 is not as strong as at sites 1 and 2. The improvements to the printing process appear to have succeeded in reducing the variability.
Excluded and Hidden Samples in Control Chart Builder The following bullets summarize the use of excluded and hidden samples in control chart builder: • Excluded subgroups are not used in the calculations of control limits, and appear on the chart as dimmed points by default. If the Show Excluded Region option is not selected, the points for the excluded subgroups do not appear in the chart, are treated as missing in Tests for Special Causes, and are not included in the count of points for Tests for Special Causes.
Chapter 3 Control Chart Builder 71 Quality and Process Methods Additional Examples of Control Chart Builder • Hidden observations are used in the calculations of control limits, but do not appear in the chart.
• Rows that are both hidden and excluded are included in the count of points for Tests for Special Causes when the Test Excluded Subgroups option is selected. An excluded row can be labeled with a special cause flag. A hidden point cannot be labeled. If the flag for a Tests for Special Causes test is on a hidden point, it will not appear in the chart.
• For partially excluded subgroups, if one or more observations within a subgroup is excluded, and at least one observation within the subgroup is included, the excluded observation is not included in the calculations of either the point statistic or the limits.
• Checks for negative and non-integer data happen on the entire data (even excluded values).
• Tests apply to all excluded subgroups only when the Test Excluded Subgroups option is selected.
Additional Examples of Control Chart Builder Note: In this section, some examples show the Control Panel while others do not. To show or hide the Control Panel, select Show Control Panel from the Control Chart Builder red triangle menu.
• “Individual Measurement and Moving Range Charts Example” • “XBar and R Chart Phase Example” • “XBar and S Charts with Varying Subgroup Sizes Example” • “Run Chart Example” • “P chart Example” • “NP chart Example” • “C chart Example” • “U chart Example” • “G chart Example” • “T chart Example” • “Three Way Control Chart Example” • “Example of Multiple Control Charts” 72 Control Chart Builder Chapter 3 Additional Examples of Control Chart Builder Quality and Process Methods Individual Measurement and Moving Range Charts Example The Pickles.jmp data in the Quality Control sample data folder contains the acid content for vats of pickles. Because the pickles are sensitive to acidity and produced in large vats, high acidity ruins an entire pickle vat. The acidity in four vats is measured each day at 1, 2, and 3 PM. The data table records day, time, and acidity measurements.
1.
Select Help > Sample Data Library and open Quality Control/Pickles.jmp.
2.
Select Analyze > Quality and Process > Control Chart Builder.
3.
Drag Acid to the Y role.
4.
Click the Control Chart Builder red triangle and select Show Limit Labels.
This option labels the control limits and averages in both charts.
Figure 3.15 Individual Measurement and Moving Range Charts for Acid The individual measurement and moving range charts monitor the acidity in each vat produced (subgroup of size 1). Vat 13 has an acidity above the upper control limit of 14.05.
You can also view a Median Moving Range chart. Continue with the following steps to change the charts to use median moving ranges.
5.
In the Limits outline, change the Sigma setting to Median Moving Range.
6.
In the Limits outline, change the Sigma setting to Median Moving Range.
Chapter 3 Control Chart Builder 73 Quality and Process Methods Additional Examples of Control Chart Builder Figure 3.16 Individual Measurement and Median Moving Range Charts for Acid The limits in the individual measurement and median moving range charts use the median moving range as the sigma, rather than the average moving range. This results in slightly narrower control limits for Acid.
XBar and R Chart Phase Example A manufacturer of medical tubing collected tube diameter data for a new prototype. The data was collected over the past 40 days of production. After the first 20 days (phase 1), some adjustments were made to the manufacturing equipment. Analyze the data to determine whether the past 20 days (phase 2) of production are in a state of control.
1.
Select Help > Sample Data Library and open Quality Control/Diameter.jmp.
2.
Select Analyze > Quality and Process > Control Chart Builder.
3.
Drag DIAMETER to the Y role.
4.
Drag DAY to the Subgroup role (at bottom).
74 Control Chart Builder Chapter 3 Additional Examples of Control Chart Builder Quality and Process Methods Figure 3.17 Control Charts for Diameter The first 20 days appear to have high variability, and in the Average chart, there are three observations that are outside of the control limits. An adjustment was made to the manufacturing equipment and new control limits were incorporated.
To compute separate control limits for each phase: 5.
Drag Phase to the Phase role.
6.
In the Average chart, right-click and select Warnings > Test Beyond Limits.
7.
In the Limits outline, select Shade Zones.
Chapter 3 Control Chart Builder 75 Quality and Process Methods Additional Examples of Control Chart Builder Figure 3.18 Control Charts for each Phase Including the Phase variable means that the control limits for phase 2 are based only on the data for phase 2. None of the phase 2 observations are outside the control limits. This is highlighted by including the zone shading on the chart. Therefore, you can conclude that the process is in control after the adjustments were made.
Filter the Control Chart by Another Variable This data table, Diameter.jmp, contains a column for the operator of the machine for each sample. You can use the Local Data Filter with Control Chart Builder to show the data for a subset of operators.
8.
Click the Control Chart Builder red triangle and deselect Show Excluded Region.
Turning off the Show Excluded Region option indicates that the subgroups that are excluded by settings in the Local Data Filter no longer appear on the horizontal axis of the control chart as you make selections in the Local Data Filter. As a result, you see only the portion of the data that are of interest.
9.
Click the Control Chart Builder red triangle and select Local Data Filter.
10. In the Local Data Filter, click on OPERATOR and click the + button.
11. In the Local Data Filter, select the bar labeled RMM.
76 Control Chart Builder Chapter 3 Additional Examples of Control Chart Builder Quality and Process Methods Figure 3.19 XBar and R Chart of Diameter for Operator RMM The XBar and R charts now show only the points for the RMM operator, as denoted by the Where() statement below the charts. The limits for both phases have been adjusted to reflect that the observations for the other three operators have been excluded.
XBar and S Charts with Varying Subgroup Sizes Example This example uses the Coating.jmp data table. This quality characteristic of interest is the Weight 2 column.
1.
Select Help > Sample Data Library and open Quality Control/Coating.jmp.
2.
Select Analyze > Quality and Process > Control Chart Builder.
3.
Drag Weight 2 to the Y role.
4.
Drag Sample to the Subgroup role (at bottom).
Chapter 3 Control Chart Builder 77 Quality and Process Methods Additional Examples of Control Chart Builder Figure 3.20 XBar and S Charts for Varying Subgroup Sizes Weight 2 has several missing values in the data, so the chart has uneven limits. Although each sample has the same number of observations, samples 1, 3, 5, and 7 each have a missing value.
Instead of viewing a line connecting the averages of each sample, you can switch to viewing box plots at each sample.
5.
In the Points outline, deselect the Show Connected Line option.
6.
In the Points outline, select the Box Plots option.
78 Control Chart Builder Chapter 3 Additional Examples of Control Chart Builder Quality and Process Methods Figure 3.21 XBar and S Chart with Box Plots Run Chart Example Run charts display a column of data as a connected series of points. This example is a Run chart for the Close variable from Stock Averages.jmp.
1.
Select Help > Sample Data Library and open Stock Averages.jmp.
2.
Select Analyze > Quality and Process > Control Chart > Run Chart.
3.
Select Close and click Y.
4.
Click OK.
Chapter 3 Control Chart Builder 79 Quality and Process Methods Additional Examples of Control Chart Builder Figure 3.22 Run Chart for Stock Averages Closing Price P chart Example The Washers.jmp sample data contains defect data for two different lot sizes from the ASTM Manual on Presentation of Data and Control Chart Analysis, American Society for Testing and Materials (1976). To view the differences between constant and variable sample sizes, you can compare charts for Lot Size and Lot Size 2.
The Washers.jmp data in the Quality Control sample data folder contains defect counts of 15 lots of 400 galvanized washers. The washers were inspected for finish defects such as rough galvanization and exposed steel. If a washer contained a finish defect, it was deemed nonconforming or defective. Thus, the defect count represents how many washers were defective for each lot of size 400. Using the Washers.jmp data table, specify a sample size variable, which would allow for varying sample sizes. This data table contains all constant sample sizes.
1.
Select Help > Sample Data Library and open Quality Control/Washers.jmp.
2.
Select Analyze > Quality and Process > Control Chart > P Control Chart.
3.
Select # defective and click Y.
4.
Select Lot Size and click n Trials.
5.
Click OK.
6.
In the Limits outline, deselect the Show Lower Limit option.
80 Control Chart Builder Chapter 3 Additional Examples of Control Chart Builder Quality and Process Methods This hides the lower limit, which is not of interest in this situation.
Figure 3.23 P chart of # defective with sample size To view the differences between constant and variable sample sizes, you can compare charts for Lot Size and Lot Size 2 by simply dragging the variables to the nTrials zone.
NP chart Example The Bottle Tops.jmp sample data contains simulated data from a bottle top manufacturing process. Sample is the sample ID number for each bottle. Status indicates whether the bottle top conformed to the design standards. In the Phase column, the first phase represents the time before the process adjustment. The second phase represents the time after the process adjustment. Notes on changes in the process are also included.
1.
Select Help > Sample Data Library and open Quality Control/Bottle Tops.jmp.
2.
Select Analyze > Quality and Process > Control Chart Builder.
3.
Drag Sample to the Subgroup role.
4.
Drag Status to the Y role.
Chapter 3 Control Chart Builder 81 Quality and Process Methods Additional Examples of Control Chart Builder Figure 3.24 NP chart of Status (Nonconforming) The original observations appear to have high variability and there are five observations (Samples 13, 15, 21, 22 and 23) that are outside of the upper control limit. Samples 15 and 23 note that new material and a new operator were introduced into the process, respectively. At the end of the phase, an adjustment was made to the manufacturing equipment. Therefore, the control limits for the entire series should not be used to assess the control during phase 2.
To compute separate control limits for each phase: 5.
Drag Phase to the Phase zone.
82 Control Chart Builder Chapter 3 Additional Examples of Control Chart Builder Quality and Process Methods Figure 3.25 NP chart by Phase Including the Phase variable means that the control limits for phase 2 are based only on the data for phase 2. None of the phase 2 observations are outside the control limits. Therefore, you can conclude that the process is in control after the adjustment.
C chart Example The Cabinet Defects.jmp sample data table contains data concerning the various defects discovered while manufacturing cabinets over two time periods.
1.
Select Help > Sample Data Library and open Quality Control/Cabinet Defects.jmp.
2.
Select Analyze > Quality and Process > Control Chart > C Control Chart.
3.
Select Type of Defect and click Y.
4.
Select Lot Number and click Subgroup.
5.
Select Date and click Phase.
6.
Click OK.
Chapter 3 Control Chart Builder 83 Quality and Process Methods Additional Examples of Control Chart Builder Figure 3.26 C chart of Type of Defect with Phases You can now view the results on the two different days. Both appear to be within limits. To examine other defect type behavior, select another defect type under the Event Chooser > Type of Defect and view the results as the limits are updated.
U chart Example The Braces.jmp data in the Quality Control sample data folder records the defect count in boxes of automobile support braces. A box of braces is one inspection unit. The number of boxes inspected (per day) is the subgroup sample size, which can vary. The U chart in Figure 3.27 is monitoring the number of brace defects per subgroup sample size. The upper and lower bounds vary according to the number of units inspected.
1.
Select Help > Sample Data Library and open Quality Control/Braces.jmp.
2.
Select Analyze > Quality and Process > Control Chart > U Control Chart.
3.
Select # defects and click Y.
4.
Select Unit size and click n Trials.
5.
Click OK.
84 Control Chart Builder Chapter 3 Additional Examples of Control Chart Builder Quality and Process Methods Figure 3.27 U chart of # Defects Because the sample sizes are not equal across subgroups, the limits are uneven. Two of the last five samples are not within the control limits.
G chart Example A G chart is an effective way to understand whether rare events are occurring more frequently than expected and warrant an intervention. See “Rare Event Control Charts”.
The Adverse Reactions.jmp sample data table contains simulated data about adverse drug events (ADEs) reported by a group of hospital patients. An ADE is any type of injury or reaction the patient suffered after taking the drug. The date of the reaction and the number of days since the last reaction were recorded.
1.
Select Help > Sample Data Library and open Quality Control/Adverse Reactions.jmp.
2.
Select Analyze > Quality and Process > Control Chart Builder.
3.
Drag Doses since Last ADE to the Y role.
4.
Drag Date of ADE to the Subgroup role.
An Individual & Moving Range chart of Doses since Last ADE appears.
5.
In the drop-down list, select Rare Event instead of Shewhart Variables.
Chapter 3 Control Chart Builder 85 Quality and Process Methods Additional Examples of Control Chart Builder A G chart of Doses since Last ADE appears, showing that the number of doses given since the last event.
Figure 3.28 G chart of Doses since Last ADE T chart Example T charts are used to measure the time that has elapsed since the last event. See “Rare Event Control Charts”.
The Fan Burnout.jmp sample data table contains simulated data for a fan manufacturing process. The first column identifies each fan that burned out. The second column identifies the number of hours between each burnout.
1.
Select Help > Sample Data Library and open Quality Control/Fan Burnout.jmp.
2.
Select Analyze > Quality and Process > Control Chart Builder.
3.
Drag Hours between Burnouts to the Y role.
4.
Drag Burnout to the Subgroup role.
86 Control Chart Builder Chapter 3 Additional Examples of Control Chart Builder Quality and Process Methods Figure 3.29 Individual and Moving Range Chart of Hours Between Burnouts 5.
In the drop-down list, select Rare Event instead of Shewhart Variables.
6.
Under Limits, change the Sigma from Negative Binomial to Weibull.
Chapter 3 Control Chart Builder 87 Quality and Process Methods Additional Examples of Control Chart Builder Figure 3.30 T chart of Hours Between Burnouts In the T chart, all points appear to be within the control limits. It is clear that the Individual & Moving Range chart was inappropriate for the analysis, as the limits were too narrow.
Three Way Control Chart Example Three way control charts are useful when there is variation between batches and variation within batches.
1.
Select Help > Sample Data Library and open Quality Control/Vial Fill Weights.jmp.
2.
Select Analyze > Quality and Process > Control Chart > Three Way Control Chart.
3.
Select Fill Weight and click Y.
4.
Select Sample and click Subgroup.
5.
Click OK.
88 Control Chart Builder Chapter 3 Additional Examples of Control Chart Builder Quality and Process Methods Figure 3.31 Three Way Control Chart for Fill Weight A Moving Range chart appears between the Range and Average charts. The limits on the Average (XBar) chart are now calculated using the moving range between each sample.
Chapter 3 Control Chart Builder 89 Quality and Process Methods Statistical Details for Control Chart Builder Example of Multiple Control Charts You can create multiple control charts in the same window using the New Y Chart option in the interactive workspace or by specifying multiple processes in a specific control chart launch window. In this example, use the I/MR control chart launch window and then save the control limits to a new data table.
1.
Select Help > Sample Data Library and open Semiconductor Capability.jmp.
2.
Select Analyze > Quality and Process > Control Chart > IMR Control Chart.
3.
Click the triangle next to Processes to show all of the processes.
4.
Select the first five processes and click Y.
5.
Click OK.
If you scroll through the report window, there are Individual and Moving Range control charts for all five selected processes. Since the processes have a Spec Limits column property, there are also Process Capability Analysis reports for each process.
6.
Click the Control Chart Builder red triangle and select Save Limits > in New Tall Table.
Figure 3.32 New Tall Table The control limits are saved to a new data table, with a row for each Y variable and a column for each statistic.
Statistical Details for Control Chart Builder Notes: • Control limits are calculated using a sigma multiplier, K, which is set to 3 by default. To change the default value of K, use the KSigma preference in File > Preferences > Platforms > Control Chart Builder.
• For sample sizes up to n = 50, JMP uses control chart constants d2(n) and d3(n) that are defined in Table 2 of Harter (1960). For samples with a sample size greater than 50, JMP 90 Control Chart Builder Chapter 3 Statistical Details for Control Chart Builder Quality and Process Methods uses the control chart constant values for sample size 50 in both the sigma and control limit calculations.
• “Control Limits for XBar and R Charts” • “Control Limits for XBar and S Charts” • “Control Limits for Individual Measurement and Moving Range Charts” • “Control Limits for P and NP Charts” • “Control Limits for U Charts and C Charts” • “Levey-Jennings Charts” • “Control Limits for G Charts” • “Control Limits for T Charts” • “Sigma Calculations for Three Way Control Charts” Control Limits for XBar and R Charts JMP generates control limits for XBar and R charts using the following formulas: LCL for XBar chart = UCL for XBar chart = LCL for R chart = UCL for R chart = Center line for R chart: By default, the center line for the ith subgroup indicates an estimate of the expected value of Ri. This value is computed as , where is an estimate of .
The standard deviation for XBar and R charts is estimated using the following formula: Xw K ˆ ni ---------– Xw K ˆ ni ---------+ max d2 ni ˆ Kd3 ni ˆ – 0 d2 ni ˆ Kd3 ni ˆ + d2 ni ˆ ˆ ˆ R1 d2 n1 ----------------- RN d2 nN ------------------+ + N -------------------------------------------------------= Chapter 3 Control Chart Builder 91 Quality and Process Methods Statistical Details for Control Chart Builder where: = weighted average of subgroup means K = the sigma multiplier and is set to 3 by default = process standard deviation ni = sample size of ith subgroup d2(n) is the expected value of the range of n independent normally distributed variables with unit standard deviation d3(n) is the standard deviation of the range of n independent normally distributed variables with unit standard deviation Ri is the range of ith subgroup N is the number of subgroups for which ni ≥2.
Control Limits for XBar and S Charts JMP generates control limits for XBar and S charts using the following formulas: LCL for XBar chart = UCL for XBar chart = LCL for S chart = UCL for S chart = Center line for S chart: By default, the center line for the ith subgroup indicates an estimate of the expected value of si. This value is computed as , where is an estimate of .
Xw Xw K ˆ ni ---------– Xw K ˆ ni ---------+ max c4 ni ˆ Kc5 ni ˆ – 0 c4 ni ˆ Kc5 ni ˆ + c4 ni ˆ ˆ 92 Control Chart Builder Chapter 3 Statistical Details for Control Chart Builder Quality and Process Methods The estimate for the standard deviation for XBar and S charts is: where: = weighted average of subgroup means K = the sigma multiplier and is set to 3 by default = process standard deviation ni = sample size of ith subgroup c4(n) is the expected value of the standard deviation of n independent normally distributed variables with unit standard deviation c5(n) is the standard error of the standard deviation of n independent normally distributed variables with unit standard deviation N is the number of subgroups for which ni ≥2 si is the sample standard deviation of the ith subgroup Control Limits for Individual Measurement and Moving Range Charts Control limits for Individual Measurement charts are computed as follows: LCL for Individual Measurement Chart = UCL for Individual Measurement Chart = Control limits for Individual Measurement charts with sigma estimated by the median moving range are computed as follows: LCL for Individual Measurement Chart = UCL for Individual Measurement Chart = Control limits for Moving Range charts are computed as follows: LCL for Moving Range Chart = ˆ s1 c4 n1 ---------------- sN c4 nN ------------------+ + N ------------------------------------------------------= Xw X K ˆ – X K ˆ + X K ˆ MMR – X K ˆ MMR + max d2 n ˆ Kd3 n ˆ – 0 Chapter 3 Control Chart Builder 93 Quality and Process Methods Statistical Details for Control Chart Builder UCL for Moving Range Chart = Control limits for Median Moving Range charts are computed as follows: LCLMMR = max(0, MMR - Kd3(n) ) UCLMMR = MMR + Kd3(n) The standard deviation for Individual Measurement and Moving Range charts is estimated as follows: The standard deviation for Individual Measurement and Moving Range charts when using the median is estimated as follows: = MMR/0.954 where: = the mean of the individual measurements K = the sigma multiplier and is set to 3 by default = the mean of the nonmissing moving ranges computed as (MR2+MR3+...+MRN)/(N-1) where MRi = |xi - xi-1|.
MMR = the median of the nonmissing moving ranges = the process standard deviation d2(n) = expected value of the range of n independent normally distributed variables with unit standard deviation.
d3(n) = standard deviation of the range of n independent normally distributed variables with unit standard deviation Note: Moving Range charts in Control Chart Builder use a range span of n = 2.
d2 n ˆ Kd3 n ˆ + ˆ MMR ˆ MMR ˆ MR d2 n --------------= ˆ MMR X MR 94 Control Chart Builder Chapter 3 Statistical Details for Control Chart Builder Quality and Process Methods Control Limits for P and NP Charts The lower and upper control limits, LCL, and UCL, respectively, are computed using the following formulas: P chart LCL = P chart UCL = NP chart LCL = NP chart UCL = where: is the average proportion of nonconforming items taken across subgroups ni is the number of items in the ith subgroup K is the sigma multiplier and is set to 3 by default Control Limits for U Charts and C Charts The lower and upper control limits, LCL, and UCL, are computed using the following formulas: U chart LCL = U chart UCL = C chart LCL = C chart UCL = max p K p 1 p – ni – 0 min p K p 1 p – ni + 1 max nip K nip 1 p – – 0 min nip K nip 1 p – + ni p p n1p1 nNpN + + n1 nn + + ----------------------------------------------X1 XN + + n1 nN + + ----------------------------------= = max u K u ni – 0 u K u ni + max niu K niu – 0 niu K niu + Chapter 3 Control Chart Builder 95 Quality and Process Methods Statistical Details for Control Chart Builder The limits vary with ni.
ui is the number of nonconformities per unit in the ith subgroup. In general, ui = ci/ni.
K is the sigma multiplier and is set to 3 by default ci is the total number of nonconformities in the ith subgroup ni is the number of inspection units in the ith subgroup is the average number of nonconformities per unit taken across subgroups. The quantity is computed as a weighted average N is the number of subgroups Levey-Jennings Charts Levey-Jennings charts show a process mean with control limits based on a long-term sigma. The control limits are placed at Ks distance from the center line, where K = 3 by default.
The standard deviation, s, for the Levey-Jennings chart is calculated the same way standard deviation is in the Distribution platform.
See Levey and Jennings (1950); Westgard (2002).
Control Limits for G Charts The negative binomial distribution is an extension of the geometric (Poisson) distribution and allows for over-dispersion relative to the Poisson. The negative binomial distribution can be used to construct both exact and approximate control limits for count data. Approximate control limits can be obtained based on a chi-square approximation to the negative binomial. All data is used as individual observations regardless of subgroup size.
Let X have a negative binomial distribution with parameters (, k). Then: u u u n1u1 nNuN + + n1 nN + + -----------------------------------------------c1 cN + + n1 nN + + --------------------------------= = s yi y – 2 N 1 – ---------------------i 1 = N = P X r P v 2 2r 1 + 1 k + --------------- 96 Control Chart Builder Chapter 3 Statistical Details for Control Chart Builder Quality and Process Methods where: is a chi-square variate with v = 2/(1+k) degrees of freedom.
Based on this approximation, approximate upper and lower control limits can be determined. For a nominal level Type 1 error probability in one direction, an approximate upper control limit is a limit UCL such that the following equation is true: Likewise, an approximate lower control limit, LCL, is a limit such that the following equation is true: Thus, an approximate level lower and upper control limits, LCL and UCL, respectively, are computed using the following formulas: where: is the upper (lower) percentile of the chi-square distribution with v = 2/(1+k) degrees of freedom. Negative lower control limits can be set to zero.
For more information about the negative binomial control limits, see Hoffman (2003).
Control Limits for T Charts The estimates of the shape and scale parameters are calculated from the data and used to obtain the percentiles of the Weibull distribution.
Note: Subgroups with a response value of zero are given a weight of zero when estimating the Weibull distribution parameters.
Define the following quantities: p1 = normalDist(-K) for Normal (0,1) 2 v P X UCL 1 P v 2 2UCL 1 + 1 k + ------------------------- – = = P X LCL 1 P v 2 2LCL 1 + 1 k + ------------------------ – = = UCL 2 v 1 – 1 k + 1 – 2 -----------------------------------------------------= LCL max 0 2 v 1 k + 1 – 2 --------------------------------------------- = 2 v 1 – 2 v Chapter 3 Control Chart Builder 97 Quality and Process Methods Statistical Details for Control Chart Builder p2 = normalDist(0) for Normal (0,1) p3 = normalDist(K) for Normal (0,1) Then the limits are calculated using the following formulas: LCL = Weibull Quantile (p1, , ) CL = Weibull Quantile (p2, , ) UCL = Weibull Quantile (p3, , ) where: is the shape parameter and is the scale parameter for the Weibull Quantile function K is the sigma multiplier and is set to 3 by default For more information about the Weibull Quantile function, see Help > Scripting Index.
Sigma Calculations for Three Way Control Charts Within Sigma Based on Average of Ranges The within sigma estimate for three way control charts that is estimated using the average of ranges can be used for the Individual on Means, Moving Range on Means and R chart.
The formula uses the following notation: Ri = range of ith subgroup ni = sample size of ith subgroup d2(ni) = expected value of the range of ni independent normally distributed variables with unit standard deviation N = number of subgroups for which ni ≥2 Within Sigma Based on Average of Unbiased Standard Deviations The within sigma estimate for three way control charts that is estimated using the average of unbiased standard deviations can be used for the Individual on Means, Moving Range on Means, and S chart.
ˆ within R1 d2 n1 ----------------- RN d2 nN ------------------+ + N -------------------------------------------------------= ˆ within s1 c4 n1 ---------------- sN c4 nN ------------------+ + N ------------------------------------------------------= 98 Control Chart Builder Chapter 3 Statistical Details for Control Chart Builder Quality and Process Methods The formula uses the following notation: si = sample standard deviation of the ith subgroup ni = sample size of ith subgroup c4(ni) = expected value of the standard deviation of ni independent normally distributed variables with unit standard deviation N = number of subgroups for which ni ≥2 Between Sigma The between sigma estimate for three way control charts is estimated using the moving range of subgroup means.
The formula uses the following notation: = the mean of the nonmissing moving ranges computed as (MR2+MR3+...+MRN)/(N-1) where MRi = |yi - yi-1|.
d2(2) = expected value of the range of two independent normally distributed variables with unit standard deviation.
, the harmonic mean of subgroup sample sizes.
Note: If between Sigma is estimated as a negative value, it is set to 0.
Between-and-Within Sigma The between-and-within sigma estimate for three way control charts is estimated using a combination of the within sigma and between sigma estimates.
ˆ between MR d2 2 -------------- 2 ˆ 2 within H ----------------------– = MR H N 1 n1 ------1 n2 ------ 1 nN -------+ + + ----------------------------------------------= ˆ between-and-within ˆ within 2 ˆ between 2 + = Chapter 4 Measurement Systems Analysis Evaluate a Continuous Measurement Process Using the EMP Method The Measurement Systems Analysis (MSA) platform assesses the precision, consistency, and bias of a measurement system. Before you can study the process itself, you need to make sure that you can accurately and precisely measure the process. If most of the variation that you see comes from the measuring process itself, then you are not reliably learning about the process. Use MSA to find out how your measurement system is performing.
This chapter covers the EMP method. The Gauge R&R method is described in “Variability Gauge Charts”.
Figure 4.1 Example of a Measurement System Analysis 100 Measurement Systems Analysis Chapter 4 Quality and Process Methods Contents Overview of Measurement Systems Analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 Example of Measurement Systems Analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 Launch the Measurement Systems Analysis Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Data Format . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 Measurement Systems Analysis Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 Average Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 Range Chart or Standard Deviation Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 EMP Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 Effective Resolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 Shift Detection Profiler . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 Bias Comparison . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 Test-Retest Error Comparison . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 Additional Example of Measurement Systems Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 Statistical Details for Measurement Systems Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 Computation of Intraclass Correlation and Probable Error . . . . . . . . . . . . . . . . . . . . . . . . . 124 Chapter 4 Measurement Systems Analysis 101 Quality and Process Methods Overview of Measurement Systems Analysis Overview of Measurement Systems Analysis The EMP (Evaluating the Measurement Process) method in the Measurement Systems Analysis platform is largely based on the methods presented in Donald J. Wheeler’s book EMP III Using Imperfect Data (2006). The EMP method provides visual information and results that are easy to interpret and helps you improve your measurement system to its full potential.
The Gauge R&R method analyzes how much of the variability is due to operator variation (reproducibility) and measurement variation (repeatability). Gauge R&R is available for many combinations of crossed and nested models, regardless of whether the model is balanced. See “Variability Gauge Charts”.
Within the Six Sigma DMAIC methodology, MSA (Measurement System Analysis) addresses the Measure phase and process behavior charts (or control charts) address the Control phase. MSA helps you predict and characterize future outcomes. You can use the information gleaned from MSA to help you interpret and configure your process behavior charts.
For more information about Control Charts, see “Control Chart Builder”.
Example of Measurement Systems Analysis In this example, three operators measured the same five parts. See how the measurement system is performing, based on how much variation is found in the measurements.
1.
Select Help > Sample Data Library and open Variability Data/Gasket.jmp.
2.
Select Analyze > Quality and Process > Measurement Systems Analysis.
3.
Assign Y to the Y, Response role.
4.
Assign Part to the Part, Sample ID role.
5.
Assign Operator to the X, Grouping role.
Notice that the MSA Method is set to EMP, the Chart Dispersion Type is set to Range, and the Model Type is set to Crossed.
6.
Click OK.
102 Measurement Systems Analysis Chapter 4 Example of Measurement Systems Analysis Quality and Process Methods Figure 4.2 MSA Initial Report The Average Chart shows the average measurements for each operator and part combination. In this example, the means of the part measurements are generally beyond the control limits. This is a desirable outcome, because it indicates that you can detect part-to-part variation.
The Range Chart shows the variability for each operator and part combination. In this example, the ranges are within the control limits. This is a desirable outcome, because it indicates that the operators are measuring parts in the same way and with similar variation.
The color coding for each part is shown in the legend below the charts.
7.
Click the red triangle next to Measurement Systems Analysis for Y and select Parallelism Plots.
Chapter 4 Measurement Systems Analysis 103 Quality and Process Methods Example of Measurement Systems Analysis Figure 4.3 Parallelism Plot for Operator and Part The Parallelism Plots chart shows the average measurements for each part by operator. Because the lines are generally parallel and there is no major crossing, you conclude that there is no interaction between operators and parts.
Tip: Interactions indicate a serious issue that requires further investigation.
8.
Click the red triangle next to Measurement Systems Analysis for Y and select EMP Results.
Figure 4.4 EMP Results Report 104 Measurement Systems Analysis Chapter 4 Example of Measurement Systems Analysis Quality and Process Methods The EMP Results report computes several statistics to help you assess and classify your measurement system. The Intraclass Correlation indicates the proportion of the total variation that you can attribute to the part.
From the EMP Results report, you can conclude the following: • The Intraclass Correlation values are close to 1, indicating that most of the variation is coming from the part instead of the measurement system.
• The classification is First Class, meaning that the strength of the process signal is weakened by less than 11%.
• There is at least a 99% chance of detecting a warning using Test 1 only.
• There is 100% chance of detecting a warning using Tests 1-4.
Note: For more information about tests and detecting process shifts, see “Shift Detection Profiler”.
There is no interaction between operators and parts, and there is very little variation in your measurements (the classification is First Class). Therefore, you conclude that the measurement system is performing quite well.
Chapter 4 Measurement Systems Analysis 105 Quality and Process Methods Launch the Measurement Systems Analysis Platform Launch the Measurement Systems Analysis Platform Launch the Measurement Systems Analysis platform by selecting Analyze > Quality and Process > Measurement Systems Analysis.
Figure 4.5 The Measurement Systems Analysis Window For more information about the options in the Select Columns red triangle menu, see Using JMP.
The Measurement Systems Analysis window contains the following features: Select Columns Lists all of the variables in your current data table. Move a selected column into a role.
MSA Method Select the method to use: EMP (Evaluating the Measurement Process) or Gauge R&R. This chapter covers the EMP method. For more information about the Gauge R&R method, see “Variability Gauge Charts”.
Chart Dispersion Type Designates the type of chart for showing variation. Select the Range option or the Standard Deviation option.
106 Measurement Systems Analysis Chapter 4 Launch the Measurement Systems Analysis Platform Quality and Process Methods Note: For the EMP method, the chart dispersion type determines how the statistics in the EMP Results report are calculated. If the Range option is selected, and you have a one factor or a two factor, balanced, crossed model, the statistics in this report are based on ranges. Otherwise, the statistics in this report are based on standard deviations.
Model Type Designates the model type: Main Variables with nominal or ordinal modeling types are treated as main effects.
Crossed The model is crossed when every level of every factor occurs with every level of every other factor.
Crossed with Two Factor Interactions The model is crossed when each level of two factors occurs with every level of the other factor.
Nested The model is nested when all levels of a factor appear within only a single level of any other factor.
Cross then Nested (3 Factors Only) The factors are crossed and then nested for 3 factors.
Nested then Crossed (3 Factors Only) The factors are nested and then crossed for 3 factors.
Options Contains the following options: Analysis Settings Sets the REML maximum iterations and convergence.
Specify Alpha Specifies the 1-alpha confidence level.
Y, Response The column of measurements.
Part, Sample, ID The column designating the part or unit.
X, Grouping The column(s) representing grouping variables.
By Identifies a column that creates a report consisting of separate analyses for each level of the variable.
Data Format To use the Measurement Systems Analysis platform, all response measurements must be in a single response column. Sometimes, responses are recorded in multiple columns, where each row is a level of a design factor and each column is a level of a different design factor. Data that are in this format must be stacked before running the Measurement Systems Analysis platform. See Using JMP.
Chapter 4 Measurement Systems Analysis 107 Quality and Process Methods Measurement Systems Analysis Platform Options Measurement Systems Analysis Platform Options Platform options appear within the red triangle menu next to Measurement Systems Analysis. Selecting an option creates the respective graph or report in the MSA report window. Deselecting an option removes the graph or report. Choose from the following options: Average Chart A plot of the average measurement values for each combination of the part and X variables. The Average Chart helps you detect product variation despite measurement variation. In an Average Chart, out of control data is desirable because it detects part-to-part variation. See “Average Chart”.
Range Chart A plot of the variability statistic for each combination of the part and X variables. Appears only if you selected Range as the Chart Dispersion Type in the launch window. The Range Chart helps you check for consistency within subgroups. In a Range Chart, data within limits is desirable, indicating homogeneity in your error. See “Range Chart or Standard Deviation Chart”.
Std Dev Chart A plot of the standard deviation statistic for each combination of the part and X variables. Appears only if you selected Standard Deviation as the Chart Dispersion Type in the launch window. The Standard Deviation Chart helps you check for consistency within subgroups. In a Standard Deviation Chart, data within limits is desirable, indicating homogeneity in your error. See “Range Chart or Standard Deviation Chart”.
Parallelism Plots An overlay plot that reflects the average measurement values for each part. If the lines are relatively not parallel or crossing, there might be an interaction between the part and X variables.
Tip: Interactions indicate a serious issue that requires further investigation. For example, interactions between parts and operators mean that operators are measuring different parts differently, on average. Therefore, measurement variability is not predictable. This issue requires further investigation to find out why the operators do not have the same pattern or profile over the parts.
EMP Results A report that computes several statistics to help you assess and classify your measurement system. See “EMP Results”.
Effective Resolution A report containing results for the resolution of a measurement system. See “Effective Resolution”.
Bias Comparison An Analysis of Means chart for testing if the X variables have different averages. See “Bias Comparison”.
Test-Retest Error Comparison An Analysis of Means for Variances or Analysis of Means Ranges chart for testing if any of the groups have different test-retest error levels. See “Test-Retest Error Comparison”.
108 Measurement Systems Analysis Chapter 4 Measurement Systems Analysis Platform Options Quality and Process Methods Shift Detection Profiler An interactive set of charts that you can adjust to see the probabilities of getting warnings on your process behavior chart. See “Shift Detection Profiler”.
Variance Components A report containing the estimates of the variance components for the given model. The calculations in this report are based on variances, not ranges. Balanced data uses the EMS method. Unbalanced data uses the REML method.
Note: This report is similar to the Variance Components report in the Variability Chart platform, except that it does not compute Bayesian variance component estimates. See “Variance Components”.
EMP Gauge R&R Results A report that partitions the variability in the measurements into part variation and measurement system variation. The calculations in this report are based on variances, not ranges. Because negative variance components are set to zero, values of zero could indicate outliers in your results.
Note: This report is similar to the Gauge R&R report in the Variability Chart platform. However, by default, the calculation for Reproducibility does not include interactions. To specify that interactions be included in the Reproducibility calculation, select the Include Interactions in Reproducibility platform preference. This preference is located in File > Preferences > Platforms > EMP Measurement Systems Analysis. For more information about Gauge R&R studies, see “About the Gauge R&R Method”.
See Using JMP for more information about the following options: Local Data Filter Shows or hides the local data filter that enables you to filter the data used in a specific report.
Redo Contains options that enable you to repeat or relaunch the analysis. In platforms that support the feature, the Automatic Recalc option immediately reflects the changes that you make to the data table in the corresponding report window.
Save Script Contains options that enable you to save a script that reproduces the report to several destinations.
Save By-Group Script Contains options that enable you to save a script that reproduces the platform report for all levels of a By variable to several destinations. Available only when a By variable is specified in the launch window.
Average Chart The red triangle menu next to Average Chart contains the following options: Show Grand Mean Draws the overall mean of the Y variable on the chart.
Chapter 4 Measurement Systems Analysis 109 Quality and Process Methods Measurement Systems Analysis Platform Options Show Connected Means Draws lines connecting all of the average measurement values.
Show Control Limits Draws lines representing the Upper Control Limit (UCL) and the Lower Control Limit (LCL) and labels those values. The control limits for the Average Chart use the same calculations as an XBar control chart. See “Control Limits for XBar and R Charts”.
Show Control Limits Shading Adds shading between the UCL and LCL.
Show Separators Draws vertical lines to delineate between the X variables.
Show Data Adds the data points to the chart.
Note: You can replace variables in the Average Chart in one of two ways: swap existing variables by dragging and dropping a variable from one axis to the other axis; or, click a variable in the Columns panel of the associated data table and drag it onto an axis.
Range Chart or Standard Deviation Chart The red triangle menu next to Range Chart or Standard Deviation Chart contains the following options: Show Average Dispersion Draws the average range or standard deviation on the chart.
Show Connected Points Draws lines connecting all of the ranges or standard deviations.
Show Control Limits Draws lines representing the Upper Control Limit (UCL) and the Lower Control Limit (LCL) and labels those values. For more information about the calculations of the limits used in the Range Chart, see “Control Limits for XBar and R Charts”. For more information about the calculations of the limits used in the Standard Deviation Chart, see “Control Limits for XBar and S Charts”.
Show Control Limits Shading Adds shading between the UCL and LCL.
Show Separators Draws vertical lines to delineate between the X variables.
Note: You can replace variables in the Range or Standard Deviation Charts in one of two ways: swap existing variables by dragging and dropping a variable from one axis to the other axis; or, click a variable in the Columns panel of the associated data table and drag it onto an axis.
110 Measurement Systems Analysis Chapter 4 Measurement Systems Analysis Platform Options Quality and Process Methods EMP Results Note: The statistics in this report are based on ranges in the following instances: if you selected EMP as the MSA Method and Range as the Chart Dispersion Type, and you have a one factor or a two factor, balanced, crossed model. Otherwise, the statistics in this report are based on variances.
The EMP Results report computes several statistics to help you assess and classify your measurement system. Using this report, you can determine the following: • How your process chart is affected.
• Which tests to set.
• How much the process signal is attenuated.
• How much the bias factors are affecting your system and reducing your potential intraclass correlation coefficient.
The EMP Results report contains the following calculations: Test-Retest Error Indicates measurement variation or repeatability (also known as within error or pure error).
Degrees of Freedom Indicates the amount of information used to estimate the within error.
Probable Error The median error for a single measurement. Indicates the resolution quality of your measurement and helps you decide how many digits to use when recording measurements. See “Effective Resolution”.
Intraclass Correlation Indicates the proportion of the total variation that you can attribute to the part. If you have very little measurement variation, this number is closer to 1.
Intraclass Correlation (no bias) Does not take bias or interaction factors into account when calculating the results.
Intraclass Correlation (with bias) Takes the bias factors (such as operator, instrument, and so on) into account when calculating the results.
Intraclass Correlation (with bias and interaction) Takes the bias and interaction factors into account when calculating the results. This calculation appears only if the model is crossed and uses standard deviation instead of range.
Bias Impact The amount by which the bias factors reduce the Intraclass Correlation.
Bias and Interaction Impact The amount by which the bias and interaction factors reduce the Intraclass Correlation. This calculation appears only if the model is crossed and uses standard deviation instead of range.
Chapter 4 Measurement Systems Analysis 111 Quality and Process Methods Measurement Systems Analysis Platform Options Classes of Process Monitors In order to understand the System and Classification parameters, you must first understand the Monitor Classification Legend.
Figure 4.6 Monitor Classification Legend This legend describes the following classifications: First, Second, Third, and Fourth Class. Each classification indicates the following: • the corresponding Intraclass Correlation values • the amount of process signal attenuation (decrease) • the chance of detecting a 3 standard error shift within 10 subgroups, using Wheeler’s test one or all four tests Wheeler (2006) identifies four detection tests known as the Western Electric Zone Tests. Within the Shift Detection Profiler, there are eight tests that you can select from. The tests that correspond to the Wheeler tests are the first, second, fifth, and sixth tests.
Tip: To prevent the legend from appearing, deselect Show Monitor Classification Legend in the EMP Measurement Systems Analysis platform preferences.
Effective Resolution The Effective Resolution report helps you determine how well your measurement increments are working. You might find that you need to add or drop digits when recording your measurements, or your current increments might be effective as is. Note the following: • The Probable Error calculates the median error of a measurement.
• The Current Measurement Increment reflects how many digits you are currently rounding to and is taken from the data as the nearest power of ten. This number is compared to the Smallest Effective Increment, Lower Bound Increment, and Largest Effective Increment. Based on that comparison, a recommendation is made.
112 Measurement Systems Analysis Chapter 4 Measurement Systems Analysis Platform Options Quality and Process Methods • Large measurement increments have less uncertainty in the last digit, but large median errors. Small measurement increments have small median errors, but more uncertainty in the last digit.
Shift Detection Profiler Use the Shift Detection Profiler to assess the sensitivity of the control chart that you use to monitor your process. The Shift Detection Profiler estimates the probability of detecting shifts in the product mean or product standard deviation. The control chart limits include sources of measurement error variation. Based on these limits, the Shift Detection Profiler estimates the Probability of Warning. This is the probability that a control chart monitoring the process mean signals a warning over the next k subgroups.
You can set the subgroup size that you want to use for your control chart. Note the following: • If the Subgroup Size equals one, the control chart is an Individual Measurement chart.
• If the Subgroup Size exceeds one, the control chart is an XBar-chart.
You can explore the effect of Subgroup Size on the control chart’s sensitivity. You can also explore the benefits of reducing bias and test-retest error.
Figure 4.7 shows the Shift Detection Profiler report for the Gasket.jmp sample data table, found in the Variability Data folder.
Chapter 4 Measurement Systems Analysis 113 Quality and Process Methods Measurement Systems Analysis Platform Options Figure 4.7 Shift Detection Profiler for Gasket.jmp Probability of Warning The Probability of Warning is the probability of detecting a change in the process. A change is defined by the Part Mean Shift and the Part Std Dev settings in the Shift Detection Profiler. The probability calculation assumes that the tests selected in the Customize and Select Tests outline are applied to the Number of Subgroups specified in the Profiler.
The control limits for the Individual Measurement chart (Subgroup Size = 1) and the XBar-chart (Subgroup Size > 1) are based on the In-Control Chart Sigma. The In-Control Sigma takes into account the bias factor (reproducibility) variation and the test-retest (repeatability) variation. These are initially set to the values obtained from your MSA study. The In-Control Chart Sigma also incorporates the In-Control Part Std Dev. Both of these values appear beneath the profiler, along with the False Alarm Probability, which is based on the In-Control Chart Sigma.
114 Measurement Systems Analysis Chapter 4 Measurement Systems Analysis Platform Options Quality and Process Methods In-Control Part Std Dev The standard deviation for the true part values, exclusive of measurement errors, for the stable process. The default value for In-Control Part Std Dev is the standard deviation of the part component estimated by the MSA analysis and found in the Variance Components report.
Often, parts for an MSA study are chosen to have specific properties and do not necessarily reflect the part-to-part variation seen in production. For this reason, you can specify the in-control part standard deviation by selecting Change In-Control Part Std Dev from the Shift Detection Profiler red triangle menu.
In-Control Chart Sigma The value of sigma used to compute control limits. This value is computed using the In-Control Part Std Dev, the Bias Factors Std Dev, and Test-Retest Std Dev specified in the Shift Detection Profiler, and the Subgroup Size. The reproducibility factors are assumed to be constant within a subgroup.
For a subgroup of size n, control limits are set at the following values: It follows that the In-Control Chart Sigma is the square root of the sum of the squares of the following terms: – In-Control Part Std Dev – Bias Factors Std Dev, as specified in the Shift Detection Profiler, multiplied by – Test-Retest Std Dev, as specified in the Shift Detection Profiler The Bias Factors Std Dev is multiplied by to account for the assumption that the reproducibility factors are constant within a subgroup.
JMP updates the In-Control Chart Sigma when you change the In-Control Part Std Dev, the Bias Factors Std Dev, the Test-Retest Std Dev, or the Subgroup Size.
False Alarm Probability The probability that the control chart tests signal a warning when no change in the part mean or standard deviation has occurred. JMP updates the False Alarm Probability when you change the Number of Subgroups or the tests in Customize and Select Tests.
For more information about the Variance Components report, see “Variance Components”.
Shift Detection Profiler Settings Number of Subgroups The number of subgroups over which the probability of a warning is computed. If the number of subgroups is set to k, the profiler gives the probability that the control chart signals at least one warning based on these k subgroups. The Number of Subgroups is set to 10 by default. Drag the vertical line in the plot to change the Number of Subgroups.
3 In-Control Chart Sigma n n n Chapter 4 Measurement Systems Analysis 115 Quality and Process Methods Measurement Systems Analysis Platform Options Part Mean Shift The shift in the part mean. By default, the profiler is set to detect a 1 sigma shift. The initial value is the standard deviation of the part component estimated by the MSA analysis and found in the Variance Components report. Drag the vertical line in the plot or click the value beneath the plot to change the Part Mean Shift.
Part Std Dev The standard deviation for the true part values, exclusive of measurement errors. The initial value for Part Std Dev is the standard deviation of the part component estimated by the MSA analysis and is found in the Variance Components report. Drag the vertical line in the plot or click the value beneath the plot to change the Part Std Dev.
Bias Factors Std Dev The standard deviation of factors related to reproducibility. Bias factors include operator and instrument. The bias factor variation does not include part and repeatability (within) variation. The initial value is derived using the reproducibility and interaction variance components estimated by the MSA analysis and is found in the Variance Components report. Drag the vertical line in the plot or click the value beneath the plot to change the Bias Factors Std Dev.
Test-Retest Std Dev The standard deviation of the test-retest, or repeatability, variation in the model. The initial value is the standard deviation of the Within component estimated by the MSA analysis and is found in the Variance Components report. Drag the vertical line in the plot or click the value beneath the plot to change the Test-Retest Std Dev.
Subgroup Size The sample size used for each subgroup. This is set to 1 by default. You can increase the sample size to investigate improvement in control chart performance. Increasing the sample size from 1 demonstrates what happens when you move from an Individual Measurement chart to an XBar-chart. Drag the vertical line in the plot to change the Subgroup Size.
Shift Detection Profiler Options The red triangle menu for the Shift Detection Profiler provides several options. Only one option is described here.
Change In-Control Part Std Dev Specify a value for the part standard deviation for the stable process. The in-control part standard deviation should reflect the variation of the true part values, exclusive of measurement errors. Enter a new value and click OK.
The In-Control Part Std Dev is originally set to the standard deviation of the part component estimated by the MSA analysis, found in the Variance Components report.
This option is useful if the parts chosen for the EMP study were not a random sample from the process.
Reset Factor Grid Displays a window for each factor allowing you to enter a specific value for the factor’s current setting, to lock that setting, and to control aspects of the grid. See Profilers.
116 Measurement Systems Analysis Chapter 4 Measurement Systems Analysis Platform Options Quality and Process Methods Factor Settings Submenu that consists of the following options: Remember Settings Adds an outline node to the report that accumulates the values of the current settings each time the Remember Settings command is invoked. Each remembered setting is preceded by a radio button that is used to reset to those settings. There are options to remove selected settings or all settings in the Remember Settings red triangle menu.
Copy Settings Script Copies the current Profiler’s settings to the clipboard.
Paste Settings Script Pastes the Profiler settings from the clipboard to a Profiler in another report.
Set Script Sets a script that is called each time a factor changes. The set script receives a list of arguments of the form: {factor1 = n1, factor2 = n2, ...} For example, to write this list to the log, first define a function: ProfileCallbackLog = Function({arg},show(arg)); Then enter ProfileCallbackLog in the Set Script dialog.
Similar functions convert the factor values to global values: ProfileCallbackAssign = Function({arg},evalList(arg)); Or access the values one at a time: ProfileCallbackAccess = Function({arg},f1=arg["factor1"];f2=arg["factor2"]); Shift Detection Profiler Legend This panel gives a brief description of four of the Shift Detection Profiler settings. See “Shift Detection Profiler Settings”.
Tip: To prevent the legend from appearing, deselect Show Shift Detection Profiler Legend in the EMP Measurement Systems Analysis platform preferences.
Customize and Select Tests In the Customize and Select Tests panel, select and customize the tests that you want to apply to the k subgroups in your control chart. The eight tests are based on Nelson (1984). For more information about the tests, see “Tests”.
Chapter 4 Measurement Systems Analysis 117 Quality and Process Methods Measurement Systems Analysis Platform Options The Shift Detection Profiler calculations take these tests into account. The Probability of Warning and False Alarm Probability values increase as you add more tests. Because the calculations are based on a quasi-random simulation, there might be a slight delay as the profiler is updated.
The Customize and Select Tests panel has the following options: Restore Default Settings If no settings have been saved to preferences, this option resets the selected tests to the first test only. The values of n are also reset to the values described in “Tests”. If settings have been saved to preferences, this option resets the selected tests and the values of n to those specified in the preferences.
Note: You can access preferences for control chart tests by selecting File > Preferences > Platforms > Control Chart Builder. Custom Tests 1 through 8 correspond to the eight tests shown in Customize and Select Tests.
Save Settings to Preferences Saves the selected tests and the values of n for use in future analyses. These preferences are added to the Control Chart Builder platform preferences.
Bias Comparison The Bias Comparison option creates an Analysis of Means chart. This chart shows the mean values for each level of the grouping variables and compares them with the overall mean. You can use this chart to see whether an operator is measuring parts too high or too low, on average.
The red triangle menu next to Analysis of Means contains the following options: Set Alpha Level Select an option from the most common alpha levels or specify any level using the Other selection. Changing the alpha level modifies the upper and lower decision limits.
Show Summary Report Shows a report containing group means and decision limits, and reports if the group mean is above the upper decision limit or below the lower decision limit.
Display Options Include the following options: Show Decision Limits Draws lines representing the Upper Decision Limit (UDL) and the Lower Decision Limit (LDL) and defines those values.
Show Decision Limit Shading Adds shading between the UDL and the LDL.
Show Center Line Draws the center line statistic that represents the average.
Point Options Changes the chart display to needles, connected points, or points.
118 Measurement Systems Analysis Chapter 4 Additional Example of Measurement Systems Analysis Quality and Process Methods Test-Retest Error Comparison The Test-Retest Error Comparison option creates a type of Analysis of Means for Variances or Analysis of Means Ranges chart. This chart shows if there are differences in the test-retest error between operators. For example, you can use this chart to see whether there is an inconsistency in how each operator is measuring. The Analysis of Mean Ranges chart is displayed when ranges are used for variance components.
• For information about the options in the red triangle menu next to Operator Variance Test, see “Bias Comparison”.
• For more information about Analysis of Means for Variances charts, see “Variance Components”.
Additional Example of Measurement Systems Analysis In this example, three operators have measured a single characteristic twice on each of six wafers. Perform a detailed analysis to find out how well the measurement system is performing.
Perform the Initial Analysis 1.
Select Help > Sample Data Library and open Variability Data/Wafer.jmp.
2.
Select Analyze > Quality and Process > Measurement Systems Analysis.
3.
Assign Y to the Y, Response role.
4.
Assign Wafer to the Part, Sample ID role.
5.
Assign Operator to the X, Grouping role.
Notice that the MSA Method is set to EMP, the Chart Dispersion Type is set to Range, and the Model Type is set to Crossed.
6.
Click OK.
Chapter 4 Measurement Systems Analysis 119 Quality and Process Methods Additional Example of Measurement Systems Analysis Figure 4.8 Average and Range Charts The Average Chart shows that some of the average part measurements fall beyond the control limits. This is desirable, indicating measurable part-to-part variation.
The Range Chart shows no points that fall beyond the control limits. This is desirable, indicating that the operator measurements are consistent within part.
Examine Interactions Take a closer look for interactions between operators and parts. Click the red triangle next to Measurement Systems Analysis for Y and select Parallelism Plots.
120 Measurement Systems Analysis Chapter 4 Additional Example of Measurement Systems Analysis Quality and Process Methods Figure 4.9 Parallelism Plot Looking at the parallelism plot by operator, you can see that the lines are relatively parallel and that there is only some minor crossing.
Examine Operator Consistency Take a closer look at the variance between operators. Click the red triangle next to Measurement Systems Analysis for Y and select Test-Retest Error Comparison.
Figure 4.10 Test-Retest Error Comparison Looking at the Test-Retest Error Comparison, you can see that none of the operators have a test-retest error that is significantly different from the overall test-retest error. The operators appear to be measuring consistently.
Chapter 4 Measurement Systems Analysis 121 Quality and Process Methods Additional Example of Measurement Systems Analysis Just to be sure, you decide to look at the Bias Comparison chart, which indicates whether an operator is measuring parts too high or too low. Click the red triangle next to Measurement Systems Analysis for Y and select Bias Comparison.
Figure 4.11 Bias Comparison Looking at the Bias Comparison chart, you make the following observations: • Operator A and Operator B have detectable measurement bias, as they are significantly different from the overall average.
• Operator A is significantly biased low.
• Operator B is significantly biased high.
• Operator C is not significantly different from the overall average.
Classify Your Measurement System Examine the EMP Results report to classify your measurement system and look for opportunities for improvement. Click the red triangle next to Measurement Systems Analysis for Y and select EMP Results.
122 Measurement Systems Analysis Chapter 4 Additional Example of Measurement Systems Analysis Quality and Process Methods Figure 4.12 EMP Results The classification is Second Class, which means that there is a better than 88% chance of detecting a three standard error shift within ten subgroups, using Test one only. You notice that the bias factors have an 11% impact on the Intraclass Correlation. In other words, if you could eliminate the bias factors, your Intraclass Correlation coefficient would improve by 11%.
Explore the Ability of a Control Chart to Detect Process Changes Use the Shift Detection Profiler to explore the probability that a control chart will be able to detect a change in your process. Click the red triangle next to Measurement Systems Analysis for Y and select Shift Detection Profiler.
Chapter 4 Measurement Systems Analysis 123 Quality and Process Methods Additional Example of Measurement Systems Analysis Figure 4.13 Shift Detection Profiler By default, the only test selected is for a point beyond the 3 sigma limits. Also note that the default Subgroup Size is 1, indicating that you are using an Individual Measurement chart.
Explore your ability to detect a shift in the mean of two part standard deviations in the 10 subgroups following the shift. Click the Part Mean Shift value of 2.1701 and change it to 4.34 (2.17 multiplied by 2). The probability of detecting a shift of twice the part standard deviation is 56.9%.
Next, see how eliminating bias affects your ability to detect the shift of two part standard deviations. Change the Bias Factors Std Dev value from 1.1256 to 0. The probability of detecting the shift increases to 67.8%.
Finally, add more tests to see how your ability to detect the two part standard deviation shift changes. In addition to the first test, select the second, fifth, and sixth tests (Wheeler’s Rules 4, 2, and 3). With these four tests and no bias variation, your probability of detecting the shift is 99.9%.
You can also explore the effect of using a control chart based on larger subgroup sizes. For subgroup sizes of two or more, the control chart is an XBar-chart. Change the Bias Factors Std Dev value back to 1.1256 and deselect all but the first test. Set the Subgroup Size in the profiler to 4. The probability of detecting the two part standard deviation shift is 98.5%.
124 Measurement Systems Analysis Chapter 4 Statistical Details for Measurement Systems Analysis Quality and Process Methods Examine Measurement Increments Finally, see how well your measurement increments are working. Click the red triangle next to Measurement Systems Analysis for Y and select Effective Resolution.
Figure 4.14 Effective Resolution The Current Measurement Increment of 0.01 is below the Lower Bound Increment of 0.09, indicating that you should adjust your future measurements to record one less digit.
Statistical Details for Measurement Systems Analysis For more information about the calculations of the limits used in the Range Chart, see “Control Limits for XBar and R Charts”. For more information about the calculations of the limits used in the Standard Deviation Chart, see “Control Limits for XBar and S Charts”.
Computation of Intraclass Correlation and Probable Error Intraclass Correlation without bias is computed as follows: Intraclass Correlation with bias is computed as follows: rpe ˆ p 2 ˆ p 2 ˆ pe 2 + ----------------------= rb ˆ p 2 ˆ p 2 ˆ b 2 ˆ pe 2 + + -----------------------------------= Chapter 4 Measurement Systems Analysis 125 Quality and Process Methods Statistical Details for Measurement Systems Analysis Intraclass Correlation with bias and interaction factors is computed as follows: Probable Error is computed as follows: Note the following: = variance estimate for pure error = variance estimate for product = variance estimate for bias factors = variance estimate for interaction factors Z0.75 = the 75% quantile of standard normal distribution rint ˆ p 2 ˆ p 2 ˆ b 2 ˆ int 2 ˆ pe 2 + + + -----------------------------------------------------= 0.75 Z ˆ pe ˆ pe 2 ˆ p 2 ˆ b 2 ˆ int 2 126 Measurement Systems Analysis Chapter 4 Statistical Details for Measurement Systems Analysis Quality and Process Methods Chapter 5 Variability Gauge Charts Evaluate a Continuous Measurement Process Using Gauge R&R Variability gauge charts analyze continuous measurements and can reveal how your measurement system is performing. You can also perform a gauge study to see measures of variation in your data.
Tip: This chapter covers only variability charts. For more information about attribute charts, see “Attribute Gauge Charts”.
Figure 5.1 Example of a Variability Chart 128 Variability Gauge Charts Chapter 5 Quality and Process Methods Contents Overview of Variability Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 Example of a Variability Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130 Launch the Variability/Attribute Gauge Chart Platform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131 Data Format . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 The Variability Gauge Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 Variability Gauge Platform Options. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 Heterogeneity of Variance Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136 Variance Components. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 About the Gauge R&R Method. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139 Gauge R&R Option . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 Discrimination Ratio. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 Misclassification Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 Bias Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 Linearity Study . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 Additional Examples of Variability Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 Example of the Heterogeneity of Variance Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 Example of the Bias Report Option . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 Statistical Details for Variability Charts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 Statistical Details for Variance Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 Statistical Details for the Discrimination Ratio. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 Statistical Details for the Misclassification Probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152 Chapter 5 Variability Gauge Charts 129 Quality and Process Methods Overview of Variability Charts Overview of Variability Charts Tip: The traditional name for a variability chart is a multi vari chart, but because that name is not well known, the more generic term variability chart is used instead.
Just as a control chart shows variation across time in a process, a variability chart shows the same type of variation across categories such as parts, operators, repetitions, and instruments. A variability chart plots the data and means for each level of grouping factors, with all plots side by side. Along with the data, you can view the mean, range, and standard deviation of the data in each category, seeing how they change across the categories. The report options are based on the assumption that the primary interest is how the mean and variance change across the categories.
Variability charts are commonly used for measurement systems analysis such as Gauge R&R. This analysis examines how much of the variability is due to operator variation (reproducibility) and measurement variation (repeatability). Gauge R&R is available for many combinations of crossed and nested models, regardless of whether the model is balanced.
130 Variability Gauge Charts Chapter 5 Example of a Variability Chart Quality and Process Methods Example of a Variability Chart Suppose that you have data containing part measurements. Three operators, Cindy, George, and Tom, each took measurements of 10 parts. They measured each part three times, making a total of 90 observations. You want to identify the variation between operators.
1.
Select Help > Sample Data Library and open Variability Data/2 Factors Crossed.jmp.
2.
Select Analyze > Quality and Process > Variability / Attribute Gauge Chart.
3.
For Chart Type, select Variability.
4.
Select Measurement and click Y, Response.
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Select Operator and click X, Grouping.
6.
Select part# and click Part, Sample ID.
7.
Click OK.
8.
Click the Variability Gauge red triangle and select Show Group Means and Connect Cell Means.
Figure 5.2 Example of a Variability Chart Chapter 5 Variability Gauge Charts 131 Quality and Process Methods Launch the Variability/Attribute Gauge Chart Platform Looking at the Std Dev chart, you can see that Cindy and George have more variation in their measurements than Tom, who appears to be measuring parts the most consistently. George seems to have the most variation in his measurements, so he might be measuring parts the most inconsistently.
Launch the Variability/Attribute Gauge Chart Platform Launch the Variability/Attribute Gauge Chart platform by selecting Analyze > Quality and Process > Variability/Attribute Gauge Chart. Set the Chart Type to Variability.
Figure 5.3 The Variability/Attribute Gauge Chart Launch Window For more information about the options in the Select Columns red triangle menu, see Using JMP.
Chart Type Choose between a variability gauge analysis (for a continuous response) or an attribute gauge analysis (for a categorical response, usually “pass” or “fail”).
Note: The content in this chapter covers only the Variability chart type. For more information about the Attribute chart type, see “Attribute Gauge Charts”.
Model Type Choose the model type (Main Effect, Crossed, Nested, and so on). See “Statistical Details for Variance Components”.
Analysis Settings Specify the method for computing variance components. See “Analysis Settings”.
Specify Alpha Specify the alpha level used by the platform.
132 Variability Gauge Charts Chapter 5 The Variability Gauge Chart Quality and Process Methods Y, Response Specify the measurement column. Specifying more than one Y column produces a separate variability chart for each response.
Standard Specify a standard or reference column that contains the “true” or known values for the measured part. Including this column enables the Bias and Linearity Study options. These options perform analysis on the differences between the observed measurement and the reference or standard value. See “Bias Report” and “Linearity Study”.
X, Grouping Specify the classification columns that group the measurements. If the factors form a nested hierarchy, specify the higher terms first. If you are doing a gauge study, specify the operator first and then the part.
Freq Identifies the data table column whose values assign a frequency to each row. Can be useful when you have summarized data.
Part, Sample ID Identifies the part or sample that is being measured.
By Identifies a column that creates a report consisting of separate analyses for each level of the variable.
Data Format To use the Variability Chart platform, all response measurements must be in a single response column. Sometimes, responses are recorded in multiple columns, where each row is a level of a design factor and each column is a level of a different design factor. Data that are in this format must be stacked before running the Variability Chart platform. See Using JMP.
The Variability Gauge Chart The variability chart and the standard deviation chart show patterns of variation. You can use these charts to identify possible groups of variation (within subgroups, between subgroups, over time). If you notice that any of these sources of variation are large, you can then work to reduce the variation for that source.
Follow the instructions in “Example of a Variability Chart” to produce the results shown in Figure 5.4.
Chapter 5 Variability Gauge Charts 133 Quality and Process Methods The Variability Gauge Chart Figure 5.4 Variability Gauge Chart The charts show the response on the y-axis and a multilevel, categorized x-axis.
In Figure 5.4, the Measurement chart shows the range of measurements for each operator by part. Each measurement appears on the chart. Maximum and minimum bars indicate the range of values for each cell, and a cell means bar indicates the median value for each combination of values. The Std Dev chart plots the standard deviation of the measurements taken on each part by operator.
You can add features to the charts, as illustrated in Figure 5.4. See “Variability Gauge Platform Options”.
To replace variables in charts, do one of the following: • Swap existing variables by dragging a variable from one axis label to the other axis label. When you drag a variable over a chart or click an axis label, the axis labels are highlighted. This indicates where to drop the variable.
• Click a variable in the Columns panel of the associated data table and drag it onto an axis label.
In other platforms, rows that are excluded in the associated data table still appear on the charts or plots. However, in variability charts, excluded rows do not appear on the charts.
points with range bars group means lines connecting cell means cell means 134 Variability Gauge Charts Chapter 5 Variability Gauge Platform Options Quality and Process Methods Variability Gauge Platform Options Use the red triangle options to modify the appearance of the chart, perform Gauge R&R analysis, and compute variance components.
Note: Figure 5.4 illustrates some of these options.
Tip: To set the default behavior of these options, select File > Preferences > Platforms > Variability Chart.
Vertical Charts Changes the layout to horizontal or vertical.
Variability Chart Shows or hides the variability chart.
Show Points Shows or hides the points for individual rows.
Show Range Bars Shows or hides the bars indicating the minimum and the maximum value of each cell.
Show Cell Means Shows or hides the mean mark for each cell.
Connect Cell Means Connects or disconnects cell means within a group of cells.
Show Separators Shows or hides the separator lines between levels of the X, Grouping variables.
Show Group Means (Available only if you have two or more X, Grouping variables or one X, Grouping variable and one Part, Sample ID variable.) Shows or hides the mean for groups of cells, represented by a horizontal solid line. A window appears, prompting you to select one of the grouping variables.
Show Grand Mean Shows or hides the overall mean, represented by a gray dotted line across the entire graph.
Show Grand Median Shows or hides the overall median, represented by a blue dotted line across the entire graph.
Show Box Plots Shows or hides box plots.
Mean Diamonds Shows or hides mean diamonds. The confidence intervals use the within-group standard deviation for each cell.
XBar Control Limits Shows or hides lines at the UCL and LCL on the variability chart. For more information about the calculations of these limits, see “Statistical Details for Control Chart Builder”.
Chapter 5 Variability Gauge Charts 135 Quality and Process Methods Variability Gauge Platform Options Points Jittered Adds some random noise to the plotted points so that coincident points do not plot on top of one another.
Show Standard Mean (Available only if you have specified a Standard variable.) Shows or hides the mean of the standard column.
Variability Summary Report Shows or hides a report that gives the mean, standard deviation, coefficient of variation (CV), standard error of the mean, lower and upper confidence intervals, and the minimum, maximum, and number of observations.
Std Dev Chart Shows or hides a separate graph that shows cell standard deviations across category cells.
Mean of Std Dev Shows or hides a line at the mean standard deviation on the Std Dev chart.
S Control Limits Shows or hides lines showing the LCL and UCL in the Std Dev chart. For more information about the calculations of these limits, see “Statistical Details for Control Chart Builder”.
Group Means of Std Dev Shows or hides the mean lines on the Std Dev chart.
Heterogeneity of Variance Tests Performs a test for comparing variances across groups. See “Heterogeneity of Variance Tests”.
Variance Components Estimates the variance components for a specific model. Variance components are computed for these models: main effects, crossed, nested, crossed then nested (three factors only), and nested then crossed (three factors only). See “Variance Components”.
Gauge Studies Contains the following options: Gauge R&R Interprets the first factors as grouping columns and the last factor as Part, and creates a gauge R&R report using the estimated variance components. (Note that there is also a Part field in the launch window). See “Gauge R&R Option”.
Discrimination Ratio Characterizes the relative usefulness of a given measurement for a specific product. It compares the total variance of the measurement with the variance of the measurement error. See “Discrimination Ratio”.
Misclassification Probabilities Shows probabilities for rejecting good parts and accepting bad parts. See “Misclassification Probabilities”.
Bias Report Shows the average difference between the observed values and the standard. A graph of the average biases and a summary table appears. This option is available only when you specify a Standard variable in the launch window. See “Bias Report”.
136 Variability Gauge Charts Chapter 5 Variability Gauge Platform Options Quality and Process Methods Linearity Study Performs a regression using the standard values as the X variable and the bias as the Y variable. This analysis examines the relationship between bias and the size of the part. Ideally, you want the slope to equal 0. A nonzero slope indicates that your gauge performs differently with different sized parts. This option is available only when you specify a Standard variable in the launch window. See “Linearity Study”.
Gauge R&R Plots Shows or hides Mean Plots (the mean response by each main effect in the model) and Std Dev plots. If the model is purely nested, the graphs appear with a nesting structure. If the model is purely crossed, interaction graphs appear. Otherwise, the graphs plot independently at each effect. For the standard deviation plots, the red lines connect for each effect.
AIAG Labels Enables you to specify that quality statistics should be labeled in accordance with the AIAG standard, which is used extensively in automotive analyses.
See Using JMP for more information about the following options: Local Data Filter Shows or hides the local data filter that enables you to filter the data used in a specific report.
Redo Contains options that enable you to repeat or relaunch the analysis. In platforms that support the feature, the Automatic Recalc option immediately reflects the changes that you make to the data table in the corresponding report window.
Save Script Contains options that enable you to save a script that reproduces the report to several destinations.
Save By-Group Script Contains options that enable you to save a script that reproduces the platform report for all levels of a By variable to several destinations. Available only when a By variable is specified in the launch window.
Heterogeneity of Variance Tests Note: See “Example of the Heterogeneity of Variance Test”.
The Heterogeneity of Variance Tests option performs a test for comparing variances across groups. The test is an Analysis of Means for Variances (ANOMV) based method. This method indicates whether any of the group standard deviations are different from the square root of the average group variance.
To be robust against non-normal data, the method uses a permutation simulation to compute decision limits. For more information about this method, see Wludyka and Sa (2004). Because the method uses simulations, the decision limits can be slightly different each time. To obtain the same results each time, press Ctrl+Shift, select the option, and then specify the same random seed.
mean weighted variance Chapter 5 Variability Gauge Charts 137 Quality and Process Methods Variability Gauge Platform Options The red triangle menus for the test reports contain the following options: Set Alpha Level Sets the alpha level for the test.
Show Summary Report Shows or hides a summary report for the test. The values in the report are the same values that are shown in the plot.
Display Options Shows or hides the decision limits, shading, center line, and needles.
Variance Components The Variance Components option models the variation from measurement to measurement. The response is assumed to be a constant mean plus random effects associated with various levels of the classification.
Note: Once you select the Variance Components option, if you did not select the Model Type in the launch window (if you selected Decide Later), you are prompted to select the model type. For more information about model types, see “Launch the Variability/Attribute Gauge Chart Platform”.
Figure 5.5 Example of the Variance Components Report The Analysis of Variance report appears only if the EMS method of variance component estimation is used. This report shows the significance of each effect in the model.
The Variance Components report shows the estimates themselves. See “Statistical Details for Variance Components”.
Analysis Settings From the launch window, click Analysis Settings to choose the method for computing variance components.
138 Variability Gauge Charts Chapter 5 Variability Gauge Platform Options Quality and Process Methods Figure 5.6 Analysis Settings Window Choose best analysis (EMS, REML, or Bayesian) Chooses the best analysis from EMS, REML, or Bayesian, using the following logic: – If the data are balanced, and if no variance components are negative, the EMS (expected mean squares) method is used to estimate the variance components.
– If the data are unbalanced, the REML (restricted maximum likelihood) method is used, unless a variance component is estimated to be negative, then the Bayesian method is used.
– If any variance component is estimated to be negative using the EMS method, the Bayesian method is used.
– If there is confounding in the variance components, then the bounded REML method is used, and any negative variance component estimates are set to zero.
Choose best analysis (EMS or REML) Chooses the best analysis from EMS or REML, using the same logic as the Choose best analysis (EMS, REML, or Bayesian) option. However, this option never uses the Bayesian method, even for negative variance components. The bounded REML method is used and any negative variance component is forced to be 0.
Use REML analysis Uses the bounded REML method, even if the data are balanced. The bounded REML method can handle unbalanced data and forces any negative variance component to be 0.
Use Bayesian analysis Uses the Bayesian method. The Bayesian method can handle unbalanced data and forces all variances components to be positive and nonzero. If there is confounding in the variance components, then the bounded REML method is used, and any negative variance component estimates are set to zero. The method implemented in JMP computes the posterior means using a modified version of Jeffreys’ prior. See Portnoy (1971) and Sahai (1974).
Maximum Iterations (Applicable only for the REML method.) For difficult problems, you might want to increase the number of iterations. Increasing this value means that JMP will try more times to find a solution in the optimization phase.
Convergence Limit (Applicable only for the REML method.) For problems where you want greater precision, you might want to change the convergence limit to be smaller. Chapter 5 Variability Gauge Charts 139 Quality and Process Methods Variability Gauge Platform Options Decreasing this value means that JMP will find the solution to a higher level of accuracy in the optimization phase. However, this can increase the time taken to find a solution. Providing a larger convergence value returns quicker results, but is less precise.
Number of Iteration Abscissas (Applicable only for the Bayesian method.) For greater accuracy, you might want to increase the number of iteration abscissas. However, this can increase the time taken to find a solution. Providing a smaller number returns quicker results, but is less precise.
Maximum Number of Function Evaluations (Applicable only for the Bayesian method.) For greater accuracy, you might want to increase the maximum number of function evaluations. However, this can increase the time taken to find a solution. Providing a smaller number returns quicker results, but is less precise.
About the Gauge R&R Method The Gauge R&R method analyzes how much of the variability in your measurement system is due to operator variation (reproducibility) and measurement variation (repeatability). Gauge R&R studies are available for many combinations of crossed and nested models, regardless of whether the model is balanced.
Tip: Alternatively, you can use the EMP method to assess your measurement system. See “Measurement Systems Analysis”.
Before performing a Gauge R&R study, you collect a random sample of parts over the entire range of part sizes from your process. Select several operators at random to measure each part several times. The variation is then attributed to the following sources: • The process variation, from one part to another. This is the ultimate variation that you want to be studying if your measurements are reliable.
• The variability inherent in making multiple measurements, that is, repeatability. In Table 5.1, this is called the within variation.
• The variability due to having different operators measure parts—that is, reproducibility.
A Gauge R&R analysis then reports the variation in terms of repeatability and reproducibility.
Table 5.1 Definition of Terms and Sums in Gauge R&R Analysis Variances Sums Term and Abbreviation Alternate Term V(Within) Repeatability (EV) Equipment Variation 140 Variability Gauge Charts Chapter 5 Variability Gauge Platform Options Quality and Process Methods A Shewhart control chart can identify processes that are going out of control over time. A variability chart can also help identify operators, instruments, or part sources that are systematically different in mean or variance.
Gauge R&R Option The Gauge R&R option shows measures of variation interpreted for a gauge study of operators and parts.
Once you select the Gauge R&R option, if you have not already selected the model type, you are prompted to do so. Then, modify the Gauge R&R specifications.
Note: The Platform preferences for Variability include the Gauge R&R Specification Dialog option. The preference is selected by default. Deselect the preference to use the spec limits that are defined in the data table.
Enter/Verify Gauge R&R Specifications The Enter/Verify Gauge R&R Specifications window contains these options: Choose tolerance entry method Choose how to enter the tolerance: Select Tolerance Interval to enter the tolerance directly, where tolerance = USL – LSL.
Select LSL and/or USL to enter the specification limits and then have JMP calculate the tolerance.
V(Operator)+V(OperatorPart) Reproducibility (AV) Appraiser Variation V(OperatorPart) Interaction (IV) Interaction Variation V(Within)+V(Operator)+V(Operator Part) Gauge R&R (RR) Measurement Variation V(Part) Part Variation (PV) Part Variation V(Within)+V(Operator)+ V(OperatorPart)+V(Part) Total Variation (TV) Total Variation Table 5.1 Definition of Terms and Sums in Gauge R&R Analysis (Continued) Variances Sums Term and Abbreviation Alternate Term Chapter 5 Variability Gauge Charts 141 Quality and Process Methods Variability Gauge Platform Options K, Sigma Multiplier K is a constant value that you choose to multiply with sigma. For example, you might type 6 so that you are looking at 6sigma or a 6 sigma process.
Tip: Modify the default value of K by selecting File > Preferences > Platforms > Variability Chart.
Tolerance Interval, USL-LSL Enter the tolerance for the process, which is the difference between the upper specification limits and the lower specification limits.
Spec Limits Enter upper and lower specification limits. See Using JMP.
Historical Mean Computes the tolerance range for one-sided specification limits, either USL-Historical Mean or Historical Mean-LSL. If you do not enter a historical mean, the grand mean is used.
Historical Sigma Enter a value that describes the variation (you might have this value from history or past experience).
The Gauge R&R Report Figure 5.7 Example of the Gauge R&R Report Note: To generate the reduced Gauge R&R report, select File > Preferences > Platforms > Variability Chart > Reduced Gauge R&R Report.
In this example, the values in the Variation column are the square roots of sums of variance components scaled by the value of k (6 in this example).
Full Gauge R&R Report Reduced Gauge R&R Report 142 Variability Gauge Charts Chapter 5 Variability Gauge Platform Options Quality and Process Methods Table 5.2 shows guidelines for measurement variation, as suggested by Barrentine (1991).
Note the following: • If you have provided a Tolerance Interval in the Enter/Verify Gauge R&R Specifications window, a % of Tolerance column appears in the Gauge R&R report. This column is computed as 100(Variation/Tolerance). Also, a Precision-to-Tolerance ratio appears at the bottom of the report. This ratio represents the proportion of the tolerance or capability interval that is lost due to gauge variability.
• If you have provided a Historical Sigma in the Enter/Verify Gauge R&R Specifications window, a % Process column appears in the Gauge R&R report. This column is defined as 100(Variation/(KHistorical Sigma)).
• The Number of Distinct Categories (NDC) is defined as (1.41(PV/RR)), rounded down to the nearest integer.
Discrimination Ratio The discrimination ratio characterizes the relative usefulness of a given measurement for a specific product. Generally, when the discrimination ratio is less than 2, the measurement cannot detect product variation, implying that the measurement process needs improvement. A discrimination ratio greater than 4 adequately detects unacceptable product variation, implying that the production process needs improvement.
See “Statistical Details for the Discrimination Ratio”.
Misclassification Probabilities Due to measurement variation, good parts can be rejected and bad parts can be accepted. This is called misclassification. Misclassification rates decrease as measurement variability decreases. When you select the Misclassification Probabilities option, you are prompted to select the model type and enter specification limits if you have not already done so.
Table 5.2 Acceptable Percent Measurement Variation < 10% excellent 11% to 20% adequate 21% to 30% marginally acceptable > 30% unacceptable Chapter 5 Variability Gauge Charts 143 Quality and Process Methods Variability Gauge Platform Options Figure 5.8 Example of the Misclassification Probabilities Report The misclassification probabilities are based on the joint probability function of Y, the measured value of the part, and X, the true value of the part. The joint probability density function used is a bivariate normal distribution. To understand the descriptions, define the following probabilities: = P[(LSL X USL) and (Y < LSL or Y > USL)] = P[(X < LSL or X > USL) and (LSL Y USL)] = P(LSL X USL) Descriptions P(Good part is falsely rejected) The conditional probability that a part is rejected given that it is a good part, or /.
P(Bad part is falsely accepted) The conditional probability that a part is accepted given that it is a bad part, or /(1-).
P(Part is good and is rejected) The joint probability that a part is good and that it is rejected, or .
P(Part is bad and is accepted) The joint probability that a part is bad and that it is accepted, or .
P(Part is good) The probability that a part is good, or .
For more information, see “Statistical Details for the Misclassification Probabilities” as well as Burdick et al. (2005).
Bias Report The Bias Report shows a graph for Overall Measurement Bias with a summary table and a graph for Measurement Bias by Standard with a summary table. The average bias, or the differences between the observed values and the standard values, appears for each level of the X variable. A t test for the bias is also given.
The Bias Report option is available only when a Standard variable is provided in the launch window.
144 Variability Gauge Charts Chapter 5 Additional Examples of Variability Charts Quality and Process Methods The Measurement Bias Report red triangle menu contains the following options: Confidence Intervals Calculates confidence intervals for the average bias for each part and places marks on the Measurement Bias Report by Standard plot.
Measurement Error Graphs Produces a graph of Measurement Error versus all grouping columns together. There are also graphs of Measurement Error by each grouping column separately.
Linearity Study The Linearity Study performs a regression analysis using the standard variable as the X variable and the bias as the Y variable. This analysis examines the relationship between bias and the size of the part. Ideally, you want to find a slope of zero. If the slope is significantly different from zero, you can conclude that there is a significant relationship between the size of the part or variable measured as a standard and the ability to measure.
The Linearity Study option is available only when a Standard variable is provided in the launch window.
The report shows the following information: • Bias summary statistics for each standard.
• An ANOVA table that tests if the slope of the line is equal to zero.
• The line parameters, including tests for the slope (linearity) and intercept (bias). The test for the intercept is useful only if the test on the slope fails to reject the hypothesis of slope = 0.
• The equation of the line appears directly beneath the graph.
The Linearity Study red triangle menu contains the following options: Set Alpha Level Changes the alpha level that is used in the bias confidence intervals.
Linearity by Groups Produces separate linearity plots for each level of the X, Grouping variables that you specified in the launch window.
Additional Examples of Variability Charts • “Example of the Heterogeneity of Variance Test” • “Example of the Bias Report Option” Chapter 5 Variability Gauge Charts 145 Quality and Process Methods Additional Examples of Variability Charts Example of the Heterogeneity of Variance Test Suppose that you have data containing part measurements. Three operators (Cindy, George, and Tom) each took measurements of 10 parts. They measured each part three times, making a total of 90 observations. You want to examine the following: • whether the variance of measurements for each operator are the same or different • whether the variance for each part is the same or different • whether the variance for each Operatorpart combination is the same or different Ideally, you want all of the variances for each of the groups to be considered the same statistically.
1.
Select Help > Sample Data Library and open Variability Data/2 Factors Crossed.jmp.
2.
Select Analyze > Quality and Process > Variability / Attribute Gauge Chart.
3.
Select Measurement and click Y, Response.
4.
Select Operator and click X, Grouping.
5.
Select part# and click Part, Sample ID.
6.
Click OK.
7.
Click the Variability Gauge red triangle and select Heterogeneity of Variance Tests.
8.
Select Crossed.
9.
Click OK.
146 Variability Gauge Charts Chapter 5 Additional Examples of Variability Charts Quality and Process Methods Figure 5.9 Heterogeneity of Variances Tests Report Note: Because the method uses simulations, the decision limits can be slightly different each time.
Chapter 5 Variability Gauge Charts 147 Quality and Process Methods Additional Examples of Variability Charts In the Operator Variance test, all three levels exceed the upper and lower decision limits. From this, you conclude that each operator has a different variability from the square root of the average group variance. You might want to examine why the variation between each operator is different.
For the part# Variance test and the interaction (Operatorpart#) Variance test, none of the levels exceed the decision limits. From this, you conclude that the variances are not statistically different from the square root of the average group variance. Each part has a similar variance to the other parts, and each Operatorpart# combination has similar variance to the other Operatorpart# combinations.
Example of the Bias Report Option Note: These data come from the Automotive Industry Action Group (2002).
Assume that as a plant supervisor, you are introducing a new measurement system into your process. As part of the Production Part Approval Process (PPAP), the bias and linearity of the measurement system needs to be evaluated. Five parts were chosen throughout the operating range of the measurement system, based on documented process variation. Each part was measured by layout inspection to determine its reference value. Each part was then measured twelve times by the lead operator. The parts were selected at random during the day. In this example, you want to examine the overall bias and the individual measurement bias (by standard).
1.
Select Help > Sample Data Library and open Variability Data/MSALinearity.jmp.
2.
Select Analyze > Quality and Process > Variability / Attribute Gauge Chart.
3.
Select Response and click Y, Response.
4.
Select Standard and click Standard.
5.
Select Part and click X, Grouping.
6.
Click OK.
7.
Click the Variability Gauge red triangle and select Gauge Studies > Bias Report.
148 Variability Gauge Charts Chapter 5 Additional Examples of Variability Charts Quality and Process Methods Figure 5.10 Measurement Bias Report The bias (Response minus Standard) is calculated for every measurement. The Overall Measurement Bias Report shows a histogram of the bias and a t test to see whether the average bias is equal to 0. You can see that the Average Bias is not zero, it is -0.0533. However, zero is contained within the confidence interval (-0.1152,0.0085), which means that the Average Bias is not significantly different from 0. Using a significance level of 0.05, you can see that the p-value is greater than 0.05, which also shows that the Average Bias is not significantly different from 0.
The Measurement Bias Report by Standard shows average bias values for each part. The bias averages are plotted on the graph along with the actual bias values for every part, so that you can see the spread. In this example, part number 1 (with a standard value of 2) is biased high and parts 4 and 5 (with standard values of 8 and 10) are biased low.
Tip: To see confidence intervals for the bias, right-click in the table and select Columns > Lower 95% and Upper 95%.
Chapter 5 Variability Gauge Charts 149 Quality and Process Methods Additional Examples of Variability Charts Example of a Linearity Study Using the same data and scenario as the Bias Report option, you can now examine the linearity to determine whether there is a significant relationship between the size of the parts and the operator’s ability to measure them.
1.
Select Help > Sample Data Library and open Variability Data/MSALinearity.jmp.
2.
Select Analyze > Quality and Process > Variability / Attribute Gauge Chart.
3.
Select Response and click Y, Response.
4.
Select Standard and click Standard.
5.
Select Part and click X, Grouping.
6.
Click OK.
7.
Click the Variability Gauge red triangle and select Gauge Studies > Linearity Study.
8.
In the window that prompts you to Specify Process Variation, type 14.9286.
The value 14.9286 is 6 times the standard deviation of the response, 2.488105.
Figure 5.11 Linearity Study Note the following: • The slope is -0.131667. This value appears as part of the equation below the graph, and also in the third table.
• The p-value associated with the test on the slope is quite small, <.0001. The t test for the slope is testing whether the bias changes with the standard value.
150 Variability Gauge Charts Chapter 5 Statistical Details for Variability Charts Quality and Process Methods Because the p-value is small, you can conclude that there is a significant linear relationship between the size of the parts and the operator’s ability to measure them. You can also see this in the graph. If the part or standard value is small, the bias is high, and vice versa.
Statistical Details for Variability Charts • “Statistical Details for Variance Components” • “Statistical Details for the Discrimination Ratio” • “Statistical Details for the Misclassification Probabilities” Statistical Details for Variance Components The exact model type that you choose depends on how the data was collected. For example, are the operators measuring the same parts (in which case you have a crossed design) or are they measuring different parts (in which case you have a nested design)? To illustrate, in a model where B is nested within A, multiple measurements are nested within both B and A, and there are na•nb•nw measurements, the following statements hold: • na random effects are due to A • na•nb random effects due to each nb B levels within A • na•nb•nw random effects due to each nw levels within B within A: .
The Zs are the random effects for each level of the classification. Each Z is assumed to have a mean of zero and to be independent from all other random terms. The variance of the response y is the sum of the variances due to each z component: .
Table 5.3 shows the supported models and what the effects in the model would be.
Table 5.3 Models Supported by the Variability Charts Platform Model Factors Effects in the Model Main Effects 1 2 unlimited A A, B and so on, for more factors yijk u Zai Zbij Zwijk + + + = Var yijk Var Zai Var Zbij Var Zwijk + + = Chapter 5 Variability Gauge Charts 151 Quality and Process Methods Statistical Details for Variability Charts Statistical Details for the Discrimination Ratio The discrimination ratio compares the total variance of the measurement, M, with the variance of the measurement error, E. The discrimination ratio is computed for all main effects, including nested main effects. The discrimination ratio, D, is computed as follows: where: P = estimated variance for a factor T = estimated total variance Crossed 1 2 3 4 unlimited A A, B, AB A, B, AB, C, AC, BC, ABC A, B, AB, C, AC, BC, ABC, D, AD, BD, ABD, CD, ACD, BCD, ABCD, and so on, for more factors Nested 1 2 3 4 unlimited A A, B(A) A, B(A), C(A,B) A, B(A), C(A,B), D(A,B,C) and so on, for more factors Crossed then Nested 3 A, B, AB, C(A,B) Nested then Crossed 3 A, B(A), C, AC, CB(A) Table 5.3 Models Supported by the Variability Charts Platform (Continued) Model Factors Effects in the Model D 2 P T P – ------------- 1 + = 152 Variability Gauge Charts Chapter 5 Statistical Details for Variability Charts Quality and Process Methods Statistical Details for the Misclassification Probabilities This section describes the computations for the probabilities in the Misclassification Probabilities report. The misclassification probabilities are based on the joint probability function of Y, the measured value of the part, and X, the true value of the part. The joint probability distribution function FYX(y, x) uses a bivariate normal distribution with mean vector [, ] and the following covariance matrix: where P is the part-to-part variation, M is the measurement variation, and is the grand mean. These quantities can be found or derived from quantities in the report window. Specifically, P + M is equal to the square of Total Variation (TV) divided by 6: (TV/6)2 and P is equal to the square of Part Variation (PV) divided by 6: (PV/6)2. The correlation YX between Y and X is defined as the square root of P/(P + M).
Next, define the following probabilities: = P[(LSL X USL) and (Y < LSL or Y > USL)] = P[(X < LSL or X > USL) and (LSL Y USL)] = P(LSL X USL) These probabilities can be expressed in terms of the joint probability distribution function FYX(y, x) and the marginal probability distribution functions for Y and X: FY(y) and FX(x): = FYX(LSL, USL) - FYX(LSL, LSL) - FYX(USL, USL) + FYX(USL, LSL) + FX(USL) - FX(LSL) = FYX(USL, LSL) - FYX(LSL, LSL) - FYX(USL, USL) + FYX(LSL, USL) + FY(USL) - FY(LSL) = FX(USL) - FX(LSL) P(Good part is falsely rejected) = / P(Bad part is falsely accepted) = /(1- P(Part is good and is rejected) = P(Part is bad and is accepted) = P(Part is good) = P M + P P P Chapter 6 Attribute Gauge Charts Evaluate a Categorical Measurement Process Using Agreement Measures Attribute charts analyze categorical measurements and can help show you measures of agreement across responses, such as raters. In attribute data, the variable of interest has a finite number of categories. Typically, data has only two possible results, such as pass or fail. You can examine aspects such as how effective raters were at classifying a part, how much they agreed with each other, and how much they agreed with themselves over the course of several ratings.
Tip: This chapter covers only attribute charts. For more information about variability charts, see “Variability Gauge Charts”.
Figure 6.1 Example of an Attribute Chart 156 Attribute Gauge Charts Chapter 6 Quality and Process Methods Contents Overview of Attribute Gauge Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 Example of an Attribute Gauge Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157 Launch the Variability/Attribute Gauge Chart Platform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 The Attribute Gauge Chart and Reports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 Agreement Reports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161 Effectiveness Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162 Attribute Gauge Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 Statistical Details for Attribute Gauge Charts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 Statistical Details for the Agreement Report. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 Chapter 6 Attribute Gauge Charts 157 Quality and Process Methods Overview of Attribute Gauge Charts Overview of Attribute Gauge Charts Before you create an attribute gauge chart, your data should be formatted using the following guidelines: • In order to compare agreement among raters, each rater in the data table must be in a separate column. These columns are then assigned to the Y, Response role in the launch window. In Figure 6.2, each rater (A, B, and C) is in a separate column.
• Responses in the different columns can be character (pass or fail) or numeric (0 or 1). In Figure 6.2, rater responses are numeric (0 for pass, 1 for fail). All response columns must have the same data type.
• Any other variables of interest that you might want to use as X, Grouping variables should appear stacked in one column each (see the Part column in Figure 6.2). You can also define a Standard column, which produces reports that compare raters with the standard. The Standard column and response columns must have the same data type.
Figure 6.2 Attribute Gauge Data Example of an Attribute Gauge Chart Suppose that you have data containing pass or fail ratings for parts. Three raters, identified as A, B, and C, each noted a 0 (pass) or a 1 (fail) for 50 parts, three times each. You want to examine how effective the raters were in correctly classifying the parts, and how well the raters agreed with each other and with themselves over the course of the ratings.
1.
Select Help > Sample Data Library and open Attribute Gauge.jmp.
2.
Select Analyze > Quality and Process > Variability / Attribute Gauge Chart.
3.
For Chart Type, select Attribute.
158 Attribute Gauge Charts Chapter 6 Example of an Attribute Gauge Chart Quality and Process Methods 4.
Select A, B, and C and click Y, Response.
5.
Select Standard and click Standard.
6.
Select Part and click X, Grouping.
7.
Click OK.
Figure 6.3 Example of an Attribute Chart The first chart (Part) shows how well the raters agreed with each other for each part. For example, here you can see that the percent agreement dropped for part 6, 12, 14, 21, 22, and so on. These parts might have been more difficult to categorize.
The second chart (Rater) shows each rater’s agreement with him or herself and the other raters for a given part, summed up over all of the parts. In this example, it looks like the performance of the raters is relatively similar. Rater C had the lowest agreement, but the difference is not major (about 89% instead of 91%).
8.
Open the Effectiveness Report and scroll down to the Conformance Report.
You can see that 0 = non-conform (fail) and a 1 = conform (pass). However in this data, it is exactly the opposite: 0 is a pass and 1 is a fail. Reverse this setting.
9.
Click the Conformance Report red triangle and select Change Conforming Category.
Chapter 6 Attribute Gauge Charts 159 Quality and Process Methods Launch the Variability/Attribute Gauge Chart Platform Launch the Variability/Attribute Gauge Chart Platform Launch the Variability/Attribute Gauge Chart platform by selecting Analyze > Quality and Process > Variability/Attribute Gauge Chart. Set the Chart Type to Attribute.
Figure 6.4 The Variability/Attribute Gauge Chart Launch Window For more information about the options in the Select Columns red triangle menu, see Using JMP.
Chart Type Choose between a variability gauge analysis (for a continuous response) or an attribute gauge analysis (for a categorical response, usually “pass” or “fail”).
Note: The content in this chapter covers only the Attribute chart type. For more information about the Variability chart type, see “Variability Gauge Charts”.
Specify Alpha Specify the alpha level used by the platform.
Y, Response Specify the columns of ratings given by each rater. You must specify more than one rating column.
Standard Specify a standard or reference column that contains the “true” or known values for the part. In the report window, an Effectiveness Report and an additional section in the Agreement Comparisons report appear, which compare the raters with the standard.
X, Grouping Specify the classification columns that group the measurements. If the factors form a nested hierarchy, specify the higher terms first.
Freq Identifies the data table column whose values assign a frequency to each row. Can be useful when you have summarized data.
160 Attribute Gauge Charts Chapter 6 The Attribute Gauge Chart and Reports Quality and Process Methods By Identifies a column that creates a report consisting of separate analyses for each level of the variable.
The Attribute Gauge Chart and Reports Attribute gauge chart plots the percent Agreement, which is a measurement of rater agreement for every part in the study. The agreement for each part is calculated by comparing the ratings for every pair of raters for all ratings of that part. See “Statistical Details for Attribute Gauge Charts”.
Follow the instructions in “Example of an Attribute Gauge Chart” to produce the results shown in Figure 6.5.
Figure 6.5 Attribute Gauge Chart The first chart in Figure 6.5 uses all X grouping variables (in this case, the Part) on the x-axis. The second chart uses all Y variables on the x-axis (typically, and in this case, the Rater).
• In the first graph, you can look for parts with a low percent Agreement value, and investigate to determine why raters do not agree about the measurement of that particular part.
Chapter 6 Attribute Gauge Charts 161 Quality and Process Methods The Attribute Gauge Chart and Reports • In the second graph, you can look for raters with a low percent Agreement value, and investigate to determine why they do not agree with the other raters or with themselves.
For information about additional options, see “Attribute Gauge Platform Options”.
Agreement Reports Note: The Kappa value is a statistic that expresses agreement. The closer the Kappa value is to 1, the more agreement there is. A Kappa value closer to 0 indicates less agreement.
The Agreement Report shows agreement summarized for each rater and overall agreement. This report is a numeric form of the data presented in the second chart in the Attribute Gauge Chart report (Figure 6.5).
The Agreement Comparisons report shows each rater compared with all other raters, using Kappa statistics. The rater is compared with the standard only if you have specified a Standard variable in the launch window.
The Agreement within Raters report shows the number of items that were inspected. The confidence intervals are score confidence intervals, as suggested by Agresti and Coull (1998). The Number Matched is the sum of the number of items inspected, where the rater agreed with him or herself on each inspection of an individual item. The Rater Score is the Number Matched divided by the Number Inspected.
The Agreement across Categories report shows the agreement in classification over that which would be expected by chance. It assesses the agreement between a fixed number of raters when classifying items.
162 Attribute Gauge Charts Chapter 6 The Attribute Gauge Chart and Reports Quality and Process Methods Figure 6.6 Agreement Reports Effectiveness Report The Effectiveness Report appears only if you have specified a Standard variable in the launch window. For a description of a Standard variable, see “Launch the Variability/Attribute Gauge Chart Platform”. This report compares every rater with the standard.
Chapter 6 Attribute Gauge Charts 163 Quality and Process Methods The Attribute Gauge Chart and Reports Figure 6.7 Effectiveness Report The Agreement Counts table shows cell counts on the number correct and incorrect for every level of the standard. In Figure 6.7, the standard variable has two levels, 0 and 1. Rater A had 45 correct responses and 3 incorrect responses for level 0, and 97 correct responses and 5 incorrect responses for level 1.
Effectiveness is defined as the number of correct decisions divided by the total number of opportunities for a decision. For example, say that rater A sampled every part three times. On the sixth part, one of the decisions did not agree (for example, pass, pass, fail). The other two decisions would still be counted as correct decisions. This definition of effectiveness is different from the MSA 3rd edition. According to MSA, all three opportunities for rater A on part six would be counted as incorrect. Including all of the inspections separately gives you more information about the overall inspection process.
In the Effectiveness table, 95% confidence intervals are given about the effectiveness. These are score confidence intervals. It has been demonstrated that score confidence intervals provide increased coverage probability, particularly where observations lie near the boundaries. See Agresti and Coull (1998).
The Misclassifications table shows the incorrect labeling. The rows represent the levels of the standard or accepted reference value. The columns contain the levels given by the raters.
164 Attribute Gauge Charts Chapter 6 Attribute Gauge Platform Options Quality and Process Methods Conformance Report The Conformance Report shows the probability of false alarms and the probability of misses. The Conformance Report appears only when the rating has two levels (such as pass or fail, or 0 or 1).
The following descriptions apply: False Alarm The part is determined to be non-conforming, when it actually is conforming.
Miss The part is determined to be conforming, when it actually is not conforming.
P(False Alarms) The number of parts that have been incorrectly judged to be nonconforming divided by the total number of parts that are judged to be conforming.
P(Miss) The number of parts that have been incorrectly judged to be conforming divided by the total number of parts that are actually nonconforming.
The Conformance Report red triangle menu contains the following options: Change Conforming Category Reverses the response category that is considered conforming.
Calculate Escape Rate Calculates the Escape Rate, which is the probability that a non-conforming part is produced and not detected. The Escape Rate is calculated as the probability that the process will produce a non-conforming part times the probability of a miss. You specify the probability that the process will produce a non-conforming part, also called the Probability of Nonconformance.
Note: Missing values are treated as a separate category in this platform. To avoid this separate category, exclude rows of missing values in the data table.
Attribute Gauge Platform Options The Attribute Gauge red triangle menu contains the following options: Attribute Gauge Charts Shows or hides the gauge attribute chart and the efficiency chart.
Show Agreement Points Shows or hides the agreement points on the charts.
Connect Agreement Points Connects the agreement points in the charts.
Agreement by Rater Confid Intervals Shows or hides the agreement by rater confidence intervals on the efficiency chart.
Chapter 6 Attribute Gauge Charts 165 Quality and Process Methods Statistical Details for Attribute Gauge Charts Show Agreement Group Means Shows or hides the agreement group means on the gauge attribute chart. This option is available when you specify more than one X, Grouping variable.
Show Agreement Grand Mean Shows or hides the overall agreement mean on the gauge attribute chart.
Show Effectiveness Points Shows or hides the effectiveness points on the charts.
Connect Effectiveness Points Draws lines between the effectiveness points in the charts.
Effectiveness by Rater Confid Intervals Shows or hides confidence intervals on the second chart in the Attribute Gauge Chart report (Figure 6.5).
Effectiveness Report Shows or hides the Effectiveness report. This report compares every rater with the standard, using the Kappa statistic.
See Using JMP for more information about the following options: Local Data Filter Shows or hides the local data filter that enables you to filter the data used in a specific report.
Redo Contains options that enable you to repeat or relaunch the analysis. In platforms that support the feature, the Automatic Recalc option immediately reflects the changes that you make to the data table in the corresponding report window.
Save Script Contains options that enable you to save a script that reproduces the report to several destinations.
Save By-Group Script Contains options that enable you to save a script that reproduces the platform report for all levels of a By variable to several destinations. Available only when a By variable is specified in the launch window.
Statistical Details for Attribute Gauge Charts For the first chart in Figure 6.5 that plots all X, Grouping variables on the x-axis, the percent Agreement is calculated as follows: % Agreement for part i number of responses for level l 2 l 1 = k Ni 2 ---------------------------------------------------------------------------------------------------= 166 Attribute Gauge Charts Chapter 6 Statistical Details for Attribute Gauge Charts Quality and Process Methods For the second chart in Figure 6.5 that plots all Y, Response variables on the x-axis, the percent Agreement is calculated as follows: Note the following: • n = number of parts (grouping variables) • ri = number of reps for part i (i = 1,...,n) • m = number of raters • k = number of levels • Ni = m x ri. Number of ratings on part i (i = 1,...,n). This includes responses for all raters, and repeat ratings on a part. For example, if part i is measured 3 times by each of 3 raters, then Ni is 3 x 3 = 9.
For example, consider the following table of data for three raters, each having three replicates for one part.
Using this table, you can make these calculations: % Agreement = % Agreement [rater A] = percent Agreement [rater B] = and Table 6.1 Three Replicates for Raters A, B, and C A B C 1 1 1 1 2 1 1 0 3 0 0 0 % Agreement for rater k number of uncounted matching levels for this rater k within part i for rep j j 1 = ri i 1 = n Ni j – j 1 = ri i 1 = n ---------------------------------------------------------------------------------------------------------------------------------= 4 2 5 2 + 9 2 -----------------------16 36 ------0.444 = = 4 3 3 + + 8 7 6 + + ---------------------10 21 ------0.476 = = Chapter 6 Attribute Gauge Charts 167 Quality and Process Methods Statistical Details for Attribute Gauge Charts % Agreement [rater C] = Statistical Details for the Agreement Report The simple Kappa coefficient is a measure of inter-rater agreement.
where: and: If you view the two response variables as two independent ratings of the n parts, the Kappa coefficient equals +1 when there is complete agreement of the raters. When the observed agreement exceeds chance agreement, the Kappa coefficient is positive, and its magnitude reflects the strength of agreement. Although unusual in practice, Kappa is negative when the observed agreement is less than the chance agreement. The minimum value of Kappa is between -1 and 0, depending on the marginal proportions.
Estimate the asymptotic variance of the simple Kappa coefficient with the following equation: where: and: The Kappas are plotted and the standard errors are also given.
4 3 2 + + 8 7 6 + + ---------------------9 21 ------0.4286 = = ˆ P0 Pe – 1 Pe – ------------------= P0 pii i = Pe pi.p.i i = var A B C – + 1 Pe – 2n --------------------------= A pii 1 pi.
p.i + 1 ˆ – – i = B 1 ˆ – 2 pij p.i pj.
+ 2 i j = C ˆ Pe 1 ˆ – – 2 = 168 Attribute Gauge Charts Chapter 6 Statistical Details for Attribute Gauge Charts Quality and Process Methods Note: The Kappa statistics in the Attribute Chart platform are shown even when the levels of the variables are unbalanced.
Categorical Kappa statistics (Fleiss 1981) are found in the Agreement Across Categories report.
Given the following assumptions: • n = number of parts (grouping variables) • m = number of raters • k = number of levels • ri = number of reps for part i (i = 1,...,n) • Ni = m x ri. Number of ratings on part i (i = 1, 2,...,n). This includes responses for all raters, and repeat ratings on a part. For example, if part i is measured 3 times by each of 2 raters, then Ni is 3 x 2 = 6.
• xij = number of ratings on part i (i = 1, 2,...,n) into level j (j = 1, 2,...,k) The individual category Kappa is defined as follows: The overall Kappa is defined as follows: The variance of and are calculated as follows: j ˆ 1 xij Ni xij – i 1 = n pjqj Ni i 1 = n Ni 1 – -----------------------------------------------------– where pj xij i 1 = n Ni i 1 = n ----------------- qj 1 pj – = = = ˆ qjpjj ˆ j 1 = k pjqj j 1 = k --------------------------= ˆ j ˆ var ˆ j 2 nN N 1 – ---------------------------= var ˆ 2 pjqj j 1 = k 2 nN N 1 – --------------------------------------------------------pjqj j 1 = k 2 pjqj qj pj – j 1 = k – = Chapter 6 Attribute Gauge Charts 169 Quality and Process Methods Statistical Details for Attribute Gauge Charts The standard errors of and are shown only when there are an equal number of ratings per part (for example, Ni = N for all i = 1,…,n).
ˆ j ˆ 170 Attribute Gauge Charts Chapter 6 Statistical Details for Attribute Gauge Charts Quality and Process Methods Chapter 7 Process Capability Measure the Variability of a Process over Time Process capability analysis, used in process control, measures how well a process is performing compared to given specification limits. A good process is one that is stable and consistently produces product that is well within specification limits. A capability index is a measure that relates process performance, summarized by process centering and variability, to specification limits.
Graphical tools such as a goal plot and box plots give you quick visual ways of identifying which process or product characteristics are within specifications. Individual detail reports display a capability report for each variable in the analysis. The analysis enables you to identify variation relative to the specifications or requirements; this enables you to achieve increasingly higher conformance values.
You can specify subgroups to compare the overall variation of the process to the within subgroup variation. You can compute capability indices for processes that produce measurements that follow various distributions. For data that follow none of the specified distributions, you can compute nonparametric capability indices.
Figure 7.1 Example of the Process Capability Platform 170 Process Capability Chapter 7 Quality and Process Methods Contents Overview of the Process Capability Platform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 Example of the Process Capability Platform with Normal Variables . . . . . . . . . . . . . . . . . . . . 173 Example of the Process Capability Platform with Nonnormal Variables. . . . . . . . . . . . . . . . . 175 Launch the Process Capability Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181 Process Selection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 Process Subgrouping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 Moving Range Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 Historical Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 Distribution Options. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 Other Specifications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 Entering Specification Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 Spec Limits Window. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 Limits Data Table. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 Spec Limits Column Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 The Process Capability Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 Goal Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 Capability Box Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 Capability Index Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 Process Capability Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 Individual Detail Reports. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 Normalized Box Plots. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 Process Performance Plot. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 Summary Reports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211 Make Goal Plot Summary Table . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 Additional Examples of the Process Capability Platform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 Process Capability for a Stable Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 Process Capability for an Unstable Process. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 Simulation of Confidence Limits for a Nonnormal Process Ppk . . . . . . . . . . . . . . . . . . . . . 220 Statistical Details for the Process Capability Platform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 Variation Statistics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 Notation for Goal Plots and Capability Box Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 Goal Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 Capability Box Plots for Processes with Missing Targets . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 Capability Indices for Normal Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 Capability Indices for Nonnormal Distributions: Percentile and Z-Score Methods . . . . . 239 Parameterizations for Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 Chapter 7 Process Capability 171 Quality and Process Methods Overview of the Process Capability Platform Overview of the Process Capability Platform The Process Capability platform provides the tools needed to measure the compliance of a process to given specifications. By default, JMP shows a Goal Plot, Capability Box Plots, and a Capability Index Plot for the variables that you fit with normal distributions. Capability indices for nonnormal variables are plotted on the Capability Index Plot. You can add normalized box plots, summary reports, and individual detail reports for the variables in your analysis.
You can supply specification limits in several ways: • in the data table, using a column property • by requesting the Spec Limits Dialog in the launch window • by loading the limits from a specification limits data table • using the Manage Spec Limits utility (Analyze > Quality and Process > Manage Spec Limits) You can specify two-sided, one-sided, or asymmetric specification limits.
Note: The Process Capability platform expands significantly on the Capability analyses that are available through Analyze > Distribution and through Analyze > Quality and Process > Control Chart.
Capability Indices A capability index is a ratio that relates the ability of a process to produce product that meets specification limits. The index relates estimates of the mean and standard deviation of the quality characteristic to the specification limits. Within estimates of capability are based on an estimate of the standard deviation constructed from within-subgroup variation. Overall estimates of capability use an estimate of standard deviation constructed from all of the process data. See “Capability Indices for Normal Distributions” and “Variation Statistics”.
Estimates of the mean or standard deviation are well-defined only if the processes related to centering or spread are stable. Therefore, interpretation of within capability indices requires that process spread is stable. Interpretation of overall capability indices requires that both process centering and spread are stable.
Capability indices constructed from small samples can be highly variable. The Process Capability platform provides confidence intervals for most capability indices. Use these to determine the range of potential values for your quality characteristic’s actual capability.
172 Process Capability Chapter 7 Overview of the Process Capability Platform Quality and Process Methods Note: When confidence intervals are not provided (for example, for nonnormal distributions) you can use the Simulate feature to construct confidence intervals. For an example, see “Simulation of Confidence Limits for a Nonnormal Process Ppk”.
Guidelines for values of capability indices can be found in Montgomery (2013). The minimum recommended value is 1.33. Six Sigma initiatives aim for much higher capability levels that correspond to extremely low rates of defective parts per million.
Capability Indices for Nonnormal Processes The Process Capability platform constructs capability indices for process measurements with the following distributions: Normal, Beta, Exponential, Gamma, Johnson, Lognormal, Mixture of 2 Normals, Mixture of 3 Normals, SHASH, and Weibull. A Best Fit option determines the best fit among these distributions and provides capability indices for this fit. The platform also provides a Nonparametric fit option that gives nonparametric estimates of capability.
For the nonnormal methods, estimates are constructed using two approaches: the ISO/Quantile method (Percentiles) and the Bothe/Z-scores method (Z-Score). For more information about these methods, see “Capability Indices for Nonnormal Distributions: Percentile and Z-Score Methods”.
Note: Process Capability analysis for individual responses is accessible through Analyze > Quality and Process > Control Chart Builder. However, nonnormal distributions are available only in the Process Capability platform.
Overall and Within Estimates of Sigma Most capability indices in the Process Capability platform can be computed based on estimates of the overall (long-term) variation and the within-subgroup (short-term) variation. If the process is stable, these two measures of variation should yield similar results since the overall and within subgroup variation should be similar. The normalized box plots and summary tables can be calculated using either the overall or the within-subgroup variation. See “Additional Examples of the Process Capability Platform” for examples of capability indices computed for stable and unstable processes.
You can specify subgroups for estimating within-subgroup variation in the launch window. You can specify a column that defines subgroups or you can select a constant subgroup size. For each of these methods, you can choose to estimate the process variation using the average of the unbiased standard deviations or using the average of the ranges. If you do not specify subgroups, the Process Capability platform constructs a within-subgroup estimate of the process variation using a moving range of subgroups of size two. Finally, you can specify a historical sigma to be used as an estimate of the process standard deviation.
Chapter 7 Process Capability 173 Quality and Process Methods Example of the Process Capability Platform with Normal Variables Capability Index Notation The Process Capability platform provides two sets of capability indices. See “Capability Indices for Normal Distributions” for more information about the calculation of the capability indices.
• Cpk, Cpl, Cpu, Cp, and Cpm. These indices are based on a within-subgroup (short-term) estimate of the process standard deviation.
• Ppk, Ppl, Ppu, Pp, and Cpm. These indices are based on an overall (long-term) estimate of the process standard deviation. Note that the process standard deviation does not exist if the process is not stable. See Montgomery (2013).
The Process Capability platform uses the appropriate AIAG notation for capability indices: Ppk labeling denotes an index constructed from an overall variation estimate and Cpk denotes an index constructed from a within-subgroup variation estimate.
Note: The AIAG (Ppk) Labeling platform preference is selected by default. You can change the reporting to use Cp notation only by deselecting this preference under Process Capability.
For more information about process capability analysis, see Montgomery (2013) and Wheeler (2004).
Example of the Process Capability Platform with Normal Variables This example uses the Semiconductor Capability.jmp sample data table. The variables represent standard measurements that a semiconductor manufacturer might make on a wafer as it is being processed. Specification limits for the variables have been entered in the data table through the Column Properties > Spec Limits property.
1.
Select Help > Sample Data Library and open Semiconductor Capability.jmp.
2.
Select Analyze > Quality and Process > Process Capability.
3.
Click the white triangle next to Processes to view all of the continuous variables.
4.
Select PNP1, PNP2, NPN2, PNP3, IVP1, PNP4, NPN3, and IVP2, and click Y, Process.
5.
Click OK.
6.
Click the Goal Plot red triangle and select Label Overall Sigma Points.
7.
Click the Capability Index Plot red triangle and select Label Overall Sigma Points.
174 Process Capability Chapter 7 Example of the Process Capability Platform with Normal Variables Quality and Process Methods Figure 7.2 Example Results for Semiconductor Capability.jmp The Goal Plot shows the spec-normalized mean shift on the x-axis and the spec-normalized standard deviation on the y-axis for each variable. The triangular region defined by the red lines in the bottom center of the plot is the goal triangle. It defines a region of capability index values. You can adjust the goal triangle using the Ppk slider to the right of the plot. When the slider is set to 1, note that PNP1, PNP3, IVP1, and IVP2 are outside of the goal triangle and possibly out of specification.
The Capability Box Plots report shows a box plot for each variable in the analysis. The values for each column are centered by their target value and scaled by the specification range. In this example, all process variables have both upper and lower specification limits, and these are symmetric about the target value. It follows that the solid green line shows where the target should be and the dashed lines represent the specification limits.
Chapter 7 Process Capability 175 Quality and Process Methods Example of the Process Capability Platform with Nonnormal Variables It appears that the majority of points for IVP1 are above its upper specification limit (USL), and the majority of points for IVP2 are less than its target. PNP2 seems to be on target with all data values inside the specification limits.
The Capability Index Plot plots the Ppk values for each variable. Four variables come from very capable processes, with Ppk values of 2 or more. Four variables have Ppk values below 1.
Example of the Process Capability Platform with Nonnormal Variables The Process Measurements.jmp data table contains measurements made on seven different processes used to construct a product. For each process, specification limits are saved as column properties. You begin by examining the distributions of your process data. You see that the distributions are not normal. Then you use the nonnormal capability features of the Process Capability platform to compute capability indices.
View the Distributions 1.
Select Help > Sample Data Library and open Process Measurements.jmp.
2.
Select Analyze > Distribution.
3.
Select all seven columns from the Select Columns list and click Y, Columns.
4.
Check the box next to Histograms Only.
5.
Click OK.
For most processes, the histograms show evidence that the theoretical distribution of measurements is skewed and does not follow a normal distribution. Therefore, for each process, you find the best fitting distributions among all of the available parametric distributions.
Perform a Capability Analysis 1.
Select Analyze > Quality and Process > Process Capability.
2.
Select all seven columns from the Columns list and click Y, Process.
3.
Select all seven columns in the Y, Process list.
4.
Open the Distribution Options panel and select Best Fit from the Distribution list.
5.
Click Set Process Distribution.
The suffix &Dist(Best Fit) is added to each variable name in the Y, Process list. The Best Fit option specifies that the best-fitting parametric distribution should be fit to each variable. The available parametric distributions are Normal, Beta, Exponential, Gamma, Johnson, 176 Process Capability Chapter 7 Example of the Process Capability Platform with Nonnormal Variables Quality and Process Methods Lognormal, Mixture of 2 Normals, Mixture of 3 Normals, SHASH, and Weibull (Figure 7.3).
6.
Open the Nonnormal Distribution Options outline. Note that the Nonnormal Capability Indices Method is set to Percentiles, the Johnson Distribution Fitting Method is set to Quantile Matching, and the Distribution Comparison Criterion is set to AICc.
Figure 7.3 Completed Launch Window The Quantile Matching method is the default method used for fitting Johnson distributions because of its stability and speed as compared to Maximum Likelihood. Note that Maximum Likelihood is used in the Distribution platform.
7.
Click OK.
8.
Click the Goal Plot red triangle and select Label Overall Sigma Points.
9.
Click the Capability Index Plot red triangle and select Label Overall Sigma Points.
Chapter 7 Process Capability 177 Quality and Process Methods Example of the Process Capability Platform with Nonnormal Variables Figure 7.4 Initial Report with Variables Labeled Note: Click a label in the plot and drag it to make the plot more interpretable. Click the right side frame of the Capability Index Plot and drag it to the right to make the labels easier to distinguish.
The Goal Plot shows only one point and it corresponds to Process 7. The Capability Box Plots report shows a single box plot for Process 7. This is because the best fit for Process 7 is a normal distribution.
10. To the right of the Capability Index Plot, set the Ppk value to 2.
178 Process Capability Chapter 7 Example of the Process Capability Platform with Nonnormal Variables Quality and Process Methods The Capability Index Plot shows Ppk values for all seven processes. Only two processes, Process 2 and Process 7, have capability values that exceed 2. Note that the best fitting nonnormal distributions are shown in parentheses to the right of the variable names in the Capability Index Plot. The best fitting distribution for Process 7 is not shown because it is a normal distribution.
11. Click the Process Capability red triangle and select Individual Detail Reports.
Because you requested Best Fit in the launch window, the Compare Distributions option has been selected from each distribution’s red triangle menu.
12. Scroll to the report entitled Process 4(Lognormal) Capability.
Chapter 7 Process Capability 179 Quality and Process Methods Example of the Process Capability Platform with Nonnormal Variables Figure 7.5 Individual Detail Report for Process 4 The title of the report for Process 4 indicates that the capability calculations are based on a lognormal fit. All of the check boxes in the Compare Distributions report, except the boxes 180 Process Capability Chapter 7 Example of the Process Capability Platform with Nonnormal Variables Quality and Process Methods for Nonparametric and Beta, are checked, indicating that these nine distributions are fit. (This is because you requested a Best Fit in the launch window.) The button that is selected in the Selected column indicates that the Lognormal distribution is the distribution that is used in the remainder of the Process 4(Lognormal) Capability report to estimate capability and nonconformance.
The Compare Distributions report enables you to compare the nine distributional fits. The Histogram - Compare Distributions report gives a visual assessment of the fit and the Comparison Details report shows fit statistics for the selected distributions. Both the plot and the fit statistics indicate that the lognormal distribution gives the best fit among the selected distributions.
The Individual Detail Report information that is shown by default includes a histogram showing the estimated best-fit distribution, a summary of the process information, capability indices based on an overall estimate of sigma, parameter estimates for the fitted lognormal distribution, and observed and expected nonconformance levels.
Chapter 7 Process Capability 181 Quality and Process Methods Launch the Process Capability Platform Launch the Process Capability Platform Launch the Process Capability Platform by selecting Analyze > Quality and Process > Process Capability. In Figure 7.6, which uses the Semiconductor Capability.jmp data table, all outlines and panels have been opened.
Figure 7.6 Process Capability Launch Window For more information about the options in the Select Columns red triangle menu, see Using JMP.
The Process Capability launch window contains the following outlines and options: • “Process Selection” • “Process Subgrouping” 182 Process Capability Chapter 7 Launch the Process Capability Platform Quality and Process Methods • “Moving Range Options” • “Historical Information” • “Distribution Options” • “Other Specifications” After you click OK in the launch window, the Spec Limits window appears unless one of the following occurs: • All of the columns contain specification limits.
• You selected No (skip columns with no spec limits) on the launch window.
The Spec Limits window also appears if you select Yes on the launch window. Otherwise, the Process Capability report window appears.
Process Selection Select the process variables to include in the capability analysis.
Y, Process Assigns the variables that you want to analyze.
Notes: • The Transform menu is not available for the Select Column list in the Process Capability launch window. Right-click a column heading in the data table and select New Formula Column to create a transform column for use in Process Capability. See Using JMP for more information about creating new formula columns.
• Reference columns for virtually joined tables are not available in the Process Capability platform.
Process Subgrouping This group of options enables you to assign each variable in the Y, Process list a subgroup ID column or a constant subgroup size.
Create Subgroups Using an ID Column 1.
Select a variable or variables in the Y, Process list.
2.
Select Subgroup ID Column from the Subgroup with options.
3.
Select a subgroup ID column in the Select Columns list.
4.
Click Nest Subgroup ID Column.
The subgroup ID column appears in brackets to the right of the variable names in the Y, Process list.
Chapter 7 Process Capability 183 Quality and Process Methods Launch the Process Capability Platform Create Subgroups Using a Constant Subgroup Size 1.
Select a variable or variables in the Y, Process list.
2.
Select Constant Subgroup Size from the Subgroup with options.
3.
Enter the subgroup size next to Set Constant Subgroup Size.
4.
Click Subgroup by Size.
The subgroup size appears in brackets to the right of the variable names in the Y, Process list.
Nest Subgroup ID Column (Available when you select Subgroup ID Column.) Assigns a column that you select from the Select Columns list to define the subgroups for the selected Y, Process columns.
Subgroup by Size (Available when you select Constant Subgroup Size.) Assigns the subgroup size that you specify in the Set Constant Subgroup Size box to define the subgroups for the selected Y, Process columns.
Set Constant Subgroup Size (Available when you select Constant Subgroup Size.) Specify the constant subgroup size for the selected Y, Process columns. You need to assign this value using Subgroup by Size.
Within-Subgroup Variation Statistic (Available when Process Subgrouping is used.) Specifies if the within-subgroup estimate of standard deviation is calculated using standard deviations or ranges.
Calculate Between-and-Within Capability (Available when Process Subgrouping is used.) Specifies that the between-and-within subgroup estimate of the standard deviation should be used in the capability analysis.
Moving Range Options Use this outline to specify which moving range statistic is used in the within sigma estimate when subgrouping is not used.
Note: When you specify subgrouping and click Calculate Between-and-Within Capability, use the Moving Range Options outline to specify which moving range statistic is used in the between sigma estimate.
Average of Moving Range Uses the mean of the moving ranges to estimate sigma. The moving range is the difference between two consecutive points.
Median of Moving Range Uses the median of the moving ranges to estimate sigma.
184 Process Capability Chapter 7 Launch the Process Capability Platform Quality and Process Methods Historical Information Use this outline to assign historically accepted values of the standard deviation to variables in the Y, Process list.
1.
Select a variable or variables in the Y, Process list.
2.
Enter a value next to Set Historical Sigma.
3.
Select Use Historical Sigma to assign that value to the selected variables.
The specified value appears in parentheses in the expression “&Sigma()” to the right of the variable names in the Y, Process list.
Note: If you set a historical sigma, then subgroup assignments for the selected process variable are no longer relevant and are removed.
Distribution Options Unless otherwise specified, all Y, Process variables are analyzed using the assumption that they follow a normal distribution. Use the Distribution Options outline to assign other distributions or calculation methods to variables in the Y, Process list and to specify options related to nonnormal calculations.
• The available distributions are the Normal, Beta, Exponential, Gamma, Johnson, Lognormal, Mixture of 2 Normals, Mixture of 3 Normals, SHASH, and Weibull distributions. Except for Johnson distributions, maximum likelihood estimation is used to fit distributions. See “Johnson Distribution Fit Method”.
• The Best Fit option determines the best fit among the available distributions and applies this fit.
• The Nonparametric option fits a distribution using kernel density estimation.
For more options related to nonnormal fits, see “Nonnormal Distribution Options”.
Specify a Distribution 1.
Select a variable or variables in the Y, Process list.
2.
Select a distribution from the Distribution list.
3.
Select Set Process Distribution to assign that distribution to the selected variables.
The specified distribution appears in parentheses in the expression “&Dist()” to the right of the variable names in the Y, Process list.
Chapter 7 Process Capability 185 Quality and Process Methods Launch the Process Capability Platform Note: If you select a distribution other than Normal, you cannot assign a Subgroup ID column or a Historical Sigma. These selections are not supported by the methods used to calculate nonnormal capability indices. See “Capability Indices for Nonnormal Distributions: Percentile and Z-Score Methods”.
Nonnormal Distribution Options Nonnormal Capability Indices Method Specifies the method used to compute capability indices for nonnormal distributions. See “Capability Indices for Nonnormal Distributions: Percentile and Z-Score Methods”.
Johnson Distribution Fit Method Specifies the method used to find the best-fitting Johnson distribution. Before estimating the parameters, the best-fitting family of distributions is determined from among the Johnson Su, Sb, and Sl families. The procedure described in Slifker and Shapiro (1980) is used to find the best-fitting family.
Quantile Matching The default method. It is more stable and faster than Maximum Likelihood. Quantile Matching Parameter estimates, assuming the best-fitting family, are obtained using a quantile-matching approach. See Slifker and Shapiro (1980).
Maximum Likelihood Parameters for the best-fitting family are determined using maximum likelihood.
Distribution Comparison Criterion (Available when a Best Fit Distribution is selected.) Specify the criterion that you want to use in determining a Best Fit. This criterion also determines the ordering of distributions in the Comparison Details report. See “Order by Comparison Criterion”.
Other Specifications By Produces a separate report for each level of the By variable. If more than one By variable is assigned, a separate report is produced for each possible combination of the levels of the By variables.
Specify Alpha Level Specifies the significance level for confidence limits.
Show Spec Limits Dialog Specifies how to handle columns that do not have specification limits.
186 Process Capability Chapter 7 Entering Specification Limits Quality and Process Methods Note: It is good practice to ensure that specification limits for all process variables are specified as Spec Limits column properties or to load specification limits from a Limits Data table (see “Limits Data Table”). Otherwise, you can specify limits interactively in the Spec Limits window that appears after you click OK in the launch window (unless you select No (skip columns with no spec limits) on the launch window).
Entering Specification Limits The lower specification limit (LSL), upper specification limit (USL), and target define the lower bound, upper bound, and target value for a quality process.
There are several ways to enter specification limits: • Enter limits in the Spec Limits window after selecting columns in the launch window. See “Spec Limits Window”.
• Import limits from a JMP data table (known as a Limits Table). See “Limits Data Table”.
• Enter limits as Spec Limits column properties in the data table. See “Spec Limits Column Property”.
• If you are creating a Process Capability report by running a JSL script, enter limits in the script. See “The Process Capability Report”.
Only one specification limit is required for a selected column. If only the USL is specified, the box plots and Goal Plot point are colored blue. If only the LSL is specified, the box plots and Goal Plot point are colored red.
Spec Limits Window After you click OK on the launch window, the Spec Limits window appears if any of the columns do not contain limits and you did not select No (skip columns with no spec limits) on the launch window. The Spec Limits window also appears if you select Yes on the launch window. Figure 7.7 shows the Spec Limits window for the Cities.jmp sample data table after selecting OZONE, CO, SO2, and NO as process variables in the launch window. Enter the known specification limits and click OK to view the Process Capability report.
You can specify process importance values for each column. Process importance values provide a mechanism to sort processes in the order that you prefer. Process importance values are used to size markers in many of the graphs in the Process Capability report.
Chapter 7 Process Capability 187 Quality and Process Methods Entering Specification Limits If you select the Show Limits option for a process and then save the specification limits to a column property, the Show as Graph Reference Lines option is selected in the saved Spec Limits column property. If you select the Show Limits option for a process and then save the specification limits to a new table, the Show Limits column in the new table contains a 1 for the process. The Select All Show Limits button selects the Show Limits option for all processes.
Figure 7.7 Spec Limits Window for Cities.jmp Limits Data Table You can also specify a limits data table with the Load spec limits from data table option from the Spec Limits window. Click the Select Data Table button and then select the appropriate data table that contains the specification limits for the analysis. After you select the appropriate limits table, the values populate the window. Click OK to view the Process Capability report.
A limits data table can be in two different formats: tall or wide. A tall limits data table has one column for the responses and the limits key words are the other columns. A wide limits data table has a column for each response with one column to label the limits keys. Either of these formats can be read using the Load spec limits from data table option.
• A tall table contains four or five columns and has one row for each process. The first column has a character data type and contains the names of the columns analyzed in the Process Capability platform. The next three columns need to be named LSL, Target, and USL. These column names can also be preceded by an underscore character. The optional final column named Show Limits specifies if the specification limits are shown as reference lines in select analysis plots.
188 Process Capability Chapter 7 Entering Specification Limits Quality and Process Methods Figure 7.8 Example of a Tall Specification Limits Table • A wide table contains three rows and one column for each column analyzed in the Process Capability platform plus a _LimitsKey column. In the _LimitsKey column, the three rows need to contain the identifiers _LSL, _Target, and _USL.
Figure 7.9 Example of a Wide Specification Limits Table The easiest way to create a limits data table is to save results computed by the Process Capability platform. The Save options in the Process Capability red triangle menu enable you to save limits from the sample values. After entering or loading the specification limits, you can do the following: • Select Save > Save Spec Limits as Column Properties to save the limits as Spec Limits column properties to the columns in the data table.
• Select Save > Save Distributions as Column Properties to save the distributions used in calculating capability as Process Capability Distribution column properties to the columns in the data table.
• Select Save > Save Spec Limits to New Table to save the limits to a new tall specification limits data table. If you have selected at least one nonnormal distribution, a column called Distribution that contains the specified distributions is also added to the limits data table.
See “Process Capability Platform Options”.
Chapter 7 Process Capability 189 Quality and Process Methods The Process Capability Report Spec Limits Column Property When you perform a capability analysis, you can use Column Properties > Spec Limits to save specification limits as a column property. The Spec Limits property applies only to numeric columns.
Some processes have one-sided specifications. Some have no target. You can enter any of these that apply: a lower specification limit, an upper specification limit, a target value, or a process importance value.
Figure 7.10 displays the Spec Limits section of the Column Properties window for OZONE in the sample data table Cities.jmp.
Figure 7.10 Spec Limits Section of the Column Properties Window Tip: Saving specification limits as a column property ensures consistency when you repeat an analysis.
The Process Capability Report By default, the Process Capability platform provides the following reports: • “Goal Plot” (provided only if at least one variable is fit with a normal distribution and shows only points for variables fit with normal distributions) 190 Process Capability Chapter 7 The Process Capability Report Quality and Process Methods • “Capability Box Plots” (provided only if at least one variable is fit with a normal distribution and shows only box plots for variables fit with normal distributions) • “Capability Index Plot” Figure 7.2 shows an example of a default Process Capability report.
Using the Process Capability red triangle menu, you can add individual detail reports, normalized box plots, and summary reports. The red triangle menu also has options for identifying out-of-spec values in your data table, creating a summary data table, changing the display order of analyzed columns, and saving out spec limits. These options are described in “Process Capability Platform Options”.
You can change the default report at File > Preferences > Platforms > Process Capability. You can also make changes to the appearance of reports produced by options by selecting the relevant Process Capability topic at File > Preferences > Platforms.
Goal Plot The Goal Plot shows, for each variable, the spec-normalized mean shift on the x-axis, and the spec-normalized standard deviation on the y-axis. It is useful for getting a quick, summary view of how the variables are conforming to specification limits. By default, the Goal Plot shows only those points for each column that are calculated using the overall sigma. Hover over each point to view the variable name and the sigma method used to calculate the point. See “Goal Plot” for more information about the calculation of the coordinates for the Goal Plot.
Note: Process variables with distributions other than Normal are not plotted on the Goal Plot.
Goal Plot Points Points on the Goal Plot correspond to columns, not rows. Selecting a point in the Goal Plot selects the corresponding column in the data table. If process importance values are specified, the goal plot points are sized by importance.
Hover over a point in the Goal Plot to view a control chart for that process. Click the control chart to launch Control Chart Builder with the corresponding control chart and capability report.
Note: A control chart is not available for a process if the unbiased pooled standard deviation is chosen as the within-group variation statistic for that process.
Chapter 7 Process Capability 191 Quality and Process Methods The Process Capability Report The points on the Goal Plot are also linked to the rows of the Goal Plot Summary Table, where each row corresponds to a column. You can select a point in the Goal Plot, right-click, and apply row states. These row states are applied to the rows of the Goal Plot Summary Table. Row states that you apply in the Goal Plot Summary Table are reflected in the Goal Plot. To see this table, select Make Goal Plot Summary Table from the Process Capability red triangle menu. See “Make Goal Plot Summary Table”.
Tip: If you hide a point in the Goal Plot, you can show the point again by changing the corresponding row state in the Goal Plot Summary Table.
Goal Plot Triangle The goal plot triangle appears in the center of the bottom of the Goal Plot. The slider to the right of the plot enables you to adjust the size of goal triangle in the plot.
By default, the Ppk slider and the value beneath it are set to Ppk = 1. This approximates a non-conformance rate of 0.0027, if the distribution is normal. The goal triangle represents the Ppk shown in the box. To change the Ppk value, move the slider or enter a number in the box.
JMP gives the Goal Plot in terms of Ppk values by default. You can change this preference at File > Preferences > Platforms > Process Capability. When the AIAG (Ppk) Labeling preference is unchecked, all of the Ppk labeling is changed to Cpk labeling, including the label of the slider to the right of the goal plot.
Goal Plot Options The Goal Plot red triangle menu has the following options: Show Within Sigma Points Shows or hides the points calculated using the within sigma estimate.
Show Within or Between-and-Within Sigma Points (Available only when Calculate Between-and-Within Capability is selected for at least one process in the launch window.) Shows or hides the points calculated using the within sigma estimate or, if specified, the between-and-within sigma estimate.
Show Overall Sigma Points Shows or hides the points calculated using the overall sigma estimate.
Shade Levels Shows or hides the Ppk level shading (Figure 7.11). When you select Shade Levels, shaded areas appear in the plot. The shaded areas depend on the relationship between p and Ppk, with p representing the value shown in the box beneath Ppk: – Points in the red area have Ppk < p.
– Points in the yellow area have p < Ppk < 2p.
192 Process Capability Chapter 7 The Process Capability Report Quality and Process Methods – Points in the green area have 2p < Ppk.
Label Within Sigma Points Shows or hides labels for points calculated using the within sigma estimate.
Label Within or Between-and-Within Sigma Points (Available only when Calculate Between-and-Within Capability is selected for at least one process in the launch window.) Shows or hides labels for points calculated using the within sigma estimate or, if specified, the between-and-within sigma estimate.
Label Overall Sigma Points Shows or hides labels for points calculated using the overall sigma estimate.
Defect Rate Contour Shows or hides a contour representing a specified defect rate.
Figure 7.11 shows the Goal Plot for the entire data set for the Semiconductor Capability.jmp sample data table after selecting Shade Levels and Show Within Sigma Points from the Goal Plot red triangle menu.
Figure 7.11 Goal Plot One-Sided or Missing Specification Limits When there is only one specification limit for a column, markers and colors are used in the following ways: • If only the upper specification limit (USL) is specified, the point on the Goal Plot is represented by a right-pointing triangle and is colored blue.
Chapter 7 Process Capability 193 Quality and Process Methods The Process Capability Report • If only the lower specification limit (LSL) is specified, the point on the Goal Plot is represented by a left-pointing triangle and is colored red.
• If at least one process has only an upper specification limit, the right half of the goal triangle is blue.
• If at least one process has only a lower specification limit, the left half of the goal triangle is red.
Processes with only an upper specification limit are represented by blue and should be compared to the blue (right) side of the goal triangle. Processes with only a lower specification limit are represented by red and should be compared to the red (left) side of the goal triangle. For more information about how the coordinates of points are calculated, see “Goal Plot”.
Capability Box Plots The Capability Box Plots show a box plot for each variable selected in the analysis. The values for each column are centered by their target value and scaled by the difference between the specification limits. If the target is not centered between the specification limits, the values are scaled by twice the minimum difference between the target and specification limits. For each process column Yj (see “Notation for Goal Plots and Capability Box Plots” for a description of the notation): For a process with a one-sided specification, see “One-Sided or Missing Specification Limits”. For the situation where no target is specified, see “Capability Box Plots for Processes with Missing Targets”.
Note: Process variables with distributions other than Normal are not plotted on the Capability Box Plot.
Figure 7.11 shows a Capability Box Plots report for eight variables in the Semiconductor Capability.jmp sample data table.
Zij Yij Tj – 2 min(Tj LSLj – USL j Tj) – ---------------------------------------------------------------------------= 194 Process Capability Chapter 7 The Process Capability Report Quality and Process Methods Figure 7.12 Capability Box Plot The plot displays dotted green lines drawn at ±0.5.
• For a process with a target that is centered between its specification limits, the dotted green lines represent the standardized specification limits.
• For a process with a target that is not centered between its specification limits, one of the dotted green lines represents the standardized specification limit for the limit closer to the target. The other dotted green line represents the same distance in the opposite direction.
This plot is useful for comparing variables with respect to their specification limits. For example, in Figure 7.12, the majority of points for IVP1 are above its USL, and the majority of its points for IVP2 are less than its target. PNP2 seems to be on target with all data points in the specification limits.
One-Sided or Missing Specification Limits When there is only one specification limit for a column, colors are used in the following ways: • If only the upper specification limit (USL) is specified, the box plot is colored blue.
• If only the lower specification limit (LSL) is specified, the box plot is colored red.
• If at least one process has only an upper specification limit, the dotted line at 0.5 is blue.
• If at least one process has only a lower specification limit, the dotted line at -0.5 is red.
Chapter 7 Process Capability 195 Quality and Process Methods The Process Capability Report Suppose that only the lower specification limit is specified and that the process target is specified. The capability box plot is based on the following values for the transformed observations. See “Notation for Goal Plots and Capability Box Plots” for a description of the notation: Suppose that only the upper specification limit is specified and that the process target is specified. The capability box plot is based on the following values for the transformed observations: For more information about how missing targets are handled with one-sided specification limits, see “Single Specification Limit and No Target”.
Capability Index Plot The Capability Index Plot shows Ppk values for all variables that you entered as Y, Process. Each variable name appears on the horizontal axis and the Ppk values appear on the vertical axis. If you fit a nonnormal distribution, the fitted distribution name appears in the plot as a parenthetical suffix to the variable name. If process importance values are specified, the points on the capability index plot are sized by importance. A horizontal line is placed at the Ppk value that is specified by the slider to the right of the plot.
Hover over a point in the Capability Index Plot to view a control chart for that process. Click the control chart to launch Control Chart Builder with the corresponding control chart and capability report.
Note: A control chart is not available for a process if the unbiased pooled standard deviation is chosen as the within-group variation statistic for that process.
Figure 7.13 shows a Capability Index Plot report for the Process Measurements.jmp sample data table. Six of the variables are fit with nonnormal distributions. Process 7 is fit with a normal distribution. Points have been labeled using the Label Overall Sigma Points option that is available in the Capability Index Plot red triangle menu.
Zij Yij Tj – 2 Tj LSLj – -------------------------------= Zij Yij Tj – 2 USLj Tj – --------------------------------= 196 Process Capability Chapter 7 The Process Capability Report Quality and Process Methods Figure 7.13 Capability Index Plot with Nonnormal Distributions Capability Index Plot Options The Capability Index Plot red triangle menu has the following options: Show Within Sigma Points Shows or hides the points calculated using the within sigma estimate.
Show Within or Between-and-Within Sigma Points (Available only when Calculate Between-and-Within Capability is selected for at least one process in the launch window.) Shows or hides the points calculated using the within sigma estimate or, if specified, the between-and-within sigma estimate.
Show Overall Sigma Points Shows or hides the points calculated using the overall sigma estimate.
Shade Levels Shows or hides the Ppk level shading. When you select Shade Levels, shaded areas appear in the plot. The shaded areas depend on the relationship between p and Ppk, with p representing the value shown in the box beneath Ppk: – Points in the red area have Ppk < p.
– Points in the yellow area have p < Ppk < 2p.
– Points in the green area have 2p < Ppk.
Chapter 7 Process Capability 197 Quality and Process Methods Process Capability Platform Options Label Within Sigma Points Shows or hides labels for points calculated using the within sigma estimate.
Label Within or Between-and-Within Sigma Points (Available only when Calculate Between-and-Within Capability is selected for at least one process in the launch window.) Shows or hides labels for points calculated using the within sigma estimate or, if specified, the between-and-within sigma estimate.
Label Overall Sigma Points Shows or hides labels for points calculated using the overall sigma estimate.
Process Capability Platform Options The Process Capability red triangle menu contains the following options: Individual Detail Reports Shows or hides individual detail reports for each variable in the analysis. See “Individual Detail Reports”.
Goal Plot Shows or hides a goal plot for the data. The Goal Plot shows the spec-normalized mean shift on the x-axis and the spec-normalized standard deviation on the y-axis for each variable. See “Goal Plot”. (Only variables for which you specify normal distributions are shown on the plot.) Capability Box Plots Shows or hides a capability box plot for each variable in the analysis. The values for each column are centered by their target value and scaled by twice the minimum difference between the target value and the specification limits. See “Capability Box Plots”. (Box plots are shown only for variables for which you specify normal distributions.) Normalized Box Plots Provides two options for plots that show normalized box plots for each process variable. Each column is standardized by subtracting its mean and dividing by an estimate of the column’s standard deviation. The box plot is constructed using quantiles for the standardized values. See “Normalized Box Plots”. (Normalized box plots are shown only for variables for which you specify normal distributions.) Within Sigma Normalized Box Plots Shows or hides a plot called Within Sigma Normalized Box Plots. The box plots are constructed using the within-subgroup estimate of standard deviation.
Within or Between-and-Within Sigma Normalized Box Plots (Available only when Calculate Between-and-Within Capability is selected for at least one process in the launch window.) Shows or hides a plot called Within or Between-and-Within Normalized Box Plots. The box plots are constructed using the within group estimate of the standard deviation or, if specified, the between-and-within estimate.
198 Process Capability Chapter 7 Process Capability Platform Options Quality and Process Methods Overall Sigma Normalized Box Plots Shows or hides a plot called Overall Sigma Normalized Box Plots. The box plots are constructed using the overall estimate of standard deviation.
Capability Index Plot Shows overall Ppk values for all variables that you entered as Y, Process. See “Capability Index Plot”.
Process Performance Plot Shows or hides a four-quadrant plot of capability versus stability. Each process that has at least one specification limit is represented by a point. See “Process Performance Plot”.
Summary Reports Provides two options for summary reports of capability indices. See “Summary Reports”.
Within Sigma Summary Report Shows or hides a summary report of capability indices calculated using the within-subgroup estimate of standard deviation. (Results are available only for variables with specified normal distributions.) Within or Between-and-Within Sigma Summary Report (Available only when Calculate Between-and-Within Capability is selected for at least one process in the launch window.) Shows or hides a summary report of capability indices calculated using the within group estimate of the standard deviation or, if specified, the between-and-within group estimate.
Overall Sigma Summary Report Shows or hides a summary report of capability indices calculated using the overall estimate of standard deviation.
Action Options The following red triangle menu options perform actions: Out of Spec Values Provides options for the cells in the data table containing values that are out of spec.
Select Out of Spec Values Selects all rows and columns in the data table that contain at least one value that does not fall within the specification limits.
Color Out of Spec Values Colors the cells in the data table that correspond to values that are out of spec. The cell is colored blue if the value is above the USL and red if the value is below the LSL.
Tip: To remove colors in specific cells, select all cells of interest. Right-click in one of the cells and select Clear Color. To remove colors in all cells, deselect Color Out of Spec Values.
Make Goal Plot Summary Table Creates a summary table for the points plotted in the Goal Plot. This table includes the variable’s name, its spec-normalized mean shift, and its Chapter 7 Process Capability 199 Quality and Process Methods Process Capability Platform Options spec-normalized standard deviation. Each variable has two rows in this table: one for each sigma type (within and overall). See “Make Goal Plot Summary Table”.
Order By Reorders the box plots, summary reports, and individual detail reports. You can reorder by Initial Order, Reverse Initial Order, Within Sigma Cpk Ascending, Within or Between-and-Within Sigma Cpk Ascending, Within Sigma Cpk Descending, Within or Between-and-Within Sigma Cpk Descending, Overall Sigma Ppk Ascending, or Overall Sigma Ppk Descending. The options that order by Within Sigma reorder plot elements only for variables with specified normal distributions.
Note: The options to order by Within or Between-and-Within Sigma are available only if Calculate Between-and-Within Capability is selected for at least one process in the launch window.
Save Provides options for saving specification limits and distributions.
Save Spec Limits as Column Properties Saves the specification limits to a column property for each variable in the analysis. If no Spec Limits column property is present, the column property is created. If a Spec Limits column property is present, the values in the column property are overwritten. See “Spec Limits Column Property”.
Save Distributions as Column Properties Saves the distribution used in calculating capability as a Process Capability Distribution column property. See Using JMP.
If a column contains the Distribution property specifying a nonnormal distribution and no Process Capability Distribution property, then the Process Capability platform applies a nonnormal fit. The Process Capability platform uses the distribution specified in the Distribution column property, or a Johnson fit if that distribution is not supported in Process Capability. If a column contains the Process Capability Distribution property, then the Process Capability platform uses the distribution specified in the Process Capability Distribution column property.
Note: If you want to use a specific distribution in the Process Capability platform, save it as a Process Capability Distribution column property.
Save Spec Limits to New Table Saves the specification limits and the setting for Show Limits for each process to a limits data table in tall format. See “Limits Data Table”.
Relaunch Dialog Opens the platform launch window and recalls the settings used to create the report.
See Using JMP for more information about the following options: Local Data Filter Shows or hides the local data filter that enables you to filter the data used in a specific report.
200 Process Capability Chapter 7 Process Capability Platform Options Quality and Process Methods Redo Contains options that enable you to repeat or relaunch the analysis. In platforms that support the feature, the Automatic Recalc option immediately reflects the changes that you make to the data table in the corresponding report window.
Save Script Contains options that enable you to save a script that reproduces the report to several destinations.
Save By-Group Script Contains options that enable you to save a script that reproduces the platform report for all levels of a By variable to several destinations. Available only when a By variable is specified in the launch window.
Individual Detail Reports The Individual Detail Reports option displays a capability report for each variable in the analysis.
Normal Distributions Figure 7.14 shows the Individual Detail Report for PNP1 from the Semiconductor Capability.jmp sample data table as described in “Example of the Process Capability Platform with Normal Variables”.
Chapter 7 Process Capability 201 Quality and Process Methods Process Capability Platform Options Figure 7.14 Individual Detail Report The Individual Details report for a variable with a normal distribution shows a histogram, process summary details, and capability and nonconformance statistics. The histogram shows the distribution of the values, the lower and upper specification limits and the process target (if they are specified), and one or two curves showing the assumed distribution. The histogram in Figure 7.14 shows two normal curves, one based on the overall estimate of standard deviation and the other based on the within-subgroup estimate.
When you fit your process with a normal distribution, the Process Summary includes the Stability Index, which is a measure of stability of the process. The stability index is defined as follows: (Overall Sigma/Within Sigma) If Calculate Between-and-Within Capability is specified for a process in the launch window, the stability index for that process is defined as follows: (Overall Sigma/Between-and-Within Sigma) A stable process has stability index near one. Higher values indicate less stability.
202 Process Capability Chapter 7 Process Capability Platform Options Quality and Process Methods Note: You can change the preferences for stability assessment type in File > Preferences > Platforms > Process Capability. This changes the stability assessment type used through the Process Capability platform.
Nonnormal Distributions Note: Capability indices based on within-subgroup variation and stability indices are not available for processes for which you have specified nonnormal distributions.
Figure 7.15 shows the Individual Detail Report for Process 1 from the Process Measurements.jmp sample data table as described in “Example of the Process Capability Platform with Nonnormal Variables”.
Figure 7.15 Individual Detail Report for Process 1 The report opens with a note summarizing the Nonnormal Distribution Options that you selected in the launch window.
Chapter 7 Process Capability 203 Quality and Process Methods Process Capability Platform Options The Individual Details report for a variable with a nonnormal distribution shows a histogram, process summary details, and capability and nonconformance statistics. The histogram shows the distribution of the values, the lower and upper specification limits and the process target (if they are specified). A curve showing the fitted distribution is superimposed on the histogram. If you selected a Nonparametric distribution, the curve shown in the histogram is the nonparametric density.
The report also shows a Parameter Estimates report if you selected a nonnormal parametric distribution or a Nonparametric Density report if you selected a Nonparametric fit. See “Parameter Estimates” and “Nonparametric Density”.
Individual Detail Report Options The outline title for each variable in the Individual Detail Reports section is of the form Capability. However, if you request nonnormal capability, the relevant distribution name is shown parenthetically in the outline title.
Each Capability report has a red triangle menu with the following options: Compare Distributions Shows or hides the control panel for comparing distributions for the process. See “Compare Distributions”.
Process Summary Shows or hides the summary statistics for the variable, including the overall sigma estimate, and, if you have specified a normal distribution, the within sigma estimate and the stability index. If you have specified Calculate Between-and-Within Capability for at least one process in the launch window, estimates for the between sigma and the between-and-within sigma are also included.
Histogram Shows or hides the histogram of the values of the variable. The histogram report includes a red triangle menu that controls the following features of the histogram: Show Spec Limits Shows or hides vertical red lines on the histogram at the specification limits for the process.
Show Target Shows or hides a vertical green line on the histogram at the process target.
Show Within Sigma Density Shows or hides an approximating normal density function on the histogram with mean given by the sample mean and standard deviation given by the within estimate of sigma.
Show Between-and-Within Sigma Density (Available only when Calculate Between-and-Within Capability is selected for at least one process in the launch window.) Shows or hides an approximating normal density function on the histogram with mean given by the sample mean and standard deviation given by the between-and-within estimate of sigma.
204 Process Capability Chapter 7 Process Capability Platform Options Quality and Process Methods Show Overall Sigma Density Shows or hides an approximating normal density function on the histogram with mean given by the sample mean and standard deviation given by the overall estimate of sigma.
Show Count Axis Shows or hides an additional axis to the right of the histogram plot showing the count of observations.
Show Density Axis Shows or hides an additional axis to the right of the histogram plot showing the density.
Capability Indices Controls display of the following capability index reports: Within Sigma Capability (Available when distribution is Normal.) Shows or hides capability indices (and confidence intervals) based on the within (short-term) sigma.
Between-and-Within Sigma Capability (Available only when distribution is Normal and Calculate Between-and-Within Capability is selected for at least one process in the launch window.) Shows or hides capability indices based on the between-and-within sigma.
Within Sigma Z Benchmark (Available when distribution is Normal.) Shows or hides Z benchmark indices based on the within (short-term) sigma.
Between-and-Within Sigma Z Benchmark (Available only when distribution is Normal and Calculate Between-and-Within Capability is selected for at least one process in the launch window.) Shows or hides Z benchmark indices based on the between-and-within sigma.
Within Sigma Target Index (Available when distribution is Normal.) Shows or hides an estimate of the target index that is based on the within (short-term) sigma.
Between-and-Within Sigma Target Index (Available only when distribution is Normal and Calculate Between-and-Within Capability is selected for at least one process in the launch window.) Shows or hides an estimate of the target index that is based on the between-and-within sigma.
Overall Sigma Capability Shows or hides capability indices (and confidence intervals) based on the overall (long-term) sigma.
Overall Sigma Z Benchmark (Available when distribution is Normal.) Shows or hides Z benchmark indices based on the overall (long-term) sigma.
Note: By default, the confidence intervals for the capability indices are constructed based on = 0.05. To change the default confidence level, select File > Preferences > Platforms > Process Capability.
Chapter 7 Process Capability 205 Quality and Process Methods Process Capability Platform Options Nonconformance Shows or hides the observed and expected percentages of observations below the LSL, above the USL, and outside of the specification limits. The Nonconformance table contains hidden columns for observed and expected PPM and counts.
Interactive Capability Plot Shows or hides the Interactive Capability Plot. The Interactive Capability Plot enables you to change the value of one or more summary statistics and see how the changes affect the capability analysis. There are Original and New reports that show the original and new summary statistics, capability indices, and expected PPM. Use the slider controls or text boxes to change the spec limits, mean, and overall sigma from the original values. You can also use the Mean Shift box to shift the mean by a factor of the original sigma. The Interactive Capability Plot report has the following red triangle menu options: Capability Shows or hides the capability indices in the Original and New reports.
Expected PPM Shows or hides the expected PPM values in the Original and New reports.
Revert to Original Values Reverts the interactive capability plot and the summary values in the New report back to the original values.
Save New Spec Limits as a Column Property Saves the new specification limits as a Spec Limits column property to the column in the original data table.
Note: The analysis is not rerun with the new specification limits unless the Auto Recalc option is turned on.
Parameter Estimates (Available when a distribution other than Normal or Nonparametric is selected.) Shows or hides the Parameter Estimates report, which gives estimates for the parameters of the selected distribution.
The estimates for all except the Johnson family distributions are obtained using maximum likelihood. For more information about Johnson family fits, see “Johnson Distribution Fit Method”.
The parameters and probability density functions for the normal, beta, exponential, gamma, Johnson, lognormal, and Weibull distributions are described in “Capability Indices for Nonnormal Distributions: Percentile and Z-Score Methods”. These are the same parameterizations used in the Distribution platform, with the exception that Process Capability does not support threshold parameters. See Basic Analysis.
Fix Parameters (Available when a distribution other than Normal or Nonparametric is selected.) Displays a window that enables you to fix one or more parameter values in a nonnormal distribution. Enter a value in the User-defined Value column for the parameters that you would like to fix. Once you click OK, the omitted parameter values 206 Process Capability Chapter 7 Process Capability Platform Options Quality and Process Methods are re-estimated given the fixed parameter values. The re-estimated parameter values appear in the Parameter Estimates report, along with a column indicating which parameters are fixed.
Nonparametric Density (Available when Nonparametric is selected as the distribution.) Shows or hides the Nonparametric Density report, which gives the kernel bandwidth used in fitting the nonparametric distribution. The kernel bandwidth is given by the following, where n is the number of observations and S is the uncorrected sample standard deviation: Compare Distributions The Compare Distributions report enables you to compare and apply various distributional fits. Note the following: • Your selected distribution is indicated in the Selected column.
• The report initially shows fit statistics for your Selected distribution and other fitted distributions in the Comparison Details report. If you selected Best Fit, the Comparison Details report initially shows statistics for all parametric fits.
• Check the distributions in the Distribution list that you want to compare.
– The probability density function for the best fitting distribution in each family that you select is superimposed on the histogram in the Histogram - Compare Distributions report.
– If the distribution is parametric, a row for that family containing fit results is added to the Comparison Details report.
– If Nonparametric is checked in the Distribution list, the Nonparametric Density report, showing the automatically selected kernel bandwidth, is added to the Compare Distributions report. See “Nonparametric Density”.
– You can change your selected distribution by selecting its radio button under Selected. The capability report is updated to show results for the selected distribution.
Figure 7.16 shows the Compare Distributions report for Process 1 in the Process Measurements.jmp sample data table. The Selected distribution, which is Lognormal, is being compared to a Normal distribution. The Comparison Details report shows fit statistics for both distributions.
To obtain probability plots, click the Compare Distributions red triangle and select Probability Plots. The points in the probability plot for the normal distribution in Figure 7.16 do not follow the line closely. This indicates a poor fit.
bandwidth 0.9S n1 5 / -----------= Chapter 7 Process Capability 207 Quality and Process Methods Process Capability Platform Options Figure 7.16 Compare Distributions with Probability Plot for Normal Compare Distributions Options The Compare Distributions red triangle menu contains the following options: Comparison Details For each distribution, gives AICc, BIC, and -2Loglikelihood values. See Fitting Linear Models. (Not available for a Nonparametric fit.) Comparison Histogram Shows or hides the Histogram report.
Probability Plots Shows or hides a report that displays probability plots for each parametric distribution that you fit (Figure 7.16). An observation’s horizontal coordinate is its observed data value. An observation’s vertical coordinate is the value of the quantile of the 208 Process Capability Chapter 7 Process Capability Platform Options Quality and Process Methods fitted distribution for the observation’s rank. For the normal distribution, the overall estimate of sigma is used in determining the fitted distribution.
The red triangle menus associated with each Probability Plot contain the following options.
Simultaneous Empirical Confidence Limits Shows or hides confidence limits that have a simultaneous 95% confidence level of containing the true probability function, given that the data come from the selected parametric family. These limits have the same estimated precision at all points. Use them to determine whether the selected parametric distribution fits the data well. See Nair (1984) and Meeker and Escobar (1998).
Simultaneous Empirical Confidence Limits Shading Shows or hides shading of the region between the Simultaneous Empirical Confidence Limits.
Parametric Fit Line Shows or hides the line that shows the predicted probabilities for the observations based on the fitted distribution.
Parametric Fit Confidence Limits Shading Shows or hides shading of the region between parametric fit confidence intervals. The parametric fit confidence limits have confidence level (1 - Alpha), where Alpha is the value that you specify in the launch window. (Available only when the parametric fit confidence limits are meaningful and when it is possible to calculate them.) When possible, the intervals are computed by expressing the parametric distribution F as a location-scale family, so that F(y) = G(z), where z = (y - )/. The approximate standard error of the fitted location-scale component at a point is computed using the delta method. Using the standard error estimate, a Wald confidence interval for z is computed for each point. The confidence interval for the cumulative distribution function F is obtained by transforming the Wald interval using G. Note that, in some cases, special accommodations are required to provide appropriate intervals near the endpoints of the interval of process measurements.
Order by Comparison Criterion Orders the distributions in the Comparison Details report according to the criterion that you select. The default ordering is by AICc, unless you selected another criterion in the Distribution Comparison Criterion panel in the launch window.
Normalized Box Plots The Normalized Box Plots options show or hide box plots that have been normalized using the specified sigma in the title. When drawing normalized box plots, JMP standardizes each column by subtracting the mean and dividing by the standard deviation. The box plots are formed for each column using these standardized values.
Chapter 7 Process Capability 209 Quality and Process Methods Process Capability Platform Options Figure 7.17 Within Sigma Normalized Box Plot Figure 7.17 shows the Within Sigma Normalized Box Plot for a selection of the process variables in the Semiconductor Capability.jmp sample data table using wafer as a subgroup variable.
The green vertical lines represent the specification limits for each variable normalized by the mean and standard deviation of each variable. The gray dotted vertical lines are drawn at ±0.5, since the data is standardized to a standard deviation of 1.
Process Performance Plot The Process Performance Plot option shows or hides a four-quadrant plot of capability versus stability. Each process that has specification limits is represented by a point. If process importance values are specified, the points are sized by importance. The horizontal coordinate of each point equals the stability index of the process and the vertical coordinate of each point equals the overall Ppk capability of the process. The plot is divided into four shaded quadrants based on the following default boundaries: • A stability index that exceeds 1.25 indicates that the process is unstable.
• A Ppk that is smaller than 1.0 indicates that the process is not capable.
210 Process Capability Chapter 7 Process Capability Platform Options Quality and Process Methods Additionally, there is a red line on the graph that indicates where the Cpk value is 1. The boundaries that define the four quadrants can be adjusted using the Ppk and Stability Index slider controls to the right of the plot. You can also set preferences for your desired Capability and Stability boundaries, as well as stability assessment type in File > Preferences > Platforms > Process Performance Plot and File > Preferences > Platforms > Process Capability.
The legend contains descriptions of the shaded regions. If any of the processes are missing a lower or upper specification limit, the legend also shows the markers used for those processes. If the markers do not appear in the legend, then all of the processes in the plot contain both lower and upper specification limits. See “One-Sided or Missing Specification Limits”.
Hover over a point in the Process Performance Plot to view a control chart for that process. Click the control chart to launch Control Chart Builder with the corresponding control chart and capability report.
Note: A control chart is not available for a process if the unbiased pooled standard deviation is selected as the within-group variation statistic for that process.
The Process Performance Plot red triangle menu contains the following option: Label Points Shows or hides labels for each point in the Process Performance Plot.
Show Within Cpk Curve Shows or hides the within Cpk curve in the Process Performance Plot.
Figure 7.18 Process Performance Plot Chapter 7 Process Capability 211 Quality and Process Methods Process Capability Platform Options Figure 7.18 shows the Process Performance Plot for a selection of the process variables in the Semiconductor Capability.jmp sample data table using wafer as a subgroup variable.
Summary Reports The Summary Report options show or hide a table that contains the following statistics for each variable: LSL, Target, USL, Sample Mean, various Sigma estimates, Stability Index, Cpk, Cpl, Cpu, Cp, Cpm, and Nonconformance statistics. If there is at least one nonmissing process importance value, an Importance column is also included in the Summary Report. These statistics are calculated using the sigma estimate specified in the report title. The columns for Stability Index, Cpk, Cpl, Cpu, and Cp are colored as green, yellow, and red to indicate adequate, marginal, and poor stability or capability. This color coding scheme matches what you would see in the Process Performance graph.
Note: You can change the preferences for stability assessment type in File > Preferences > Platforms > Process Capability. This changes the stability assessment type used through the Process Capability platform.
Figure 7.19 shows a subset of columns for both summary reports as described in “Example of the Process Capability Platform with Normal Variables”. The following optional columns are available for this report: • Confidence intervals for Cpk, Cpl, Cpu, CP, and Cpm • Expected and observed PPM statistics (outside, below LSL, above USL) Note: The expected PPM statistics are the percentages you would expect to see based on the distribution chosen. By default, the distribution is normal. The observed PPM statistics are the percentages based on the actual data.
• Sample standard deviation • The sample size (N), the minimum, and the maximum.
• Target Index Note: Target Index is only available in the Within Sigma Capability Summary report.
To reveal these optional columns, right-click the report and select the column names from the Columns submenu.
Note that the report (based on overall sigma) shows the overall capability indices Ppk, Ppl, Ppu, and Pp instead of the within capability indices Cpk, Cpl, Cpu, and Cp. The labeling of the overall capability indices depends on the setting of the AIAG (Ppk) Labeling preference.
212 Process Capability Chapter 7 Process Capability Platform Options Quality and Process Methods Figure 7.19 Within Sigma and Overall Sigma Capability Summary Reports Make Goal Plot Summary Table The Make Goal Plot Summary Table option produces a summary data table that includes each variable’s name, its spec-normalized mean shift (Spec-Standardized Mean), and its spec-normalized standard deviation (Spec-Standardized Std Dev). For each variable, there is a row for each of the sigma types.
Note: If a variable is fit with a distribution other than normal, the name of the fitted distribution is appended parenthetically to the variable name. The Spec-Standardized Mean and Spec-Standardized Std Dev values are not provided for nonnormal variables.
The points in the Goal Plot are linked to the rows in the Goal Plot Summary Table. If you apply row states to a point in the Goal Plot, you can change the corresponding row states in the Goal Plot Summary Table. Conversely, if you apply row states in the Goal Plot Summary Table, they are reflected on the Goal Plot.
Figure 7.20 shows the Goal Plot Summary Table for the Semiconductor Capability.jmp sample data table as described in “Example of the Process Capability Platform with Normal Variables”.
Chapter 7 Process Capability 213 Quality and Process Methods Additional Examples of the Process Capability Platform Figure 7.20 Summary Table Additional Examples of the Process Capability Platform • “Process Capability for a Stable Process” • “Process Capability for an Unstable Process” • “Simulation of Confidence Limits for a Nonnormal Process Ppk” Process Capability for a Stable Process In this example, you verify the assumptions that enable you to estimate PPM defective rates based on a capability analysis. You access Process Capability through Control Chart Builder and then directly. The data consist of 22 subgroups of size five. There are six missing readings, with three in each of two consecutive subgroups.
Process Capability through Control Chart Builder You can use Control Chart Builder to check process stability and the normality assumption for your process characteristic. You can also obtain Process Capability information directly within Control Chart Builder.
1.
Select Help > Sample Data Library and open Quality Control/Clips2.jmp.
2.
Select Analyze > Quality and Process > Control Chart Builder.
3.
Drag Date to the Subgroup zone.
214 Process Capability Chapter 7 Additional Examples of the Process Capability Platform Quality and Process Methods 4.
Drag Gap to the Y zone.
Figure 7.21 XBar and R Control Chart for Gap The control chart indicates that Gap is stable over time. Because Gap has the Spec Limits column property, a Process Capability Analysis report appears to the right of the control chart.
Figure 7.22 Histogram in Process Capability Analysis for Gap The histogram and fitted normal blue curve suggest that the distribution of Gap is approximately normal. Although the process is stable, the distribution of Gap is shifted to the right of the specification range.
Chapter 7 Process Capability 215 Quality and Process Methods Additional Examples of the Process Capability Platform The Process Summary report shows the specification limits that are saved to the Spec Limits column property. It also shows that the estimate of sigma calculated from within-subgroup variation (Within Sigma) does not differ greatly from the overall estimate given by the sample standard deviation (Overall Sigma). Consequently, the Stability Index is near one (0.979268). This is expected because the process is stable.
5.
Right-click in the body of the Nonconformance report and select Expected Within PPM from the Columns submenu.
Figure 7.23 Capability Indices and Nonconformance Report The Cpk value calculated using subgroup variation is 0.966, indicating that the process is not very capable. The Cpl value suggests good performance, but this is because the process is shifted away from the lower specification limit. Defective parts generally result from large values of Gap.
Note that the confidence interval for Cpk is wide; it ranges from 0.805 to 1.128. This occurs even though there are 104 observations. Capability indices are surprisingly variable, due to the fact that they are ratios. It is easy to reach incorrect conclusions based on the point estimate of a capability index.
The estimates of out-of-specification product given in the Nonconformance report provide a direct measure of process performance. The PPM values in the Nonconformance report indicate that Gap hardly ever falls below the lower specification limit (1.4 parts per million). However, the number of parts for which Gap falls above the upper specification limit is 1869.0 parts per million.
For an uncentered process, the Cp value indicates potential capability if the process were adjusted to be centered. If this process were adjusted to be centered at the target value of 14.8, then its capability would be 1.264, with a confidence interval from 1.071 to 1.457.
Process Capability Platform Now that you have verified stability and normality for Gap, you can obtain additional information in the Process Capability platform.
1.
Select Analyze > Quality and Process > Process Capability.
216 Process Capability Chapter 7 Additional Examples of the Process Capability Platform Quality and Process Methods 2.
Select Gap and click Y, Process.
3.
Open the Process Subgrouping outline.
4.
Select Date in the Select Columns list and Gap in the Roles list.
5.
Click Nest Subgroup ID Column.
By default, the Within-Subgroup Variation Statistic selection is set to Average of Unbiased Standard Deviations. In the Control Chart Builder example (“Process Capability through Control Chart Builder”), subgroup ranges were used.
6.
Click OK.
Figure 7.24 Goal Plot and Box Plot for Gap The Goal Plot shows the Ppk index for Gap as being essentially equal to 1. The box plot shows that most values fall within specifications, but the preponderance of data values are shifted to the right within the specification range.
7.
Click the Process Capability red triangle and select Individual Detail Reports.
The report is the one obtained using Control Chart Builder, except that the Within Sigma is based on average standard deviations rather than average ranges. See “Histogram in Chapter 7 Process Capability 217 Quality and Process Methods Additional Examples of the Process Capability Platform Process Capability Analysis for Gap” and “Capability Indices and Nonconformance Report”.
Process Capability for an Unstable Process This example shows a case where the overall variation differs from the within variation because the process is not stable. It uses the Coating.jmp data table from the Quality Control folder of Sample Data (taken from the ASTM Manual on Presentation of Data and Control Chart Analysis). The process variable of interest is the Weight column, grouped into subgroups by the Sample column.
1.
Select Help > Sample Data Library and open Quality Control/Coating.jmp.
2.
Select Analyze > Quality and Process > Process Capability.
3.
Select Weight and click Y, Process.
4.
Open the Process Subgrouping outline.
5.
Select Sample in the Select Columns list on the left.
6.
Select Weight in the Cast Selected Columns into Roles list on the right.
7.
Click Nest Subgroup ID Column.
8.
Click OK.
9.
Enter 16 for LSL, 20 for Target, and 24 for USL in the Spec Limits window.
10. Click OK.
11. Click the Goal Plot red triangle and select Show Within Sigma Points.
12. Click the Process Capability red triangle and select Individual Detail Reports.
218 Process Capability Chapter 7 Additional Examples of the Process Capability Platform Quality and Process Methods Figure 7.25 Process Capability Report for Coating.jmp Data Chapter 7 Process Capability 219 Quality and Process Methods Additional Examples of the Process Capability Platform Figure 7.25 shows the resulting Process Capability report. The Goal Plot shows two points that represent the mean shift and standard deviation standardized to the specification limits. There is a legend next to the Goal Plot that identifies the two points. The Overall Sigma point is calculated using the overall sample standard deviation. The Within Sigma point is calculated using a within-subgroup estimate of the standard deviation.
The point calculated using Overall Sigma is outside the goal triangle corresponding to a Ppk of 1. This indicates that the variable Weight results in non-conforming product.
However, the point calculated using Within Sigma is inside the goal triangle. This indicates that, if the process were stable, Weight values would have a high probability of falling within the specification limits.
13. Hover cursor over one of the points in the Goal Plot.
14. Click the control chart.
Figure 7.26 XBar and R Chart for Weight The control chart indicates that the Weight measurements are unstable. The process is affected by special causes and is unpredictable. This makes the interpretation of capability indices and nonconformance estimates highly questionable. Even estimates based on Overall Sigma are questionable, because the process is not predictable.
220 Process Capability Chapter 7 Additional Examples of the Process Capability Platform Quality and Process Methods The histogram in Figure 7.25 shows the distribution of the Weight values with normal density curves using both sigma estimates superimposed over the histogram. The normal curve that uses the Overall Sigma estimate is flatter and wider than the normal curve that uses the Within Sigma estimate. This normal curve is more dispersed because the estimate of Overall Sigma is inflated by the special causes that make the process unstable. If the process were stable, the narrower normal curve would reflect process behavior.
You can also compare the Cpk estimate (1.142) to the Ppk estimate (0.814). The fact that Ppk is much smaller than Cpk is additional evidence that this is an unpredictable process. The Cpk estimate is a forecast of the capability that you would achieve by bringing the process to a stable state.
Note: The Individual Detail Reports Cutoff preference determines whether the Individual Reports appear by default. If the preference is enabled, the Individual Reports appear by default if the number of process variables is less than or equal to the number specified in the preference. You can change this preference in Preferences > Platforms > Process Capability.
Simulation of Confidence Limits for a Nonnormal Process Ppk In this example, you first perform a capability analysis for the three nonnormal variables in Tablet Measurements.jmp. You then use Simulate to find confidence limits for the nonconformance percentage for the variable Purity.
Nonnormal Capability Analysis If you prefer not to follow the steps below, you can obtain the results in this section by running the Process Capability table script in Tablet Measurements.jmp.
1.
Select Help > Sample Data Library and open Tablet Measurements.jmp.
2.
Select Analyze > Quality and Process > Process Capability.
3.
Select Weight, Thickness, and Purity and click Y, Process.
4.
Select Weight, Thickness, and Purity in the Cast Selected Columns into Roles list on the right.
5.
Open the Distribution Options outline.
6.
From the Distribution list, select Best Fit.
7.
Click Set Process Distribution.
The &Dist(Best Fit) suffix is added to each column name in the list on the right.
8.
Click OK.
A Capability Index Plot appears, showing the Ppk values. Note that only the Thickness variable appears above the line that denotes Ppk = 1. Purity is nearly on the line. Although Chapter 7 Process Capability 221 Quality and Process Methods Additional Examples of the Process Capability Platform the number of measurements, 250, seems large, the estimated Ppk value is still quite variable. For this reason, you construct a confidence interval for the true Purity Ppk value.
Note: Because a Goal Plot is not shown, you can conclude that a normal distribution fit was not the best fit for any of the three variables.
9.
Click the Process Capability red triangle and select Individual Detail Reports.
The best fits are different for each process.
– Weight: Lognormal – Thickness: SHASH – Purity: Weibull Construct the Simulation Column To use the Simulate utility to estimate Ppk confidence limits, you need to construct a simulation formula that reflects the fitted Weibull distribution. If you prefer not to follow the steps below, you can obtain the results in this section by running the Add Simulation Column table script.
1.
Scroll to the Purity (Weibull) Capability report and find the Parameter Estimates report.
Figure 7.27 Weibull Parameter Estimates for Purity These are the parameter estimates for the best fitting distribution, which is Weibull.
2.
In the Tablet Measurements.jmp sample data table, select Cols > New Columns.
3.
Next to Column Name, enter Simulated Purity.
4.
From the Column Properties list, select Formula.
5.
In the formula editor, select Random > Random Weibull.
6.
The placeholder for beta is selected. Click the insertion element (^).
222 Process Capability Chapter 7 Additional Examples of the Process Capability Platform Quality and Process Methods Figure 7.28 Formula Editor for Simulated Purity Column This adds a placeholder for the parameter alpha.
7.
In the Process Capability report, under Purity (Weibull) Capability, right-click in the Parameter Estimates report table and select Make into Data Table.
8.
Copy the entry in Row 2 in the Estimate column (1589.7167836).
9.
In the formula editor window, select the placeholder for beta in the Random Weibull formula and paste 1589.7167836 into the placeholder for beta.
10. In the data table that you created from the Parameter Estimates report, copy the entry in Row 1 in the Estimate column (99.918708989).
11. In the formula editor window, select the placeholder for alpha in the Random Weibull formula and paste 99.918708989 into the placeholder for alpha.
Figure 7.29 Completed Formula Window 12. Click OK in the formula editor window.
13. In the Tablet Measurements.jmp data table, right-click the Simulated Purity column and select Column Properties > Spec Limits.
14. Next to Lower Spec Limit, type 99.5.
15. Click OK in the New Column window.
Chapter 7 Process Capability 223 Quality and Process Methods Additional Examples of the Process Capability Platform The Simulated Purity column contains a formula that simulates values from the best-fitting distribution.
Simulate Confidence Intervals for Purity Ppk and Expected Percent Nonconforming When you use Simulate, the entire analysis is run the number of times that you specify. To shorten the computing time, you can minimize the computational burden by running only the required analysis. In this example, because you are interested only in Purity with a fitted Weibull distribution, you perform only this analysis before running Simulate.
Note: If you do not care about computing time, you can use the same report you created in the previous section and start with step 7.
1.
In the Process Capability report, click the Process Capability red triangle and select Relaunch Dialog.
2.
(Optional) Close the Process Capability report.
3.
In the launch window, from the Cast Selected Columns into Roles list, select Weight&Dist(Lognormal) and Thickness&Dist(SHASH).
4.
Click Remove.
5.
Click OK.
6.
Click the Process Capability red triangle and select Individual Detail Reports.
Both Ppk and Ppl values are provided, but they are identical because Purity has only a lower specification limit.
7.
In the Overall Sigma Capability report, right-click the Estimate column and select Simulate.
In the Column to Switch Out list, make sure Purity is selected. In the Column to Switch In list, make sure Simulated Purity is selected.
8.
Next to Number of Samples, type 500.
Note: The next step is not required. However, it ensures that you obtain exactly the simulated values shown in this example.
9.
(Optional) Next to Random Seed, type 12345.
10. Click OK.
The calculation might take several seconds. A data table entitled Process Capability Simulate Results (Estimate) appears. The Ppk and Ppl columns in this table each contain 500 values calculated based on the Simulated Purity formula. The first row, which is excluded, contains the values for Purity obtained in your original analysis. Because Purity has only a lower specification limit, the Ppk values are identical to the Ppl values.
224 Process Capability Chapter 7 Additional Examples of the Process Capability Platform Quality and Process Methods 11. In the Process Capability Simulate Results (Estimate) data table, click the green triangle next to the Distribution script.
Figure 7.30 Distribution of Simulated Ppk Values for Purity Chapter 7 Process Capability 225 Quality and Process Methods Additional Examples of the Process Capability Platform Note: Your values may vary slightly from what is shown, depending on the decimal precision of the parameters in the Simulated Purity column formula.
Two Distribution reports are shown, one for Ppk and one for Ppl. But Purity has only a lower specification limit, so that the Ppk and Ppl values are identical. For this reason, the Distribution reports are identical.
The Simulation Results report shows that a 95% confidence interval for Ppk is 0.909 to 1.145. This indicate that the true Ppk value could be above 1.0, which would place Purity above the Ppk = 1 line in the Capability Index Plot you constructed in “Nonnormal Capability Analysis”.
12. In the Process Capability report, right-click the Expected Overall % column in the Nonconformance report and select Simulate.
In the Column to Switch Out list, make sure Purity is selected. In the Column to Switch In list, make sure Simulated Purity is selected.
13. Next to Number of Samples, enter 500.
14. (Optional) Next to Random Seed, enter 12345.
15. Click OK.
The calculation might take several seconds. A data table entitled Process Capability Simulate Results (Expected Overall %) appears. Because Purity has only a lower specification limit, the Below LSL values are identical to the Total Outside values.
16. In the Process Capability Simulate Results (Expected Overall %) data table, click the green triangle next to the Distribution script.
226 Process Capability Chapter 7 Additional Examples of the Process Capability Platform Quality and Process Methods Figure 7.31 Distribution of Simulated Total Outside Values for Purity Chapter 7 Process Capability 227 Quality and Process Methods Statistical Details for the Process Capability Platform Note: Your values may vary slightly from what is shown, depending on the decimal precision of the parameters in the Simulated Purity column formula.
Again, two identical Distribution reports appear. The Simulation Results report shows that a 95% confidence interval for the Expected Overall % nonconforming is 0.055 to 0.238.
Statistical Details for the Process Capability Platform • “Variation Statistics” • “Notation for Goal Plots and Capability Box Plots” • “Goal Plot” • “Capability Box Plots for Processes with Missing Targets” • “Capability Indices for Normal Distributions” • “Capability Indices for Nonnormal Distributions: Percentile and Z-Score Methods” • “Parameterizations for Distributions” Variation Statistics Denote the standard deviation of the process by. The Process Capability platform provides two types of capability indices. The Ppk indices are based on an estimate of that uses all of the data in a way that does not depend on subgroups. This overall estimate can reflect special cause as well as common cause variation. The Cpk indices are based on an estimate that attempts to capture only common cause variation. The Cpk indices are constructed using within-subgroup or between-and-within-subgroup estimates of . In this way, they attempt to reflect the true process standard deviation. When a process is not stable, the different estimates of can differ markedly.
Overall Sigma The overall sigma does not depend on subgroups. The overall estimate of is calculated as follows: The formula uses the following notation: N = number of nonmissing values in the entire data set ˆ 1 N 1 – -------------yi y – 2 i 1 = N = 228 Process Capability Chapter 7 Statistical Details for the Process Capability Platform Quality and Process Methods yi = value of the ith observation = mean of nonmissing values in the entire data set Caution: When the process is stable, the Overall Sigma estimates the process standard deviation. If the process is not stable, the overall estimate of is of questionable value, since the process standard deviation is unknown.
Estimates of Sigma Based on Within-Subgroup Variation An estimate of that is based on within-subgroup variation can be constructed in one of the following ways: • Within sigma estimated by average of ranges • Within sigma estimated by average of unbiased standard deviations • Within sigma estimated by moving range • Within sigma estimated by unbiased pooled standard deviation If you specify a subgroup ID column or a constant subgroup size on the launch window, you can specify your preferred within-subgroup variation statistic. See “Launch the Process Capability Platform”. If you do not specify a subgroup ID column, a constant subgroup size, or a historical sigma, JMP estimates the within sigma using the third method (moving range of subgroups of size two).
Within Sigma Based on Average of Ranges Within sigma estimated by the average of ranges is the same as the estimate of the standard deviation of an XBar and R chart: The formula uses the following notation: Ri = range of ith subgroup ni = sample size of ith subgroup d2(ni) = expected value of the range of ni independent normally distributed variables with unit standard deviation N = number of subgroups for which ni ≥2 y ˆ R1 d2 n1 ----------------- RN d2 nN ------------------+ + N -------------------------------------------------------= Chapter 7 Process Capability 229 Quality and Process Methods Statistical Details for the Process Capability Platform Within Sigma Based on Average of Unbiased Standard Deviations Within sigma estimated by the average of unbiased standard deviations is the same as the estimate for the standard deviation in an XBar and S chart: The formula uses the following notation: ni = sample size of ith subgroup c4(ni) = expected value of the standard deviation of ni independent normally distributed variables with unit standard deviation N = number of subgroups for which ni ≥2 si = sample standard deviation of the ith subgroup Within Sigma Based on Average Moving Range Within sigma estimated by average moving range is the same as the estimate for the standard deviation for Individual Measurement and Moving Range charts: The formula uses the following notation: = the mean of the nonmissing moving ranges computed as (MR2+MR3+...+MRN)/(N-1) where MRi = |yi - yi-1|.
d2(2) = expected value of the range of two independent normally distributed variables with unit standard deviation.
Within Sigma Based on Median Moving Range Within sigma estimated by median moving range: The formula uses the following notation: MMR = the median of the nonmissing moving ranges computed as Median(MR2, MR3,..., MRN) where MRi = |yi - yi-1|.
ˆ s1 c4 n1 ---------------- sN c4 nN ------------------+ + N ------------------------------------------------------= ˆ MR d2 2 --------------= MR ˆ MMR 0.954 ----------------= 230 Process Capability Chapter 7 Statistical Details for the Process Capability Platform Quality and Process Methods Within Sigma Based on Unbiased Pooled Standard Deviation Within sigma estimated by the unbiased pooled standard deviation: The formula uses the following notation: ni = sample size of ith subgroup n = n1 + + nN, the total sample size c4(n) = expected value of the standard deviation of n independent normally distributed variables with unit standard deviation N = number of subgroups for which ni ≥2 si = sample standard deviation of the ith subgroup Estimate of Sigma Based on Between Group Variation Between Sigma Based on Moving Range The estimate of that is based on between-subgroup variation is estimated by the moving range of subgroup means: The formula uses the following notation: = the mean of the nonmissing moving ranges computed as (MR2+MR3+...+MRN)/(N-1) where MRi = |yi - yi-1|.
d2(2) = expected value of the range of two independent normally distributed variables with unit standard deviation.
2 within = the specified within sigma estimate.
, the harmonic mean of subgroup sample sizes.
ˆ n1 1 – s1 2 nN 1 – sN 2 + + c4 n n1 n2 N – + + -----------------------------------------------------------------------------= ˆ MR d2 2 -------------- 2 ˆ 2 Within ˆ H -----------------------– = MR H N 1 n1 ------1 n2 ------ 1 nN -------+ + + ----------------------------------------------= Chapter 7 Process Capability 231 Quality and Process Methods Statistical Details for the Process Capability Platform Estimate of Sigma Based on Between and Within Group Variation Between-and-Within Sigma The estimate of sigma that is based on the combined between and within group variation is defined as follows: Notation for Goal Plots and Capability Box Plots The formulas for the Goal Plot and Capability Box Plots use the following notation: Yij = ith observation for process j = mean of the observations on process j SD(Yj) = standard deviation of the observations on process j Tj = target value for process j LSLj = lower specification limit for process j USLj = upper specification limit for process j Goal Plot This section provides details about the calculation of the mean shift and standard deviation standardized to specification quantities plotted in the Goal Plot. This section uses the notation defined in “Notation for Goal Plots and Capability Box Plots”.
The mean shift and the standard deviation standardized to the specification limits for the jth column are defined as follows: Spec-Standardized Mean = Spec-Standardized Std Dev = Note: If either LSLj or USLj is missing, twice the distance from the target to the nonmissing specification limit is used in the denominators of the Goal Plot coordinates.
ˆ ˆ within 2 ˆ between 2 + = Yj Yj Tj – 2 min(Tj LSLj – USL j Tj) – ---------------------------------------------------------------------------SD Yj 2 min(Tj LSLj – USL j Tj) – ---------------------------------------------------------------------------232 Process Capability Chapter 7 Statistical Details for the Process Capability Platform Quality and Process Methods Goal Plot Points for Processes with Missing Targets Suppose that the process has both a lower and an upper specification limit but no target. Then the formulas given in “Goal Plot” are used, replacing Tj with the midpoint of the two specification limits.
Suppose that the process has only one specification limit and no target. To obtain (x,y) coordinates for a point on the Goal Plot, the capability indices of the process are used. (See “Capability Indices for Normal Distributions” for definitions in terms of the theoretical mean and standard deviation.) For sample observations, the following relationships hold: If a process has two specification limits and a target at the midpoint of the limits, then the (x,y) coordinates for the point on the Goal Plot satisfy these relationships: To obtain coordinates when there is only one specification limit and no target, these relationships are used. To identify a unique point requires an assumption about the slope of the line from the origin on which the points fall. A slope of 0.5 is assumed for the case of an upper specification limit and of -0.5 for a lower specification limit. When capability values are equal to one and the Ppk slider for the goal plot triangle is set to 1, these slopes place the points on the goal plot triangle lines.
Consider the case of only an upper specification limit and no target. Using the assumption that the (x,y) coordinates fall on a line from the origin with slope 0.5, solving for x and y gives the following coordinates: Consider the case of only a lower specification limit and no target. Using the assumption that the (x,y) coordinates fall on a line from the origin with slope -0.5, solving for x and y gives the following coordinates: Cpu USLj Yj – 3SD Yj ------------------------= Cpl Yj LSLj – 3SD Yj -----------------------= Cpu 0.5 x – 3y = Cpl 0.5 x + 3y = x 1 = 3Cpu 2 + y 1 = 6Cpu 4 + x 1 – = 3Cpl 2 + y 1 = 6Cpl 4 + Chapter 7 Process Capability 233 Quality and Process Methods Statistical Details for the Process Capability Platform Note: If either Cpu or Cpl is less than -0.6, then it is set to -0.6 in the formulas above. At the value -2/3, the denominator for x assumes the value 0. Bounding the capability values at -0.6 prevents the denominator from assuming the value 0 or switching signs.
Capability Box Plots for Processes with Missing Targets A column with no target can have both upper and lower specification limits, or only a single specification limit. This section uses the notation defined in “Notation for Goal Plots and Capability Box Plots”.
Two Specification Limits and No Target When no target is specified for the jth column, the capability box plot is based on the following values for the transformed observations: Single Specification Limit and No Target Suppose that only the lower specification limit is specified. (The case where only the upper specification limit is specified in a similar way.) When no target is specified for the jth column, the capability box plot is based on the following values for the transformed observations: Note: When a column has only one specification limit and no target value, and the sample mean falls outside the specification interval, no capability box plot for that column is plotted.
Capability Indices for Normal Distributions This section provides details about the calculation of capability indices for normal data.
Zij Yij LSLj USLj + 2 – USLj LSLj – ---------------------------------------------------------= Zij Yij Yj – 2 Yj LSLj – --------------------------------= 234 Process Capability Chapter 7 Statistical Details for the Process Capability Platform Quality and Process Methods For a process characteristic with mean and standard deviation , the population-based capability indices are defined as follows. For sample observations, the parameters are replaced by their estimates: Cp = Cpl = Cpu = Cpk = Cpm = Target Index = The formulas use the following notation: LSL = Lower specification limit USL = Upper specification limit T = Target value For estimates of Within Sigma capability, is estimated using the subgrouping method that you specified. For estimates of Overall Sigma capability, is estimated using the sample standard deviation. If either of the specification limits is missing, the capability indices containing the missing specification limit are reported as missing.
Note: With the default AIAG (Ppk) Labeling, the indices based on Overall Sigma are denoted by Pp, Ppl, Ppu, and Ppk. The labeling for the index Cpm does not change when Overall Sigma is used. The formulas in this section are defined using Cp labels.
Confidence Intervals for Capability Indices Confidence intervals for capability indices are available only for processes with normal distributions. Confidence intervals are calculated for both Within and Overall Sigma capability and are shown in the Individuals Detail Reports.
USL LSL – 6 ---------------------------- LSL – 3 --------------------USL – 3 ---------------------min( Cpl, Cpu ) min T LSL – USL T – 31 T – ------------- 2 + -------------------------------------------------------------3 Cp Cpk – Chapter 7 Process Capability 235 Quality and Process Methods Statistical Details for the Process Capability Platform Cp The 100(1 - )% confidence interval for Cp is calculated as follows: where is the estimated value for Cp is the (/2)th quantile of a chi-square distribution with df degrees of freedom df is the degrees of freedom N is the number of observations m is the number of subgroups For Overall Sigma capability, the degrees of freedom is equal to N - 1.
For Within Sigma capability, the degrees of freedom depends on the subgrouping and the within sigma estimation method.
• For Within Sigma capability with unbalanced subgroups, the degrees of freedom calculation is the same regardless of the within sigma estimation method. The degrees of freedom is equal to N - m.
• For Within Sigma capability with balanced subgroups of size n = N/m, the degrees of freedom calculation depends on the within sigma estimation method.
– When Within Sigma is estimated by the average of the unbiased standard deviations, the degrees of freedom is equal to f (N - m). The scale factor f, which varies between 0.875 and 1, is defined as follows: where (n) is the gamma function evaluated at n.
For more information, see Bissell (1990).
– When Within Sigma is estimated by the average of ranges, the degrees of freedom is calculated as df = 1/A - (3/16) A + (3/64) A2 + 0.25. A is defined as follows: Cp ˆ 2 df 2 df -------------------- Cp ˆ 1 2 – df 2 df ---------------------------- Cp ˆ 2 df 2 f 1 2 n 1 – n 1 – 2 -----------------n 1 – 2 ------------ n 2 --- ---------------------- 2 1 – -----------------------------------------------------------------------------------= 236 Process Capability Chapter 7 Statistical Details for the Process Capability Platform Quality and Process Methods d2(n) is the expected value of the range of n independent normally distributed variables with unit standard deviation d3(n) is the standard deviation of the range of n independent normally distributed variables with unit standard deviation For more information, see David (1951).
– When Within Sigma is estimated by the unbiased pooled standard deviations, the degrees of freedom is equal to N - m.
• For Within Sigma capability with no subgroups, the degrees of freedom calculation depends on the within sigma estimation method.
– When Within Sigma is estimated by the average moving ranges, the degrees of freedom is calculated as 0.62 (N - 1).
– When Within Sigma is estimated by the median moving ranges, the degrees of freedom is calculated as 0.32 (N - 1).
For more information, see Wheeler (2004, p. 82).
Cpk The 100(1 - )% confidence interval for Cpk is calculated as follows: where is the estimated value for Cpk is the (1 - /2)th quantile of a standard normal distribution df is the degrees of freedom N is the number of observations m is the number of subgroups For Overall Sigma capability, the degrees of freedom is equal to N - 1.
For Within Sigma capability, the degrees of freedom depends on the subgrouping and the within sigma estimation method.
A 2d3 n 2 m d 2 n 2 --------------------------= C ˆ pk 1 1 – 1 2 – 1 9NC ˆ pk 2 ------------------1 2df --------+ – C ˆ pk 1 1 – 1 2 – 1 9NC ˆ pk 2 ------------------1 2df --------+ + C ˆ pk 1 – 1 2 – Chapter 7 Process Capability 237 Quality and Process Methods Statistical Details for the Process Capability Platform • For Within Sigma capability with unbalanced subgroups, the degrees of freedom calculation is the same regardless of the within sigma estimation method. The degrees of freedom is equal to N - m.
• For Within Sigma capability with balanced subgroups of size n = N/m, the degrees of freedom calculation depends on the within sigma estimation method.
– When Within Sigma is estimated by the average of the unbiased standard deviations, the degrees of freedom is equal to f (N - m). The scale factor f, which varies between 0.875 and 1, is defined as follows: where (n) is the gamma function evaluated at n.
For more information, see Bissell (1990).
– When Within Sigma is estimated by the average of ranges, the degrees of freedom is calculated as df = 1/A - (3/16) A + (3/64) A2 + 0.25. A is defined as follows: d2(n) is the expected value of the range of n independent normally distributed variables with unit standard deviation d3(n) is the standard deviation of the range of n independent normally distributed variables with unit standard deviation For more information, see David (1951).
– When Within Sigma is estimated by the unbiased pooled standard deviations, the degrees of freedom is equal to N - m.
• For Within Sigma capability with no subgroups, the degrees of freedom calculation depends on the within sigma estimation method.
– When Within Sigma is estimated by the average moving ranges, the degrees of freedom is calculated as 0.62 (N - 1).
– When Within Sigma is estimated by the median moving ranges, the degrees of freedom is calculated as 0.32 (N - 1).
For more information, see Wheeler (2004, p. 82).
f 1 2 n 1 – n 1 – 2 -----------------n 1 – 2 ------------ n 2 --- ---------------------- 2 1 – -----------------------------------------------------------------------------------= A 2d3 n 2 m d 2 n 2 --------------------------= 238 Process Capability Chapter 7 Statistical Details for the Process Capability Platform Quality and Process Methods Cpm Note: The confidence interval for Cpm is computed only when the target value is centered between the lower and upper specification limits.
The 100(1 - )% confidence interval for Cpm is calculated as follows: where is the estimated value for Cpm is the (/2)th quantile of a chi-square distribution with degrees of freedom N is the number of observations is the mean of the observations T is the target value s is the sigma estimate For Overall Sigma capability, s is the Overall Sigma estimate. For Within Sigma capability, s is replaced by the Within Sigma estimate.
Tip: For more information on confidence intervals for Cp, Cpk, and Cpm, see Pearn and Kotz (2006).
Cpl and Cpu Lower and upper confidence limits for Cpl and Cpu are computed using the method of Chou et al. (1990).
The 100(1 - )% confidence limits for Cpl (denoted by CPLL and CPLU) satisfy the following equations: where Cpm ˆ 2 2 ------------------ Cpm ˆ 1 2 – 2 -------------------------- Cpm ˆ 2 2 N 1 x T – s ------------ 2 + 2 1 2 x T – s ------------ 2 + --------------------------------------------= x Pr tn 1 – L 3C ˆ pl n 2 = L 3CPLL n = Chapter 7 Process Capability 239 Quality and Process Methods Statistical Details for the Process Capability Platform where where tn-1() has a non-central t-distribution with n - 1 degrees of freedom and noncentrality parameter is the estimated value for Cpl The 100(1 - )% confidence limits for Cpu (denoted by CPUL and CPUU) satisfy the following equations: where where where tn-1() has a non-central t-distribution with n - 1 degrees of freedom and noncentrality parameter is the estimated value for Cpu Capability Indices for Nonnormal Distributions: Percentile and Z-Score Methods This section describes how capability indices are calculated for nonnormal distributions. Two methods are described: the Percentile (also known as ISO/Quantile) method and the Z-Score (also known as Bothe/Z-scores) method. When you select a distribution for a nonnormal process variable, you can fit a parametric distribution or a nonparametric distribution. You can use either the Percentile or the Z-Score methods to calculate capability indices for the process variable of interest. However, unless you have a very large amount of data, a nonparametric fit might not accurately reflect behavior in the tails of the distribution.
Note: For both the Percentile and the Z-Score methods, if the data are normally distributed, the capability formulas reduce to the formulas for normality-based capability indices.
The descriptions of the two methods use the following notation: LSL = Lower specification limit USL = Upper specification limit Pr tn 1 – U 3C ˆ pl n 2 = U 3CPLU n = C ˆ pl Pr tn 1 – L 3C ˆ pu n 2 = L 3CPUL n = Pr tn 1 – U 3C ˆ pu n 2 = U 3CPUU n = C ˆ pu 240 Process Capability Chapter 7 Statistical Details for the Process Capability Platform Quality and Process Methods T = Target value Percentile (ISO/Quantile) Method The percentile method replaces the mean in the standard capability formulas with the median of the fitted distribution and the 6 range of values with the corresponding percentile range. The method is described in AIAG (2005).
Denote the 100th percentile of the fitted distribution by P. Then Percentile method capability indices are defined as follows: Z-Score (Bothe/Z-Scores) Method The Z-Score method transforms the specification limits to values that have the same probabilities on a standard normal scale. It computes capability measures that correspond to a normal distribution with the same risk levels as the fitted nonnormal distribution.
Let F denote the fitted distribution for a process variable with lower and upper specification limits given by LSL and USL. Equivalent standard normal specification limits are defined as follows: Ppk min P0.5 LSL – P0.5 P0.00135 – -------------------------------------- USL P0.5 – P0.99865 P0.5 – --------------------------------------, = Ppl P0.5 LSL – P0.5 P0.00135 – --------------------------------------= Ppu USL P0.5 – P0.99865 P0.5 – --------------------------------------= Pp USL LSL – P0.99865 P0.00135 – ------------------------------------------------= Cpm min T LSL – P0.5 P0.00135 – -------------------------------------- USL T – P0.99865 P0.5 – --------------------------------------, 1 T – ------------- 2 + ------------------------------------------------------------------------------------------------= LSLF 1 – F LSL = USLF 1 – F USL = Chapter 7 Process Capability 241 Quality and Process Methods Statistical Details for the Process Capability Platform Then the Z-Score method capability indices are defined as follows: Note: Because Cpm is a target-based measure, it cannot be calculated using the Z-Scores method.
Note: For very capable data, F(LSL) or F(USL) can be so close to zero or one, respectively, that LSLF or USLF cannot be computed. In these cases, JMP automatically switches from the Z-Score method to the Percentile method by default. This gives more meaningful capability indices. To turn off this default setting, select File > Preferences > Platforms > Process Capability.
Parameterizations for Distributions This section gives the density functions f for the distributions used in the Process Capability platform. It also gives expected values and variances for all but the Johnson and SHASH distributions.
Normal , , , E(X) = Var(X) = 2 Beta , , , E(X) = Ppk min LSLF – 3 USLF 3 , = Ppl LSLF – 3 = Ppu USLF 3 = Pp USLF LSLF – 6 = f x 1 2 --------------1 22 ---------– x – 2 exp = – x – 0 f x 1 B ------------------x 1 – 1 x – 1 – = 0 x 1 0 0 + -------------242 Process Capability Chapter 7 Statistical Details for the Process Capability Platform Quality and Process Methods Var(X) = where B(·) is the Beta function.
Exponential , , E(X) = Var(X) = Gamma , , , E(X) = Var(X) = where (·) is the gamma function.
Johnson Johnson Su , , , Johnson Sb , Johnson Sl , , where (·) is the standard normal probability density function.
+ 2 1 + + -------------------------------------------------f x 1 ---x – exp = x 0 0 f x 1 --------------------x 1 – x – exp = x 0 0 0 f x --- 1 x – ------------ 2 + 1 – 2 x – ------------ sinh 1 – + = – x 0 0 f x x – x – – -------------------------- ln + x – x – – ------------------------------------------------ = x + 0 f x x – --------------- x – ------------ ln + = x 0 if 1 = x 0 if 1 – = Chapter 7 Process Capability 243 Quality and Process Methods Statistical Details for the Process Capability Platform Lognormal , , , E(X) = Var(X) = Mixture of Normals The Mixture of 2 Normals and Mixture of 3 Normals options for Distribution share the following parameterization: E(X) = Var(X) = where i, i, and i are the respective mean, standard deviation, and proportion for the ith group, and (·) is the standard normal probability density function. For the Mixture of 2 Normals, k is equal to 2. For the Mixture of 3 Normals distribution, k is equal to 3. A separate mean, standard deviation, and proportion of the whole is estimated for each group in the mixture.
SHASH , , , 0 < where (·) is the standard normal pdf f x 1 2 --------------x log – – 2 22 -------------------------------------exp x -----------------------------------------------------= x 0 – 0 2 2 + exp 2 2 + exp 2 2 + exp – f x i i i i i ----- x i – i -------------- i 1 = k = i i 1 = k i i i 2 i 2 + i 1 = k i i 1 = k i 2 – f x w cosh 2 x – 2 + -------------------------------------sinh w = – x 244 Process Capability Chapter 7 Statistical Details for the Process Capability Platform Quality and Process Methods Note: When = 0 and = 1, the SHASH distribution is equivalent to the normal distribution with location and scale .
Weibull , , E(X) = Var(X) = where (·) is the gamma function.
w x – ------------ sinh–1 + = f x -------x 1 – x --- – exp = 0 0 1 1 ---+ 2 1 2 ---+ 2 1 1 ---+ – Chapter 8 CUSUM Control Charts Create Tabular CUSUM Control Charts with Decision Limits Cumulative Sum (CUSUM) control charts enable you to detect small shifts in a process. They are useful in detecting shifts that occur over time, such as a gradual drift, and that are not necessarily accompanied by a sudden shift. The CUSUM Control Chart platform creates a CUSUM chart with decision limits, similar to a Shewhart chart. This chart is also called a tabular CUSUM chart. To create a V-mask cumulative sum control chart, see “V-Mask CUSUM Control Charts”.
The CUSUM Control Chart platform also provides information about average run length (ARL). The average run length is the average number of samples or observations that can be expected to occur before an out-of-control signal occurs. You can use the average run length to assess the performance of a CUSUM chart, given specific parameters and assuming constant variance.
Figure 8.1 CUSUM Control Chart 246 CUSUM Control Charts Chapter 8 Quality and Process Methods Contents Overview of the CUSUM Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 Example of CUSUM Control Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 Launch the CUSUM Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 249 The CUSUM Control Chart Platform Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 Control Panel for CUSUM Control Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 CUSUM Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 CUSUM Control Chart Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252 Average Run Length (ARL) Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254 Additional Examples of CUSUM Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 Example of the Data Units Option in a CUSUM Control Chart . . . . . . . . . . . . . . . . . . . . . 255 Example of CUSUM Chart with Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 Statistical Details for the CUSUM Control Chart Platform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 Statistical Details for CUSUM Control Chart Construction . . . . . . . . . . . . . . . . . . . . . . . . . 258 Statistical Details for Shift Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 Statistical Details for Average Run Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 Chapter 8 CUSUM Control Charts 247 Quality and Process Methods Overview of the CUSUM Control Chart Platform Overview of the CUSUM Control Chart Platform A tabular CUSUM chart consists of two one-sided decision limits charts superimposed on one chart. The chart contains decision limits that signal when the process is out of control and places a shift line on the chart where the shift is suspected to have occurred. To use the CUSUM Control Chart platform, you must determine the smallest change in the mean that you consider important. You can view the CUSUM control chart in standard deviation units or in data units. For more information about tabular CUSUM charts, see Woodall and Adams (1998) and Montgomery (2013).
Another form of a cumulative sum control chart is the V-mask chart. To create a V-mask CUSUM chart, see “V-Mask CUSUM Control Charts”.
Note: The summary results in the CUSUM Control Chart platform do not always match the summary results in the V-mask CUSUM platform. Specifically, the summary results for a two-sided V-mask CUSUM chart do not match those from a CUSUM Control Chart with both Upper Side and Lower Side options selected. However, the one-sided summary reports from the CUSUM Control Chart platform and the V-mask CUSUM platform do match.
Example of CUSUM Control Chart You want to detect small shifts in the temperature of an engine. The data table contains temperature measurements from the engine thermostat.
1.
Select Help > Sample Data Library and open Quality Control/Engine Temperature Sensor.jmp.
2.
Select Analyze > Quality and Process > Control Chart > CUSUM Control Chart.
3.
Select Y and click Y.
4.
Click OK.
5.
In the Target box, type 100.
6.
In the Sigma box, type 10.
248 CUSUM Control Charts Chapter 8 Example of CUSUM Control Chart Quality and Process Methods Figure 8.2 CUSUM Control Chart Report The vertical line on the CUSUM Chart indicates that a shift in the temperature measurements started around sample 26.
Note: You can compare this result to the Individual Moving Range control chart by running the IMR Chart table script in Engine Temperature Sensor.jmp. The IMR chart does not trigger any of the Nelson tests.
Example of a One-Sided CUSUM Control Chart Continuing the previous example, suppose that you care only about increasing temperature changes. To change the CUSUM control chart in Figure 8.2 to a one-sided chart, deselect the Lower Side check box. When you do that, the points for the negative cumulative sums are removed from the chart. You are left with a CUSUM control chart that contains only the positive cumulative sum points.
Chapter 8 CUSUM Control Charts 249 Quality and Process Methods Launch the CUSUM Control Chart Platform Figure 8.3 One-Sided CUSUM Control Chart Report Launch the CUSUM Control Chart Platform Launch the CUSUM Control Chart platform by selecting Analyze > Quality and Process > Control Chart > CUSUM Control Chart.
Figure 8.4 The CUSUM Control Chart Launch Window For more information about the options in the Select Columns red triangle menu, see Using JMP.
The CUSUM Control Chart platform launch window contains the following options: Y Identifies the variables that you want to chart.
Note: The rows of the data table must be sorted in the order in which the observations were collected.
250 CUSUM Control Charts Chapter 8 The CUSUM Control Chart Platform Report Quality and Process Methods X Identifies a subgroup variable. The horizontal axis of the CUSUM chart is labeled by the subgroup variable. If a value of this column is present more than once, the average response at each X value is plotted on the CUSUM chart.
By Produces a separate report for each level of the By variable. If more than one By variable is assigned, a separate report is produced for each possible combination of the levels of the By variables.
Show Excluded Region (Applicable only when an X variable is specified.) Specifies that subgroups that are entirely excluded are shown on the horizontal axis in the CUSUM control chart.
Data Units Specifies that data units be used in the report rather than standard deviation units. By default, the chart and parameters are shown in standard deviation units. However, if you select the Data Units option in the launch window, the chart and parameters are shown in the units of the data column that is being analyzed.
When you use standard deviation units, values for the h and k parameters do not depend on the process standard deviation. This can be an advantage.
The CUSUM Control Chart Platform Report By default, the CUSUM Control Chart platform produces a report that contains a parameter control panel and a CUSUM chart.
Figure 8.5 CUSUM Control Chart Report Chapter 8 CUSUM Control Charts 251 Quality and Process Methods The CUSUM Control Chart Platform Report Control Panel for CUSUM Control Chart The Control Panel report contains the current values for the chart parameters. The current values are in boxes that enable you to update the parameter values. There are also boxes for Upper Side and Lower Side. If you have specified the Data Units option in the launch window, this setting is denoted in the Control Panel below the check boxes.
The following options appear in the Control Panel report: Target The known value of the mean. This is the value of the center line in the chart. By default, this parameter is set to the Target value in the Spec Limits column property for the Y column. If the Y column does not have a Target value in the Spec Limits column property, this parameter is set to the overall average of the Y column.
Note: To use the overall average of the Y column as the value of the center line even if the Y column has a Target value in the Spec Limits column property, select the Use Process Mean for Center Line platform preference. This preference is located in File > Preferences > Platforms > CUSUM Control Chart.
Sigma The known value of the standard deviation. By default, this parameter is set to the average moving range of the Y column. If there is an X variable, the Sigma parameter is set to the average moving range of the summary data.
Head Start The value of the cumulative sums before the first sample. Starting the cumulative sums at a nonzero value increases the sensitivity of the CUSUM chart near the beginning of the samples. This parameter is also known as the fast initial response (FIR) value. By default, this parameter is set to 0.
h or H The value of the parameter that defines the limits. If the Data Units option was not selected in the launch window, this is the h parameter. If the Data Units option was selected in the launch window, this is the H parameter. Note that H is equal to h times Sigma. By default, h is equal to 5 and H is equal to 5 times Sigma.
k or K The value of the parameter that defines the smallest change in the mean that is valuable to detect. If the Data Units option was not selected in the launch window, this is the k parameter. If the Data Units option was selected in the launch window, this is the K parameter. Note that K is equal to k times Sigma. By default, k is equal to 0.5 and K is equal to one half of Sigma.
Upper Side Shows or hides the positive values for the cumulative sum on the chart. These values are the C+ values.
Lower Side Shows or hides the negative values for the cumulative sum on the chart. These values are the C- values.
252 CUSUM Control Charts Chapter 8 CUSUM Control Chart Platform Options Quality and Process Methods Using Data Units The presence of this text indicates that the Data Units option was selected in the launch window and that the values in the CUSUM chart are centered but not standardized.
CUSUM Chart The CUSUM Chart report contains the cumulative sum control chart with decision limits that are determined by the current values of the chart parameters. The samples (or subgroups if you specified an X variable) are denoted on the horizontal axis. The vertical axis denotes centered values of the positive and negative values for the cumulative sum. If the Data Units option was not selected in the launch window, the vertical axis denotes cumulative sums for standardized response values. If the Data Units option was selected in the launch window, the vertical axis denotes cumulative sums for unstandardized response values.
CUSUM Control Chart Platform Options The CUSUM Control Chart red triangle menu contains the following options: Show Limits Shows or hides the upper and lower decision limits in the CUSUM Chart.
Show Center Line Shows or hides the center line in the CUSUM Chart.
Show Shift Lines (Available only when there is a shift detected in the data.) Shows or hides the vertical lines in the CUSUM Chart that designate shifts. Shift lines are drawn at the start of a shift.
– A positive shift occurs when the value of C+ exceeds the upper limit on the chart. The start of the shift is defined as the first point after the most recent zero value for C+.
– A negative shift occurs when the value of C- falls below the lower limit on the chart. The start of the shift is defined as the first point after the most recent zero value for C-.
Show ARL Shows or hides the Average Run Length (ARL) report. See “Average Run Length (ARL) Report”.
ARL Profiler Shows or hides a profiler of average run length versus the parameters h and k. If you have specified the Data Units option in the launch window, the profiler plots average run length versus the parameters H and K.
The average run length (ARL) for a specified shift is the average number of runs expected before an out-of-control signal occurs. For example, the ARL at 0 represents the average Chapter 8 CUSUM Control Charts 253 Quality and Process Methods CUSUM Control Chart Platform Options number of runs expected before a false-alarm signal occurs when the process is in control. When the process is in control, the shift size is 0.
The ARL Profiler enables you to explore how various settings of the parameters affect the performance of the corresponding CUSUM chart. As the parameters in the Control Panel report are updated, the ARL Profiler is updated as well. An ideal CUSUM chart has a high ARL(0) value and a low ARL() value, where is the size shift that is of interest.
The ARL Profiler depends on the settings of the Upper Side and Lower Side options in the Control Panel report: – If both the Upper Side and Lower Side options are selected, the profiler represents the average run length for crossing either the upper or lower decision limits on the CUSUM chart.
– If only the Upper Side option is selected, the profiler represents the average run length for the upper decision limit on the CUSUM chart.
– If only the Lower Side option is selected, the profiler represents the average run length for the lower decision limit on the CUSUM chart.
For more information about the options in the red triangle menu next to ARL Profiler, see Profilers.
Control Panel Shows or hides a report of the current values of the parameters. This report enables you to change the parameter values as well as the sidedness of the CUSUM chart.
Parameters Report Shows or hides a report of the current values of the parameters.
Test Beyond Limits Shows or hides a red circle around any point that is above the upper limit or below the lower limit in the CUSUM chart.
Save Summaries Creates a new data table that contains statistics for each subgroup in the CUSUM chart. The following statistics are saved to the new data table: the subgroup number and size, the subgroup sample mean, an indicator of shift starts, a value that indicates each interval between shift starts, the upper and lower cumulative sums and corresponding consecutive run counts, and the LCL and UCL values.
Tune Chart Shows or hides a control that enables you to set an acceptable range for the Y variable. Adjust the minimum and maximum values of the acceptable range and click Done. At this point, the CUSUM chart updates based on a new value of the k parameter.
– If you specified the Data Units option in the launch window, the imputed k parameter is the average of the minimum and maximum values.
254 CUSUM Control Charts Chapter 8 CUSUM Control Chart Platform Options Quality and Process Methods – If the Data Units option is not specified, the imputed K parameter is the average of the minimum and maximum values divided by the Sigma parameter.
Setting the acceptable range for the Y variable enables you to set the practical significance of the CUSUM chart. This is particularly helpful when the testing interval is more frequent, which can result in a much shorter practical average run length.
Reset to Defaults Resets all parameters back to their default values.
Alarm Script Enables you to write and run a script that indicates when the data fail special causes tests. Results can be written to a file, written to the log or sent in an email. There is an option to include an explanation of why the test failed.
As an Alarm Script is invoked, the following variables are available, both in the issued script and in subsequent JSL scripts: qc_col is the name of the column qc_test is the test that failed qc_sample is the sample number Tip: After an alarm script is specified, the alarm script is invoked when the Test Beyond Limits option is turned on.
See the Scripting Guide for more information about writing custom Alarm Scripts.
See Using JMP for more information about the following options: Redo Contains options that enable you to repeat or relaunch the analysis. In platforms that support the feature, the Automatic Recalc option immediately reflects the changes that you make to the data table in the corresponding report window.
Save Script Contains options that enable you to save a script that reproduces the report to several destinations.
Save By-Group Script Contains options that enable you to save a script that reproduces the platform report for all levels of a By variable to several destinations. Available only when a By variable is specified in the launch window.
Average Run Length (ARL) Report The Average Run Length (ARL) report contains a table and a graph of ARL values. The average run length (ARL) for a specified shift is the average number of runs expected before an out-of-control signal occurs. For example, the ARL at 0 represents the average number of runs expected before seeing a false-alarm signal when the process is in control. When the process is in control, the shift size is 0.
Chapter 8 CUSUM Control Charts 255 Quality and Process Methods Additional Examples of CUSUM Control Charts The table and graph in the ARL report enable you to explore how various settings of the parameters affect the performance of the corresponding CUSUM chart. As the h and k parameters in the Control Panel report are updated, the ARL report is updated as well. An ideal CUSUM chart has a high ARL(0) value and a low ARL() value, where is the size shift that is of interest.
The Average Run Length (ARL) report depends on the settings of the Upper Side and Lower Side options in the Control Panel report. If only one option is selected, the ARL report uses calculations for the corresponding one-sided CUSUM chart. If both options are selected, the ARL report uses calculations for the two-sided CUSUM chart. Note that the two-sided ARL values are related to the positive and negative one-sided ARL values by the following equation: ARL Table The ARL Table shows the average run length for shifts () between zero and three at 0.25 increments. If the Data Units option is specified, the shift is represented by 2K/Sigma2. If the Data Units option is not specified, the shift is represented by 2k/Sigma.
ARL Graph The ARL Graph shows the average run length for shifts () between 0 and 3. This graph contains the same data points as the ARL Table to the left of the ARL Graph.
Additional Examples of CUSUM Control Charts • “Example of the Data Units Option in a CUSUM Control Chart” • “Example of CUSUM Chart with Subgroups” Example of the Data Units Option in a CUSUM Control Chart This example uses the Data Units option and reproduces the analysis in “Example of CUSUM Control Chart”. You want to detect small shifts in the temperature of an engine. The data table contains temperature measurements from the engine thermostat.
1.
Select Help > Sample Data Library and open Quality Control/Engine Temperature Sensor.jmp.
2.
Select Analyze > Quality and Process > Control Chart > CUSUM Control Chart.
3.
Select Y and click Y.
1 ARL ------------1 Positive ARL -----------------------------------1 Negative ARL --------------------------------------+ = 256 CUSUM Control Charts Chapter 8 Additional Examples of CUSUM Control Charts Quality and Process Methods 4.
Select the box next to Data Units.
5.
Click OK.
6.
In the Target box, type 100.
7.
In the Sigma box, type 10.
Note that the options below Head Start are H and K, instead of h and k. These parameters are now specified in units of the data column, rather than in standard deviation units.
Figure 8.6 CUSUM Control Chart Report Like in the example using sigma units, the vertical line on the CUSUM Chart indicates that a shift in the temperature measurements started around sample 26.
Example of CUSUM Chart with Subgroups A machine fills 8-ounce cans of two-cycle engine oil additive. The filling process is believed to be in statistical control. The process is set so that the average weight of a filled can (0) is 8.10 ounces. Previous analysis shows that the standard deviation of fill weights (0) is 0.05 ounces.
Subgroup samples of four cans are selected and weighed every hour for twelve hours. Each observation in the Oil1 Cusum.jmp data table contains one value of weight and its associated value of hour. The observations are sorted so that the values of hour are in increasing order. You want to be able to detect a 2 shift in the process.
1.
Select Help > Sample Data Library and open Quality Control/Oil1 Cusum.jmp.
2.
Select Analyze > Quality and Process > Control Chart > CUSUM Control Chart.
3.
Select weight and click Y.
4.
Select hour and click X.
Chapter 8 CUSUM Control Charts 257 Quality and Process Methods Additional Examples of CUSUM Control Charts 5.
Click OK.
6.
In the Target box, type 8.1.
This is the target mean for the process.
7.
In the Sigma box, type 0.05.
This is the known standard deviation for the process.
8.
In the h box, type 2.
This defines the decision limits to be 2 standard deviations in each direction.
Figure 8.7 CUSUM Control Chart with Subgroups The CUSUM Chart does not show any points outside of the upper or lower decision limits. There is no evidence that a shift in the process has occurred.
Note: Montgomery (2013) states that “only if there is some significant economy of scale or some other valid reason for taking samples of size greater than one should subgroups of size greater than one be used with the CUSUM.” The use of rational subgroups in the tabular CUSUM chart does not always improve the performance of the chart.
258 CUSUM Control Charts Chapter 8 Statistical Details for the CUSUM Control Chart Platform Quality and Process Methods Statistical Details for the CUSUM Control Chart Platform • “Statistical Details for CUSUM Control Chart Construction” • “Statistical Details for Shift Detection” • “Statistical Details for Average Run Length” Statistical Details for CUSUM Control Chart Construction This section defines the statistics that are used in the construction of the CUSUM Chart. Some of these statistics are also saved in the data table that is created by the Save Summaries command.
One-Sided Upper and Lower Cumulative Sums The definitions of C+ and C- depend on the setting of the Data Units option.
Note: In the Save Summaries data table, C+ and C- are labeled Upper Cumulative Sum and Lower Cumulative Sum, respectively.
Cumulative Sums in Standardized Units If the Data Units option is not selected, C+ and C- for each step are defined as follows: where: xi is the value of the response at the ith step T is the target of the process is the standard deviation of the process k is the reference value, in units of standard deviations If a value is specified for Head Start, that value is used as the initial C+ value and the negative of that value is used as the initial C- value. Otherwise, the initial values of C+ and C- are zero.
Ci + max 0 xi T – --------------k – Ci 1 – + + = Ci -min 0 xi T – --------------k Ci 1 – -+ + = Chapter 8 CUSUM Control Charts 259 Quality and Process Methods Statistical Details for the CUSUM Control Chart Platform Cumulative Sums in Data Units If the Data Units option is selected, C+ and C- for each step are defined as follows: where: xi is the value of the response at the ith step T is the target of the process is the standard deviation of the process K is the reference value, in units of the data If a value is specified for Head Start, that value is used as the initial C+ value and the negative of that value is used as the initial C- value. Otherwise, the initial values of C+ and C- are zero.
Counters for Positive and Negative Runs N+ at each step is the number of steps since the most recent zero value for C+. N- at each step is the number of steps since the most recent zero value for C-.
Note: In the Save Summaries data table, N+ and N- are labeled Positive Runs and Negative Runs, respectively.
Statistical Details for Shift Detection A positive shift occurs when the value of C+ exceeds the upper limit on the chart. The start of the shift is defined as the first point after the most recent zero value for C+.
A negative shift occurs when the value of C- exceeds the lower limit on the chart. The start of the shift is defined as the first point after the most recent zero value for C-.
Statistical Details for Average Run Length The one-sided average run length (ARL) values are calculated using the integral equation method (with 24 Gaussian points) described by Goel and Wu (1971). If the Head Start value is greater than 0, the values are calculated according to the method in Appendix A.1 of Lucas and Crosier (1982).
Ci + max 0 xi T – K – Ci 1 – + + ( ) = Ci -min 0 xi T – K Ci 1 – -+ + ( ) = 260 CUSUM Control Charts Chapter 8 Statistical Details for the CUSUM Control Chart Platform Quality and Process Methods Note that the two-sided ARL values are related to the positive and negative one-sided ARL values by the following equation: Lucas and Crosier (1982) describe the properties of a Head Start value for CUSUM charts in which the initial CUSUM S0 is set to a nonzero value. This is sometimes referred to as a fast initial response (FIR) feature. Average run length calculations given by them show that the FIR feature has little effect when the process is in control and that it leads to a faster response to an initial out-of-control condition than a standard CUSUM chart.
1 ARL ------------1 Positive ARL -----------------------------------1 Negative ARL --------------------------------------+ = Chapter 9 EWMA Control Charts Create Exponentially Weighted Control Charts Exponentially weighted moving average (EWMA) charts can be used to detect small shifts in a process. Each point on an EWMA chart is the weighted average of all the previous subgroup means, including the mean of the present subgroup sample. The weights decrease exponentially going backward in time.
Figure 9.1 EWMA Control Chart Report 262 EWMA Control Charts Chapter 9 Quality and Process Methods Contents Overview of the EWMA Control Chart Platform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 Example of the EWMA Control Chart Platform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 Launch the EWMA Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 The EWMA Control Chart Platform Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 Control Panel for EWMA Control Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 266 EWMA Chart Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267 The EWMA Control Chart Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 Average Run Length (ARL) Report for EWMA Control Charts. . . . . . . . . . . . . . . . . . . . . . 272 Additional Example of the EWMA Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 Statistical Details for the EWMA Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 Chapter 9 EWMA Control Charts 263 Quality and Process Methods Overview of the EWMA Control Chart Platform Overview of the EWMA Control Chart Platform Exponentially weighted moving average (EWMA) charts can be used to detect small shifts in a process. Each point on an EWMA chart is the weighted average of all the previous subgroup means, including the mean of the present subgroup sample. The weights decrease exponentially going backward in time. For more information about exponentially weighted moving average charts, see Box et al. (2009) and Montgomery (2013).
The EWMA Control Chart platform pairs an EWMA chart with an X chart and a residual chart. If you do not specify a Subgroup variable, the X chart is an individual measurements chart. If you specify a Subgroup variable and at least one subgroup size is greater than 1, the X chart is an XBar chart.
Example of the EWMA Control Chart Platform In the sample data table Clips1.jmp, the measure of interest is the gap between the ends of manufactured metal clips. To monitor the process for a change in the average gap, subgroup samples of five clips have been measured daily.
1.
Select Help > Sample Data Library and open Quality Control/Clips1.jmp.
2.
Select Analyze > Quality and Process > Control Chart > EWMA Control Chart.
3.
Select Gap and click Y.
4.
Select Sample and click Subgroup.
5.
Click OK.
264 EWMA Control Charts Chapter 9 Example of the EWMA Control Chart Platform Quality and Process Methods Figure 9.2 EWMA Control Chart Purple vertical lines in the EWMA chart denote shifts. Shift starts are detected at samples 4 and 17.
Chapter 9 EWMA Control Charts 265 Quality and Process Methods Launch the EWMA Control Chart Platform Launch the EWMA Control Chart Platform Launch the EWMA Control Chart platform by selecting Analyze > Quality and Process > Control Chart > EWMA Control Chart.
Figure 9.3 EWMA Control Chart Launch Window For more information about the options in the Select Columns red triangle menu, see Using JMP.
The EWMA Control Chart platform launch window contains the following options: Y Identifies the variables that you want to chart.
Note: If you do not specify a Subgroup variable, the rows of the data table must be sorted in the order in which the observations were collected.
Subgroup Identifies a subgroup variable. The horizontal axis of the EWMA chart is labeled by the subgroup variable.
By Produces a separate report for each level of the By variable. If more than one By variable is assigned, a separate report is produced for each possible combination of the levels of the By variables.
Center Data Specifies that the data are centered by subtracting the target from each observation.
Show Excluded Region (Applicable only when a Subgroup variable is specified.) Specifies that subgroups that are entirely excluded in the data table are shown in the EWMA control chart.
266 EWMA Control Charts Chapter 9 The EWMA Control Chart Platform Report Quality and Process Methods The EWMA Control Chart Platform Report By default, the EWMA Control Chart platform produces a report that contains a parameter control panel, an EWMA chart, an X chart, and a residuals chart.
Figure 9.4 EWMA Control Chart Report Control Panel for EWMA Control Chart The Control Panel contains the current values for the chart parameters. The current values are in boxes that enable you to update the parameter values.
Chapter 9 EWMA Control Charts 267 Quality and Process Methods The EWMA Control Chart Platform Report The following options appear in the Control Panel: Target The known value of the mean. This is the value of the center line in the chart. By default, this parameter is set to the Target value in the Spec Limits column property for the Y column. If the Y column does not have a Target value in the Spec Limits column property, this parameter is set to the overall average of the Y column.
The Target value cannot be edited in these situations: – If you specified the Center Data option in the launch window.
– If the Y variable has a Spec Limits column property that contains a value for Target. However, you can select the Use Overall Mean for Target option in the red triangle menu to switch between the Target value in the Spec Limits column property and the overall mean of the Y variable.
Sigma The known value of the standard deviation. By default, this parameter is set to the average moving range of the Y column. If there is a Subgroup variable, the Sigma parameter is set to the average of the moving ranges of the subgroup means.
Note: The default Sigma value calculation does not include excluded rows.
Lambda The value of the smoothing constant for weighting prior samples. By default, this parameter is set to 0.2.
Using Centered Data The presence of this text indicates that the Centered Data option was selected in the launch window and that the values in the EWMA chart and X chart are centered around the Target value.
EWMA Chart Report The EWMA Chart report contains three charts: an EWMA chart, an X chart, and a residuals chart. For more information about the interpretation of these charts, see Box et al. (2009).
EWMA Chart The EWMA chart is an exponentially weighted moving average (EWMA) chart with decision limits that are determined by the current values of the chart parameters: • If you specified a Subgroup variable and at least one subgroup size is greater than 1, the horizontal axis denotes the subgroups.
• Otherwise, the horizontal axis denotes the samples.
268 EWMA Control Charts Chapter 9 The EWMA Control Chart Platform Report Quality and Process Methods In both cases, the vertical axis denotes the exponentially weighted moving average. If you specified the Center Data option in the launch window, the vertical axis denotes the exponentially weighted moving average with the target value subtracted from it. Each sample or subgroup has a single point on the chart. There is one additional point that is a forecast point, which is shown in blue.
Note: If the last sample (or subgroup) is both hidden and excluded, the line connecting the last sample (or subgroup) to the forecast point is not drawn.
X Chart The X chart is a Shewhart control chart of the observations: • If you specified a Subgroup variable and at least one subgroup size is greater than 1, the X chart is an XBar chart of the mean values. The horizontal axis denotes the subgroups. The vertical axis denotes the subgroup means. If you selected the Center Data option in the launch window, the vertical axis denotes the subgroup means with the target value subtracted from them. Each subgroup has a single point on the chart.
• Otherwise, the X chart is an individual measurements chart of the values. The horizontal axis denotes the samples. The vertical axis denotes the measurements. If you selected the Center Data option in the launch window, the vertical axis denotes the measurements with the target value subtracted from them. Each sample has a single point on the chart.
For more information about the limits on the X chart, see “Statistical Details for Control Chart Builder”.
Residuals Chart The residuals chart enables you to visually check for autocorrelation: • If you specified a Subgroup variable and at least one subgroup size is greater than 1, the residuals chart is a chart of the differences between each subgroup mean and the EWMA value for the previous subgroup. The ith residual is calculated as ri = Xi - EWMAi-1 where Xi denotes the ith subgroup mean and EWMAi-1 denotes the (i-1)th EWMA value.
• Otherwise, the residuals chart is a chart of the differences between each sample value and the EWMA value for the previous sample. The ith residual is calculated as ri = Xi - EWMAi-1 where Xi denotes the ith sample value and EWMAi-1 denotes the (i-1)th EWMA value.
The limits on the residuals chart are ±3ResidSigma, where ResidSigma is the standard deviation of the residuals.
Chapter 9 EWMA Control Charts 269 Quality and Process Methods The EWMA Control Chart Platform Options The EWMA Control Chart Platform Options The EWMA Control Chart red triangle menu contains the following options: Show Limits Shows or hides the upper and lower decision limits in the EWMA chart, X chart, and residuals chart.
Show Center Line Shows or hides the center line in the EWMA chart, X chart, and residuals chart.
Show Shift Lines Shows or hides the vertical lines in the EWMA chart that designate shifts. Shift lines are drawn at the start of a shift. A shift start is defined as the first point after the EWMA value crosses the center line in a particular direction.
– A positive shift occurs when the EWMA value exceeds the upper limit on the chart. The start of the shift is defined as the first point after the most recent EWMA value below the Target line.
– A negative shift occurs when the EWMA value falls below the lower limit on the chart. The start of the shift is defined as the first point after the most recent EWMA value above the Target line.
Figure 9.5 Shift Lines Example In this example, the point for sample 7 is above the upper limit. Looking back from sample 7, the most recent point below the Target line is sample 3. Therefore, sample 4 is the first point above the Target line since the most recent point below the Target line and is denoted as the start of the positive shift.
Similarly, the point for sample 20 is below the lower limit and the most recent point above the Target line is sample 16. Therefore, sample 17 is denoted as the start of the negative shift.
Test Beyond Limits Shows or hides a red circle around any point that is above the upper limit or below the lower limit in the EWMA and X charts.
270 EWMA Control Charts Chapter 9 The EWMA Control Chart Platform Options Quality and Process Methods Show ARL Shows or hides the Average Run Length (ARL) report. See “Average Run Length (ARL) Report for EWMA Control Charts”.
Control Panel Shows or hides a report of the current values of the parameters. This report enables you to change the parameter values in the EWMA chart.
Parameters Report Shows or hides a report of the current values of the parameters.
Constant Limits Specifies that the EWMA chart limits are calculated using an asymptotic expression so that the limits on the EWMA chart are constant.
Caution: The Constant Limits option has no effect when the sample sizes are not equal across subgroups.
Save Summaries Creates a new data table that contains statistics for each subgroup in the EWMA chart. The following statistics are saved to the new data table: the subgroup number, the subgroup label, the subgroup size, the subgroup mean, an indicator of shift starts, a value that indicates each interval between shift starts, the exponentially weighted moving average of each subgroup, the number of positive and negative consecutive run counts, the LCL and UCL values, and test failure indicators. The forecast value is saved in the last row of the summary table.
Reset to Defaults Resets all parameters to their default values. If a Lambda value has been specified in the Lambda platform preference, the Lambda value is reset to the value that is specified in the platform preference.
Note: When the Y variable has a Spec Limits column property that contains a value for Target, the Reset to Defaults option sets the Target to the Target value in the Spec Limits column property.
Restart EWMA After Empty Subgroup Specifies how calculations for the moving average and limits are handled when there are empty subgroups. A subgroup can be empty if all the observations for the subgroup are missing values or are in excluded rows. If this option is selected, the calculations restart in the first nonmissing subgroup that follows an empty subgroup. The restart of the calculations resets the moving average to the overall mean. If this option is not selected, the EWMA calculations continue with the most recent nonmissing subgroup moving average.
Use Overall Mean for Target (Available only when the Y variable has a Spec Limits column property that contains a value for Target.) Sets the Target in the EWMA chart to the overall mean of the Y variable. If this option is not selected, the Target in the EWMA chart is set to the Target value in the Spec Limits column property.
Overlay Charts Specifies whether the individual location values are overlaid on the EWMA chart. When this option is selected, the Location chart is no longer shown. Instead, the Chapter 9 EWMA Control Charts 271 Quality and Process Methods The EWMA Control Chart Platform Options points from the Location chart appear on the EWMA chart as unconnected gray Xs. The limits from the Location chart do not appear on the EWMA chart, unless the Show X Limits on Overlay Charts option is selected.
Show X Limits on Overlay Charts (Available only when the Overlay Charts option is selected.) Shows or hides the limits from the Location chart on the EWMA chart when the Overlay Charts option is selected. When the Location chart limits are shown on the EWMA chart, they appear as dashed lines.
Note: The Test Beyond Limits option is applied to the Location chart values in the Overlay Chart only when the Show X Limits on Overlay Charts option is selected.
Lambda Slider Shows or hides a slider control that enables you to change the value of the Lambda parameter interactively.
Show X Chart Shows or hides the Location chart below the EWMA chart.
Note: When the Overlay Charts option is selected, the Show X Chart option shows or hides the individual location values and limits that are overlaid on the EWMA chart.
Show Residuals Chart Shows or hides a chart of residuals.
Alarm Script Enables you to write and run a script that indicates when the data fail special causes tests. Results can be written to a file, written to the log or sent in an email. There is an option to include an explanation of why the test failed.
As an Alarm Script is invoked, the following variables are available, both in the issued script and in subsequent JSL scripts: qc_col is the name of the column qc_test is the test that failed qc_sample is the sample number Tip: After an alarm script is specified, the alarm script is invoked when the Test Beyond Limits option is turned on.
See the Scripting Guide for more information about writing custom Alarm Scripts.
See Using JMP for more information about the following options: Redo Contains options that enable you to repeat or relaunch the analysis. In platforms that support the feature, the Automatic Recalc option immediately reflects the changes that you make to the data table in the corresponding report window.
Save Script Contains options that enable you to save a script that reproduces the report to several destinations.
272 EWMA Control Charts Chapter 9 The EWMA Control Chart Platform Options Quality and Process Methods Save By-Group Script Contains options that enable you to save a script that reproduces the platform report for all levels of a By variable to several destinations. Available only when a By variable is specified in the launch window.
Average Run Length (ARL) Report for EWMA Control Charts The Average Run Length (ARL) report contains a table and a graph of ARL values. The average run length (ARL) for a specified shift is the average number of runs expected before an out-of-control signal occurs. For example, the ARL at 0 represents the average number of runs expected before seeing a false-alarm signal when the process is in control. When the process is in control, the shift size is 0.
The table and graph in the ARL report enable you to explore how various settings of the Lambda parameter affect the performance of the corresponding EWMA chart. The table and graph also enable you to compare the performance of the EWMA chart with a Shewhart chart, such as the X chart in the EWMA Chart report. The Shewhart ARL column is equivalent to the EWMA ARL column when Lambda is set to 1.
The value of EWMA ARL at 0 depends on the setting of the Constant Limits option: • If the Constant Limits option is selected, the process is assumed to have been in control long enough that the effect of the starting value is negligible. In this case, also referred to as a steady state of the EWMA chart, the value of EWMA ARL(0) is calculated using the method described in Crowder (1987).
• If the Constant Limits option is not selected, the value of EWMA ARL(0) is calculated using the method described in Knoth (2004). This situation is also referred to as a zero state of the EWMA chart.
ARL Report The ARL Report shows the average run length for standardized shifts () between 0 and 3 at 0.25 increments. The shift is represented by sqrt(n)RawShift/Sigma, where n is the constant sample size in each subgroup. For the multiplier used in calculating control limits, K = 3 is assumed. This table contains ARL values for the EWMA chart as well as a Shewhart chart.
ARL Graph The ARL Graph shows the average run length for standardized shifts () between 0 and 3. This graph contains the same data points as the ARL Table to the left of the ARL Graph. The solid line corresponds to the EWMA ARL values, and the dashed line corresponds to the Shewhart ARL values.
Chapter 9 EWMA Control Charts 273 Quality and Process Methods Additional Example of the EWMA Control Chart Platform Additional Example of the EWMA Control Chart Platform In this example, you create a chart that overlays the subgroup points and limits on the EWMA control chart. This enables you to see the EWMA points in the context of the subgroup points. The Quench.jmp sample data table contains simulated observations of temperature at the time of magnet quench. Magnet quench occurs when part of the superconducting coil enters a resistive state and increases in temperature. When this happens, containment of the fusion reactor is at risk. Immediately shutting down power to the magnet is required to prevent a breach.
1.
Select Help > Sample Data Library and open Quality Control/Quench.jmp.
2.
Select Analyze > Quality and Process > Control Chart > EWMA Control Chart.
3.
Select Temp and click Y.
4.
Select Time stamp and click Subgroup.
5.
Click OK.
6.
Click the EWMA Control Chart red triangle and select Test Beyond Limits.
274 EWMA Control Charts Chapter 9 Additional Example of the EWMA Control Chart Platform Quality and Process Methods Figure 9.6 EWMA Chart with Points Beyond Limits Highlighted Note that three observations are outside the limits in the EWMA chart and one observation is outside the limits in the X chart. The observations outside of the limits are indicated by a red circle. The purple vertical line in the EWMA chart denotes a shift upward around 1:16 AM.
7.
Click the EWMA Control Chart red triangle and select Overlay Charts.
The X chart limits are not shown on the overlay chart by default, so the observation outside the X limits is not circled until you turn on the X limits for the overlay chart.
8.
Click the EWMA Control Chart red triangle and select Show X Limits on Overlay Charts.
Chapter 9 EWMA Control Charts 275 Quality and Process Methods Additional Example of the EWMA Control Chart Platform Figure 9.7 EWMA Chart with Overlaid X Chart The solid lines and circles in the top chart represent the limits and points for the EWMA chart. The dashed lines and x markers represent the limits and points for the X chart. Note that the observations outside the EWMA limits and the X limits are still circled.
9.
Click the EWMA Control Chart red triangle and select Constant Limits.
276 EWMA Control Charts Chapter 9 Statistical Details for the EWMA Control Chart Platform Quality and Process Methods Figure 9.8 EWMA Chart with Constant Limits Because the time span of the data is so long, the asymptotic limits differ from the original EWMA limits only for the first few subgroups. In this case, you might decide that you can use constant (asymptotic) limits instead of the EWMA limits.
Statistical Details for the EWMA Control Chart Platform This section defines the statistics that are used in the construction of the EWMA chart. Some of these statistics are also saved in the data table that is created by the Save Summaries command.
The ith point on the EWMA chart is calculated as follows: EWMAi = Xi + (1 - )EWMAi-1 where: = Lambda parameter Xi = ith sample value (or subgroup mean) Chapter 9 EWMA Control Charts 277 Quality and Process Methods Statistical Details for the EWMA Control Chart Platform EWMAi-1 = (i-1)th EWMA value When i = 1, define EWMA0 as the Target value.
Note: When the Restart EWMA after Empty Subgroup option is selected, the EWMAi-1 value following an empty subgroup is the Target value. When the Restart EWMA after Empty Subgroups option is not selected, the EWMAi-1 value following an empty subgroup is the EWMA value for the most recent non-empty subgroup.
The computation of the control limits on the EWMA chart is determined by the setting of the Constant Limits option.
EWMA Limits When the Constant Limits option is not selected and the subgroup sizes are not equal, the EWMA control limits are computed as follows: LCL = UCL = where: T = Target value K = the sigma multiplier and is set to 3 by default = Sigma value i = the number of the sample (or subgroup) ni = the size of subgroup i When the Constant Limits option is not selected and the subgroup sizes are equal, the formulas for the EWMA control limits simplify as follows: LCL = UCL = T K 1 – 2 i j – nj ---------------------------------j 1 = i – T K 1 – 2 i j – nj ---------------------------------j 1 = i + T K n 2 – -------------------- 1 1 – 2i – – T K n 2 – -------------------- 1 1 – 2i – + 278 EWMA Control Charts Chapter 9 Statistical Details for the EWMA Control Chart Platform Quality and Process Methods where: T = Target value K = the sigma multiplier and is set to 3 by default = Sigma value i = the number of the sample (or subgroup) n = the size of each subgroup (or 1 if no subgroup is specified) Constant Limits When the Constant Limits option is selected, the EWMA control limits are computed as follows: LCL = UCL = where: T = Target value K = the sigma multiplier and is set to 3 by default = Sigma value n = the subgroup size (or 1 if no subgroup is specified) For more information about constructing exponentially weighted moving average charts, see Montgomery (2013).
T K n 2 – --------------------– T K n 2 – --------------------+ Chapter 10 Multivariate Control Charts Monitor Multiple Process Characteristics Simultaneously Multivariate control charts are used to monitor two or more interrelated process variables. Where univariate control charts are used to monitor a single independent process characteristic, multivariate control charts are necessary when process variables are correlated. The Multivariate Control Chart platform enables you to build Hotelling T2 charts. You can use the platform to determine whether a process is stable as well as to monitor a process as new data are collected.
For monitoring and diagnosing complex processes, see “Model Driven Multivariate Control Charts”.
Figure 10.1 Example of a Multivariate Control Chart 280 Multivariate Control Charts Chapter 10 Quality and Process Methods Contents Overview of Multivariate Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 Example of a Multivariate Control Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281 Launch the Multivariate Control Chart Platform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 The Multivariate Control Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 Multivariate Control Chart Platform Options. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 286 T Square Partitioned . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 Change Point Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 288 Principal Components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 Additional Examples of Multivariate Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 Example of Monitoring a Process Using Sub-Grouped Data . . . . . . . . . . . . . . . . . . . . . . . . 289 Example of T Square Partitioned . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 Example of Change Point Detection. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294 Statistical Details for Multivariate Control Charts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 Statistical Details for Individual Observations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296 Statistical Details for Observations in Rational Subgroups . . . . . . . . . . . . . . . . . . . . . . . . . 297 Statistical Details for Change Point Detection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 Chapter 10 Multivariate Control Charts 281 Quality and Process Methods Overview of Multivariate Control Charts Overview of Multivariate Control Charts Multivariate control charts are used to monitor two or more interrelated process variables. Where univariate control charts are used to monitor a single independent process characteristic, multivariate control charts are necessary when process variables are correlated. A Hotelling T2 chart, or just T2 chart for short, is one type of multivariate control chart. A T2 chart can detect shifts in the mean or the relationship between several interrelated variables. The observations can either be individual observations of the process variables or they can be grouped into rational subgroups.
You can construct a multivariate control chart using current or historical data. The control chart is said to be a Phase I chart if it is constructed using current data; the control chart is said to be a Phase II chart if it is constructed using target statistics from a historical data set. In Phase I, you check that the process is stable and establish a historical data set from which to calculate target statistics for the process. In Phase II, the multivariate control chart uses the target statistics from Phase I in order to monitor new process observations.
To construct a Phase II multivariate control chart, first identify a period of time during which the process is stable and capable.
1.
Develop a Phase I control chart to verify that the process is stable over this period.
The data used in Phase I provides a historical data set.
2.
Save the target statistics for this historical data set.
3.
Monitor the on-going process using a Phase II control chart based on the target statistics that were saved in step 2.
Example of a Multivariate Control Chart This example illustrates constructing a control chart for data that are not sub-grouped. The data are measurements on a steam turbine engine. For an example that uses sub-grouped data, “Example of Monitoring a Process Using Sub-Grouped Data”.
Step 1: Determine Whether the Process Is Stable 1.
Select Help > Sample Data Library and open Quality Control/Steam Turbine Historical.jmp.
2.
Select Analyze > Quality and Process > Control Chart > Multivariate Control Chart.
3.
Select all of the columns and click Y, Columns.
4.
Click OK.
282 Multivariate Control Charts Chapter 10 Example of a Multivariate Control Chart Quality and Process Methods Figure 10.2 Initial Multivariate Control Chart The process seems to be in reasonable statistical control, because there is only one out-of-control point. Therefore, it is appropriate to create targets based on this data.
Step 2: Save Target Statistics 1.
Click the red triangle next to Multivariate Control Chart and select Save Target Statistics.
This creates a new data table containing target statistics for the process.
Figure 10.3 Target Statistics for Steam Turbine Data 2.
Save the new data table as Steam Turbine Targets.jmp.
Now that target statistics have been established, create the multivariate control chart that monitors the process.
Step 3: Monitor the Process 1.
Select Help > Sample Data Library and open Quality Control/Steam Turbine Current.jmp.
Chapter 10 Multivariate Control Charts 283 Quality and Process Methods Example of a Multivariate Control Chart This sample data table contains recent observations from the process.
2.
Select Analyze > Quality and Process > Control Chart > Multivariate Control Chart.
3.
Select all of the columns and click Y, Columns.
4.
Click Get Targets.
5.
Open the Steam Turbine Targets.jmp table that you saved.
6.
Click OK.
The default alpha level is set to 0.05. Change it to 0.001.
7.
Click the red triangle next to Multivariate Control Chart and select Set Alpha Level > Other.
8.
Type 0.001 and click OK.
Figure 10.4 Steam Turbine Control Chart Figure 10.4 shows out-of-control conditions occurring at observations 2, 3, 4, 5, and 8. This result implies that these observations do not conform to the historical data from Steam Turbine Historical.jmp, and that the process should be further investigated. To find an assignable cause, you might want to examine individual univariate control charts or perform another univariate procedure.
284 Multivariate Control Charts Chapter 10 Launch the Multivariate Control Chart Platform Quality and Process Methods Launch the Multivariate Control Chart Platform Launch the Multivariate Control Chart platform by selecting Analyze > Quality And Process > Control Chart > Multivariate Control Chart.
Figure 10.5 The Multivariate Control Chart Launch Window For more information about the options in the Select Columns red triangle menu, see Using JMP.
The Multivariate Control Chart platform launch window contains the following options: Y, Columns Specify the columns to be analyzed.
Subgroup Enter a column with sub-grouped data. Hierarchically, this group is nested within Group.
Group Enter a column that specifies group membership at the highest hierarchical level.
Weight Identifies the data table column whose variables assign weight (such as importance or influence) to the data.
Freq Identifies the data table column whose values assign a frequency to each row. Can be useful when your data table contains summarized data.
By Identifies a column that creates a report consisting of separate analyses for each level of the variable.
Get Targets Click to select a JMP table that contains historical targets for the process.
Chapter 10 Multivariate Control Charts 285 Quality and Process Methods The Multivariate Control Chart The Multivariate Control Chart Use the multivariate control chart to quickly identify shifts in your process and to monitor your process for special cause indications.
Follow the instructions in “Example of a Multivariate Control Chart” to produce the results shown in Figure 10.6.
Figure 10.6 Multivariate Control Chart Tip: For information about additional options, see “Multivariate Control Chart Platform Options”.
286 Multivariate Control Charts Chapter 10 Multivariate Control Chart Platform Options Quality and Process Methods The multivariate control chart plots Hotelling’s T2 statistic. The calculation for the control limit differs based on whether targets have been specified. To understand how the T2 statistic and the UCL (Upper Control Limit) are calculated, see “Statistical Details for Multivariate Control Charts”. For more information about control limits, see Tracy et al. (1992).
In this example, the Principal Components reports for both data sets indicate that the first eigenvalue, corresponding to the first principal component, explains about 95% of the total variation in the variables. The values in both Eigenvectors tables indicate that the first principal component is driven primarily by the variables Fuel and Steam Flow. You can use this information to construct a potentially more sensitive control chart based only on this first component. For more information about the Principal Components reports, see “Principal Components”.
Multivariate Control Chart Platform Options The Multivariate Control Chart red triangle menu contains the following options: T Square Chart Shows the T2 chart. Hotelling’s T2 chart is a multivariate extension of the XBar chart that takes correlation into account.
T Square Partitioned Constructs multivariate control charts based on the principal components of Y. Specify the number of major principal components for T2. See “T Square Partitioned”.
Set Alpha Level Set the level used to calculate the control limit. The default is =0.05.
Show Covariance Shows the Covariance report. Covariance is a measure of the linear relationship between two variables. When a subgroup variable is specified, the Pooled Covariance report is shown.
Show Correlation Shows the Correlation report. When a subgroup variable is specified, the Pooled Correlation report is shown.
Show Inverse Covariance Shows the Inverse Covariance report. If the inverse covariance is singular, a generalized inverse of the covariance matrix is reported. When a subgroup variable is specified, the Pooled Inverse Covariance report is shown.
Show Inverse Correlation Shows the Inverse Correlation report. If the inverse correlation is singular, a generalized inverse of the correlation matrix is reported. When a subgroup variable is specified, the Pooled Inverse Correlation report is shown.
Show Means Shows the Group Means report, which contains the means for each group.
Save T Square Creates a new column in the data table containing T2 values.
Chapter 10 Multivariate Control Charts 287 Quality and Process Methods Multivariate Control Chart Platform Options Save T Square Formula Creates a new column in the data table. Stores a formula in the column that calculates the T2 values.
Save Target Statistics Creates a new data table containing target statistics for the process. Target statistics include: sample size, the number of samples, mean, standard deviation, and any correlations.
Change Point Detection (Not available for sub-grouped data.) Shows a Change Point Detection plot of test statistics by row number and indicates the row number where the change point appears. See “Change Point Detection”.
Principal Components Shows reports showing eigenvalues and their corresponding eigenvectors. Principal components help you understand which of the many variables you might be monitoring are primarily responsible for the variation in your process. See “Principal Components”.
Save Principal Components Creates new columns in the data table that contain the scaled principal components.
T Square Partitioned If you are monitoring a large number of correlated process characteristics, you can use the T Square Partitioned option to construct a control chart based on principal components. If a small number of principal components explains a large portion of the variation in your measurements, then a multivariate control chart based on these big components might be more sensitive than a chart based on your original, higher-dimensional data.
The T Square Partitioned option is also useful when your covariance matrix is ill-conditioned. When this is the case, components with small eigenvalues, explaining very little variation, can have a large, and misleading, impact on T2. It is useful to separate out these less important components when studying process behavior.
Once you select the T Square Partitioned option, you need to decide how many major principal components to use.
The option creates two multivariate control charts: T Square with Big Principal Components and T Square with Small Principal Components. Suppose that you enter r as the number of major components when you first select the option. The chart with Big Principal Components is based on the r principal components corresponding to the r largest eigenvalues. These are the r components that explain the largest amount of variation, as shown in the Percent and Cum Percent columns in the Principal Components: on Covariances reports. The chart with Small Principal Components is based on the remaining principal components.
288 Multivariate Control Charts Chapter 10 Multivariate Control Chart Platform Options Quality and Process Methods For a given subgroup, its T2 value in the Big Principal Components chart and its T2 value in the Small Principal Components chart sum to its overall T2 statistic presented in the T2 with All Principal Components report. For more information about how the partitioned T2 values are calculated, see Kourti and MacGregor (1996).
Tip: Consider using the Model Driven Multivariate Control Chart platform for decomposition of the T2 statistic. See “Model Driven Multivariate Control Charts”.
Change Point Detection When the data set consists of multivariate individual observations, a control chart can be developed to detect a shift in the mean vector, the covariance matrix, or both. This method partitions the data and calculates likelihood ratio test statistics for a shift. The statistic that is plotted on the control chart is an observation’s likelihood ratio test statistic divided by the product of the following: • Its approximate expected value assuming no shift.
• An approximate value for an upper control limit.
Division by the approximate upper control limit allows the points to be plotted against an effective upper control limit of 1. A Change Point Detection plot readily shows the change point for a shift occurring at the maximized value of the control chart statistic. The Change Point Detection implementation in JMP is based on Sullivan and Woodall (2000) and is described in “Statistical Details for Change Point Detection”.
Note: The Change Point Detection method is designed to show a single shift in the data. Detect multiple shifts by recursive application of this method.
Note the following about the Change Point Detection plot: • Values above 1.0 indicate a possible shift in the data.
• Control chart statistics for the Change Point Detection plot are obtained by dividing the likelihood ratio statistic of interest (either a mean vector or a covariance matrix) by a normalizing factor.
• The change point of the data occurs for the observation having the maximum test statistic value for the Change Point Detection plot.
Note the following about the scatterplot matrix: • This plot shows the shift in the sample mean vector.
• In the “Example of Change Point Detection”, data are divided into two groups. The first 24 observations are classified as the first group. The remaining observations are classified as the second group.
Chapter 10 Multivariate Control Charts 289 Quality and Process Methods Additional Examples of Multivariate Control Charts Principal Components The Principal Components report contain the following information: Eigenvalue Eigenvalues for the covariance matrix.
Percent Percent variation explained by the corresponding eigenvector. Also shows an accompanying bar chart.
Cum Percent Cumulative percent variation explained by eigenvectors corresponding to the eigenvalues.
ChiSquare Provides a test of whether the correlation remaining in the data is of a random nature. This is a Bartlett test of sphericity. When this test rejects the null hypothesis, this implies that there is structure remaining in the data that is associated with this eigenvalue.
DF Degrees of freedom associated with the Chi-square test.
Prob > ChiSq p-value for the test.
Eigenvectors Table of eigenvectors corresponding to the eigenvalues. Note that each eigenvector is divided by the square root of its corresponding eigenvalue.
For more information about principal components, see Multivariate Methods.
Additional Examples of Multivariate Control Charts • “Example of Monitoring a Process Using Sub-Grouped Data” • “Example of T Square Partitioned” • “Example of Change Point Detection” Example of Monitoring a Process Using Sub-Grouped Data The workflow for monitoring a multivariate process with sub-grouped data is similar to the one for individual data. See “Example of a Multivariate Control Chart”. You create an initial control chart to save target statistics and then use these targets to monitor the process.
Step 1: Determine Whether the Process Is Stable 1.
Select Help > Sample Data Library and open Quality Control/Aluminum Pins Historical.jmp.
2.
Select Analyze > Quality and Process > Control Chart > Multivariate Control Chart.
3.
Select all of the Diameter and Length columns and click Y, Columns.
290 Multivariate Control Charts Chapter 10 Additional Examples of Multivariate Control Charts Quality and Process Methods 4.
Select subgroup and click Subgroup.
5.
Click OK.
Figure 10.7 Multivariate Control Chart for Sub-Grouped Data, Step 1 The process appears to be in statistical control, making it appropriate to create targets using this data.
Step 2: Save Target Statistics 1.
Click the red triangle next to Multivariate Control Chart and select Save Target Statistics.
This creates a new data table containing target statistics for the process.
2.
Save the new data table as Aluminum Pins Targets.jmp.
Now that target statistics have been established, create the multivariate control chart for process monitoring.
Step 3: Monitor the Process 1.
Select Help > Sample Data Library and open Quality Control/Aluminum Pins Current.jmp.
This sample data table contains recent observations from the process.
2.
Select Analyze > Quality and Process > Control Chart > Multivariate Control Chart.
3.
Select all of the Diameter and Length columns and click Y, Columns.
4.
Select subgroup and click Subgroup.
5.
Click Get Targets.
6.
Open the Aluminum Pins Targets.jmp table that you saved.
7.
Click OK.
Chapter 10 Multivariate Control Charts 291 Quality and Process Methods Additional Examples of Multivariate Control Charts 8.
Click the red triangle next to Multivariate Control Chart and select Show Means.
The Show Means option gives the means for each subgroup. You can then observe which groups are most dissimilar from each other.
Figure 10.8 Multivariate Control Chart for Sub-Grouped Data, Step 3 292 Multivariate Control Charts Chapter 10 Additional Examples of Multivariate Control Charts Quality and Process Methods Figure 10.8 shows indications of instability at subgroups 4-7, 9-11, 18, and 20. This result implies that these observations do not conform to the historical data from Aluminum Pins Historical.jmp, and that the process should be further investigated. To determine why the process was out of control at these points, you might want to examine individual univariate control charts or perform another univariate procedure.
An alternative method to monitoring this process is based on the big principal components. In this example, for the historical data, the first three principal components account for about 98% of the variation. Based on this, you might construct a chart for the first three principal components. Then you would monitor current data using those three components. The control limits for the chart used in monitoring the process should be based on the corresponding chart for the historical data.
Example of T Square Partitioned Use T Square Partitioned to separate out the more important components from the less important components when studying process behavior. In this example, the coating on each of 50 bars was measured at 12 uniformly spaced locations across the bar. You want to examine the variation in the measurements and determine whether the causes of variation need to be investigated further.
1.
Select Help > Sample Data Library and open Quality Control/Thickness.jmp.
2.
Select Analyze > Quality and Process > Control Chart > Multivariate Control Chart.
3.
Select all of the Thickness columns and click Y, Columns.
4.
Click OK.
The current alpha level is set to 0.05, which corresponds to a 5% false alarm rate. You want to set the false alarm rate to 1%.
5.
To change the alpha level, click the red triangle next to Multivariate Control Chart, select Set Alpha Level, and choose 0.01.
Chapter 10 Multivariate Control Charts 293 Quality and Process Methods Additional Examples of Multivariate Control Charts Figure 10.9 Initial Multivariate Control Chart for Thickness.jmp The overall control chart in Figure 10.9 suggests that special causes affected bars 1, 2, 4, 5, and 22. Looking at the Principal Components report, you can see that almost 95% of the variation in the 12 thickness measurements is explained by the first principal component. You want to study the variation associated with this principal component further.
6.
Click the red triangle Multivariate Control Chart and select T Square Partitioned.
7.
Accept the default value of 1 principal component by clicking OK.
294 Multivariate Control Charts Chapter 10 Additional Examples of Multivariate Control Charts Quality and Process Methods Figure 10.10 T Square Partitioned Control Charts In contrast to the Principal Components report, the T Square with Big Principal Components chart, which reflects variation for only the first component, shows no evidence of special causes. The T Square with Small Principal Components chart shows that the special cause indications reside in the remaining smaller components. These smaller components do not explain much variation, and likely represent random noise. Therefore, you might conclude that the variation in the thickness measurements is not a major cause for concern.
Example of Change Point Detection Use change point detection to find the point at which a shift occurs in your data.
1.
Select Help > Sample Data Library and open Quality Control/Gravel.jmp.
2.
Select Analyze > Quality and Process > Control Chart > Multivariate Control Chart.
3.
Select Large and Medium and click Y, Columns.
4.
Click OK.
Chapter 10 Multivariate Control Charts 295 Quality and Process Methods Statistical Details for Multivariate Control Charts 5.
Click the red triangle next to Multivariate Control Chart and select Change Point Detection.
Figure 10.11 Change Point Detection for Gravel.jmp Tip: You might need to drag the axes to see the density ellipses for the two groups, depending on your data.
In the Change Point Detection plot, values above 1.0 indicate a possible shift in the data. At least one shift is apparent; the change point occurs at observation 24 and the shift occurs immediately after observation 24. The 95% prediction regions for the two groups have approximately the same size, shape, and orientation, visually indicating that the sample covariance matrices are similar.
Statistical Details for Multivariate Control Charts • “Statistical Details for Individual Observations” • “Statistical Details for Observations in Rational Subgroups” • “Statistical Details for Change Point Detection” 296 Multivariate Control Charts Chapter 10 Statistical Details for Multivariate Control Charts Quality and Process Methods Statistical Details for Individual Observations Consider measurements that are not sub-grouped, that is, where the natural subgroup size is n = 1. Denote the number of observations by m and the number of variables measured by p. A T2 statistic is calculated and plotted for each observation. The calculation of the T2 statistic and upper control limit (UCL) depends on the source of the target statistics. In a Phase I chart, the limits are based on the same data that is being plotted on the control chart. In a Phase II chart, the limits are based on target statistics that were calculated from a historical data set. For more information about T2 statistic and control limit calculations for Hotelling T2 control charts, see Montgomery (2013).
Calculations for Phase I Control Charts In Phase I control charts, the T2 statistic for the ith observation is defined as follows: where: Yi is the column vector of p measurements for the ith observation is the column vector of sample means of the p variables S-1 is the inverse of the sample covariance matrix The Ti 2 value for each of the i observations are the points plotted on the multivariate control chart.
When computing Phase I control limits, the UCL is based on the beta distribution. Specifically, the upper control limit (UCL) is defined as follows: where: p = number of variables m = number of observations = (1–)th quantile of a Beta distribution Ti 2 Yi Y – 'S 1 – Yi Y – = Y UCL m 1 – 2 m --------------------- 1 p 2 -- m p – 1 – 2 ----------------------; ; – = 1 p 2 -- m p – 1 – 2 ----------------------; ; – p 2 -- m p – 1 – 2 ---------------------- Chapter 10 Multivariate Control Charts 297 Quality and Process Methods Statistical Details for Multivariate Control Charts Calculations for Phase II Control Charts In Phase II control charts, define the historical data set as X. Then the T2 statistic for the ith observation is defined as follows: where: Yi is the column vector of p measurements for the ith observation is the column vector of sample means of the p variables, calculated from the historical data set SX -1 is the inverse of the sample covariance matrix, calculated from the historical data set The Ti 2 value for each of the i observations are the points plotted on the multivariate control chart.
When computing Phase II control limits, new observations are independent of the historical data set. In this case, the upper control limit (UCL) is a function of the F distribution and partially depends on the number of observations in the historical data set from which the targets are calculated. The UCL is defined as follows: where: p = number of variables m = number of observations in the historical data set = (1–)th quantile of an F distribution Statistical Details for Observations in Rational Subgroups Consider the case where p variables are monitored and m subgroups of size n > 1 are obtained. A T2 statistic is calculated and plotted for each subgroup. The calculation of the T2 statistic and upper control limit (UCL) depends on the source of the target statistics. In a Phase I chart, the limits are based on the same data that is being plotted on the control chart. In a Phase II chart, the limits are based on target statistics that were calculated from a historical data set. For more information about T2 statistic and control limit calculations for Hotelling T2 control charts, see Montgomery (2013).
Ti 2 Yi X – 'SX 1 – Yi X – = X UCL p m 1 + m 1 – m m p – -----------------------------------------F 1 – p m p – if m 100 p m 1 – m p – ---------------------F 1 – p m p – if m 100 = F 1 – p m p – p m p – 298 Multivariate Control Charts Chapter 10 Statistical Details for Multivariate Control Charts Quality and Process Methods Calculations for Phase I Control Charts For Phase I control charts, the T2 statistic for the jth subgroup is defined as follows: where: is the mean of the n column vectors of p measurements for the jth subgroup is the mean of the subgroup means Sj is the sample covariance matrix for the n observations in the jth subgroup is the pooled covariance matrix, calculated as the mean of the within-subgroup covariance matrices The Phase I upper control limit (UCL) is defined as follows: where: p = number of variables n = sample size for each subgroup m = number of subgroups = (1–)th quantile of an F distribution Calculations for Phase II Control Charts In Phase II control charts, define the historical data set from which the target statistics are calculated as X. Then the T2 statistic for the jth subgroup is defined as follows: where: is the mean of the n column vectors of p measurements for the jth subgroup Tj 2 Yj Y – 'Sp 1 – Yj Y – = Yj Y 1 m ----Yj j 1 = m = Sp 1 m ----Sj j 1 = m = UCL p m 1 – n 1 – mn m – p – 1 + ---------------------------------------F 1 – p mn m p – 1 + – = F 1 – p mn m p – 1 + – p mn m p – 1 + – Tj 2 Yj X – 'Sp 1 – Yj X – = Yj Chapter 10 Multivariate Control Charts 299 Quality and Process Methods Statistical Details for Multivariate Control Charts is the mean of the n column vectors of p measurements for the kth subgroup from the historical data set is the overall mean of the observations Sk is the sample covariance matrix for the n observations in the kth subgroup from the historical data set is the pooled covariance matrix, calculated as the mean of the within-subgroup covariance matrices The Phase II upper control limit (UCL) is defined as follows: where: p = number of variables n = subgroup sample size m = number of subgroups in the historical data set = (1–)th quantile of an F distribution Additivity of Test Statistics for Observations in Rational Subgroups When a sample of mn independent normal observations is grouped into m rational subgroups each of size n, define T2 M as the distance between the mean of the jth subgroup and the target value. (T2 M is equivalent to T2 in the previous sections for observations in rational subgroups.) You can also calculate T2 statistics related to the internal variability in each subgroup and the overall variability around the target value. The components of the T2 statistic are additive, much like sums of squares. Specifically, the following relationship is true for each of the m subgroups: In all of the following definitions, Sp is defined as it is in the previous sections, depending on whether the control chart is a Phase I or a Phase II control chart. Also, define as for Phase I control charts and as for Phase II control charts.
Xk X 1 m ----Xk k 1 = m = Sp 1 m ----Sk k 1 = m = UCL p m 1 + n 1 – mn m – p – 1 + ---------------------------------------F 1 – p mn m p – 1 + – = F 1 – p mn m p – 1 + – p mn m p – 1 + – Yj TAj 2 TMj 2 TDj 2 + = Y X 300 Multivariate Control Charts Chapter 10 Statistical Details for Multivariate Control Charts Quality and Process Methods The distance from the target value for the jth subgroup is defined as follows: The internal variability for the jth subgroup is defined as follows: where Yji is the ith column vector of p measurements for the jth subgroup.
The overall variability for the jth subgroup is defined as follows: where Yji is the ith column vector of p measurements for the jth subgroup.
Note: When you select the Save T Square or Save T Square Formula options from the Multivariate Control Chart red triangle menu, the three values saved in each row correspond to one value of i in the three definitions above.
Statistical Details for Change Point Detection This discussion follows the development in Sullivan and Woodall (2000).
Assumptions Denote a multivariate distribution of dimension p with mean vector i and covariance matrix i by Np(i,i). Suppose that the xi are m (where m > p) independent observations from such a distribution: If the process is stable, the means i and the covariance matrices i equal a common value so that the xi have a Np(, ) distribution.
Suppose that a single change occurs in either the mean vector or the covariance matrix, or both, between the m1 and m1+1 observations. Then the following conditions hold: • Observations 1 through m1 have the same mean vector and the same covariance matrix (a,a).
• Observations m1 + 1 to m have the same mean vector and covariance matrix (b,b).
• One of the following occurs: TMj 2 n Yj – 'SP 1 – Yj – = TDj 2 Yji Yj – 'SP 1 – Yji Yj – i 1 = n = TAj 2 Yji – 'SP 1 – Yji – i 1 = n = xi Np i i , , i 1 m , = Chapter 10 Multivariate Control Charts 301 Quality and Process Methods Statistical Details for Multivariate Control Charts – If the change affects the mean, a b.
– If the change affects the covariance matrix, a b.
– If the change affects both the mean and the covariance matrix, a b and a b.
Overview A likelihood ratio test approach is used to identify changes in one or both of the mean vector and covariance matrix. The likelihood ratio test statistic is used to compute a control chart statistic that has an approximate upper control limit of 1. The control chart statistic is plotted for all possible m1 values. If any observation’s control chart statistic exceeds the upper control limit of 1, this is an indication that a shift occurred. Assuming that exactly one shift occurs, that shift is considered to begin immediately after the observation with the maximum control chart statistic value.
Likelihood Ratio Test Statistic The maximum value of twice the log-likelihood function for the first m1 observations is defined as follows: The equation for l1 uses the following notation: • S1 is the maximum likelihood estimate of the covariance matrix for the first m1 observations.
• k1 = Min[p,m1-1] is the rank of the p x p matrix S1.
• The notation denotes the generalized determinant of the matrix S1, which is defined as the product of its k1 positive eigenvalues j: The generalized determinant is equal to the ordinary determinant when S1 has full rank.
Denote the maximum of twice the log-likelihood function for the subsequent m2 = m - m1 observations by l2, and the maximum of twice the log-likelihood function for all m observations by l0. Both l2 and l0 are given by expressions similar to that given for l1.
The likelihood ratio test statistic compares the sum l1 + l2 to l0. The sum l1 + l2 is twice the log-likelihood that assumes a possible shift at m1. The value l0 is twice the log-likelihood that assumes no shift. If l0 is substantially smaller than l1 + l2, the process is assumed to be unstable.
l1 m1k1 2 log – m1 S1 k1 log m1k1 – – = S1 k1 S1 k1 j j 1 = k1 = 302 Multivariate Control Charts Chapter 10 Statistical Details for Multivariate Control Charts Quality and Process Methods The likelihood ratio test statistic for a test of whether a change begins at observation m1 + 1 is defined as follows: The distribution of the likelihood ratio test statistic is asymptotically chi-square distributed with p(p + 3)/2 degrees of freedom. Large log-likelihood ratio values indicate that the process is unstable.
The Control Chart Statistic Simulations indicate that the expected value of lrt[m1] varies based on the observation’s location in the series, and, in particular, depends on p and m. See Sullivan and Woodall (2000).
Approximating formulas for the expected value of lrt[m1] are derived by simulation. To reduce the dependence of the expected value on p, lrt[m1] is divided by its asymptotic expected value, p(p + 3)/2.
The formulas for the approximated expected value of lrt[m1] divided by p(p+3)/2 are defined as follows: where and For p = 2, the value of ev[m,p,m1] when m1 or m2 = 2 is 1.3505.
Note: The formulas above are not accurate for p > 12 or m < (2p + 4). In such cases, simulation should be used to obtain approximate expected values.
lrt m1 l1 l2 l0 – + = m1 p k1 – m2 p k2 – + 1 2 log + = m S m1 S1 k1 m2 S2 k2 log – log – log + ev m p m1 , , ap m1bp, if m1 p 1 + + ap m m1 – bp, if m m – 1 p 1 + + 1 m 2p – 1 – m1 p – m p – m1 – -----------------------------------------------------, otherwise + = ap 0.08684 p 14.69 – p 2.036 – p 2 – --------------------------------------------------------------------------– = bp 0.1228 p 1.839 – p 2 – --------------------------------------------= Chapter 10 Multivariate Control Charts 303 Quality and Process Methods Statistical Details for Multivariate Control Charts An approximate upper control limit that yields a false out-of-control signal with probability approximately 0.05, assuming that the process is stable, is calculated as follows: Note that this formula depends on m and p.
The control chart statistic is defined to be twice the log of the likelihood ratio test statistic divided by p(p + 3), divided by its approximate expected value, and also divided by the approximate value of the control limit. Because of the division by the approximate value of the UCL, the control chart statistic can be plotted against an upper control limit of 1. The approximate control chart statistic is calculated as follows: UCL m p , 3.338 2.115 p 0.8819 p log 2 0.1382 p log 3 – + log – + 0.6389 0.3518 p 0.01784 p log 3 m .
log + log – y ˆ m1 2lrt m1 p p 3 + ev m p m1 , , UCL m p [ , ] ----------------------------------------------------------------------------------= 304 Multivariate Control Charts Chapter 10 Statistical Details for Multivariate Control Charts Quality and Process Methods Chapter 11 Model Driven Multivariate Control Charts Monitor and Diagnosis a Complex Process Model-driven multivariate control charts are used to monitor parameters for multiple processes in a single control chart. The Model Driven Multivariate Control Chart (MDMVCC) platform enables you to build a control chart based on principal components or partial least squares models. For a set of continuous variables, the MDMVCC platform uses principal components to build the control chart. For saved principal components or partial least squares score functions, the MDMVCC platform builds a control chart based on the provided models. Use the MDMVCC platform to interactively explore and understand the underlying components that lead to out-of-control signals.
Figure 11.1 Model-driven Multiple Control Chart 306 Model Driven Multivariate Control Charts Chapter 11 Quality and Process Methods Contents Overview of Model Driven Multivariate Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 Example of Model Driven Multivariate Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 Launch the Model Driven Multivariate Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . 310 Data Format . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 The Model Driven Multivariate Control Chart Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312 Model Driven Multivariate Control Chart Platform Options. . . . . . . . . . . . . . . . . . . . . . . . . . . 312 Plot Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 Score Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 Additional Examples of the Model Driven Multivariate Control Chart Platform . . . . . . . . . 315 Example of an MDMVCC with Historical Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 316 Example of an MDMVCC with a PLS Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 Statistical Details for the Model Driven Multivariate Control Chart Platform . . . . . . . . . . . . 318 Monitoring Statistics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 Contributions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 Score Plot Group Comparisons. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 Score Plot Loadings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 Chapter 11 Model Driven Multivariate Control Charts 307 Quality and Process Methods Overview of Model Driven Multivariate Control Charts Overview of Model Driven Multivariate Control Charts The Model Driven Multivariate Control Chart (MDMVCC) platform has two primary functions: monitoring and diagnosing.
• Use multivariate control charts to monitor a multivariate process.
• You can interactively drill down to investigate the contributions of individual variables to the overall signal to diagnosis the process.
For more information about multivariate control charts, see Kourti and MacGregor (1996).
Example of Model Driven Multivariate Control Charts This example uses the Steam Turbine Historical.jmp sample data table that contains process variables from a steam turbine system. You want to build a control chart for the six monitored variables.
1.
Select Help > Sample Data Library and open Quality Control/Steam Turbine Historical.jmp.
2.
Select Analyze > Quality and Process > Model Driven Multivariate Control Chart.
3.
Select all six columns, click Process, and click OK.
Figure 11.2 Steam Turbine Report Note that the process shifts after sample 16.
308 Model Driven Multivariate Control Charts Chapter 11 Example of Model Driven Multivariate Control Charts Quality and Process Methods 4.
Select the sample 17 data point. Right-click and select Rows > Row Label.
5.
Hover over the sample 17 data point to view the T2 contribution proportion plot for that point. Click on the plot to open the plot in the report window.
Figure 11.3 Contribution Proportion Plot for Sample 17 Note that Cool Temp contributes 40% of the T2 value. The Cool Temp bar is green indicating that sample 17 is within the univariate control limits for Cool Temp. Steam Flow and MW each contribute about 20% of the T2 value. They are both red, which indicates that sample 17 is outside of the univariate control limits for each variable. Steam Temp has a zero contribution to the T2 value. In this example, you found variables where the multivariate out-of-control sample could be traced to an out-of-control univariate variable. However, that is not always the case. In multivariate process control you may observe an out-of-control point on the T2 chart but find that the sample is in-control at the univariate level for all variables.
6.
Hover over the Steam Flow bar in the contribution proportion plot to see a univariate control chart for Steam Flow. Click on the chart to open in a new report window.
Chapter 11 Model Driven Multivariate Control Charts 309 Quality and Process Methods Example of Model Driven Multivariate Control Charts Figure 11.4 Individual Chart for Steam Flow The individual chart indicates that the steam flow might have experienced an upset around sample 17. When control limits are specified in the Controls Limits column property, the individual chart respects those limits.
7.
In the PCA Model Driven Multivariate Control Chart report window, Click the T2 for 3 Principal Components red triangle and select Contribution Proportion Heat Map.
Figure 11.5 Contribution Proportion Heat Map 310 Model Driven Multivariate Control Charts Chapter 11 Launch the Model Driven Multivariate Control Chart Platform Quality and Process Methods The contribution proportion heat map shows that there is a shift in the contribution proportions for rows 16, 17, and 18 and again at row 23 as compared to other rows. Generally, Steam Temp, Cool Temp, and Pressure contribute the most to the T2 value for each row.
Launch the Model Driven Multivariate Control Chart Platform Launch the Model Driven Multivariate Control Chart (MDMVCC) platform by selecting Analyze > Quality and Process > Model Driven Multivariate Control Chart.
Figure 11.6 The Model Driven Multivariate Control Chart Launch Window For more information about the options in the Select Columns red triangle menu, see Using JMP.
The Model Driven Multivariate Control Chart platform launch window contains the following options: Process Assigns the process columns. See “Data Format”.
Time ID Assigns a column that is used to identify samples. If no Time ID is assigned, the row number identifies the observations. If the Time ID column is a time, the time identifies each sample. Otherwise, the numeric value of the Time ID identifies each sample.
By Produces a separate report for each level of the By variable. If more than one By variable is assigned, a separate report is produced for each possible combination of the By variables.
Historical Data End at Row Specifies a row number to indicate where historical data end. This enables you to calculate chart limits based on historical data. Both historical and current data are plotted on the charts. Historical data are also known as Phase I data, and current data are also known as Phase II data.
Chapter 11 Model Driven Multivariate Control Charts 311 Quality and Process Methods Launch the Model Driven Multivariate Control Chart Platform Data Format The MDMVCC platform accepts data in the following three allowable data formats: Raw Data Use continuous process data to build a control chart that is based on the principal components of the data. The default dimension of the control chart is based on the number of principal components that account for 85% of the process variation. This number is based on the cumulative percent of the principal component eigenvalues.
Principal Components Use principal component columns that were previously saved from a principal component analysis (PCA). The default dimension of the control chart is the number of components specified as process variables.
Partial Least Squares Score Data Use score columns that were previously saved from a partial least squares (PLS) analysis to build a control chart that is based on the score columns. The default dimension of the control chart is the number of scores specified as process variables.
Notes: • PCA or PLS models built with a frequency or weight column are not supported.
• PCA or PLS models built with historical data must use the same number of historical data rows as specified in the MDMVCC launch window.
• PCA models built from within the Multivariate platform are not supported.
312 Model Driven Multivariate Control Charts Chapter 11 The Model Driven Multivariate Control Chart Report Quality and Process Methods The Model Driven Multivariate Control Chart Report The initial Model Driven Multivariate Control Chart Report shows a T2 control chart. The hover labels on the chart are themselves charts. Click the hover label charts to open larger versions of the charts. Depending on the chart, they open in a separate report window or in the Diagnosis the Process section of the MDMVCC report. You can use the graphlets to interactively drill down into the data.
Figure 11.7 MDMVCC Report with a Hover Graphlet Model Driven Multivariate Control Chart Platform Options Show History Summary Statistics Shows or hides summary statistics for rows designated as historical data or all rows if historical data rows are not specified. Summary statistics include univariate means and standard deviations for process variables. For PCA-based charts, the eigenvalues and eigenvectors are displayed. For charts based on PLS scores, the standard deviation of scores and the score loadings are displayed.
Monitor the Process Show Monitoring Plots Shows or hides the selected process monitoring plots.
Set Component Enables you to set the number of components for the T2, DModX, or SPE plots. The number of components can range from one up to the number of valid eigenvectors for PCA driven analysis or from one up to the number of PLS model factors for PLS driven analysis.
Set Level Enables you to adjust the alpha level that is used for all control chart limits.
Chapter 11 Model Driven Multivariate Control Charts 313 Quality and Process Methods Model Driven Multivariate Control Chart Platform Options T2 Plot Shows or hides a T2 plot. The T2 statistic is a summary of multivariate variation that measures how far away an observation is from the center of a PCA or PLS model.
Normalized DModX Plot Shows or hides a plot of the normalized DModX values. DModX measures the distance of each observation to the PCA or PLS model. A high DModX value indicates an observation that deviates from the underlying correlation structure of the data.
Squared Prediction Error Plot Shows or hides the squared prediction error (SPE) plot. SPE measures the sum of squared of the residuals from the PCA or PLS model. A high SPE value indicates an observation that deviates from the underlying correlation structure of the data.
Score Plot Shows or hides a score plot of principal components or partial least squares factors. See “Score Plot”.
Diagnose the Process (Available when at least one diagnostic plot is active.) Shows or hides diagnostic plots.
See Using JMP for more information about the following options: Redo Contains options that enable you to repeat or relaunch the analysis. In platforms that support the feature, the Automatic Recalc option immediately reflects the changes that you make to the data table in the corresponding report window.
Save Script Contains options that enable you to save a script that reproduces the report to several destinations.
Save By-Group Script Contains options that enable you to save a script that reproduces the platform report for all levels of a By variable to several destinations. Available only when a By variable is specified in the launch window.
Plot Options The following options apply to the T2, Normalized DModX, Squared Prediction Error, and Score Plots. The plots, when selected, appear in the Diagnosis the Process section of the report window.
Show Limit Summaries (Not available for the Score Plot.) Shows or hides control chart limits and summary data in a report table below the chart.
Contribution Heat Map Shows or hides a heat map that is colored by the variable contributions of each observation.
Contribution Proportion Heat Map Shows or hides a heat map that is colored by the variable contributions expressed as proportions of the overall value of each observation.
314 Model Driven Multivariate Control Charts Chapter 11 Model Driven Multivariate Control Chart Platform Options Quality and Process Methods Contribution Plot for Selected Samples (Available only when one or more points are selected.) Shows or hides a bar chart of the individual component contributions to the overall value for each selected sample.
Contribution Proportion Plot for Selected Samples (Available only when one or more points are selected.) Shows or hides a bar chart of the individual component contributions expressed as proportions of contribution of the overall value for each selected sample.
Mean Contribution Proportion Plot for Selected Samples (Available only when two or more points are selected.) Shows or hides a bar chart of the average of the individual component contributions to the overall value for each selected sample.
Note: Bars are green if the value or mean value is within 3 sigma of the mean. Bars are red if the value or mean value is beyond 3 sigma of the mean. When control limits are specified in the Controls Limits column property, the bar coloring respects those limits.
The following options are available for contribution plots: Sort Bars Enables you to sort the bars from largest to smallest contribution or from largest to smallest average contribution for multiple plots.
Label Bars Enables you to label the bars by the value, the column name, or to remove labels (No Label).
Control Charts for Selected Items Shows a control chart for the selected columns. When control limits are specified in the Controls Limits column property, the individual chart respects those limits.
Scatterplot Matrix (Available when two or more bars are selected.) Shows a scatterplot matrix for the selected variables.
Remove Plot Removes the plot from the report window.
Normalized Score Plot for Selected Samples (Available only for a Score Plot when one or more points are selected.) Shows or hides a bar chart of normalized scores for each sample selected.
Score Ellipse Coverage (Available only for Score Plots with two components.) Adds an ellipse with the specified coverage to the score plot. The ellipse is based on historical data. When both Phase I and Phase II data are present, there is an ellipse for each phase and the Phase II ellipse is dashed.
Connect Points (Available only for Score Plots.) Connects data points in the score plot.
Show Loadings (Available only for Score Plots.) Shows the PCA loadings on the score plot using biplot arrows.
Save Columns For each plot there are three options to save values to the data table: Chapter 11 Model Driven Multivariate Control Charts 315 Quality and Process Methods Additional Examples of the Model Driven Multivariate Control Chart Platform Save Values Saves values (T2, Normalized DModX, SPE, or scores) to a new data table column.
Save Contributions Saves contributions to new data table columns.
Save Contribution Proportions Saves contribution proportions to new data table columns.
Score Plot The Score Plot displays a plot of principal components or partial least squares factors. Use the controls below the plot to change the components shown in the Score Plot.
Use the buttons to the right of the plot to assign and compare the relative contributions between two groups of points. Relative contributions show how two or more samples differ from each other. Relative contributions show what changes in the underlying process variables contribute to differences in groups of samples. One use is to investigate the differences between an in-control sample and an out-of-control sample.
Group A is the reference group and Group B the comparator group. Each group can consist of one or more points. To assign the reference group, select one or more points and then click Group A. To assign the comparator group, select one or more points and then click Group B. To display the Relative Score Contribution Plot, click Compare.
Figure 11.8 Score Plot with Relative Contribution Plot for Row 17 Relative to Row 24 Additional Examples of the Model Driven Multivariate Control Chart Platform • “Example of an MDMVCC with Historical Data” • “Example of an MDMVCC with a PLS Model” 316 Model Driven Multivariate Control Charts Chapter 11 Additional Examples of the Model Driven Multivariate Control Chart Platform Quality and Process Methods Example of an MDMVCC with Historical Data This example demonstrates the use of historical data to set the monitoring limits for current data.
1.
Select Help > Sample Data Library and open Quality Control/Flight Delays.jmp.
2.
Select Analyze > Quality and Process > Model Driven Multivariate Control Chart.
3.
Select the AA through WN and click Process.
4.
Select Flight Date and click Time ID.
5.
Enter 16 for Historical Data End at Row.
6.
Click OK.
Figure 11.9 T2 Chart for Historical and Current Data Note that there are two sets of limits. One set applies to the historical data. A second set of limits applies to the current data. For more information about how the historical data is used to calculate the two sets of limits, see “Limits”.
Tip: Turn on Automatic Recalc to enable the chart to automatically update as you add additional observations to the data table. The Automatic Recalc option is under redo when you click the PCA Model Driven Multivariate Control Chart red triangle.
Chapter 11 Model Driven Multivariate Control Charts 317 Quality and Process Methods Additional Examples of the Model Driven Multivariate Control Chart Platform Example of an MDMVCC with a PLS Model This example demonstrates the use of a PLS model for monitoring a multivariate process. Consider a process with 14 inputs and 5 quality variables. You have a PLS model that explains the process and you want to use this model to monitor the process for deviations.
1.
Select Help > Sample Data Library and open Polyethylene Process.jmp.
This data table contains 14 process variables and 5 quality or output variables. The first 100 rows are historical data used to build a PLS model to describe the process. These rows are colored blue. The remaining 239 rows are data collected since the model was built.
The partial least squares model finds 4 score functions that describe the process. These functions are saved to the data table in columns X Score 1 Formula to X Score 4 Formula. To build the PLS model, use the table script Set current data as excluded to exclude the 239 rows of data collected after the historical data. Then use the Fit Partial Least Squares table script to build the PLS model to relate the quality variable to the process variables.
2.
Select Analyze > Quality and Process > Model Driven Multivariate Control Chart.
3.
Select the X Score 1 Formula through X Score 4 Formula and click Process.
4.
Set the Historical Data End at Row to 100.
5.
Click OK.
Figure 11.10 T2 Chart The blue data points represent the historical data. The black data points are data points collected after the control chart was established. The process experienced an upset that begins around sample number 326.
6.
Hover over the sample data points that are above the upper control limit to view contribution plots.
318 Model Driven Multivariate Control Charts Chapter 11 Statistical Details for the Model Driven Multivariate Control Chart Platform Quality and Process Methods 7.
Select the first cluster of data points above the upper control limit. Click the T2 for 4 Factors red triangle and select Mean Contribution Proportion Plot for Selected Samples.
8.
Click the red triangle next to T2 Mean Contribution Proportion Plot for Selected Samples and select Sort Bars.
Figure 11.11 Mean Contribution Proportion Plot Notice that the contributions plot is in terms of the PLS model input variables. It appears that z2 and Tmax2 are causing the process upset. Tmax2 and z2 are related. Tmax2 is a reactor temperature and z2 is the location of the Tmax2 temperature.
Note: The descriptions of the factors are recorded in the Notes column property.
Statistical Details for the Model Driven Multivariate Control Chart Platform Monitoring Statistics T2 The value for each of the i observations is plotted on the T2 control chart. For historical and current data, the T2 values for a PCA or PLS model with k components are defined as: where: Ti 2 Ti 2 ti TSk 1 – ti = Chapter 11 Model Driven Multivariate Control Charts 319 Quality and Process Methods Statistical Details for the Model Driven Multivariate Control Chart Platform ti = the vector of k scores for the ith observation Sk = the diagonal sample covariance matrix of the k scores for historical observations For PCA models, Sk is the diagonal eigenvalue matrix.
The mean of each of the k historical score vectors is 0 when the data is centered during the data preprocessing step. This step occurs in PCA on correlations or covariances and in PLS with centering. For preprocessing options where X is not centered, the data is assumed to have been centered by the user, so the mean of each of the k score vectors is 0. For more information about Hotelling’s T2, see Montgomery (2013).
SPE For both PCA and PLS models, the preprocessed X matrix can be decomposed as: where Tk = (t1,...,tk) is the k dimensional score matrix and Pk = (p1,...,pk) is a matrix with the first k eigenvectors for PCA models or the loading matrix for PLS models. The squared prediction error of this PCA or PLS model is used for the SPE control chart.
The SPEi value for each of the i observations is plotted on the SPE control chart. The squared prediction error is defined as: where ei = the residual vector for observation i p = number of variables DModX The DModXi value for each of the i observations is plotted on the DModX control chart. The normalized distance to model (DModX) is defined as: where X TkPk T E + = SPEi ei Tei eij 2 j 1 = p = = DModXi SPEi df1 S i 1 = n PEi df2 ----------------------------------------------eij 2 df1 j 1 = p eij 2 df2 j 1 = p i 1 = n ---------------------------------------------eij 2 j 1 = p d -----------------= = = 320 Model Driven Multivariate Control Charts Chapter 11 Statistical Details for the Model Driven Multivariate Control Chart Platform Quality and Process Methods eij = the residual for observation i and variable j df1 = p−k df2 = (n−k−1)(p−k) if the data is centered and (n−k)(p−k) if the data is not centered n = number of historical data observations k = number of PCA/PLS components p = number of variables Note: DModXi is equal to SPEi scaled by 1/d.
Limits All data are treated as historical data when the number of historical rows is not specified in the launch window. See “Launch the Model Driven Multivariate Control Chart Platform”.
T2 The upper control limit (UCL) for historical data is based on the Beta distribution and defined as: where: n = number of historical data observations k = number of PCA or PLS components = (1−)th quantile of a Beta distribution.
The UCL for current data is based on the F distribution and defined as: where: n = number of historical data observations k = number of PCA or PLS components F(1−; k; n−k) = (1−)th quantile of an F(k; n−k) distribution.
UCL n 1 – 2 n --------------------1 k 2 -- n k – 1 – 2 ---------------------; ; – = 1 k 2 -- n k – 1 – 2 ---------------------; ; – k 2 -- n k – 1 – 2 ---------------------; UCL k n 1 + n 1 – n n k – --------------------------------------F 1 k n k – ; ; – = Chapter 11 Model Driven Multivariate Control Charts 321 Quality and Process Methods Statistical Details for the Model Driven Multivariate Control Chart Platform DModX For PCA and PLS models, the UCL is based on the F distribution. The DModX UCL for PCA models is defined as: where: df1 = p−k df2 = (n−k−1)(p−k) if the data is centered and (n−k)(p−k) if the data is not centered n = number of historical data observations k = number of PCA components p = number of variables F(1−; n−p−1; p−k) = (1−)th quantile of a F(n−p−1; p−k) distribution.
The DModX UCL for PLS models is defined as: where: = historical sample mean of SPE = historical sample variance of SPE n = number of historical data observations F(1−; h; nh) = (1−)th quantile of an F(h; nh) distribution.
SPE The SPE UCL for PCA models is defined as: where: UCL F 1 df1 df2 ; ; – = UCL F 1 h nh ; ; – = h 2 ˆ SPE 2 ˆ SPE 2 = ˆ SPE ˆ SPE 2 UCL 1 1 2h0 1 h0 – 1 2 --------------------------------– z1 – 22h0 2 1 2 1 --------------------------------------------+ = 1 h0 322 Model Driven Multivariate Control Charts Chapter 11 Statistical Details for the Model Driven Multivariate Control Chart Platform Quality and Process Methods a = the ath eigenvalue k = number of PCA components z1− = (1−)th quantile of the standard normal distribution For more information about SPE control limits for PCA models, see Jackson and Mudholkar (1979).
For PLS models, the UCL is based on the chi-square distribution and defined as: where = historical sample mean of SPE = historical sample variance of SPE 2(1−; h) = (1−)th quantile of an 2(h) distribution The g and h parameters are estimated by the method of moments. For more information about SPE control limits for PLS models, see Nomikos (1995).
h0 1 213 32 2 – = 1 a a 1 = k = 2 a 2 a 1 = k = 3 a 3 a 1 = k = UCL g1 h ; – 2 = g ˆ SPE 2 2 ˆ SPE = h 2 ˆ SPE 2 ˆ SPE 2 = ˆ SPE ˆ SPE 2 Chapter 11 Model Driven Multivariate Control Charts 323 Quality and Process Methods Statistical Details for the Model Driven Multivariate Control Chart Platform Contributions T2 The T2 contributions for a PCA or PLS model with p variables and k components are calculated as: where: ti = the vector of k scores for the ith observation Sk = the diagonal sample covariance matrix of the k scores for historical observations. For PCA models, Sk is the diagonal eigenvalue matrix.
sa = the ath diagonal element of Sk rja = the jth element of the ath eigenvector for PCA models and the ath column of the Rk loading matrix for PLS models. Rk is the matrix used to relate the score matrix, Tk to the X matrix, such that Tk=XRk.
xij = the value of the jth variable for the ith observation.
Note: The p terms in the last sum are the variable contributions.
The contribution of each variable is the sum of its contribution to each score, weighted by the normalized score value. A variable is considered to have a large contribution to if there is a large normalized score value, and the variable contribution is large.
the contribution proportion of variable j is defined as: Ti 2 ti TSk 1 – ti = tia 2 sa ------a 1 = k = tia sa ------a 1 = k rjaxij j 1 = p = tia sa ------rjaxij a 1 = k j 1 = p = Ti 2 Cont Ti 2 j j 1 = p Ti 2 = 324 Model Driven Multivariate Control Charts Chapter 11 Statistical Details for the Model Driven Multivariate Control Chart Platform Quality and Process Methods Note: When computing T2 contribution proportions, JMP zeros out negative contributions. Negative contributions are possible due to the interaction of variables during the projection of X in PCA and PLS models. The negative contributions are zeroed in order to identify the variable contributions that represent a large proportion of the total positive contributions.
For more information about PCA contributions and negative contributions, see Kourti and MacGregor (1996). For more information about PLS contributions, see Li et al. (2009).
DModX For PCA and PLS models, the contribution of variable j to DModXi is defined as: Note that since the contribution proportion of variable j is defined as: SPE For PCA and PLS models, the contribution of variable j to SPEi is defined as: Note that since the contribution proportion of variable j is defined as: ContProp Ti 2 j Cont Ti 2 j Cont Ti 2 j j 1 = p --------------------------------------= Cont DModXi j eij d -------= DModXi eij 2 d -----j 1 = p = ContProp DModXi j eij 2 d DModXi -----------------------= Cont SPEi j eij = SPEi eij 2 j 1 = p = Chapter 11 Model Driven Multivariate Control Charts 325 Quality and Process Methods Statistical Details for the Model Driven Multivariate Control Chart Platform Score Contributions The score contribution computation is the same as T2 contributions but are computed only for the dimensions selected in the score plot.
Score Plot Group Comparisons For the score plot group comparisons, the relative score contribution for variable j is the difference between the average contribution in group B and the average contribution in group A: where Ti = the ith row of the score matrix with columns corresponding to the dimensions shown in the score plot.
A = the set of observations in group A B = the set of observations in group B na = the number of observations in group A nb = the number of observations in group B.
Score Plot Loadings The loadings shown on the score plot are based on PCA eigenvectors or PLS X score loadings (R matrix). These loadings are scaled by the maximum absolute value of scores. The scaling is performed in order to graph the loadings on the score plot. The loadings illustrate each variable’s approximate influence on each score.
ContProp SPEi j eij 2 SPEi -------------= Cont Ti j nb ----------------------------i B Cont Ti j na ----------------------------i A – 326 Model Driven Multivariate Control Charts Chapter 11 Statistical Details for the Model Driven Multivariate Control Chart Platform Quality and Process Methods Chapter 12 Legacy Control Charts Create Variable and Attribute Control Charts A control chart is a graphical and analytic tool for monitoring process variation. The natural variation in a process can be quantified using a set of control limits. Control limits help distinguish common-cause variation from special-cause variation. Typically, action is taken to identify and eliminate special-cause variation. It is also important to quantify the common-cause variation in a process, as this determines the capability of a process.
The legacy control chart platforms in JMP provide a variety of control charts, as well as runs charts, V-Mask CUSUM charts, and weighted moving average charts. To support process improvement initiatives, most of the control chart options display separate control charts for different phases of a project on the same chart.
Figure 12.1 Control Chart Example 328 Legacy Control Charts Chapter 12 Quality and Process Methods Contents Example of a Legacy Control Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 329 Legacy Control Chart Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 Control Charts for Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 Control Charts for Attributes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 Launch a Legacy Control Chart Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334 Process Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 Limits Specifications. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 Specified Statistics. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 Legacy Control Chart Reports. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340 V-Mask CUSUM Chart Reports . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 Interpret a Two-Sided V-Mask CUSUM Chart. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 Interpret a One-Sided CUSUM Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343 Legacy Control Chart Platform Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344 Window Options for Legacy Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344 Chart Options for Legacy Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 346 Chart Options for V-Mask CUSUM Control Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349 Saving and Retrieving Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349 Excluded, Hidden, and Deleted Samples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 Additional Examples of the Control Chart Platform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354 Presummarize Chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 354 V-Mask CUSUM Chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356 One-Sided CUSUM Chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 358 UWMA Chart Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 Statistical Details for the Control Chart Platform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 Control Limits for Median Moving Range Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 Statistical Details for Capability Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 360 Statistical Details for V-Mask CUSUM Control Charts. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367 Statistical Details for Weighted Moving Average Charts . . . . . . . . . . . . . . . . . . . . . . . . . . . 370 Chapter 12 Legacy Control Charts 329 Quality and Process Methods Example of a Legacy Control Chart Example of a Legacy Control Chart This example uses the Coating.jmp sample data table in the Quality Control sample data folder (taken from the ASTM Manual on Presentation of Data and Control Chart Analysis). The quality characteristic of interest is the Weight column. A subgroup sample of four is chosen.
1.
Select Help > Sample Data Library and open Quality Control/Coating.jmp.
2.
Select Analyze > Quality And Process > Legacy Control Charts > XBar.
Note the selected chart types of XBar and R.
3.
Select Weight and click Process.
4.
Select Sample and click Sample Label.
5.
Click OK.
Figure 12.2 Variables Charts for Coating Data An XBar chart and an R chart for the process are shown in Figure 12.2. Sample six indicates that the process is not in statistical control. To check the sample values, click the sample six summary point on either control chart. The corresponding rows highlight in the data table.
{ 330 Legacy Control Charts Chapter 12 Legacy Control Chart Types Quality and Process Methods Note: If an S chart is chosen with the XBar chart, then the limits for the XBar chart are based on the standard deviation. Otherwise, the limits for the XBar chart are based on the range.
Legacy Control Chart Types Legacy control charts are broadly classified into two categories: • “Control Charts for Variables”— IR, XBar, Runs Chart, Levey-Jennings, Presummarize, CUSUM, UWMA, and EWMA.
• “Control Charts for Attributes”— P, NP, C, and U.
Control Charts for Variables Control charts for variables are classified according to the subgroup summary statistic plotted on the chart: • The IR selection provides additional chart types: – Individual Measurement charts display individual measurements. These charts are appropriate when only one measurement is available for each subgroup sample.
– Moving Range charts display moving ranges of two or more successive measurements. See “Moving Range (Average) Charts”.
• XBar charts display subgroup means (averages). This selection provides additional chart types: – R charts display subgroup ranges (maximum – minimum).
– S charts display subgroup standard deviations.
For quality characteristics measured on a continuous scale, a typical analysis shows both the process mean and its variability with a mean chart aligned above its corresponding R or S chart.
• Runs Chart displays data as a connected series of points. Runs charts can also plot the group means when the Sample Label role is used, either on the window or through a script.
• Levey-Jennings charts show a process mean with control limits based on a long-term sigma. The control limits are placed at 3s distance from the center line. The standard deviation, s, for the Levey-Jennings chart is calculated the same way standard deviation is in the Distribution platform.
• Presummarize charts display subgroup means and standard deviations. See “Presummarize Charts”.
Chapter 12 Legacy Control Charts 331 Quality and Process Methods Legacy Control Chart Types • CUSUM charts show cumulative sums of subgroup or individual measurements from a target value. See “V-Mask CUSUM Control Charts”.
• UWMA charts show a uniformly weighted moving average of a specified number of measurements. See “Uniformly Weighted Moving Average Charts”.
• EWMA charts show an exponentially weighted moving average of all measurements with a specified weight. See “Exponentially Weighted Moving Average Charts”.
Moving Range (Average) Charts In a Moving Average chart, the quantities that are averaged can be individual observations instead of subgroup means. However, a Moving Average chart for individual measurements is not the same as a control chart for individual measurements or moving ranges with individual measurements plotted.
Moving Range (Average) charts display moving ranges of two or more successive measurements. Moving ranges are computed for the number of consecutive measurements that you enter in the Range Span box. The default range span is 2. Because moving ranges are correlated, these charts should be interpreted with care.
A Median Moving Range chart is also available. If you choose a Median Moving Range chart and an Individual Measurement chart, the limits on the Individual Measurement chart use the Median Moving Range as the sigma, rather than the Average Moving Range.
Presummarize Charts If your data consist of repeated measurements of the same process unit, you can combine these into one measurement for the unit. Pre-summarizing is not recommended unless the data have repeated measurements on each process or measurement unit.
Presummarize summarizes the process column into sample means or standard deviations, based either on the sample size or sample label chosen. Then it charts the summarized data based on the options chosen in the launch window. You can also append a capability analysis by checking the appropriate box in the launch window.
The Presummarize launch window has the following options for chart types: • Individual on Group Means • Individual on Group Std Devs • Moving Range on Group Means • Moving Range on Group Std Devs • Median Moving Range on Group Means • Median Moving Range on Group Std Devs There is also an option for setting the range span that is used for the moving range chart types.
332 Legacy Control Charts Chapter 12 Legacy Control Chart Types Quality and Process Methods V-Mask CUSUM Control Charts V-Mask Cumulative Sum (CUSUM) control charts show cumulative sums of subgroup or individual measurements from a target value. V-Mask CUSUM charts can help you decide whether a process is in a state of statistical control by detecting small, sustained shifts in the process mean. In comparison, standard Shewhart control charts can detect sudden and large changes in measurement, such as a two or three sigma shift, but they are less effective at spotting smaller changes, such as a one sigma shift.
The CUSUM menu selection has options for V-mask cumulative sum charts. In addition to KSigma, you also specify: • The vertical distance h between the origin for the V-mask and the upper or lower arm of the V-mask for a two-sided chart. For a one-sided chart, H is the decision interval. Choose H as a multiple of the standard error.
• The reference value k, where k is greater than zero.
Another form of a cumulative sum control chart is the tabular CUSUM chart. To create a tabular CUSUM chart, see “CUSUM Control Charts”. The tabular CUSUM chart is recommended over the V-mask chart for a variety of reasons, including the following: • The V-mask must be moved with each observation, not simply placed on the last observation.
• The cumulative sums in the V-mask procedure can end up a long way from the center of the graph, even for an on-target process.
Caution: Montgomery (2013) strongly “advises against using the V-mask procedure.” Uniformly Weighted Moving Average Charts Each point on a Uniformly Weighted Moving Average (UWMA) chart is the average of the w most recent subgroup means, including the present subgroup mean. When you obtain a new subgroup sample, the next moving average is computed by dropping the oldest of the previous w subgroup means and including the newest subgroup mean. The constant, w, is called the span of the moving average.
In addition to KSigma and Alpha, in the UWMA launch window you also specify: • The Moving Average Span, or w, which indicates how many subgroups to include to form the moving average. The larger the Moving Average Span (w), the smoother the UWMA line, and the less it reflects the magnitude of shifts. This means that larger values of w guard against smaller shifts. See “Control Limits for UWMA Charts”.
Chapter 12 Legacy Control Charts 333 Quality and Process Methods Legacy Control Chart Types Exponentially Weighted Moving Average Charts Each point on an Exponentially Weighted Moving Average (EWMA) chart is the weighted average of all the previous subgroup means, including the mean of the present subgroup sample. The weights decrease exponentially going backward in time.
Note: An Exponentially Weighted Moving Average (EWMA) chart can also be called a Geometric Moving Average (GMA) chart.
In addition to KSigma and Alpha, in the EWMA launch window you also specify: • A Weight parameter, which is the weight (0 < weight ≤1) assigned to the present subgroup sample mean. Small values of Weight are used to guard against small shifts. See “Control Limits for EWMA Charts”.
Tip: See “EWMA Control Charts” for the newer EWMA Control Charts platform.
Control Charts for Attributes In the previous types of charts, measurement data was the process variable. This type of data is often continuous, and the charts are based on theory for continuous data. Another type of data is count data, where the variable of interest is a discrete count of the number of defects or blemishes per subgroup. For discrete count data, attribute charts are applicable, as they are based on binomial and Poisson models. Because the counts are measured per subgroup, it is important when comparing charts to determine whether you have a similar number of items in the subgroups between the charts. Attribute charts, like variables charts, are classified according to the subgroup sample statistic plotted on the chart.
Determining Which Attribute Chart to Use Each item is judged as either conforming or non-conforming: p-chart Shows the proportion of defective items.
np-chart Shows the number of defective items.
The number of defects is counted for each item: c-chart Shows the number of defects.
u-chart Shows the proportion of defects.
334 Legacy Control Charts Chapter 12 Launch a Legacy Control Chart Platform Quality and Process Methods For attribute charts, specify the column containing the defect count or defective proportion as the Process variable. The data are interpreted as counts, unless the column contains non-integer values between 0 and 1.
• P charts display the proportion of nonconforming (defective) items in subgroup samples, which can vary in size. Since each subgroup for a P chart consists of Ni items, and an item is judged as either conforming or nonconforming, the maximum number of nonconforming items in a subgroup is Ni.
• NP charts display the number of nonconforming (defective) items in subgroup samples. Because each subgroup for an NP chart consists of Ni items, and an item is judged as either conforming or nonconforming, the maximum number of nonconforming items in subgroup i is Ni.
Note: To use the Sigma column property for P or NP charts, the value needs to be equal to the proportion. JMP calculates the sigma as a function of the proportion and the sample sizes.
• C charts display the number of nonconformities (defects) in a subgroup sample that usually, but does not necessarily, consists of one inspection unit.
Caution: For a C chart, if you do not specify a Sample Size or Constant Size, then the Sample Label is used as the sample size.
• U charts display the proportion of nonconformities (defects) in each subgroup sample that can have a varying number of inspection units.
Caution: For a U chart, if you do not specify a Unit Size or Constant Size, then the Sample Label is used as the unit size.
Launch a Legacy Control Chart Platform When you launch a legacy control chart platform by selecting Analyze > Quality And Process > Legacy Control Charts, a launch window similar to Figure 12.3 appears. The specific controls vary depending on which type of chart you select. Initially, the window shows the following types of information: • “Process Information” for measurement variable selection • Chart type information (for more information, see “Legacy Control Chart Types”) • “Limits Specifications” • “Specified Statistics” Chapter 12 Legacy Control Charts 335 Quality and Process Methods Launch a Legacy Control Chart Platform Figure 12.3 XBar Control Chart Launch Window For more information about the options in the Select Columns red triangle menu, see Using JMP.
Process Information The launch window displays a list of columns in the current data table. Here, you specify the variables to be analyzed and the subgroup sample size.
Process The Process role selects variables for charting: • For variables charts, specify measurements as the process.
• For attribute charts, specify the defect count or defective proportion as the process. The data are interpreted as counts, unless it contains non-integer values between 0 and 1.
Note: The rows of the data table must be sorted in the order in which the observations were collected. Even if there is a Sample Label variable specified, you still must sort the observations accordingly.
limits specification process information add capability analysis to the report enter or remove known statistics chart type information 336 Legacy Control Charts Chapter 12 Launch a Legacy Control Chart Platform Quality and Process Methods Sample Label The Sample Label role enables you to specify a variable whose values label the horizontal axis and can also identify unequal subgroup sizes. If you do not specify a sample label variable, the samples are identified by their subgroup sample number.
• If the sample subgroups are the same size, select the Sample Size Constant option and enter the size in the text box. If you entered a Sample Label variable, its values are used to label the horizontal axis. The sample size is used in the calculation of the limits regardless of whether the samples have missing values.
• If the sample subgroups have an unequal number of rows or have missing values and you have a column identifying each sample, select the Sample Grouped by Sample Label option and enter the sample identifying column as the sample label.
For attribute charts (P, NP, C, and U charts), this variable is the subgroup sample size. Additional options appear on the launch window, including Sample Size, Constant Size, or Unit Size, depending on your selection. In variables charts, it identifies the sample. When the chart type is IR, a Range Span text box appears. The range span specifies the number of consecutive measurements from which the moving ranges are computed.
Notes: • The rows of the data table must be sorted in the order in which the observations were collected. Even if there is a Sample Label variable specified, you still must sort the observations accordingly.
• The non-integer part of the value for Constant Size is truncated. If you have a constant non-integer subgroup sample size, you must specify a column of constant values.
Chapter 12 Legacy Control Charts 337 Quality and Process Methods Launch a Legacy Control Chart Platform The illustration in Figure 12.4 shows an XBar chart for a process with unequal subgroup sample sizes, using the Coating.jmp sample data from the Quality Control sample data folder.
Figure 12.4 Variables Charts with Unequal Subgroup Sample Sizes Phase The Phase role enables you to specify a column identifying different phases, or sections. A phase is a group of consecutive observations in the data table. For example, phases might correspond to time periods during which a new process is brought into production and then put through successive changes. Phases generate, for each level of the specified Phase variable, a new sigma, set of limits, zones, and resulting tests.
On the window for XBar, R, S, IR, P, NP, C, U, Presummarize, and Levey-Jennings charts, a Phase variable button appears. If a phase variable is specified, the phase variable is examined, row by row, to identify to which phase each row belongs. Saving to a limits file reveals the sigma and specific limits calculated for each phase.
By The By role identifies a variable to produce a separate analysis for each value that appears in the column.
338 Legacy Control Charts Chapter 12 Launch a Legacy Control Chart Platform Quality and Process Methods Limits Specifications You can specify computations for control limits by entering a value for k (K Sigma), or by entering a probability for (Alpha), or by retrieving a limits value from the process columnsʹ properties or a previously created Limits Table. Limits Tables and the Get Limits button are discussed in the section “Saving and Retrieving Limits”. There must be a specification of either K Sigma or Alpha. The window default for K Sigma is 3.
KSigma The KSigma parameter option enables specification of control limits in terms of a multiple of the sample standard error. KSigma specifies control limits at k sample standard errors above and below the expected value, which shows as the center line. To specify k, the number of sigmas, click the radio button for KSigma and enter a positive k value into the text box. The usual choice for k is 3, which is three standard deviations. The examples shown in Figure 12.5 compare the XBar chart for the Coating.jmp data with control lines drawn with KSigma = 3 and KSigma = 4.
Figure 12.5 K Sigma =3 (left) and K Sigma=4 (right) Control Limits Alpha The Alpha parameter option specifies control limits (also called probability limits) in terms of the probability that a single subgroup statistic exceeds its control limits, assuming that the process is in control. To specify alpha, click the Alpha radio button and enter the desired probability. Reasonable choices for are 0.01 or 0.001. For XBar charts under the assumption of normality and known in-control parameters, the Alpha value equivalent to a KSigma of 3 is 0.0027.
Chapter 12 Legacy Control Charts 339 Quality and Process Methods Launch a Legacy Control Chart Platform Specified Statistics After specifying a process variable, if you click the Specify Stats (when available) button on a Control Chart launch window, a tab with editable fields is appended to the bottom of the window. This lets you enter historical statistics (that is, statistics obtained from historical data) for the process variable. The Control Chart platform uses those entries to construct control charts. The example here shows 1 as the standard deviation of the process variable and 20 as the mean measurement.
Figure 12.6 Example of Specify Stats Note: When the mean is user-specified, it is labeled in the plot as 0.
If you check the Capability option on a Control Chart launch window (Figure 12.3), a window appears as the platform is launched asking for specification limits. The standard deviation for the control chart selected is sent to the window and appears as a Specified Sigma value, which is the default option. After entering the specification limits and clicking OK, capability output appears in the same window next to the control chart. For information about how the capability indices are computed, see “Capability Indices for Normal Distributions”.
340 Legacy Control Charts Chapter 12 Legacy Control Chart Reports Quality and Process Methods Legacy Control Chart Reports The analysis produces a chart that can be used to determine whether a process is in a state of statistical control. The report varies depending on the type of chart that you select. Figure 12.7 displays the parts of a simple control chart. Control charts update dynamically as data is added or changed in the data table.
Figure 12.7 Example of a Control Chart Note: Any rows that are excluded in the data table are also hidden in Runs charts, P charts, U charts, and C charts.
Control charts have the following characteristics: • Each point plotted on the chart represents an individual process measurement or summary statistic. In Figure 12.7, the points represent the average for a sample of measurements.
Subgroups should be chosen rationally, that is, they should be chosen to maximize the probability of seeing a true process change between subgroups. Often, this requires knowledge of the process to determine the most effective grouping strategy. See Wheeler (2004); Woodall and Adams (1998).
• The vertical axis of a control chart is scaled in the same units as the summary statistic.
• The horizontal axis of a control chart identifies the subgroup samples and is time ordered. Observing the process over time is important in assessing if the process is changing.
• The green line is the center line, or the average of the data. The center line indicates the average (expected) value of the summary statistic when the process is in statistical control. Measurements should appear equally on both sides of the center line. If not, this is possible evidence that the process average is changing.
• The two red lines are the upper and lower control limits, labeled UCL and LCL. These limits give the range of variation to be expected in the summary statistic when the process is in statistical control. If the process is exhibiting only routine variation, then all the points Subgroup Sample Axis LCL Center Line UCL Out of Control Point Measurement Axis Chapter 12 Legacy Control Charts 341 Quality and Process Methods Legacy Control Chart Reports should fall randomly in that range. In Figure 12.7, one measurement is above the upper control limit. This is evidence that the measurement could have been influenced by a special cause, or is possibly a defect.
• A point outside the control limits (or the V-mask of a CUSUM chart) signals the presence of a special cause of variation.
Options within each platform create control charts that can be updated dynamically as samples are received and recorded or added to the data table.
When a control chart signals abnormal variation, action should be taken to return the process to a state of statistical control if the process degraded. If the abnormal variation indicates an improvement in the process, the causes of the variation should be studied and implemented.
When you double-click the horizontal or vertical axis, the appropriate Axis Specification window appears for you to specify the format, axis values, number of ticks, gridline, reference lines, and other options to display on the axis.
For example, the Pickles.jmp data lists measurements taken each day for three days. In Figure 12.8, by default, the horizontal axis is labeled at every other tick. Sometimes this gives redundant labels, as shown to the left in Figure 12.8. If you specify a label at an increment of eight, the horizontal axis is labeled once for each day, as shown in the chart on the right.
Figure 12.8 Example of Labeled x Axis Tick Marks Tip: For information about warnings and rules, see “Tests” and “Westgard Rules”.
342 Legacy Control Charts Chapter 12 V-Mask CUSUM Chart Reports Quality and Process Methods V-Mask CUSUM Chart Reports Interpret a Two-Sided V-Mask CUSUM Chart Note: See also “V-Mask CUSUM Chart Example”.
To interpret a two-sided CUSUM chart, compare the points with limits that compose a V-mask. A V-mask is a shape in the form of a V on its side that is superimposed on the graph of the cumulative sums. The V-mask is formed by plotting V-shaped limits. The origin of a V-mask is the most recently plotted point, and the arms extended backward on the horizontal axis, as in Figure 12.9. As data are collected, the cumulative sum sequence is updated and the origin is relocated at the newest point.
Figure 12.9 V-Mask for a Two-Sided CUSUM Chart Shifts in the process mean are visually easy to detect on a CUSUM chart because they produce a change in the slope of the plotted points. The point where the slope changes is the point where the shift occurs. A condition is out-of-control if one or more of the points previously plotted crosses the upper or lower arm of the V-mask. Points crossing the lower arm signal an increasing process mean, and points crossing the upper arm signal a downward shift.
d h, the rise in the arm corresponding to the distance (d) from origin to vertex k, the rise in the arm corresponding to one sampling unit upper arm vertex 1 unit lower arm Chapter 12 Legacy Control Charts 343 Quality and Process Methods V-Mask CUSUM Chart Reports There are important differences between CUSUM charts and Shewhart charts: • A Shewhart control chart plots points based on information from a single subgroup sample. In CUSUM charts, each point is based on information from all samples taken up to and including the current subgroup.
• On a Shewhart control chart, horizontal control limits define whether a point signals an out-of-control condition. On a CUSUM chart, the limits can be either in the form of a V-mask or a horizontal decision interval.
• The control limits on a Shewhart control chart are commonly specified as 3 limits. On a CUSUM chart, the limits are determined from average run length.
A CUSUM chart is more efficient for detecting small shifts in the process mean. Lucas (1976) states that a V-mask detects a 1 shift about four times as fast as a Shewhart control chart.
Interpret a One-Sided CUSUM Chart Use a one-sided CUSUM chart to identify data approaching or exceeding the side of interest.
Figure 12.10 Example of a One-Sided CUSUM Chart The decision interval or horizontal line is set at the H value that you entered in the launch window. In this example, it is 0.25. Any values exceeding the decision interval of 0.25 indicate a shift or out-of-control condition. In this example, observation 4 appears to be where a shift occurred. Also note that no V-mask appears for one-sided CUSUM charts.
344 Legacy Control Charts Chapter 12 Legacy Control Chart Platform Options Quality and Process Methods Legacy Control Chart Platform Options Legacy control charts have red triangle menus that affect various parts of the platform: • The menu on the top-most title bar affects the whole platform window. Its items vary with the type of chart that you select. See “Window Options for Legacy Control Charts”.
• There is a menu of items on the chart type title bar with options that affect each chart individually. See “Chart Options for Legacy Control Charts”.
Window Options for Legacy Control Charts The red triangle menu on the window title bar lists options that affect the report window. If you request XBar and R at the same time, you can check each chart type to show or hide it. The specific options that are available depend on the type of control chart you request. Unavailable options show as grayed menu items.
Show Limits Legend Shows or hides the Avg, UCL, and LCL values to the right of the chart.
Connect Through Missing Connects points when some samples have missing values. In Figure 12.11, the left chart has no missing points. The middle chart has samples 2, 11, 19, and 27 missing with the points not connected. The right chart appears if you select the Connect Through Missing option, which is the default.
Figure 12.11 Example of Connected through Missing Option Use Median For Runs Charts, when you select the Show Center Line option in the individual Runs Chart red triangle menu, a line is drawn through the center value of the column. The center line is determined by the Use Median setting of the main Runs Chart red triangle menu. When Use Median is selected, the median is used as the center line. Otherwise, the mean is used. When saving limits to a file, both the overall mean and median are saved.
No missing points Missing points are not connected Missing points are connected Chapter 12 Legacy Control Charts 345 Quality and Process Methods Legacy Control Chart Platform Options Capability (Not available when a Phase variable is specified.) Performs a Capability Analysis for your data. A pop-up window is first shown, where you can enter the Lower Spec Limit, Target, and Upper Spec Limit values for the process variable.
Figure 12.12 Capability Analysis Window An example of a capability analysis report is shown in Figure 12.13 for Coating.jmp when the Lower Spec Limit is set as 16.5, the Target is set to 21.5, and the Upper Spec Limit is set to 23.
Figure 12.13 Capability Analysis Report for Coating.jmp For additional information, see “Statistical Details for Capability Analysis”.
Save Sigma Saves the computed value of sigma as a column property in the process variable column in the JMP data table.
Save Limits Saves the control limits in one of the following ways: 346 Legacy Control Charts Chapter 12 Legacy Control Chart Platform Options Quality and Process Methods in Column Saves control limits as a column property in the existing data table for the response variable. If the limits are constant, LCL, Avg, and UCL values for each chart type in the report are saved. This option is not available with phase charts. In addition, the option has no effect if the sample sizes are not constant for each chart.
in New Table Saves the standard deviation and mean for each chart into a new data table. If the limits are constant, the LCL, Avg, and UCL for each chart are saved as well. If there are phases, a new set of values is saved for each phase. You can use this data table to use the limits later. In the Control Chart launch window, click Get Limits and then select the saved data table. See the section “Saving and Retrieving Limits”.
Save Summaries Creates a new data table that contains the sample label, sample sizes, the statistic being plotted, the center line, and the control limits. The specific statistics included in the table depend on the type of chart.
Alarm Script Enables you to write and run a script that indicates when the data fail special causes tests. Results can be written to the log or spoken. See “Tests” of this guide. See the Scripting Guide for more information about writing custom Alarm Scripts.
See Using JMP for more information about the following options: Redo Contains options that enable you to repeat or relaunch the analysis. In platforms that support the feature, the Automatic Recalc option immediately reflects the changes that you make to the data table in the corresponding report window.
Save Script Contains options that enable you to save a script that reproduces the report to several destinations.
Save By-Group Script Contains options that enable you to save a script that reproduces the platform report for all levels of a By variable to several destinations. Available only when a By variable is specified in the launch window.
Chart Options for Legacy Control Charts The red triangle menu of chart options appears when you click the icon next to the chart name. Some options are also available under Chart Options when you right-click the chart. Not all of the options below are available for all control chart types.
Box Plots Superimposes box plots on the subgroup means plotted in a Mean chart. The box plot shows the subgroup maximum, minimum, 75th percentile, 25th percentile, and median. Markers for subgroup means show unless you deselect the Show Points option. The control limits displayed apply only to the subgroup mean. The Box Plots option is available only for charts. It is most appropriate for larger subgroup sample sizes (more than 10 samples in a subgroup).
Needle Connects plotted points to the center line with a vertical line segment.
X Chapter 12 Legacy Control Charts 347 Quality and Process Methods Legacy Control Chart Platform Options Connect Points Shows or hides the line that connects the data points.
Show Points Shows or hides the points representing summary statistics. Initially, the points show. You can use this option to suppress the markers denoting subgroup means when the Box Plots option is in effect.
Connect Color Displays the JMP color palette for you to choose the color of the line segments used to connect points.
Center Line Color Displays the JMP color palette for you to choose the color of the line segments used to draw the center line.
Limits Color Displays the JMP color palette for you to choose the color of the line segments used in the upper and lower limits lines.
Line Width Enables you to select the width of the control lines. Options are Thin, Medium, or Thick.
Point Marker Enables you to select the marker used on the chart.
Show Center Line Shows or hides the center line in the control chart.
Show Control Limits Shows or hides the chart control limits and their legends.
Limits Precision Sets the decimal limit for labels.
Tests Shows a submenu that enables you to choose which tests to mark on the chart when the test is positive. Tests apply only for charts whose limits are 3 limits. Tests 1 to 4 apply to Mean, Individual, and attribute charts. Tests 5 to 8 apply to Mean charts, Presummarize, and Individual Measurement charts only. If tests do not apply to a chart, the Tests option is dimmed. When sample sizes are unequal, the Test options are grayed out. If the samples change while the chart is open and they become equally sized, and the zone or test option is selected, the zones or tests are applied immediately and appear on the chart. These special tests are also referred to as the Western Electric Rules. For more information about special causes tests, see “Tests”.
Westgard Rules Westgard rules are control rules that help you decide whether a process is in or out of control. The different tests are abbreviated with the decision rule for the particular test. See the text and chart in “Westgard Rules”.
Test Beyond Limits Flags as a “” any point that is beyond the limits. This test works on all charts with limits, regardless of the sample size being constant, and regardless of the size of k or the width of the limits. For example, if you had unequal sample sizes, and wanted to flag any points beyond the limits of an R chart, you could use this command.
Show Zones Shows or hides the zone lines. The zones are labeled A, B, and C as shown here in the Mean plot for weight in the Coating.jmp sample data. Control Chart tests use the 348 Legacy Control Charts Chapter 12 Legacy Control Chart Platform Options Quality and Process Methods zone lines as boundaries. The seven zone lines are set one sigma apart, centered on the center line.
Figure 12.14 Show Zones Shade Zones Shows or hides the default green, yellow, and red colors for the three zone areas and the area outside the zones. Green represents the area one sigma from the center line, yellow represents the area two and three sigmas from the center line, and red represents the area beyond three sigmas. Shades can be shown with or without the zone lines.
Tip: To change the colors used to shade the zones, right-click in the control chart and select Customize. In the Customize Graph window, you can specify colors for each of the three zones.
Figure 12.15 Shade Zones OC Curve (Available when subgroup sizes are equal.) Opens a new window that contains the operating characteristic (OC) curve, using all the calculated values directly from the active control chart. See “Operating Characteristic Curves Utility”.
Chapter 12 Legacy Control Charts 349 Quality and Process Methods Saving and Retrieving Limits Chart Options for V-Mask CUSUM Control Charts Mask Color (Available only when the Show V Mask option is selected.) Enables you to select a line color for the V-mask.
Show Shift Shows or hides the shift that you entered in the launch window.
Show V Mask Shows or hides the V-mask based on the statistics that you specified in the CUSUM Control Charts launch window.
Show Parameters Shows or hides a report that summarizes the CUSUM charting parameters.
Show ARL Shows or hides the average run length (ARL) information. The average run length is the expected number of samples taken before an out-of-control condition is signaled: – ARL (Delta), sometimes denoted ARL1, is the average run length for detecting a shift in the size of the specified Delta.
– ARL(0), sometimes denoted ARL0, is the in-control average run length for the specified parameters (Montgomery 2013).
Saving and Retrieving Limits JMP can use previously established control limits for control charts: • Upper and lower control limits, and a center line value.
• Parameters for computing limits such as a mean and standard deviation.
The control limits or limit parameter values must be either in a JMP data table, referred to as the Limits Table, or stored as a column property in the process column. When you specify the Control Chart command, you can retrieve the Limits Table with the Get Limits button on the Control Chart launch window.
Tip: To add specification limits to several columns at once, see “Manage Spec Limits Utility”.
The easiest way to create a Limits Table is to save results computed by the Control Chart platform. The Save Limits command in the red triangle menu for each control chart automatically saves limits from the sample values. The type of data saved in the table varies according to the type of control chart in the analysis window. You can also use values from any source and create your own Limits Table.
All Limits Tables must have: • A column of special keywords that identify each row.
350 Legacy Control Charts Chapter 12 Saving and Retrieving Limits Quality and Process Methods • A column for each of the variables whose values are the known standard parameters or limits. This column name must be the same as the corresponding process variable name in the data table to be analyzed by the Control Chart platform.
The following table describes the limit keywords and their associated control chart for both legacy control charts and charts created with Control Chart Builder.
Table 12.1 Limits Table Keys with Appropriate Charts and Meanings Keywords For Charts Meaning _KSigma All except Control Chart Builder and V-Mask CUSUM multiples of the standard deviation of the statistics to calculate the control limits; set to missing if the limits are in terms of the alpha level _Alpha All except Control Chart Builder Type I error probability used to calculate the control limits _Range Span IM, MR, MMR number of consecutive measurements for which moving ranges are computed. Not applicable in the Control Chart Builder platform, where the range span is always equal to 2.
_Sample Size All except Levey-Jennings and Presummarize subgroup size _Std Dev XBar, R, S, IM, MR, G, T, V-Mask CUSUM, and Levey-Jennings known process standard deviation _U C, U known average number of nonconformities per unit _P NP, P known value of average proportion nonconforming _LCL, _UCL XBar, IM, P, NP, C, U, G, T, and Levey-Jennings lower and upper control limit for Mean Chart, Individual Measurement chart, or any attribute or rare event chart _AvgR R, MR average range or average moving range Chapter 12 Legacy Control Charts 351 Quality and Process Methods Saving and Retrieving Limits You can save limits in a new data table or as properties of the response column. When you save control limits using the in New Table command, the limit keywords written to the table depend on the current chart types displayed.
Figure 12.16 shows examples of control limits saved to a data table using Coating.jmp. The rows with values _Mean, _LCL, and _UCL are for the Individual Measurement chart. The values with the R suffix (_AvgR, _LCLR, and _UCLR) are for the Moving Range chart. If you create these charts again using this Limits Table, the Control Chart platform identifies the appropriate limits from keywords in the _LimitsKey column.
_LCLR, _UCLR R, MR lower control limit for R or MR chart upper control limit for R or MR chart _AvgS, _LCLS, _UCLS S Chart average standard deviation, upper and lower control limits for S chart _AvgR_PreMeans _AvgR_PreStdDev _LCLR_PreMeans _LCLR_PreStdDev _UCLR_PreMeans _UCLR_PreStdDev _Avg_PreMeans _Avg_PreStdDev _LCL_PreMeans _LCL_PreStdDev _UCL_PreMeans _UCL_PreStdDev IM, MR Mean, upper, and lower control limits based on pre-summarized group means or standard deviations.
_Data Units _Two Sided _Headstart _Beta _Delta V-Mask CUSUM specifications for V-Mask CUSUM chart Table 12.1 Limits Table Keys with Appropriate Charts and Meanings (Continued) Keywords For Charts Meaning 352 Legacy Control Charts Chapter 12 Excluded, Hidden, and Deleted Samples Quality and Process Methods Figure 12.16 Example of Saving Limits in a Data Table Note that values for _KSigma, _Alpha, and _Range Span can be specified in the Control Chart Launch window. JMP always looks at the values from the window first. Values specified in the window take precedence over those in an active Limits Table.
Rows with unknown keywords and rows marked with the excluded row state are ignored. Except for _Range Span, _KSigma, _Alpha, and _Sample Size, any needed values not specified are estimated from the data.
Excluded, Hidden, and Deleted Samples The following table summarizes the effects of various conditions on samples and subgroups: Table 12.2 Excluded, Hidden, and Deleted Samples All rows of the sample are excluded before creating the chart.
Sample is not included in the calculation of the limits, but it appears on the graph.
Sample is excluded after creating the chart.
Sample is included in the calculation of the limits, and it appears in the graph. Nothing changes on the output by excluding a sample with the graph open.
Chapter 12 Legacy Control Charts 353 Quality and Process Methods Excluded, Hidden, and Deleted Samples Some additional notes: • Hide and Exclude operate only on the row state of the first observation in the sample. For example, if the second observation in the sample is hidden, but the first observation is not hidden, the sample still appears on the chart.
Note: Excluded rows in Presummarize charts are excluded from calculations, regardless of which position they are within a sample.
• An exception to the exclude/hide rule: Both hidden and excluded rows are included in the count of points for Tests for Special Causes. An excluded row can be labeled with a special cause flag. A hidden point cannot be labeled. If the flag for a Tests for Special Causes is on a hidden point, it will not appear in the chart.
• Because of the specific rules in place (Table 12.2), the control charts do not support the Automatic Recalc script.
Sample is hidden before creating the chart.
Sample is included in the calculation of the limits, but does not appear on the graph.
Sample is hidden after creating the chart.
Sample is included in the calculation of the limits, but does not appear on the graph. The sample marker disappears from the graph, the sample label still appears on the axis, but limits remain the same.
All rows of the sample are both excluded and hidden before creating the chart.
Sample is not included in the calculation of the limits, and it does not appear on the graph.
All rows of the sample are both excluded and hidden after creating the chart.
Sample is included in the calculation of the limits, but does not appear on the graph. The sample marker disappears from the graph, the sample label still appears on the axis, but limits remain the same.
Data set is subsetted with Sample deleted before creating chart.
Sample is not included in the calculation of the limits, the axis does not include a value for the sample, and the sample marker does not appear on the graph.
Data set is subsetted with Sample deleted after creating chart.
Sample is not included in the calculation of the limits, and does not appear on the graph. The sample marker disappears from the graph, the sample label is removed from the axis, the graph shifts, and the limits change.
Table 12.2 Excluded, Hidden, and Deleted Samples (Continued) 354 Legacy Control Charts Chapter 12 Additional Examples of the Control Chart Platform Quality and Process Methods Additional Examples of the Control Chart Platform • “Presummarize Chart Example” • “V-Mask CUSUM Chart Example” • “One-Sided CUSUM Chart Example” • “UWMA Chart Example” Presummarize Chart Example This example uses the Coating.jmp data table.
1.
Select Help > Sample Data Library and open Quality Control/Coating.jmp.
2.
Select Analyze > Quality and Process > Legacy Control Charts > Presummarize.
3.
Select Weight and click Process.
4.
Select Sample and click Sample Label.
5.
Select both Individual on Group Means and Moving Range on Group Means. The Sample Grouped by Sample Label button is automatically selected when you choose a Sample Label variable.
When using Presummarize charts, you can select either On Group Means options or On Group Std Devs options or both. Each option creates two charts (an Individual Measurement, also known as an X chart, and a Moving Range chart) if both IR chart types are selected.
The On Group Means options compute each sample mean and then plot the means and create an Individual Measurement and a Moving Range chart on the means.
The On Group Std Devs options compute each sample standard deviation and plot the standard deviations as individual points. Individual Measurement and Moving Range charts for the standard deviations then appear.
6.
Click OK.
Chapter 12 Legacy Control Charts 355 Quality and Process Methods Additional Examples of the Control Chart Platform Figure 12.17 Example of Charting Presummarized Data Although the points for XBar and S charts are the same as the Individual on Group Means and Individual on Group Std Devs charts, the limits are different because they are computed as Individual charts.
Another way to generate the presummarized charts, with the Coating.jmp data table: 1.
Choose Tables > Summary.
2.
Select Sample and click Group.
3.
Select Weight, and then click Statistics > Mean and Statistics > Std Dev.
4.
Click OK.
5.
Select Analyze > Quality and Process > Legacy Control Charts > IR.
6.
Select Mean(Weight) and Std Dev(Weight) and click Process.
7.
Click OK.
The resulting charts match the presummarized charts.
356 Legacy Control Charts Chapter 12 Additional Examples of the Control Chart Platform Quality and Process Methods V-Mask CUSUM Chart Example A machine fills 8-ounce cans of two-cycle engine oil additive. The filling process is believed to be in statistical control. The process is set so that the average weight of a filled can (0) is 8.10 ounces. Previous analysis shows that the standard deviation of fill weights (0) is 0.05 ounces.
Subgroup samples of four cans are selected and weighed every hour for twelve hours. Each observation in the Oil1 Cusum.jmp data table contains one value of weight and its associated value of hour. The observations are sorted so that the values of hour are in increasing order.
1.
Select Help > Sample Data Library and open Quality Control/Oil1 Cusum.jmp.
2.
Select Analyze > Quality And Process > Legacy Control Charts > CUSUM.
3.
Select weight and click Process.
4.
Select hour and click Sample Label.
5.
Select the Two Sided check box if it is not already checked.
6.
In the Parameters area, click the H button and type 2.
7.
Click Specify Stats.
8.
Type 8.1 next to Target.
8.1 is the average weight in ounces of a filled can. This is the target mean.
9.
Type 1 next to Delta.
1 is the absolute value of the smallest shift to be detected as a multiple of the process standard deviation or of the standard error.
10. Type 0.05 next to Sigma.
0.05 is the known standard deviation of fill weights (0) in ounces.
Chapter 12 Legacy Control Charts 357 Quality and Process Methods Additional Examples of the Control Chart Platform Figure 12.18 Completed Launch Window 11. Click OK.
Figure 12.19 Two-Sided CUSUM Chart for Oil1 Cusum.jmp Data You can interpret the chart by comparing the points with the V-mask. The right edge of the V-mask is centered at the most recent point (the 12th hour). Because none of the points cross the arms of the V-mask, there is no evidence that a shift in the process has occurred. See “V-Mask CUSUM Chart Reports”.
358 Legacy Control Charts Chapter 12 Additional Examples of the Control Chart Platform Quality and Process Methods One-Sided CUSUM Chart Example Consider the data used in “V-Mask CUSUM Chart Example”, where the machine fills 8-ounce cans of engine oil. In order to cut costs, the manufacturer is now concerned about significant over-filling (and not so concerned about under-filling). Use a one-sided CUSUM chart to identify any instances of over-filling. Anything that is 0.25 ounces beyond the mean of 8.1 is considered a problem.
1.
Select Help > Sample Data Library and open Quality Control/Oil1 Cusum.jmp.
2.
Select Analyze > Quality And Process > Legacy Control Charts > CUSUM.
3.
Deselect Two Sided.
4.
Select weight and click Process.
5.
Select hour and click Sample Label.
6.
Click H and type 0.25.
7.
Click Specify Stats.
8.
Type 8.1 next to Target.
8.1 is the average weight in ounces of a filled can. This is the target mean.
9.
Type 1 next to Delta.
1 is the absolute value of the smallest shift to be detected as a multiple of the process standard deviation or of the standard error.
10. Type 0.05 next to Sigma.
0.05 is the known standard deviation of fill weights (0) in ounces.
11. Click OK.
Figure 12.20 One-Sided CUSUM Chart for Oil1 Cusum.jmp Data Chapter 12 Legacy Control Charts 359 Quality and Process Methods Additional Examples of the Control Chart Platform The decision interval is set at the H value that you entered (0.25). You can see that at the fourth hour, some significant over-filling occurred.
UWMA Chart Example In the sample data table Clips1.jmp, the measure of interest is the gap between the ends of manufactured metal clips. To monitor the process for a change in the average gap, subgroup samples of five clips are selected daily. A UWMA chart with a moving average span of three is examined.
1.
Select Help > Sample Data Library and open Quality Control/Clips1.jmp.
2.
Select Analyze > Quality and Process > Legacy Control Charts > UWMA.
3.
Select Gap and click Process.
4.
Select Sample and click Sample Label.
5.
Change the Moving Average Span to 3.
6.
Click OK.
Figure 12.21 UWMA Charts for the Clips1 data The point for the first day is the mean of the five subgroup sample values for that day. The plotted point for the second day is the average of subgroup sample means for the first and second days. The points for the remaining days are the average of subsample means for each day and the two previous days.
The average clip gap appears to be decreasing, but no sample point falls outside the 3 limits.
360 Legacy Control Charts Chapter 12 Statistical Details for the Control Chart Platform Quality and Process Methods Statistical Details for the Control Chart Platform • “Control Limits for Median Moving Range Charts” • “Statistical Details for Capability Analysis” • “Statistical Details for V-Mask CUSUM Control Charts” • “Statistical Details for Weighted Moving Average Charts” Note: For more information about other types of charts (such as XBar and R charts, P and NP charts, and more) see the “Statistical Details for Control Chart Builder”.
Control Limits for Median Moving Range Charts Control limits for Median Moving Range charts are computed as follows: LCLMMR = max(0, MMR - kd3(n) ) UCLMMR = MMR + kd3(n) where: MMR is the median of the nonmissing moving ranges. = MMR/0.954 d3(n) is the standard deviation of the range of n independent normally distributed variables with unit standard deviation.
Statistical Details for Capability Analysis This section contains details about the computation of the statistics in the Capability Analysis report.
Variation Statistics All capability analyses use the same formulas. Options differ in how sigma ()is computed: Long Term Sigma Uses the overall sigma. This option is used for Ppk statistics, and computes sigma as follows: ˆ ˆ ˆ Chapter 12 Legacy Control Charts 361 Quality and Process Methods Statistical Details for the Control Chart Platform Note: By default, the capability indices in the Long Term Sigma report use the Cp labeling that is used in the other sigma reports. To use Ppk labeling in the Long Term Sigma report, select the File > Preferences > Platforms > Distribution > PpK Capability Labeling preference.
Control Chart Sigma Uses a sigma that is determined by the control chart settings.
– If you specify a value for Sigma using the Specify Stats button in the control launch window, the specified value is used for computing capability indices.
– In an IR chart that uses the Moving Range (Average) option, the value for sigma is computed as follows: where: is the average of the moving ranges.
d2(n) is the expected value of the range of n independent normally distributed variables with unit standard deviation, where n is the value of the Range Span option.
– In an IR chart that uses the Median Moving Range option, the value for sigma is computed as follows: where: MMR is the median of the nonmissing moving ranges.
d4(n) is the median of the range of n independent normally distributed variables with unit standard deviation, where n is the value of the Range Span option.
– In an XBar chart that uses the R option, the value for sigma is computed as follows: where: Ri = range of ith subgroup ˆ xi x – 2 n 1 – ---------------------i 1 = n = ˆ R d2 n --------------= R ˆ MMR d4 n ----------------= ˆ R1 d2 n1 ----------------- RN d2 nN ------------------+ + N -------------------------------------------------------= 362 Legacy Control Charts Chapter 12 Statistical Details for the Control Chart Platform Quality and Process Methods ni = sample size of ith subgroup d2(ni) = expected value of the range of ni independent normally distributed variables with unit standard deviation N = number of subgroups for which ni ≥2 – In an XBar chart that uses the S option, the value for sigma is computed as follows: where: ni = sample size of ith subgroup c4(ni) = expected value of the standard deviation of ni independent normally distributed variables with unit standard deviation N = number of subgroups for which ni ≥2 si = sample standard deviation of the ith subgroup Capability Indices for Normal Distributions This section provides details about the calculation of capability indices for normal data.
For a process characteristic with mean and standard deviation , the population-based capability indices are defined as follows: Cp = Cpl = Cpu = Cpk = Cpm = where: LSL is the lower specification limit.
ˆ s1 c4 n1 ---------------- sN c4 nN ------------------+ + N ------------------------------------------------------= USL LSL – 6 ---------------------------- LSL – 3 --------------------USL – 3 ---------------------min( Cpl, Cpu ) min T LSL – USL T – 31 T – ------------- 2 + -------------------------------------------------------------Chapter 12 Legacy Control Charts 363 Quality and Process Methods Statistical Details for the Control Chart Platform USL is the upper specification limit.
T is the target value.
For sample-based capability indices, the parameters are replaced by their estimates. The estimate for uses the method that you specified in the Capability Analysis window. See “Variation Statistics”.
If either of the specification limits is missing, the capability indices containing the missing specification limit are reported as missing.
Tip: A capability index of 1.33 is often considered to be the minimum value that is acceptable. For a normal distribution, a capability index of 1.33 corresponds to an expected number of nonconforming units of about 6 per 100,000.
Confidence Intervals for Capability Indices Note: Confidence intervals for capability indices appear only in the Long Term Sigma report.
The 100(1 - )% confidence interval for Cp is calculated as follows: where: is the estimated value for Cp. is the (/2)th quantile of a chi-square distribution with n - 1 degrees of freedom.
n is the number of observations.
The 100(1 - )% confidence interval for Cpk is calculated as follows: where: is the estimated value for Cpk.
CP ˆ 2 n 1 – 2 n 1 – --------------------------- CP ˆ 1 2 – n 1 – 2 n 1 – ----------------------------------- CP ˆ 2 n 1 – 2 C ˆ pk 1 1 – 1 2 – 1 9nC ˆ pk 2 -----------------1 2 n 1 – --------------------+ – C ˆ pk 1 1 – 1 2 – 1 9nC ˆ pk 2 -----------------1 2 n 1 – --------------------+ + C ˆ pk 364 Legacy Control Charts Chapter 12 Statistical Details for the Control Chart Platform Quality and Process Methods is the (1 - /2)th quantile of a standard normal distribution.
n is the number of observations.
The 100(1 - )% confidence interval for CPM is calculated as follows: where: is the estimated value for CPM. is the (/2)th quantile of a chi-square distribution with degrees of freedom.
n is the number of observations. is the mean of the observations.
T is the target value.
s is the long-term sigma estimate.
Note: The confidence interval for CPM is computed only when the target value is centered between the lower and upper specification limits.
Lower and upper confidence limits for CPL and CPU are computed using the method of Chou et al. (1990).
The 100(1 - )% confidence limits for CPL (denoted by CPLL and CPLU) satisfy the following equations: where where where: tn-1() has a non-central t-distribution with n - 1 degrees of freedom and noncentrality 1 – 1 2 – CPM ˆ 2 2 ------------------ CPM ˆ 1 2 – 2 -------------------------- CPM ˆ 2 2 n 1 x T – s ------------ 2 + 2 1 2 x T – s ------------ 2 + ------------------------------------------= x Pr tn 1 – L 3C ˆ pl n 2 = L 3CPLL n = Pr tn 1 – U 3C ˆ pl n 2 = U 3CPLU n = Chapter 12 Legacy Control Charts 365 Quality and Process Methods Statistical Details for the Control Chart Platform parameter is the estimated value for Cpl.
The 100(1 - )% confidence limits for CPU (denoted by CPUL and CPUU) satisfy the following equations: where where where: tn-1() has a non-central t-distribution with n - 1 degrees of freedom and noncentrality parameter is the estimated value for Cpu.
Capability Indices for Nonnormal Distributions This section describes how capability indices are calculated for nonnormal distributions. These generalized capability indices are defined as follows: Cp = Cpk = Cpm = Cpl = Cpu = C ˆ pl Pr tn 1 – L 3C ˆ pu n 2 = L 3CPUL n = Pr tn 1 – U 3C ˆ pu n 2 = U 3CPUU n = C ˆ pu USL LSL – P0.99865 P0.00135 – ------------------------------------------------min Cpl Cpu min T LSL – P0.5 P0.00135 – -------------------------------------USL T – P0.99865 P0.5 – -------------------------------------, 1 T – ------------- 2 + ----------------------------------------------------------------------------------------------P0.5 LSL – P0.5 P0.00135 – -------------------------------------USL P0.5 – P0.99865 P0.5 – -------------------------------------366 Legacy Control Charts Chapter 12 Statistical Details for the Control Chart Platform Quality and Process Methods where: LSL is the lower specification limit.
USL is the upper specification limit.
T is the target value.
P is the 100th percentile of the fitted distribution.
For the calculation of Cpm, and are estimated using the expected value and square root of the variance of the fitted distribution. For more information about the relationship between the parameters in the Parameter Estimates report and the expected value and variance of the fitted distributions, see Basic Analysis.
Sigma Quality Statistics The Sigma Quality statistics for each Portion (Below LSL, Above USL, and Total Outside) are calculated as follows: where: Pct is the value in the Percent column of the report. is the (1 - Pct/100)th quantile of a standard normal distribution.
Note: Even though the Percent Below LSL and Percent Above USL sum to the Percent Total Outside value, the Sigma Quality Below LSL and Sigma Quality Above USL values do not sum to the Sigma Quality Total Outside value. This is because calculating Sigma Quality involves finding normal distribution quantiles, and is therefore not additive.
Benchmark Z Statistics Benchmark Z statistics are available only for capability analyses based on the normal distribution. The Benchmark Z statistics are calculated as follows: Z Bench = Z LSL = = 3 Cpl Z USL = = 3 Cpu Sigma Quality 1 – 1 Pct 100 – 1.5 + = 1 – 1 Pct 100 – 1 – 1 P LSL – P USL – LSL – --------------------USL – ---------------------Chapter 12 Legacy Control Charts 367 Quality and Process Methods Statistical Details for the Control Chart Platform where: LSL is the lower specification limit.
USL is the upper specification limit.
is the sample mean.
is the sample standard deviation.
-1 1 - P(LSL) - P(USL) is the (1 - P(LSL) -P(USL))th quantile of a standard normal distribution.
P(LSL) = Prob(X < LSL) = 1 - (Z LSL).
P(USL) = Prob(X > USL) = 1 - (Z USL).
is the standard normal cumulative distribution function.
Statistical Details for V-Mask CUSUM Control Charts The following notation is used in these formulas: • denotes the mean of the population, also referred to as the process mean or the process level.
• 0 denotes the target mean (or goal) for the population. Sometimes, the symbol is used for 0. See the American Society for Quality Statistics Division (2004). You can provide 0 as the Target in the Known Statistics for CUSUM Chart area on the launch window.
• denotes the population standard deviation. denotes an estimate of .
• 0 denotes a known standard deviation. You can provide o as the Sigma in the Known Statistics for CUSUM Chart area on the launch window.
• n denotes the nominal sample size for the CUSUM chart.
• denotes the shift in to be detected, expressed as a multiple of the standard deviation. You can provide as the Delta in the Known Statistics for CUSUM Chart area on the launch window.
• denotes the shift in to be detected, expressed in data units. If the sample size n is constant across subgroups, then the following computation applies: You can provide as the Shift in the Known Statistics for CUSUM Chart area on the launch window.
Note: Some authors use the symbol D instead of .
X0 ˆ X n = = 368 Legacy Control Charts Chapter 12 Statistical Details for the Control Chart Platform Quality and Process Methods One-Sided CUSUM Charts Positive Shifts If the shift to be detected is positive, the CUSUM for the tth subgroup is computed as follows: St = max(0, St – 1+ (zt – k)) t = 1, 2,..., n, where S0 = 0, zt is defined as for two-sided charts, and the parameter k, termed the reference value, is positive. If the parameter k is not specified in the launch window, k is set to /2. The CUSUM St is referred to as an upper cumulative sum. St can be computed as follows: The sequence St cumulates deviations in the subgroup means greater than k standard errors from 0. If St exceeds a positive value h (referred to as the decision interval), a shift or out-of-control condition is signaled.
Negative Shifts If the shift to be detected is negative, the CUSUM for the tth subgroup is computed as follows: St = max(0, St – 1 – (zt + k)) t = 1, 2,..., n, where S0 = 0, zt is defined as for two-sided charts, and the parameter k, termed the reference value, is positive. If the parameter k is not specified in the launch window, k is set to /2. The CUSUM St is referred to as a lower cumulative sum. St can be computed as follows: The sequence St cumulates the absolute value of deviations in the subgroup means less than k standard errors from 0. If St exceeds a positive value h (referred to as the decision interval), a shift or out-of-control condition is signaled.
Note that St is always positive and h is always positive, regardless of whether is positive or negative. For charts designed to detect a negative shift, some authors define a reflected version of St for which a shift is signaled when St is less than a negative limit.
Lucas and Crosier (1982) describe the properties of a fast initial response (FIR) feature for CUSUM charts in which the initial CUSUM S0 is set to a “head start” value. Average run length calculations given by them show that the FIR feature has little effect when the process is in control and that it leads to a faster response to an initial out-of-control condition than a standard CUSUM chart. You can provide a Head Start value in the Known Statistics for CUSUM Chart area on the launch window.
max 0 St 1 – Xi 0 kXt + – Xt -----------------------------------------+ max 0 St 1 – Xi 0 kXt – – Xt -----------------------------------------– Chapter 12 Legacy Control Charts 369 Quality and Process Methods Statistical Details for the Control Chart Platform Constant Sample Sizes When the subgroup sample sizes are constant (= n), it might be preferable to compute CUSUMs that are scaled in the same units as the data. CUSUMs are then computed as follows: where > 0 where < 0. In either case, the parameter k is rescaled to . If the parameter k is not specified in the launch window, kʹ is set to /2. A shift is signaled if St exceeds . Some authors use the symbol H for hʹ.
Two-Sided CUSUM Charts If the CUSUM chart is two-sided, the cumulative sum St plotted for the tth subgroup is defined as follows: St = St - 1 +zt t = 1, 2,..., n. Here S0=0, and the term zt is calculated as follows: where is the tth subgroup average, and nt is the tth subgroup sample size. If the subgroup samples consist of individual measurements xt, the term zt simplifies to the following computation: zt = (xt – 0)/ The first equation can be rewritten as follows: where the sequence St cumulates standardized deviations of the subgroup averages from the target mean 0.
In many applications, the subgroup sample sizes ni are constant (ni = n), and the equation for St can be simplified: St max 0 St 1 – Xt 0 k n + – + = St max 0 St 1 – Xt 0 k n – – – = k' k n = h' h n = zt Xt 0 – nt = Xt St zi i 1 = t Xi 0 – Xi i 1 = t = = St 1 X Xi 0 – i 1 = t n Xi 0 – i 1 = t = = 370 Legacy Control Charts Chapter 12 Statistical Details for the Control Chart Platform Quality and Process Methods In some applications, it might be preferable to compute St as follows: which is scaled in the same units as the data. In this case, the procedure rescales the V-mask parameters h and k to and , respectively. Some authors use the symbols F for kʹ and H for hʹ.
If the process is in control and the mean is at or near the target 0, the random walk model applies. Therefore, the points might wander away from zero, but they will not exhibit a large trend since positive and negative displacements from 0 tend to cancel each other. If shifts in the positive direction, the points exhibit an upward trend, and if shifts in the negative direction, the points exhibit a downward trend.
Statistical Details for Weighted Moving Average Charts Control Limits for UWMA Charts Control limits for UWMA charts are computed for each subgroup i as follows: LCLi = UCLi = where: w is the span parameter (number of terms in moving average) ni is the sample size of the ith subgroup k is the number of standard deviations is the weighted average of the subgroup means is the estimated process standard deviation St Xi 0 – i 1 = t = h' h n = k' k n = 0 k ˆ min i w ------------------------1 ni -----1 ni 1 – ------------- 1 n1 max i w 0 – + ------------------------------------------+ + + – 0 k ˆ min i w ------------------------1 ni -----1 ni 1 – ------------- 1 n1 max i w 0 – + ------------------------------------------+ + + + 0 ˆ Chapter 12 Legacy Control Charts 371 Quality and Process Methods Statistical Details for the Control Chart Platform Control Limits for EWMA Charts Control limits for EWMA charts are computed as follows: LCL = UCL = where: r is the EWMA weight parameter (0 < r 1) xij is the jth measurement in the ith subgroup, with j = 1, 2, 3,..., ni ni is the sample size of the ith subgroup k is the number of standard deviations is the weighted average of the subgroup means is the estimated process standard deviation 0 k ˆ r 1 r – 2j ni j – --------------------j 0 = i 1 – – 0 k ˆ r 1 r – 2j ni j – --------------------j 0 = i 1 – + 0 ˆ 372 Legacy Control Charts Chapter 12 Statistical Details for the Control Chart Platform Quality and Process Methods Chapter 13 Pareto Plots Focus Improvement Efforts on the Vital Few Improve the statistical quality of your process or operation using Pareto plots. A Pareto plot is a chart that shows severity (frequency) of problems in a quality-related process or operation. Pareto plots help you decide which problems to solve first by highlighting the frequency and severity of problems.
Figure 13.1 Pareto Plot Examples 374 Pareto Plots Chapter 13 Quality and Process Methods Contents Overview of the Pareto Plot Platform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375 Example of the Pareto Plot Platform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375 Launch the Pareto Plot Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378 The Pareto Plot Report . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379 Pareto Plot Platform Options. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 Causes Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382 Additional Examples of the Pareto Plot Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382 Threshold of Combined Causes Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 Using a Constant Size across Groups Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384 Using a Non-Constant Sample Size across Groups Example. . . . . . . . . . . . . . . . . . . . . . . . 386 One-Way Comparative Pareto Plot Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 387 Two-Way Comparative Pareto Plot Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389 Statistical Details for the Pareto Plot Platform . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390 Likelihood Ratio Chi-Square Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390 Chapter 13 Pareto Plots 375 Quality and Process Methods Overview of the Pareto Plot Platform Overview of the Pareto Plot Platform The Pareto Plot platform produces charts to display the relative frequency or severity of problems in a quality-related process or operation. The Pareto plot is displayed initially as a bar chart that shows the classification of problems arranged in decreasing order. The column whose values are the cause of a problem is assigned the Y role and is called the process variable.
You can also generate a comparative Pareto plot, which combines two or more Pareto plots for the same process variable. The single display shows plots for each value in a column assigned the X role, or combination of levels from two X variables. Columns assigned the X role are called classification variables.
The Pareto plot can chart a single Y (process) variable with no X classification variables, with a single X, or with two X variables. The Pareto function does not distinguish between numeric and character variables or between modeling types. You can switch between a bar chart and a pie chart. All values are treated as discrete, and bars or wedges represent either counts or percentages.
Example of the Pareto Plot Platform This example uses the Failure.jmp sample data table, which contains failure data and a frequency column. It lists causes of failure during the fabrication of integrated circuits and the number of times each type of defect occurred. From the analysis, you can determine which factors contribute most toward process failure.
1.
Select Help > Sample Data Library and open Quality Control/Failure.jmp.
2.
Select Analyze > Quality and Process > Pareto Plot.
3.
Select failure and click Y, Cause.
This column lists the causes of failure. It is the variable that you want to inspect.
4.
Select N and click Freq.
This column list the number of times that each type of failure occurred.
5.
Click OK.
376 Pareto Plots Chapter 13 Example of the Pareto Plot Platform Quality and Process Methods Figure 13.2 Pareto Plot Report Window The left axis represents the count of failures, and the right axis represents the percent of failures in each category. The bars are in decreasing order with the most frequently occurring failure to the left. The curve indicates the cumulative failures from left to right.
6.
Click the Pareto Plot red triangle and select Label Cum Percent Points.
Note that Contamination accounts for approximately 45% of the failures. The point above the Oxide Defect bar shows that Contamination and Oxide Defect together account for approximately 71% of the failures.
7.
Click the Pareto Plot red triangle and deselect Label Cum Percent Points and Show Cum Percent Curve.
8.
Click the label for the y-axis labeled N and rename it Count.
9.
Double-click the y-axis to display the Y Axis Settings window.
– In the Maximum field, type 15.
– In the Increment field, type 2.
– In the Axis Label Row panel, select Grid Lines for the Major grid line.
– Click OK.
10. Click the Pareto Plot red triangle and select Category Legend.
Chapter 13 Pareto Plots 377 Quality and Process Methods Example of the Pareto Plot Platform Figure 13.3 Pareto Plot with Display Options Figure 13.3 shows the counts of different types of failures and has a category legend. The vertical count axis is rescaled and has grid lines at the major tick marks.
11. To view the data as a pie chart, click the Pareto Plot red triangle and select Pie Chart.
Figure 13.4 Pareto Plot as a Pie Chart Contamination and Oxide Defect clearly represent the majority of the failures.
378 Pareto Plots Chapter 13 Launch the Pareto Plot Platform Quality and Process Methods Launch the Pareto Plot Platform Launch the Pareto Plot platform by selecting Analyze > Quality and Process > Pareto Plot.
Figure 13.5 The Pareto Plot Launch Window For more information about the options in the Select Columns red triangle menu, see Using JMP.
The Pareto Plot launch window contains the following options: Y, Cause Identifies the column whose values are the cause of a problem. It is called the process variable and is the variable that you want to inspect.
X, Grouping Identifies the grouping factor. The grouping variable produces one Pareto plot window with side-by-side plots for each value. You can have no grouping variable, one grouping variable (see “One-Way Comparative Pareto Plot Example”), or two grouping variables (see “Two-Way Comparative Pareto Plot Example”).
Weight Assigns a variable to give the observations different weights.
Freq Identifies the column whose values hold the frequencies.
By Identifies a variable to produce a separate analysis for each value that appears in the column.
Threshold of Combined Causes Enables you to specify a threshold for combining causes by specifying a minimum rate or count. Select the option and then select Tail % or Count and enter the threshold value. The Tail percent option combines smaller count groups against the percentage specified of the total (combined small groups count/total group count). The Count option enables you to specify a specific count threshold. For an example, see “Threshold of Combined Causes Example”.
Per Unit Analysis Enables you to compare defect rates across groups. JMP calculates the defect rate as well as 95% confidence intervals of the defect rate. Select the option and then select Constant or Value in Freq Column and enter the sample size value or cause code, respectively. The Constant option enables you to specify a constant sample size on the Chapter 13 Pareto Plots 379 Quality and Process Methods The Pareto Plot Report launch window. The Value In Freq Column option enables you to specify a unique sample size for a group through a special cause code to designate the rows as cause rows.
Although causes are allowed to be combined in Pareto plots, the calculations for these analyses do not change correspondingly.
For examples, see “Using a Constant Size across Groups Example” and “Using a Non-Constant Sample Size across Groups Example”.
The Pareto Plot Report The Pareto plot combines a bar chart displaying percentages of variables in the data with a line graph showing cumulative percentages of the variables.
Figure 13.6 Pareto Plot Example The Pareto plot can chart a single Y (process) variable with no X classification variables, with a single X, or with two X variables. The Pareto plot does not distinguish between numeric and character variables or between modeling types. All values are treated as discrete, and bars represent either counts or percentages. The following list describes the arrangement of the Pareto plot: • A Y variable with no X classification variables produces a single chart with a bar for each value of the Y variable. For an example, see “Example of the Pareto Plot Platform”.
• A Y variable with one X classification variable produces a row of Pareto plots. There is a plot for each level of the X variable with bars for each Y level. These plots are referred to as 380 Pareto Plots Chapter 13 Pareto Plot Platform Options Quality and Process Methods the cells of a comparative Pareto plot. There is a cell for each level of the X (classification) variable. Because there is only one X variable, this is called a one-way comparative Pareto plot. For an example, see “One-Way Comparative Pareto Plot Example”.
• A Y variable with two X variables produces rows and columns of Pareto plots. There is a row for each level of the first X variable and a column for each level of the second X variable. Because there are two X variables, this is called a two-way comparative Pareto plot. The rows have a Pareto plot for each value of the first X variable, as described previously. The upper left cell is called the key cell. Its bars are arranged in descending order. The bars in the other cells are in the same order as the key cell. You can reorder the rows and columns of cells. The cell that moves to the upper left corner becomes the new key cell and the bars in all other cells rearrange accordingly. For an example, see “Two-Way Comparative Pareto Plot Example”.
• Each bar is the color for which the rows for that Y level are assigned in the associated data table. Otherwise, a single color is used for all of the bars whose Y levels do not have rows with an assigned color. If the rows for a Y level have different colors, the bar for that Y level is the color of the first row for that Y level in the data table.
You can change the type of scale and arrangement of bars and convert the bars into a pie chart using the options in the Pareto Plot red triangle menu. See “Pareto Plot Platform Options”.
Pareto Plot Platform Options The Pareto Plot red triangle menu contains options that customize the appearance of the plots. It also has options in the Causes submenu that affect individual bars within a Pareto plot. The following commands affect the appearance of the Pareto plot as a whole: Percent Scale Shows or hides the count and percent left vertical axis display.
N Legend Shows or hides the total sample size in the plot area.
Category Legend Shows or hides labeled bars and a separate category legend.
Pie Chart Shows or hides the bar chart and pie chart representation.
Reorder Horizontal, Reorder Vertical Reorders grouped Pareto plots when there is one or more grouping variables.
Ungroup Plots Splits up a group of Pareto charts into separate plots.
Count Analysis Performs defect per unit analyses. Enables you to compare defect rates and perform ratio tests across and within groups: Per Unit Rates Compares defect rates across groups. If a sample size is specified, Defects Per Unit (DPU) and Parts Per Million (PPM) columns are added to the report.
Chapter 13 Pareto Plots 381 Quality and Process Methods Pareto Plot Platform Options Test Rate Within Groups Performs a likelihood ratio Chi-square test to determine whether the rates across causes are the same within a group. See “Statistical Details for the Pareto Plot Platform”.
Test Rates Across Groups Performs a likelihood ratio Chi-square test to determine whether the rate for each cause is the same across groups. See “Statistical Details for the Pareto Plot Platform”.
Show Cum Percent Curve Shows or hides the cumulative percent curve above the bars and the cumulative percent axis on the vertical right axis.
Show Cum Percent Axis Shows or hides the cumulative percent axis on the vertical right axis.
Show Cum Percent Points Shows or hides the points on the cumulative percent curve.
Label Cum Percent Points Shows or hides the labels on the points on the cumulative curve.
Cum Percent Curve Color Changes the color of the cumulative percent curve.
Causes Has options that affect one or more individual chart bars. See “Causes Options”, for a description of these options.
See Using JMP for more information about the following options: Local Data Filter Shows or hides the local data filter that enables you to filter the data used in a specific report.
Redo Contains options that enable you to repeat or relaunch the analysis. In platforms that support the feature, the Automatic Recalc option immediately reflects the changes that you make to the data table in the corresponding report window.
Save Script Contains options that enable you to save a script that reproduces the report to several destinations.
Save By-Group Script Contains options that enable you to save a script that reproduces the platform report for all levels of a By variable to several destinations. Available only when a By variable is specified in the launch window.
382 Pareto Plots Chapter 13 Additional Examples of the Pareto Plot Platform Quality and Process Methods Causes Options You can highlight a bar by clicking on it. Use Control-click to select multiple bars that are not contiguous. When you select bars, you can access the commands on the red triangle menu that affect Pareto plot bars. They are found on the Causes submenu on the red triangle menu. These options are also available with a right-click anywhere in the plot area. The following options apply to highlighted bars instead of to the chart as a whole: Combine Causes Combines selected (highlighted) bars. You can select either Selected, Last Causes, First Causes or select from a list of variables.
Figure 13.7 Combine Causes Window Separate Causes Separates selected bars into their original component bars.
Move to First Moves one or more highlighted bars to the left (first) position.
Move to Last Moves one or more highlighted bars to the right (last) position.
Colors Shows the colors palette for coloring one or more highlighted bars.
Markers Shows the markers palette for assigning a marker to the points on the cumulative percent curve, when the Show Cum Percent Points command is in effect.
Label Displays the bar value at the top of all highlighted bars.
Additional Examples of the Pareto Plot Platform • “Threshold of Combined Causes Example” • “Using a Constant Size across Groups Example” • “Using a Non-Constant Sample Size across Groups Example” • “One-Way Comparative Pareto Plot Example” • “Two-Way Comparative Pareto Plot Example” Chapter 13 Pareto Plots 383 Quality and Process Methods Additional Examples of the Pareto Plot Platform Threshold of Combined Causes Example This example uses the Failure.jmp sample data table, which contains failure data and a frequency column. It lists causes of failure during the fabrication of integrated circuits and the number of times each type of defect occurred. A threshold value of 2 is specified for this example.
1.
Select Help > Sample Data Library and open Quality Control/Failure.jmp.
2.
Select Analyze > Quality and Process > Pareto Plot.
3.
Select failure and click Y, Cause.
4.
Select N and click Freq.
5.
Select Threshold of Combined Causes and then select Count.
6.
Enter 2 as the threshold value.
7.
Click OK.
Figure 13.8 Pareto Plot with a Threshold Count of 2 Figure 13.8 displays the plot after specifying a count of 2. All causes with counts 2 or fewer are combined into the final bar labeled 4 Others.
8.
To separate the combined bars into original categories as shown in Figure 13.9, select Causes > Separate Causes.
384 Pareto Plots Chapter 13 Additional Examples of the Pareto Plot Platform Quality and Process Methods Figure 13.9 Pareto Plot with Separated Causes Using a Constant Size across Groups Example This example uses the Failures.jmp sample data table, which contains failure data and a frequency column. It lists causes of failure during the fabrication of integrated circuits and the number of times each type of defect occurred for two processes. A constant sample size of 1000 is specified for this example.
1.
Select Help > Sample Data Library and open Quality Control/Failures.jmp.
2.
Select Analyze > Quality and Process > Pareto Plot.
3.
Select Causes and click Y, Cause.
4.
Select Process and click X, Grouping.
5.
Select Count and click Freq.
6.
Select Per Unit Analysis and then select Constant.
7.
Enter 1000 in Sample Size.
8.
Click OK.
Chapter 13 Pareto Plots 385 Quality and Process Methods Additional Examples of the Pareto Plot Platform Figure 13.10 Pareto Plot Report Window Process A indicates Contamination as the top failure while Process B indicates Oxide Defect as the leading failure.
9.
Click the Pareto Plot red triangle and select Count Analysis > Test Rates Across Groups.
Figure 13.11 Test Rates across Groups Results Note that the DPU for Contamination across groups (Process A and B) is around 0.06.
386 Pareto Plots Chapter 13 Additional Examples of the Pareto Plot Platform Quality and Process Methods Using a Non-Constant Sample Size across Groups Example This example uses the Failuressize.jmp sample data table, which contains failure data and a frequency column. It lists causes of failure during the fabrication of integrated circuits and the number of times each type of defect occurred for two processes. Among the other causes (Oxide Defect, Silicon Defect, and so on) is a cause labeled size. Specifying size as the cause code designates the rows as size rows.
1.
Select Help > Sample Data Library and open Quality Control/Failuressize.jmp.
2.
Select Analyze > Quality and Process > Pareto Plot.
3.
Select Causes and click Y, Cause.
4.
Select Process and click X, Grouping.
5.
Select Count and click Freq.
6.
Select Per Unit Analysis and then select Value in Freq Column.
7.
Enter size in Cause Code.
8.
Click OK.
Figure 13.12 Pareto Plot Report Window 9.
Click the Pareto Plot red triangle and select Count Analysis > Per Unit Rates and Count Analysis > Test Rates Across Groups.
Chapter 13 Pareto Plots 387 Quality and Process Methods Additional Examples of the Pareto Plot Platform Figure 13.13 Per Unit Rates and Test Rates across Groups Results Note that the sample size of 101 is used to calculate the DPU for the causes in group A. However, the sample size of 145 is used to calculate the DPU for the causes in group B.
If there are two group variables (say, Day and Process), Per Unit Rates lists DPU or rates for every combination of Day and Process for each cause. However, Test Rate Across Groups only tests overall differences between groups.
One-Way Comparative Pareto Plot Example This example uses the Failure2.jmp sample data table. This table records failures in a sample of capacitors manufactured before cleaning a tube in the diffusion furnace and in a sample manufactured after cleaning the furnace. For each type of failure, the variable clean identifies the samples with the values “before” or “after.” 1.
Select Help > Sample Data Library and open Quality Control/Failure2.jmp.
2.
Select Analyze > Quality and Process > Pareto Plot.
3.
Select failure and click Y, Cause.
4.
Select clean and click X, Grouping.
5.
Select N and click Freq.
6.
Click OK.
388 Pareto Plots Chapter 13 Additional Examples of the Pareto Plot Platform Quality and Process Methods Figure 13.14 displays the side-by-side plots for each value of the variable, clean.
Figure 13.14 One-way Comparative Pareto Plot The horizontal and vertical axes are scaled identically for both plots. The bars in the first plot are in descending order of the y-axis values and determine the order for all cells.
7.
Rearrange the order of the plots by clicking the title (after) in the first tile and dragging it to the title of the next tile (before).
A comparison of these two plots shows a reduction in oxide defects after cleaning. However, the plots are easier to interpret when presented as the before-and-after plot shown in Figure 13.15. Note that the order of the causes changes to reflect the order based on the first cell.
Chapter 13 Pareto Plots 389 Quality and Process Methods Additional Examples of the Pareto Plot Platform Figure 13.15 One-way Comparative Pareto Plot with Reordered Cells Two-Way Comparative Pareto Plot Example This example uses the Failure3.jmp sample data table. The data monitors production samples before and after a furnace cleaning for three days for a capacitor manufacturing process. The data table has a column called date with values OCT 1, OCT 2, and OCT 3.
1.
Select Help > Sample Data Library and open Quality Control/Failure3.jmp.
2.
Select Analyze > Quality and Process > Pareto Plot.
3.
Select failure and click Y, Cause.
4.
Select clean and date and click X, Grouping.
5.
Select N and click Freq.
6.
Click OK.
Figure 13.16 displays the Pareto plot with a two-way layout of plots that show each level of both X variables. The upper left cell is called the key cell. Its bars are arranged in descending order. The bars in the other cells are in the same order as the key cell.
7.
Click Contamination and Metallization in the key cell and the bars for the corresponding categories highlight in all other cells.
390 Pareto Plots Chapter 13 Statistical Details for the Pareto Plot Platform Quality and Process Methods Figure 13.16 Two-way Comparative Pareto Plot The Pareto plot illustrates highlighting the vital few. In each cell of the two-way comparative plot, the bars representing the two most frequently occurring problems are selected. Contamination and Metallization are the two vital categories in all cells. After furnace cleaning, Contamination is less of a problem.
Statistical Details for the Pareto Plot Platform Likelihood Ratio Chi-Square Test Notation The likelihood ratio Chi-square test statistic computed in the Pareto Plot platform uses the following notation: • nij is the count for Cause i in Group j.
Chapter 13 Pareto Plots 391 Quality and Process Methods Statistical Details for the Pareto Plot Platform • Ej is the expected count for Group j. This is the mean count of each group, across causes.
• Ei is the expected count for Cause i. This is the mean count of each cause, across groups.
Likelihood Ratio Chi-Square Test Statistic within Groups Likelihood Ratio Chi-Square Test Statistic across Groups Gj 2 2 nij nij Ej ln i 1 = K = Gi 2 2 nij nij Ei ln j 1 = J = 392 Pareto Plots Chapter 13 Statistical Details for the Pareto Plot Platform Quality and Process Methods Chapter 14 Cause-and-Effect Diagrams Identify Root Causes Use the Diagram platform to construct cause-and-effect diagrams, also known as Ishikawa charts or fishbone charts. Use these diagrams to: • Organize the causes of an effect (sources of a problem) • Brainstorm • Identify variables in preparation for further experimentation Figure 14.1 Example of a Cause-and-Effect Diagram 394 Cause-and-Effect Diagrams Chapter 14 Quality and Process Methods Contents Overview of Cause-and-Effect Diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395 Example of a Cause-and-Effect Diagram. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395 Prepare the Data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396 Launch the Diagram Platform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396 The Cause-and-Effect Diagram. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397 Right-Click Menus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397 Cause and Effect Diagram Menu Options. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400 Save the Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 Save the Diagram as a Data Table. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 Save the Diagram as a Journal. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 Save the Diagram as a Script . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402 Chapter 14 Cause-and-Effect Diagrams 395 Quality and Process Methods Overview of Cause-and-Effect Diagrams Overview of Cause-and-Effect Diagrams Use the Diagram platform to construct cause-and-effect diagrams, also known as Ishikawa charts or fishbone charts. Use these diagrams to: • Organize the causes of an effect (sources of a problem) • Brainstorm • Identify variables in preparation for further experimentation Example of a Cause-and-Effect Diagram You have data about defects in a circuit board. You want to examine the major factors and possible causes of the defects in a diagram.
1.
Select Help > Sample Data Library and open Ishikawa.jmp.
2.
Select Analyze > Quality and Process > Diagram.
3.
Select Parent and click X, Parent.
4.
Select Child and click Y, Child.
5.
Click OK.
Figure 14.2 Ishikawa.jmp Diagram The major factors are Inspection, Solder process, Raw card, Components, and Component insertion. From each major factor, possible causes branch off, such as Inspection, Measurement, and Test coverage for the Inspection factor.
You can focus on one area at a time to further examine the possible causes or sources of variation for each major factor.
396 Cause-and-Effect Diagrams Chapter 14 Prepare the Data Quality and Process Methods Prepare the Data Before you produce the diagram, begin with your data in two columns of a data table.
Figure 14.3 Example of the Ishikawa.jmp Data Table Notice that the Parent value Defects in circuit board (the effect) has five major factors, listed in the Child column. One of these major factors is Inspection, which has its own causes listed in the Child column. Parent values have children, and children can have their own children (and therefore be listed in both the Parent and Child columns.) Launch the Diagram Platform Launch the Diagram platform by selecting Analyze > Quality And Process > Diagram.
Figure 14.4 The Diagram Launch Window Chapter 14 Cause-and-Effect Diagrams 397 Quality and Process Methods The Cause-and-Effect Diagram For more information about the options in the Select Columns red triangle menu, see Using JMP.
Tip: To create a basic diagram that is not based on a data table, leave the Y, Child, and X, Parent fields empty and click OK. Then edit the nodes using the options in the right-click menu. See “Right-Click Menus”.
Y, Child Represents the child factors contributing to the parent factors.
X, Parent Represents the parent factors (including the effect) that have child factors.
Label Includes the text from the Label columns in the nodes of the diagram.
By Produces separate diagrams for each value of the By variable.
The Cause-and-Effect Diagram In Figure 14.5, the effect or problem, Defects in circuit board, appears on the right as the center line. The major contributing factors appear at the end of the branches (Inspection, Solder process, Raw Card, and so on.) Possible causes branch off each major factor.
Figure 14.5 Cause-and-Effect Diagram Right-Click Menus Right-click a highlighted node to modify text, insert new nodes, change the diagram type, and more. Note the following: • Right-click a title to change the font and color, positioning, visibility, or formatting.
• Click and highlight a node to rename it.
• Click and drag a node to move it.
effect major factor possible causes 398 Cause-and-Effect Diagrams Chapter 14 The Cause-and-Effect Diagram Quality and Process Methods Text Menu The Text menu contains the following options: Font Select the font of the text or numeric characters.
Color Select the color of the text or numeric characters.
Rotate Left, Rotate Right, Horizontal Rotates the text or numbers to be horizontal, 90 degrees left, or 90 degrees right.
Insert Menu Use the Insert menu to insert items onto existing nodes. The Insert menu contains the following options: Before Inserts a new node to the right of the highlighted node. For example, Figure 14.6 inserts Child 1.5 before Child 2.
Figure 14.6 Insert Before After Inserts a new node to the left of the highlighted node. For example, Figure 14.7 inserts Child 3 after Child 2.
Figure 14.7 Insert After Above Inserts a new node at a level above the current node. For example, Figure 14.8 inserts Grandparent at a level above Parent.
Figure 14.8 Insert Above Chapter 14 Cause-and-Effect Diagrams 399 Quality and Process Methods The Cause-and-Effect Diagram Below Inserts a new node at a level below the current node. For example, Figure 14.9 inserts Grandchild at a level below Child 2.
Figure 14.9 Insert Below Move Menu Use the Move menu to move nodes or branches. The Move menu contains the following options: First Moves the highlighted node to the first position under its parent.
Last Moves the highlighted node to the last position under its parent.
Other Side Moves the highlighted node to the opposite side of its parent line.
Force Left Makes all horizontally drawn elements appear to the left of their parent.
Force Right Makes all horizontally drawn elements appear to the right of their parent.
Force Up Makes all vertically drawn elements appear above their parent.
Force Down Makes all vertically drawn elements appear below their parent.
Force Alternate Draws children on alternate sides of the parent line.
400 Cause-and-Effect Diagrams Chapter 14 The Cause-and-Effect Diagram Quality and Process Methods Figure 14.10 Force Options Other Menu Options The right-click menu for a highlighted node also contains these options: Change Type Changes the entire chart type to Fishbone, Hierarchy, or Nested.
Uneditable Disables all other commands except Move and Change Type.
Text Wrap Width Specifies the width of labels where text wrapping occurs.
Make Into Data Table Converts the currently highlighted node into a data table. Convert the all nodes by highlighting the whole diagram (effect).
Close Shows the highlighted node.
Delete Deletes the highlighted node and all of its children.
Cause and Effect Diagram Menu Options The Cause and Effect Diagram red triangle menu contains the following options: See Using JMP for more information about the following options: Redo Contains options that enable you to repeat or relaunch the analysis. In platforms that support the feature, the Automatic Recalc option immediately reflects the changes that you make to the data table in the corresponding report window.
Force Left Force Alternate Force Right Force Up Force Down Chapter 14 Cause-and-Effect Diagrams 401 Quality and Process Methods Save the Diagram Save Script Contains options that enable you to save a script that reproduces the report to several destinations.
Save By-Group Script Contains options that enable you to save a script that reproduces the platform report for all levels of a By variable to several destinations. Available only when a By variable is specified in the launch window.
Save the Diagram There are different ways to save your diagram. Choose from one of the following: • save the diagram as a data table • save the diagram as a journal • save the diagram as a script Save the Diagram as a Data Table Note the following about this approach: • If you have other processes that need to update the data table, this can be a good approach to choose.
• Very little customization is available, because the data table cannot represent the customization.
To save the diagram as a data table: 1.
Highlight the entire diagram.
2.
Right-click and select Make Into Data Table.
3.
Save the new data table.
Save the Diagram as a Journal Note the following about this approach: • This option can be a good choice for impromptu work. For example, you can manually build the diagram, save it as a journal, then reopen the journal later and continue building and editing the diagram.
• Any customization exists only in the journal, and the journal is not connected to the data table.
402 Cause-and-Effect Diagrams Chapter 14 Save the Diagram Quality and Process Methods To save the diagram as a journal: 1.
Highlight the entire diagram.
2.
Right-click and select Edit > Journal.
3.
Save the new journal.
Save the Diagram as a Script Note the following about this approach: • If you have other processes that need to update the data table, this can be a good approach to choose.
• If you created the diagram from a data table, a simple script appears that relaunches against the data table with no customization.
• If you created the diagram without using a data table (or from a journal), a more complex script appears that contains all the commands needed to add and customize each area of the diagram.
To save the diagram as a script: 1.
Click the red triangle next to Cause and Effect Diagram and select Save Script > To Script Window.
2.
Save the new script.
Chapter 15 Quality Utilities Manage Specification Limits and Create OC Curves This chapter covers utilities in the Analyze > Quality and Process menu. Specifically, the Manage Spec Limits utility and OC Curves utility are discussed.
404 Quality Utilities Chapter 15 Quality and Process Methods Contents Manage Spec Limits Utility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405 Example of the Manage Spec Limits Utility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405 Manage Spec Limits Options. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 408 Operating Characteristic Curves Utility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 Launch the OC Curves Utility. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 OC Curves for Control Chart Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 OC Curves for Acceptance Sampling Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 Chapter 15 Quality Utilities 405 Quality and Process Methods Manage Spec Limits Utility Manage Spec Limits Utility The Manage Spec Limits utility enables you to quickly add or edit many specification limits for several columns at once. The specification limits are then used in any future analyses. You can also specify importance values for each process and indicate whether limits should appear in graphs as reference lines.
Example of the Manage Spec Limits Utility 1.
Select Help > Sample Data Library and open Cities.jmp.
2.
Select Analyze > Quality and Process > Manage Spec Limits.
3.
Specify the columns that you want to set specification limits on. For this example, select OZONE, CO, SO2, and NO, and click Process Variables.
Figure 15.1 Specify Columns 4.
Click OK.
5.
Add your specification limits. You can do this by loading existing limits from a JMP data table (Load from Limits Table) or by entering limits manually. For this example, enter the following limits manually: – OZONE: LSL 0.12, USL 0.2 – CO: LSL 6, USL 12 – SO2: LSL 0.015, USL 0.06 – NO: LSL 0.02, USL 0.04 6.
Click the red triangle next to Manage Spec Limits and select Show Limits All.
406 Quality Utilities Chapter 15 Manage Spec Limits Utility Quality and Process Methods Specification limits for all columns will appear in graphs for any future analyses. If you want to show the specification limits only for individual columns, check the Show Limits box next to those columns.
Figure 15.2 Set Specification Limits 7.
Choose how you want to save the specification limits. For this example, click Save to Column Properties. This saves them as column properties in the corresponding data table. You could also save them to a new data table (tall or wide format).
In the Cities.jmp data table Columns panel, notice that asterisks indicating the Spec Limits column property appear next to OZONE, CO, SO2, and NO.
8.
To see values that are outside the limits in the data table, click the red triangle next to Manage Spec Limits and select Color Out of Spec Values. Go to the Cities.jmp data table, and you can see that any values that are outside the limits are now colored.
9.
Now, you can run any analysis. For this example, select Analyze > Distribution.
10. Select OZONE, CO, SO2, and NO, and click Y, Columns.
11. Click OK.
Chapter 15 Quality Utilities 407 Quality and Process Methods Manage Spec Limits Utility Figure 15.3 Specification Limits for OZONE in Distribution The specification limits that you added to the OZONE column appear in the histogram. Because the column contains a Spec Limits column property, the Distribution report also contains a Capability Analysis report.
408 Quality Utilities Chapter 15 Manage Spec Limits Utility Quality and Process Methods Manage Spec Limits Options In the window where you set specification limits, there are buttons to save and load specification limits, and options in the Manage Spec Limits red triangle menu.
Report Table Columns The table at the top of the Manage Spec Limits report contains a row for each process column specified in the launch window. This table enables you to specify specification limits as well as the following options for each column: Show Limits Specifies that the Show as Graph Reference Lines option is selected in the Spec Limits column property for the specified column.
Process Importance Specifies the process importance value for each column. Process importance values provide a mechanism to sort processes in the order that you prefer. Process importance values are used to size markers in many platform graphs.
Units Specifies the units for each column.
Buttons Load from Limits Table Loads specification limits from a JMP data table.
Save to Column Properties Saves the specification limits as column properties in the associated data table.
Save to Tall Spec Limits Table Saves the specification limits to a new data table in tall format.
Save to Wide Spec Limits Table Saves the specification limits to a new data table in wide format.
Red Triangle Options Show Limits All Selects boxes under Show Limits for all of the columns. If Show Limits is selected for a column, the Show as Graph Reference Lines option is selected in the Spec Limits column property. The Show as Graph Reference Lines option displays the specification limits and target that you specify as reference lines in select analysis plots.
Note: If all boxes under Show Limits are selected, the Show Limits All option deselects all of the boxes under Show Limits.
Round Decimals Sets the number of decimal places to which you want the specification limits rounded.
Chapter 15 Quality Utilities 409 Quality and Process Methods Operating Characteristic Curves Utility Color Out of Spec Values Colors any values in the data table that are outside the specification limits for the columns.
See Using JMP for more information about the following options: Redo Contains options that enable you to repeat or relaunch the analysis. In platforms that support the feature, the Automatic Recalc option immediately reflects the changes that you make to the data table in the corresponding report window.
Save Script Contains options that enable you to save a script that reproduces the report to several destinations.
Operating Characteristic Curves Utility The Operating Characteristic (OC) Curves utility enables you to construct OC curves for control charts and attribute acceptance sampling plans. OC curves are available for XBar, P, NP, C, and U charts. For specified control charts, the OC curve shows the probability of failing to detect a shift of a particular size. In addition, there are OC curves for single and double attribute acceptance sampling plans. For a specified acceptance sampling plan, the OC curve shows how the probability of accepting a lot changes with the lot quality.
Note: The OC curves for control charts are two-sided curves. They are drawn for negative and positive shifts. Often OC curves display one curve for the absolute shift.
Launch the OC Curves Utility Launch the Operating Characteristic Curves by selecting Analyze > Quality and Process > OC Curves. Select the OC curve of interest and click OK to launch.
Figure 15.4 OC Curves Launch Window 410 Quality Utilities Chapter 15 Operating Characteristic Curves Utility Quality and Process Methods OC Curves for Control Chart Options For control charts, you can use OC curves to explore how chart parameter settings impact , which is the probability of failing to detect a specified shift of interest. Use the text boxes, sliders, or the LCL, Shift, and UCL handles on the OC curve to set and adjust chart parameters. The chart parameters are as follows: Lower Control Limit Specifies the lower control limit from your control chart.
Upper Control Limit Specifics the upper control limit from your control chart.
Sample Size (Not available for C charts.) Specifies the sample size used for your control chart measure.
Sigma (Available only for XBar charts.) Specifies your control chart sigma.
Shift Specifies the shift to detect.
Beta Specifies the probability of failing to detect the specified shift given the control chart specifications. Beta updates as the specifications are changed.
OC Curves for Acceptance Sampling Options For attributes acceptance sampling, you can use OC curves to explore how sampling plans and assumed product quality impact the probability of accepting a lot. Use the text boxes, sliders, or fraction defective handle on the OC curve to set and adjust your sampling plan.
Single Sampling OC Curve Options Sampling Type Enables you to select Lot Sampling or Binomial Sampling.
Lot Sampling Enables you to specify and explore an acceptance plan based on a fixed lot size.
Binomial Sampling Enables you to specify and explore an acceptance plan for a continuous process or other situation where the binomial distribution is appropriate.
Lot size (N) (Available only for Lot Sampling.) Specifies the size of the lot that you are sampling from.
Sample Size (n) Specifies the number of units for inspection.
Acceptable failures (c1) specifies the number of allowable failures. If the number of observed defects is greater than c1, then the lot is rejected.
Fraction Defective Specifies the expected fraction defective in the lot.
Chapter 15 Quality Utilities 411 Quality and Process Methods Operating Characteristic Curves Utility Probability of Acceptance Specifies the probability of accepting the lot given the sampling plan as defined. The Probability of Acceptance updates as the specifications are changed, but you can also adjust this value directly. When you adjust the Probability of Acceptance directly, the Fraction Defective value is updated.
Double Sampling OC Curve Options First Sample Contains the following settings for the first sample: Number inspected (n1) Specifies the number of units inspected in the first sample.
Acceptable failures (c1) specifies the number of allowable failures in the first sample.
Second Sample Contains the following settings for the second sample: Number inspected (n2) Specifies the number of units inspected in the second sample.
Acceptable failures (c1+c2) specifies the total number of allowable failures.
Fraction Defective Specifies the expected fraction defective in the lot.
Probability of Acceptance Specifies the probability of accepting the lot given the sampling plan as defined. The Probability of Acceptance updates as the specifications are changed, but you can also adjust this value directly. When you adjust the Probability of Acceptance directly, the Fraction Defective value is updated.
412 Quality Utilities Chapter 15 Operating Characteristic Curves Utility Quality and Process Methods Appendix A References Agresti, A., and Coull, B. (1998). “Approximate is Better Than ‘Exact’ for Interval Estimation of Binomial Proportions.” The American Statistician 52:119–126.
American Society for Quality Statistics Division (2004). Glossary and Tables for Statistical Quality Control. 4th ed. Milwaukee, WI: ASQC Quality Press.
American Society for Testing and Materials (1976). ASTM Manual on Presentation of Data and Control Chart Analysis. Philadelphia: ASTM.
Automotive Industry Action Group (2002). Measurement Systems Analysis Reference Manual. 3rd ed. Troy, MI: Automotive Industry Action Group.
AIAG (2005). Statistical Process Control. 2nd ed. Troy, MI: Automotive Industry Action Group.
Barrentine, L. B. (1991). Concepts for R&R Studies. Milwaukee, WI: ASQC Quality Press.
Bissell, A. F. (1990). “How Reliable Is Your Capability Index?” Journal of the Royal Statistical Society, Series C 39:331–340.
Box, G. E. P, Luceño, A., and Paniagua-Quinones, M. d. C. (2009). Statistical Control by Monitoring and Adjustment. 2nd ed. New York: John Wiley & Sons.
Burdick, P. K., Borror, C. M., and Montgomery, D. C. (2005). Design and Analysis of Gauge R&R Studies: Making Decisions with Confidence Intervals in Random Mixed ANOVA Models. Philadelphia, PA and Alexandria, VA: SIAM and ASA.
Chou, Y.-M., Owen, D. B., and Borrego, S. A. (1990). “Lower confidence limits on process capability indices.” Journal of Quality Technology 22:223–229.
Crowder, S. V. (1987). “Average Run Lengths of Exponentially Weighted Moving Average Charts.” Journal of Quality Technology 29:401–408.
David, H. A. (1951). “Further Applications of Range to the Analysis of Variance.” Biometrika 38:393–407.
Fleiss, J. L. (1981). Statistical Methods for Rates and Proportions. New York: John Wiley & Sons.
Goel, A. L., and Wu, S. M. (1971). “Determination of A.R.L. and a Contour Nomogram for Cusum Charts to Control Normal Mean.” Technometrics 13:221–230.
Harter, H. L. (1960). “Tables of Range and Studentized Range.” Annals of Mathematical Statistics 31:1122–1147.
Hoffman, D. (2003). “Negative Binomial Control Limits for Count Data with Extra-Poisson Variation.” Pharmaceutical Statistics 2:127–132.
Jackson, J. E., and Mudholkar, G. S. (1979). “Quality Control Procedures for Residuals Associated with Principal Component Analysis.” Annals of Mathematical Statistics 25:290– 302.
414 References Appendix A Quality and Process Methods Knoth, S. (2004). “Fast Initial Response Features for EWMA Control Charts.” Statistical Papers 46:47–64.
Kourti, T., and MacGregor, J. F. (1996). “Multivariate SPC Methods for Process and Product Monitoring.” Journal of Quality Technology 28:409–428.
Levey, S., and Jennings, E. R. (1950). “The Use of Control Charts in the Clinical Laboratory.” American Journal of Clinical Pathology 20:1059–1066.
Li, G., Qin, S., Ji, Y., Zhou, D.(2009). “Total PLS Based Contribution Plots for Fault Diagnosis.” Acta Automatica Sinica 35:759 - 765.
Lucas, J. M. (1976). “The Design and Use of V–Mask Control Schemes.” Journal of Quality Technology 8:1–12.
Lucas, J. M., and Crosier, R. B. (1982). “Fast Initial Response for CUSUM Quality Control Schemes: Give Your CUSUM a Head Start.” Technometrics 24:199–205.
Meeker, W. Q., and Escobar, L. A. (1998). Statistical Methods for Reliability Data. New York: John Wiley & Sons.
Montgomery, D. C. (2013). Introduction to Statistical Quality Control. 7th ed. New York: John Wiley & Sons.
Nair, V. N. (1984). “Confidence Bands for Survival Functions with Censored Data: A Comparative Study.” Technometrics 26:265–275.
Nelson, L. (1984). “The Shewhart Control Chart—Tests for Special Causes.” Journal of Quality Technology 15:237–239.
Nelson, L. (1985). “Interpreting Shewhart X Control Charts.” Journal of Quality Technology 17:114–116.
Nomikos, P., and MacGregor, J. F., (1995). “Multivariate SPC Charts for Monitoring Bach Processes.” Technometrics 37:41–59.
Pearn, W. L., and Kotz, S. (2006). Encyclopedia and Handbook of Process Capability Indices. Vol. 12 of Series on Quality, Reliability, and Engineering Statistics. New Jersey: World Scientific.
Portnoy, S. (1971). “Formal Bayes Estimation with Application to a Random Effects Model.” The Annals of Mathematical Statistics 42:1379–1402.
Sahai, H. (1974). “Some Formal Bayes Estimators of Variance Components in the Balanced Three-Stage Nested Random Effects Model.” Communication in Statistics – Simulation and Computation 3:233–242.
Slifker, J. F., and Shapiro, S. S. (1980). “The Johnson System: Selection and Parameter Estimation.” Technometrics 22:239–246.
Sullivan, J. H., and Woodall, W. H. (2000). “Change-point detection of mean vector or covariance matrix shifts using multivariate individual observations.” IIE Transactions 32:537–549.
Tracy, N. D., Young, J. C., and Mason, R. L. (1992). “Multivariate Control Charts for Individual Observations.” Journal of Quality Technology 24:88–95.
Westgard, J. O. (2002). Basic QC Practices. 2nd ed. Madison, WI: Westgard QC.
Appendix A References 415 Quality and Process Methods Wheeler, D. J. (2004). Advanced Topics in Statistical Process Control. 2nd ed. Knoxville, TN: SPC Press.
Wheeler, D. J. (2006). EMP III Using Imperfect Data. Knoxville, TN: SPC Press.
Wludyka, P., and Sa, P. (2004). “A robust I-Sample analysis of means type randomization test for variances for unbalanced designs.” Journal of Statistical Computation and Simulation 74:701–726.
Woodall, W. H., and Adams, B. M. (1998). Handbook of Statistical Methods for Engineers and Scientists. 2nd ed. Chapter 7. New York: McGraw-Hill.
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In probability theory and statistics, two of the most important concepts we rely on are expectation and variance - tools that help us understand the average outcome of a random variable and its uncertainty. Sometimes, it is hard to compute these quantities directly. In some cases, we can use the law of total expectation and the law of total variance to break down the problem into simpler parts.
This post aims to make these laws clearer by visualizing them. It assumes familiarity with random variables, expectation, and variance.
Law of Total Expectation
The Law of Total Expectation states that the expected value of a random variable X can be decomposed into the expectation of its conditional expectation given another random variable Y:
E[X]=E[E[X∣Y]] But what does this mean intuitively?
Let’s consider a simple example. Suppose we have two random variables X and Y. X represents the height of a person, and Y represents a group that the person belongs to. There are three groups: “short (1)” “medium (2)”, and “tall (3)”.
The expected value can be computed by averaging the heights of all people. But we can also computing averages within each group and taking weighted averages of the group means. This is what the law of total expectation tells us.
Law of Total Variance
The Law of Total Variance states that the variance of a random variable X can be decomposed into the variance of its conditional expectation given another random variable Y, plus the expectation of the conditional variance given Y:
Var[X]=Var[E[X∣Y]]+E[Var[X∣Y]]
Let’s consider the same example as before. The variance of the heights can be computed by averaging the squared differences from the mean height. But we can also compute it by averaging the variances within each group and adding the variance among the group means.
The variance measures the spread of the data. The spread of the data is visualized by the horizontal lines, and the conditional averages are shown by the vertical lines. However, note that horizontal lines are used only for visualization purposes, and they should not be interpreted literally.
Conclusion
The laws of total expectation and total variance are powerful tools that help us break down complex problems into simpler parts. I hope that these visualizations have made these laws more intuitive and easier to understand. |
6847 | https://emergent-scientist.edp-open.org/articles/emsci/full_html/2019/01/emsci180004/emsci180004.html | Motion of rain drops on a car side window | Emergent Scientist
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| Issue | | Emergent Scientist Volume3, 2019 IPT 2018 |
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| Article Number | | 3 |
| Number of page(s) | | 9 |
| Section | | Physics |
| DOI | | |
| Published online | | 01 April 2019 |
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Abstract
1 Introduction
2 Methods
3 Results
4 Discussion
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Acknowledgments
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Emergent Scientist 3, 3 (2019)
Research Article
Motion of rain drops on a car side window
Julie André 1, Clément Brochet 1, Quentin Louis 1, Amaury Barral 1, Anthony Guillen 1, Fang-Ting Goh 1, Angel Prieto 1 and Thibault Guillet 2
1 Ecole polytechnique, 91128 Palaiseau, France
2 LadHyX, UMR 7636, CNRS, Ecole polytechnique, 91128 Palaiseau, France
e-mail: julie.andre.2016@polytechnique.org
Received: 24 June 2018
Accepted: 11 February 2019
Abstract
The present paper studies the initial motion of rain drops on a car side window. An explanation to the common observation of drops going up the window when a car reaches a given speed is provided, which takes into account the dimensions of the drops and the wind speed. We also discuss the importance of the actual window geometry, the state of the surface, and the drop size distribution. This work may thus be used as a basis for complete study of a population of drops against a car window.
Key words: Drop / trajectory / car / wind / adhesion
© J. André et al., Published by EDP Sciences, 2019
This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1 Introduction
You will have probably already noticed a puzzling phenomenon while driving your car in the rain: on the side-windows of a moving car, rain drops may adopt a wide range of behaviors and trajectories, as shown in Figure 1, some of them stay at rest, while others tend to move either downwards or upwards. The purpose of this study is to identify those different regimes of behaviors and to explain the observed trajectories.
The motion of rain drops on a car is of tremendous interest, since if they stick on the glass they limit the visibility through the window, which may trigger safety issues. This subject is therefore greatly interesting the automobile field. Another subject, important for visibility issues and quite close to the dynamic of drops on a car window, is the formation and stability of thin films and rivulets, such as the localized canals observed in Figure 1. This subject has already been closely studied [1–4] and will not be consider in this paper.
The dynamic of drops on a plane surface has been investigated usually in simpler conditions: the influence of gravity and drag force [6,7] has been separately looked at. A thorough study of the parameters involved in the drag force can be found in . The specific problem of a car window also involves aerodynamic effects such as side vortices, which have been meticulously studied in . The subject concerning drops on a car side window has been studied in [9,10], where emphasis was put on removing drops from the glass to enhance visibility. However, none of those studies have tackled the determination of the drops' trajectories.
In this paper, we will focus on the trajectories of rain drops once they are on the side windows, where all the forces mentioned above must be taken into account. We first present our method to model the dynamics of a drop undergoing gravity, drag forces and capillarity forces (due to the interaction of the drop with the surface). Experimental results are then provided and enable us to validate the models and explain the different regimes observed on a car window. We finally discuss how our model could be used as a basis to study a whole population of drops on a car window, where the geometry of the car, surface treatment and drop merging play crucial roles.
Fig. 1
Picture of rain drops on the side window of a moving car. The drops have different trajectories; some of them are going up, other have almost horizontal directions. Merging between drops occur, which makes the different paths quite erratic. The wetting of the window makes it easier to take some predefined paths (canals), as we can see on the top part on the window.
2 Methods
Hypothesis: we consider a single drop holding on a car side window. We assume that the car is driven with a constant speed on a straight line (so that it can be considered as an inertial frame of reference) and that we have access to the local wind on the car window (value and direction).
We then intend to find:
the conditions needed to dislodge the drop;
the direction taken by the drop once dislodged;
the conditions forcing its trajectory to have an upward component.
2.1 Forces acting on a drop
Different forces are acting on this drop: the weight P→, the adhesion force of the window F→a d, the drag force F→w exerted by the wind, and the shear force within the drop. In our situation, the shear force within the drop when it deforms is negligible, since the drop deforms slowly and can therefore be considered as quasi-static. As long as the drop is at rest, there is a mechanical equilibrium in the plane of the window. Thus, one has: (P→⋅t→)t→+F→w+F→ad=0→, where t→ denotes a unit vector tangential to the plane, directed downwards. Figure 2 sums up the situation.
Fig. 2
Schematic view of the three forces applying on a drop, holding on a tilted plane of inclination α and under a wind of given direction β w on the plane. t→ is an unit vector tangential to the plane. The sum Σ→ of the two external forces (weight P s i n αt→ and drag force F→w) is represented in blue. The angle between Σ→ and t→ is defined as β.
2.2 Dislodging a drop
To dislodge the drop, the adhesion force must be overtaken by the two external forces. This gives the general condition of sliding:||P sin α t→+F→w||≥F ad(1)where α denotes the inclination angle of the plane. We began by considering two simpler cases where only one external force out of the two plays a role: first, the weight, then the drag force, which can both lead to the dislodgement of the drop.
2.3 Influence of the weight: theoretical prediction
The following part models the competition between weight and adhesion forces, when the wind speed is null. The expression of weight is well known: P = Ω ρ water g with Ω the volume of the drop, ρ water the density of water and g the gravity. To derive a model for the adhesion force, we need to explain why a drop can hold on an inclined plane.
While a still drop on a horizontal plane has a perfectly symmetric shape, it deforms and becomes asymmetric if one inclines the plane: the contact angle varies along the perimeter of the drop, see Figure 3. MacDougall and Ockrent and then Furmidge have elaborated a simple model to explain this: the drop can deform to minimize its energy when submitted to an external force (such as weight or wind drag) and to capillarity forces sticking it on the surface; the asymmetry of contact angles enables the drop to hold on the tilted plane. The drop's deformation is limited. When the contact angle at the front reaches the maximum angle θ a then the contact line advances: θ a is thus defined as the advancing angle. On the other hand, when the contact angle at the back reaches the minimum angle θ r, the contact line retracts, therefore θ r is defined as the receding angle. The drop will start sliding when θ r and θ a are both reached on two opposite sides of the drop, see Figure 3.
According to Furmidge, the maximum adhesion force can be modeled at the first order as F a d=γ(cos θ r−cos θ a)D, where γ is the surface tension of water and D is the diameter of the drop. This maximum adhesion force is reached when the drop is the most asymmetric, and after that, it starts sliding.
Hence, just before sliding the mechanical equilibrium of the drop on a plane with maximum inclination α s reads: P sin α s = F ad. This gives us a relation between this inclination and the drop's volume Ω:ρ Ω g s i n α s=γ(c o s θ r−c o s θ a)D(2)
Fig. 3
Schematic representations of : (1) a drop of diameter D on a horizontal plane, (2) its deformation under gravity, when the plane which is progressively inclined and (3) the moment when the drop begins to slide. θ a is the advancing angle and θ r the receding angle.
2.4 Influence of the weight: experimental setup
For the first set of experiments, we used an acrylic board (made of polymethyl methacrylate) as a support for the drops, which we cleaned with ethanol to make the experiment reproducible. With a μ l-pipette, we injected a drop of known volume Ω (ranging from 15 μ L to 100 μ L) on the plane. By taking a picture of it from above, we had access to the drop diameter D (approximately between 4 mm and 10 mm). Then, we inclined the plane slowly (doing so, the projection of the weight on the plane increases) up to the angle α s where the drop slides. The angle of sliding α s may theoretically vary from 0° (superhydrophobic behavior) to 90° (the drop is stuck on the surface whatever the angle α).
2.5 Dislodging a drop with drag force: theoretical prediction
In the following part, we investigate the influence of the wind blowing on a horizontal plane, where a single drop of volume Ω is lying. Equilibrium is obtained similarly to the situation where the drop is undergoing gravity: adhesion forces equalize the drag force up to a maximum deformation of the drop. When the drag force increases beyond the maximum adhesion force, the drop begins to slide in the direction imposed by the wind. We now want to obtain a consistent estimate of the drag force.
We first consider a spot at a distance x from the edge of the plane, under a wind blowing horizontally. The Reynolds number Re(x)=x.u ν a with ν a the kinematic viscosity of air, at x ≈ 10 cm from the edge of the plane and with an air velocity u=10 m/s is about 10 4. For such high Reynolds number, the boundary layer thickness δ(x) follows with good approximation the law established by Prandtl : δ(x)=ν a x/u. This yields δ (x) ≈0.5 mm at a distance x ≈ 10 cm from the edge of the plane. If a drop of diameter D ≈ 1 cm sits on this spot, as we have D/δ>> 1, we can consider the local wind speed u on the drop to be uniform.
As the Reynolds number of the flow around the drop R e D D.u ν a≈10 4 is great compared to 1, the expression of the drag force is thus F w=1 2 C x ρ a S u 2 (as can be found in ), with C x the horizontal drag coefficient, independent of Re, linked to the drop geometry, ρ a the air volume weight and S the surface of the drop projected on the plane perpendicular to the direction of the wind.
Finally, the equilibrium of the adhesion force and the drag force gives :1 2 C x ρ a S u s 2=γ(cos θ r−cos θ a)D.(3)
From which we can deduce the speed of the wind u s needed to force a given still drop to slide on a horizontal plane:u s 2=2 γ(cos θ r−cos θ a)D C x ρ a S.(4)
2.6 Dislodging a drop with drag force: experimental setup
For this second set of experiments, we used the same acrylic board as described above, kept horizontal. We injected drops of given volume Ω and measured diameter D (respectively between 5 μ L and 100 μ L and from 3 mm to 11 mm) on the plane. We then switched on the wind blower and increased the wind speed until the drops began to slide. Wind speeds ranged from 5 m s−1 to a maximum value of 25 m s−1. The wind blower had a precision of 0.5 m s−1, see Figure 4.
Fig. 4
Schematic view of the experiment to dislodge a horizontal drop with drag force. We use a wind tunnel, to ensure that the air flow is laminar and to shrink wind speed variability. The wind could blow from 1 to 30 m s-1.
2.7 Direction of the drop once dislodged: theoretical prediction
Now we assume that the drop has been dislodged under the combined action of gravity and wind, as shown in Figure 2. The adhesion force is a reaction force having the same direction that the sum of the external forces Σ→=P→+F→w. The drop thus follows the direction of the resulting global force Σ→+F→ad, which is in the direction of Σ→. To have access to the angle β taken by the drop, we can project Σ→ on the plane, which gives:cos β=Σ→.t→||Σ→||=P sin α+F w cos β w(P sin α)2+2 P sin α.F w cos β w+F w 2.(5)
Limit cases of this general formula are easy to check: a situation with no wind (F w = 0) gives cos β = 1, which means the drop follows the weight; a situation with a horizontal window (α = 0) gives cos β = cos β w which means the drop slides in the direction of the wind.
The drop will then move upwards if 90° ≤ β ≤ 180 ° which means c o s β≤0. Therefore, the condition for an upward trajectory given by equation (6) is expressed as follows:P sin α+F w cos β w≤0.(6)
Since α is taken between 0° and 90°, this condition yields cos β w ≤ 0, which corresponds to the intuitive condition that the wind must have a vertical component directed upwards.
2.8 Direction of the drop once dislodged: experimental setup
The acrylic board was placed vertically in front of the wind blower, the inclination was therefore α = 90°. The direction of the wind was kept horizontal throughout the experiment (β w = 90°). We put a drop of known volume Ω, small enough so that it held on the vertical plane: a first round of trials was performed with 25 μ L drops, and another one with 20 μ l drops. Colored water was used to increase the contrast. We suddenly let the wind blow at a given speed u, strong enough to dislodge the drop (from 12 to 30 m s−1). Taking a video in front of the plane, we had access to the trajectory of the drop, more precisely to the angle β the trajectory of the drop made with the vertical. One time-lapse photography obtained from the video can be seen in Figure 5.
Equation (5) becomes in the particular case of this setup cos β=P P 2+F w 2=P||Σ→|| or in its equivalent form:tan β=F w P.(7)
Fig. 5
(Top) schematic view of the experiment performed in order to measure the direction taken by a drop when dislodged by an horizontal wind. The support is maintained vertical. (Bottom) time-lapse photography of a drop in motion on a vertical plane under an horizontal wind.β w is the angle the wind makes with the vertical, while β is the angle which the sliding drop takes.
2.9 Global condition
Finally, two conditions need to be fulfilled to observe a drop going up on a car window:
the drop is dislodged when: ||P sin α t→+F→w||≥F ad
the drop goes up when: P sin α + F w cos β w ≤ 0
3 Results
3.1 Adhesion force
The experimental data for the angle α s needed to dislodge a given drop are shown on Figure 6.
We noticed a high variability of results. Some drops seem to stay stuck on some irregularities on the surface, which enables them to stay on a higher inclination. This effect will be discussed later on.
We observed two main regimes: for a drop of volume greater than about 17±1 μ l, the drop slides when the angle α s≤ 90 ° is reached. On the contrary, a smaller drop holds even when the plane reaches the vertical position. Such a drop does not slide at all under its own weight, we thus cannot define an angle of sliding α s; therefore, this situation is not represented on the graph. However, we draw a vertical line to enhance the change in regime at the limit size.
For drops big enough to slide, we indeed see a clear dependency of the inclination α s on the volume Ω of the drops, as predicted in equation (2): bigger drops begin to slide at smaller inclination angles.
Fig. 6
(Top) a plot of α s, the inclination needed to dislodge a drop of water, is plotted against the volume Ω of the drop, put on a plane of acrylic. Dashed lines indicate limits for definition of α s, which cannot exceed 90° (in this case the drop will not fall under its weight). Thus for volumes on the left of the vertical dashed line, α s is not defined.(Middle) a plot of us, the minimum wind speed needed to dislodge a drop, is plotted against the diameter D of the drop lying on a plane of acrylic. (Bottom) the direction, taken by a drop of volume put on a vertical plane in front of a blower, is plotted against the speed u of the wind. Two values of the volume are used (blue dots: 20 µ l, and green crosses: 25 µ l).
3.2 Drag force on the drop
Studying the minimum air speed needed to dislodge a given drop of volume Ω, we observe that bigger drops tend to be dislodged under weaker wind speeds, see Figure 6. For each volume data point, we tested three drops. As for the experiment on adhesion forces, one may notice the large uncertainties on the graph, which stem from the distribution of adhesive forces created by surface imperfections.
3.3 On a vertical plane − both weight and wind
Figure 6 shows the experimental data for the direction β taken by a drop of volume Ω put on a vertical plane in front of the blower, plotted against the wind speed u. As foreseen in Methods with equation (7), the stronger the wind blows, the more the drop tends to follow the direction of the wind (that is β comes closer to 90°).
4 Discussion
4.1 Scaling laws
In the following paragraphs, we want to simplify the equations of the first section into scaling laws, hoping it will help understand the dependence of minimum wind speed u s, maximum inclination α s and direction of motion β regarding the different parameters. We therefore need a relation between the volume Ω, the projected surface S and the diameter D of a drop lying on a plane.
We assess in first approximation that the drops are small enough to be considered as half-spheres of diameter D. Thus, the volume is expected proportional to D 3 and the projected surface as D 2. We poured drops of known volume and measured the diameter by taking pictures from above. Figure 7 shows the relation we obtained between the diameter and the volume of the drop.
The experimental data are in good agreement with the approximation Ω = aD 3 (with a is a dimensionless geometrical constant), as long as the drop is not too big. When the drop is about five times the capillary length L c ≈ 2.7 mm, the drop spreading under its own weight is not negligible. Its width becomes much smaller than its radius, the model of the half sphere is no longer valid. This explains why the points of Ω=100 μ l are far from the fit Ω = aD 3. Our simple approximation of half-spherical drops is thus accurate for volumes smaller than 50 μ l, that we used in our experiments. In the following models, we now assess that:Ω∝D 3(8)
Fig. 7
Experimental law between the volume Ω and the diameter D of a drop defined as the diameter seen from above the drop as it is lying on a horizontal floor. Ω ∝ D 3 is a consistent model for small enough drops. Fit does not take into account the two last points, where drops are so big that they spread under their own weight.
4.2 Interpretation of results in scaling laws
4.2.1 Adhesion force
We can rewrite the equation (2) obtained in Section 2 in a more convenient way:s i n α s=γ ρ g(c o s θ r−c o s θ a)D Ω
The factor c o s θ r−c o s θ a is only due to the interaction between water and the material of the plane used for the experiment. It is considered as a fixed constant during our experiment, since we used the same plane each time. By using the relation (8) between the volume and the diameter, and denoting L c γ/ρ g the capillary length, we obtain the following scaling law:sin α s∼L c 2 D Ω∼L c 2 Ω−2/3(9)
Plotting s i n α s depending on Ω−2 3, as done in Figure 8 shows a linear relationship as expected. This scaling law (9) indicates that the expression of the adhesion force as a force proportional to the diameter of the drop is relevant.
Fig. 8
(Top) scaling of s i n α s (α s the sliding angle for a drop) as a function of L c 2 Ω−2 3, with L c the capillary length for water under normal conditions of pressure and temperature. Values of Ω range from 15 µ l to 100 µ l. A linear fit (dashed line) to experimental data is expected, according to our theory. (Middle) experimental law for the relation between wind speed u s and drop volume Ω on an horizontal plane, law stemming from the competition between the drag force and the adhesion force. Linear correlation (red line) is consistent with our theoretical model. (Bottom) experimental results on the angle β taken by drops of given volume Ω under an horizontal wind u. Drops small enough are set on a vertical plane (we took Ω=20 μ l or 25 μ l so they can stay DESPITE GRAVITY). Then the wind blower is switched on, and blows horizontally with a speed u bigger than u s(Ω), the minimum value needed to dislodge the drop. As our model predicts, there is a linear law between t a n β and the ratio u 2 Ω−1/3/g.
4.2.2 Drag force
Once again, we can simplify the theoretical equation (3) obtained in Section 2 into a scaling law. As an expression for the speed of the wind u s needed to dislodge a drop of volume Ω on an horizontal plane, we obtain:u s 2∼D S∝Ω−1/3.(10)
We can now plot our data again according to equation (10) and check that we have a linear relation between u s 2 and Ω−1/3. Figure 8 shows that our data are consistent with a linear law. Consequently, our expression for the drag force proportional to the projected surface of the drop is validated.
4.2.3 Direction β under an horizontal wind
In order to check the relations obtained on t a n β on a vertical plane with a strong horizontal wind, equation (7) can be simplified. Using the above determined expressions for drag force and weight in scaling laws, it gives :tan β=F w P=C x ρ a 2 ρ water g S u 2 Ω=1 gu 2 Ω 1/3(11)where g=2 ρ water C x ρ a g is a reduced acceleration. Drag coefficients for drops are given in and estimated to 0.1–0.2 for a drop of water on a plane surface. For a drag coefficient of C x = 0.15 we have g≈1.3×10 5 m s−2. We thus plot t a n β according to the ratio u 2 Ω−1/3/g in Figure 8.
The relation is indeed linear, as predicted by the scaling law (11). This scaling law is one more argument in favor of the global relationship (5) which gives the direction of the drop depending on its volume and the speed of the wind.
4.2.4 On the complete model
The individual motion of a drop deposited on a plane seems to be rather well described by the above presented theory: wind and gravity compete with the adhesion force to put the drop in movement. The initial direction of the drop is then given by a global relation (11) between the angle of motion β and the ratio u 2/(Ω 1/3 g).
For a given value of a horizontal wind speed, say u=50 k m/h, drops under 3 m m diameter will not detach, drops between 3 m m and 4.5 m m diameter will have an initial direction closer to the wind (β> 45 °), while drops over 4.5 m m diameter will be more sensitive to gravity (β< 45 °). For u=100 k m/h, the limit for being more sensitive to the wind is D<17 m m, which is the case of most drops on a car window.
For a drop of fixed volume (say 20 μ l), the trajectory will mostly follow the wind (β> 45 °) only if u>42 k m/h. These figures are obtained after the fit given in Figure 8, with numerical values on an acrylic board.
From there, we can deduce properties of the actual motion of rain drops on the side window of a car. Two major trends may be mentioned:
Dependence on the drop volume: the general trend resulting from equations (9) and (10) is that smaller drops tend to be more attached to the window than bigger ones. Bigger drops are easier to dislodge than smaller one, and their trajectory is more susceptible to the influence of weight. Indeed the weight scales in D 3, so it grows quicker than the drag force which scales in D 2 and than the adhesion force (proportional to D). The volume of the drop thus has a critical role in the problem in so far as it determines both the transition between the static and dynamical behaviors and the direction taken by the drop when put in motion.
Dependence on the car speed: we conducted experiments to assess the importance of the speed u of the wind actually pushing the drop. This parameter is critical, on an equal footage with the volume, as it influences both the detaching and the direction taken by the drop. The speed u grows with the speed of the car on a real situation, yet a more precise correlation would require advanced numerical simulation of the vortices against the car window, which is far from our reach.
4.3 Dead ends
One may wish to extend the model to the complete description of a population of drops on a car window. Such a description should satisfyingly address the following points, which our model cannot exactly account for.
4.3.1 Influence of surface nature and treatment
When writing equation (2), we regard the factor c o s θ r−c o s θ a as a dimensionless constant, depending only on the chemical interactions between the surface and the drop, which does not play any role in the following scaling laws [Eqs. (9) and (3)]. Yet of course this factor's scale has a prominent role to play in the motion of the drop. Indeed, equation (2) states the existence of a minimal volume for water drops to detach under gravity, depending on surface nature and treatment.
We conducted experiments and verified our model with acrylic. It is noteworthy that the same law holds for other surfaces. We checked it out experimentally with normal glass, as can be seen in Figure 9. This enables us to extend the expression of adhesion force to any kind of ideal surface, from regular glass to car glass, or to glass treated to be hydrophobic, as may be the case for a luxury car. Only the coefficients of proportionality may change from one surface to another, as long as imperfections are neglected.
Complex models of the contact angles have been developed, namely by de Gennes and Cox , which quantitatively relate the contact angles to the properties of the fluid and the size of the contact line of the drop with the surface. However, these variations have no impact on the physics we presented.
The wear of the surface may also play a significant role on the behavior of the drops. As mentioned in Section 3, we noted that imperfections such as millimeter sized scratches on the surface may force the drops to stick under bigger external constraints (weight or drag). Indeed the geometry of the contact zone between the drop and the plane is modified by the presence of the scar, and the surface of contact is usually enhanced. Since there is no perfectly smooth surface, the size of imperfections contact line with the drop is a crucial parameter for pinning and depinning mechanisms, as has been studied by Paxton and Varanasi in .
Fig. 9
Experimental results for the inclination α s needed to dislodge a drop of volume Ω, put on an initially horizontal plane of glass that we slowly inclined. Ω ranges from 15 µ l to 100 µ l. This graph shows that for a glass surface, the scaling law between s i n α s and L c Ω−2/3 is once again reasonably valid (as it was for a plane of acrylic). What changes, compared to the previous experimental results with a plane of acrylic, is the slope of the fit. This slope is indeed linked to the contact angles between water and the plane, which are different for a plane of glass and acrylic.
4.3.2 Drops size distribution and merging
We have focused our study on the motion of an individual drop on a car window. For a better understanding of the global phenomena showed in Figure 1, it would be necessary to take into account the influence of the other drops. The presence of a drop on the path of a first drop of comparable size will impact its direction: the two drops will touch and merge. As the resulting drop has grown in size, the weight takes more importance compared to wind drag (it scales in D 3, so grows quicker than the drag force which scales in D 2): the drop's direction tends to turn downwards. Furthermore, the trajectory of drops go suddenly up and down (even if only on a few mm) while they merge with surrounding drops, which creates erratic-looking behaviors. A model for the dynamics of merging drops can be found in .
It would be interesting to have access to a size distribution of drops on a window. Though models for the size of drops in a rainfall exist since long (see for example ), this is unfortunately no easy task. We can however give some simple idea: the quicker the car moves, the stronger the local wind, and the smaller the drops which stay stuck on the window. Furthermore, drops tend to splash out more violently when they impact the car at higher speeds, which may also increase the proportion of small drops on the window for high car velocities.
Besides, some parts of the window are wet, covered by a thin layer of water or by a rivulet. It is energetically interesting for the drop to go on such a wet surface, as the interfaces between glass/water and water/air are already existing. As mentioned in the introduction, we have not tackled these aspects in this paper.
4.3.3 Air flow around the car
Cars have very different shapes, and this has a great influence on the air flow, and therefore on the local wind on the size window. This is why the global behavior of drops on one car may be different from drops on another car, and totally differs from drops on trucks.
The influence of the car geometry has been studied by Gilliéron and Kourta . He has shown that on conventional cars, a vortex forms at the junction between the windscreen and the side window. Whatever the car model considered, this vortex always has the same direction of rotation, which yields an upside component of the wind speed on the side window. This is the reason why some of the drops can be seen going up against gravity: there are indeed regions of the window with strong upwards wind.
As the photo of an experiment illustrates in Figure 10, the wind direction is not uniform on a window. Its value depends on the position of the drop of the window as well, and on the car model considered.
The main difficulty encountered in developing our theory further was to accurately estimate the wind along the car window. We first tried to measure the dependence between the magnitude of the upward wind speed and the magnitude of the car speed using an anemometer. We obtained the expected coarse correlation between the two: the wind speed increases when the car moves faster. However, the dimensions of the anemometer (a few decimeters) are not the least negligible given the Reynolds number of the flow. Therefore, the presence of the device itself modifies the geometry of the flow. This prevents us from getting accurate measures on the upward flow against the car, all the more so as we were forced to open the window on a small length to hold the anemometer. We then tried to simulate Navier–Stokes equations with a Computing Fluid Dynamic software, but we were not able to obtain enough precision so as to predict the formation of vortices studied in the literature .
One more step which could account for the erratic trajectories of a population of drops would be to take the drop-induced flow modification (as studied experimentally in ) into account while doing simulations. However, this is once again far from our reach.
Fig. 10
Photography from an experiment we conducted in a real car, driving at 90 km/h on a straight road. Little cotton threads had been previously stuck on the outside surface of the window. They show the direction of the local wind. On the picture, orange arrows are added for a better visualization of the direction of the threads.
5 Conclusion
The present work gives the basis for further study of the motion of a population of drops on the side window of a car. We developed a simple model taking into account the weight, drag force and contact force undergone by a single drop sticking on a plane. Applying the equations of motion yielded consistent experimental results regarding the initial direction taken by the drop. We explicitly showed the dependence of the motion on the main parameters of the problem: the volume and diameter of the drop – that we treated as a single parameter in first approximation – and the speed of the wind against the window.
We showed that three distinct regimes exist for fixed condition of wind speed. On an acrylic board for instance, if u=50 km/h, small drops (D<3 m m) will stick to the surface and would not move, medium size drops (3<D<4.5 m m) will mostly follow the wind and large drops (D>4.5 m m) will be more sensitive to gravity than wind.
We also showed there is a variability of the detaching parameters due to the surface state of the window.
Being given a map of the wind on the window, the size and initial position of a single drop, our model should be able to predict its trajectory, assuming quasistatic motion, through point-to-point approximation. The complete problem of raindrop motion will be the object of future research: taking into account the size distribution of drops on the window, the local inclination of the plane on which each drop stands and the complete velocity field of the wind against the car will enable further studies to explore the movement of a whole population of drops on the side window. This would allow to optimize both surface treatment and window geometry to evacuate drops faster.
Acknowledgments
This work is based on one of the problems proposed in the 2018 edition of the International Physicist' Tournament (IPT), a scientific competition open to students. The problem in question was stated as follows : ``When a car moves with high speed in rain sometimes the drops on its side window walk up but not down. Explain the phenomenon and find the conditions for it to occur (size of the drops and the car speed for example). What determines the drop trajectory and how does it depend on the important parameters?". Following the guidelines imposed by the IPT, we investigated the motion of rain drops on a car side window from the point of view of this final question, and focused on the direction taken by the drops once they undergo a strong wind coupled to gravity.
We are grateful to Pierre Lecointre, Christophe Clanet, from the LadHyX, Guilhem Gallot from the LOB, and Fabian Cadiz from the PMC at Ecole polytechnique, for the help they provided during the project and then for the fruitful commentaries and corrections they brought to the core of this article. We also thank Pierre Gilliéron, from Renault, as well as Barbara Brudieu and Alban Sauret, from Saint-Gobain, for their helping us on understanding better automobile aerodynamics and glass wetting.
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Cite this article as: Julie André, Clément Brochet, Quentin Louis, Amaury Barral, Anthony Guillen, Fang-Ting Goh, Angel Prieto, Thibault Guillet, Motion of rain drops on a car side window, Emergent Scientist 3, 3 (2019)
All Figures
Fig. 1
Picture of rain drops on the side window of a moving car. The drops have different trajectories; some of them are going up, other have almost horizontal directions. Merging between drops occur, which makes the different paths quite erratic. The wetting of the window makes it easier to take some predefined paths (canals), as we can see on the top part on the window.
In the text
Fig. 2
Schematic view of the three forces applying on a drop, holding on a tilted plane of inclination α and under a wind of given direction β w on the plane. t→ is an unit vector tangential to the plane. The sum Σ→ of the two external forces (weight P s i n αt→ and drag force F→w) is represented in blue. The angle between Σ→ and t→ is defined as β.
In the text
Fig. 3
Schematic representations of : (1) a drop of diameter D on a horizontal plane, (2) its deformation under gravity, when the plane which is progressively inclined and (3) the moment when the drop begins to slide. θ a is the advancing angle and θ r the receding angle.
In the text
Fig. 4
Schematic view of the experiment to dislodge a horizontal drop with drag force. We use a wind tunnel, to ensure that the air flow is laminar and to shrink wind speed variability. The wind could blow from 1 to 30 m s-1.
In the text
Fig. 5
(Top) schematic view of the experiment performed in order to measure the direction taken by a drop when dislodged by an horizontal wind. The support is maintained vertical. (Bottom) time-lapse photography of a drop in motion on a vertical plane under an horizontal wind.β w is the angle the wind makes with the vertical, while β is the angle which the sliding drop takes.
In the text
Fig. 6
(Top) a plot of α s, the inclination needed to dislodge a drop of water, is plotted against the volume Ω of the drop, put on a plane of acrylic. Dashed lines indicate limits for definition of α s, which cannot exceed 90° (in this case the drop will not fall under its weight). Thus for volumes on the left of the vertical dashed line, α s is not defined.(Middle) a plot of us, the minimum wind speed needed to dislodge a drop, is plotted against the diameter D of the drop lying on a plane of acrylic. (Bottom) the direction, taken by a drop of volume put on a vertical plane in front of a blower, is plotted against the speed u of the wind. Two values of the volume are used (blue dots: 20 µ l, and green crosses: 25 µ l).
In the text
Fig. 7
Experimental law between the volume Ω and the diameter D of a drop defined as the diameter seen from above the drop as it is lying on a horizontal floor. Ω ∝ D 3 is a consistent model for small enough drops. Fit does not take into account the two last points, where drops are so big that they spread under their own weight.
In the text
Fig. 8
(Top) scaling of s i n α s (α s the sliding angle for a drop) as a function of L c 2 Ω−2 3, with L c the capillary length for water under normal conditions of pressure and temperature. Values of Ω range from 15 µ l to 100 µ l. A linear fit (dashed line) to experimental data is expected, according to our theory. (Middle) experimental law for the relation between wind speed u s and drop volume Ω on an horizontal plane, law stemming from the competition between the drag force and the adhesion force. Linear correlation (red line) is consistent with our theoretical model. (Bottom) experimental results on the angle β taken by drops of given volume Ω under an horizontal wind u. Drops small enough are set on a vertical plane (we took Ω=20 μ l or 25 μ l so they can stay DESPITE GRAVITY). Then the wind blower is switched on, and blows horizontally with a speed u bigger than u s(Ω), the minimum value needed to dislodge the drop. As our model predicts, there is a linear law between t a n β and the ratio u 2 Ω−1/3/g.
In the text
Fig. 9
Experimental results for the inclination α s needed to dislodge a drop of volume Ω, put on an initially horizontal plane of glass that we slowly inclined. Ω ranges from 15 µ l to 100 µ l. This graph shows that for a glass surface, the scaling law between s i n α s and L c Ω−2/3 is once again reasonably valid (as it was for a plane of acrylic). What changes, compared to the previous experimental results with a plane of acrylic, is the slope of the fit. This slope is indeed linked to the contact angles between water and the plane, which are different for a plane of glass and acrylic.
In the text
Fig. 10
Photography from an experiment we conducted in a real car, driving at 90 km/h on a straight road. Little cotton threads had been previously stuck on the outside surface of the window. They show the direction of the local wind. On the picture, orange arrows are added for a better visualization of the direction of the threads.
In the text
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6848 | https://jeffjar.me/fourier/lec8.html | Scaling & Derivatives
Home
Introduction
Introduction
Fourier Series
Periodicity
Monsieur Fourier
Finding Coefficients
Interpretation
Hot Rings
Orthogonality
Fourier Transforms
Motivation
Inversion&Examples
Duality&Symmetry
Scaling&Derivatives
ConvolutionScaling & Derivatives
You know, one of the issues with recorded classes is that people don't bother to show up to class! I swear, on the first day of class, there were a lot more people sitting here than there are today. I don't blame them; it's super convenient to view lectures on your own time…
‘‘Did I make some comment about not getting paid?’’
Today we'll do some more derivations about properties of the Fourier transform. Just like last time, we'll talk about the sort of logic behind these sort of arguments; and again, we'll leave the rigor police off-duty.
Properties of F.T.
Linearity
The F.T. is a linear transformation; that is, the F.T. of a sum is the sum of the F.T.; and if you multiply a function by a constant, it's that same constant times the F.T>. The whole point of this princple of superposition is that if you can write a system as a sum of different components, you can just sum of the F.T. of each of the parts. Life is really nice when things are linear.
Now this is a sort of property we take for granted. The more interesting property is the Shift theorem.
The Shift therom
What happens to the F.T. if the time is delayed by constant ?
We want to find the fourier transfrom of , which is
Notational note
there's a problem with notation here! We mean the ‘‘fourier transform of the shifted function’’, but on the LHS we're also evaluating the F.T.'s function at the point ….. Notation is a genuine problem in the subject! And it can get in the way of understanding things! (On the homework, we'll talk about scaling and shifting operators….which is nice, since you don't have to explicitly say the argument of ….but it's also confusing since it introduces extra things). Anyways, we should just remember to be careful about what variables mean so we don't go wrong.
To evaluate the integral here, we do a change of variables (as usual) to , so that the argument of becomes . And then there's an extra factor of up in the exponent, so we just have a total phase shift of which we can pull out of the integral.
The net result of the shifting operation is that we end up with a phase shift in the spectrum.
Often, we'll hear this said as ‘‘a time shift in time corresponds to a phase shift in spectrum’’.
Notice that the magnitude of the spectrum doesn't change when you shift the time; only its phase changes.
Again, these properties aren't terribly difficult to derive, but they're fundamental and important to know, and highly applicable! The logic we use in them is pretty important to know…simple things, but we should do them with ‘‘gusto and confidence’’!
Stretch theorem or Similarity theorem
Now the question is, ‘‘what happens when you take the Fourier transform of a stretched sigal’’, , where ?
To figure this out, we do the same thing. Again, we switch our variable of integration to so that the argument inside is a simple , but this time around, we have to be a bit careful….
goes to …
the in the exponent goes to , and we can slide the one over to the , so that now we just have a in place of .
If we take , we end up with
The subtle thing here is what happens when is negative! Then the limits of integration swap around when we change variables to , and we'll need to introduce an overall minus sign to put them back to and . So this time the result looks like
with a minus sign!
And to combine the two cases into one equation, we put an absolute value around the a outside the Fourier transform.
There's a number of comments to make here:
‘‘This is a fine example of a reciprocal relationship’’ – which as you remember, help us organize our understadning and build our intution. In this case, if we scale the time by a factor , then the frequency and the magnitude of the F.T. have both been scaled by , the reciprocal of !
One way to interpret this is graphically. (Graphs aren't always our most useful tool, but in this case they can help us!) Remember that when we scale , the graph gets squeezed along the x-axis if , and it's stretched if .
There's a silly way to remember what this scaling operation does to …does it squeeze it or stretch it?….we went through a fun argument in class, and it's also in the course reader…
And also, remember that even though the F.T> of a function is generally complex, we can still plot it on a graph by considering its magnitdue.
This means that…
Stretching and squeezing
If , then:
the time-domain function is squeezed in time
the frequency-domain function is stretched in time.
If , then:
the time-domain function is stretched in time
the frequency-domain function is squeezed in time.
The consequence is that you can't have a function that's concentrated both in time and in frequency! If you try to squeeze it in the time domain, you stretch it in the freq domain, and vice versa if you try the other way.
This is reminiscent of the uncertainty principle in quantum mechanics! Hehehe.
The limiting extreme case of this principle is the delta function. It's infinitely concentrated in the time domain, but when you take the fourier transform, it has a flat and uniform frequency for all frequency components.
There's different sorts of transforms in signal processing such as wavelet transforms where you can concentrate them in both the frequency and time domains.
Philosophically, the stretching theorem tells us a pretty important property of Fourier Transforms:
You can't squeeze a function in both the time domain and the frequency domain.
As a side note, in higher dimensions, the generalization of the scaling theorem leads to a more general idea of what a reciprocal means.
Is this related to reciprocal lattice and Miller planes in the field of crystallography? I've always been so confused by the whole idea…
Some examples
Then we went through a few examples to gain some intution. These examples look much better with actual graphs to stare at, but the best I can do is to describe in words what we did in calss.
The square wave.
Remember that the FT of the square wave is the sinc function which was defined as . If we apply the scaling theorem, it says:
The FT of the stretched rectangle function is a squeezed sinc function. It's squeezed both in the x and y axes when you graph it –
‘‘Not hard, but you've got to get it in your gut!’’
The Derivative Theorem
This is true magic! It's not clear at all that the Fourier Transform has anything to do with derivatives!! But remarkably, something magical happens when you mix in differentiation and Fourier transforming.
Fir'st let's consider what happens when you take the derivative of the FT of a function. What is ?
We expand out the definition…
We pull the derivative inside the integral.
the rigor police are now rolling aroudn in their graves! There's a lot of conditions and religions and such that need to apply for us to do this.
physicsits do this all the time with impunity!
Now the derivative hits the exponential and pulls down a factor of as well as some constants in front of it.
The end result is that we end up with the FT of something times . Wow!
Multiplication by time corresponds to differentian in frequency domain!
The dual statement
If we ask a similar but related question, we end up with a nice symmetric result as well. Instead of finding the derivative of the Fourier Transfrom, this time we want to find the FT of a derivative.
This time, the ‘‘proof’’ had a sort of different flavor, where we first wrote the derivative in terms of a difference quotient (from the first few weeks of calculus class) and then plugged that into the Fourier transform.
We use linearity to separate out the terms, and we have the FT of two different terms – one with , the other one with .
We apply the shift theorem to the first term, so we get an overall phase shift of on the first term
Both terms have a fourier transfrom of , so we factor it out. We're left with the expression
And then we can simplify that fraction in the limit . In class, we performed the limit by recognizing it as the derivative of a function . Personally, I would do a physicisty-thing and just expand the exponential in terms of a Taylor series, and then drop off all higher-order terms .
No matter how you do it, the fraction turns into as . So our end result is…
Wow! The fourier transform can turn derivatives into multiplication!
This is very useful if we want to solve differential equations; the Fourier xfm lets us turn them into algebraic equations, which can be much easier to solve. Think to the Laplace transform in your ODE eclass.
And remember, there's no reason a priori to expect the Fourier transform – which isa complicated operation – to turn multiplication into derivatives – which are also another complicated operation! This is a quite magical relationship. And we'll put it to good use soon.
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6849 | https://www.youtube.com/watch?v=FIMK50NkvoA | Combinatorics. Section 13.1: Coloring regions with two colors
Henry Adams
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Posted: 25 Nov 2018
This video accompanies Math 301, Introduction to Combinatorial Theory, taught at Colorado State University in Fall 2018 by Henry Adams. The course is based on the book "Discrete Mathematics: Elementary and Beyond" by Lovasz, Pelikan, and Vesztergombi.
For our class webpage, please see
For our class notes, please see
Transcript:
I would like to give it a brief review of section thirteen point one coloring regions in the plane with two colors the main theorem in this section is thirteen point one point one and it states that if you draw circles in the plane then you can color the resulting regions with only two colors and furthermore satisfy the property that adjacent regions are colored differently in the picture here we have six circles drawn and we've colored the regions in red and blue and adjacent regions are always colored one red one blue for the proof of this theorem we'll proceed by induction on n the number of circles and our base case is when we've drawn N equals one circle in the plane we have two resulting regions inside the circle and outside the circle we need to draw these two regions with different colors well that's easy use the color red for inside the circle and the color blue for outside now in the inductive step we'll assume that the theorem is true whenever you have n minus 1 circles drawn in the plane and our task is to prove that remains true for n circles so pretend we're given n circles drawn in the plane in the figure on the top left and is equal to 5 and we'll remove one of these circles call that circle see that we remove giving us the drawing of n minus 1 circles here four circles on the top right by our inductive assumption we can color these remaining regions formed by n minus 1 circles with our two colors and now what we'll do as drawn on the bottom here is we'll add in our last remaining circle and we'll reverse the color inside this remaining circle so if we had a portion that was colored with a dark color now inside our newly added circle colored light or if we had a region that was previously colored with our light color now inside this newly added circle color it dark and we want to verify that this new coloring preserves the property that adjacent regions are colored differently so there's a couple arcs two cases of arcs that we need to check so first consider an arc that's outside the newly added circle see well the colorings on either side of this arc haven't switched and so this arc has one color on each side next consider an arc that's inside the newly added circle see well before we had light and dark and now that we've switched the color inside the circle we have dark and light so we had both colors on either side of this arc and now we've just reversed the color so we've maintained the property that one color still is on either side of this arc finally consider arcs that live a long circle C as such as this arc here well before we had the lighter color on either side of this arc but now that we've added in the circle and reverse the colors inside we have one color on each side of this arc and that's the end of our proof by induction we've assumed that we can color regions with formed by n minus 1 circles and we've used that to prove that we can color the regions formed by n circles using only two colors and therefore by induction the theorem is true for all n all numbers of circles let me end this section with an aside in elementary school we would draw checkerboard patterns like this and then on top of this draw shapes like circles and triangles and rectangles and then color these pictures as you see here in this sort of fashion that creates a 3d effect it looks like the circle is on top of the triangle which is on top of the underlying checkerboard grid and the other side I want to make here is that you can use a similar proof technique to prove that any such drawing can be drawn with only two colors the way you do this is that you cover the background grid alternating light and dark colors and then as you add inductively add on one more shape or one much more shape you repeat the inductive argument that we just gave for circles in the plane thanks |
6850 | https://www.youtube.com/watch?v=wMAWbnbvtio | DISEÑO DE EJES POR TEORÍAS DE ENERGÍA DE DISTORSIÓN Y ESFUERZO CORTANTE MÁXIMO | EJER. 5-27, SHIGLEY
PROFE JN El canal del ingeniero
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Posted: 28 Apr 2022
Este video enseña cómo calcular el diámetro para un eje sometido a cargas estáticas y por las teorías de ENERGÍA DE DISTORSIÓN y las de ESFUERZO CORTANTE MÁXIMO
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Transcript:
[Música] e [Música] hola bienvenidos a mi canal este es un canal especializado en temas de ingeniería para esta ocasión vamos a trabajar diseño de ejes vamos a hacer dos ejercicios completos en la descripción del vídeo les voy a dejar los pdf de los ejercicios para que lo tengan como material de estudio quédate hasta el final del vídeo y si te gusta el vídeo por favor no se te olvide regálame like y comparte a los compañeros y amigos empecemos con el vídeo el ejercicio 527 nos plantea la figura es una representación esquemática de un contra eje que sostiene dos poleas con bandas en vez ahí está el eje y están las dos poleas me dice que con bandas en ve en cada polea las tensiones en las bandas son paralelas ahí se muestra atención 2 para ver la contención 1 y la carga de 50 paralela con la 270 para la polea da considere que la atención de la banda cuando está suelta es igual al 15% de la atención que se presenta en el lado apretado entonces me están diciendo que una atención pongamos la atención uno es en el lado apretado entonces en el lado suelto será el 15% del valor de esta atención bueno para esta aplicación se seleccionará 10 n s 10 180 estirado en frío con un diámetro uniforme es decir nuestro eje no va a tener escalones va a tener un diámetro uniforme para un análisis estático ojo vamos a trabajar esto análisis estático con un factor de seguridad de 3 determine el diámetro mínimo preferido y me piden acá en el 527 use la teoría de la energía distorsión y en el 528 que es el mismo problema me dice que resuelven 527 pero usando la teoría del esfuerzo cortante máximo pues vamos a hacer los dos problemas como siempre lo primero que hacemos es sacar la información del problema entonces aquí he sacado ya la información miremos lo primero que saqué es que la atención 2 es el 15 por ciento de la atención 1 como yo sé que debo expresar la atención 2 en función de la atención 1 y no la atención 1 en función de la atención 2 entonces dicen como a mí me dan como información estos 270 y estos 50 quiere decir que esta polea está recibiendo torque en este sentido ese es el sentido del torque de la polea en ve por qué por qué hacia dónde va la fuerza mayor min la fuerza mayor es hacia acá entonces la polea vaya girando así si este torque en be lleva a esta dirección que si ustedes aplican la mano derecha a la regla de la mano derecha ese torque iría entrando ese sería un momento negativo a un torque negativo y orientando entonces para que haya equilibrio porque esto del equilibrio en los apoyos se supone que no hay pérdidas de torque porque estos son apoyos a rodamientos entre no haber pérdida de torque quiere decir que el torque en esta pelea debe ser exactamente este mismo torque en b pero en sentido opuesto y entonces si va en sentido opuesto pues el torque tendrá que ser así este será el torque en la polea si ustedes siguen la mano la regla de la mano derecha pues el dedo pulgar apuntará hacia afuera hacia el eje x y este es el torque de la pole na sí si ustedes se dan cuenta los dos torpes son contrarios tienen que ser contrarios ahora cómo tienen que ser contrarios quiere decir que la mayor fuerza para que este torque se en esta condición la mayor fuerza tiene que ser t 1 y entonces en ese caso de 2 debe ser la menor o sea el 15% de tf1 el 15% de 0-15 entonces este 12 015 de tf1 esto quiere decir que el valor que yo ponga acá lo multiplicó por ser un 15 pues él o el 0 15 lo hace más pequeño o sea que te 2 es más pequeña que te 1 y en este caso la rotación será en este sentido bueno una vez que tenemos claro eso ya sabemos la relación entre t 2 y tf1 entonces lo otro es que aquí me dicen que esté a 0 es una 0 n s g 10 180 si yo me había la apéndice del libro del libro de chicles de la octava edición la tablada 20 entonces aquí encuentro el acero mil g 10 180 que en la numeración sae y sidi y sae es el mismo 10 18 aquí está pero aquí está estirado en caliente y estirado en frío recuerde estirado en periodos cold round porque aquí me dicen estirado en frío desierto y acá estirado en frío entonces como está ahí en los dos estirado en caliente estirado en frío entonces me voy por el estirado en frío acá y de acá tengo información importante que es la resistencia a la fluencia la resistencia de la prudencia 370 mega pascal es porque ese es una información que voy a necesitar más adelante ahora el problema también me dice que tome un factor de seguridad de 3 y el bid en calcular el diámetro por energía distorsión y por cortante máximo bueno ya hemos organizado la información del problema ahora vamos con lo primero que se hace en este tipo de ejercicios que es la parte estática para la parte estática tengo que calcular de entrada la atención 1 y la tensión 2 para eso voy a hacer sumatoria de momentos respecto al eje pero el eje que está sobre mi eje es el eje x entonces vamos a hacer sumatoria de momentos alrededor de x como estamos en el equilibrio igual a 0 ojo que esté mal el equilibrio no quiere decir que esto no pueda rotar esto rota pero rota de tal manera que el torque entrada tiene que ser el mismo torque de salida o sea la sumatoria de todos los toques tienen que dar 0 y eso es lo que vamos a hacer entonces el primer momento fíjense que esta tensión de 270 produce un giro así y esta tensión de 50 produce un giro en sentido opuesto las dos fuerzas están a la misma distancia que es el radio o sea que aquí voy a tener dos torques pero uno en este sentido que sería el sentido negativo mire hasta 200 de 30 me da un torque negativo y esta me da un torque positivo ahora como este es más grande pues va a quedar el torque negativo ahí bueno pero para sacar el torque el resultante ahí sencillamente lo que hago es tomar los 270 newton restar los 50 newton y multiplicar esto por el radio si el diámetro aquí me dice que son 300 milímetros quiere decir que el radio son 150 milímetros osea 0.150 metros y aquí lo que estoy haciendo es sacando el torque mi fuerza por distancia fuerza por distancia en los doctor que fuerza por distancia 11 en un sentido el otro en el otro sentido pero ya dijimos que el resultante resultante de esos dos como esta es la fuerza mayor va en este sentido y aplicando la regla de la mano derecha es hacer el sentido negativo luego este torque es negativo ahora a este torque le vamos a sumar a este torque que ya dijimos que era en este sentido porque en b no le vamos a sumar el torque na perderéis recuerde el torque na va a ir en este sentido quiere decir que esta tensión 1 es la mayor entonces aquí vamos a colocar el torque es igual a la tensión uno menos el torque en el sentido opuesto que me lo genera la atención 2 pero de una vez la atención 2000 y la tensión 12 015 la atención uno entonces menos 0 15 de la atención 1 y ahí ya tengo la fuerza resultante eso multiplicó por él radio si el diámetro son 250 el radio serán 125 es aquí por 0,125 metros y ahí está lo mismo no fuerza por distancia fuerza por distancia este porque en un sentido este torque en otro sentido el mayor ojo el mayor va en dirección de t uno tiene que ser contrario al torque en b que es este torque nada cierto si seguimos la regla de la mano derecha ya dijimos que será positivo por eso aquí colocamos el positivo listo como sólo están esas dos poleas sobre las cuales no entra la torque y en la otra sal saldrá del torque entonces ya la suma de eso tiene que darme igual a cero y ahora aquí simplemente despejó la única incógnita pues este uno este primer producto aquí me da menos 33 ahora de estos dos que son términos semejantes vamos a dejar uno solo vamos a sumar los entonces esto será 1 t 1 - 0 15 t 1 esto me dará 0 85 0 85 de 1 finalmente multiplicamos este es el 85 de 1 por 0 125 y esto me da 0 punto 10 625 de 1 ahora este menos 33 pasaría al otro lado como 33 positivo entonces igual a 33 esto estará en newton-metro despejamos el t 1 estos en los 10 625 son metros cierto entonces el t 1 será 33 newton por metro dividido entre 0.10 625 metros cancelamos y piensen que ya me queda el t 1 bueno de uno me queda 310 punto 5 88 bueno si aproximamos esto nos queda 310 puntos 6 newton nos quedan newton y ahí está la atención 1 la atención 2 pues sencillo no arriba ya dijimos la atención 12 es el 15% de la atención 1 sea 0 15 porque 1 luego de 2 x 0 15 y nos da aproximadamente igual a 46.6 newton ahí están las dos tensiones calculadas las dos tensiones ahora nos vamos con las reacciones en los apoyos una vez calculadas las fuerzas tenemos todas las fuerzas externas vamos con las reacciones aquí en los dos apoyos y aquí con las reacciones en los apoyos cuatro reacciones porque sólo cuatro dicen que estas cargas están en dirección del eje z estas contienen componentes hacia z y hacia allí porque están en el plano z pero no hay componentes en x no hay fuerzas en x si no hay fuerzas en x pues no hay reacciones aquí en x entonces solo dibuje esas 4 reacción es bueno porque estás bajando porque cuando saque los componentes de t1 y t2 como ambas van subiendo pues las reacciones tendrán que ir bajando estas como van para la derecha pues coloque está para la izquierda o bueno para la lgct a negativo y está también la coloque hacia el eje z negativo de igual manera si tengo alguna en dirección opuesta pues el signo cuando las calcula y me lo va a decir entonces ahora vamos a hacer lo de siempre equilibrio sumatoria de fuerzas en x igual a cero pero como no hay fuerzas en x entonces no hay ecuación luego sumatoria de fuerzas en que igual a cero fuerzas en jesse hay entonces cuántas fuerzas ahí en que ésta está oye esta esta serie ambas negativas porque van bajando 15 pero están las componentes de t1 y t2 porque viese si tomamos las componentes vertical la componente vertical de t 2 será ésta miren esta que será de 2 en que la componente vertical de t 1 ésta será de 1 engine y ambas van para arriba cierto cuánto vale en esos componentes pues si usted se da cuenta esta la lleva acá y fíjese forma este triángulo miralo acá este triángulo esta componente será el opuesto al ángulo de 45 grados yo recuerde será la hipotenusa por el seno de 45 grados entonces de una vez acá de 2 seno de 45 grados positiva porque iría subiendo más de 10 a 45 grados positiva porque iría subiendo menos 6 - oye que son las dos reacciones aquí en los apoyos verticales listo como sólo hay estas cuatro cerramos la ecuación igualando a cero ahora como yo tengo aquí hoy o en contra cada que te uno aquí este t 1 son 300 10.6 newton y t 2 son 46.6 newton de la reemplazó las multiplicó por 0 45 cada una de ellas sacó el resultado de esto estas dos las pasó al otro lado para que me queden positivas y me queda la ecuación si soy igual a 250 y 2.578 newton que es el resultado de esto estas 2 al otro lado positivas bien la primera ecuación ahora hay más fuerzas acá pero en la dirección zeta entonces hacemos sumatoria de fuerzas en z igual a 0 porque hasta acá pues tengo dos incógnitas que no los puedo solucionar miremos la fuerza del zeta entonces tengo 270 y 50 en cada para ambas van para z positivo esto me da 320 desde una vez aquí vamos colocando 320 en sentido positivo 320 320 minutos ahora quien va negativa menos 11 está y menos ez pero me faltan las componentes de la atención 2 y de la atención 1 porque estas dos fuerzas fish en miremos los componentes ambas tienen componentes en z físicas dos componentes bueno ya dijimos que la de ella era con seno pues entonces la de z que es el otro cateto pues será con coseno entonces ambas con coseno en este ejercicio no importarían si se equivocan porque el seno y el coseno de 45 valen lo mismo pero hagámoslo como es entonces aquí está de una vez reemplace el valor 22 cierto mire la quite 2 coseno de 45 y el valor de tf1 este 1 por coseno de 45 ambas con coseno porque son los catetos adyacentes listo cerramos igualando a 0 porque no hay más fuerzas ahora reducimos este estreno está con esta con esta bueno estas negativas ambas componentes porque fíjense que ambos vectores miren las dos componentes van hacia el eje z negativo millón de está z positivo se está positiva hacia este lado de acá y esas van para el lado opuesto bueno y aquí está nuestra ecuación reducida entonces estos tres meses 67.4 21 estas dos negativas pasan al otro lado positivas y me queda la ecuación número 2 hasta ahí dice que llevo cuatro incógnitas que son obviamente las cuatro reacciones ahora sólo tengo dos ecuaciones luego debo sacar más ecuaciones entonces como ya hicimos toda la sumatoria de fuerzas ahora me quedan la sumatoria momentos vamos a hacer sumatoria de momentos el punto más conveniente pues será aquí el origen o entonces sumatoria el momento respecto al origen para que si yo hago sumatoria de momento respecto al origen y sé que estas dos se pierden como incógnita porque están ahí en el punto donde voy a hacer la rotación quiere decir que las incógnitas van a hacerse y sz y entonces voy a tener una ecuación que es una tercera ecuación bueno otras dos ecuaciones porque hay realmente van a salir dos ecuaciones con dos incógnitas y ahí ya podré completar todo el sistema para resolverlo recuerden que acá como estamos en 3-d el momento lo vamos a hacer r cruz efe como víctor es nada entonces empecemos bueno coloquemos acaso materia de momentos igual a cero r pensemos el primer ojo salimos del origen entonces mi primer radio vector reposición será del origen a donde encuentro fuerzas acá donde encuentro estas dos componentes de una vez lo vamos a escribir lr 300 milímetros o sea 03 metros y piense que hoy navegando en y positiva no vengo para acá y positiva entonces una vez vamos a escribir 0.3 n s slr cruz las fuerzas pero las fuerzas pero fíjense que voy a trabajar es con las componentes o sea voy a sumarle una vez t 2 z y t 1 zeta la suma de esos dos valores cierto que ya sabemos que sale 46 6% o 45 y la otra por con 145 y la otra 316 por cos en el 45 la suma de sus dos mil 252 entonces aquí de una vez 250 y 2.5 78 newton en qué dirección dicen que estaba hacia menos que estés aquí en menos que esté al menos ahí está en dirección menos acá porque dicen ambas van hacia z negativo listo ahora estas 2 pero piensen que estas dos van subiendo y la suma de estas dos me da las mismas suma de estas dos por ser el ángulo de 45 grados luego entonces aquí como van subiendo colocaré más 250 y 2.5 78 pero subiendo hacia jota positivo y cierro bueno ahora seguimos el otro momento me lo genera la resultante de estas dos 270 y 50 320 entonces voy a colocar una de 320 acá en el eje entonces será la distancia desde hoy hasta nuestra polea que será esta suma 300 y 400 700 milímetros y me desplazaré mire esos 700 los vengo caminando así o sea 700 en y pasado a metros 0.7 en positivo cruz la fuerza entonces la fuerza ya dijimos que eran 320 newton en qué dirección en z positivo no o sea en k aquí este es el otro momento ahí está la suma de estos 2 320 hacia z positivo que es acá listo y nos faltaría estas dos acá cz entonces desde o hasta donde está el apoyo en c estarán 700 y 850 entonces pasado a metros más 0,850 en y positivo no porque si venimos desde o hasta donde están vamos caminando en positivo 850 milímetros y ahí encuentro las dos que las colocó con signo no se lle negativa y 60 negativa entonces acá será cruz 6 como es negativa será en menos j ahí está en menos j - cc está y estará en menos acá o cc está en menos acá cierro y como no hay más fuerza cierto no hay más momentos ahí solamente de esas fuerzas y esos momentos estoy cierro la ecuación igualando a cero ahora hacemos los productos cruz entonces acá bueno voy a apoyar en el reloj cierto por lo que los vectores no están completos entonces empecemos 03 por menos 2 52 58 inicialmente me da menos 75 773 pero cuando el producto cruz y cruz acá entonces venimos acá y cruz acá nos da menos jota no y con el menos de acá me daría más luego el siguiente producto que me da exactamente el mismo resultado pero ahora el producto vectorial y cruzó está entonces y cruzó está recuerde acá el siguiente producto 0.7 por 320 esto a 224 inicialmente positivo luego algo del producto y cruz acá entonces y cruz acá me da menos j entonces por eso aquí negativo el siguiente producto entonces será el 0 80 150 2 085 por 6 inicialmente más x menos menos negativo y hago el producto y cruzo está eso me da acá positivo positivo por negativo pues sigue siendo negativo y el último producto inicialmente negativo 0 850 por menos cc está cierto y cuando el producto vectorial y cruz acá pues aquí esta cruz camera - j y que en el menos que tenía me da más ahí están los productos lo pueden hacer así ayudados con el reloj que es mucho más rápido que hacer todos los determinantes ahora me dicen que nos quedaron términos en jota términos en jota términos en gota y los otros son términos en cada y términos en que entonces se paró no el separar es que yo hago sumatoria de momentos en que las gotas son los djs ya está igualado a cero entonces que tenemos 75.773 esa está en j el otro que está en jota es esta de menos 224 y la otra jota es este +0 85 cc está + 0 85 cc está cierro la ecuación igualando a 0 y esta será mi ecuación número 3 mi ecuación número 4 pues haremos sumatoria de momentos lo otro es en kano o sea sumatorio de momento respecto a z igual a cero tomo lo de acá dos lógicas lo que está en rojo los 75.773 bueno estos son newton-metro y lo otro que está en que es menos 0 85 6 cierto y cierro la ecuación igualando a 0 y esta es mi ecuación número 4 y ahora si tengo cuatro ocasiones cuatro incógnitas solucionamos entonces de la 4 pues se puede despejar directamente serie porque sólo es la única incógnita que tengo ahí pues paso a paso todo esto al otro lado positivo devuelvo al 085 a dividir y tengo que seguir 89 puntos 145 en la 3 física la única en comités ez el resto de estos dos esto me dará negativo pasa al otro lado positivo lo viven 100 85 y 60 media 174,3 85 newton ambas positivas ahora como ya tengo 6 ez entonces me devuelvo a la ecuación 1 y 2 fi se reemplazó a 15 y en esta reemplazo cz para calcular oye y ó z y al reemplazar y calcular obtengo que oye y objeta son entonces ahí está si reemplazamos tenemos que oye va a tomar un valor de 163 punto 433 y ó z dice que hace tan nos dio negativa no quiere decir que os estaba en el sentido opuesto a como lo habíamos considerado dicen que miremos la k o z que es ésta la habíamos considerado entrando realmente o se estaba en el sentido opuesto desde una vez aquí puedo cambiarle el sentido ahí está y es lógico no porque te 1 727 160 pies en ambas van hacia la izquierda pues está dividir a la derecha para contrarrestarlas listo ahí está el sentido real de océ está listo ahora una vez que ya tenemos resuelta toda la parte estática es decir calculas todas las reacciones entonces nos vamos a producir oa calcular los momentos en la polea y los momentos en la polea para seleccionar el mayor de los dos para eso voy a tomar estas reacciones que acaba de encontrar y estas cargas y voy a dibujar por planos voy a dejar las fuerzas presentes en el plano de x y las fuerzas presentes en el plano zx entonces aquí ya las tengo llevadas felices y antes de continuar ya te suscribiste esperas suscribirte al canal activa la campanita para que cada que su hombre ejercicio de estos de otros temas de beneficios sea el primero en ver el material y recuerda somos una comunidad si te gusta mi trabajo obviamente que al botón unirse vuélvete miembro de esta comunidad desde 2.5 dólares muy económico para ti apoya es mi trabajo me beneficia a mí y te beneficias bueno continuemos con el vídeo fuerzas presentes en el plano de x y las fuerzas presentes en el plano zx y esas son las fuerzas aquí en las poleas y en los apoyos aquí fuerza en la polea y las componentes de fuerza en la polea y en los apoyos bueno aquí la se ha aproximado al primer decimal voy a trabajar con la aproximación al primer decimal entonces dicen que está de 163,4 es la reacción o lleno la que encontramos acá minella la reacción oye pero aproximada hay al primer decimal y entonces ahí está dibujada fi se oye bajando que están en el plano de zeta entonces ahí está en el plano de zeta oye bajando ésta es la suma de las dos componentes recuerden que la suma de las dos componentes no sabe a 2 252 6 a que me refiero a estos dos componentes que van subiendo embistes sumadas me dan igual que estas dos componentes no entonces esa suma de 250 y 2.6 van subiendo y aquí la reacción de serie que nos dio que bueno que le habíamos dibujado bajando fíjese seguía aquí esta serie dibujada bajando y sigue ahí nos dio 89 145 entonces la aproximamos a 89 puntos newton y lo mismo hacemos en el plano zx fíjense en que ésta lo habíamos calculado y nos había dado negativa no - 106 974 la próxima menos 107 pero entonces ya la dibujo en la dirección correcta missing hacia el zeta positivo listo y esta que es la suma de las otras dos componentes estoy refiriendo a estas componentes las que están en verde dicen que van hacia z negativo aquí me dice está positivo es hacia abajo van a cc está negativo está la de 320 que es la suma de estas dos mil 320 dicen que esta queda en el plano zx ahí está en el plano zx o xz y la reacción cc está bueno y ahí está entonces ahora calculamos momentos en la polea y momento en la polea b para calcular los momentos aquí es sencillo por tratarse de cargas puntuales si no fueran cargas puntuales me tocaba generar las ecuaciones y hacer los diagramas pero como son cargas puntuales entonces yo voy a partir aquí antes de la polea mi eje y voy a tom o bueno quedan dos dos partes del eje cierto si partimos acá nos quedará o bien esto o bien esto trabajo con cualquiera de las dos secciones y usted ya lo sabe si trabajo con esta sección que es la más corta el momento respecto a ser a estos 163.4 por la distancia de esta carga al punto a que son los 300 milímetros volvemos lo arriba son estos 300 milímetros o sea que el primer momento 163 newton por 300 milímetros es 03 metros pero ese es el momento en este plano ahora el momento en este plano entonces lo mismo hago un corte acá cierto por ser carga puntual entonces yo tomo cualquiera de las dos secciones fijo puede tomar esta y entonces el momento será la carga de 107 por la distancia acá la distancia al punto a que son también 300 milímetros entonces aquí el momento respecto a bueno pero aquí de una dehesa no tenemos los éste es en el plano xy y éste es en el plano x y serán los 107 newton por los 300 milímetros o sea 03 metros listo y ahora el momento resultante respecto a recuerden que como están en planos perpendiculares el momento resultante será la raíz cuadrada de cada componente este es una componente y esta es la otra componente raíz cuadrada de cada componente al cuadrado bueno y ahí está si hago las operaciones esto me da 48 9 esto me dio 32 1 hago pitágoras y eso me a aproximadamente cierto 58,6 newton-metro este es el momento presente dentro de la polea da ahora voy cálculo el momento presente dentro de la polea ve exactamente de la misma manera solo que corto acá miren si corto acá por qué debo cortar antes o aquí en la polea de entonces se me generan al cortar me quedan dos porciones me queda o bien esta porción cierto en la cual piense que trabajaría con la carga de 89 2 por la distancia que son estos 150 milímetros o bien tendrían que sacar la resultante de estas dos y por la distancia obviamente me queda más fácil está igual agua acá corto también acá fíjese entre bs trabajo con alguna de las dos secciones obviamente la más sencilla de trabajar ese es tan cierto y entonces el momento acá será la carga de 174 por esta distancia a los 150 milímetros y ahí ya tendré mis momentos bueno y ahí está calculado físico el momento en el plano de 500 dijimos la carga de 89,2 por 0 15 250 milímetros y el otro momento la carga de 174 por la distancia que es también 0 15 metros listo y nuevamente igual que lo que hicimos arriba el momento resultante como estos momento están en planos perpendiculares tendré que aplicar pitágoras y será la raíz cuadrada de 13.38 estos son newton-metro no esto al cuadrado más el otro momento perpendicular que es el de 26.16 newton-metro esto al cuadrado le repito esto por tratarse de cargas puntuales y entonces aquí si hago las operaciones me da que el momento en b tiene un valor de 29 38 newton-metro ahora de estos dos debo seleccionar el máximo entonces el máximo es el que yo encuentro acá en la polea ahora esos momentos hacen que mi eje se curve que lo llamamos que se flexiona la flexión genera esfuerzos normales normales quiere decir perpendiculares perpendiculares a quien a la sección transversal que dice que si es un eje su sección transversal es un círculo acá entonces en la sección transversal en una cara o la atención en la otra compresión debido al momento pero la mayor va a estar acá o la mayor cantidad esfuerzo va a estar acá sobre la pole aena y la dirección del esfuerzo normal es perpendicular a la sección o sea a lo largo del eje quiere decir en dirección del eje y luego entonces calculemos el esfuerzo normal a lo largo del eje x que recuerden que por momentos lector se calcula como m por c sobre y listo para el caso de un eje será el momento si es un eje recuerde que tiene que ser la distancia del eje neutro al punto más alejado en el eje o sea el radio entonces será m por el radio sobre el momento de inercia de un círculo pero recordemos que el radio es igual al diámetro sobre 2 porque aquí me piden calcular el diámetro antes de votar y trabajar en función del diámetro y no del radio y el momento de inercia un círculo espí cuartos radio a la 4 pero como voy a trabajar en función de el diámetro no del radio entonces será el diámetro sobre dos llama el radio y esther a la 4 si hacemos el desarrollo matemático tendríamos m sobre 1 sobre pico actos de sobre 2 sobre de a la 4 sobre 2 a las 4 16 si hacemos producto extremos y medios entonces multiplicaría 16 por 4 64 y esto por m y por ley entonces tendría 64 por md sobre en los medios quien quedaría 2 por 1 2 por 2 2 pide a la 4 pues aquí 2 p de a la 4 listo pero esto lo puedo simplificar porque 64 y en 232 entonces será 32 por el momento stade también puedo cancelar una de de arriba con una de las cuatro que tenemos abajo y entonces nos queda sobre pi por de a la 3 ahora como mi hija también está sometido a torsión porque piense que estos son tork es cierto estos son toques estos también son torpes entonces me está sometido a torsión la torsión genera esfuerzos cortantes que recordemos los esfuerzos cortantes los vamos a calcular aquí esta cortante tal de porsche sobre j así se calculan los esfuerzos cortantes en un eje pero como este c es el radio y yo no necesito trabajar en función del radio necesito trabajar en función del diámetro entonces tendré torque radio es diámetro sobre 2 sobre el jota que es el momento polar de un círculo que es para medios radio a la 4 pero el radio el diámetro sobre 2 / diámetro sobre 2 a la 4 y aquí tendré torque por diámetro sobre 2 y aquí tendré y medios diámetro a la 42 a la 416 hacemos la oreja nuevamente y entonces que tendré 16 por los 32 por torque por diámetro 32 por torque por diámetro sobre la oreja en medio 2 por people diámetro la 42 por people diámetro ahora 4 canceló 32 días en 2 16 veces el torque este diámetro cancela a uno de los cuatro abajo antes me cantares abajo antes me queda diámetro a la 3 y me quede el pi y podemos decir que para el caso de un eje el esfuerzo cortante que genera el torque lo podemos calcular como 10 veces el torque sobre mi diámetro al cubo y esta también se conoce como una fórmula para calcular el esfuerzo cortante de un eje en función del diámetro ahora lo que hacemos es reemplazar cierto entonces el 16 lo dejamos igual el torque el torque fíjese que sólo hay dos porque es el torque entrará y el torque de salida cualquiera de los dos dedos lo mismo y de ser opuesto no recuerden que aquí están la sumatoria y los torques michelle y recuerden que ése torque a 33 y el otro también da 33 porque hayamos t1 necesario para que 33 entonces aquí por 33 newton por metro sobre el diámetro del cubo nuevamente hago 16 por 33 y diop y eso me da 168 puntos 0 67 sobre el diámetro al cubo y ahí está el esfuerzo importante en función del diámetro entre ya tengo el esfuerzo normal y el esfuerzo cortante lo siguiente es si voy a utilizar la teoría en la energía de distorsión necesitaré calcular el esfuerzo dos meses recuerde que el esfuerzo normal de von mises se puede calcular de dos maneras o a partir de los esfuerzos principales o a partir de estos esfuerzos que acabamos de calcular que son los esfuerzos en cada punto entonces vamos a calcularlo a partir de estos esfuerzos para no tener que calcular esfuerzos principales y si es el caso del esfuerzo 2006 es el esfuerzo en x al cuadrado más el esfuerzo en el cuadrado pero no tenemos el forn sánchez solamente tenemos esfuerzo en x + 3 veces el cortante lado x este también al cuadrado entonces así vamos a calcular el esfuerzo dos meses a partir del esfuerzo normal y del esfuerzo cortante que encontramos luego si reemplazamos que nos queda el esfuerzo normal nos dio 500 96.8 94 sobre el diámetro a la 3 esto al cuadrado más tres veces el cortante que nos dio 168 puntos 0 67 sobre el diámetro a la 3 este también al cuadrado y ahí está si hago las operaciones de el numerador y luego esto al cuadrado me a este valor de a la 3a al cuadro me da de a la 6 acá l oeste 168 al cuadrado y el resultado por tres me a este valor y esto sobre de a las seis listo ahora como éste suma de fracciones si tiene el mismo denominador pues sencillamente ese es el mismo denominador y sumó los numeradores y esto me da 441.000 0 21.98 los humanos numeradores sobre de a las 6 ahora sacó raíz cuadrada el numerador sacó raíz cuadrada al denominador y entonces ahí ya tengo el esfuerzo dos meses en función del diámetro no ese esfuerzo 2006 en función del diámetro me dará no me dará me da 664 sobre el diámetro al cubo listo está el esfuerzo 2006 recuerden que el esfuerzo 2006 lo necesito para aplicar la ecuación de energía distorsión que dice que el factor de seguridad para un elemento frágil sometido a carga estática es igual al esfuerzo de influencia sobre el esfuerzo o misses y con esta información ya puedo calcular el diámetro porque sencillamente el esfuerzo de florencia y el factor de seguridad miren no van en el problema te recuerden aquí me dicen factor de seguridad 3 y como me dicen el material de recuerden que aquí hemos sacado el esfuerzo influencia del material 370 mega pascal es entonces si yo vengo y reemplazo en mi ecuación acá me va a quedar factor de civilidad 3 igual a esfuerzo influencia recordemos que esto es 370 politizarlas en newton sobre metros cuadrados o sea 350 70 megas pascal es sobre 1 la colocó sobre el esfuerzo 2006 que recuerde que nos dio en función del diámetro 664 sobre diámetro a la 3 listo hacemos producto extremos producto de medios y que vamos a obtener entonces vamos a obtener en el producto de visitaremos 370 por vencer a 6 por el diámetro al cubo y en el denominador 664 por 1 664 ahora si despejó el diámetro acá porque pasó al 664 multiplicará 3 y el 370 por utilizar las 6 a dividir entonces y saco raíz cúbica no porque este diámetro está a la 33 saco raíz cúbica ese diámetro me va a dar 0 0 175 metros que si lo pasó a milímetros vamos a tener aproximadamente 17 puntos 5 milímetros ese debe ser el diámetro de nuestro eje hemos calculado el diámetro del edificio y esto ha sido por energía de distorsión recuerden que estamos haciendo dos problemas el 527 que era ahí dice decía calculando lo por la energía de distorsión y ahora el 528 que lo vamos a calcular por el esfuerzo cortante máximo entonces aquí está ejercicio 528 recuerden que por la teoría del esfuerzo cortante máximo me dice que el factor de seguridad es el esfuerzo influencia sobre 2 cierto o 05 el esfuerzo afluencia como usted lo quieran trabajar sobre el cortante máximo donde el cortante máximo recuerde que es el radio del círculo de moore que se calcula como el sol son x menos el esfuerzo en 10 sobre 2 esto al cuadrado más el cortante quilla al cuadrado ahora como el esfuerzo en que es cero cierto entonces lo primero que voy a cortar a calcular es el esfuerzo cortante máximo entonces esfuerzo cortante máximo es igual a la raíz cuadrada de el esfuerzo en x 500 596 1500 96.8 estos son newton-metro me quedaría sobre dos y eso por el diámetro de la tres más el cortante recuerden que el constante nos dio 168 168 punto 1 sobre el diámetro de la 3 esto también va a que al cuadrado bueno alargamos esto acá y eso es el cortante máximo luego si dividido esto en dos y elevó esto al cuadrado entonces a 89.000 142.5 sobre de a la 6 no quería las tres salas desde a las 6 768 punto 1 al cuadrado media 28 mil 52 57.6 sobre d a la 6 también como tiene el mismo denominador entonces sumó los numeradores dejó quieto el denominador porque son fracciones homogéneas y entonces tendríamos que el cortante máximo va a ser igual a 117.300 punto 11 sobre diámetro de 6 sacó raíz cuadrada el numerador sacó raíz cuadrada el denominador entonces que tendría esto me da 340 y 2.5 esta raíz cuadrada sobre la raíz cuadrada diámetro de 6 diámetro la 3 ya tengo aquí el cortante máximo bueno como ya tengo el cortante máximo entre ya vengo y reemplazo acá en mi ecuación de la teoría del esfuerzo cortante máximo factor de seguridad es igual a entonces el esfuerzo el fluence 370 por 10 a las 6 mega pascal es cierto 10 a la señito sobre metro cuadrado que son los pascales y estos sobre el cortante máximo desde aquí colocamos sobre 340 y 2.5 sobre de ala 3 bueno lo de arriba nos toca colocarlo esta expresión sobre 1 para hacer productos extremos productos de medios y que nos quedara 3 igual a producto en los extremos me queda el 370 por 10 a las 6 bueno esto son hitos sobre metro cuadrado sobre este diámetro de la actriz esto sobre 340 y 2.5 y de acá dice que lo único que tengo que hacer es despejar este diámetro que está a la 3 no entonces éste que está dividiendo lo mandamos a multiplicar al 3 éste lo mandamos a dividir y sacamos raíz cúbica luego me queda que el diámetro es igual a la raíz cúbica de 3 por 340 y 2.5 sobre 370 por 10 a las 6 hacemos esa operación cuando yo saco raíz cúbica kimi a 17.7 por 10 a la menos tres metros que si lo pasamos a milímetros entonces nos da un diámetro aproximado de 17.7 y ahí tenemos nuevamente calculado el diámetro pero a partir de la teoría de esfuerzo cortante máximo ahora podemos hacer la comparación ese diámetro nos debe 17 7 y el otro diámetro unos 17 5 muy aproximados éste nos dio 2 milímetros más grande dos décimas de milímetro más grande porque este a más grande porque es que la teoría del esfuerzo cortante máximo es más conservadora entonces siempre me da valores mayores dimensiones mayores es más conservadora que la de la teoría de la energía distorsión de un mix el yankee y esto ha sido todo en nuestro vídeo los invito compartan este material a compañeros y amigos de like al vídeo recomienden el canal y en la descripción del vídeo les dejo el pdf de los dos ejercicios para que lo descarguen y lo tengan como material de estudio y los espero en el próximo vídeo haremos diseño de ejes pero por teorías de diseño para materiales frágiles |
6851 | https://www.wikihow.it/Calcolare-la-Resistenza-Totale-nei-Circuiti | Accedi
Come Calcolare la Resistenza Totale nei Circuiti
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Questo articolo è stato co-redatto da Marvin Woo. Marvin Woo è un elettricista abilitato e titolare della ditta Woo's Electrical & Appliance, con sede a East O'ahu (Hawaii). Con oltre 20 anni di esperienza, è specializzato in risoluzione di problemi e manutenzione di impianti elettrici residenziali. Possiede l'abilitazione e l'assicurazione necessaria per eseguire lavori elettrici nello stato delle Hawaii.
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Esistono due modi per collegare fra loro i componenti elettrici. I circuiti in serie sono composti da elementi collegati l'uno di seguito all'altro, mentre quelli in parallelo sono formati da pezzi disposti su rami paralleli. La maniera in cui le resistenze sono collegate determina come queste contribuiscono a formare la resistenza totale del circuito.
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Circuito in Serie
Circuito in Parallelo
Circuito Combinato
Formule che Utilizzano la Potenza
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6852 | https://rafamarino.com/posts/n-cubes/ | # Rafael Marino
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Elements of Hyperdimensional Cubes
Table of Contents
Summary
A review of the building blocks of lower-dimensional cubes and the functions that describe their transformations into higher dimensions. We’ll begin by working with points, segments, squares, and cubes. By observing these lower-level objects we’ll be able to derive generalized formulas valid for the n-cube. These equations will describe precisely all constituent elements of any hypercube.
Introduction
One way of describing the processes involved in going from lower to higher dimensions is through motion. When a point moves to another point, this movement traces out a segment in a new dimension. In turn, a segment sweeps out to create a square, a square a cube, a cube a tesseract (Figure 1), etc. It can go on indefinitely.
Figure 1. Point, segment, square, cube and tesseract1
Note that although the cube2 is tacitly understood as the 3-cube, I will also be making use of standard n-cube notation for all dimensions, both lower and higher. Thus an edge becomes a 1-cube, and a penteract a 5-cube. This is a covenient standardization scheme, specially considering that individualized names can become a bit unwieldy for more complex figures (see Table 1).
Table 1. Nomenclature—Zero to Six-Dimensional Cubes
| Dim | n-cube | Names |
---
| 0D | 0-cube | Point, Vertex |
| 1D | 1-cube | Segment, Edge, -gon |
| 2D | 2-cube | Square, Face, Tetragon |
| 3D | 3-cube | Cube, Cell, Hexahedron |
| 4D | 4-cube | Tesseract, 8-Cell, Tetracube, Octachoron |
| 5D | 5-cube | Penteract, Deca-5-tope, Decateron |
| 6D | 6-cube | Hexeract, Dodeca-6-tope, Dodecapeton |
The Building Blocks—Vertices & Edges
Let’s define some foundational concepts.
Point—the lowest level of abstraction in geometry, it represents location, with no extent3.
Vertex—a point where two or more edges meet; they are [at] the corners of a hypercube. All vertices are points, but not all points are vertices.
Edge—a segment connecting 2 vertices.
( V )—the set of all vertices.
( E )—the set of all edges.
As dimensionality ( (n) ) grows, ( V ) and ( E ) also grow, thereby reshaping the hypercube. Our goal is to understand and describe this transformation mathematically for all dimensions. Fortunately for us, these mechanics can be formulated explicitly at lower dimensions and then extrapolated into higher dimensions.
First, let’s look at vertices. To move to dimension ( n + 1 ) from dimension ( n ), the number of vertices doubles. Segments have 2 vertices, squares have 4, cubes have 8, and so on. That is to say, vertices grow at a rate of ( 2^n ). Consequently, each dimension has twice the number of vertices as its predecessor, ( |V|_{n} = 2|V|_{n-1} ).
Then we have edges. Edges can be thought of as the traces left behind by the shifting of vertices into a higher dimension (Figure 2). In other words, the vertices of dimension ( n ) travel to dimension ( n + 1 ), building a new set of edges in the process. Let’s find out how many exactly.
Figure 2. A Point Sweeps out into 3D Space4
The simplest way to derive the edge count in dimension ( n ) is probably by observing the following invariant: the number of edges in ( n ) is always ( n ), times the number of vertices in the previous dimension (which we already generalized)—that is, ( n \times 2^{n-1} ). Thus, 2-cubes have ( 2 \times 2^{1} = 4 ) edges, 3-cubes have ( 3 \times 2^{2} = 12 ) edges, and so on. Note how this formula depends exclusively on ( n ), unlike other methods5.
Table 2. Vertices and Edges for an N-cube
| ( n ) | Vertices | Edges | Squares |
--- --- |
| 0 (Point) | 1 | 0 | 0 |
| 1 (Edge) | 2 | 1 | 0 |
| 2 (Square) | 4 | 4 | 1 |
| 3 (Cube) | 8 | 12 | ? |
| … | … | … | … |
| n-cube | ( 2^n ) | ( n \times 2^{n-1} ) | ? |
We are now able to determine mechanically the vertex and edge counts of any n-cube (Table 2). That completes our understanding of the elements of the square, but there are still missing elements in all higher-dimensional cubes, ( n \ge 3 ). For example, we know that a cube also has square faces, and we have no formula for the counting of squares.
In general, any ( n )-cube is built by composing all its lower-level ( (n-k) )-cubes, with ( k \in { 1, …, n } ). It follows, by example, that a full accounting of the 7-cube requires knowing how many [0-6]-cubes it contains. But this structure seems a bit daunting to conceive, hence our next task is to devise a general formula that yields the precise count of k-cubes in an n-cube directly.
Generalizing to N Dimensions
By now we have repeated the shifting procedure a few times, and some patterns have begun to emerge. Namely, the way to construct the next dimensional cube is always by duplicating the cube in the current dimension and moving it perpendicularly to itself6. Due to the reflective nature of this procedure, the resulting higher-dimensional cubes are inherently and increasingly symmetric.
(…) grouping the faces of an object is particularly effective when the object possesses a great deal of symmetry, as does the hypercube. A segment possesses one symmetry, obtained by interchanging its endpoints. A square has a much larger number of symmetries: we can rotate the square into itself by one, two, or three quarter-turns about its center, and we can reflect the square across either of its diagonals, or across the horizontal or vertical lines through its center. The even larger group of symmetries of the cube enables us to move any vertex to any other vertex (…) The collection of symmetries is one of the most important examples of an algebraic structure known as a group . —Banchoff7
The key observation to obtain a general formulation begins with the vertex. All vertices of an n-cube are constructed identically, therefore an entire n-cube can be generated by looking at a single n-vertex.
For instance, in a tesseract each vertex projects 4 edges—which means that at each vertex there are as many 3-cubes as ways of choosing subsets of 3 edges out of 4. Mathematically, this is expressed by the binomial coefficient , ( nCk = 4C3 = 4 ) ways per vertex. Multiplied by 16 vertices in the tesseract for a total of 64. But each cube uses only 8 vertices, so we divide 64 by 8 to arrive at the fact that a tesseract contains 8 cubes (Figure 3).
Figure 3. A 2D Representation of a Tesseract’s Eight 3D-Cubes8
Note. There are eight cubes in a tesseract---this explains the Greek octa-choron, or "eight rooms". Of the eight, two are accounted for by the displaced copies (blue), and the other six by shifting the six squares of the original copy (orange).
Let’s generalize. In dimension ( n ) each vertex projects ( n ) edges. At each vertex we can determine in how many ways we can choose bundles of ( k ) out of these ( n ) edges (( nCk )). Then, in order to create k-dimensional cubes, this coefficient is multiplied by the number of vertices in the ( n )-cube ( (2^n) ) and divided by the number of vertices in the ( k )-cube ( (2^k) ); thereby obtaining the quantity of k-cubes contained in an n-cube ( (Q) ). After subtracting the powers we get:
[ Q(k, n) = nCk \times 2^{(n-k)} \hspace{1cm} (1) ]
Turns out, (1) is the only equation we will need to take full inventory of all elements of a hypercube. In Table 3, we used it to crunch the numbers up to the 9-cube. The formulas we derived before, for vertices and edges, are actually two special cases of (1)9. Note also two intuitive notions confirmed by the table: a) any n-cube is self-contained (diagonal of ones); b) any n-cube contains zero ( (n + k) )-cubes, for ( k > 0 ).
Table 3. Statistics of All Elements in 0-9D Cubes
| k=0 | k=1 | k=2 | k=3 | k=4 | k=5 | k=6 | k=7 | k=8 | k=9 |
--- --- --- --- --- |
| n=0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| n=1 | 2 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| n=2 | 4 | 4 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
| n=3 | 8 | 12 | 6 | 1 | 0 | 0 | 0 | 0 | 0 | 0 |
| n=4 | 16 | 32 | 24 | 8 | 1 | 0 | 0 | 0 | 0 | 0 |
| n=5 | 32 | 80 | 80 | 40 | 10 | 1 | 0 | 0 | 0 | 0 |
| n=6 | 64 | 192 | 240 | 160 | 60 | 12 | 1 | 0 | 0 | 0 |
| n=7 | 128 | 448 | 672 | 560 | 280 | 84 | 14 | 1 | 0 | 0 |
| n=8 | 256 | 1024 | 1792 | 1792 | 1120 | 448 | 112 | 16 | 1 | 0 |
| n=9 | 512 | 2304 | 4608 | 5376 | 4032 | 2016 | 672 | 144 | 18 | 1 |
The N-Cube In Code
The code below implements the N_cube class in Python. The method named all_elements_matrix was used to generate the contents of Table 3.
import numpy as np from math import comb class N_cube: """Compute the counts of all lower-dimensional elements of an n-cube""" def kcubes_in_ncube(self, k = 2, n = 3): """Return the number of k-cubes in an n-cube.""" k, n = int(k), int(n) k_cubes = 2 (n - k) comb(n, k) return int(k_cubes) def all_elements_matrix(self, n = 5): """Compute the counts of all (n-k)-cubes""" ks, ns = np. arange(n), np. arange(n) n_statistics = [self. kcubes_in_ncube(k, n) for k in ks for n in ns] n_statistics = np. array(n_statistics) n_statistics = n_statistics. reshape(n, n, order = "F") return n_statistics def elements(self, n): """Return element counts for a single n-cube""" elements = self. all_elements_matrix(n + 1) n_cube_elements = elements[n] return n_cube_elements
```
instantiate and call n_cube = N_cube() n_cube. kcubes_in_ncube(k = 2, n = 3) # 6 squares in a cube n_cube. all_elements_matrix(10) # see Table 3 n_cube. elements(3) # array([ 8, 12, 6, 1])
```
Notes
Wikipedia – Tesseract . By user Nerd‐Boy1392 (Creative Commons BY-SA 3.0) ↩︎
From French cube (13c.) and directly from Latin cubus, from Greek kybos “a six-sided die,” used metaphorically of dice-like blocks of any sort (…) – Etymology.com ↩︎
“That which has no part” – Euclid ↩︎
Physics Insights.org – Hypercubes ↩︎
For example, here’s another way to derive the edge count that depends on more than ( n ): we know that as we shift into the next dimension, ( n + 1 ), ( |V| ) exactly doubles; necessarily, ( |E| ) also doubles. So now we have a lower bound of ( 2|E|_{n} ) and we still need to add the new edges—the traces that connect these two dimensions. This number is identical to the number of vertices they need to connect, i.e. ( |V|_{n} ). Therefore, the total number of edges ( |E|_{n+1} ), could also be determined by the formula: ( 2|E|_{n} + |V|_{n} ). However, this formula is recursive on ( |E| ) which makes it less convenient if the number of edges in ( n ) is unknown. ↩︎
For instance, from 2D to 3D, after the square is duplicated we have 2 squares, plus the 4 squares outlined by the shifting of the copy, for a total of 6 squares (this is why the cube is also called hexa-hedron, or “six faces”). ↩︎
Banchoff – Beyond3D – Counting the Faces of Higher-Dimensional Cubes ↩︎
Instructable Workshop – Tesseract Model .—Color scheme my own. ↩︎
For vertices, ( 2^{(n-k)} \times nCk = 2^{(n-0)} \times nC0 = 2^n \times 1 = 2^n) and. For edges, ( 2^{(n-1)} \times nC1 = 2^{(n-1)} \times n ). ↩︎ |
6853 | https://math.stackexchange.com/questions/944353/what-is-the-solution-of-m-in-the-following-equation | linear algebra - what is the solution of m in the following equation? - Mathematics Stack Exchange
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what is the solution of m in the following equation?
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I have a equation (−1)2 m(−1)2 m=1 1 so for this m m has to be integer but if i write it as ((−1)2)m((−1)2)m=1 , m m can be rational as well. so which solution is right and why? thanks for your help
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asked Sep 24, 2014 at 13:47
avz2611avz2611
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Usually m m denotes an integer. Are you looking for rational solutions?Angelo Rendina –Angelo Rendina 2014-09-24 13:51:14 +00:00 Commented Sep 24, 2014 at 13:51
well if i take 2 inside all real values of m are accepted but thats not the case when 2 is not taken inside , which feels a bit contradictory to me avz2611 –avz2611 2014-09-24 13:53:11 +00:00 Commented Sep 24, 2014 at 13:53
It's the old mess due to radicals. If you bring 2 inside you are changing equation.Angelo Rendina –Angelo Rendina 2014-09-24 13:55:17 +00:00 Commented Sep 24, 2014 at 13:55
Intuitively, (−1)(−1) raised to any even power is 1 1, so we have all integers as a solution to m m.Daccache –Daccache 2014-09-24 13:55:35 +00:00 Commented Sep 24, 2014 at 13:55
1 It is. The problem is exactly that you are squaring the −1−1.Angelo Rendina –Angelo Rendina 2014-09-24 14:04:55 +00:00 Commented Sep 24, 2014 at 14:04
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As a start consider m=1 2 m=1 2 - then
(−1)2 m=(−1)1=−1≠1(−1)2 m=(−1)1=−1≠1
In general, we do not allow rational powers of negative numbers for this very reason - if x x is negative and a,b∈Q a,b∈Q then in general
x a b≠(x a)b x a b≠(x a)b
Some textbooks allow rational powers of negative numbers if the denominator is odd - in this case you are correct, and any m∈Q m∈Q with odd denominator will be a solution.
The occurs because we define x 1 2 x 1 2 to be the positive values y y satisfying y 2=x y 2=x. But −y−y will also satisfy this equation. For example,
x 2−−√=x x 2=x
is only true if x x is non-negative.
In addition, it is not immediately clear how to define something like (−1)1 4(−1)1 4. It could be any of ±1 2–√(1±i)±1 2(1±i) and there is no obvious way to choose which of these we should assign to be (−1)1 4(−1)1 4.
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edited Sep 24, 2014 at 14:03
answered Sep 24, 2014 at 13:53
Mathmo123Mathmo123
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oh well i did not know that , but why does this happen ?avz2611 –avz2611 2014-09-24 13:56:01 +00:00 Commented Sep 24, 2014 at 13:56
In short because (−1)1 2(−1)1 2 is not a real number. For a similar problem, you would expect x 2−−√x 2 to be x x, but this is not true if x x is negative Mathmo123 –Mathmo123 2014-09-24 14:01:29 +00:00 Commented Sep 24, 2014 at 14:01
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Obviously, 2 m 2 m must be an exponent that you accept as valid for a negative base.
Irrational numbers are ruled out.
The case of integers depends on the parity. Let use denote as e e an even integer and o o an odd one: (−1)e=1(−1)e=1 and (−1)o=−1(−1)o=−1.
The case of integer inverses is clear as well: (−1)1/e(−1)1/e is not defined and (−1)1/o=−1(−1)1/o=−1.
The case of rational needs more care as (−1)p/q(−1)p/q could be interpreted both as ((−1)p)1/q((−1)p)1/q or ((−1)1/q)p((−1)1/q)p.
Anyway,
((−1)o)1/o′=(−1)1/o′=−1((−1)o)1/o′=(−1)1/o′=−1 and ((−1)1/o′)o=(−1)o=−1((−1)1/o′)o=(−1)o=−1, so (−1)o/o′=−1(−1)o/o′=−1,
((−1)e)1/o=1 1/o=1((−1)e)1/o=1 1/o=1 and ((−1)1/o)e=(−1)e=1((−1)1/o)e=(−1)e=1, so (−1)e/o=1(−1)e/o=1,
((−1)o)1/e=(−1)1/e((−1)o)1/e=(−1)1/e and ((−1)1/e)o((−1)1/e)o are both undefined and so is (−1)o/e(−1)o/e.
In this sense, raising a negative to a rational power is a well defined operation.
Now it makes sense to accept
2 m=e o,2 m=e o,
or
m=i o,m=i o,
where i i is any integer.
As has been said by others, we may not use ((−1)a)b)=(−1)a b((−1)a)b)=(−1)a b, so that (−1)2 m=1(−1)2 m=1 and ((−1)2)m=1((−1)2)m=1 are two different equations, with different solutions.
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edited Sep 24, 2014 at 15:19
answered Sep 24, 2014 at 14:20
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6854 | https://math.libretexts.org/Bookshelves/PreAlgebra/Prealgebra_1e_(OpenStax)/08%3A_Solving_Linear_Equations/8.04%3A_Solve_Equations_with_Variables_and_Constants_on_Both_Sides_(Part_1) | 8.4.1
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8.4: Solve Equations with Variables and Constants on Both Sides (Part 1)
Last updated
: May 1, 2022
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8.3: Solve Equations Using the Division and Multiplication Properties of Equality
8.5: Solve Equations with Variables and Constants on Both Sides (Part 2)
Page ID
: 5021
OpenStax
OpenStax
( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
Solve an equation with constants on both sides
Solve an equation with variables on both sides
Solve an equation with variables and constants on both sides
Solve equations using a general strategy
be prepared!
Before you get started, take this readiness quiz.
Simplify: 4y − 9 + 9. If you missed this problem, review Example 2.3.10.
Solve: y + 12 = 16. If you missed this problem, review Example 2.5.4.
Solve: −3y = 63. If you missed this problem, review Example 3.9.6.
Solve an Equation with Constants on Both Sides
You may have noticed that in all the equations we have solved so far, all the variable terms were on only one side of the equation with the constants on the other side. This does not happen all the time—so now we’ll see how to solve equations where the variable terms and/or constant terms are on both sides of the equation.
Our strategy will involve choosing one side of the equation to be the variable side, and the other side of the equation to be the constant side. Then, we will use the Subtraction and Addition Properties of Equality, step by step, to get all the variable terms together on one side of the equation and the constant terms together on the other side.
By doing this, we will transform the equation that started with variables and constants on both sides into the form ax = b. We already know how to solve equations of this form by using the Division or Multiplication Properties of Equality.
Example 8.4.18.4.1:
Solve: 4x + 6 = −14.
Solution
In this equation, the variable is only on the left side. It makes sense to call the left side the variable side. Therefore, the right side will be the constant side. We’ll write the labels above the equation to help us remember what goes where.
| | |
--- |
| Since the left side is the variable side, the 6 is out of place. We must "undo" adding 6 by subtracting 6, and to keep the equality we must subtract 6 from both sides. Use the Subtraction Property of Equality. | 4x+6−6=−14−6 4x+6−6=−14−6(8.4.1) |
| Simplify. | 4x=−20 4x=−20(8.4.2) |
| Now all the x's are on the left and the constant on the right. | |
| Use the Division Property of Equality. | 4x4=−204 4x4=−204(8.4.3) |
| Simplify. | x=−5 x=−5(8.4.4) |
| Check: Let x = −5. | 4x+6=−144(−5)+6=−14−20+6=−14−14=−14✓ 4x+64(−5)+6−20+6−14=−14=−14=−14=−14✓ |
Exercise 8.4.18.4.1:
Solve: 3x + 4 = −8.
Answer
: x = -4
Exercise 8.4.28.4.2:
Solve: 5a + 3 = −37.
Answer
: a = -8
Example 8.4.28.4.2:
Solve: 2y − 7 = 15.
Solution
Notice that the variable is only on the left side of the equation, so this will be the variable side and the right side will be the constant side. Since the left side is the variable side, the 7 is out of place. It is subtracted from the 2y, so to ‘undo’ subtraction, add 7 to both sides.
| | |
--- |
| Add 7 to both sides. | 2y−7+7=15+7 2y−7+7=15+7(8.4.5) |
| Simplify. | 2y=22 2y=22(8.4.6) |
| The variables are now on one side and the constants on the other. | |
| Divide both sides by 2. | 2y2=222 2y2=222(8.4.7) |
| Simplify. | y=11 y=11(8.4.8) |
| Check: Substitute: y = 11. | 2y−7=152⋅11−7?=1522−7?=1515=15✓ 2y−72⋅11−722−715=15=?15=?15=15✓ |
Exercise 8.4.38.4.3:
Solve: 5y − 9 = 16.
Answer
: y = 5
Exercise 8.4.48.4.4:
Solve: 3m − 8 = 19.
Answer
: m = 9
Solve an Equation with Variables on Both Sides
What if there are variables on both sides of the equation? We will start like we did above—choosing a variable side and a constant side, and then use the Subtraction and Addition Properties of Equality to collect all variables on one side and all constants on the other side. Remember, what you do to the left side of the equation, you must do to the right side too.
Example 8.4.38.4.3:
Solve: 5x = 4x + 7.
Solution
Here the variable, x, is on both sides, but the constants appear only on the right side, so let’s make the right side the “constant” side. Then the left side will be the “variable” side.
| | |
--- |
| We don't want any variables on the right, so subtract the 4x. | 5x−4x=4x−4x+7 5x−4x=4x−4x+7(8.4.9) |
| Simplify. | x=7 x=7(8.4.10) |
| We have all the variables on one side and the constants on the other. We have solved the equation. | |
| Check: Substitute 7 for x. | 5x=4x+75(7)?=4(7)+735?=28+735=35✓ 5x5(7)3535=4x+7=?4(7)+7=?28+7=35✓ |
Exercise 8.4.58.4.5:
Solve: 6n = 5n + 10.
Answer
: n = 10
Exercise 8.4.68.4.6:
Solve: −6c = −7c + 1.
Answer
: c = 1
Example 8.4.48.4.4:
Solve: 5y − 8 = 7y.
Solution
The only constant, −8, is on the left side of the equation and variable, y, is on both sides. Let’s leave the constant on the left and collect the variables to the right.
| | |
--- |
| Subtract 5y from both sides. | 5y−5y−8=7y−5y 5y−5y−8=7y−5y(8.4.11) |
| Simplify. | −8=2y −8=2y(8.4.12) |
| We have the variables on the right and the constants on the left. Divide both sides by 2. | −82=2y2 −82=2y2(8.4.13) |
| Simplify. | −4=y −4=y(8.4.14) |
| Rewrite with the variable on the left. | y=−4 y=−4(8.4.15) |
| Check: Let y = −4. | 5y−8=7y5(−4)−8?=7(−4)−20−8?=−28−28=−28✓ 5y−85(−4)−8−20−8−28=7y=?7(−4)=?−28=−28✓ |
Exercise 8.4.78.4.7:
Solve: 3p − 14 = 5p.
Answer
: p = -7
Exercise 8.4.88.4.8:
Solve: 8m + 9 = 5m.
Answer
: m = -3
Example 8.4.58.4.5:
Solve: 7x = − x + 24.
Solution
The only constant, 24, is on the right, so let the left side be the variable side.
| | |
--- |
| Remove the −x from the right side by adding x to both sides. | 7x+x=−x+x+24 7x+x=−x+x+24(8.4.16) |
| Simplify. | 8x=24 8x=24(8.4.17) |
| All the variables are on the left and the constants are on the right. Divide both sides by 8. | 8x8=248 8x8=248(8.4.18) |
| Simplify. | x=3 x=3(8.4.19) |
| Check: Substitute x = 3. | 7x=−x+247(3)?=−(3)+2421=21✓ 7x7(3)21=−x+24=?−(3)+24=21✓ |
Exercise 8.4.98.4.9:
Solve: 12j = −4j + 32.
Answer
: j = 2
Exercise 8.4.108.4.10:
Solve: 8h = −4h + 12.
Answer
: h = 1
Solve Equations with Variables and Constants on Both Sides
The next example will be the first to have variables and constants on both sides of the equation. As we did before, we’ll collect the variable terms to one side and the constants to the other side.
Example 8.4.68.4.6:
Solve: 7x + 5 = 6x + 2.
Solution
Start by choosing which side will be the variable side and which side will be the constant side. The variable terms are 7x and 6x. Since 7 is greater than 6, make the left side the variable side and so the right side will be the constant side.
| | |
--- |
| Collect the variable terms to the left side by subtracting 6x from both sides. | 7x−6x+5=6x−6x+2 7x−6x+5=6x−6x+2(8.4.20) |
| Simplify. | x+5=2 x+5=2(8.4.21) |
| Now, collect the constants to the right side by subtracting 5 from both sides. | x+5−5=2−5 x+5−5=2−5(8.4.22) |
| Simplify. | x=−3 x=−3(8.4.23) |
| The solution is x = −3. | |
| Check: Let x = −3. | 7x+5=6x+27(−3)+5?=6(−3)+2−21+5?=−18+2−16=−16✓ 7x+57(−3)+5−21+5−16=6x+2=?6(−3)+2=?−18+2=−16✓ |
Exercise 8.4.118.4.11:
Solve: 12x + 8 = 6x + 2.
Answer
: x = -1
Exercise 8.4.128.4.12:
Solve: 9y + 4 = 7y + 12.
Answer
: y = 4
We’ll summarize the steps we took so you can easily refer to them.
HOW TO: SOLVE AN EQUATION WITH VARIABLES AND CONSTANTS ON BOTH SIDES
Step 1. Choose one side to be the variable side and then the other will be the constant side.
Step 2. Collect the variable terms to the variable side, using the Addition or Subtraction Property of Equality.
Step 3. Collect the constants to the other side, using the Addition or Subtraction Property of Equality.
Step 4. Make the coefficient of the variable 1, using the Multiplication or Division Property of Equality.
Step 5. Check the solution by substituting it into the original equation.
It is a good idea to make the variable side the one in which the variable has the larger coefficient. This usually makes the arithmetic easier.
Example 8.4.78.4.7:
Solve: 6n − 2 = −3n + 7.
Solution
We have 6n on the left and −3n on the right. Since 6 > − 3, make the left side the “variable” side.
| | |
--- |
| We don't want variables on the right side—add 3n to both sides to leave only constants on the right. | 6n+3n−2=−3n+3n+7 6n+3n−2=−3n+3n+7(8.4.24) |
| Combine like terms. | 9n−2=7 9n−2=7(8.4.25) |
| We don't want any constants on the left side, so add 2 to both sides. | 9n−2+2=7+2 9n−2+2=7+2(8.4.26) |
| Simplify. | 9n=9 9n=9(8.4.27) |
| The variable term is on the left and the constant term is on the right. To get the coefficient of n to be one, divide both sides by 9. | 9n9=99 9n9=99(8.4.28) |
| Simplify. | n=1 n=1(8.4.29) |
| Check: Substitute 1 for n. | 6n−2=−3n+76(1)−2?=+74=4✓ 6n−26(1)−24=−3n+7=?+7=4✓ |
Exercise 8.4.138.4.13:
Solve: 8q − 5 = −4q + 7.
Answer
: q = 1
Exercise 8.4.148.4.14:
Solve: 7n − 3 = n + 3.
Answer
: n = 1
Example 8.4.88.4.8:
Solve: 2a − 7 = 5a + 8.
Solution
This equation has 2a on the left and 5a on the right. Since 5 > 2, make the right side the variable side and the left side the constant side.
| | |
--- |
| Subtract 2a from both sides to remove the variable term from the left. | 2a−2a−7=5a−2a+8 2a−2a−7=5a−2a+8(8.4.30) |
| Combine like terms. | −7=3a+8 −7=3a+8(8.4.31) |
| Subtract 8 from both sides to remove the constant from the right. | −7−8=3a+8−8 −7−8=3a+8−8(8.4.32) |
| Simplify. | −15=3a −15=3a(8.4.33) |
| Divide both sides by 3 to make 1 the coefficient of a. | −153=3a3 −153=3a3(8.4.34) |
| Simplify. | −5=a −5=a(8.4.35) |
| Check: Let a = −5. | 2a−7=5a+82(−5)−7?=5(−5)+8−10−7?=−25+8−17=−17✓ 2a−72(−5)−7−10−7−17=5a+8=?5(−5)+8=?−25+8=−17✓ |
Note that we could have made the left side the variable side instead of the right side, but it would have led to a negative coefficient on the variable term. While we could work with the negative, there is less chance of error when working with positives. The strategy outlined above helps avoid the negatives!
Exercise 8.4.158.4.15:
Solve: 2a − 2 = 6a + 18.
Answer
: a = -5
Exercise 8.4.168.4.16:
Solve: 4k − 1 = 7k + 17.
Answer
: k = -6
To solve an equation with fractions, we still follow the same steps to get the solution.
Example 8.4.98.4.9:
Solve: 3232x + 5 = 1212x − 3.
Solution
Since 32>1232>12, make the left side the variable side and the right side the constant side.
| | |
--- |
| Subtract 1212x from both sides. | 32x−12x+5=12x12x−3 32x−12x+5=12x12x−3(8.4.36) |
| Combine like terms. | x+5=−3 x+5=−3(8.4.37) |
| Subtract 5 from both sides. | x+5−5=−3−5 x+5−5=−3−5(8.4.38) |
| Simplify. | x=−8 x=−8(8.4.39) |
| Check: Let x = −8. | 32x+5=12x−332(−8)+5?=12(−8)−3−12+5?=−4−3−7=−7✓ 32x+532(−8)+5−12+5−7=12x−3=?12(−8)−3=?−4−3=−7✓ |
Exercise 8.4.178.4.17:
Solve: 7878x - 12 = −18−18x − 2.
Answer
: x = 10
Exercise 8.4.188.4.18:
Solve: 7676y + 11 = 1616y + 8.
Answer
: y = -3
We follow the same steps when the equation has decimals, too.
Example 8.4.108.4.10:
Solve: 3.4x + 4 = 1.6x − 5.
Solution
Since 3.4 > 1.6, make the left side the variable side and the right side the constant side.
| | |
--- |
| Subtract 1.6x from both sides. | 3.4x−1.6x+4=1.6x−1.6x−5 3.4x−1.6x+4=1.6x−1.6x−5(8.4.40) |
| Combine like terms. | 1.8x+4=−5 1.8x+4=−5(8.4.41) |
| Subtract 4 from both sides. | 1.8x+4−4=−5−4 1.8x+4−4=−5−4(8.4.42) |
| Simplify. | 1.8x=−9 1.8x=−9(8.4.43) |
| Use the Division Property of Equality. | 1.8x1.8=−91.8 1.8x1.8=−91.8(8.4.44) |
| Simplify. | x=−5 x=−5(8.4.45) |
| Check: Let x = −5. | 3.4x+4=1.6x−53.4(−5)+4?=1.6(−5)−5−17+4?=−8−5−13=−13✓ 3.4x+43.4(−5)+4−17+4−13=1.6x−5=?1.6(−5)−5=?−8−5=−13✓ |
Exercise 8.4.198.4.19:
Solve: 2.8x + 12 = −1.4x − 9.
Answer
: x = -5
Exercise 8.4.20:
Solve: 3.6y + 8 = 1.2y − 4.
Answer
: y = -5
Contributors and Attributions
Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Commons Attribution License v4.0 "Download for free at
8.3: Solve Equations Using the Division and Multiplication Properties of Equality
8.5: Solve Equations with Variables and Constants on Both Sides (Part 2) |
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This algebra video tutorial explains how to find the slope and y-intercept of a line in slope intercept form and in standard form.
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Writing Equations Given a Point & Slope:
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Transcript:
in this video we're going to talk about how to find the slope and Y intercept from a linear equation now it's important to be familiar with the slope intercept form of a linear equation and here it is M represents the slope m is a number in front of x b represents the y intercept so if we were to write the slope intercept equation right below this one we can clearly see that M which is the number front of X that's two and B is the constant uh that's three so in this problem the slope which represents m is 2 and the Y intercept which is equal to B that's three so that's a quick and simple way of how you could find the slope and a y intercept from the equation however not all examples are simple as this so we're going to cover a lot of examples in this video go ahead and try these two examples let's say Y is equal to 34 x - 5 and also Y is = 8 - 4x so feel free to pause the video and try those examples if you want to so for number two we can see that the slope is just a number in front of X so the slope is going to be 3 over 4 and the Y intercept is the number other than this it's the constant without the X so it's -5 now for the next example it's not written in slope intercept form but we can change that if we were to switch to 8 with the -4x and write it this way we get -4x + 8 keep in mind there's a plus in front of the eight you just don't see it immediately so writing it this way we can see that the slope is -4 and the Y intercept is 8 if it's not given to you in slope intercept form you need to adjust the equation and put it in slope intercept form once in slope intercept form you could find the slope and the Y intercept go ahead and try these two examples let's say we have y is equal to 5 - x and also Y is equal to -7x so for the fourth example I'm going to reverse or switch to five in the negative X so this ISX + 5 now you might be wondering what is the slope here we don't really have a number if you don't see a number there this assume it's always a one so this is --1x + 5 that's the same ASX + 5 therefore the slope for this line is -1 and the Y intercept is five so that's it for number four now what about number five all we have is just a -7x it doesn't appear to have any B value in a situation like this you can rewrite this equation as follows so Y is equal to -7x + 0 -7x + 0 is the same as -7x and writing it this way you can clearly see that the slope is -7 but the Y intercept is is zero so that's all you need to do for that particular example now what about these two let's say if we have y is equal to 3 and X is equal to 4 what is the slope and Y intercept for these two lines I want you to think about that for a moment now for y equal 3 we can rewrite this um like this we can say it's Y is equal to 0x + 3 so this is in slope intercept form 0x + 3 is equal to 3 these two have the same value so we can clearly see that m is zero and B is three so for this problem the slope is zero and the Y intercept is three now what about for xal 4 because you can't really put that in slope intercept form so what do you do to get the answer there for xal 4 you need to graph it why equal three this is a horizontal line at three horizontal lines they have a slope of zero because they don't go up they don't go down they're just flat and you could see the horizontal line it touches the Y AIS this is the x- axis this is the y- axis it touches the y- AIS at three so that's why the Y intercept is three because that's where it touches the y- axis now for a vertical line x = 4 the situation is different so this is how we can graph xals 4 it looks like this for vertical lines there is no slope if you try to calculate the slope of a vertical line it's going to be undefined if you're wondering how you can calculate the slope of a horizontal or vertical line you could use this formula m is equal to Y2 - y1 / X2 - X1 so for the horizontal line if you pick any two points let's say this is y1 Y2 in both cases y1 and Y2 they'll have a value of three regardless of what x 1 and X2 is and so it's always going to be zero all the Y values on a horizontal line will have the same y value of three because Y is three it's always three anywhere on this horizontal line so this is always going to be 3 - 3 for this line giving you a slope of zero well for a vertical line X is always the same so these two points may have different V y values this may be y1 Y 2 but the x value is always four and so if you were to calculate the slope using this formula it will be Y2 - y1 but on the bottom X2 and X1 they're going to be four and so you're going to get a zero on the bottom and whenever you have a zero in the denominator of a fraction it's going to be undefined so for a vertical line the slope is always undefined now what about the Y intercept now this vertical line will it ever touch the y- axis the answer is no because it's parallel to the y- axis so it will never touch the y- axis therefore for the Y intercept there's none there is no y intercept so whenever you see an equation like this x equals some number the slope is going to be undefined and the Y intercept is just going to be nothing there's no Y intercept by the way for those of you who want access to more video related content feel free to check out the links in the description if you click on this more button you're going to see other videos relating to the video that you're currently watching and these links are separated by chapter and of course you could check out my website video.net where you'll get access to my video playlists final exam videos and also test prep videos so feel free to take a look at that when you get a chance so let's say if we have y = -6 and x = -2 so whenever you have y is equal to some number whether it's a positive or negative number the Y intercept is going to be whatever y equals and the slope is always going to be zero for this kind of line when you whenever you see x equals a number whether it's positive or negative the slope is going to be nothing and I mean the Y intercept is nothing but the slope is going to be undefined it's always going to be that way based on the explanation U that I gave you you for the last two examples now what would you do if you get a linear equation in this form so right now this linear equation is in standard form and in order to determine the slope and the Y intercept we need to put it in slope intercept form yal MX plus b so the way to convert it from standard form to slope intercept form is you need to solve for y in other words you need to get y by itself on one side of the equation so first we need to get rid of this two in front of Y and this term we we need to move it to the other side so what I want to do is get rid of the two first I'm going to divide everything by two so -4 / 2 is -2 2 / 2 is 1 so 1 Y is the same as just WR in y and 6 / 2 is three so we have this now I'm going to take this term move it to the other side on the left side it's -2X when I move it to the other side it will be positive 2x the other way to do this is to add 2x to both sides if you add 2x to both sides this will cancel here and we'll get Y is = 2x + 3 as you can see if you simply move it it changes from -2X on the left to positive 2x on the right and we can see that the slope is 2 and the Y intercept is three so that's it for this example let's try a somewhat similar problem let's say we have 3x - 5 Y is equal 8 go ahead and find a slope and a y intercept for this line so first I want to get rid of the number in front of Y so I'm going to divide both sides by5 so I'm going to have -3 over 5x these two will cancel so I'm just going to get + y5 /5 is POS 1 and then this equals -8 over 5 now I'm going to take this term move it to the other side so I'm going to get Y is equal to it's negative on the left but it's going to be positive on the right so POS 3 over 5x and then minus 8 over 5 so this is my M value m is a number in front of x m is 3 over 5 and this is my B value B is going to be -8 over5 so now I have the slope and the Y intercept now let's work on one more example let's say I have -2 7x + 3 7 Y is equal to 4 s so what should we do to convert this standard equation into its slope intercept form what would you do now just like before I want to get rid of the 3 over 7 in front of the Y but if I divide everything by 2 7 that can make the math a bit complicated so because I have fractions and I want to get rid of the fractions I'm going to multiply everything by seven doing this all the sevens will cancel with this seven so I'm going to get -2x + 3 Y is = to 4 next I'm going to get rid of the 3 in front of the Y by dividing everything by 3 so I'm going to get -2 over 3x 3 over 3 is 1 so this is just going to be + y is equal to 43 then I'm going to take this term move it to the other side and I'll get Y is equal to positive 2 over 3x + 43 and so I could see that the slope is POS 2 over3 and the Y intercept is 4 over 3 so that's how you could find the slope and the Y intercept from a linear equation all you have to do is convert it to slope intercept form and identify your m and B values so that's it for this video uh thanks for watching for those of you who want more videos on algebra linear equations feel free to check out the links in the description section below thanks for watching |
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Detail of Bayeux Tapestry: William at the battle of Hastings. Image by Myrabella, 2013, licensed under Creative Commons.
egregious
June 15, 2022
A flock of white sheep with a single black sheep standing in their midst
15 June 2022
The adjective egregious is often used in negative contexts, meaning flagrant, outrageous, or offensive. But it can also have a positive connotation, meaning exceptional, eminent, or especially worthy. The etymology is quite straightforward; it’s a direct borrowing from Latin, but the etymology is of interest because of the meaning of its root and when and how it was borrowed into English.
The Latin egregius means distinguished or excellent; it has no negative connotation—that connotation would develop in English usage. The root greg means flock, as in sheep or goats, and is also the root of gregarious, which literally means associated with the flock. Egregious, on the other hand, literally means outside of or distinguished from the flock.
Egregious is also an example of what is called an inkhorn term. These were words imported into English from Latin or Greek in the Early Modern Era, that is the sixteenth and early seventeenth centuries. Inkhorn words were learned terms, associated with the classroom and ostentatious displays of education and erudition. As a result, most never entered common usage. Some, however, did pass into common use, and egregious is one of these.
The word appears in English in the mid sixteenth century in a translation of Polydore Vergil’s history of England. Here it is used in a passage about the Roman Emperor Constantine, who prior to becoming Emperor had led military campaigns in Britain:
This manne, as we have seyde beefore, after hee hadd geven the overthrowe to Maxentius and seased Italie into his handdes, proceaded to Rome, unto whome shortelie repaired Sylvester Bisshoppe, of singuler and ægregius holliness, and with facilitee perswaded himme to deserve well of the Christian religion, whoe of his owne accorde all readie hadd good affiaunce therein; farthermore, beefore that he went to Rome (as it is crediblie thought) hee was soe instructed of his own moother Helena, that goinge towards battayle he used the sygne of the crosse as a defence.
In the above passage egregious is being used in its positive sense, but the negative sense appears at about the same time. Here is an example of the adverb egregiously being used negatively in a 1553 translation of one of John Calvin’s homilies:
Be cause thei wold not seme to fight agai[n]st god w[ith]out swerd or buckler / thei bring and obiect the autoritie of this or [that] ma[n]n / as though the absolution of any one manne may exempt and deliv[er] them [that] thei be not co[n]demned of god. I wil not saie [that] thei lye egregiously / whe[n] thei alege soch men as thei do for the defence of their cause.
And here is an example of the word’s early use from Christopher Marlowe’s 1590 play Tamburlaine:
Egregious Viceroyes of these Eastern parts
Plac'd by the issue of great Baiazeth:
And sacred Lord the mighty Calapine:
Who liues in Egypt, prisoner to that slaue,
Which kept his father in an yron cage:
Now haue we martcht from faire Natolia
So, egregious is egregious in that the ovine etymology is not obvious to most English speakers today and in that it is an inkhorn term that managed to survive in common use.
Discuss this post
Sources:
Calvin, John. Certain Homilies. Wesel, 1553, sig. Dviii. Early English Books Online.
Ellis, Henry, ed. Polydore Vergil’s English History, vol. 1 (c.1550). London: John Bowyer Nicols and Son for the Camden Society, 1846, 93. HathiTrust Digital Archive.
Marlowe, Christopher. Tamburlaine, part 2. London: Richard Ihones, 1590, 2.1.1, sig. F3.
Oxford English Dictionary, third edition, June 2014, s.v. egregious, adj., egregiously, adv.
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6857 | https://www.quora.com/What-is-the-relation-between-zero-covariance-and-independence-What-are-examples-in-science-of-variables-which-are-not-independent-but-have-zero-covariance | Something went wrong. Wait a moment and try again.
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What is the relation between zero covariance and independence? What are examples in science of variables which are not independent but have zero covariance?
Keith Winstein
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Benjamin Golub
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Justin Rising
, PhD in statistics · Author has 391 answers and 2M answer views
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11y
In general terms, correlation and covariance measure whether two random variables have a linear relationship. Statistical independence is about whether the variables have any relationship at all; i.e. whether knowing something about one tells you anything about the other.
Here's a simple example.
Random variable A takes on the values {1, -1} with equal probability.
Random variable B depends on A as follows:
If A = 1, then B takes on the values {1000, -1000} with equal probability.
If A = -1, then B = 0.
A and B are uncorrelated, that is, they have zero covariance: both variables have zero mean, and
In general terms, correlation and covariance measure whether two random variables have a linear relationship. Statistical independence is about whether the variables have any relationship at all; i.e. whether knowing something about one tells you anything about the other.
Here's a simple example.
Random variable A takes on the values {1, -1} with equal probability.
Random variable B depends on A as follows:
If A = 1, then B takes on the values {1000, -1000} with equal probability.
If A = -1, then B = 0.
A and B are uncorrelated, that is, they have zero covariance: both variables have zero mean, and the expected value of A ⋅ B = 1000 - 1000 + 0 - 0 = 0.
But the two variables are clearly dependent; without knowing B, A could be +1 or -1 with equal probability. But if you know B, then you also know A exactly! And if you know A, you know B's absolute value exactly.
( discusses this a bit as well.)
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Justin Rising
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11y
There's a general class of multivariate distributions known as the spherically symmetric distributions which have this property. In brief, a distribution over Rn with a density f is spherically symmetric if the density has the property that f(→x)=g(||→x||2) for some scalar function g. The standard multivariate Gaussian distribution is a simple example, but so are multivariate generalizations of the student's T-distribution.
If a random vector →X has a spherically symmetric distribution and its covariance matrix exists, then that covariance mat
There's a general class of multivariate distributions known as the spherically symmetric distributions which have this property. In brief, a distribution over Rn with a density f is spherically symmetric if the density has the property that f(→x)=g(||→x||2) for some scalar function g. The standard multivariate Gaussian distribution is a simple example, but so are multivariate generalizations of the student's T-distribution.
If a random vector →X has a spherically symmetric distribution and its covariance matrix exists, then that covariance matrix is of the form cIn, where In is the identity matrix of the appropriate dimension and c is some constant. As a result, every spherical symmetric distribution has uncorrelated components. However, the multivariate Gaussian distribution is the only spherically distribution that has independent components.
There's a nice writeup in which has this result as its Theorem 1 (although the proof is not given). It also has a table of the common spherically symmetric distributions, most of which are new unless you specialize in multivariate statistics.
Aaron Brown
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In addition to the previous replies, I’d add that the distinction becomes increasingly important with more variables.
Suppose you have ten binary variables, each with 10% chance of being 1 and 90% chance of being zero.
If they are independent, the chance that all will be 1 is 0.1^{10} and the chance that all will be 0 is 0.9^{10}=0.3487.
Zero covariance just means that the chance of any pair of them being 1 is 0.1^2=0.01. With two variables, that’s the same as independence. But with ten it is quite different.
Imagine you have one hundred slips of paper in a hat, and you’re going to draw one out to
In addition to the previous replies, I’d add that the distinction becomes increasingly important with more variables.
Suppose you have ten binary variables, each with 10% chance of being 1 and 90% chance of being zero.
If they are independent, the chance that all will be 1 is 0.1^{10} and the chance that all will be 0 is 0.9^{10}=0.3487.
Zero covariance just means that the chance of any pair of them being 1 is 0.1^2=0.01. With two variables, that’s the same as independence. But with ten it is quite different.
Imagine you have one hundred slips of paper in a hat, and you’re going to draw one out to determine which of the ten variables is going to be 1.
You have to write each variable on ten slips of paper to have 0.1 probability of 1.
You have to write each pair of numbers on one slip of paper to have 0.01 probability that they are both one.
But any assignment that meets these two conditions means zero covariance. You could put all variables on one slip of paper, each variable alone on nine slips, and leave nine slips blank. Now the probability of ten 1’s is 0.01 instead of 0.0000000001. The probability of zero 1’s is 0.09 instead of 0.3487.
Another assignment is to put every pair of names on one slip each, 45 slips in total. You put each variable alone on another slip. leaving 45 slips blank. Now the probability of ten 1’s is zero—also the probability of three or more 1’s. The chance of zero 1’s is 45%.
There are 287 different ways to put variables on slips that have zero covariance, most of them leading to multivariate distributions very far from independence.
Another warning is that in this example, every pair of variables is independent; but all then variables can be very far from independent.
Related questions
Why doesn't 0 covariance imply independence if the converse is true?
Does independence mean zero covariance?
Why can we say that two independent random variables have zero covariance if they both take on only non-negative values?
What is an example of two independent events with zero covariance?
Are two data sets independent from each other if their covariance is zero?
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1y
The concepts of covariance and independence are fundamental in statistics and probability theory, and they describe different relationships between random variables.
Covariance and Independence
Covariance:
Covariance is a measure of the degree to which two random variables change together. If the covariance is zero, it indicates that there is no linear relationship between the two variables. However, it does not imply that the variables are independent.
Independence:
Two random variables are independent if the occurrence of one does not affect the probability of the occurrence of the ot
The concepts of covariance and independence are fundamental in statistics and probability theory, and they describe different relationships between random variables.
Covariance and Independence
Covariance:
Covariance is a measure of the degree to which two random variables change together. If the covariance is zero, it indicates that there is no linear relationship between the two variables. However, it does not imply that the variables are independent.
Independence:
Two random variables are independent if the occurrence of one does not affect the probability of the occurrence of the other. In mathematical terms, two variables X and Y are independent if:
P(X∩Y)=P(X)⋅P(Y)
Independence implies that there is no relationship (linear or otherwise) between the variables.
Zero Covariance vs. Independence
Zero Covariance does not imply independence. Two variables can have zero covariance but still be dependent in a non-linear way.
Independence implies zero covariance, but the converse is not true.
Examples of Zero Covariance with Dependence
Non-linear Relationships:
Consider a random variable X that is uniformly distributed between −1 and 1, and let Y=X2.
Here, X and Y have zero covariance because the average of X (which is zero) and the average of Y (which is 1/3) do not correlate linearly, but Y is clearly dependent on X since Y is a function of X.
Circular Relationships:
If you take X to be the angle of a point on a unit circle (in radians) and Y=sin(X), the covariance between X (which is uniformly distributed over [0,2π]) and Y is zero. However, Y is dependent on X because knowing X directly gives you Y.
Conclusion
In summary, while zero covariance indicates no linear relationship between two variables, it does not rule out the possibility of a non-linear dependence. Examples from science, such as those involving polynomial relationships or circular functions, illustrate how variables can exhibit zero covariance while still being dependent.
Vandad Ghiassi
Asset allocator, Strategist
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Author has 150 answers and 181.6K answer views
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11y
x and x^2 are dependent with zero covariance over [-1, 1], so are real interest rates and their volatility, dependant but with zero correlation from time to time and on some windows (though IR up usually means vol up). x and abs(x) is another example.
More technically covariance applies to certain classes of random variables and the more the variable resembles Gaussian variables, the more its significance is close to what intuition tells you. But if you take variables such that there is not a mean... then it still can make sense to measure covariance if variables are cointegrated, ie a linea
x and x^2 are dependent with zero covariance over [-1, 1], so are real interest rates and their volatility, dependant but with zero correlation from time to time and on some windows (though IR up usually means vol up). x and abs(x) is another example.
More technically covariance applies to certain classes of random variables and the more the variable resembles Gaussian variables, the more its significance is close to what intuition tells you. But if you take variables such that there is not a mean... then it still can make sense to measure covariance if variables are cointegrated, ie a linear combination of the two, seen through some filters (Arima and like) is a "white noise"
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Aaron Brown
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Justin Rising
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4y
Related
Why doesn't 0 covariance imply independence if the converse is true?
There are two main reasons.
The first, and less important one, is that covariance is a linear measure. You can have non-linear dependence that results in zero covariance, but is still dependence.
For example, suppose you run an experiment where people are allowed to choose an amount to bet, and you do two independent fair coin flip bets. The covariance between the two bet outcomes will be zero. But the second bet outcome will always be equal to either the same as the first, or the opposite of the first. Knowing the first bet outcome tells you a lot about the second.
That is, if a person wins $10
There are two main reasons.
The first, and less important one, is that covariance is a linear measure. You can have non-linear dependence that results in zero covariance, but is still dependence.
For example, suppose you run an experiment where people are allowed to choose an amount to bet, and you do two independent fair coin flip bets. The covariance between the two bet outcomes will be zero. But the second bet outcome will always be equal to either the same as the first, or the opposite of the first. Knowing the first bet outcome tells you a lot about the second.
That is, if a person wins $10 on the first coin flip, her expected value on the second coin flip is $0, the same as someone who wins $100, or loses $1, or any other outcome on the second flip. But there are only two possible results after winning $10 on the first flip, win $10 on the second, or lose $10 on the second.
The second and more important reason is covariance is only a pairwise concept, while dependence can involve more than two variables. Suppose you have ten bonds, each with a 10% chance of defaulting, and defaults are uncorrelated. If the bond defaults were independent, not just uncorrelated, the chance of all ten bonds defaulting would be 10−10 and the chance of zero defaults would be 0.910≈0.35.
But all zero covariance means is that the probability of any two bonds defaulting is 1% (the product of the individual probabilities).
Imagine putting 100 slips of paper into a hat, and drawing one out to decide which bonds default. Each bond must be written on 10 slips of paper (10% chance of default) and each pair of bonds on one slip (1% chance of both defaulting).
There are many ways to satisfy these conditions. For one, write all 10 bonds on one slip, put each bond individually on 9 other slips (90 total) and leave 9 slips blank. Now the chance of 10 defaults in 1%, and the chance of zero defaults is 9%, both quite different from the case of independence.
Another thing you could do is write each pair of bonds on one slip (45 slips in all), each bond alone on one other slip (10 slips total) and have 45 blank slips. Now the chance of 10 defaults—or more than 2 defaults—is zero and the chance of none is 45%.
Related questions
(X1, X2) follows a singular normal distribution. If their covariance is 0, are they independent? Why?
Does zero spearman's rho imply zero covariance (covariance, intuition, spearman rho, statistics)?
Can you provide examples of independent and dependent variables that can be used to calculate covariance in statistics?
Can you calculate covariance for more than 2 variables?
If two independent, normally distributed variables have zero correlation, does their covariance also have to be zero?
Anish Sarkar
Professor
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8y
Originally Answered: If the covariance of two random variables is zero, then are they statistically independent? Is this true?
·
No. Zero covariance of two random variables imply independence if they have at most two point support or jointly bi-variate normally distiibuted.
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4y
Related
Why doesn't 0 covariance imply independence if the converse is true?
In layman’s terms, X and Y are independent from each other means
even if you know everything about X, it does not help you know better about Y,
while zero covariance means,
even if you know everything about X,
but if you are only allowed to use linear models
, it does not help you know better about Y.
The two concepts are confusing sometimes, because we intuitively understand the term “covariance” or “correlation” as “the two things being somehow related with each other”. But the number of different ways in which two things can be related with each other is limitless. The lack of a linear relations
In layman’s terms, X and Y are independent from each other means
even if you know everything about X, it does not help you know better about Y,
while zero covariance means,
even if you know everything about X,
but if you are only allowed to use linear models
, it does not help you know better about Y.
The two concepts are confusing sometimes, because we intuitively understand the term “covariance” or “correlation” as “the two things being somehow related with each other”. But the number of different ways in which two things can be related with each other is limitless. The lack of a linear relationship between the two does not tell you much about whether they are somehow related with each other.
Pratheep Kumar Reddy Yaddala
8y
Related
Does independence mean zero covariance?
Yes.
How?
Linear relationship between A and B is measured using Covariance(A, B).
Statistical Independence between A and B : No relation at all between A and B.
If we know that A and B are independent which means no relationship exists. It implies that no linear relationship exists as linear relationship is also a type of relationship. Hence Covariance(A, B) = 0.
But if we know that Covariance(A, B) = 0, it means that there is no linear relationship. Some any other relationship may exist. Hence Covariance(A, B) does not guarantee that A and B are independent.
Thanks
pratheep
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Maurice Dupre
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1y
Related
Is the covariance always equal to zero when two random variables are uncorrelated but not necessarily independent?
QUESTION:
Is the covariance always equal to zero when two random variables are uncorrelated but not necessarily independent?
ANSWER:
If Cov(X,Y) is the covariance of X and Y, SD(W) is standard deviation for any W and R(X,Y) is the correlation of X and Y, then
Cov(X,Y)=[R(X,Y)][SD(X)][SD(Y)],
so if X and Y are both having non zero SD, then uncorrelation (R=0) implies covariance zero.
For X and Y to be independent would require that X^m and Y^n be uncorrelated (or zero covariance) for any positive integers m and n. So if XX=X and YY=Y, then uncorrelation and independence are the same. Notice WW=W simp
QUESTION:
Is the covariance always equal to zero when two random variables are uncorrelated but not necessarily independent?
ANSWER:
If Cov(X,Y) is the covariance of X and Y, SD(W) is standard deviation for any W and R(X,Y) is the correlation of X and Y, then
Cov(X,Y)=[R(X,Y)][SD(X)][SD(Y)],
so if X and Y are both having non zero SD, then uncorrelation (R=0) implies covariance zero.
For X and Y to be independent would require that X^m and Y^n be uncorrelated (or zero covariance) for any positive integers m and n. So if XX=X and YY=Y, then uncorrelation and independence are the same. Notice WW=W simply says the only possible values of W are 0 and 1, so W is simply an indicator for some statement. Thus for statements, independence and uncorrelation are the same.
Dennis Clason
Ph.D. in Statistics (academic discipline), Kansas State University (Graduated 1987)
·
Author has 559 answers and 566.1K answer views
·
1y
Related
Can the covariance of two uncorrelated random variables be proven to be zero? Can we conclude that they are independent based on this?
Well, since the correlation is defined as
ρ=Cov(X,Y)√Var(X)⋅Var(Y)
with the restriction that X and Y cannot be degenerate random variables, ρ=0 if and only Cov(X,Y)=0.
It’s also fairly well-known that ρ=0 implies independence characterizes the multivariate Normal distribution. So the answer to your second question is, “Only if your random variables are jointly Normal.”
Ph.D. in Mathematics, University of Manchester (Graduated 1982)
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Author has 79 answers and 64.2K answer views
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4y
Related
Why doesn't 0 covariance imply independence if the converse is true?
My favourite counterexample is to let
Y
=
X
2
where, for example,
X
can be any integer value from
−
3
to
3
, with equal probability. You don't have to do any calculations to see that:
X
and
Y
are dependent (as
Y
=
X
2
)
2. Cov
(
X
,
Y
)
=
0
(by the symmetry of the scatter diagram in
x
=
0
)
The thing is, of course, that covariance only measures linear correlation.
Aaron Brown
MBA in Finance & Statistics (academic discipline), The University of Chicago Booth School of Business (Graduated 1982)
·
Author has 25.4K answers and 101.9M answer views
·
1y
Related
Can two variables have no correlation and still be considered correlated? If not, can you provide examples of this?
Variables can have a zero correlation correlation but not be independent, which means we might say in English that they are correlated.
For example, suppose X has a standard Gaussian distribution. Y=-X has a zero correlation coefficient with X, yet if you know X, you know Y exactly.
Another example is a set of variables can have all pairwise correlation coefficients zero, but strong higher-order dependence.
For example, suppose there are 100 equally likely scenarios for ten events. In one scenario all ten events occur. Each of the ten events occurs alone in nine scenarios, and in nine scenarios n
Variables can have a zero correlation correlation but not be independent, which means we might say in English that they are correlated.
For example, suppose X has a standard Gaussian distribution. Y=-X has a zero correlation coefficient with X, yet if you know X, you know Y exactly.
Another example is a set of variables can have all pairwise correlation coefficients zero, but strong higher-order dependence.
For example, suppose there are 100 equally likely scenarios for ten events. In one scenario all ten events occur. Each of the ten events occurs alone in nine scenarios, and in nine scenarios no event occurs.
Each of the events occurs in 10 scenarios, so has a 10% chance of occuring.
Each pair of events occurs in 1 scenario, so has a 1% chance. Since the chance of the pair occurring is the product of the individual event probabilities, they have zero correlation coefficient.
But no one would say the events are collectively uncorrelated. We know that either zero, one or ten events happen, never 2 to 9 events. If we know two events happened, we know all ten did. If we know any event did not happen, we know either zero or one event did. So knowing some events gives us a lot of information about others.
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Family of subsets of [n][n] and intersections
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Let F F be a family of subsets of [n]={1,…,n}[n]={1,…,n} such that for all A,B,C∈F A,B,C∈F at most 3 3 out of 8 8 triples A∩B∩C A∩B∩C, A∩B c∩C A∩B c∩C, A∩B∩C c A∩B∩C c, A∩B c∩C c A∩B c∩C c, A c∩B∩C A c∩B∩C, A c∩B c∩C A c∩B c∩C, A c∩B∩C c A c∩B∩C c and A c∩B c∩C c A c∩B c∩C c are non-empty (here X c X c is the complement of X X). Prove that the size |F||F| is bounded from above by a constant independent of n n.
Apart from considering the contrapositive (i.e. prove that if |F|>C 0|F|>C 0 then it contains A,B,C A,B,C with at least four non-empty intersection triples) I do not know what to do. I tried some extreme examples (such as F F to consist of many pairwise disjoint sets) but do not see how to connect them for the general case.
Any help appreciated!
combinatorics
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edited Aug 3, 2022 at 20:17
DesmondMilesDesmondMiles
asked Jun 3, 2020 at 10:49
DesmondMilesDesmondMiles
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2 HINT: The statement is very strong - it said there is a finite max size for |F||F| even if n→∞n→∞. Try to construct such F F and maximize its size, to see how big you can get |F||F| to be. You should find that the max size you can achieve is a very small integer, and that should give you an idea how to prove it.antkam –antkam 2020-06-04 03:33:35 +00:00 Commented Jun 4, 2020 at 3:33
@antkam Yep, makes sense that C 0 C 0 is small, but could you give some more directions on this? As I said, I cannot connect different particular examples :(DesmondMiles –DesmondMiles 2020-06-04 08:48:02 +00:00 Commented Jun 4, 2020 at 8:48
1 OK another HINT: I suggest playing with n=2 n=2. There are only 4 4 subsets of . How many of these can you fit into F F while satisfying the given property? Then try n=3 n=3 and the 8 8 subsets of . How many of these can you fit into F F while satisfying the given property?antkam –antkam 2020-06-04 13:10:23 +00:00 Commented Jun 4, 2020 at 13:10
1 Ok for n=2 n=2 the answer is 4 4, for n=3 n=3 I see that in general I cannot have three pairwise disjoint non-empty subsets whose union is not [n][n] (or the complements of three such subsets) as these would give 4 intersection triples; so I can take at most ∅∅, {1,2,3}{1,2,3} and at most 2 2 from {1},{2},{3}{1},{2},{3} and {1,2},{1,3},{2,3}{1,2},{1,3},{2,3} and I think that ∅∅, {1,2,3}{1,2,3}, {1},{2}{1},{2}, {1,2},{1,3}{1,2},{1,3} are fine. Anyway, it was tedious to verify for n=3 n=3, no general reasons apart from the above observation come up to my mind...DesmondMiles –DesmondMiles 2020-06-04 14:11:23 +00:00 Commented Jun 4, 2020 at 14:11
2 Probably we can consider some large graph (or hypergraph) and then apply the Ramsey's theorem somehow, but I'm not sure.richrow –richrow 2020-06-05 13:28:30 +00:00 Commented Jun 5, 2020 at 13:28
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My answer is inspired by Ewan Delanoy’s that. He bounded size of F F in terms of Ramsey numbers, showing that |F|≤2+R(2,R(4,4,3))=2+R(4,4,3)|F|≤2+R(2,R(4,4,3))=2+R(4,4,3). The best known bounds for R(4,4,3)R(4,4,3) are 55≤R(4,4,3)≤77 55≤R(4,4,3)≤77, see [R, p. 39]. We shall show that |F|≤14|F|≤14.
Following Ewan Delanoy, given a subset A A of [n][n] we put A+=A A+=A, A−=A c A−=A c and let A±A± denotes A A or A c A c. Subsets A A and B B of [n][n] are independent, if all four intersections A±∩B±A±∩B± are nonempty. Assume that |F|≥3|F|≥3.
Lemma. The family F F does not contain independent members.
Proof. Suppose to the contrary that A,B∈F A,B∈F are independent and C C be an arbitrary member of C C distinct from A A and B B. For any choice ∗,∗∗∗,∗∗ of signs in ±±, a set A∗∩B∗∗A∗∩B∗∗ is nonempty, so at least one of sets A∗∩B∗∗∩C+A∗∩B∗∗∩C+ and A∗∩B∗∗∩C−A∗∩B∗∗∩C− is non-empty. So the family of sets of the form A±∩B±∩C±A±∩B±∩C± has at least four non-empty members, a contradiction. □◻
Let a family F∗F∗ is obtained from family F∖{∅,[n]}F∖{∅,[n]} by replacing each member A A of F F with |A|>n/2|A|>n/2 by A c A c. Then |F∗|≥|F|/2−1|F∗|≥|F|/2−1 and F∗F∗ satisfies the question condition. Since A c∩B c A c∩B c is non-empty for each A,B∈F∗A,B∈F∗, Lemma implies that any members of F∗F∗ are disjoint or one is contained in the other. It follows that any member of F∗F∗ contains a minimal element and minimal elements of F∗F∗ are pairwise disjoint. If F∗F∗ contains four minimal members A A, B B, C C, and D D then sets A∩B c∩C c=A A∩B c∩C c=A, A c∩B∩C c=B A c∩B∩C c=B, A c∩B c∩C=C A c∩B c∩C=C, and A c∩B c∩C c⊃D A c∩B c∩C c⊃D are non-empty, a contradiction. Thus F F contains at most three minimal elements. Let A A be any of them. Suppose to the contrary that there exist distinct elements B⊃A B⊃A and C⊃A C⊃A of F∗∖{A}F∗∖{A}. Since B∩C⊃A B∩C⊃A is non-empty, it follows that B⊂C B⊂C or C⊂B C⊂B. Anyway, sets A c∩B c∩C c A c∩B c∩C c, A c∩B c∩C A c∩B c∩C, A c∩B∩C A c∩B∩C, and A∩B∩C A∩B∩C are non-empty, a contradiction. Thus each minimal member of F∗F∗ is contained in at most one other member. Thus |F∗|≤3⋅2=6|F∗|≤3⋅2=6, and |F|≤2(|F∗|+1)≤14|F|≤2(|F∗|+1)≤14.
References
[R] Stanisław P. Radziszowski, Small Ramsey numbers. Dynamic Surveys. Electronic Journal of Combinatorics, revision #15: March 3, 2017.
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edited Aug 3, 2022 at 20:25
darij grinberg
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Here is a proof that |F|≤2+R(2,R(4,4,3))|F|≤2+R(2,R(4,4,3)), where R R denotes Ramsey numbers.
Let E E denote the subets of [n][n] that are neither empty or full, and F′=F∩E=F∖{∅,[n]}F′=F∩E=F∖{∅,[n]}. It will be convenient to use the notation A−A− for A c A c and define A+=A A+=A ; then A±A± means "A A or its complement". Say also that A,B⊆[n]A,B⊆[n] are independent if all four intersections A±∩B±A±∩B± are nonempty.
Step 1.F F does not contain three mutually independent subsets.
Suppose that A 1,A 2,A 3 A 1,A 2,A 3 are mutually independent subsets in F F. Let s 1 s 1 and s 2 s 2 be signs in ±±. By independence, A s 1 1∩A s 2 2 A 1 s 1∩A 2 s 2 is nonempty, so at least one of A s 1 1∩A s 2 2∩A−3 A 1 s 1∩A 2 s 2∩A 3− and A s 1 1∩A s 2 2∩A+3 A 1 s 1∩A 2 s 2∩A 3+ is non-empty. When (s 1,s 2)(s 1,s 2) varies over its four possible values, this already gives us four non-empty triple intersections, which contradicts the hypothesis on F F. QED
Step 2.F′F′ does not contain an increasing sequence of three subsets.
Indeed, if A 1⊂A 2⊂A 3 A 1⊂A 2⊂A 3 is an increasing sequence of three subsets in E E, then the following four intersections are nonempty : A−1∩A−2∩A−3 A 1−∩A 2−∩A 3−, A−1∩A−2∩A+3 A 1−∩A 2−∩A 3+, A−1∩A+2∩A+3 A 1−∩A 2+∩A 3+, A+1∩A+2∩A+3 A 1+∩A 2+∩A 3+. QED
Step 3.F′F′ does not contain four mutually disjoint subsets, or four subsets whose complements are mutually disjoint.
Indeed, if A k(1≤k≤4)A k(1≤k≤4) are four mutually disjoint subsets in E E, then the following four intersections are nonempty : A+1∩A−2∩A−3 A 1+∩A 2−∩A 3−, A−1∩A+2∩A−3 A 1−∩A 2+∩A 3−, A+1∩A−2∩A−3 A 1+∩A 2−∩A 3−, A−1∩A−2∩A−3 A 1−∩A 2−∩A 3−. QED
Step 4.|F′|<R(2,R(4,4,3))|F′|<R(2,R(4,4,3)).
Suppose by contradiction that |F′|≥R(2,R(4,4,3))|F′|≥R(2,R(4,4,3)). Then, by step 1, there is a F′′⊆F′F″⊆F′ or cardinality at least R(4,4,3)R(4,4,3) such that any A,B∈F′′A,B∈F″ are never independent ; so there must be signs s A,s B s A,s B such that A s A∩B s B=∅A s A∩B s B=∅. Color a pair (A,B)(A,B) blue if (s A,s B)=(−,−)(s A,s B)=(−,−), red if (s A,s B)=(+,+)(s A,s B)=(+,+), and yellow otherwise. Then, we must have either a blue or red four-set (ruled out by step 3), or a yellow three-set (ruled out by step 2). This finishes the proof.
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answered Jun 5, 2020 at 13:35
Ewan DelanoyEwan Delanoy
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In step 1, do you mean F′F′? Perhaps I misunderstand the term "mutual independence" when it comes to sets -- but when it comes to probability the events ∅∅ and Ω Ω are considered mutually independent with everything else.antkam –antkam 2020-06-05 14:52:34 +00:00 Commented Jun 5, 2020 at 14:52
@antkam 1) No, step 1 happens to be the only step where you can work on F F directly rather than F′F′. 2) Indeed, "independence" here has a meaning different from the probabilistic one.Ewan Delanoy –Ewan Delanoy 2020-06-05 14:58:06 +00:00 Commented Jun 5, 2020 at 14:58
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6859 | https://www-backup.salemstate.edu/is-hbr-a-weak-acid | Unveiling the Acid Strength: Is HBR a Weak Acid? - Salem State Vault
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Home/Members/Unveiling the Acid Strength: Is HBR a Weak Acid?
Unveiling the Acid Strength: Is HBR a Weak Acid?
Ashley
July 27, 2025 8 min read
The concept of acid strength is a fundamental aspect of chemistry, particularly in the realm of inorganic and organic chemistry. It is defined as the ability of an acid to donate a proton (H+ ion), which is a critical process in various chemical reactions. Hydrobromic acid (HBr) is one of the many acids that have been extensively studied, and its acid strength has been a topic of interest among chemists and researchers. In this article, we will delve into the properties of HBr and examine its acid strength to determine if it is indeed a weak acid.
Table of Contents
Understanding Acid Strength
Properties of Hydrobromic Acid
Determining the Acid Strength of HBr
Factors Influencing Acid Strength
Key Points
Hydrobromic acid (HBr) is a strong acid due to its complete dissociation in water.
The acid strength of HBr is influenced by its bond strength and the stability of its conjugate base.
The pKa value of HBr is approximately -9, indicating its strong acidic nature.
HBr is commonly used in various chemical reactions, including the production of pharmaceuticals and dyes.
The handling and storage of HBr require special precautions due to its corrosive and toxic properties.
Understanding Acid Strength
Acid strength is typically measured by the pKa value, which is the negative logarithm of the acid dissociation constant (Ka). A lower pKa value indicates a stronger acid, as it is more likely to donate a proton. The acid dissociation constant is a measure of the equilibrium between the acid and its conjugate base. In the case of HBr, its acid dissociation constant is relatively high, indicating a strong tendency to donate a proton.
Properties of Hydrobromic Acid
Hydrobromic acid (HBr) is a hydrohalic acid, consisting of hydrogen and bromine atoms. It is a colorless, corrosive, and toxic liquid with a characteristic pungent odor. HBr is highly soluble in water and is commonly used in various chemical reactions, including the production of pharmaceuticals, dyes, and other organic compounds. Its chemical properties make it an essential reagent in many industrial and laboratory applications.
| Property | Value |
--- |
| Molecular Weight | 80.91 g/mol |
| Density | 1.49 g/cm³ |
| Boiling Point | 122°C |
| pKa Value | -9 |
💡 The pKa value of HBr is a critical factor in determining its acid strength. A pKa value of -9 indicates that HBr is a strong acid, as it is highly likely to donate a proton in aqueous solutions.
Determining the Acid Strength of HBr
To determine the acid strength of HBr, we can examine its acid dissociation constant (Ka) and its pKa value. The Ka value of HBr is approximately 1.3 × 10^9, indicating a strong tendency to donate a proton. The pKa value, which is the negative logarithm of the Ka value, is approximately -9. This value is significantly lower than that of weak acids, such as acetic acid (pKa = 4.76), indicating that HBr is indeed a strong acid.
Factors Influencing Acid Strength
The acid strength of HBr is influenced by several factors, including its bond strength and the stability of its conjugate base. The bond strength between the hydrogen and bromine atoms is relatively weak, making it easier for the acid to donate a proton. Additionally, the conjugate base of HBr, Br-, is a relatively stable anion, which further contributes to the acid’s strong acidic nature.
What is the primary use of hydrobromic acid?
+
Hydrobromic acid is commonly used in various chemical reactions, including the production of pharmaceuticals, dyes, and other organic compounds.
What are the hazards associated with handling hydrobromic acid?
+
Hydrobromic acid is a corrosive and toxic substance, requiring special precautions for handling and storage. It can cause severe burns, respiratory problems, and other health issues if not handled properly.
How does the pKa value of hydrobromic acid compare to other acids?
+
The pKa value of hydrobromic acid is approximately -9, indicating that it is a strong acid. This value is significantly lower than that of weak acids, such as acetic acid (pKa = 4.76), and is comparable to other strong acids, such as hydrochloric acid (pKa = -7).
In conclusion, the acid strength of hydrobromic acid is a critical factor in its chemical properties and applications. With a pKa value of approximately -9, HBr is indeed a strong acid, making it a valuable reagent in various chemical reactions. Its corrosive and toxic properties require special precautions for handling and storage, and its uses are diverse, ranging from the production of pharmaceuticals to the synthesis of dyes and other organic compounds. By understanding the acid strength of HBr, we can better appreciate its importance in chemistry and its potential applications in various fields.
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二次函数最值求法
Raymond Fang
从来如此，便对么？
19 人赞同了该文章
Upd in 2022/12/19：第一题题目有误，应当是最小值，而非最大值
Upd in 2024/02/24：typo
二次函数最值能有多难？今天让我们好好聊一聊有关二次函数最值的问题解法。
1.二次函数求顶点——二次函数最值中第一关
对于任意一个二次函数 y=a x 2+b x+c(a≠0)y=ax^2+bx+c(a\neq0)对于任意一个二次函数 y=a x 2+b x+c(a≠0)y=ax^2+bx+c \ (a\ne0) ，我们都可以通过二次函数配方法或者公式法来求出顶点坐标。当 a<0 a<0 时，开口朝下；当 a>0 a>0 时，开口朝上。通过配方法可得：
y=a x 2+b x+c=a(x 2+b a x)+c=a(x 2+2⋅b 2 a x+b 2 4 a 2−b 2 4 a 2)+c=a(x+b 2 a)2−b 2 4 a+c=a(x+b 2 a)2+4 a c−b 2 4 a\begin{align} y&=ax^2+bx+c \ &=a\Big(x^2+\frac{b}{a}x\Big)+c \ &=a\Big(x^2+2 \cdot \frac{b}{2a}x+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}\Big)+c \ &=a\Big(x+\frac{b}{2a}\Big)^2-\frac{b^2}{4a}+c \ &=a\Big(x+\frac{b}{2a}\Big)^2+\frac{4ac-b^2}{4a} \end{align}
所以说，顶点坐标 (−b 2 a,4 a c−b 2 4 a)\Big(-\frac{b}{2a},\frac{4ac-b^2}{4a}\Big) 。这是进入二次函数求最值最先应当掌握的知识点。
2.当二次函数出现在分母——二次函数最值第二关
这里可以给出一道题目：
已知函数 y=3 x 2+6 x+5 1 2 x 2+x+1，求 y 的最小值。\text{已知函数 }y=\frac{3x^2+6x+5}{\frac{1}{2}x^2+x+1} \text{，求 }y \text{的最小值。}
不难发现，分子分母中 x x 的二次项系数和一次项系数比例相同，但常数项比例不同，所以说可以将分母化成常数，这样这道题就可以解出来了。具体过程如下：
y=3 x 2+6 x+5 1 2 x 2+x+1=6(1 2 x 2+x+1)−1 1 2 x 2+x+1=6−1 1 2 x 2+x+1 要让 y 尽可能小，就让 1 2 x 2+x+1 尽可能小（注意负号），∵1 2 x 2+x+1=1 2(x+1)2+1 2∴1 2 x 2+x+1 最 小 值 为 1 2∴y min=4\begin{align} y&=\frac{3x^2+6x+5}{\frac{1}{2}x^2+x+1} \ &=\frac{6(\frac{1}{2}x^2+x+1)-1}{\frac{1}{2}x^2+x+1} \ &=6-\frac{1}{\frac{1}{2}x^2+x+1} \ &\text{要让 }y\text{ 尽可能小，就让 }\frac{1}{2}x^2+x+1 \text{尽可能小（注意负号），} \ \because \ &\frac{1}{2}x^2+x+1=\frac{1}{2}(x+1)^2+\frac{1}{2} \ \therefore \ &\frac{1}{2}x^2+x+1 \ 最小值为\frac{1}{2} \ \therefore \ &y_{\text{min}}=4 \end{align}
但是，假如说，这题稍微改一下，比如三对系数比例均不相同，或者需要求 y y 的最大值，这个方法就不适用了。所以说这不是通法，我们一般不用（至少我不用）。
3.判别式法——二次函数最值第三关
更熟悉的应当是这个方法——判别式法。这可以说是绝大部分题都是这么做的。我们要先建立一元二次方程，然后通过计算判别式大于或大于等于0来建立不等式。
以上面那题为例， y=\frac{3x^2+6x+5}{\frac{1}{2}x^2+x+1} 可以将分母移到左边，得：
y\Big(\frac1 2x^2+x+1\Big)=3x^2+6x+5 \ x^2\Big(\frac 1 2y-3\Big)+x(y-6)+(y-5)=0 \ \Delta=(y-6)^2-4 \cdot \Big(\frac1 2y-3\Big)(y-5)\geq0 \ -y^2+10y-24\geq0 \ (y-5)^2\leq1 \ -1\leq y-5\leq1 \ 4\leq y\leq 6
也就是说， y_{\text{min}}=4,\ y_{\text{max}}=6 。通过 \Delta\geq0 计算出二次函数最值是通法。假若说，分母是一次函数，也可以按照同样方法算出函数取值范围。
4.韦达定理+判别式法
韦达定理描述：
对于一元n次方程 \ a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0=0 \ 的根分别为x_1,x_2,\cdots,x_n，那么有\ \begin{equation} \left{ \begin{aligned} &x_1+x_2+\cdots+x_n=-\frac{a_{n-1}}{a_n} \ &x_1x_2+x_1x_3+\cdots+x_{n-1}x_n=\frac{a_{n-2}}{a_n} \ &x_1x_2x_3+x_1x_2x_4+\cdots+x_{n-2}x_{n-1}x_n=-\frac{a_{n-3}}{a_n} \ &\cdots \ &x_1x_2\cdots x_n=(-1)^n\frac{a_0}{a_n} \end{aligned} \right. \end{equation} \ 特别地，当 n=2 时，则有：\ 对于一元二次方程\ ax^2+bx+c=0\ 的根x_1,x_2，有\ \left{ \begin{aligned} &x_1+x_2=-\frac b a \ &x_1x_2=\frac c a \end{aligned} \right.
韦达定理的逆定理：
若实数a,b满足 \ \left{ \begin{aligned} &a+b=m\ &ab=n \end{aligned} \right. \ 则a,b为一元二次方程x^2-mx+n=0的两个根。
主要东西在下面的那个方程组。我们可以通过它解决很多问题。先上例题：
已知实数a,b满足 a^2+ab+b^2=1，求a^2-ab+b^2的取值范围。
这道题很简短。我们可以先设 t=a^2-ab+b^2 ，用 t 表示出 a+b 和 ab ，建立一元二次方程，就可以把 t 的取值范围求出来了。
具体过程：
令t=a^2-ab+b^2，则\ ab=\frac{1-t}{2}, a^2+b^2=\frac{1+t}2 \ 所以 \ \begin{aligned} a+b&=\pm\sqrt{a^2+b^2+2ab} \ &=\pm\sqrt{\frac{1+t}{2}+1-t} \ &=\pm\sqrt{\frac{3-t}{2}} \end{aligned} \ 所以a,b是一元二次方程x^2\mp\sqrt{\frac{3-t}2}x+\frac{1-t}2=0的两个根，且\frac{3-t}2\geq0\ 所以 \ \left{ \begin{aligned} &\Delta=\frac{3-t}2-4\cdot \frac{1-t}2\geq0 \ &\frac{3-t}2\geq0 \end{aligned} \right. \ 解得\frac1 3\leq t\leq 3。\ 所以\frac1 3\leq a^2-ab+b^2 \leq 3。
韦达定理非常好用，它十分灵活。它十分适用于十分复杂的压轴题。下面是另一道例题：
已知有正整数m,n，且m\neq2。对于任意实数t，满足|mt+3|\leq|2t+n|，求m,n的值。
在此就不写思路解析了，直接上过程：
\begin{aligned} &\because |mt+3|\leq|2t+n| \ &\therefore (mt+3)^2\leq(2t+n)^2 \ &t^2(m^2-4)+t(6m-4n)+(9-n^2)\geq0 \ &\therefore \left{ \begin{aligned} &m^2-4>0 \ &(6m-4n)^2-4(m^2-4)(9-n^2)\leq0 \end{aligned} \right. \ &\therefore \left{ \begin{aligned} &m>2 \ &mn=6 \end{aligned} \right. \ &\because m,n为正整数 \ &\therefore \left{ \begin{aligned} m=3 \ n=2 \end{aligned} \right. 或 \left{ \begin{aligned} m=6 \ n=1 \end{aligned} \right. \end{aligned}
注意，第四行 m^2-4>0 是因为上述表达式恒大于等于0，所以开口朝上。
有关这样的题目还有很多，我在此附上3道（第三道难题预警）：
1.已知实数x,y,z满足x+y+z=0，且xyz=2，求|x|+|y|+|z|的最\color{red}{小}值。 \ 2.已知实数x,y,z满足x+y+z=5，且xy+yz+zx=3，求z的取值范围。\ 3.已知a,b,c都是正整数，且二次函数y=ax^2+bx+c交x轴于A(x_1,0) \ B(x_2,0) 两点。\若A,B与原点的距离都小于1，求a+b+c的最小值。
以上就是有关二次函数求最值的方法。本人发的第一篇文章，烦多多指点，有误尽请指明。
编辑于 2024-02-23 16:31・湖南
二元函数
二次函数
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言当
作者您好，第三个判别式法在得到4≤y≤6时，将y代入原函数化简会得到1=0的结果，这也说明当y＝3时无解，应该是闭区间。（或者在使用韦达定理前先验证y=6也会得到同样结论）
2023-09-16 · 陕西
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言当
打错了，应该是开区间
2023-09-16 · 陕西
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十分受用
2023-01-27 · 江西
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2023-01-10 · 湖南
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6861 | https://bingweb.binghamton.edu/~suzuki/QuantumMechanicsII/4-9_Spherical_harmonics.pdf | Spherical Harmonics in Quantum mechanics Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: March 03, 2023) Here we discuss the properties of spherical harmonics. Spherical harmonics Dirac delta function Recurrence relation Associated Legendre functions Parity Time reversal operator SphericalPlot3D ContourPlot3D Series expansion 1. Formulation The relation between the spherical coordinates and Cartesian coordinates are schematically shown below. Fig.1 Spherical coordinates. 3 2 2 ( )( sin ) sin dV d dr rd r d r dr d d r drd r . The spherical harmonics: , , , ( , ) m l l m l m Y n , where m l m m l Lz , , ˆ ℏ , or ˆ , , , z L l m i l m m l m n n n ℏ ℏ , , n . The closure relation: 1 ˆ , , d , or ˆ 1 d n n where the solid angle, dsindd The and dependence of m l, n is given by 2 2 2 2 2 2 1 1 ˆ [ (sin )] ( , ) sin sin ( 1) ( , ) m l m l L lm Y l l Y n ℏ ℏ (1) ˆ , ( , ) ( , ) m m z l l L lm l m Y mY i i n n ℏ ℏ ℏ (2) Equation (2) shows that im m l m l e Y ) ( ) , ( . It is required that the eigenfunction must be single valued e ime im (2), which means that m = 0, ±1, ±2, (integers). Equation (1) can be rewritten as 0 ) ( )] 1 ( sin ) (sin sin 1 [ 2 2 m l l l m d d d d . When we change the variable; cos x ( 1 x ). sin d dx d d d d dx dx Then we have the differential equation (the Legendre differential equation) as 0 ) ( ] 1 ) 1 ( [ ) ( ) 1 ( 2 2 2 x P x m l l x P dx d x dx d m l m l . where ) (x Pm l is the associated Legendre function; ) (cos ) ( m l lm m l P C . The orhogonality relation ' , ' , , ' , ' m m l l m l m l leads to ' , ' , ' ' ', ' , sin ( , ) ( , ) m m l l m m l l d l m l m d d Y Y n n . Note that this differential equation (Sturm-Liouville type) can be solved by using the series expansion method (see later). To obtain the form ofY l m(,), we may start with m = l. 0 , ˆ l m l L , or ˆ , ( cot ) , 0 i L l m l i e i l m l n n ℏ . Since , ( , ) ( ) l l il l l l m l Y e n ( d dlcot)l l() 0, or Y l l(,) C le ilsin l where Cl is a normalization constant. 4 )!
1 2 ( !
2 ) 1 ( l l C l l l . The result for m≥0 is l m l m l m im l l m l d d e m l m l l l Y 2 ) (sin ) (cos sin 1 )!
( )!
( 4 ) 1 2 ( !
2 ) 1 ( ) , ( , and we define Y l m(,) ( 0 m ) by Y l m(,) (1) m[Y l m(,)] , for 0 m , or ) , ( ) 1 ( )] , ( [ m l m m l Y Y . where somewhat peculiar choice of sign is conventional. The definition of ( , ) m l Y with 0 m arises from the property of time-reversal operator; ˆ , ( 1) , m l m l m , or ˆ , ( 1) , m l m l m where ˆ is the time-reversal operator. We note that ˆ , , ( 1) , , ( 1) , , m m l m l m l m or Y l m(,) (1) m[Y l m(,)] , for 0 m , since ˆ , , , , l m l m from the definition (see the Time reversal operator in the lecture note of Quantum Mechanics, Graduate course), Angular momentum in spherical coordinate is ( ) 1 1 ( ) sin 1 ( ) sin r r i i r r r r i L r e e e e e e ℏ ℏ ℏ The angular momentum Lx, Ly, and Lz (Cartesian components) can be described by 1 [ ( sin cos ) (cos cos cos sin sin ) ] sin y y z i x x L e e e e e ℏ . or ) cos cot (sin ℏ i Lx , ) sin cot cos ( ℏ i Ly , ℏ i Lz . We define L+ and L- as ) cot ( i e i iL L L i y x ℏ , and ) cot ( i e i iL L L i y x ℏ . 2. Dirac delta function The Dirac delta function can be described by ) ' ( ) ' ( 1 ) ' ( 2 n n r r r r r since 3 '2 2 1 ' ( ') ( ') ' ' ( ') ' ' ( ') 1 d r r r dr d r d r r r n n n n Here, we note that 0 0 0 ' ( ') , , ' ( ', ') ( , ) 2 1 ( ') 4 l l m l l m m l l l m l l l l m l m Y Y l P n n n n n n n n , where we use the addition theorem 4 ( ') ( , ) ( ', ') 2 1 l m m l l l m l P Y Y l n n . x y z r r' q q' ff' Fig.2 Angle such that ' ' cos ' rr r r n n In summary, the Dirac delta function is expressed by 2 2 0 1 ( ') ( ') ( ') 1 2 1 ( ') ( ') 4 l l r r r l r r P r r r n n n n This formula will be useful in the theory of scattering from a spherical potential. ((Note)) ) (x P l is the l-th Legendre polynomial which is defined by the Rodrigues formula 3. Associate Legendre function ) , ( m l Y can be also expressed by ) (cos )!
( )!
( 4 1 2 ) 1 ( ) , ( m l im m m l P e m l m l l Y , where ) (cos m l P is the associated Legendre function. l m l m l m l l m m m m l x dx d x l x P dx d x x P ) 1 ( ) 1 ( !
2 1 ) ( ) 1 ( ) ( 2 2 / 2 2 / 2 for m0. ) ( )!
( )!
( ) 1 ( ) ( x P m l m l x P m l m m l The Rodrigues' formula: l l l l l l x dx d l x P x P ) 1 ( !
2 1 ) ( ) ( 2 0 , 1 ) 1 ( l P , l l P ) 1 ( ) 1 ( , and ) ( ) ( ) 1 2 ( ) ( ) 1 ( 1 1 x lP x xP l x P l l l l where 1 ) ( 0 x P , x x P ) ( 1 , ) 1 3 ( 2 1 ) ( 2 2 x x P , ) 3 5 ( 2 1 ) ( 3 3 x x x P , ) 3 30 35 ( 8 1 ) ( 2 4 4 x x x P , ) 15 70 63 ( 8 1 ) ( 3 5 5 x x x x P 4. Parity operator ˆ Fig.3 Space reflection on parity. . . n n ˆ , or n n ˆ For n to -n, we have , and, m l Y Y m l m l l m l l m l , ) 1 ( ) , ( ) 1 ( ) , ( , , ˆ n n n , or m l m l l , ) 1 ( , ˆ . ((Note-1)) Note that L L ˆ ˆ ˆ ˆ , or ˆ ˆ ˆ ˆ L L , 0 , ˆ ˆ 0 , ˆ ˆ l L l L . Here we suppose that 0 , l is the state with either even or odd parity 0 , 0 , ˆ l p l e . Then we have 0 , ˆ 0 , ˆ ˆ 0 , ˆ ˆ l L p l L l L e . This implies that 0 , ˆ l L has also the same parity as the state 0 , l . Repeating this procedure, we can find that the parity of the state m l, is the same as that of the state 0 , l . The problem is reduced to the determination of the parity of the state 0 , l . ˆ ,0 ,0 ,0 e l l p l n n n . Here 0 2 ( 1) 2 1 ,0 ( , ) (sin ) 2 !
4 (cos ) l l l l l l l d l Y l d n 2 0 ( 1) 2 1 ,0 ( 1) (sin ) ( 1) ( , ) 2 !
4 (cos ) l l l l l l l l l d l Y l d n when . Therefore, we have m l m l l , ) 1 ( , ˆ . ((Note-2)) ((Binney)) We start with l il l l l e C l m l Y sin , , ) , ( We note that l m l e e e C Y l m l l m l l l il il l il l l l , , ) 1 ( ) ( sin ) ( sin ) , ( , , , ˆ , ) ( or l m l l m l l , ) 1 ( , ˆ . In other words, l m l , has an even parity if l is an even number and odd parity if l is an odd number. The ladder operators y x L i L L ˆ ˆ ˆ are even parity operators; 0 ] ˆ , ˆ [ L . We use the relation 1 , ) 1 )( ( , ˆ m l m l m l m l L , 1 , ) 1 )( ( , ˆ m l m l m l m l L . We apply the parity operator to 1 , l m l . l m l L l l m l L l l m l L l l m l l , ) 1 ( ˆ 2 1 , ˆ ˆ 2 1 , ˆ 2 1 ˆ 1 , ˆ or 1 , ) 1 ( , ˆ 2 1 ) 1 ( 1 , ˆ l m l l m l L l l m l l l So that 1 , l m l has the same parity as 1 , l m l . Since all the m l, for a given l can be obtained by repeated application of L ˆ to l m l , , it follows that they all have the same parity, (-1)l. 5. Determination of the parity with the use of Mathematica We evaluate the parity ) , ( ) , ( m l m l Y Y , as a function of (l, m). m l m l , ) 1 ( , ˆ ℓ . ((Mathematica-1)) Table of { , , l m the parity}; the parity is assumed to be ( 1)l . We make sure of this prediction using the Mathematica. Clear "Global" ; P L , m : SphericalHarmonicY L, m, , SphericalHarmonicY L, m, , Simplify; Column Table L, m, P L, m , L, 0, 5 , m, L, L, 1 , Left 0, 0, 1 1, 1, 1 , 1, 0, 1 , 1, 1, 1 2, 2, 1 , 2, 1, 1 , 2, 0, 1 , 2, 1, 1 , 2, 2, 1 3, 3, 1 , 3, 2, 1 , 3, 1, 1 , 3, 0, 1 , 3, 1, 1 , 3, 2, 1 , 3, 3, 1 4, 4, 1 , 4, 3, 1 , 4, 2, 1 , 4, 1, 1 , 4, 0, 1 , 4, 1, 1 , 4, 2, 1 , 4, 3, 1 , 4, 4, 1 5, 5, 1 , 5, 4, 1 , 5, 3, 1 , 5, 2, 1 , 5, 1, 1 , 5, 0, 1 , 5, 1, 1 , 5, 2, 1 , 5, 3, 1 , 5, 4, 1 , 5, 5, 1 6. Recurrence relation For m = 0, l l l l l l d d l l Y 2 0 ) (sin ) (cos 4 1 2 !
2 ) 1 ( ) , ( which can be written in the form ) (cos 4 1 2 ) , ( 0 l l P l Y , where l l l l l l d d l P 2 ) (sin ) (cos !
2 ) 1 ( ) (cos , or P l(x) (1)l 2 ll!
dl dx l (1 x 2) l. Pl(x) is the l-th order Legendre polynomial. It has l zeros in the interval (-1≤x≤1). Note that Pl(1)=1. P l(x) (1) l P l(x) . (i) L ˆ Through the repeat action of ˆ L , we construct , , , ( , ) m l l m l m Y n , with m = l, l-1, l-2,....., ˆ , , 1 ( )( 1) 1 ( )( cot ) , ( )( 1) 1 ( cot ) , ( )( 1) i i L l m l m l m l m i e i l m l m l m e i l m l m l m n n n n ℏ ℏ ℏ (recurrence relation) (ii) ˆ L Through the repeat action of ˆ L , we construct , , , ( , ) m l l m l m Y n with m = 0,1, 2,....., l: ˆ , , 1 ( )( 1) 1 ( )( cot ) , ( )( 1) i L l m l m l m l m i e i l m l m l m n n n ℏ ℏ ℏ Here we use 0 2 1 , 0 ( , ) (cos ) 4 l l l l m Y P n . where ) (cos l P is the Legendre polynomial. We can find the exact expressions for the spherical harmonics using the above recurrence relation, using the Mathematica. We can make sure that ( , ) ( 1) [ ( , )] m m m l l Y Y for 0 m . ((Mathematica-2)) 7. Useful formula (summary) (i) , ( ) ( , ) m m l l l m Y Y n n , , , ( , ) m l l m l m Y n n (ii) Orthogonality , ' .
' 2 ' ' 0 0 ', ' , '.
' , sin ( , ) ( , ) l l m m m m l l l m l m d l m l m d d Y Y n n where d d d sin (iii) ,0 2 1 , ( 0, ) 4 m z l m l l m Y e (iv) ) (cos 4 1 2 ) , ( 0 , 0 l l P l Y m l n (v) ˆ z R n e , ˆ z R n e where ) ( ˆ ) ( ˆ ˆ y z R R R is the rotation operator ' ',0 ' ( ) ,0 , , [ ( , )] ˆ , ˆ , , ' , ' 2 1 ˆ , , ' 4 2 1 ˆ , ,0 4 2 1 ˆ ( ) 4 m z z m m m l m l m l m Y l m R l m R l m l m l m R l m l m R l D R n n e e ℓ ℓ ℓ ℓ or ) ( 0 , )] , ( [ 1 2 4 0 , ˆ , ) ˆ ( m l m Y l R m l R D ℓ ℓ . (vi) ˆ ˆ ˆ , , , z z R l m e R R l m l m n e ' ( ) ', ' ˆ , , ˆ , ' , ' , ˆ , ' ( ) z l m l l l m m m l R l m l m l m l m R l m l m D R n e n n or ( ) ,0 ', ' 2 1 ˆ , , ' ( ) 4 l l z m m m m l l l m l m D R e n . We also have ' ' ( ) , ' ' ( ) ',0 , ' ' ( ) ,0 ˆ , , ˆ , ' , ' , ˆ , ' , , ' ˆ , ' ( ) 2 1 ˆ ( ) 4 2 1 ˆ ( ) 4 z l z m l l z m l l l z m m m l l l m m m m l l m l m R l m l m l m R l m l m l m R l m l m D R l D R l D R n e e e e since ,0 2 1 , ( 0, ) 4 m z l m l l m Y e In summary, we get ( ) ,0 2 1 ˆ ( ) , 4 l m l D R l m n , or ( ) ,0 4 4 ˆ ( ) , ( ) 2 1 2 1 l m m l D R l m Y l l n n . (vii) 0 0 0 ' ( ') , , ' ( ', ') ( , ) 2 1 ( ') 4 l l m l l m m l l l m l l l l m l m Y Y l P n n n n n n n n , where we use the addition theorem 4 ( ') ( , ) ( ', ') 2 1 l m m l l l m l P Y Y l n n . 8. Calculation of ) ( 0 , 0 , ˆ , ) ˆ ( l R m l R D l m ) ( 0 , )] , ( [ 1 2 4 0 , ˆ , ) ˆ ( m l m Y l R m l R D ℓ ℓ where ) ( ˆ ) ( ˆ ˆ y z R R R . (a) l = 1 sin 2 1 0 , 1 ˆ 1 , 1 i e m l R m l cos 0 , 1 ˆ 0 , 1 m l R m l sin 2 1 0 , 1 ˆ 1 , 1 i e m l R m l where sin 2 3 2 1 ) , ( 1 1 i e Y cos 3 2 1 ) , ( 0 1 Y sin 2 3 2 1 ) , ( 1 1 i e Y (b) l = 2 2 2 sin 2 3 2 1 0 , 2 ˆ 2 , 2 i e m l R m l cos sin 2 3 0 , 2 ˆ 1 , 2 i e m l R m l ) 2 cos 3 1 ( 4 1 0 , 2 ˆ 0 , 2 m l R m l cos sin 2 3 0 , 2 ˆ 1 , 2 i e m l R m l 2 2 sin 2 3 2 1 0 , 2 ˆ 2 , 2 i e m l R m l where 2 2 2 2 sin 2 15 4 1 ) , ( i e Y cos sin 2 15 2 1 ) , ( 1 2 i e Y ) 1 cos 3 ( 5 4 1 ) , ( 2 0 2 Y cos sin 2 15 2 1 ) , ( 1 2 i e Y 2 2 2 2 sin 2 15 4 1 ) , ( i e Y (c) l = 3 3 3 sin 5 4 1 0 , 3 ˆ 3 , 3 i e m l R m l . cos sin 2 15 2 1 0 , 3 ˆ 2 , 3 2 2i e m l R m l . ) 3 sin 5 (sin 3 16 1 0 , 3 ˆ 1 , 3 i e m l R m l . ) 3 cos 5 cos 3 ( 8 1 0 , 3 ˆ 0 , 3 m l R m l . ) 3 sin 5 (sin 3 16 1 0 , 3 ˆ 1 , 3 i e m l R m l . cos sin 2 15 2 1 0 , 3 ˆ 2 , 3 2 2i e m l R m l . 3 3 sin 5 4 1 0 , 3 ˆ 3 , 3 i e m l R m l . where 3 3 3 3 sin 35 8 1 ) , ( i e Y . cos sin 2 105 4 1 ) , ( 2 2 2 3 i e Y . ) 1 cos 5 ( sin 21 8 1 ) , ( 2 1 3 i e Y . ) cos 3 cos 5 ( 7 4 1 ) , ( 3 0 3 Y . ) 1 cos 5 ( sin 21 8 1 ) , ( 2 1 3 i e Y . cos sin 2 105 4 1 ) , ( 2 2 2 3 i e Y . 3 3 3 3 sin 35 8 1 ) , ( i e Y . (d) l = 4 4 4 sin 2 35 8 1 0 , 4 ˆ 4 , 4 i e m l R m l . cos sin 35 4 1 0 , 4 ˆ 3 , 4 3 3 i e m l R m l . 2 2 sin )] 2 cos( 7 5 [ 2 5 8 1 0 , 4 ˆ 2 , 4 i e m l R m l . )] 4 sin( 7 ) 2 sin( 2 [ 5 32 1 0 , 4 ˆ 1 , 4 i e m l R m l . )] 4 cos( 35 ) 2 cos( 20 9 [ 64 1 0 , 4 ˆ 0 , 4 m l R m l . )] 4 sin( 7 ) 2 sin( 2 [ 5 32 1 0 , 4 ˆ 1 , 4 i e m l R m l . 2 2 sin )] 2 cos( 7 5 [ 2 5 8 1 0 , 4 ˆ 2 , 4 i e m l R m l cos sin 35 4 1 0 , 4 ˆ 3 , 4 3 3 i e m l R m l 4 4 sin 2 35 8 1 0 , 4 ˆ 4 , 4 i e m l R m l where 4 4 4 4 sin 2 35 16 3 ) , ( i e Y cos sin 35 8 3 ) , ( 3 3 3 4 i e Y ) 1 cos 7 ( sin 2 5 8 3 ) , ( 2 2 2 2 4 i e Y ) cos 3 cos 7 ( sin 5 8 3 ) , ( 3 1 4 i e Y ) 3 cos 30 cos 35 ( 1 16 3 ) , ( 2 4 0 4 Y ) cos 3 cos 7 ( sin 5 8 3 ) , ( 3 1 4 i e Y ) 1 cos 7 ( sin 2 5 8 3 ) , ( 2 2 2 2 4 i e Y cos sin 35 8 3 ) , ( 3 3 3 4 i e Y 4 4 4 4 sin 2 35 16 3 ) , ( i e Y See the Table of spherical harmonics (in detail) 9. Addition theorem Let 1 1 1 1 1 ˆ( , ) ( , ) z z R n e e with the geometrical rotation defined by 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 cos 0 sin sin sin cos sin cos cos sin sin cos cos ) ( ) ( ) , ( y z 1 0 0 0 cos sin 0 sin cos ) ( 1 1 1 1 1 z , cos 0 sin 0 1 0 sin 0 cos ) ( 1 y The unit vector n1 is defined by 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 ( , ) cos cos sin sin cos 0 cos sin cos sin sin 0 sin 0 cos 1 sin cos sin sin cos z n e Another ket 2 n is defined by 2 2 2 2 2 ˆ( , ) ( , ) z z R n e e where the geometrical rotation matrix is given by 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 cos 0 sin sin sin cos sin cos cos sin sin cos cos ) ( ) ( ) , ( y z Note that 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ( , ) cos cos sin sin cos 0 cos sin cos sin sin 0 sin 0 cos 1 sin cos sin sin cos z n e Then we have the inner product as 1 2 1 2 1 2 1 2 cos cos sin sin cos( ) n n We assume a new ket defined by 2 1 1 1 2 1 1 1 ) , ( ) , ( ˆ ' n n n R , and 2 1 1 ˆ ' ( , ) R n n Note that the matrix ) , ( 1 1 1 is given by 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 cos sin sin cos sin 0 cos sin sin sin cos cos cos ) , ( and 1 1 1 2 1 2 1 2 1 2 2 1 2 1 2 1 2 1 2 ' ( , ) sin cos cos sin cos( ) sin sin( ) cos cos sin sin cos( ) ( , ) sin cos sin sin cos z n n e Since 1 2 1 2 1 2 ' cos cos sin sin cos( ) z n e it is found that 1 2 ' cos z n n n e In other words, is the angle between ' n and z e and it is also the angle between 1 n and 2 n . Fig.4 Three vectors 1 1 1 ( , ) z n e . 2 2 2 ( , ) z n e . 1 1 1 2 ' ( , ) ( , ) z n n e . The Mathematica is used for this figure. 60 1 . 45 2 . 30 1 . 50 2 . Without Mathematica, it is hard for me to get the line for the unit vector n’. Using the closure relation, we get ' 1 1 2 2 ' ' 1 1 2 1 1 2 , ) , ( ˆ ' , ) , ( , ) , ( ˆ ' , ' , , ) , ( ˆ ' ) , ( m m l m m l m l R m l Y m l R m l m l m l R lm Y n n n (1) where m l R m l D l m m , ) , ( ˆ ' , ) , ( 1 1 1 1 ) ( ' Equation (1) relates spherical harmonics in three different directions. The most useful case is m = 0; ) ( ) , ( 1 ) ( 0 ' ' 1 1 ) ( 0 ' 1 l m m im l m m d e D where 1 1 ' ' 1 ) ( 0 ' )] , ( [ 1 2 4 ) ( 1 m l im l m m Y l e d and ) (cos )] , ( [ 1 2 4 ) ( 1 1 1 ' 0 ' 1 ) ( 0 , 0 ' l m l l m m P Y l d Then we have m m l m l m m l m l im im m l m m im m l m l m m m l m l Y Y l Y Y e e l d e Y D Y Y ) , ( )] , ( [ 1 2 4 ) , ( )] , ( [ 1 2 4 ) ( ) , ( ) , ( ) , ( ) , ( 2 2 1 1 ' 2 2 ' 1 1 ' ' ' ' 1 ) ( 0 ' ' 2 2 ' ' 1 1 ) ( 0 ' 2 2 ' 0 1 1 1 which leads to the addition theorem for the spherical harmonics m m l m l l Y Y l P ) , ( )] , ( [ 1 2 4 ) (cos 2 2 1 1 (a) ˆ ' R n n , ˆ ' R n n , ˆ ' R n n Using the closure relation, we get m l R m l m l m l R m , ˆ ' , ' , , ˆ ' and ' ˆ ˆ ' , ' , ' , ' , m R l m l m l m R l m n n Noting that ˆ ' R n n , we get ' ˆ , ' , ' , ' , m l m l m l m R l m n n , or ' ( ) ', ' ˆ ( ) ( ') ( ) m m l l l m m m Y Y D R n n . (b) ˆ ' R n n ,, ˆ ' R n n Using the closure relation, we get m l R m l m l m l R m , ˆ ' , ' , , ˆ ' and m l R m l m l m l R m , ˆ ' , ' , ' , ˆ ' ' n n Noting that ˆ ' R n n , we get ' ˆ , ' , ' , ' , m l m l m l m R l m n n or ' ( ) ', ' ˆ ( ) ( ') ( ) m m l l l m m m Y Y D R n n 10. SphericalPlot3D of 2 ) , ( m l Y (i) 0 , 0 m l n 2 1 ) , ( 0 0 Y l = 0, m = 0 Fig.5 SphericalPlot3D of 2 0 0 ) , ( Y (ii) m l , 1 n (m = 1, 0, -1) r iy x e Y i ) ( 2 3 2 1 sin 2 3 2 1 ) , ( 1 1 , r z Y 4 3 cos 4 3 ) , ( 0 1 , r iy x e Y i ) ( 2 3 2 1 sin 2 3 2 1 ) , ( 1 1 . l = 1, m = ±1 l = 1, m = 0 Fig.6 SphericalPlot3D of 2 1 ) , ( m Y with m = 1, 0, -1. (ii) m l n , 2 (m = 2, 1, 0, -1, -2) 2 2 2 2 2 2 ) ( 2 15 4 1 sin 2 15 4 1 ) , ( r iy x e Y i , 2 1 2 ) ( 2 15 2 1 cos sin 2 15 2 1 ) , ( r z iy x e Y i , ) 3 ( 5 4 1 ) 1 cos 3 ( 5 4 1 ] 2 cos 3 1 [ 5 8 1 ) , ( 2 2 2 2 0 2 r r z Y , 2 1 2 ) ( 2 15 2 1 cos sin 2 15 2 1 ) , ( r z iy x e Y i , 2 2 2 2 2 2 ) ( 2 15 4 1 sin 2 15 4 1 ) , ( r iy x e Y i . l = 2, m = ±2 l = 2, m = ±1 l = 2, m = 0 Fig.7 SphericalPlot3D of 2 2 ) , ( m Y with m = 2, 1, 0, -1, -2. (iii) m l , 3 n (m = 3, 2, 1, 0, -1, -2, -3) 3 3 3 3 3 3 ) ( 35 8 1 sin 35 8 1 ) , ( r iy x e Y i , 3 2 2 2 2 3 ) ( 2 105 4 1 cos sin 2 105 4 1 ) , ( r iy x z e Y i , ) )( 5 ( 21 8 1 sin ) 1 cos 5 ( 21 8 1 sin )] 2 cos( 5 3 [ 21 16 1 ) , ( 2 2 2 2 1 3 iy x r r z e e Y i i , 3 2 2 2 0 3 ) 3 5 ( 7 16 4 ) 3 cos 5 ( cos 7 16 4 ) 3 cos( 5 cos 3 [ 7 16 1 ) , ( r r z z Y , ) )( 5 ( 21 8 1 sin ) 1 cos 5 ( 21 8 1 sin )] 2 cos( 5 3 [ 21 16 1 ) , ( 2 2 2 2 1 3 iy x r r z e e Y i i , 3 2 2 2 2 3 ) ( 2 105 4 1 cos sin 2 105 4 1 ) , ( r iy x z e Y i , 3 3 3 3 3 3 ) ( 35 8 1 sin 35 8 1 ) , ( r iy x e Y i . l = 3, m = ±3 l = 3, m = ±2 l = 3, m = ±1 l = 3, m = 0 Fig.8 SphericalPlot3D of 2 3 ) , ( m Y with m = 3, 2, 1, 0, -1, -2, -3 ((Mathematica-3, SphericalPlot3D)) _______________________________________________________________________ l = 0 (m =0): SphericalPlot3D Fig.9 _____________________________________________________________________ l = 1 (m =1, 0): SphericalPlot3D Fig.10 ________________________________________________________________________ l = 2 (m =2, 1, 0): SphericalPlot3D Fig.11 ________________________________________________________________________ l = 3 (m = 3, 2, 1, 0): SphericalPlot3D Fig.12 ________________________________________________________________________ l = 4 (m =4, 3, 2, 1, 0): SphericalPlot3D Fig.13 ____________________________________________________________ l = 6, m = 6, 5, 4, 3, 2, 1, 0: SphericalPlot3D Fig.14 11. Contour plot of )] , ( Re[ m l Y on the Bloch sphere Here we show a contour plot of )] , ( Re[ m l Y for several spherical harmonics, where the contours of )] , ( Re[ m l Y =constant are drawn on the unit sphere (the Bloch sphere). Since the spherical harmonics are functions on the unit sphere, the figures show a series of balls with contours drawn on them. We show the plot contours on which the squares of the real part of the spherical harmonics is constant. The contours on which 0 )] , ( Re[ m l Y are denoted by red lines, while the contours on which 0 )] , ( Re[ m l Y are denoted by blue lines. For large l, )] , ( Re[ l l Y is significantly non-zero only where 1 sin , i.e., around the equator. As m decreases from m = l to m = 0, the region of the unit sphere in which )] , ( Re[ l l Y is significantly non-zero gradually spreads from the equator towards the poles. For large l the phase of )] , ( Re[ l l Y changes rapidly with . As m decreases, the change of the phase of )] , ( Re[ l l Y becomes smaller. ((Mathematica-4)) (i) m l , 3 state (a) l = 3, m = 0 (b) l = 3, m = 1 (c) l = 3, m = 2 (d) l = 3, m = 3 (ii) m l , 15 state (a) l = 15, m = 2 (b) l = 15, m = 7 (c) l = 15, m = 15 Fig.15 Contour plot of )] , ( Re[ m l Y on the Bloch sphere 12. Rotational motion We define the unit: Kaiser (cm-1) as ] [ ) ( 10 23984 .
1 2 4 eV cm c ck E ℏ ℏ ℏ This is the relation between energy (erg) and Kaiser (cm-1). When one discusses the optical spectrum, the unit (Kaiser, cm-1) is conventionally use. The Hamiltonian for the rotation is given by z z y z x x I L I L I L H 2 ˆ 2 ˆ 2 ˆ ˆ 2 2 2 where Ix, Iy, and Iz are the moment of inertia. (a) I I I I z y x (isotropic) 2 ˆ ˆ 2 H I L . m l, is the eigenket of 2 ˆ \ L with the eigenvalue ) 1 ( 2 l l ℏ ; 2 2 ˆ , ( 1) , l m l l l m L ℏ . Then we have m l l E m l I l l m l H , ) ( , 2 ) 1 ( , ˆ 2 ℏ . where m = l, l-1,...., -l+1, and -l. The energy level is given by I l l l E 2 ) 1 ( ) ( 2 ℏ . The degeneracy of the energy level is (2l + 1). Note that ) 1 ( 2 ) 1 ( 2 ) 2 )( 1 ( ) ( ) 1 ( 2 2 2 l I I l l I l l l E l E ℏ ℏ ℏ . The wave function is given by , , .
( , ) m l l m l m Y n . (b) y x I I , but x z I I 2 2 2 2 2 2 ˆ ˆ ˆ ˆ ˆ ˆ ˆ 2 2 2 2 x z z z z x z x z L L L L L H I I I I L . Since 2 ˆ ˆ [ , ] 0 z L L , m l, is the simultaneous eigenket of 2 ˆ L and z L ˆ with the eigenvalue ) 1 ( 2 l l ℏ and m ℏ , respectively. 2 2 ˆ , ( 1) , l m l l l m L ℏ , m l m m l Lz , , ˆ ℏ . Then we have m l m l E m l H , ) , ( , ˆ where z x I m I m l l m l E 2 2 ] ) 1 ( [ ) , ( 2 2 2 2 ℏ ℏ We note that , ( , ) m l l m Y n . 13. Spherical harmonics in the Cartesian coordinate Using the relation given by cos sin r x , sin sin r y , cos r z , the spherical harmonics can be expressed as follows, ) , ( 3 4 )] , ( ) , ( [ 3 2 )] , ( ) , ( [ 3 2 0 1 1 1 1 1 1 1 1 1 Y r z Y Y i r y Y Y r x ) , ( 3 4 2 ) , ( 3 4 2 1 1 1 1 Y r iy x Y r iy x ) , ( 15 2 4 ) ( ) , ( 15 2 2 ) ( ) , ( 5 4 ) ( 2 ) , ( 15 2 2 ) ( ) , ( 15 2 4 ) ( 2 2 2 2 1 2 2 0 2 2 2 2 2 1 2 2 2 2 2 2 Y r iy x Y r iy x z Y r y x z Y r iy x z Y r iy x )] , ( ) , ( [ 15 2 )] , ( ) , ( [ 15 2 )] , ( ) , ( [ 15 2 2 2 2 2 2 1 2 1 2 2 1 2 1 2 2 Y Y i r xy Y Y i r yz Y Y r zx ((Mathematica-5)) Spherical harmonics using the Cartesian co-ordinate (x, y, z) 14. Example-1 ((Sakurai 3-15)) The wave function of a particle subjected to a spherically symmetrical potential V(r) is given by ) ( ) ( ) ( r f z y x r (a) Is ) (r an eigenfunction of L2? If so, what is the l-value? If not, what are the possible values of l we may obtain when L2 is measured? (b) What are the probabilities for the particle to be found in various m states? Noting that ) , ( 3 4 )] , ( ) , ( [ 3 2 )] , ( ) , ( [ 3 2 0 1 1 1 1 1 1 1 1 1 Y r z Y Y i r y Y Y r x we have )] , ( 2 ) , ( ) 1 ( ) , ( ) 1 [( 3 2 0 1 1 1 1 1 Y Y i Y i r z y x Then, ) (r can be rewritten as ) ( )] , ( 2 ) , ( ) 1 ( ) , ( ) 1 [( 3 2 ) , , ( 0 1 1 1 1 1 r rf Y Y i Y i z y x This implies that ] 1 , 1 ) 1 ( 0 , 1 2 1 , 1 ) 1 ( [ 6 1 i i So we get ) 1 ( 2 2 l l ℏ L with l = 1. 3 1 ) , 1 ( 2 m m l P , which is independent of m. 15. Example-2 A particle moving in a potential is described by the wave packet )] ( exp[ ) ( ) ( 2 2 2 2 z y x zx yz xy r What is the probability that a measurement of L2 and Lz yields the results 6ℏ2 and ℏ, respectively? ((Solution)) We note that )] , ( ) , ( [ 15 2 )] , ( ) , ( [ 15 2 )] , ( ) , ( [ 15 2 2 2 2 2 2 1 2 1 2 2 1 2 1 2 2 Y Y i r xy Y Y i r yz Y Y r zx Then, we have )] , ( ) , ( [ 15 2 )] , ( ) , ( [ 15 2 1 2 1 2 1 2 1 2 2 Y Y i Y Y r zx yz xy )] , ( ) , ( [ 15 2 2 2 2 2 Y Y i or ] ) , ( ) , ( ) , ( ) 1 ( ) , ( ) 1 [( 15 2 2 2 2 2 1 2 1 2 2 i iY iY Y i Y i r zx yz xy . Thus, the wave function can be rewritten as 2 2 2 2 2 2 1 2 1 2 )] , ( ) , ( ) , ( ) 1 ( ) , ( ) 1 [( 15 2 ) , , ( r e r iY iY Y i Y i z y x , or ] 2 , 2 1 , 2 ) 1 ( 1 , 2 ) 1 ( 2 , 2 [ 6 1 i i i i . The probability that a measurement of L2 and Lz yields the results 6ℏ2 and ℏ, respectively (l = 2, m = 1) is 3 1 6 1 1 , 2 2 2 i P 16. The Series expansion of the Legendre differential equation We note that 0 ] ˆ , ˆ [ L , and 0 ] ˆ , ˆ [ 2 L , where ˆ is the parity operator and L ˆ is the orbital angular momentum. Then we have a simultaneous eigenket of ˆ and 2 ˆ L such that m l m l .
, ˆ , ) 1 ( , ˆ 2 2 l l m l ℏ L . We obtain m l m l m l , , , ˆ , , ˆ , . This means that under the parity operation, we get cos cos( ) cos x x , Here we consider the case of m = 0. The function ( ) y x is defined as ( ) , , 0 y x l m , where cos x . The function ( ) y x is either even or odd function of x , ( ) ( ) y x y x . depending on the value of l. The function ( ) y x satisfies the differential equation 2 (1 ) "( ) 2 '( ) ( ) 0 x y x xy x y x . where is the eigenvalue to be determined. So it is reasonable to assume that 2 2 0 0 ( ) (2 ) (2 ) p k k p k k y x x a k x a k x , where 0 ) 0 ( a and the index p is expected to be either 0 (even function) or 1 (odd function). We solve the second order differential equation by using the series expansion. Then we have 0 ) 0 ( ) 1 ( a p p , 0 ) 2 ( ) 2 )( 1 ( ) 0 ( )] 1 ( [ a p p a p p , 0 ) 4 ( ) 4 )( 3 ( ) 2 ( )] 3 )( 2 ( [ a p p a p p , 0 ) 6 ( ) 6 )( 5 ( ) 4 ( )] 5 )( 4 ( [ a p p a p p , 0 ) 8 ( ) 8 )( 7 ( ) 6 ( )] 7 )( 6 ( [ a p p a p p , 0 ) 10 ( ) 10 )( 9 ( ) 8 ( )] 9 )( 8 ( [ a p p a p p , …………………………………………………………… From the first equation, we have 0 p , or 1 p . In general, we have the recursion relation ) 2 ( ) 2 2 )( 1 2 ( ) 1 2 )( 2 ( ) 2 2 ( k a p k p k p k p k k a When ) 1 2 )( 2 ( p k p k , we have 0 ...
) 4 2 ( ) 2 2 ( k a k a Then the series terminates. The solution is just the Legendre polynomial. (i) 0 p We have ) 2 ( ) 2 2 )( 1 2 ( ) 1 2 )( 2 ( ) 2 2 ( k a k k k k k a with 0 ) 2 ( 2 ) 0 ( a a 0 ) 4 ( 12 ) 2 ( ) 6 ( a a 0 ) 6 ( 30 ) 4 ( ) 20 ( a a 0 ) 8 ( 56 ) 6 ( ) 42 ( a a 0 ) 10 ( 90 ) 8 ( ) 72 ( a a (ii) 1 p ) 2 ( ) 3 2 )( 2 2 ( ) 2 2 )( 1 2 ( ) 2 2 ( k a k k k k k a with 0 ) 2 ( 6 ) 0 ( ) 2 ( a a 0 ) 4 ( 20 ) 2 ( ) 12 ( a a 0 ) 6 ( 42 ) 4 ( ) 30 ( a a 0 ) 8 ( 72 ) 6 ( ) 56 ( a a 0 ) 10 ( 110 ) 8 ( ) 90 ( a a where ) 1 ( l l ((Mathematica-6)) Series expansion method Clear@"Global∗"D; eq1 = I1 −ξ2M D@ϕ@ξD, 8ξ, 2<D −2 ξ D@ϕ@ξD, ξD + L HL + 1L ϕ@ξD; DSolve@eq1 0, ϕ@ξD, ξD êê Simplify 88ϕ@ξD →C@1D LegendreP@L, ξD + C@2D LegendreQ@L, ξD<< eq2 = I1 −ξ2M D@ϕ@ξD, 8ξ, 2<D −2 ξ D@ϕ@ξD, ξD + λ ϕ@ξD; f1@x_D := x p ‚ k=0 6 a@2 kD x2 k; rule1 = 8ϕ →Hf1@ D &L<; eq3 = ξ2−p eq2 ê. rule1 êê Simplify; list1 = Table@8 n, Coefficient@eq3, ξ, 2 nD<, 8n, 0, 5<D êê FullSimplify; f2@x_D := x p ‚ m=−3 3 a@ 2 k + 2 mD x2 k+2 m; rule2 = 8ϕ →Hf2@ D &L<; eq4 = ξ8−p−2 k Heq2 ê. rule2L êê FullSimplify; list2 = Table@8 n, Coefficient@eq4, ξ, 2 nD<, 8n, 2, 6> seq13 = seq12 ê. p →0 ::a@2 + 2 kD → I2 + 6 k + 4 k2 −λM a@2 kD H2 + 2 kL H3 + 2 kL >> list1 ê. p →0 êê TableForm 0 0 1 λ a@0D + 2 a@2D 2 −H6 −λL a@2D + 12 a@4D 3 −H20 −λL a@4D + 30 a@6D 4 −H42 −λL a@6D + 56 a@8D 5 −H72 −λL a@8D + 90 a@10D list1 ê. p →1 êê TableForm 0 0 1 −H2 −λL a@0D + 6 a@2D 2 −H12 −λL a@2D + 20 a@4D 3 −H30 −λL a@4D + 42 a@6D 4 −H56 −λL a@6D + 72 a@8D 5 −H90 −λL a@8D + 110 a@10D (b) Legendre function determined from the series expansion (1) l = 0 0 , 0 ) 2 ( a ) 0 ( ) ( 0 a P (2) l = 1 2 , 0 ) 2 ( a ) 0 ( ) ( 1 a P (3) l = 2 6 , 0 ) 4 ( a , ) 0 ( 3 ) 2 ( a a ) 3 1 )( 0 ( ) ( 2 2 a P (4) l = 3 12 , 0 ) 4 ( a , ) 0 ( 3 5 ) 2 ( a a ) 5 3 ( 3 ) 0 ( ) 3 5 1 ( ) 0 ( ) ( 3 2 3 a a P (5) l = 4 20 , 0 ) 6 ( a , ) 0 ( 10 ) 2 ( a a , ) 0 ( 3 35 ) 4 ( a a ] 35 30 3 [ 3 ) 0 ( ) 3 35 10 1 )( 0 ( ) ( 4 2 4 2 4 a a P Note that the value of a(0) can be determined from the normalization such that 1 2 2 ) ( ) ( ' , 1 1 ' l P P d l l l l (c). Legendre function (Mathematica) ((Mathematica)) Clear@"Global`∗"D; Table@8L, LegendreP@L, ξD<, 8L, 0, 10<D êê TableForm 0 1 1 ξ 2 1 2 I−1 + 3 ξ2M 3 1 2 I−3 ξ + 5 ξ3M 4 1 8 I3 −30 ξ2 + 35 ξ4M 5 1 8 I15 ξ −70 ξ3 + 63 ξ5M 6 1 16 I−5 + 105 ξ2 −315 ξ4 + 231 ξ6M 7 1 16 I−35 ξ + 315 ξ3 −693 ξ5 + 429 ξ7M 8 1 128 I35 −1260 ξ2 + 6930 ξ4 −12 012 ξ6 + 6435 ξ8M 9 1 128 I315 ξ −4620 ξ3 + 18018 ξ5 −25 740 ξ7 + 12 155 ξ9M 10 1 256 I−63 + 3465 ξ2 −30030 ξ4 + 90 090 ξ6 −109 395 ξ8 + 46 189 ξ10M ____________ REFERENCE A.R. Edmonds, Angular Momentum in Quantum Mechanics (Princeton University, 1957). M.E. Rose, Elementary Theory of Angular Momentum (Dover, 1995). J. Binney and D. Skinner, The Physics of Quantum Mechanics (Oxford, 2014). E. Merzbacher, Quantum Mechanics, 3rd edition (John Wiley & Sons, New York, 1998). A. Goswami, Quantum Mechanics, 2nd edition (Waveland Press, 2003). D.H. McIntyre, Quantum Mechanics A Paradigms Approach (Pearson Education, 2012). APPENDIX: Recursion formula ˆ ˆ ˆ x y L L iL , ˆ ˆ ˆ x y L L iL . ˆ , ( )( 1) , 1 L l m l m l m l m ℏ . ˆ , ( )( 1) , 1 L l m l m l m l m ℏ . (a) Method-I: the use of ˆ L Suppose that , l m l n is given exactly. We determine the exact form of , l m n in the following way. ˆ ,0 ( 1) ,1 L l l l l ℏ ˆ ,1 ( 1)( 2) ,2 L l l l l ℏ ˆ ,2 ( 2)( 3) ,3 L l l l l ℏ ………………………………………………………………………. ˆ , 2 ( 2)( 1) , 1 L l m l m l m l m ℏ ˆ , 1 ( 1)( ) , L l m l m l m l m ℏ By multiplying terms of both sides separately, we have ˆ ( ) ,0 ( 1)....( 1) ( 1)( 2)....( ) , m m L l l l l m l l l m l m ℏ or 1 ( )! ˆ , ( ) ,0 ( )!
m m l m l m L l l m ℏ or 1 ( )!
, ( ) ,0 ( )!
m m l m l m L l l m n n ℏ where L is the differential operator and is given by ( cot ) i L e i ℏ In Mathematica, this differential operator is given by ( [#, ] cot [#, ])& i L e D i D ℏ (b) Method-II: the use of ˆ L Suppose that , l m l n is given exactly. We determine the exact form of , l m n in the following way. ˆ , 2 1 , 1 L l l l l l ℏ , ˆ , 1 (2 1) 2 , 2 L l l l l l ℏ , ˆ , 2 (2 2) 3 , 3 L l l l l l ℏ ………………………………………………………………………………. ˆ , 2 ( 2)( 1) , 1 L l m l m l m l m ℏ ˆ , 1 ( 1)( ) , L l m l m l m l m ℏ By multiplying these terms of both sides separately, we have (2 )!( )!
ˆ ( ) , , ( )!
l m l m l l m L l l l m l m ℏ or 1 ( )!
ˆ , ( ) , (2 )!( )!
l m l m l m l m L l l l l m ℏ or 1 ( )!
, ( ) , (2 )!( )!
l m l m l m l m L l l l l m n n ℏ where L is the differential operator ( cot ) i L e i ℏ In Mathematics, the differential operator is given by ( [#, ] cot [#, ])& i L e D i D ℏ ((Note)) ( cot ) i L e i ℏ , ( cot ) i L e i ℏ , with (sin cot cos ) x L i ℏ . ( cos cot sin ) y L i ℏ . z L i ℏ . (a) Differential operator: ( cot ) i L e i L ℏ . For the operators in quantum mechanics ˆ ˆ ˆ ˆ x y L L iL L , ˆ ˆ ˆ ˆ x y L L iL L . ˆ ˆ ˆ [ ] L L L L L L n n n n n n ˆ ˆ ˆ [ ] L L L L L L n n n n n n Note that ˆ L L n n , ˆ L L n n . (b) ˆ ˆ ˆ [ ( cot ) ( )] ( cot ) ( ) ( ) i i L L L e i e i L L n n n n n n n ℏ ℏ ˆ ˆ ˆ [ ( cot ) ( )] ( cot ) ( ) ( ) i i L L L e i e i L L n n n n n n n ℏ ℏ ((Mathematica)) |
6862 | https://www.forbes.com/councils/forbesbusinesscouncil/2022/11/28/the-trusted-advisor-role-do-you-have-it/ | Flash Sale: Less than $1/week
The Trusted Advisor Role—Do You Have It?
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The Trusted Advisor Role—Do You Have It?
ByNorma Watenpaugh,
Former Forbes Councils Member.
for
Forbes Business Council
COUNCIL POST
Expertise from Forbes Councils members, operated under license. Opinions expressed are those of the author.
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Nov 28, 2022, 09:45am EST
CEO of PhoenixCG specializing in collaborative business ecosystems to accelerate innovation and open new markets.
In the world of ecosystem partnering, everyone clamors to be "the trusted advisor," often without fully understanding what that means. Not every partner in your ecosystem can lay claim to that role—a role that, as it turns out, can be multifaceted. It’s possible to claim that role at a strategic level in advising clients about their business, at an operational level in managing the day-to-day or even in niche specialties. Few can claim all of the above.
The Heavy Lift Of Advising
Providing business strategy or transformation consulting is typically the most influential trusted advisor role—one that is provided by advisory and consulting firms and is also very much related to the role of building the customer value proposition and business case for transformational change. This is a heavy load to carry.
These services address the heart of the customer business model. For example, if your client is a bank, then how can you help their customers feel financially secure? If your client is a hospital, how can you help them improve patient outcomes and avoid litigation by reducing medical mistakes? For a manufacturer, how can you help them navigate supply chain disruption? Essentially, how do you help clients be more competitive, more profitable and able to provide better customer experiences and outcomes? How do you help them not be overrun by disruption, or even to be the disruptor in the market themselves?
Once the vision for transformation is set, it's then time to architect the change through people and processes and to specify the technology and infrastructure needed to achieve the anticipated outcomes that lead to acquisition, implementation and operation.
It can be very frustrating to product-oriented partners that the sale and recurring revenue happen toward the end of the transformation process. The key to success for product-oriented companies is to be sure your products offer a better value proposition in achieving the anticipated outcome so that you get spec’d into that transformation strategy early on.
Understanding The "Trust" Part Of Trusted Advisor
Partners who can bring deep industry expertise are particularly valuable in providing a compelling customer experience. These advisors know and understand the drivers and disrupters in the customer’s industry. They know what problems they are facing that are unique to their business, as well as how others have overcome those problems. This understanding is what creates the "trust" part of a trusted advisor.
At the onset of the pandemic, for example, partners who knew the healthcare industry were in a prime position to deliver digital health systems as the demand for remote health services soared. Organizations that just knew how to build a portal were at a great disadvantage to those who knew and understood the healthcare industry.
Because today’s technology implementations for digital transformation are complex and comprised of multiple vendors and services, strong relationships among these partners are crucial. Customers want an integrated, worry-free solution, and it is incumbent on partners to provide it.
When there are issues in integration or performance, you must work together to resolve them instead of leaving your customers with a never-ending cycle of finger-pointing and blame. This reinforces the role of the trusted advisor who takes responsibility for orchestrating vendors of the component products and services to ensure problems get solved without dragging the customer into the middle of it all.
Customers don’t want to buy a bag of parts (some assembly required). Customers are seeking business outcomes and expect a complete solution to deliver those outcomes. This, of course, requires collaboration and integration of products and services around the customer value proposition. And customers need to have confidence that their trusted advisor can deliver. Without that, the customer is left holding the bag (of parts).
Forbes Business Council is the foremost growth and networking organization for business owners and leaders. Do I qualify?
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6863 | https://brainly.com/question/20863509 | [FREE] What is the hypotenuse from the following Pythagorean triple? 25 in., 15 in., 20 in. - brainly.com
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What is the hypotenuse from the following Pythagorean triple?
25 in., 15 in., 20 in.
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The hypotenuse of the Pythagorean triple 25 in, 15 in, 20 in is 25 in
The Pythagoras theorem is used to find the sides of a right angle triangle. The formula below is used to find the sides of a right angle triangle(Pythagoras's theorem) is represented below
c² = a² + b²
where
c = hypotenuse
a and b are either opposite or adjacent sides
A right angle triangle is a triangle that has one of its angles as 90°.
A right angle triangle has the following sides :
Hypotenuse
Opposite side
Adjacent side
The longest side of a right angle triangle is the hypotenuse. Base on the Pythagorean triple, the hypotenuse is 25 inches because its the longest side.
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The hypotenuse of the Pythagorean triple 25 in., 15 in., and 20 in. is 25 in. This is the longest side of the right triangle. The lengths are confirmed by the Pythagorean theorem, demonstrating that 625 = 400 + 225.
Explanation
To determine the hypotenuse from the Pythagorean triple of 25 in., 15 in., and 20 in., we first identify the longest side of the triangle. In the context of a right triangle, the hypotenuse is defined as the side opposite the right angle and is always the longest side.
In this case, we have the following lengths:
25 in.
15 in.
20 in.
Among these sides, 25 in. is the longest side. Therefore, according to the Pythagorean theorem, which states that
c 2=a 2+b 2 where c represents the hypotenuse and a and b represent the other two sides, the hypotenuse of the triangle is indeed 25 in.
To verify, we can check the relationship:
2 5 2=2 0 2+1 5 2 625=400+225 625=625 This confirms that the Pythagorean triple is valid and that the hypotenuse is confidently 25 inches.
Examples & Evidence
For example, if you have a right triangle with legs of 3 in. and 4 in., you can find the hypotenuse using the same method: c 2=3 2+4 2 resulting in c=5 in.
The definition of the Pythagorean theorem and the verification calculation provided confirms that our answer is accurate.
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Hypotenuse is the longest side in the right angle triangle. It is proven by the fact that the square of hypotenus is the sum of the squares of the two other sides.
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6864 | https://brilliant.org/wiki/triangles-circumcenter/ | Circumcenter
Xuming Liang,
Sandeep Bhardwaj,
Andrew Ellinor,
and
Anuj Shikarkhane
Ken Jennison
Brilliant Mathematics
Worranat Pakornrat
Karthik Sharma
Calvin Lin
Jimin Khim
contributed
Contents
Basic Properties
Worked Examples
Applying Properties of the Circumcenter
Recognizing Concyclic Points
See Also
Basic Properties
Let's look into the basic properties of circumcircles and some relevant examples. We can derive angle-related properties using the properties of angle in a circle (inscribed angle). Writing down the properties, however, may be a bit tricky since the circumcenter O can lie either outside or inside triangle ABC.
Properties
O is the intersection of perpendicular bisectors of the sides of triangle ABC.
Connecting the circumcenter O to the vertices of triangle ABC gives three isosceles triangles OAB,OBC,OCA.
R=2sinAa=4Sabc follows from the extended sine rule, where S denotes the area of the triangle ABC.
Location of circumcenter differs for the acute, obtuse, and right-angled triangles. This can be deduced from the central angle property:
If ∠B is acute, then ∠BOC=2∠A.
If ∠B is right, then O lies on the midpoint of AC.
If ∠B is obtuse, then O lies on the opposite side of AC from B and ∠BOC=2∠A.
Moreover, the alternate segment theorem states that the angle formed by the tangent to the circle and one of the sides of the triangle is equal to the angle between the other two sides:
In the next section we will learn how these applications can be used for solving problems.
Worked Examples
Now that we've gone through the properties possessed by a circumcircle and its circumcenter, we will apply them to work out some examples and problems which should make you more confident in their usage.
What is the circumcenter of triangle ABC with vertices A=(1,4),B=(−2,3),C=(5,2)?
By definition, a circumcenter is the center of the circle in which a triangle is inscribed. For this problem, let O=(a,b) be the circumcenter of △ABC. Then, since the distances to O from the vertices are all equal, we have AO=BO=CO.
From the first equality, we have
AO2(a−1)2+(b−4)2−2a+1−8b+163a+b=BO2=(a+2)2+(b−3)2=4a+4−6b+9=2.(1)
Similarly, from the second equality, we have
BO2(a+2)2+(b−3)24a+4−6b+97a−b=CO2=(a−5)2+(b−2)2=−10a+25−4b+4=8.(2)
Taking (1)+(2) gives a=1, which in turn gives b=−1.
Therefore, the circumcenter of triangle ABC is O=(1,−1). □
We can use other properties of the circumcircle to help us find the circumcenter to solve this problem. For example, you can use the fact that it is the intersection of the perpendicular bisectors.
What is the radius of the circumcircle for the equilateral triangle with side length 2?
Let O be the circumcenter, and D the foot of the perpendicular from A to BC. Since OB=OC, this tells us that O lies on the perpendicular bisector of BC, and thus O lies on AD. In this diagram, we have OA=OB=OC=R, the radius of the circumcircle. Since BD=CD=1, the height of triangle ABC is AD=3. Then, from the right triangle OBD, we obtain OD2+BD2=OB2, or (3−R)2+12=R2. Expanding this and solving for R, we obtain 3−23R+R2+1=R2⟹4=23R⟹R=323. □
We have worked out the above examples so that you can get familiar with using the properties above in problem solving. Now it's your turn to try a couple of problems on your own and make the properties worth learning.
Triangle ABC is an isosceles triangle with AB=BC and ∠ABC=123∘. Point D is the midpoint of AC, point E is the foot of the perpendicular from D to BC, and point F is the midpoint of DE. The intersection point of AE and BF is G. What is the measure (in degrees) of ∠BGA?
The correct answer is: 90
Let H be the midpoint of AB. Since BD⊥AC, hence H is the circumcenter of BAD, which we will call Γ. Since DH∥BC⊥DE, this shows that DE is tangential to Γ. Let BF intersect Γ at G′.
Then, FE2=FD2=FG′⋅FB by power of a point. This shows that FEG′ and FBE are similar triangles, and thus FG′E is a right angle. Since AG′B is also a right angle, it shows that A,G′,E are collinear, and thus that G′ is the intersection of BF and AE, and thus G=G′. Hence, ∠AGB=90∘.
In ΔABC,
AB2+BC2+AC2=cosAsinBsinC+sinAcosBsinC+sinAsinBcosC.
If the area of the circumcircle of ΔABC can be represented as baπ, where a and b are coprime positive integers, what is the value of a+b?
Note: This problem is posed by Kshitij J.
The correct answer is: 9
By the Law of Cosines and Law of Sines,
sinAsinBcosC=4R2BC⋅AC⋅2⋅BC⋅ACBC2+AC2−AB2=8R2BC2+AC2−AB2,
where R is the radius of the circumcircle. Therefore,
cosAsinBsinC+sinAcosBsinC+sinAsinBcosC=8R2AB2+AC2−BC2+8R2AB2+BC2−AC2+8R2BC2+AC2−AB2=8R2BC2+AC2+AB2.
Since this is equal to BC2+AC2+AB2, we have 8R2=1. Therefore,
R2=81and πR2=8π,
implying a=1,b=8, and a+b=9.
Find the area of the circumcircle of an isosceles triangle ABC above, where a=b and ∠A=15∘, provided that the perimeter of the triangle is 25 cm.
Let R be the circumradius of △ABC. Then, from the extended sine rule, we have
sinAa=sinBb=sinCc⇒R=2R=2sinAa.
Since we are given a=b and a+b+c=25, it follows that c=25−2a, which implies
sinAa=sinCc=sinC25−2a.
Since it is given that ∠A=15∘ and ∠C=150∘, solving further gives a=1+4sin15∘50sin15∘ and R=1+4sin15∘25.
Hence the area of the circumcircle is πR2=π(1+4sin15∘25)2=474.005. □
Find the area of rhombus ABCD given that the radii of the circumcircles of triangles ABD and ACD are 12.5 and 25, respectively.
The radius of the circumcircle of triangle ABD is (2sinAd1)=12.5.(1)
The radius of the circumcircle of triangle ACD is (2sin(180∘−A)d2)=25.(2)
Taking (2)÷(1), we get d1d2=2⟹d2=2d1.(3)
The area of the rhombus is 2d1d2. Using the formula for the radius of the circumcircle of a triangle, we can write the area of △ABC as
A=4Rabc=(4)(25)s2d2,
where the side length of the rhombus s can be written as 21d12+d22. Since twice the area of this triangle is equal to the area of the rhombus, we have
2d1d2=(8)(25)(d12+d22)d2⟹20d1=d12.(since d2=2d1)
This means d1=20 and d2=40. Therefore, the area of the rhombus is 400. □
Applying Properties of the Circumcenter
In this section, we will try problems that can be greatly simplified by applying properties of the circumcenter, even though the problems themselves don't directly involve the circumcenter. Since the circumcenter is a rich structure that interrelates angles and lengths, using it correctly in a problem (e.g. International Mathematical Olympiad, or IMO) can be very powerful. For this reason, it is important to know how to spot circumcenters and the appropriate moment to use them.
O is the circumcenter of ABC if and only if
AO=BO=CO;
BO=CO and ∠BOC=2∠A when ∠A is acute and A,O are on the same side of BC;
BO=CO and ∠BOC=2(180−∠A) when ∠A is obtuse and A,O are on opposite sides of BC.
Let's see some of these criteria in action.
Quadrilateral ABCD satisfies ∠A=76∘,∠B=72∘,∠C=142∘, and AD=AB. Find ∠ACB−∠ACD.
Since AD=AB and the angles of the quadrilateral are given, it does not hurt to see if A is a circumcenter.
By the definition of a quadrilateral we know A,C are on opposite sides of BD. In addition,
2(180∘−∠BCD)=2(180∘−142∘)=76∘=∠BAD.
Then, from criterion number 3 above, we know that A is the circumcenter of △BCD.
Therefore, ∠ACB=∠ABC=72∘,∠ACD=∠ADC=360∘−76∘−72∘−142∘=70∘, and our answer is 72∘−70∘=2∘. □
Note: As one can see, realizing that A is the circumcenter of △BCD immediately made the problem clear. It is fine to say that the circumcenter is the hidden configuration in this problem, and revealing it is the key. Now imagine approaching this with trigonometry. Although considering it is natural, using trig here will definitely be a pain.
(Indian National Mathematical Olympiad 2015)
Let ABC be a right-angled triangle with ∠B=90∘. Let BD be the altitude from B on to AC. Let P,Q and I be the incenters of triangles ABD,CBD and ABC, respectively. Show that the circumcenter of triangle PIQ lies on the hypotenuse AC.
Suppose O is the circumcenter of triangle PIQ. Since ∠PIQ=∠AIC=135∘, by the angle properties of cirumcenter ∠POQ=2(180−135)=90∘. Note that ∠PDQ=90∘. Hence D,O,P,Q are concyclic.
Since PO=QO, DO externally bisects ∠PDQ. However, AC is also the external bisector of ∠PDQ. In conclusion, O must lie on AC. □
In △ABC, let D be the midpoint of BC. If ∠ADB=45∘ and ∠ACD=30∘, what is ∠BAD in degrees?
The correct answer is: 30
The following is an advanced example that requires you to recognize the circumcenter and utilize it to perform crucial angle chasing.
△ABC is a right triangle with ∠B=90∘. D is a point inside △ABC such that AD=AB, and E is on AC satisfying DE⊥BD. Also, F is the point where the circumcircle of △CDE intersects BC.
Prove that BE=EF.
We want to show that ∠EBF=∠EFB. Since ∠EFB=180∘−∠EFC=180∘−∠EDC, it suffices to show that ∠EBC+∠EDC=180∘.
We almost want to say they are opposite angles of a cyclic quad, but B,D lie on the same side of EC, so that won't be possible. This motivates the construction of D′ as the reflection of D in the line EC, and it suffices to show that BED′C is a cyclic quad. There are many different approaches that we could use, and we will show that ∠ECB=∠ED′B. We pick this angle because ∠ACB is fixed in this question, while D,E,D′ are variable points.
Now, observe that AB=AD=AD′, so A is the circumcenter of △BDD′. This means that ∠BD′D=21∠BAD. This motivates constructing M as the midpoint of BD, so that we get ∠BD′D=∠BAM. Hence, it suffices to show that ∠ED′B+∠BD′D=∠ACB+∠BAM.
Observe that AM∥ED since they are both perpendicular to BD. We then have ∠ACB+∠BAM=90∘−∠MAC=90∘−DEC=∠EDD′. This is equal to ∠ED′D due to isosceles triangle ED′D. Hence we are done. □
Recognizing Concyclic Points
Our first problem can be solved by constructing similar triangles and proceeded by angle chasing. See below for a more direct proof that arose from recognizing concyclic points.
Three circles with the same radius intersect at the center of a bigger circle with twice the radius.
Lines are drawn from the intersection points where two small circles meet on the big circle such that they form a triangle, as shown above left. The triangle is then divided into 3 quadrilaterals by drawing the lines from the center of the big circle to the intersection points of the small circles: red, blue, and green regions.
Alternatively, the triangle can be divided into 3 smaller triangles by simply drawing the lines from the center to the vertices: yellow, purple, and cyan regions in the diagram on the right.
Suppose that the ratio of the area of the red region, the area of the dark blue region, and the area of the green region is 11:12:13.
Then, if the ratio of area of the yellow region, the area of the purple region, and the area of the light blue region is a:b:c, where gcd(a,b,c)=1, compute 3a+b+c.
The correct answer is: 6
Two circles ω1 and ω2 intersect at points A and B. The tangent to ω1 passing through A intersects ω2 at X. The tangent to ω2 passing through A intersects ω1 at Y. Let O be the circumcenter of △XAY. Then what is the measure of ∠OBA in degrees?
The correct answer is: 90
Let M be the midpoint of side AC of acute triangle ABC with AB>BC, and let Ω be the circumcircle of △ABC. The tangents to Ω at points A and C meet at P, and BP and AC intersect at S. Also, let AD be the altitude of △ABP, and ω the circumcircle of △CSD. Given that ω and Ω intersect at K=C, prove that ∠CKM=90∘.
Since AP=CP, we have ∠AMP=90∘=∠ADP, which implies A,P,D,M are concyclic. Since C,K,D,S are given to be concyclic, by the alternate segment theorem ∠KAP=∠ACK=∠KDP⟹K∈⊙(APDM). Applying the properties of cyclic quadrilaterals and alternate segment theorem again, we have
∠MKP=180∘−∠CAP=∠AKC⟹∠CKM=∠AKC−∠AKM=∠MKP−∠AKM=∠AKP=90∘. □
Believe it or not, a lot of IMO problems can be solved simply by some clever constructions of cyclic points. Here's an example from 2014:
[IMO 2014]
Convex quadrilateral ABCD has ∠ABC=∠CDA=90∘. Point H is the foot of the perpendicular from A to BD. Points S and T lie on sides AB and AD, respectively, such that H lies inside △SCT and
∠CHS−∠CSB=90∘,∠THC−∠DTC=90∘.
Prove that line BD is tangent to the circumcircle of △TSH.
First of all, points S,T are defined in an unconventional manner, so let's try to tackle them first. Rearranging the angle equality gives
∠CHS=90+∠CSB=180−∠SCB.
Seeing this reminds us of opposite angles in a cyclic quadrilateral summing to 180∘, and is enough to motivate us to reflect C over AB,AD to obtain E,F with cyclic CHSE,CHTF, respectively.
Now there are several ways to show that the problem is equivalent to proving CH⊥ST, one of which will be presented at the end.
Project C onto SH,TH to get J,K, respectively. Right away we observe cyclic CJSB,CKTD. Notice ∠CHK=∠TFC=∠TCF=∠TKD, and since CKH is a right triangle at K, KD bisects CH. Similarly we can show that BJ bisects CH, implying BJ,DK intersection on the midpoint of HC, denoted by M, or the circumcenter of CJHK. It is well known that the reflection of C over the perpendicular bisector of BD lies on AH, and hence M lies on perpendicular bisector of BD, which is sufficient to conclude that BDJK is an isosceles trapezoid, a cyclic quadrilateral.
Consider the radical axis of ⊙CJSB,⊙CKTD, which is perpendicular to ST. We then know BJ∩DK=M lies on it by radical axis theorem. Consequently line CMH is the radical axis, establishing that CH⊥ST. Hence
∠TSH=∠90∘−∠JHC=∠HCJ=∠HKJ=∠THD.
By the alternate segment theorem, converse BD is tangent to the circumcircle of TSH, and we are done. □
In the next example, a solution with a clear intention to create a cyclic figure is showcased. The technique is very similar to the one used in the proof example in the last section.
[2011 IMO SL6]
Let ABC be a triangle with AB=AC and let D be the midpoint of AC. The angle bisector of ∠BAC intersects the circle through D,B and C at the point E inside the triangle ABC. The line BD intersects the circle through A,E and B in two points B and F. The lines AF and BE meet at a point I, and the lines CI and BD meet at a point K. Show that I is the incentre of triangle KAB.
From cyclic and symmetry, we can show ∠EBD=∠ECD=∠ABE, which implies BI bisects ∠KBA. By the properties of incenter, I is the incenter of KAB⟺∠AIC=∠ABI+90∘. Note that
∠AFD=180∘−∠BAE=2∠BEC=2∠BDC,
which implies
∠DAF=∠BDC−∠AFD=∠AFD⟹AD=DF=DC⟹∠AFC=90∘.
So now it suffices to show ∠ABI=∠AIC−90∘=∠ICF.
Reflect B over AF to obtain B′. Now C,B′ are on the same side of IF, so we just need to show I,C,B′,F are concyclic, which is equivalent to ∠IB′C=∠IFC=90∘. Since IB′ internally bisects ∠FB′A by symmetry, we want to prove B′C externally bisects ∠FB′A, or 180∘−∠FB′C=∠AB′C=∠ACB′ since AB′=AB=AC. The last equivalence amounts to proving FB′∣∣AC, which is true by 180∘−∠AFB′=∠AFD=∠CAF. □
See Also
Intersection of Circles
Two circles in a plane intersect in zero, one, two, or infinitely many points. The latter case occurs only in the case of two identical circles. The first case (zero points of intersection) occurs whenever the distance between the centers of the circles is greater than the sum of the radii, or the distance between the centers is less than the absolute value of the difference in their radii. Here are all the cases, explained graphically:
Incenter and Incircle of Triangles
An incircle is an inscribed circle of a polygon, i.e. a circle at which each side of the polygon is a tangent. The center I of the incircle is called the Incenter and the radius r of the circle is called the inradius. The incenter is the point of concurrence (where they cross) of the triangle’s angle bisectors.
While an incircle does not necessarily exist for arbitrary polygons, it exists and is moreover unique for triangles and regular polygons.
Cite as:
Circumcenter.
Brilliant.org.
Retrieved
00:10, August 13, 2025,
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6865 | https://math.stackexchange.com/questions/3850365/applying-the-polynomial-remainder-theorem-does-this-logic-hold-for-working-out | Applying the polynomial remainder theorem - does this logic hold for working out factors? - Mathematics Stack Exchange
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Applying the polynomial remainder theorem - does this logic hold for working out factors?
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Let's say I have some function of x x and an unknown integer d d, given by
f(x)=x 2+d 2+d x f(x)=x 2+d 2+d x
and I want to see if it's divisible by something like x−d x−d;as I understand it, we could apply the polynomial remainder theorem and find the remainder, r=f(d)=3 d 2 r=f(d)=3 d 2. Now, I believe I can then argue that this is zero if d=0 d=0, which is obviously true. But I also can see that this isn't sufficient alone; for example, if x−d=1 x−d=1, then x−d x−d will divide always.
My question is thus, can I say that x−d x−d is a factor of f(x)f(x) only when EITHER d=0 d=0 or x−d=1 x−d=1, or are there another situations I'm missing?
polynomials
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asked Oct 3, 2020 at 19:32
DRGDRG
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2 It doesn't make sense to say that x−d=1 x−d=1; you're considering this function as x x ranges over all of its possible values, and d d is unknown but fixed.Qiaochu Yuan –Qiaochu Yuan 2020-10-03 19:35:16 +00:00 Commented Oct 3, 2020 at 19:35
1 What Qiaochu Yuan commented. However, I upvoted your query, for your good presentation and work, and for stretching your intuition by questioning everything.user2661923 –user2661923 2020-10-03 19:38:26 +00:00 Commented Oct 3, 2020 at 19:38
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When we speak about division over polynomials, it means p(x)|q(x)p(x)|q(x) for any x x, as @QiaochuYuan said. If you consider x−d=1 x−d=1, it means x x is no more free variable and is x=d+1 x=d+1, so f(x)=f(d+1)f(x)=f(d+1), and therefore both become constants, and you are speak about division over constant integers! Notice in polynomials, there is no necessity to have integer coefficients for speaking about division.
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answered Oct 3, 2020 at 20:58
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Local currency needed for Iceland and Norway
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Local currency needed for Iceland and Norway
By knuf,
February 23 in Viking Ocean
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knuf
Posted February 23
knuf
Members
3
September 5, 2016
New Jersey
#1
Posted February 23
We are taking a 7 day Viking Iceland cruise with a 4 day Norway extension. We have no idea how much locsl currency is needed for tipping. We'll be doing tours (arranged through Viking) every day. We will have transportation as well as tour guides.
Any help is greatly appreciated!
2cruisecats
Posted February 23
2cruisecats
Members
351
March 15, 2004
Ottawa, Canada
#2
Posted February 23 (edited)
We spent a few weeks in Iceland last summer prior to boarding the Viking Iceland cruise and have been to Norway several times. Neither country has a tipping culture but if you do want to tip, use the amount of dollars suggested by Viking & convert to local currency. Also, in Iceland, you may see some stores with a "no cash" sign -- e-wallet and credit cards widely accepted.
Edited February 23 by 2cruisecats
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1
Redtravel
Posted February 23
Redtravel
Members
7.6k
July 10, 2004
#3
Posted February 23
No need for local currency. Use credit cards.
3
1
kag3120
Posted February 23
kag3120
Members
45
December 27, 2020
Atlanta
#4
Posted February 23
We were pretty surprised by how cashless Norway was, probably more so than any country we’ve visited. Credit cards rule there. Don’t know about Iceland though
1
pavementends
Posted February 23
pavementends
Members
417
May 10, 2021
#5
Posted February 23
Both Iceland and Norway have their own currency (imagine that). It doesn't make sense to change money just to tip guides. I suspect that guides will be happy to receive euros, maybe a little less happy with dollars. For purchases (if any) use credit cards.
1
Xyzzy123
Posted February 23
Xyzzy123
Members
42
September 28, 2022
#6
Posted February 23
Iceland, at least, is almost completely cashless. Everything (and I mean everything) is tap-to-pay credit cards. They also aren't a tipping culture, so I wouldn't worry to much about tipping your guides. Maybe a few dollars or euros if you really want, but no need to stock up on local currency.
4
Baron Barracuda
Posted February 23
Baron Barracuda
Members
3.4k
August 17, 2011
RAMSEY, NJ
#7
Posted February 23
We often pick up small amounts of local currency and at end of cruise use whatever's unspent for extra tips to crew.
2
sharkster77
Posted February 23
sharkster77
Members
2.1k
March 11, 2015
Massachusetts
#8
Posted February 23
Agree with @Xyzzy123-- we spent 9 days in Iceland and had brought with us the equivalent of $200 USD in Icelandic krona. We ended up spending none of it, and used it to tip our tour director and bus driver (it was a bus tour).
Going into COVID Europe was IMO more cashless than the US, and the pandemic helped push the US along to get more caught up. Plus, things are so expensive in Iceland, that using a credit card makes sense.
1
janetcbl
Posted February 24
janetcbl
Members
1.5k
December 23, 2007
#9
Posted February 24 (edited)
In Iceland even the restrooms only use credit cards. No need for cash in either of these countries.
Edited February 24 by janetcbl
2
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OneSixtyToOne
Posted February 24
OneSixtyToOne
Rare
Members
3.7k
July 2, 2015
San Diego
#10
Posted February 24
We were on the Iceland - Norway cruise last August and the only place we used cash was the Faroe Islands and that was because we had Danish Krone leftover from a 2019 trip to Copenhagen. We had a small amount of change coming and the store clerk was shocked when I told her to keep it.
1
GonzoWCS
Posted February 24
GonzoWCS
Members
3.1k
December 30, 2013
Gulf Breeze, Florida
#11
Posted February 24
As stated above Zero cash needed in Iceland. Credit Cards rule!
Cheers
1
AroundWithMAPTravels
Posted February 24
AroundWithMAPTravels
Members
470
March 31, 2022
East Coast
#12
Posted February 24
On 2/23/2025 at 7:44 PM, Xyzzy123 said:
Iceland, at least, is almost completely cashless. Everything (and I mean everything) is tap-to-pay credit cards. They also aren't a tipping culture, so I wouldn't worry to much about tipping your guides. Maybe a few dollars or euros if you really want, but no need to stock up on local currency.
Expand
In the Harpa center, where many tour buses meet, it is cashless to the extent that instead of a euro coin needed for use of rest room, you must tap your credit card instead. No coins accepted. (Beautifully clean restrooms, I might add)
but honestly, if you are spending multiple days there, no need to change money. If you do choose to tip a guide (not necessary), a euro or two is probably fine… USD maybe, but euro more preferred, I think.
1
Heidi13
Posted February 24
Heidi13
Members
13.7k
June 10, 2014
British Columbia
#13
Posted February 24
On 2/24/2025 at 2:20 AM, AroundWithMAPTravels said:
In the Harpa center, where many tour buses meet, it is cashless to the extent that instead of a euro coin needed for use of rest room, you must tap your credit card instead. No coins accepted. (Beautifully clean restrooms, I might add)
but honestly, if you are spending multiple days there, no need to change money. If you do choose to tip a guide (not necessary), a euro or two is probably fine… USD maybe, but euro more preferred, I think.
Expand
Neither Iceland or Norway use the Euro and they also don't us US Dollars.
Norway uses the Norwegian Knone and Iceland the Krona. So neither Euros or US Dollars are legal tender in either country.
8
1
mahasamatman
Posted March 1
mahasamatman
Members
3.8k
July 16, 2007
Coquitlam BC
#14
Posted March 1
On 2/24/2025 at 4:53 AM, Heidi13 said:
So neither Euros or US Dollars are legal tender in either country.
You're not paying for goods or services, so no need to be restricted when tipping. They won't reject either one, though Euros would definitely be more appreciated than USD.
2
duquephart
Posted March 1
duquephart
Members
2.5k
January 30, 2019
Land of 10,000 Lakes
#15
Posted March 1
On 3/1/2025 at 9:40 PM, mahasamatman said:
You're not paying for goods or services, so no need to be restricted when tipping. They won't reject either one, though Euros would definitely be more appreciated than USD.
Tipping whom?
1
Heidi13
Posted March 1
Heidi13
Members
13.7k
June 10, 2014
British Columbia
#16
Posted March 1
On 3/1/2025 at 9:40 PM, mahasamatman said:
You're not paying for goods or services, so no need to be restricted when tipping. They won't reject either one, though Euros would definitely be more appreciated than USD.
Correct, you are not purchasing a goods or service, but by providing a tip, you are thanking the guide for providing exemplary service. Why would you provide them with a foreign currency they can't spend at home, and make them go to a bank to exchange.
I received tips many years ago, when carrying bags around the Old Course at St Andrews. In my experience, one nationality often provided foreign currency. It was always accepted with a smile, but as we trudged our way to the Post Office/Bank to exchange it, we may have made some non-complimentary comments. So yes, tips will be accepted in most countries, but when providing foreign currency, you may be the recipient of ridicule.
Sorry, couldn't agree Euros are more appreciated, as other than the airport or hotels, they probably don't accept Euros or US $. Iceland also does not have a tipping culture.
4
knuf
Posted March 2
knuf
Members
3
September 5, 2016
New Jersey
Author
#17
Posted March 2
I agree. When we did our Rhine River Cruise last year, we tipped in local currency. We plan to do the same in Iceland and Norway. I guess we'll just need to estimate how much we'll need.
1
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uktog
Posted March 2
uktog
Rare
Members
21.7k
August 4, 2004
The Kingdom of Fife
#18
Posted March 2
How would you feel if you were a hard up server in the US and the tip you got was Norwegian kroner?
In Europe not only do they have to trail to the bank they also have to fill in complex documentation
4
2
knuf
Posted March 2
knuf
Members
3
September 5, 2016
New Jersey
Author
#19
Posted March 2
Agree with you 💯
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aungrl
Posted March 3
aungrl
Rare
Members
738
March 22, 2018
coastal Georgia
#20
Posted March 3
It's more difficult for us, but we also always tip in local currency, when and where tips are warranted. And that's what we do; we estimate in advance the amount we'll need, and then add a bit of a cushion. In many years of travel, we've almost always been a little on the plus side, so generally, the last person "in country" got a bit more generous tip to use up the last of the currency.
3
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6867 | https://ocw.mit.edu/courses/8-02t-electricity-and-magnetism-spring-2005/pages/lecture-notes/ | search
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8.02T | Spring 2005 | Undergraduate
Electricity and Magnetism
Lecture Notes
The course notes were written by John Belcher, Peter Dourmashkin, and Sen-Ben Liao.
The TEAL classroom includes the opportunity for students to use the Personal Response System (PRS). Questions are posed to the class to stimulate discussion and indicate how concepts are going over. Students “vote” on answers electronically and their answers are tallied.
| WEEK # | SES # | TOPICS | COURSE NOTES | PRESENTATIONS | PRS |
--- --- --- |
| 1 | 1 | Hour 1 Why Physics? Why Studio Physics? (and How?) Vector and Scalar Fields Hour 2 Gravitational Fields Electric Fields | Chapter 1: Sections 1.1 - 1.6; 1.8 (PDF) Chapter 2 (PDF) | (PDF) | (PDF) |
| 2 | Hour 1 Review: Electric Fields Charge Dipoles Hour 2 Continuous Charge Distributions | Chapter 1: Section 1.6 (PDF) Chapter 2 (PDF) | (PDF) | (PDF) |
| 3 | Problem Solving Session 1: Line and Surface Integrals | | | |
| 2 | 4 | Hour 1 Working In Groups Experiment 1: Visualizations Hour 2 Electric Potential | Chapter 3: Sections 3.1 - 3.5 (PDF) | (PDF) | (PDF) |
| 5 | Hour 1 Gauss’ Law Hour 2 Gauss’ Law | Chapter 4 (PDF) | (PDF) | (PDF) |
| 6 | Problem Solving Session 2: Electric Field of Continuous Charge Distributions | | | |
| 3 | 7 | Hour 1 Conductors and Insulators Experiment 2: Electrostatic Force Hour 2 Capacitors | Chapter 4: Sections 4.3 - 4.4 (PDF) Chapter 5 (PDF - 1.3 MB) | (PDF) | (PDF) |
| 8 | Hour 1 Last Time: Conductors Experiment 3: Faraday Ice Pail Hour 2 Capacitors and Dielectrics | Chapter 4: Sections 4.3 - 4.4 (PDF) Chapter 5 (PDF - 1.3 MB) | (PDF) | (PDF) |
| 9 | Problem Solving Session 3: Gauss’ Law | Chapter 4 (PDF) | | |
| 4 | 10 | Hour 1 DC Circuits Hour 2 Kirchhoff’s Loop Rules | Chapter 6 (PDF) Chapter 7: Sections 7.1 through 7.4 (PDF) | (PDF - 1.2 MB) | (PDF) |
| 11 | Problem Solving Session 4: Capacitance | Chapter 5 (PDF - 1.3 MB) | | |
| 5 | 12 | Hour 1 Working with Circuits Experiment 4: Part I: Measuring V, I, R Hour 2 RC Circuits Experiment 4: Part II: RC Circuits | Chapter 7 (PDF) | (PDF) | (PDF) |
| 13 | Hour 1 Concept Review/Overview PRS Questions - Possible Exam Questions Hour 2 Sample Exam | | (PDF) | (PDF) |
| 6 | 14 | Hour 1 Magnetic Fields Experiment 5: Magnetic Fields Hour 2 Charges moving in B Fields Exam Review | Chapter 8 (PDF) | (PDF) | (PDF) |
| 15 | Hour 1 Magnetic Force Experiment 6: Magnetic Force Hour 2 Creating B Fields: Biot-Savart | Chapter 9: Sections 9.1 - 9.2 (PDF - 1.9 MB) | (PDF - 1.1 MB) | (PDF) |
| 16 | Problem Solving Session 5: Magnetic Torque and Moments | Chapter 8: Sections 8.3 - 8.4 (PDF) Chapter 9: Sections 9.1 - 9.2 (PDF - 1.9 MB) | | |
| 7 | 17 | Hour 1 Dipoles and Magnetic Fields Hour 2 Experiment 7: Dipoles in B Fields | Chapter 8: Section 8.4 (PDF) Chapter 9: Sections 9.1 - 9.2, 9.5 (PDF - 1.9 MB) | (PDF) | (PDF) |
| 18 | Hour 1 Levitation Experiment 8: Magnetic Forces Hour 2 Ampere’s Law | Chapter 9 (PDF - 1.9 MB) | (PDF - 1.6 MB) | (PDF) |
| 19 | Problem Solving Session 6: Ampere’s Law | Chapter 9: Sections 9.3 - 9.4; 9.10.2, 9.11.6, 9.11.7 (PDF - 1.9 MB) | | |
| 8 | 20 | Hour 1 Faraday’s Law Hour 2 Faraday’s Law: Applications | Chapter 10 (PDF) | (PDF) | (PDF) |
| 21 | Hour 1 Experiment 9: Faraday’s Law Hour 2 Faraday’s Law Transformers Magnetic Materials | Chapter 10 (PDF) Chapter 11: Section 11.1 (PDF - 1.0 MB) | (PDF) | (PDF) |
| 22 | Problem Solving Session 7: Faraday’s Law | Chapter 10 (PDF) | | |
| 9 | 23 | Hour 1 Concept Review/Overview PRS Questions - Possible Exam Questions Hour 2 Sample Exam | | (PDF) | (PDF) |
| 24 | Hour 1 Inductance and LR Circuits Hour 2 Energy in Inductors | Chapter 11: Sections 11.1 - 11.4 (PDF - 1.0 MB) | (PDF) | (PDF - 1.3 MB) |
| 10 | 25 | Hour 1 Experiment 10: Part I: Measuring L LC Circuits Hour 2 Experiment 10: Part II: LRC Circuit | Chapter 11: Sections 11.5 - 11.6 (PDF - 1.0 MB) | (PDF) | (PDF) |
| 26 | Hour 1 Driven Harmonic Motion (RLC) Hour 2 Experiment 11: Driven RLC Circuit | Chapter 12 (PDF) | (PDF) | (PDF) |
| 27 | Problem Solving Session 8: Driven RLC Circuits | Chapter 12 (PDF) | | |
| 11 | 28 | Hour 1 Displacement Current Maxwell’s Equations Hour 2 Electromagnetic Waves | Chapter 13 (PDF) | (PDF) | (PDF) |
| 29 | Problem Solving Session 9: Displacement Current, Poynting | Chapter 13 (PDF) | | |
| 12 | 30 | Hour 1 Traveling and Standing Waves Hour 2 Electromagnetic (EM) Waves | Chapter 13 (PDF) | (PDF) | (PDF) |
| 31 | Hour 1 Concept Review/Overview PRS Questions - Possible Exam Questions Hour 2 Sample Exam | | (PDF) | |
| 13 | 32 | Hour 1 Generating Electromagnetic Waves Plane EM Waves Electric Dipole EM Waves Hour 2 Experiment 12: Microwaves Review Exam 3 Results | Chapter 13 (PDF) | (PDF - 2.9 MB) | (PDF) |
| 33 | Hour 1 Hour 2 Experiment 13: Interference | Chapter 14 (PDF) | (PDF - 1.4 MB) | (PDF) |
| 34 | Problem Solving Session 10: Interference | | | |
| 14 | 35 | Hour 1 The Structure of Space and Time Hour 2 The Structure of Space and Time | | | |
| 36 | Hour 1 Concept Review/Overview PRS Questions - Possible Exam Questions Hour 2 Sample Exam | | (PDF) | |
Course Info
Instructors
Bruce Knuteson
Prof. Eric Hudson
Dr. George Stephans
Prof. John Belcher
Prof. John Joannopoulos
Prof. Michael Feld
Dr. Peter Dourmashkin
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| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- ---
| Two-Digit Divisor in Long Division This is a complete lesson with examples and exercises about two-digit divisor in long division, meant for initial teaching in 5th grade. The first exercises have grids to complete the division, and space for students to write the multiplication table of the divisor in the margin. Then there are conversion problems between inches/feet and ounces/pounds, because those are solved with division. For extra tips and practice problems, check out IXL's dividing by two-digit numbers lesson! | | | Often, it is helpful to write the multiplication table of the divisor before you divide. | | Example 1. The division is by 16. Here is the multiplication table of 16: 3 × 16 = 484 × 16 = 645 × 16 = 806 × 16 = 96 7 × 16 = 112 8 × 16 = 128 9 × 16 = 144 | | | | | --- | | | 0 3 | | 16 | ) | 5 5 6 8 | 16 goes into 5 zero times, so we look at 55. How many times does 16 go into 55? Check in the table on the left. We see it goes into 55 three times. | | | | | --- | | | 0 3 4 | | 16 | ) | 5 5 6 8 | | 4 8 | | | | 7 6 | Now, how many times does 16 go into 76? From the table we can see that it is four times. | | | | | --- | | | 0 3 4 8 | | 16 | ) | 5 5 6 8 | | 4 8 | | | | 7 6 | | | | - 6 4 | | | | 1 2 8 | | | | - 1 2 8 | | | | 0 | Lastly, 16 goes into 128 exactly 8 times, and the division is over. | | Example 2. We are dividing by 32. Here is the multiplication table of 32: 3 × 32 = 964 × 32 = 1285 × 32 = 1606 × 32 = 1927 × 32 = 2248 × 32 = 2569 × 32 = 288 | | | | | --- | | | 0 1 | | 32 | ) | 4 7 0 7 | | 3 2 | | | | 1 5 | 32 goes into 47 once. | | | | | --- | | | 0 1 4 | | 32 | ) | 4 7 0 7 | | 3 2 | | | | 1 5 0 | | 1 2 8 | | | | 2 2 | 32 goes into 150 four times. | | | | | --- | | | 0 1 4 7 | | 32 | ) | 4 7 0 7 | | 3 2 | | | | 1 5 0 | | 1 2 8 | | | | 2 2 7 | | | | - 2 2 4 | | | | 3 | 32 goes into 224 seven times. Notice there is a remainder. | 1. Divide. First write a multiplication table for the divisor. Check each answer by multiplying. | | | | | | --- --- | | | | | | --- --- | | | | | Table of 21: 2 × 21 = 3 × 21 = 4 × 21 = 5 × 21 = 6 × 21 = 7 × 21 = 8 × 21 = 9 × 21 = | | | | | | | | 2. Divide. First write a multiplication table for the divisor. Check each answer by multiplying. | | | | | | | --- --- --- | | | | | | | | --- --- | a. | | | | Table of 15: 2 × 15 = 3 × 15 = 4 × 15 = 5 × 15 = 6 × 15 = 7 × 15 = 8 × 15 = 9 × 15 = | | | | | | | | | | | | | | | --- --- | b. | | | | Table of 12: 2 × 12 = 3 × 12 = 4 × 12 = 5 × 12 = 6 × 12 = 7 × 12 = 8 × 12 = 9 × 12 = | | | | | | | | | | | | | | --- --- | | c. | | | | Table of 25: 2 × 25 = 3 × 25 = 4 × 25 = 5 × 25 = 6 × 25 = 7 × 25 = 8 × 25 = 9 × 25 = | | | | | | | | | | --- --- | | d. | | | | Table of 16: 2 × 16 = 3 × 16 = 4 × 16 = 5 × 16 = 6 × 16 = 7 × 16 = 8 × 16 = 9 × 16 = | | | | 3. Divide. Check each answer by multiplying. | | | | | | --- --- | | | | | | --- --- | | a. | | | | Table of 12: 2 × 12 = 3 × 12 = 4 × 12 = 5 × 12 = 6 × 12 = 7 × 12 = 8 × 12 = 9 × 12 = | | | | | | | | | | --- --- | | b. | | | | Table of 22: 2 × 22 = 3 × 22 = 4 × 22 = 5 × 22 = 6 × 22 = 7 × 22 = 8 × 22 = 9 × 22 = | | | | | | | | | | --- --- | | c. | | | | Table of 14: 2 × 14 = 3 × 14 = 4 × 14 = 5 × 14 = 6 × 14 = 7 × 14 = 8 × 14 = 9 × 14 = | | | | | | | | | | --- --- | | d. | | | | Table of 51: 2 × 51 = 3 × 51 = 4 × 51 = 5 × 51 = 6 × 51 = 7 × 51 = 8 × 51 = 9 × 51 = | | | | 4. Mental math! If 20 goes into 800 forty times, then 20 goes into 820 one time more, or 41 times. In each box, use the top problem to help you solve the bottom problem. | | | | --- | a. 800 ÷ 20 = 820 ÷ 20 = | b. 700 ÷ 50 = 750 ÷ 50 = | c. 150 ÷ 15 = 300 ÷ 15 = | | d. 480 ÷ 40 = 520 ÷ 40 = | e. 600 ÷ 30 = 690 ÷ 30 = | f. 1,200 ÷ 60 = 1,320 ÷ 60 = | 5. a. How many inches are in one foot? b. Convert 245 inches into feet and inches. c. Convert 387 inches into feet and inches. 6. a. How many ounces are in one pound? b. Convert 163 ounces into pounds and ounces. c. Convert 473 ounces into pounds and ounces. 7. A newborn baby gains weight at approximately one ounce per day. Suppose that the baby gained weight at that rate for a FULL YEAR. (In reality, babies don’t; their growth rate slows down.) How many pounds and ounces would the baby gain in a year? See also Long division worksheets Create an unlimited supply of worksheets for long division (grades 4-6), including with 2-digit and 3-digit divisors. The worksheets can be made in html or PDF format - both are easy to print. You can also customize them using the generator. This lesson is taken from Maria Miller's book Math Mammoth Multiplication & Division 3, and posted at www.HomeschoolMath.net with permission from the author. Copyright © Maria Miller. Math Lessons menu | | | | --- | Place Value Grade 1+ Using a 100-bead abacus in elementary math + Teaching tens and ones + Practicing with two-digit numbers + Counting in groups of ten + Skip-counting practice (0-100) + Comparing 2-digit numbers + Cents and dimes Grade 2+ Three-digit numbers + Comparing 3-digit numbers Grade 3+ Place value with thousands + Comparing 4-digit numbers + Rounding & estimating + Rounding to the nearest 100 Grade 4+ Place value - big numbers | Add & subtract lessons Grade 1+ Missing addend concept (0-10) + Addition facts when the sum is 6 + Addition & subtraction connection Grade 2+ Fact families & basic addition/subtraction facts + Sums that go over over the next ten + Add/subtract whole tens (0-100) + Add a 2-digit number and a single-digit number mentally + Add 2-digit numbers mentally + Regrouping in addition + Regrouping twice in addition + Regrouping or borrowing in subtraction Grade 3+ Mental subtraction strategies + Rounding & estimating | Multiplication Grade 3+ Multiplication concept as repeated addition + Multiplication on number line + Commutative + Multiply by zero + Word problems + Order of operations + Structured drill for multiplication tables + Drilling tables of 2, 3, 5, or 10 + Drilling tables of 4, 11, 9 Grade 4+ Multiplying by whole tens & hundreds + Distributive property + Partial products - the easy way + Partial products - video lesson + Multiplication algorithm + Multiplication Algorithm — Two-Digit Multiplier + Scales problems - video lesson + Estimation when multiplying | | Division Grade 3+ Division as making groups + Division/multiplication connection + Division is repeated subtraction + Zero in division + Division that is not exact (remainder) + Divisibility Grade 4+ How to teach long division + Long division as repeated subtraction + Why long division works + Zero in dividend + Remainder & long division + Two-digit divisor + Review of division topics Divisibility+ Divisibility within 0-1000 + Divisibility rules + Prime factorization 1 + Prime factorization 2 + Sieve of Eratosthenes | Fraction Lessons + Understanding fractions + Finding fractional parts with division + Mixed numbers + Fractions to mixed numbers and vv. + Adding like fractions + Equivalent fractions + Adding unlike fractions 1 + Adding unlike fractions 2: Finding the common denominator + Adding mixed numbers + Subtracting mixed numbers + Subtracting mixed numbers 2 + Measuring in inches + Comparing fractions + Simplifying fractions + Multiply fractions by whole numbers + Multiply fractions by fractions + Multiplication and area + Simplify before multiplying + Dividing fractions by whole numbers + Dividing fractions: fitting the divisor + Dividing fractions: reciprocal numbers + Dividing fractions: using the shortcut | Geometry Lessons + Lines, rays, and angles + Measuring angles + Parallel & perpendicular + Acute, obtuse, and right triangles + Angle sum of a triangle + Equilateral & isosceles triangles + Circles + Symmetry + Altitude of a triangle + Polygons + Perimeter + Area of rectangles + Area of right triangles + Area of parallelograms + Area of triangles + Area versus Perimeter + Angles in Polygons (PDF) + Review: Area of Polygons (PDF) + Surface Area (PDF) | | Decimals Lessons + Decimals videos + Decimals (1 decimal digit) + Decimal place value (1 decimal digit) + Decimals (2 decimal digits) + Decimal place value (2 decimal digits) + Decimals (3 decimal digits) + Add & subtract (1 decimal digit) + Add & subtract (2 decimal digits) + Add and subtract decimals — practice + Comparing decimals + Multiply a decimal by a whole number + Multiply decimals by decimals + Divide decimals—mental math + Divide decimals by decimals + Multiply and divide decimals by 10, 100, and 1000 + Decimals review lesson | Percents Lessons + How to teach proportions + Percent - the basic concept + Percentage of a number—mental math + How to calculate a percentage of a number + How to calculate percentages + Basics of percent of change | General + Four habits of highly effective math teaching + Why are math word problems SO difficult for children? Hint: it has to do with a "recipe" that many math lessons follow. + The do's and don'ts of teaching problem solving in mathAdvice on how you can teach problem solving in elementary, middle, and high school math. + How to set up algebraic equations to match word problems Students often have problems setting up an equation for a word problem in algebra. To do that, they need to see the RELATIONSHIP between the different quantities in the problem. This article explains some of those relationships. + Seven reasons behind math anxiety and how to prevent it + Mental math "mathemagic" with Arthur Benjamin (video) + Keeping math skills sharp in the summer + Geometric vanish puzzles + Science resources Short reviews of the various science resources and curricula I have used with my own children. | | | |
6869 | http://www.math.hawaii.edu/~ramsey/Probability/TwoDice.html | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | |
--- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- |
| | | Rolling Two Dice When rolling two dice, distinguish between them in some way: a first one and second one, a left and a right, a red and a green, etc. Let (a,b) denote a possible outcome of rolling the two die, with a the number on the top of the first die and b the number on the top of the second die. Note that each of a and b can be any of the integers from 1 through 6. Here is a listing of all the joint possibilities for (a,b): | | | | | | | --- --- --- | | (1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) | | (2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) | | (3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) | | (4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) | | (5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) | | (6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) | Note that there are 36 possibilities for (a,b). This total number of possibilities can be obtained from the multiplication principle: there are 6 possibilities for a, and for each outcome for a, there are 6 possibilities for b. So, the total number of joint outcomes (a,b) is 6 times 6 which is 36. The set of all possible outcomes for (a,b) is called the sample space of this probability experiment. With the sample space now identified, formal probability theory requires that we identify the possible events. These are always subsets of the sample space, and must form a sigma-algebra. In an example such as this, where the sample space is finite because it has only 36 different outcomes, it is perhaps easiest to simply declare ALL subsets of the sample space to be possible events. That will be a sigma-algebra and avoids what might otherwise be an annoying technical difficulty. We make that declaration with this example of two dice. With the above declaration, the outcomes where the sum of the two dice is equal to 5 form an event. If we call this event E, we have E={(1,4),(2,3),(3,2),(4,1)}. Note that we have listed all the ways a first die and second die add up to 5 when we look at their top faces. Consider next the probability of E, P(E). Here we need more information. If the two dice are fair and independent, each possibility (a,b) is equally likely. Because there are 36 possibilities in all, and the sum of their probabilities must equal 1, each singleton event {(a,b)} is assigned probability equal to 1/36. Because E is composed of 4 such distinct singleton events, P(E)=4/36= 1/9. In general, when the two dice are fair and independent, the probability of any event is the number of elements in the event divided by 36. What if the dice aren't fair, or aren't independent of each other? Then each outcome {(a,b)} is assigned a probability (a number in [0,1]) whose sum over all 36 outcomes is equal to 1. These probabilities aren't all equal, and must be estimated by experiment or inferred from other hypotheses about how the dice are related and and how likely each number is on each of the dice. Then the probability of an event such as E is the sum of the probabilities of the singleton events {(a,b)} that make up E. | | | Go to the home page for Tom Ramsey | | | Go to the home page for the UHM Department of Mathemati cs | | | Your comments and questions are welcome. Please email them ramsey@math.hawaii.edu. | | |
6870 | https://commons.wikimedia.org/wiki/Category:Apollo_11 | Category:Apollo 11 - Wikimedia Commons
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Category:Apollo 11
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Apollo 11 (July 16-24, 1969) was the United States spaceflight mission that was the first time that humans landed on the moon. The NASA crew consisted of Neil Armstrong (commander), Edwin "Buzz" Aldrin (lunar module commander), and Michael Collins (command module pilot). The lunar module, "Eagle", landed on July 20. Armstrong stepped on the lunar surface on July 21, becoming the first person to step on the moon. Nineteen (19) minutes later, Aldrin joined him. Together, they explored the site, which they named "Tranquility Base". They returned to the command module "Columbia" 21 hours later.
| v t e Apollo program |
| | Uncrewed Apollo-Saturn IB and Saturn V | AS-201 AS-203 AS-202 Apollo 4 Apollo 5 Apollo 6 |
| Manned missions | Apollo 1 Apollo 7 Apollo 8 Apollo 9 Apollo 10 Apollo 11 Apollo 12 Apollo 13 Apollo 14 Apollo 15 Apollo 16 Apollo 17 |
| | number: | (eleven) |
| Apollo program missions: | 1·4·5·6·7·8·9·10·11·12·13·14·15·16·17 |
English: This category is part of the category scheme space exploration.
Français : Cette catégorie fait partie du schéma des catégories concernant l'exploration de l'espace.
日本語： このカテゴリーは category scheme space exploration の一部です。
Apolo 11; Tungllendingin (1969); اپولو ۱۱; Apollo 11; Apollo 11; اپولو ۱۱; Аполо 11; Apollo 11; اپالو 11; Apollo 11; Apollo 11; Аполлон-11; Аполлон-11; Απόλλων 11; 阿波羅11號; 阿波罗11号; อะพอลโล 11; 아폴로 11호; Аполлон 11; Apollo 11; Аполо 11; Apollo 11; Apollo 11; অ্যাপোলো ১১; Apollo 11; Apollo 11; Apollo 11; এপ'ল' ১১; 阿波罗11号; Apollo 11; אפאלא 11; अपोलो ११; Apollo 11; Apollo 11; ଆପୋଲୋ ୧୧; Apollo 11; Apollo 11; Apolo 11; אפולו 11; Apollo 11; Apollo 11; Apollo 11; Аполлон 11; Apollo 11; Apollo 11; Apollon 11; Apollo 11; Apollo 11; ಅಪೋಲೊ ೧೧; Apollo 11; Apollo 11; أبولو 11; Apollo 11; اپولو 11; အပိုလို ၁၁; 阿波羅11號; Apollo–11; አፖሎ ፲፩; Apollo 11; Apollo 11; Apollo 11; Apollo 11; Apollo 11; Apollo 11; Apollo 11; Apollo XI; Apollo 11; آپولو ۱۱; 阿波罗11号; Apollo 11; აპოლო 11; アポロ11号; Apollo 11; Apollo 11; Apollo 11; Apollo 11; ඇපලෝ 11; Apollo 11; 阿波罗11号; अपोलो ११; అపోలో 11; Apollo 11; 阿波罗11号; Apollo 11; Apollo 11; அப்பல்லோ 11; Apollo 11; Аполлон-11; એપોલો ૧૧; Apollo 11; 阿波羅11號; Apollo 11; Ապոլո 11; 阿波羅11號; Apollo 11; Апалён-11; آپولو 11; Apollo 11; Apollo 11; Apollo 11; 阿波罗11号; Apollo 11; Apollo 11; Apollo 11; Apollo 11; Apollo 11; Apollo XI; Apollo 11; അപ്പോളോ 11; 阿波羅11號; Апалон-11; Аполлон 11; Apollo 11; Apollo 11; Apollo 11; Apollo 11; Apollo 11; ئەپۆلۆ ١١; primera misión espacial que permitió a los humanos llegar a la superficie lunar; az amerikai Apollo-program első holdra szállása; સૌપ્રથમ ચંદ્ર પર સમાનવ ઉતરાણ કરતું અભિયાન હતું; primer vuelu espacial que llevó humanos a la Lluna; primera missió espacial que va posar una persona a la superfície de la Lluna; erste bemannte Raumfahrtmission mit Mondlandung; ulja e parë në Hënë me ekuipazh (1969); اولین ماموریت فضایی همراه با فضانوردان برای فرود روی ماه; 人類首次登陸月球的載人太空飛行任務; første bemandede månelanding; prima misiune în care omul a pășit pe Lună, parte a programului spațial american Apollo; 史上初めて人類を月に着陸させることに成功したアポロ宇宙船; första bemannade rymdfärden som landade på månen; המשימה המאוישת השישית בתוכנית אפולו והראשונה שהנחיתה אדם על הירח; 人類首次登陸月球的太空任務; प्रथम सफल चंद्रमा मिशन; ensimmäinen Kuuhun laskeutunut miehitetty avaruuslento; Usona kosmoveturilo, per kiu homoj unufoje surluniĝis; првиот екипажен вселенски лет што слетал на Месечината; missione del programma spaziale statunitense Apollo; প্রথম মনুষ্যবাহী মহাকাশ অভিযান; 11e mission du programme spatial américain Apollo (premiers pas de l'homme sur la Lune); misia v rámci amerického vesmírneho programu Apollo; місія касьмічнай праграмы „Апалён“; එක්සත් ජනපද ඇපලෝ වැඩසටහනේ පළමු සඳ ගොඩබෑම සහ පස්වන මිනිසුන් සහිත ගමන; американский пилотируемый космический корабль серии «Аполлон»; first Moon landing and fifth crewed flight of the United States Apollo program; 1969년 발사된 유인 우주선; missioon, mis sooritas esimese mehitatud maandumise Kuu pinnale; മനുഷ്യനെ ആദ്യമായി ചന്ദ്രനിൽ ഇറക്കിയ ബഹിരാകാശ ദൗത്യം; missão do Programa Apollo; Ay'a ilk insanlı iniş (1969); ruimtevaartmissie; Amerikaanse ruimtesending waarmee die eerste mense op die maan geland het; прва свемирска мисија која је одвела људе на Месец; ପ୍ରଥମ ମାନବଧାରୀ ଯାନ ଯାହା ଚନ୍ଦ୍ର ପୃଷ୍ଠରେ ଅବତରଣ କରିଥିଲା ।; chuyến bay đầu tiên đưa con người đặt chân lên bề mặt Mặt Trăng; Әлемдегі ең бірінші айға қону миссиясы; Космічний корабель; โครงการการบินอวกาศขององค์การนาซา; załogowa misja kosmiczna zwieńczona lądowaniem na Księżycu; første romferd som landet mennesker på månen; 人類首次登陸月球的太空任務; misi pandaratan manusia partamo di Bulan; mise programu Apollo, která jako první dostala člověka na povrch Měsíce; ಚಂದ್ರನಲ್ಲಿಗೆ ಮಾನವನನ್ನು ಕೊಂಡೊಯ್ದ ಮೊದಲ ಚಾಂದ್ರನೌಕೆ; misi pendaratan manusia pertama di Bulan; misión espacial tripulada de Estados Unidos, a primeira en levar persoas á Lúa; أول رحلة مأهولة إلى سطح القمر; Διαστημική πτήση που προσεδάφισε τους πρώτους δύο ανθρώπους στη Σελήνη.; y chied tarlheim er yn Eayst as y queiggoo mishoon lesh skimmee 'sy chlaare Americaanagh Apollo; Apollo XI; Alunizaje del Apollo XI; Apolo XI; Llegada del hombre a la Luna; Primer viaje a la luna; Eagle; Apollo 11; Apollo-11; Apolló 11; አፖሎ 11; Apolo 11; Apollo XI; Высадка американцев на Луну; Высадка на луну; Аполлон 11; Первая высадка человека на Луну; Erste Mondlandung; Apollo-11; اپولو 11; آپولو 11; اپولو ۱۱; Аполо-11; Apollo 11 kontrol odası; Apollo XI; Månlandningen; Аполон 11; Apollo 11; Аполон-11; Аполлон 11; Apollo11; அப்போலோ 11; அப்போலோ - 11; Apollo XI; এপোলো ১১; Appolo 11; Apollo XI; אַפאָלאָ 11; Apollo XI; 아폴로11호; ยานอะพอลโล 11; primo sbarco umano sulla luna; Apollo XI; Адамның біріншірет айға қонуы; Apolo XI; Apolo 11; Apollo XI; Apollo 11; Apollon-11; Apollo XI; Apol·lo 11; Eerste maanlanding; 太陽神11; Apollo 11; 太阳神11号; 太陽神11號; 阿波羅11號任務; აპოლონ 11; Apollo XI; Apollo ١١; أبولو ١١; ابولو ١١; Apollo 11; ابولو 11; Apollo 11; Απόλλο 11; Appolo 11
Apollo 11Collapse first Moon landing and fifth crewed flight of the United States Apollo program
Crew of Apollo 11. From left: Neil Armstrong, Michael Collins, and Buzz Aldrin.
Duration: 4 minutes and 30 seconds.4:30 Subtitles available.CC
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Spoken text audioEnglishⓘ
Instance of human spaceflight
Moon landing (1969–1969)
artificial satellite
Part of Apollo space program
Location United States
Operator National Aeronautics and Space Administration
Start point Kennedy Space Center Launch Complex 39A
Space launch vehicle Saturn V (SA-506)
UTC date of spacecraft launch 16 July 1969 (13:32:00)
Type of orbit lunar orbit
Significant event rocket launch (1969, Kennedy Space Center Launch Complex 39A)
docking and berthing of spacecraft (Apollo 11 Command and Service Module, Eagle, 1969–1969)
orbital activity (lunar orbit, Apollo 11 Command and Service Module, 1969–1969)
Moon landing (Mare Tranquillitatis, 77,901 second, Eagle, 1969–1969)
extra-vehicular activity (Mare Tranquillitatis, 9,060 second, 1)
docking and berthing of spacecraft (Apollo 11 Command and Service Module, Eagle, 1969–1969)
splashdown (Columbia, Pacific Ocean, 13°19′N 169°9′W, 1969)
Mass 49,734.6 kg (launch weight)
4,931.9 kg (landing)
Duration 703,115 s
Follows Apollo 10
Followed by Apollo 12
Different from Apollo 11 Cave
| Collapse Authority file |
| Q43653 ISNI: 0000000123370045 VIAF cluster ID: 145407613 GND ID: 4356607-8 Library of Congress authority ID: n79129542 Bibliothèque nationale de France ID: 12510596r National Library of Israel ID (old): 002083590 National Library of Israel J9U ID: 987007594503605171 |
| Reasonator Scholia Wikidocumentaries PetScan statistics WikiMap Locator tool KML file Search depicted |
Subcategories
This category has the following 37 subcategories, out of 37 total.
Audio files of Apollo 11(8 F)
Videos of Apollo 11(1 C, 46 F)
A
Anniversaries of the First Moon Walk(8 C)
Apollo 11 (2019 film)(1 C, 2 F)
Apollo 11 artist concept(39 F)
Apollo 11 goodwill messages(7 F)
Apollo 11 lunar plaque(12 F)
Apollo 11 moon landing in newspapers(12 F)
Apollo 11 training(2 C, 83 F)
Apollo 11 in art(4 C, 7 F)
Astronaut photography during Apollo 11(13 C, 8 F)
B
Buzz Aldrin's Apollo space suit at the National Air and Space Museum(2 F)
C
Command Module, Apollo 11 (NASA & Smithsonian Institution)(269 F)
Crew of Apollo 11(3 C, 87 F)
CSM-107(2 C, 11 F)
E
Lunar extra-vehicular activity of Apollo 11(2 C, 44 F)
F
F-1 rocket engine (Cincinnati Museum Center at Union Terminal Museum of Natural History)(4 F)
Figures from Apollo 11 Preliminary Science Report(114 F)
Apollo 11 Flight Plan(181 F)
G
Jack Garman(2 F)
H
Hatch, Crew, Apollo 11 (NASA & Smithsonian Institution)(1 C, 596 F)
K
Apollo 11 at Kennedy Space Center(1 C, 96 F)
L
Statio Tranquillitatis(3 C, 26 F)
LM-5 Eagle(5 C, 13 F)
Logo of Apollo 11(7 F)
Apollo 11 lunar samples(2 C, 24 F)
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Apollo 11 at the Manned Spacecraft Center(1 C, 47 F)
Apollo 11 Mobile Quarantine Facility(32 F)
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Neil Armstrong's Apollo 11 space suit(18 F)
Neil Armstrong's Apollo space suit at the Armstrong Air and Space Museum(3 F)
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Recovery of the Apollo 11 mission(51 F)
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S-IC-6(6 F)
S-II-6(2 F)
S-IVB-506(19 F)
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Apollo 11 on television(29 F)
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Apollo 11 in the White House(9 F)
Apollo 11 world tour(2 C, 15 F)
Pages in category "Apollo 11"
The following 2 pages are in this category, out of 2 total.
Apollo 11
A
Apollo 11 flight (artist concept)
Gallery Slideshow Category Slideshow
Media in category "Apollo 11"
The following 200 files are in this category, out of 226 total.
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The Arm-co-operator, December 1962 - DPLA - d441dfbfd11f30be3ca6a6def89b4673 (page 13).jpg 1,380 × 1,810; 548 KB
Apollo Lunar Landing Mission Symposium - Apollo Earth Return Abort Capabilities - Fig 1.jpg 843 × 568; 59 KB
Apollo Lunar Landing Mission Symposium - Apollo Earth Return Abort Capabilities - Fig 2.jpg 878 × 515; 51 KB
LunarLandingMIssionSymposium1966 1978075303.pdf 1,270 × 1,650, 698 pages; 17.27 MB
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Buzz Aldrin (left) and Mike Collins on arrival at Patrick Air Force Base in a T-38.jpg 2,244 × 2,244; 561 KB
Apollo 11 Earth Orbit Chart 1.jpg 6,204 × 1,776; 6.83 MB
Alunizaje seguido desde el Trust Joyero.jpg 960 × 627; 137 KB
Apollo 11 insignia.png 5,380 × 5,432; 17.99 MB
Gold Olive Branch Left on the Moon by Neil Armstrong - GPN-2002-000070.jpg 3,000 × 2,912; 4.05 MB
John Hirasaki, detail (crop of NASA photo S69-45487).jpg 1,092 × 1,093; 322 KB
John K. Hirasaki decontaminating Apollo 11 (NASA photo S69-45487).jpg 4,105 × 4,121; 2.29 MB
Apollo11FlightPlan.jpg 770 × 598; 206 KB
Apollo 11 pre-flight press conference (48292441401).jpg 3,810 × 3,050; 4.24 MB
Neil Armstrong reviewing Apollo 11 flight plan (48293562382).jpg 2,963 × 3,763; 2.5 MB
Presentator Henk Terlingen met een model van de maanraket Apollo-11, Bestanddeelnr 922-6256.jpg 2,434 × 3,667; 1.29 MB
Regisseur Rudolf Spoor (links) en presentator Henk Terlingen met een model van d, Bestanddeelnr 922-6259.jpg 3,601 × 2,425; 1.33 MB
Buzz Aldrin reviewing Apollo 11 landing site imagery (48293465046).jpg 2,575 × 3,725; 3.65 MB
Buzz Aldrin (left) and Neil Armstrong at supper in crew quarters on the night before launch.jpg 2,847 × 2,220; 1.12 MB
Deke Slayton (on stool at left), Buzz Aldrin, Neil Armstrong, and Michael Collins during the last pre-flight press conference.jpg 3,000 × 2,427; 3.02 MB
Invite to Apollo 11 launch.png 1,456 × 864; 1.62 MB
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In the event of moon disaster.djvu 3,017 × 3,938, 2 pages; 9 KB
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Apollo 11. Television clip of Buzz descending the ladder and stepping onto the moon, 1094228.ogv 1 min 15 s, 320 × 240; 4.35 MB
Buzz Aldrin’s Apollo 11 Insurance Cover.jpg 300 × 168; 20 KB
President Richard Nixon Speaking by Telephone to the Apollo 11 Astronauts on the Moon.jpg 3,093 × 2,068; 1.28 MB
Apollo 11 lunar module cropped and shaded.jpg 960 × 540; 22 KB
Land on the Moon 7 21 1969.jpg 2,260 × 2,839; 3.31 MB
Customs and Immigration form signed by Apollo 11 astronauts after returning from the Moon.jpg 850 × 1,355; 188 KB
Huntsville Von Braun (36096240466).jpg 3,000 × 2,294; 2.81 MB
PPK of Micheal Collins.jpg 1,335 × 2,000; 2.74 MB
Henk Terlingen (l) en Rudolf Spoor, Bestanddeelnr 922-6727.jpg 3,667 × 2,413; 1.64 MB
Apollo 11 photo map.png 2,363 × 2,363; 221 KB
Proefnemingen voor lancering van Apollo II met Aldrin, Collins en Armstrong, USA, Bestanddeelnr 922-7055.jpg 2,637 × 3,406; 1.2 MB
Proefnemingen voor lancering van Apollo II met Aldrin, Collins en Armstrong, USA, Bestanddeelnr 922-7057.jpg 2,710 × 3,440; 1.23 MB
Proefnemingen voor lancering van Apollo II met Aldrin, Collins en Armstrong, USA, Bestanddeelnr 922-7058.jpg 3,406 × 2,715; 1.21 MB
First On The Moon - A Voyage with Neil Armstrong, Michael Collins, Edwin Aldrin Jr. (the Crew of Apollo 11).png 205 × 300; 94 KB
Michael Collins' razor and shaving cream.jpg 1,440 × 1,182; 81 KB
Tricot 1996 - Man on the Moon.jpg 1,254 × 1,449; 308 KB
Gateway to space 2016, Budapest, Apollo 11 hand casts.jpg 2,592 × 1,944; 1.24 MB
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Ruimtepak in Artis Zoo (2130202677).jpg 2,592 × 3,872; 3.85 MB
Apollo 11 Tapes Interview - DPLA - 059e6d6c57b553ee92c1231b8ebff8fa.jpg 3,832 × 2,550; 5.33 MB
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Apollo 11 Patch (3726741770).jpg 948 × 948; 393 KB
To the Moon and Beyond - Cover (3725935771).jpg 1,239 × 1,600; 844 KB
To the Moon and Beyond - Front Page (3726742496).jpg 1,230 × 1,600; 506 KB
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Apollo 11 Tapes Report.pdf 1,275 × 1,650, 20 pages; 564 KB
Hands of Apollo 11.jpg 3,744 × 5,616; 8.64 MB
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Apollo 11 Engines Recovered by Bezos Expeditions (8574459399).jpg 1,012 × 721; 446 KB
1969 Lunar Landing (9316776255).jpg 5,010 × 3,247; 1.78 MB
Media Viewer-Apollo 11.png 1,592 × 871; 486 KB
Lunar Television Camera for Apollo 11 Moon Landing, Westinghouse, identical to the model used on the moon - National Electronics Museum - DSC00569.JPG 3,554 × 3,320; 4.63 MB
Apollo11 Landscape 3D (14532843260).jpg 1,522 × 1,560; 2.17 MB
Exposição Relíquias do Mundo no Shopping Iguatemi em Ribeirão Preto. Revestimento interno Kapton foil da nave Eagle na missão Apollo 11, primeira missão a levar astronautas na Lua. Neil Armstrong e Edw - panoramio.jpg 3,456 × 4,608; 2.75 MB
Stanley Kubrick The Exhibition - TIFF - Shining (15675616924).jpg 5,030 × 3,353; 5.56 MB
Apollo 11 LM interior 20Jul69 (NASA via RJF) (18166925900).jpg 1,000 × 808; 131 KB
Jack King's Apollo 11 Launch Commentary o4OBKOlgmfo.webm 21 s, 1,280 × 720; 2.98 MB
The Neil Armstrong artifacts in the Smithsonian Restoration Labs - Flickr - jurvetson.jpg 3,840 × 3,216; 2.22 MB
Nixon Presidential Library & Museum (30873158416).jpg 2,736 × 1,824; 2.38 MB
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Buzz Aldrin's Spare Apollo Meal (35214203144).jpg 4,032 × 3,647; 3.7 MB
Apollo 11 Sample Return Case at Space Center Houston 2017.jpg 5,312 × 2,988; 4.74 MB
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20180706 Michael Collins Quarantine Suit Johnson Space Center.jpg 2,194 × 4,936; 8.28 MB
Apollo space suit.jpg 1,919 × 3,203; 1.6 MB
Apollo 11 Timeline.jpg 1,439 × 496; 74 KB
Apollo 11 Rare NASA Films from the National Archives.webm 2 min 46 s, 1,920 × 1,080; 57.88 MB
Apollo 11 artifacts, National Air and Space Museum, April 2019.jpg 3,456 × 2,304; 2.99 MB
Apollo 11 Moon Landing (49595207591).jpg 1,920 × 1,080; 1.06 MB
Apollo 50th - launch on NASA TV and Real Time Apollo (50136630638).jpg 4,032 × 3,024; 3.03 MB
Moone Exhibition at NMM - Apollo 11 flight plan (50137397597).jpg 5,472 × 3,648; 3.83 MB
Moon Exhibition at NMM - Apollo Camera (50137398317).jpg 3,648 × 5,472; 3.48 MB
A giant leap for mankind.png 1,146 × 500; 22 KB
La Luna del Apollo 11 - 20.07.1969.png 635 × 563; 48 KB
Apollo 11 Moon. 50 years after first step (annotated). (48364423771).jpg 5,472 × 3,648; 1.65 MB
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Space Program.jpg 1,471 × 1,960; 1.83 MB
Apollo 11 video console.jpg 2,651 × 1,490; 413 KB
Apollo 50th - fun with AR (50136623363).jpg 1,125 × 2,334; 524 KB
Apollo 11 50th - More Augmented reality fun (50137867186).png 736 × 1,595; 1.41 MB
Legacy of Branko Pešić in Adligat, signatures of the astronauts of Apollo 11.jpg 1,200 × 1,600; 208 KB
Chandrayaan-2 Apollo.jpg 755 × 572; 218 KB
028 Festa Major de Sants 2022 (Barcelona), carrer d'Alcolea, part de dalt.jpg 3,120 × 4,160; 4.34 MB
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430-L1-S1 640 (cropped).jpg 738 × 951; 276 KB
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5-shot Panorama (A11pan5454-58) taken by Buzz Aldrin just after the landing of Apollo 11 on the Moon. This is a contingency shot in case of a NO-STAY call.webp 1,186 × 479; 42 KB
A11 5847-8 dmh.webp 1,467 × 921; 235 KB
Apollo 10 and 11 crews photographed during Apollo 10 debriefing (S69-35504).jpg 4,009 × 2,638; 1.37 MB
Apollo 10 and 11 crews photographed during Apollo 10 debriefing (S69-35507).jpg 3,955 × 2,626; 1.55 MB
Apollo 11 - Space Particle Alert Network (SPAN) - Preparations - Mission Control Center (MCC) - MSC (S69-39753).jpg 4,734 × 3,685; 2.58 MB
Apollo 11 crewmen during first postflight debriefing (S69-40205).jpg 3,750 × 2,448; 1.28 MB
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Apollo 11 Flight Plan Photocopies of Materials - NARA - 117677443 (page 1).jpg 1,706 × 2,128; 86 KB
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Apollo 11 Ladder Mini-Pan.webp 3,670 × 2,523; 4.22 MB
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Apollo 11 Panorama Station 2.jpg 8,123 × 646; 3.46 MB
APOLLO 11 PATCH - NARA - 17450904.jpg 6,213 × 4,869; 23.94 MB
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Apollo 11 Rollout (624109main 1969-05-20-2 full).jpg 2,617 × 2,065; 625 KB
Apollo 11 Saturn V F-1 Engine Injector Plate (52742983218).jpg 3,683 × 2,832; 2.64 MB
Apollo 11 spacecraft Command Module hoisted aboard U S S Hornet (S69-21294).jpg 4,158 × 4,213; 2.98 MB
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Apollo 11, Science Museum, Winchester, Londýn.jpg 3,000 × 4,000; 2.36 MB
Apollo Lunar Module Scale Model - NARA - 6922350 (page 1).jpg 3,264 × 2,448; 954 KB
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Apollo Rd Sign.jpg 4,080 × 3,072; 2.68 MB
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Apollo XI Crewmen - Dining - Crew Reception Area - Lunar Receiving Lab (LRL) - MSC (S69-40307).jpg 4,548 × 3,498; 1.6 MB
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Carl Albert with Apollo 11 astronauts, including Edwin E. Buzz Aldrin, Neil Armstrong, and Michael Collins, and his son David. Hale Boggs is on the right.jpg 4,831 × 3,737; 2.44 MB
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CSIRO ScienceImage 2124 The Parkes Control Room Apollo 11 Mission.jpg 740 × 945; 435 KB
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6871 | https://fiveable.me/key-terms/hs-speech-debate/standard-deviation | Standard Deviation - (Speech and Debate) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Standard Deviation
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key term - Standard Deviation
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Definition
Standard deviation is a statistical measure that quantifies the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, while a high standard deviation suggests that the values are spread out over a wider range. This concept is crucial when analyzing both quantitative and qualitative data, as it helps to understand how consistent or variable the data points are in relation to the average.
5 Must Know Facts For Your Next Test
Standard deviation is commonly used in research to determine how representative a sample mean is of the population mean, allowing for more informed conclusions.
In practice, standard deviation helps researchers and analysts gauge whether their data sets show significant variability or consistency, impacting decision-making processes.
Calculating standard deviation involves finding the difference between each data point and the mean, squaring those differences, averaging them, and then taking the square root.
Standard deviation can be influenced by outliers, which can skew the perception of data variability; therefore, understanding its context is essential.
In qualitative data analysis, while standard deviation is less frequently applied directly, it can be helpful for quantifying responses or ratings in surveys to assess variability in opinions.
Review Questions
How does standard deviation help in understanding the consistency of a data set?
Standard deviation provides insight into how closely data points cluster around the mean. A low standard deviation indicates that most values are similar to the average, suggesting high consistency among data points. Conversely, a high standard deviation reveals greater spread among values, indicating variability. This understanding is crucial for interpreting data patterns and making informed decisions based on those patterns.
What impact do outliers have on standard deviation and how should they be considered when analyzing data?
Outliers can significantly affect standard deviation because they can increase the spread of data points from the mean, leading to a misleadingly high standard deviation. When analyzing data, it's important to identify and consider outliers since they can distort our understanding of variability. Researchers might choose to exclude outliers or conduct further analysis to determine if they represent genuine values or anomalies that should be addressed.
Evaluate the role of standard deviation in both quantitative and qualitative data analysis and its implications for research outcomes.
Standard deviation plays a pivotal role in quantitative data analysis by providing a clear measure of variability around the mean, enabling researchers to assess reliability and consistency of findings. In qualitative research, while its application might be less direct, it still aids in understanding variability in responses or attitudes when converted into numerical formats through rating scales. The implications for research outcomes are profound; accurately interpreting standard deviation ensures that conclusions drawn reflect true patterns within the data rather than being skewed by misleading variances.
Related terms
Mean:
The average of a set of values, calculated by adding all the numbers together and dividing by the count of values.
Variance:A statistical measure that represents the degree of spread in a set of data points; it is the square of the standard deviation.
Normal Distribution:A probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean.
"Standard Deviation" also found in:
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6872 | https://www.dictionary.com/browse/cosset | Daily Crossword
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View synonyms for cosset
cosset
[kos-it]
verb (used with object)
to treat as a pet; pamper; coddle.
noun
a lamb brought up without its dam; pet lamb.
any pet.
cosset
/ ˈkɒsɪt/
verb
to pamper; coddle; pet
noun
any pet animal, esp a lamb
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Other Word Forms
uncosseted adjective
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Word History and Origins
Origin ofcosset1
First recorded in 1570–80; Middle English; derivative verb use of the noun cosset “a lamb raised as a pet”; of uncertain etymology, but possibly Middle English cot-sēte “cottage dweller, cottager,” from Old English cot- sǣta
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Word History and Origins
Origin ofcosset1
C16: of unknown origin
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Example Sentences
Examples are provided to illustrate real-world usage of words in context. Any opinions expressed do not reflect the views of Dictionary.com.
Is there a benefit to being ignored by the industry — you don’t end up cosseted or overly corporate — at least as far as your stand-up goes?
FromLos Angeles Times
It’s hard to imagine the last time Trump, who cossets himself with an entourage of fawning sycophants, has been spoken to in that fashion.
FromLos Angeles Times
The Supreme Court’s so-called conservative justices have variously been vetted, feted and cosseted by the Federalist Society, along with and other well-heeled enablers, to encourage unstinting partisanship.
FromSalon
“But I seem to have been cosseted in some way that I couldn’t get to the writers’ room.
FromLos Angeles Times
Trump is a hero to many in the South and many Western states because, cossetted New York rich-boy though he may be, Republican voters see him as a rough-and-ready fellow rebel.
FromSalon
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6873 | https://www.ecdc.europa.eu/en/pneumococcal-disease/facts | Skip to main content
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Pneumococcal diseases are symptomatic infections caused by the bacterium Streptococcus pneumoniae (S. pneumoniae), commonly referred to as pneumococci. The term invasive pneumococcal disease (IPD) is used for more severe and invasive pneumococcal infections, such as bacteraemia, sepsis, meningitis and osteomyelitis, in which the bacterium can be isolated from normally sterile sites. Pneumococcal infections and IPDs are major causes of communicable disease morbidity and mortality in Europe and globally, with the highest burden of disease found in young children and the elderly. A large proportion of IPD is vaccine preventable.
The pathogen
Streptococcus pneumoniae is a Gram-positive bacterium that appears in pairs—diplococci—when examined in the microscope.
The polysaccharide capsule is an important virulence factor that protects the organism from phagocytosis.
Encapsulated strains are almost exclusively the cause of invasive infections and non-encapsulated strains only rarely cause disease.
Streptococcus pneumoniae is classified into serotypes based on the polysaccharide capsule antigens. The more than 90 immunologically distinct serotypes are numbered and structurally related serotypes are grouped together and labelled alphabetically (e.g., 6A, 6B).
Some serotypes possess distinct epidemiological properties and 23 serotypes account for most of pneumococcal bacteraemia and meningitis worldwide.
Clinical features and sequelae
Streptococcus pneumoniae is a common cause of otitis media, sinusitis, conjunctivitis and community-acquired pneumonia in addition to causing more severe IPDs such as meningitis, osteomyelitis and sepsis.
Other, less frequent infections caused by S. pneumoniae include periorbital cellulitis, osteomyelitis, endocarditis, pericarditis, peritonitis, pyogenic arthritis, soft tissue infections and neonatal septicaemia.
Clinical signs and symptoms do not allow for pneumococcal infections to be distinguished from other bacterial infections in the absence of laboratory tests.
Streptococcus pneumoniae is the most commonly isolated bacterial pathogen in acute otitis media (AOM) in both children and adults. The disease is frequently preceded by a viral respiratory tract infection.
Streptococcus pneumoniae conjunctivitis is a characteristically acute and painless infection with occasional pruritus. On inspection, the conjunctiva is thickened and injected and there is purulent, sometimes profuse, discharge. The infection is often self-limiting but topical antibiotic treatment may be required.
Streptococcus pneumoniae is the most common cause of community acquired bacterial pneumonia. Pneumococcal pneumonia is frequently preceded by a viral respiratory tract infection and typically presents abruptly with chills and high fever often followed by productive cough and pleuritic pain. The clinical picture is often less specific in infants and children who, more often than adults, will have bacteraemia on presentation. Symptoms are often less florid in the elderly and in immunocompromised patients. Pleural effusion is the most common complication of pneumococcal pneumonia. Lung abscesses and pericarditis resulting from local extension of the lung infection are relatively rare but well recognised complications of pneumococcal pneumonia. Bacteraemia is found as often as in 15–30% of patients with pneumococcal pneumonia. Hospital mortality has been estimated at 15%. Prognostic factors include age, underlying diseases, extent and complications of infection and the timeliness of effective antibiotic therapy.
Streptococcus pneumoniae is the most frequent aetiology of mastoiditis, as well as of the feared complications of mastoiditis; meningitis and brain abscesses.
Invasive pneumococcal disease is defined as the isolation of S. pneumoniae from blood or another normally sterile site. Invasive pneumococcal disease is therefore not one condition but a group of pneumococcal infections in which the pathogen has penetrated the body’s barrier defence and invaded normally sterile sites.
Streptococcus pneumoniae is the most common cause of bacterial meningitis in adults. Invasion of the meninges is usually via the bloodstream but can result from direct entry following a skull fracture. Presentation may be acute or sub-acute with fever, irritability, confusion and seizures. Signs of meningism are often prominent, and patients may present with focal neurological deficits and cranial nerve palsies. The case fatality ratio is as high as 10–30% and pneumococcal meningitis is associated with a higher risk of both death and permanent disability than other bacterial meningitis.
Bacteraemia is the most common manifestation of IPD. A large proportion (up to 90%) of culture confirmed IPD is the result of a primary lung infection as bacteraemia is common with pneumococcal pneumonia. Bacteraemia can present without localised symptoms and signs although, particularly in adults, there is often a focus from which the infection has spread. Bacteraemia may result in the seeding of pneumococcal infection to the meninges, peritoneum, bones, joints or lungs.
Sepsis refers to the clinical systemic manifestations of a severe infection and is associated with tachycardia, low blood pressure and circulatory collapse. It is therefore a more severe condition than bacteraemia which can be transient or occult.
Streptococcus pneumoniae is a common cause of osteomyelitis with or without contiguous joint infection, but a less common cause of soft tissue infections.
Epidemiology
Pneumococcal infections affect people of all ages but children younger than two years of age and adults aged 65 years and older are at higher risk.
A limited number of S. pneumoniae serotypes are responsible for most serious pneumococcal infections in both adults and children around the world.
The prevalence and distribution of invasive S. pneumoniae serotypes differs across populations and geographical areas.
The incidence of pneumococcal diseases peaks in the winter months in temperate climates. The seasonality is attributed to multiple factors including lower humidity, indoor crowding, associated viral infections, cold weather and air pollution.
Streptococcus pneumoniae is the leading cause of community-acquired pneumonia and the incidence is estimated at one per one thousand adults per year.
The introduction of Haemophilus influenzae type b (Hib) vaccine in the 1990s dramatically reduced invasive Hib disease in many European countries and led to S. pneumoniae becoming the leading cause of bacterial meningitis and sepsis in young children.
Routine immunisation with the 7-valent Pneumococcal Conjugate Vaccine (PCV 7) has in recent years profoundly changed the epidemiology of IPD in many European countries. Studies from the US, which in 2000 was the first country to introduce PCV 7, have documented reductions in the incidence of IPD caused by vaccine serotypes of 94% and overall reductions of IPD incidence of 75% in children below 5 years of age.
Immunisation against pneumococcal disease reduces carriage rates of vaccine serotypes. This leads to herd immunity against vaccine serotypes as transmission of vaccine strains from colonised to susceptible individuals goes down. Substantial reductions in the incidence of IPD and pneumonia have been demonstrated in unimmunised children and adults as a result of routine PCV 7 immunisation.
Otitis media is a leading indication for antibiotic treatment in Europe and pneumococcal vaccination has been shown to reduce the risk of otitis media.
The reported incidence of IPD in Europe ranges from 0.4 cases per 100 000 population to 20 cases per 100 000 population but it should be noted that surveillance strategies for IPD are heterogeneous across Europe making it difficult to compare data. The large variations in the reported IPD incidence across Europe are likely to reflect both true differences as well as differences in diagnostic and surveillance practices..
IPD cases are reported to the European Community network under Decision No 2119/98/EC and epidemiological updates are published annually in the Annual epidemiological report on communicable diseases in Europe.
Most cases of serious pneumococcal disease (pneumonia, septicaemia and meningitis) are sporadic, but outbreaks have been described in closed settings such as long-term care facilities, hospitals and households.
Transmission
Transmission of S. pneumoniae is from person to person by respiratory droplets.
The incubation period is uncertain but assumed to be around 1–3 days.
The infectious period is not known but is presumed to last until discharges from mouth and nose no longer contain pneumococci in significant numbers, and once effective antimicrobial treatment has been started, patients are thought to remain contagious for less than 24 hours.
The human nasopharynx is the only known reservoir for S. pneumoniae.
Asymptomatic nasopharyngeal colonization is common and ranges from 20 to 40% in children and from 5 to 10% in adults.
Colonisation with S. pneumoniae occurs early in life and is generally acquired at about six months of age, although there is considerable variation between populations.
The outcome of colonisation depends on the virulence of the specific serotype and on the host’s immune response. Susceptibility is increased by processes affecting the integrity of the lower respiratory tract including influenza, chronic lung disease or exposure to irritants, such as cigarette smoke.
There is a marked variation in the propensity of different S. pneumoniae serotypes to colonise the nasopharynx. For example serotypes 1, 3 and 46 are rarely found in the nasopharynx, even in populations in which they comprise a high proportion of IPD isolates, while other serotypes commonly identified in carriage studies rarely cause invasive disease.
Host risk factors for pneumococcal infection include young age, old age, diabetes, smoking, chronic lung disease, alcohol abuse, functional asplenia, sickle cell disease, leukaemia, multiple myeloma, HIV infection, and other immunodeficiencies.
The organism may spread locally from the nasopharynx to the sinuses or middle ear cavity, causing sinusitis or otitis media. The conjunctiva is infected via respiratory droplets or by direct contact. The primary focus of infection is not always obvious and in one study only 50% of adults with pneumococcal septic arthritis had another focus of pneumococcal infection.
Daycare centres provide an environment that facilitates transmission and outbreaks have also been reported among adults in closed settings such as nursing homes, military camps, prisons and shelters for the homeless.
Children are frequently the route of entry of the bacterium into households and the risk of transmission within a household is associated with family size.
Prevention
Childhood immunisation against S. pneumoniae is the most effective public health measure for preventing IPD both among vaccine recipients (direct effect), and among unimmunised populations (indirect ‘herd’ effect).
All EU/EEA countries but one include as part of their childhood immunisation schedule vaccination against pneumococcal invasive disease.
There are two principal types of pneumococcal vaccines currently in use: pneumococcal polysaccharide vaccine (PPV) and pneumococcal conjugate vaccines (PCV):
PPV-23 contains purified capsular polysaccharide from the 23 serotypes that most commonly cause IPD. It is poorly immunogenic in children younger than two years of age and does not reduce pneumococcal carriage. The vaccine induces a T-cell independent response and there is no booster effect from repeated immunisations.
PCV differ in their composition and target 10, 13, 15 or 20 seroptypes.
Details on vaccine used in respective EU/EEA Member States are found in the ECDC vaccine scheduler.
Management and treatment
Pneumococcal infections are treated with antibiotics and the choice of antibiotic should reflect local resistance patterns and national treatment guidelines.
Note: The information contained in this factsheet is intended for the purpose of general information and should not be used as a substitute for the individual expertise and judgement of healthcare professionals.
Page last updated
28 Nov 2023
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6874 | https://www.youtube.com/watch?v=z7vvUaqu5As | Paramagnetic vs Diamagnetic - Paired vs Unpaired Electrons - Electron Configuration
The Organic Chemistry Tutor
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This chemistry video tutorial focuses on paramagnetism and diamagnetism. It shows you how to identify if an element is paramagnetic or diamagnetic by writing the ground state electron configuration using noble gas notation and determining if the element has unpaired electrons by drawing the orbital diagram. Elements with unpaired electrons are paramagnetic which are weakly attracted to an external magnetic field. Elements with paired electrons only are diamagnetic and are weakly repelled by an external magnetic field.
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Electron Configuration of Ions:
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Paired & Unpaired Electrons:
Aufbau's Principle & Hund's Rule:
Paramagnetic & Diamagnetic Elements:
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Transcript:
In this video, we're going to focus on identifying substances as being paramagnetic or diiamagnetic. What you need to know is that if it's paramagnetic, it contains unpaired electrons. Now, if it doesn't have any unpaired electrons, it's diamagnetic. So, let's go through some examples. Magnesium is it paramagnetic or is it diamagnetic? So what we need to do is write the electron configuration for magnesium. So the first energy level only has the 1s suble. The second energy level has the 2s and 2p suble. For the third level it's 3s, 3 p, 3d. And for the fourth, 4s, 4 p, 4d, 4f. Now, the s suble can only hold two electrons, and the p suble can hold six. D can hold up to 10, and f can hold up to 14 electrons. The atomic number for magnesium, if you look at the periodic table, it's 12. So we're going to write the electron configuration until the exponents add up to 12. So we're going to start with the 1s orbital. It's going to be uh 1 s2 because s can hold up to two electrons. Then after 1 s we have the 2s suble. So 2 s2. After 2s we have 2 p then 3s. So it's going to be 2 p6. If we add the exponents right now we have a total of 10. We only need two more to get to 12. So we can stop at 3 s2. This is the electron configuration for magnesium. It's 1 s2 2 s2 2 p6 3s2. So now what we're going to do is we're going to draw the orbital diagram. So here's the electron configuration. And first we're going to have the 1s level, 2s level. P has three orbitals. S only has one. So this is 1 S, 2 S, 2 P, 3 S. Now notice that the energy levels or the orbitals are completely filled. There's no unpaired electrons. Therefore, since all of the electrons are paired, magnesium is diamagnetic. A substance that is diamagnetic is weakly repelled by an external magnetic field. So let's go ahead and try another example. What about manganesees? Is it paramagnetic or is it diiamagnetic? So feel free to pause the video and try this example yourself. So the first thing we need to do is write the electron configuration. So let's go ahead and let's do that. Now the atomic number for manganese is 25. So we're going to start with the 1s suble. S can hold up to two electrons. After 1 s we have 2 s. So this is going to be 2 s2. Then after that it's 2 p then 3s. P can hold up to six electrons. So it's 2 P6 and then 3 S2. After 3S we have 3 P then 4S. So 3 P6 4 S2. So 2 + 2 + 6 that's 10. And two and six and two that's also 10. So we have a total of 20. We need five more to get to 25. Now, the 3d suble can hold up to 10 electrons, but since we only need five more, we're going to stop at 3d5. So, that is the ground state electron configuration for manganesees. So, now we can draw the orbital diagram. Now, we know everything is going to be filled except the last um suble. So, let's focus on the 3D suble. Now according to Hun's rule, whenever you have orbitals or degenerate orbitals, that's orbitals of the same energy, you need to add the electrons one at a time. So you can't add them like this. You have to add them one at a time with parallel spins. So if the first one is facing up, the second arrow has to be facing up. So that's the main idea behind Huns rule. And as you can see, we have five unpaired electrons, which means that manganese is paramagnetic. So it's weakly attracted to an external magnetic field. So what about this uh element nickel? Is it paramagnetic or is it diamagnetic? Feel free to pause the video and try this example yourself. So nickel has an atomic number of 28. And so if we write the electron configuration, it's going to be starting with 1s. We have 1 S2 and then 2 S2 and then 2 P6 3 S2 and then after 3S2 it's uh 3 P6 4 S2 and then after that we need to go to the 3D suble. Right? And now if you add all the exponents, 2 + 2 + 6 + 2 + 6 + 2, that's 20. We need eight more. So we're going to stop at 3d8. So let's draw the orbital diagram for this element. So this is going to be 1 S, 2 S, 2 P, 3 S, 3 P and then 4 S. And then the D suble has five orbitals. Now according to the offball principle, we need to fill the lower energy levels first before filling the higher energy levels. So you got to start at the bottom. And then according to Hun's rule, once you have orbitals of the same energy, like these three 2p orbitals, you have to add the electrons one at a time with parallel spins. So now we need to fill the 3D suble. So 1 2 3 4 5 6 7 8. So nickel has two unpaired electrons which makes it a paramagnetic. Now how many paired electrons does it have? A quick way that you can find the number of paired electrons instead of counting them is notice that the total number of electrons in a neutral atom of nickel is 28. So 28 minus the two unpaired electrons will give you the number that is paired. So we have 26 paired electrons, two unpaired electrons. But overall the substance is still paramagnetic. So for the sake of practice, let's try one final example. Let's use zinc. Zinc has an atomic number of 30. So go ahead and try this example and see if you can get the answer. Determine if it's paramagnetic or if it's diamagnetic. So let's begin. So we have 1 S, 2 S, 2 P, 3 S, 3 P, 3 D, 4 S, 4 P, 4 D, 4 F. So just like before, we're going to start with 1 S. So we have 1 S2 and then 2 S2 and then after 2S we have the 2P suble. So it goes up to 2 P6 then 3 S2. After a while you can easily commit this to memory. So then this is going to be 3 P6 4 S2. So we have a total of 20. We need 10 more. Now 3D can hold up to 10. So we're going to stop at 3d10. This is the electron configuration for zinc. By the way, make sure you know the exceptions. The exceptions are chromium, malibdinum, copper, silver, gold. When you're writing the electron configuration for those elements, an S electron is going to jump into the d orbital. So, make sure you're aware of that. Now, this is the ground state electron configuration. Do you know how to write the electron configuration using noble gas notation? The noble gas that's before zinc is argon. Argon has an atomic number of 18. So therefore, argon covers the electron configuration up to 3 p6. If you add 2 2 6 2 and 6, you get 18. So the electron configuration of zinc using noble gas notation, it's argon 4s2 3d10. So now let's uh fill the uh orbital diagram. So here's the 4s suble and here is the 3D suble. So the 4s energy level is going to be completely filled and the same is true for the 3D suble. So as you can see all of the electrons are paired in zinc. So therefore zinc is diamagnetic. It has 30 paired electrons, no unpaired electrons. So that is it for this video. You know how to write the electron configuration um ground state and use a noble gas notation. And now you know how to identify if an element is paramagnetic or diiamagnetic. So thanks for watching this video and have a great day. |
6875 | https://www.mathway.com/popular-problems/Algebra/246321 | Find the x and y Intercepts 2x-3y=5 | Mathway
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Find the x and y Intercepts 2x-3y=5
2 x−3 y=5 2 x-3 y=5
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Find the x-intercepts.
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2 x+0=5 2 x+0=5
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2 x=5 2 x=5
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Divide each term in 2 x=5 2 x=5 by 2 2 and simplify.
Tap for more steps...
Step 1.2.2.1
Divide each term in 2 x=5 2 x=5 by 2 2.
2 x 2=5 2 2 x 2=5 2
Step 1.2.2.2
Simplify the left side.
Tap for more steps...
Step 1.2.2.2.1
Cancel the common factor of 2 2.
Tap for more steps...
Step 1.2.2.2.1.1
Cancel the common factor.
2 x 2=5 2 2 x 2=5 2
Step 1.2.2.2.1.2
Divide x x by 1 1.
x=5 2 x=5 2
x=5 2 x=5 2
x=5 2 x=5 2
x=5 2 x=5 2
x=5 2 x=5 2
Step 1.3
x-intercept(s) in point form.
x-intercept(s): (5 2,0)(5 2,0)
x-intercept(s): (5 2,0)(5 2,0)
Step 2
Find the y-intercepts.
Tap for more steps...
Step 2.1
To find the y-intercept(s), substitute in 0 0 for x x and solve for y y.
2(0)−3 y=5 2(0)-3 y=5
Step 2.2
Solve the equation.
Tap for more steps...
Step 2.2.1
Simplify 2(0)−3 y 2(0)-3 y.
Tap for more steps...
Step 2.2.1.1
Multiply 2 2 by 0 0.
0−3 y=5 0-3 y=5
Step 2.2.1.2
Subtract 3 y 3 y from 0 0.
−3 y=5-3 y=5
−3 y=5-3 y=5
Step 2.2.2
Divide each term in −3 y=5-3 y=5 by −3-3 and simplify.
Tap for more steps...
Step 2.2.2.1
Divide each term in −3 y=5-3 y=5 by −3-3.
−3 y−3=5−3-3 y-3=5-3
Step 2.2.2.2
Simplify the left side.
Tap for more steps...
Step 2.2.2.2.1
Cancel the common factor of −3-3.
Tap for more steps...
Step 2.2.2.2.1.1
Cancel the common factor.
−3 y−3=5−3-3 y-3=5-3
Step 2.2.2.2.1.2
Divide y y by 1 1.
y=5−3 y=5-3
y=5−3 y=5-3
y=5−3 y=5-3
Step 2.2.2.3
Simplify the right side.
Tap for more steps...
Step 2.2.2.3.1
Move the negative in front of the fraction.
y=−5 3 y=-5 3
y=−5 3 y=-5 3
y=−5 3 y=-5 3
y=−5 3 y=-5 3
Step 2.3
y-intercept(s) in point form.
y-intercept(s): (0,−5 3)(0,-5 3)
y-intercept(s): (0,−5 3)(0,-5 3)
Step 3
List the intersections.
x-intercept(s): (5 2,0)(5 2,0)
y-intercept(s): (0,−5 3)(0,-5 3)
Step 4
2 x−3 y=5 2 x-3 y=5
and,,x x
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6876 | http://staff.ustc.edu.cn/~wangzuoq/Courses/20S-Topology/Notes/Lec11.pdf | Topology (H) Lecture 10 Lecturer: Zuoqin Wang Time: April 27, 2020 TIETZE EXTENSION THEOREM Last time we learned: • Separation axioms: (T1), (T2), (T3), (T4) – The only “(Ti) = ⇒(Tj)” (1 ≤i ̸= j ≤4) is (T2) = ⇒(T1) – (T1) + (T4) = ⇒(T3) , (T1) + (T3) = ⇒(T2) – compact + (T3) = ⇒(T4) , compact + (T2) = ⇒(T3) – Lindel¨ of + (T3) = ⇒(T4) • Urysohn’s lemma: (X, T ) is (T4) if and only if disjoint closed subsets can be separated by a continuous map.
– A, B disjoint closed sets in (T4) space = ⇒∃continuous f : X →[0, 1] such that f(A) = 0, f(B) = 1.
– In a (T4) space, A = f −1(0) for some continuous function f : X →[0, 1] ⇐ ⇒A is a closed Gδ set.
1. Tietze Extension Theorem Although the function we get in Urysohn’s lemma looks too special, they can be used as building blocks to construct more complicated continuous functions with certain properties, as we have seen in the proof of Urysohn’s metrization theorem and in the proof of the variant of Urysohn’s lemma. In this section we will give another application of Urysohn’s lemma, Tietze extension theorem,1 which can be viewed as a generalization of Urysohn’s lemma (although they are in fact equivalent), and thus is directly applicable to more situations.
We start with a trivial definition.
Definition 1.1. Let A ⊂X be a subset. We say a map ˜ f : X →Y is an extension of a map f : A →Y if ˜ f = f on A.
In analysis, it is always important to extend a given function from a smaller domain to a larger domain, while keeping some properties, e.g. continuity (or smoothness), 1According to wikipedia, the theorem was first proved by Brouwer and Lebesgue for the special case of the theorem when X is Rn, and then was extended by Tietze to all metric spaces. The current version for normal space was proved by Urysohn.
1 2 TIETZE EXTENSION THEOREM boundedness. As a result, Tietze extension theorem is one of the most useful theorems in topology.
Theorem 1.2 (Tietze Extension Theorem). A topological space (X, T ) is normal if and only if for any closed set A ⊂X and given any continuous function f : A →[0, 1], there exists a continuous function e f : X →[0, 1] which is an extension of f.
Proof.
(⇐) Let A, B be disjoint closed sets in X. Then A ∪B is closed in X, and f : A ∪B →[0, 1], f(x) = ¨ 0, x ∈A 1, x ∈B is a continuous function on A ∪B. By assumption, f can be extended to a continuous function e f : X →[0, 1] such that e f = f on A ∪B. So by Urysohn’s lemma, X is (T4).
(⇒) The idea Consider the “restriction map” R : C(X, [0, 1]) →C(A, [0, 1]), g 7→g|A, where C(X, [0, 1]) means the space of all continuous maps f : X → [0, 1]. Then we want to prove that R is surjective, i.e. we want to solve the equation Rg = f.
We will apply a standard trick in analysis: First find an approximate solution.
Then iteratively find better and better approximations.
Finally prove the sequence of approximate solutions converges to a true solution.
Now we realize the idea. For simplicity, we replace [0, 1] by [−1, 1].
Step 1 [Construction an approximate solution] First we approximate the function f by f1 : A →R, f1(x) = 8 < : 1/3, if f(x) ⩾1/3, f(x), if |f(x)| ⩽1/3, -1/3, if f(x) ⩽−1/3.
By construction we have |f(x) −f1(x)| ⩽2 3, ∀x ∈A.
Then we use Urysohn’s lemma to find a continuous function g : X →R s.t.
Rg ≈f1.
TIETZE EXTENSION THEOREM 3 There is a very obvious candidate for such a function g: since A1 := x ∈A f(x) ⩽1 3 and B1 := x ∈A f(x) ⩽−1 3 are both closed in X, there exists a continuous g : X →[−1 3, 1 3] s.t.
g(x) = 1 3 on A1, g(x) = −1 3 on B1.
It’s easy to see that g(x) also satisfies |f(x) −Rg(x)| ⩽2 3, ∀x ∈A.
Step 2 [Do iteration] Write f = f1. According to Step 1, we have obtained a continuous function g1 : X →[−1 3, 1 3] s.t.
|f1(x) −Rg1(x)| ⩽2 3, ∀x ∈A.
Repeat Step 1 with f replaced by f2 = f1 −Rg1, we can construct a continuous function g2 : X →[−1 3 · 2 3, 1 3 · 2 3] s.t.
|f2(x) −Rg2(x)| ⩽(2 3)2, ∀x ∈A.
Iteratively we can find a sequence of continuous functions gn : X → −1 3(2 3)n−1, 1 3(2 3)n−1 s.t. if we denote fn+1 = fn −Rgn, then |fn(x) −Rgn(x)| ⩽(2 3)n, ∀x ∈A.
Step 3 [Converges to a solution] Let e f(x) = ∞ X n=1 gn(x).
Since each gn is continuous on X, and |gn(x)| ⩽1 3(2 3)n−1, we see the series converges uniformly and thus e f is continuous on X, and | e f(x)| ⩽1, ∀x ∈X.
4 TIETZE EXTENSION THEOREM Finally for ∀x ∈A, we have f(x) − N X n=1 gn(x) = |f1 −g1 −· · · −gN| = |f2 −g2 −· · · −gN| = · · · = |fN −gN| ⩽(2 3)N, So f(x) = e f(x) for x ∈A.
□ 2. Various generalizations Obviously in the statement of Tietze extension theorem, we can replace the target space [0, 1] by any closed interval [a, b]: We only need to compose the functions we get with the linear transform t 7→a + t(b −a) and its inverse transform.
A not-that-obvious extension is: we can replace [0, 1] by R.
Theorem 2.1 (Tietze extension theorem for unbounded functions). Suppose X is normal and A ⊂X is closed.
Then any continuous function f : A →R can be extended to a continuous function e f : X →R.
Proof. Composing f with the function arctan(x), we get a continuous function f1 := arctan ◦f : A →(−π 2 , π 2 ).
By Tietze extension theorem, we can extend f1 to a continuous function f f1 : X →[−π 2 , π 2 ].
Let B = f f1 −1(±π 2 ).
Then B is closed in X, and B ∩A = ∅. By Urysohn’s lemma, there exists a continuous function g : X →[0, 1] s.t.
g(A) = 1 and g(B) = 0.
Define h(x) = f f1(x)g(x).
TIETZE EXTENSION THEOREM 5 Then h is a continuous function mapping X into (−π 2, π 2). Finally we let e f(x) = tan h(x).
Then e f : X →R is continuous, and e f(x) = tan h(x) = tan f f1(x) = tan f1(x) = x, ∀x ∈A.
□ Remark. One can also extend continuous vector-valued functions f : A →[0, 1]n, f : A →Rn, or f : A →[0, 1]S to continuous vector-valued functions on X, i.e. to e f : X →[0, 1]n, e f : X →Rn, or e f : X →[0, 1]S, where S is an arbitrary set. To do so, one only need to extend each component of f separately. [Note: A map to the product space is continuous if and only if all its components are continuous.] On the other hand, for a topological space Y , in general one can’t expect to extend any continuous function f : A →Y to a continuous function e f : X →Y .
• To extend a function f : {0, 1} →Y to a continuous map e f : [0, 1] →Y, a necessary condition is: f(0) and f(1) should lie in the same “path component” of Y .
• To extend a continuous map f : S1 →Y to a continuous map e f : D →Y , where D is the unit disc in the plane, one need require the image f(S1) to be “contractible” in Y . In particular, we will see that the identity map f : S1 →S1, x 7→x can not be extended to a continuous map e f : D →S1. [Brouwer fixed point theorem] We will study these connectivity phenomena in the second half of this course.
Remark. One may pose extra assumptions on the extension. For example, one can extend smooth functions to smooth functions, which is known as Whitney extension theorem. One can also require the extension to preserve other properties like Lips-chitz/H¨ older continuity (for metric space), or boundedness (See PSet).
To apply Urysohn’e lemma or Tietze extension theorem, one need to assume that the source space is normal. One class of normal spaces that appears in many applica-tions are those topological spaces that are both compact and Hausdorff. However, in some other applications, compactness is too strong, and we only have “local compact-ness” and Hausdorff. Recall [PSet 3-2-2(c)]: 6 TIETZE EXTENSION THEOREM Definition 2.2. A topological space (X, T ) is called locally compact if for any x ∈X, there exists a compact set Kx and an open set Ux such that x ∈Ux ⊂Kx.
Notation. A locally compact Hausdorffspace will be abbreviated as LCH.
Example.
• Any compact Hausdorffspace is LCH.
• Rn is LCH.
• More generally, any locally Euclidian Hausdorffspace is LCH.
– We say a topological space is locally Euclidian if for any x ∈X, there exists a neighborhood U of x which is homeomorphic to an open ball in Euclidian space.
• Let K be any compact Hausdorffspace. Let p be any point such that X = K \ {p} is non-compact. Then Fact 1. X is a non-compact LCH.
As a subspace of a Hausdorffspace, X is Hausdorff.
To see X is locally compact, for any x ∈X ⊂K, since K is Hausdorffwe can find disjoint open sets U, V in K such that x ∈U, p ∈V . Now K \ V is a closed subset in compact space K, and thus is compact in K. Since K \ V ⊂X, it is also compact in X. By definition it is a compact neighborhood of x in X, since U ⊂K \ V .
Conversely, Fact 2. Any non-compact LCH arises this way: According to PSet3-1-3(b), any non-compact topological space admits a one-point compactification X∗= X ∪{∞} 2. In the case X is a non-compact LCH, one can easily check that the X∗is a compact Hausdorff space. In particular, this shows that any LCH can be realized an an open subspace of some compact Hausdorffspace.
Remark. One can prove that the space Qp, i.e. the completion of Q under the p-adic metric, is locally compact, and thus LCH. As a result, analysis on LCH is very useful in p-adic analysis.
• Q ⊂R is NOT locally compact. [Why?] • The Sorgenfrey line (R, TSorgenfrey) is NOT locally compact. [Why?] 2The one-point compactification is also known as the Alexandrov compactification, and sometimes denoted by αX.
TIETZE EXTENSION THEOREM 7 LCHs appear widely in analysis. For LCHs, people also want to apply Urysohn’s lemma or Tietze extension theorem to construct continuous functions with specific properties. Unfortunately, not all LCH’s are normal, so that we can’t extend all con-tinuous functions defined on closed sets. [Note: normal is a necessary and sufficient condition in Urysohn’e lemma as well as in Tietze extension theorem.] However, we can prove that a weaker version holds for LCH, which is one of the most useful results in the analysis on LCH: Theorem 2.3 (Urysohn’s Lemma, LCH version). Let X be a LCH, and K, F be disjoint subsets in X with K compact and F closed. Then there exists a continuous function f : X →[0, 1] such that f(K) = 1 and f(F) = 0.
Proof. The proof is based on the following separation property in LCH. [You can com-pare this with the equivalent characterization of (T4) that we proved last time.] Lemma 2.4. Let X be a LCH, K be a compact set in X, and U be an open set in X such that K ⊂U. Then there exists an open set V such that V is compact, and K ⊂V ⊂V ⊂U.
We will leave the proof of the lemma as an exercise. Applying the lemma to the pair K ⊂F c, we can find an open set V with V compact, such that K ⊂V ⊂V ⊂F c.
Applying the lemma again, but now to K ⊂V , we will find an open set U with U compact, such that K ⊂U ⊂U ⊂V.
Since V is a compact Hausdorffspace, it is normal. According to Urysohn’s lemma, one can find a continuous function g : V →[0, 1] such that g(K) = 1, g(V \ U) = 0.
We can extend g to a continuous function f : X →[0, 1] by simply “extend by 0”, i.e.
f(x) = ¨ g(x), x ∈V , 0, x ∈X \ V .
Then f is continuous on both V and X \ U. It follows that f is continuous on X.
Obviously f(K) = 1 and f(F) = 0.
□ In a very similar way, one can prove the following LCH version of Tietze extension theorem. We will leave the proof as an exercise as well: Theorem 2.5 (Tietze extension theorem, LCH version). Let X be a LCH, and K be a compact subset. Then any continuous function f : K →R can be extended to a compactly supported continuous function f : X →R.
8 TIETZE EXTENSION THEOREM 3. Various applications Tietze extension theorem has many applications. For example, in real analysis, Tietze extension theorem was used to produce a sequence of continuous functions that approximates (a.e.) a given measurable function. In what follows we give more applications of Tietze extension theorem.
Application 1: Yet another characterization of compactness for metric spaces.
We have given several different characterization of compactness for metric spaces.
Here is another one: Proposition 3.1. A metric space (X, d) is compact if and only if any continuous function f : X →R is bounded.
Proof. If (X, d) is compact, then by the extreme value theorem, any continuous function f : X →R is bounded.
To prove the converse, we argue by contradiction. Suppose (X, d) is non-compact, then there exists A = {x1, x2, · · · } such that A′ = ∅. It follows that A is closed and each xn is isolated in A. So the function f : A →R, f(xn) = n is continuous on A. By Tietze extension theorem, there exists a continuous function e f : X →R such that e f = f on A. Obviously e f is an unbounded continuous function on X, a contradiction.
□ Application 2: From the Cantor set to space-filling curves Our second application is concerned with the Cantor set C. Recall C = [0, 1] \ ∞ [ n=1 3n−1−1 [ k=0 (3k + 1 3n , 3k + 2 3n ).
As we mentioned at the beginning of Lecture 7, one way to understand C is via the ternary representation of real numbers, i.e. regard C as the image of the map g : {0, 1}N →[0, 1], a = (a1, a2, · · · ) 7→ ∞ X k=1 2 3k ak.
We have checked in PSet 3-2-1(a)) that the map g is a homeomorphism from ({0, 1}N, Tproduct) onto the Cantor set C.
On the other hand, the map h : {0, 1}N →[0, 1]2, a = (a1, a2, · · · ) 7→ ∞ X k=1 a2k−1 2k , ∞ X k=1 a2k 2k !
is continuous and surjective: TIETZE EXTENSION THEOREM 9 To check continuity, one only need to check continuity of each compo-nent, which can be done easily via sub-base. The surjectivity is just another way to say that each real number has a binary representation.
[Note: h can’t be injective, otherwise as a bijective continuous map from a compact space to a Hausdorffspace, it will has to be a homeomor-phism, which is absurd.] As a consequence, we get a continuous surjective map h ◦g−1 : C →[0, 1]2.
Since C is closed in [0, 1], Tietze extension theorem indicates that there exists a con-tinuous surjective map f : [0, 1] →[0, 1]2.
Definition 3.2. Any continuous surjective map from [0, 1] to [0, 1]2 is called a Peano Curve or a space-filling curve.
Remarks.
(1) By using a very similar argument, one can easily construct surjective continuous map f : [0, 1] →[0, 1]n or even surjective continuous map f : [0, 1] →[0, 1]N.
(2) Of course our argument is an “non-constructive” proof of the existence Peano curve. In literature there are also many “constructive proofs” which iteratively construct such a curve.
10 TIETZE EXTENSION THEOREM (3) The space-filling curves are not just theoretic monsters. They have many prac-tical applications in real life. For example, it is used in storing multidimensional data into computer (which is arranged linearly), e.g. Google maps, so that when you move a little bit on the map, you only move a little bit in the memory, that is why we require continuity of the function.
Application 3: The Stone-ˇ Cech compactification of LCHs.
Let X be a non-compact LCH. We have seen how to compactify X to a compact Hausdorffspace via the Alexandrov compactification (i.e. the-one point compactifica-tion). Geometrically, the Alexandrov compactification αX of X is a compactification that glue all “open ends” of X together, which is not as good as we want in many ap-plications. Here is another widely used way to compactify X, via continuous functions.
As usual we will denote by C(X, [0, 1]) the space of all continuous functions f : X →[0, 1]. Consider the “huge product space” Q = [0, 1]C(X,[0,1]), equipped with the product topology. Since compactness and Hausdorffare both pro-ductive [PSet 4-2-2], Q is a compact Hausdorffspace.
Theorem 3.3. Let X be a LCH. Then the map θ defined by θ : X →Q, x 7→(f(x))f∈C(X,[0,1]).
is a homeomorphism to its image θ(X).
Proof. For simplicity we denote βX = θ(X).
It is compact since it is a closed subset in the compact space Q. It is Hausdorffsince it is a subspace of the Hausdorffspace Q.
Before we prove the theorem, first note that any f ∈C(X, [0, 1]) induces a map πf : Q →[0, 1] is continuous since it is just the coordinate map. We denote ˜ f = πf|βX : βX →[0, 1].
Then ˜ f is continuous and satisfies ˜ f ◦θ = f. Since βX is the closure of θ(X), such a continuous function ˜ f : βX →[0, 1] is unique. Moreover, by our construction, if suppf ⊂K for some compact set K, then ˜ f = 0 on βX \ θ(K).
To prove θ : X →θ(X) is a homeomorphism, it is enough to prove f is continuous, bijective and open.
• θ is continuous since each component of θ is continuous.
TIETZE EXTENSION THEOREM 11 • θ is injective (and thus bijective onto θ(X)): For any x ̸= y, by Urysohn lemma one can find continuous function f ∈C(X, [0, 1]) such that f(x) ̸= f(y). Then the continuous function ˜ f satisfies ˜ f(θ(x)) ̸= ˜ f(θ(y)). So θ(x) ̸= θ(y).
• θ is open: For any open set U in X, and any point y = θ(x) ∈θ(U), by locally compactness one can find a compact set K ⊂U such that x ∈Int(K) ⊂K. [To get such K, one can apply Lemma 2.4 to the pair {x} ⊂U. The set V we get there can be taken as K here.] According to LCH version of Urysohn’s lemma, there exists f ∈C(X, [0, 1]) such that f(x) = 1 and supp(f) ⊂K. Consider the function ˜ f : βX →[0, 1] as constructed above. Then ˜ f = 0 on βX \ θ(U).
It follows that the open set V = ˜ f −1((0, +∞)) ⊂βX is contained in θ(U). Clearly we have y ∈V since f(x) = 1.
□ The closure of the image, βX := θ(X), is a compactification of X. It is known as the Stone-ˇ Cech compactification of X.
In the proof, we also get the following important property of the Stone-ˇ Cech com-pactification.
Proposition 3.4. Let X be a non-compact LCH. Then any bounded continuous func-tion f : X →R can be “extended” uniquely to a continuous function ˜ f : βX →R.
Remark. So a space may have many different compactifications. We can view each compactification X as a pair (ι, X), where ι : X →X is a homeomorphism from X to its image ι(X). We say a compactification (ι, X) of X is finer than another compactification (ι′, X ′), (and say (ι′, X ′) is coarser than (ι, X)), if three exists a continuous map g : X ′ →X such that ι = π ◦ι′. One can prove that for non-compact LCHs, the Alexandrov compactification αX is the coarsest compactification, while the Stone-ˇ Cech compactification is the finest compactification.
Application 4: Partition of unity.
Our last application is to “Partition of unity”, which we will have more to say next time. Today we will only prove a simple version: Theorem 3.5 (“Partition of unity”). Let Kα be closed subsets in a normal space X such that [ α Kα = X.
Let Uα be open neighbourhoods of Kα’s which is locally finite, i.e.
∀x ∈X, ∃open Ux ∋x s.t. Ux ∩Uα ̸= ∅for at most finitely many α’s.
12 TIETZE EXTENSION THEOREM Then there exist continuous functions fα : X →[0, 1] such that fα > 0 on Kα.
fα = 0 on U c α.
P α fα(x) = 1, ∀x ∈X.
Proof. By Tietze extension theorem (in fact, Urysohn’s lemma), there exist continuous functions gα : X →[0, 1] such that gα = 1 on Kα, gα = 0 on U c α.
Define g(x) = X α gα(x).
Then on open set Ux, g is a finite sum of continuous functions. So g is well-defined and is continuous on each Ux, and hence g is well-defined and is continuous on the whole of X. Moreover, g(x) ≥1, ∀x since S α Kα = X. Now we set fα(x) = gα(x) g(x) .
It is easy to check that fα’s are what we need.
□ Remark. Next time we will discuss topological conditions which guarantee the local finiteness assumption above, as well as applications of “partition of unity”. |
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Lutjanus guttatus, Spotted rose snapper : fisheries, gamefish
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Lutjanusguttatus (Steindachner, 1869)
Spotted rose snapper
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Lutjanus guttatus
Juvenile Picture by Allen, G.R.
Classification / Names Common names | Synonyms | Catalog of Fishes(genus, species) | ITIS | CoL | WoRMS | Cloffa
Teleostei (teleosts) >Eupercaria/misc (Various families in series Eupercaria) >Lutjanidae (Snappers) > Lutjaninae
Etymology: Lutjanus:Malay, ikan lutjan, name of a fish.
More on author: Steindachner.
Environment: milieu / climate zone / depth range / distribution range Ecology
Marine; reef-associated; depth range 10 - 60 m (Ref. 91172). Tropical; 33°N - 12°S, 115°W - 78°W (Ref. 55)
Distribution Countries | FAO areas | Ecosystems | Occurrences | Point map | Introductions | Faunafri
Eastern Pacific: Mexico to Peru.
Length at first maturity / Size / Weight / Age
Maturity: L m26.4, range 17 - 18 cm
Max length : 105 cm TL male/unsexed; (Ref. 106806); max. published weight: 1.3 kg (Ref. 40637)
Short description Identification keys | Morphology | Morphometrics
Dorsalspines (total): 10; Dorsalsoft rays (total): 12 - 13; Analspines: 3; Analsoft rays: 8. Preopercular notch and knob weak. Scale rows on back rising obliquely above the lateral line. Pale crimson on side, often with silvery sheen of horizontal rows of bluish spots; belly golden yellow. Head with bluish spots and irregular broken lines, especially across cheek. A large blackish blotch on the upper back below the posterior of the dorsal spines.
Body shape(shape guide): fusiform / normal;Cross section: oval.
Biology Glossary (e.g. epibenthic)
Adults are found over hard bottoms in inshore reef areas. Generally solitary or in small groups but may occasionally form big schools (Ref. 9313). Juveniles inhabit estuaries and mouths of rivers (Ref. 9313). Carnivorous, feed on invertebrates and fish (Ref. 9313). Marketed fresh or frozen (Ref. 9313).
Life cycle and mating behavior Maturity | Reproduction | Spawning | Eggs | Fecundity | Larvae
Main reference Upload your references | References | Coordinator | Collaborators
Allen, G.R., 1985. FAO Species Catalogue. Vol. 6. Snappers of the world. An annotated and illustrated catalogue of lutjanid species known to date. FAO Fish. Synop. 125(6):208 p. Rome: FAO. (Ref. 55)
IUCN Red List Status (Ref. 130435: Version 2025-1)
Least Concern (LC); Date assessed: 21 May 2007
CITES
Not Evaluated
CMS (Ref. 116361)
Not Evaluated
Threat to humans
Harmless
Human uses
Fisheries: commercial; gamefish: yes
FAO - Aquaculture systems: production; Fisheries: landings; Publication: search | FishSource | Sea Around Us
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Estimates based on models
Preferred temperature (Ref. 123201): 22.3 - 29.1, mean 26.2 °C (based on 214 cells).
Phylogenetic diversity index (Ref. 82804):PD 50 = 0.5000 [Uniqueness, from 0.5 = low to 2.0 = high].
Bayesian length-weight: a=0.01349 (0.00851 - 0.02138), b=2.91 (2.78 - 3.04), in cm total length, based on LWR estimates for this species & Genus-body shape (Ref. 93245).
Trophic level (Ref. 69278):4.0 ±0.2 se; based on diet studies.
Resilience (Ref. 120179):Medium, minimum population doubling time 1.4 - 4.4 years (K = 0.19; Fec = 66,400).
Prior r = 0.57, 95% CL = 0.37 - 0.85, Based on 1 data-limited stock assessment.
Fishing Vulnerability (Ref. 59153):Low to moderate vulnerability (35 of 100). 🛈
Price category (Ref. 80766):High.
Nutrients (Ref. 124155):Calcium = 28.4 [17.6, 41.8] mg/100g; Iron = 0.327 [0.204, 0.502] mg/100g; Protein = 18.8 [17.3, 20.2] %; Omega3 = 0.127 [0.086, 0.183] g/100g; Selenium = 71.2 [46.4, 111.1] μg/100g; VitaminA = 88 [17, 326] μg/100g; Zinc = 0.362 [0.282, 0.513] mg/100g (wet weight);
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6878 | https://julioandrade.weebly.com/uploads/4/0/3/2/40324815/notes2015v35.pdf | ECM3704 Number Theory 2015–2016 Henri Johnston H.Johnston@exeter.ac.uk 10th December 2015 Abstract These notes are based on the lecture notes and handouts of Dr Robin Chapman who gave this course in 2013–2014.
The lectures were originally typed up by Oliver Bond who sat in on the course that year. The notes have since been completely rewritten, but consider-able thanks is clearly due to both Robin and Oliver. All errors are my own; please do email me if you find any.
1 Divisibility and primes 1.1 Divisibility Definition 1.1. We recall the definitions of the following sets.
(i) N (the natural numbers) is defined to be {1, 2, 3, . . .} (note 0 / ∈N).
(ii) Z (the integers) is defined to be {. . . , −3, −2, −1, 0, 1, 2, 3, . . .}.
(iii) Q (the rational numbers) is defined to be {a b | a, b ∈Z, b ̸= 0}.
Remark 1.2. (Z, +, ×) is a commutative ring with 1 and (Q, +, ×) is a field.
Definition 1.3. Given a, b ∈Z, we say that ‘a divides b’ or ‘a is a divisor of b’ or ‘b is a multiple of a’ or ‘a is a factor of b’ if and only if there exists c ∈Z such that b = ac. (If a ̸= 0 this means b a ∈Z.) The notation ‘a | b’ means ‘a divides b’.
Proposition 1.4. Let a, b, c, n, x, y ∈Z. Divisibility has the following prop-erties: (i) a | a (reflexive property), (ii) a | b and b | c implies a | c (transitive property), (iii) a | b and a | c implies a | (xb ± yc) (linearity property), 1 (iv) a | b implies an | bn (multiplication property), (v) an | bn and n ̸= 0 implies a | b (cancellation property), (vi) 1 | n (1 divides every integer), (vii) n | 0 (every integer divides 0), (viii) 0 | n implies n = 0 (zero divides only zero), (ix) a | b and b ̸= 0 implies |a| ≤|b| (comparison property), (x) a | b and b | a implies |a| = |b|, i.e., a = ±b.
Proof. Checking the properties is straightforward. We check (iii) and leave the others as an exercise. Since a | b and a | c there exist m, n ∈Z such that b = an and c = am. For any x, y ∈Z we have xb ± yc = xan ± yam = a(xn ± ym).
So we have found an integer q = xn ± ym such that xb ± yc = aq. Thus a | (xb ± yc), as desired.
1.2 The Division Algorithm Well-Ordering Principle (WOP). Every non-empty subset of N ∪{0} contains a least element.
Theorem 1.5 (The Division Algorithm). Given a ∈Z, b ∈N, there exist unique integers q and r satisfying a = bq + r and 0 ≤r < b.
Proof. We first establish the existence of such a pair of integers q and r.
Define S := {a −xb | x ∈Z and a −xb ≥0}. Note that S ̸= ∅since: • if a ≥0, by choosing m = 0, we get a −mb = a ≥0; • if a < 0, by choosing m = a, we get a −mb = a −ab = (−a)(b −1) ≥0 since −a > 0 and b > 0.
Hence S is a non-empty subset of N ∪{0} and so by the Well-Ordering Principle S contains a least element r ≥0. Since r ∈S we have there exists q ∈Z such that a −qb = r, so a = qb + r. It remains to show that r < b.
Assume for a contradiction that r ≥b and let r1 = r −b ≥0. Then a = qb + r = qb + (r1 + b) = (q + 1)b + r1 and so a −(q + 1)b = r1 ∈S and is smaller than r: a contradiction. Hence q and r satisfy the required properties.
We now show that the pair q, r is unique. Assume that there is another pair of integers q′, r′ such that a = q′b + r′ with 0 ≤r′ < b. Then from a = qb + r = q′b + r′ we have (q −q′)b = r′ −r. If q = q′ then we must have r = r′ and we are done. Suppose for a contradiction that q ̸= q′. Then b ≤|q −q′||b| = |r′ −r|.
2 However, since 0 ≤r, r′ < b we must have |r′ −r| < b, which gives a contradiction.
1.3 The Greatest Common Divisor Theorem 1.6. Let a, b ∈Z. Then there exists a unique d ∈N ∪{0} and (non-unique) x, y ∈Z such that (i) d | a and d | b, (ii) if e ∈Z, e | a and e | b then e | d, (iii) d = ax + by.
Proof. If a = b = 0, then it is easy to check that we must have d = 0. So suppose that a and b are not both zero. Let S = {am + bn | m, n ∈Z and am + bn > 0}.
Now a2+b2 > 0 so S is a non-empty subset of N. Hence by the Well-Ordering Principle, S has a minimal element d > 0 and we can write d = ax + by for some x, y ∈Z.
By the Division Algorithm, a = qd + r for some q, r ∈Z with 0 ≤r < d.
Suppose for a contradiction that r ̸= 0. Then 0 < r = a −qd = a −q(ax + by) = (1 −qx)a −qby.
Hence r ∈S. But r < d, contradicting the minimality of d in S. So we must have r = 0, i.e., d | a. The same argument also shows that d | b.
Suppose e ∈Z, e | a and e | b. Then e divides any linear combination of a and b, so in particular, e | d.
Suppose that e ∈N∪{0} also satisfies (i) & (ii). Then e | d and d | e and so d = ±e. But d, e ≥0 so we have d = e. Thus d is unique.
Corollary 1.7. If a, b ∈Z then there exists a unique d ∈N ∪{0} such that (i) d | a and d | b, (ii) if e ∈Z, e | a and e | b then e | d.
Proof. The existence of such a d is given by Theorem 1.6. In the proof of uniqueness in Theorem 1.6, we only used properties (i) & (ii).
Definition 1.8. Let a, b ∈Z.
Then the d of Corollary 1.7 is called the greatest common divisor of a and b and is written gcd(a, b). (Note this is the same d as in Theorem 1.6.) This is sometimes also referred to as the highest common factor and written as hcf(a, b). If gcd(a, b) = 1 then a and b are said to be coprime or relatively prime.
3 Combining Theorem 1.6, Corollary 1.7 and Definition 1.8, we have: Bezout’s Identity. Given a, b ∈Z there exist (non-unique) x, y ∈Z such that gcd(a, b) = ax + by.
Proposition 1.9. Let a, b, c ∈Z. The gcd has the following properties: (i) gcd(a, b) = gcd(b, a) (commutative law), (ii) gcd(a, gcd(b, c)) = gcd(gcd(a, b), c) (associative law), (iii) gcd(ac, bc) = |c| gcd(a, b) (distributive law), (iv) gcd(a, 1) = gcd(1, a) = 1, (v) gcd(a, 0) = gcd(0, a) = |a|, (vi) c | gcd(a, b) if and only if c | a and c | b, (vii) gcd(a + cb, b) = gcd(a, b).
Proof. Checking properties (i),(ii),(iv),(v) & (vi) is straightforward and is left as an exercise. (For property (vi) use Bezout’s Identity and the linearity property of divisibility).
We prove (iii). Let d = gcd(a, b) and let e = gcd(ac, bc). We wish to show that e = |c|d. By property (vi), cd | e = gcd(ac, bc) since cd | ac and cd | bc.
By Bezout’s Identity, there exist x, y ∈Z such that d = ax + by. Then cd = acx + bcy.
But e | ac and e | bc and so by linearity of divisibility we have e | cd.
Therefore |e| = |cd|, i.e., e = |c|d.
Finally, we prove (vii). Let e = gcd(a + bc, b) and f = gcd(a, b). Then e | (a + bc) and e | b. Thus by linearity of divisibility e | a. Hence e | a and e | b so by property (vi), we have e | f. Similarly, f | a and f | b so again by linearity of divisibility f | (a + bc). Thus f | (a + bc) and f | b and so again by property (vi), we have f | e. Therefore e | f and f | e and f, e ≥0 so we conclude that e = f.
Remark 1.10. Note that gcd(a, b) = 0 if and only if a = b = 0. Otherwise gcd(a, b) ≥1.
Theorem 1.11 (Euclid’s Lemma). Let a, b, c ∈Z. If a | bc and gcd(a, b) = 1 then a | c.
Proof. Suppose that a | bc and gcd(a, b) = 1. By Bezout’s Identity there exist x, y ∈Z such that 1 = ax + by. Hence c = acx + bcy. But a | acx and a | bcy, so a | c by the linearity property of divisibility.
Theorem 1.12 (Solubility of linear equations in the integers). Let a, b, c ∈Z.
The equation ax + by = c is soluble with x, y ∈Z if and only if gcd(a, b) | c.
4 Proof. Let d = gcd(a, b). Then d | a and d | b so if there exist x, y ∈Z such that c = ax+by then d | c by linearity of divisibility. Now suppose that d | c.
Then we can write c = qd for some q ∈Z. By Bezout’s Identity there exist x′, y′ ∈Z such that d = ax′ + by′. Hence c = qd = aqx′ + bqy′ and so x = qx′ and y = qy′ gives a suitable solution.
1.4 Euclid’s Algorithm Theorem 1.13 (Euclid’s Algorithm). Let a, b ∈N with a ≥b > 0 and b ∤a.
Let r0 = a, r1 = b and apply the Division Algorithm repeatedly to obtain a set of remainders r2, r3, . . . , rn, rn+1 defined successively by the relations r0 = r1q1 + r2, 0 < r2 < r1, r1 = r2q2 + r3, 0 < r3 < r2, .
.
.
rn−2 = rn−1qn−1 + rn, 0 < rn < rn−1, rn−1 = rnqn + rn+1, rn+1 = 0.
Then the last non-zero remainder, rn, is equal to gcd(a, b).
Proof. There is a stage at which rn+1 = 0 because the ri are strictly de-creasing and non-negative. Recall from Proposition 1.9 (vii) that for any x, y, z ∈Z we have gcd(x, y) = gcd(x + zy, y). In particular, we have gcd(ri, ri+1) = gcd(ri+1qi+1 + ri+2, ri+1) = gcd(ri+2, ri+1) = gcd(ri+1, ri+2).
Applying this result repeatedly gives gcd(a, b) = gcd(r0, r1) = gcd(r1, r2) = gcd(r2, r3) = · · · = gcd(rn−1, rn) = rn where the last equality is because rn | rn−1.
Remark 1.14. One can also use Euclid’s Algorithm to find x, y ∈Z such that gcd(a, b) = ax + by by ‘working backwards’.
Example 1.15. Work out the greatest common divisor of 841 and 160 and express it as a linear combination of 841 and 160: 841 = 160 × 5 + 41 160 = 41 × 3 + 37 41 = 37 × 1 + 4 37 = 4 × 9 + 1 4 = 1 × 4 + 0.
5 Hence gcd(841, 160) = 1 (i.e. they are coprime) and working backwards gives: 1 = 37 × 1 −4 × 9 = 37 × 1 −(41 −37) × 9 = 37 × 10 −41 × 9 = (160 −3 × 41) × 10 −41 × 9 = 160 × 10 −41 × 39 = 160 × 10 −(841 −160 × 5) × 39 = −39 × 841 + 205 × 160.
Note that such a solution is not unique. For example, we will also have 1 = (160 −39) × 841 + (205 −841) × 160 = 121 × 841 −636 × 160.
1.5 The Extended Euclidean Algorithm Instead of performing Euclid’s Algorithm to compute gcd(a, b) and then ‘working backwards’ to compute x, y ∈Z such that gcd(a, b) = ax + by, one can instead compute x, y during the course of performing Euclid’s Al-gorithm. This is known as the Extended Euclidean Algorithm.
The sequences of quotients qi and remainders ri are defined as in Theorem 1.13. We also define sequences of integers xi, yi such that ri = axi+byi. Recall that we defined rn to be the last non-zero remainder and that rn = gcd(a, b).
Therefore we have gcd(a, b) = rn = axn +byn and so we set (x, y) := (xn, yn).
So how do we explicitly find xi and yi? Recall that r0 = a and r1 = b.
Thus r0 = 1 × a + 0 × b and r1 = 0 × a + 1 × b, and so we set (x0, y0) := (1, 0) and (x1, y1) := (0, 1). Now assume that i ≥2 and that xj, yj are known for j < i. Then ri−2 = ri−1qi−1 + ri and so we have ri = ri−2 −ri−1qi−1 = (axi−2 + byi−2) −(axi−1 + byi−1)qi−1 = a(xi−2 −xi−1qi−1) + b(yi−2 −yi−1qi−1).
Thus we set xi := xi−2 −xi−1qi−1 and yi := yi−2 −yi−1qi−1. In other words, for i ≥2 we define (xi, yi) recursively by (xi, yi) := (xi−2, yi−2) −qi−1(xi−1, yi−1).
6 Example 1.16. We compute gcd(841, 160) and express it as a linear combin-ation of 841 and 160 using the Extended Euclidean Algorithm.
i ri−2 ri−1 qi−1 ri xi yi 0 841 1 0 1 160 0 1 2 841 = 160 × 5 + 41 1 −5 3 160 = 41 × 3 + 37 −3 16 4 41 = 37 × 1 + 4 4 −21 5 37 = 4 × 9 + 1 −39 205 6 4 = 1 × 4 + 0 Therefore gcd(841, 160) = 1 = 841 × (−39) + 160 × 205.
1.6 Primes Definition 1.17. Prime and composite numbers in N: (i) A number p ∈N with p > 1 is prime if and only if its only divisors are 1 and p (i.e. if n ∈N and n | p then n = 1 or n = p).
(ii) A number n ∈N with n > 1 is composite if and only if it is not prime (i.e. n = ab for some a, b ∈N with a, b > 1).
Note that n = 1 is neither prime nor composite.
Proposition 1.18. If n ∈N with n > 1 then n has a prime factor.
Proof. We use strong induction, i.e., we prove that if for all m ∈N with 1 < m < n, m has a prime factor, then n has a prime factor.
Case (i): if n is prime, then n is a prime factor of n.
Case (ii): If n is composite then n = ab where a, b ∈N with a, b > 1. So 1 < a < n. By the induction hypothesis, there is a prime p with p | a. Hence p | a and a | n, so by the transitivity property of divisibility p | n.
Proposition 1.19. If n ∈N with n > 1 then we can write n = p1p2 · · · pk where k ∈N and p1, . . . , pk are (not necessarily distinct) primes.
Proof. If n is prime then the result is clear. So suppose that n is composite.
Then by Proposition 1.18 n has a prime factor, i.e., n = p1n1 where p1 is prime and n1 ∈N with n1 > 1. If n1 is prime, we are done. If n1 is composite, it has a prime factor p2 and we can write n1 = p2n2 where n2 ∈N with n2 > 1. If n2 is prime, we are done, otherwise we take out another prime factor and keep on going. The process does eventually terminate since n > n1 > n2 > · · · > 1. Hence after at most n steps we obtain a prime factorisation of n.
7 Example 1.20. We have 666 = 3 × 222 = 3 × 2 × 111 = 3 × 2 × 3 × 37.
Theorem 1.21. There are infinitely many primes.
Euclid’s proof. For a contradiction, assume {p1, p2, . . . , pn} is a complete list of primes. Consider N := 1 + p1p2 . . . pn ∈N. Then N > 1 so by Proposition 1.18, N has a prime factor p. However, every prime is supposedly one of p1, . . . , pn, so p = pi for some i. Then p = pi | (p1 . . . pn), so p | (N −1).
However, we also have p | N and we can write 1 = N −(N −1), so p | 1, which is a contradiction.
1.7 The Fundamental Theorem of Arithmetic Lemma 1.22. Let n ∈Z. If a prime p does not divide n then gcd(p, n) = 1.
Proof. Let d = gcd(p, n). Then d | p so by definition of prime either d = 1 or d = p. But d | n so d ̸= p because p ∤n. Hence d = 1.
Theorem 1.23 (Euclid’s Lemma for Primes). Let a, b ∈Z and let p a be prime. If p | ab then p | a or p | b.
Proof. Assume that p | ab and that p ∤a. We shall prove that p | b. By Lemma 1.22, gcd(p, a) = 1 so by Euclid’s Lemma (Theorem 1.11), p | b.
Remark 1.24. Euclid’s Lemma for Primes immediately generalises to several factors: if p is prime and p | a1a2 . . . ak then p | aj for some j.
Definition 1.25. Let n ∈N and let p be a prime. Then vp(n) := max{k ∈N ∪{0} : pk | n}.
(Note that this set is always non-empty as it must contain 0; moreover, it is clear that it is bounded above.) In other words, k is the unique non-negative integer such that pk | n but pk+1 ∤n. Equivalently, vp(n) = k if and only if n = pkn′ where n′ ∈N and p ∤n′. (For the right to left implication, use the cancellation property of divisibility.) Example 1.26. The following example illustrates the definition of vp(n).
• v2(720) = 4 because 720 16 = 45 is odd, so 24 | 720 but 25 ∤720.
• v3(720) = 2 because 3 ∤80 = 720 9 , so 32 | 720 but 33 ∤720.
• v5(720) = 1 because 5 ∤144 = 720 5 , so 5 | 720 but 52 ∤720.
• If p ≥7 then vp(720) = 0 because p ∤720.
Lemma 1.27. Let n, m ∈N and let p be a prime. Then vp(mn) = vp(m) + vp(n).
8 Proof. Let k = vp(m) and ℓ= vp(n). Then we can write m = pkm′ where p ∤m′ and similarly n = pℓn′ where p ∤n′. Then mn = pk+ℓm′n′. By Euclid’s Lemma for Primes, p ∤m′n′. Therefore vp(mn) = k + ℓ.
Theorem 1.28 (The Fundamental Theorem of Arithmetic). Let n ∈N with n > 1. Then (i) (Existence) The number n can be written as a product of primes.
(ii) (Uniqueness) Suppose that n = p1 · · · pr = q1 · · · qs where each pi and qj is prime. Assume further that p1 ≤p2 ≤· · · ≤pr and q1 ≤q2 ≤· · · ≤qs.
Then r = s and pi = qi for all i.
Proof. The existence of a factorisation into primes is just Proposition 1.19.
Thus it remains to show uniqueness. Let ℓbe any prime. Then by Lemma 1.27 we have vℓ(n) = vℓ(p1 · · · pr) = vℓ(p1) + · · · + vℓ(pr).
However, vℓ(pi) = 1 if ℓ= pi, 0 if ℓ̸= pi.
Therefore vℓ(n) = # of i for which ℓ= pi = # of times ℓappears in the factorisation n = p1 · · · pr.
Similarly, vℓ(n) = # of times ℓappears in the factorisation n = q1 · · · qs.
Thus every prime ℓappears the same number of times in each factorisation, giving the desired result.
Remark 1.29. Another way of interpreting this result is to say that for n ∈N n = p vp1(n) 1 p vp2(n) 2 · · · pvpr(n) r where p1, . . . , pr are the distinct prime factors of n. Note that we take the empty product to be 1, which covers the case n = 1.
Lemma 1.30. Let n = Qr i=1 pai i where each ai ∈N ∪{0} and the pi’s are distinct primes. The set of positive divisors of n is the set of numbers of the form Qr i=1 pci i where 0 ≤ci ≤ai for i = 1, . . . , r.
Proof. Exercise.
9 2 Modular Arithmetic 2.1 Congruences Definition 2.1. Suppose a, b ∈Z and n ∈N. We write a ≡b mod n (or a ≡b (mod n)), and say ‘a is congruent to b mod n’, if and only if n | (a−b).
If n ∤(a −b) we write a ̸≡b mod n and say that ‘a and b are incongruent mod n’.
Remark 2.2. In particular, a ≡0 mod m if and only if m | a.
Examples 2.3.
(i) 4 ≡30 mod 13 since 13 | (4 −30) = −26.
(ii) 17 ̸≡−17 mod 4 since 17 −(−17) = 34 but 4 ∤34.
(iii) n is even if and only if n ≡0 mod 2.
(iv) n is odd if and only if n ≡1 mod 2.
(v) a ≡b mod 1 for every a, b ∈Z.
Proposition 2.4. Let n ∈N. Being congruent mod n is an equivalence relation, i.e., the relation is: (i) Reflexive: For all a ∈Z we have a ≡a mod n.
(ii) Symmetric: Let a, b ∈Z. If a ≡b mod n then b ≡a mod n.
(iii) Transitive: Let a, b, c ∈Z.
If a ≡b mod n and b ≡c mod n then a ≡c mod n.
Proof. The proof follows at once from the following properties of divisibility: (i) n | 0.
(ii) If n | (a −b) then n | (b −a).
(iii) If n | (a −b) and n | (b −c) then n | (a −b) + (b −c) = (a −c).
Proposition 2.5. Congruences respect addition, subtraction and multiplic-ation. Let n ∈N and let a, b, α, β ∈Z. Suppose that a ≡α mod n and b ≡β mod n. Then (i) a + b ≡α + β mod n, (ii) a −b ≡α −β mod n, and (iii) ab ≡αβ mod n.
Moreover, if f(x) ∈Z[x] then f(a) ≡f(α) mod n.
Proof. We will check that ab ≡αβ mod n; the rest is an exercise. Since a ≡α mod n, we have n | (a −α) and so a = α + ns for some s ∈Z.
Similarly, b = β + nt for some t ∈Z. Hence ab = (α + ns)(β + nt) = αβ + n(sβ + tα + nst) and so n | (ab −αβ). Therefore ab ≡αβ mod n, as required.
10 Example 2.6. Let n ∈N and write n in decimal notation n = k X i=0 ai × 10i where 0 ≤ai ≤9 and ai ∈N ∪{0} for all i.
Define f(x) by f(x) = k X i=0 aixi.
Then, since 10 ≡−1 mod 11, we see that n = f(10) ≡f(−1) mod 11, whence 11 | n ⇐ ⇒11 | f(−1) ⇐ ⇒11 | (a0 −a1 + a2 −a3 + . . . + (−1)kak).
This gives an easy way to test integers for divisibility by 11.
Example 2.7. Does the equation x2 −3y2 = 2 have a solution with x, y ∈Z?
Let x, y ∈Z. Note that x2 −3y2 ≡x2 mod 3. Now x ≡0, 1 or 2 mod 3, so x2 ≡0, 1 or 4 mod 3. But 4 ≡1 mod 3 so in fact x2 ≡0 or 1 mod 3. Hence x2 −3y2 ≡x2 ̸≡2 mod 3 and so x2 −3y2 ̸= 2.
Theorem 2.8. There are infinitely many primes p with p ≡3 mod 4.
Proof. If p is a prime then p ≡0, 1, 2 or 3 mod 4. But p ̸≡0 mod 4 because 4 ∤p. If p ≡2 mod 4 then p = 4k + 2 for some k ∈Z so 2 | p and so in fact p = 2. Therefore there are three types of primes: (i) p = 2, (ii) p ≡1 mod 4, (iii) p ≡3 mod 4.
Let N ∈N.
It suffices to show that there exists a type (iii) prime with p > N. Let M = 4(N!) −1. Then M ≥3 and so by the existence statement of Fundamental Theorem of Arithmetic (i.e. Proposition 1.19) M has a prime factorisation M = p1p2 · · · pk. If p is a prime such that p ≤N then M ≡ −1 mod p so p ∤M. Hence pj > N for all j. Moreover, pj ̸= 2 for all j because M is odd. Therefore for each j we have pj ≡1 or 3 mod 4. If pj ≡3 mod 4 for any j then we are done. If this is not the case then pj ≡1 mod 4 for all j, and so M ≡1 × · · · × 1 ≡1 mod 4; but by definition of M we have M ≡−1 ≡3 mod 4 - contradiction!
Remark 2.9. Congruences do not respect division. For example, 4 ≡14 mod 10 but 2 ̸≡7 mod 10.
Proposition 2.10. Let a, b, s ∈Z and d, n ∈N.
(i) If a ≡b mod n and d | n then a ≡b mod d.
(ii) Suppose s ̸= 0. Then a ≡b mod n if and only if as ≡bs mod ns.
11 Proof. (i) follows from the transitivity property of divisibility; (ii) follows from the multiplication and cancellation properties.
Theorem 2.11 (Cancellation Law for Congruences). Let a, b, c ∈Z and n ∈N. Let d = gcd(c, n). Then ac ≡bc mod n ⇐ ⇒a ≡b mod n d. In particular, if n and c are coprime, then ac ≡bc mod n ⇐ ⇒a ≡b mod n.
Proof. Since d = gcd(c, n), we may write n = dn′ and c = dc′ where n′, c′ ∈Z.
Suppose ac ≡bc mod n. Then n | c(a −b) and hence the cancellation property of divisibility gives n′ | c′(a −b). However, gcd(n′, c′) = 1 and so n′ | (a −b) by Euclid’s Lemma. Thus a ≡b mod n′.
Suppose conversely that a ≡b mod n′. Then n′ | (a−b) and so n | d(a−b).
But d | c so d(a−b) | c(a−b) and thus n | c(a−b) by the transitive property of divisibility. Thus ac ≡bc mod n.
Proposition 2.12. Let a, m, n ∈Z. If m and n are coprime and if m | a and n | a then mn | a.
Proof. Since m | a we can write a = mc for some c ∈Z. Now n | a = mc and gcd(m, n) = 1 so by Euclid’s Lemma, n | c. Thus by the multiplicative property of divisibility, mn | mc = a.
Corollary 2.13. Let m, n ∈N be coprime and let a, b ∈Z. If a ≡b mod m and a ≡b mod n then a ≡b mod mn.
Proof. We have n | (a −b) and m | (a −b). Since m and n are coprime we therefore have mn | (a −b) by Proposition 2.12, i.e., a ≡b mod mn.
2.2 Residue classes and complete residue systems Proposition 2.14. Let a, b ∈Z and n ∈N. If a ≡b mod n and |b −a| < n then a = b.
Proof. Since n | (a −b), by the comparison property of divisibility we have n ≤|a −b| unless a −b = 0.
Recall Proposition 2.4 that congruence modn is an equivalence relation.
Definition 2.15. Consider a fixed modulus n ∈N. For a ∈Z we write [a]n for the equivalence class of a mod n. Thus [a]n = {x ∈Z | x ≡a mod n} = {a + qn | q ∈Z}.
This is called the residue class of a modulo n.
12 Example 2.16. Consider the case n = 2. Then 2 = {x ∈Z | x ≡0 mod 2} = {the even integers}, 2 = {x ∈Z | x ≡1 mod 2} = {the odd integers}.
To understand the results in this section, it is often helpful to think of the case n = 2 and odd and even integers.
Proposition 2.17. Let n ∈N. The n residue classes n, n, . . . , [n −1]n are disjoint and their union is the set of all integers. In other words, every x ∈Z is congruent modulo n to precisely one element of {0, 1, 2, . . . , n −1}.
Proof. The integers 0, 1, 2, . . . , n−1 are incongruent modulo n by Proposition 2.14. Hence the residue classes n, n, . . . , [n −1]n are distinct and thus disjoint (recall that distinct equivalences classes are always disjoint). Now every integer x must be in exactly one of these classes because by the Division Algorithm we can write x = qn + r where 0 ≤r < n, so x ≡r mod n and hence x ∈[r]n.
Definition 2.18. Let n ∈N. If S is a subset of Z containing exactly one element of each residue class modulo n we say that S is a complete residue system modulo n.
Remark 2.19. Proposition 2.17 says that S = {0, 1, 2, . . . , n−1} is a complete residue system modulo n. Note that if S is any complete residue system modulo n we must have |S| = n because Proposition 2.17 shows that there are precisely n residue classes modulo n. In fact, any set consisting of n integers, incongruent modulo n, is a complete residue system modulo n.
Examples 2.20. Let n ∈N. Then {0, 1, . . . , n −1}, {1, . . . , n}, {1, n + 2, 2n + 3, 3n + 4, . . . , n2}, {x ∈Z | −n/2 < x ≤n/2}, are all complete residue systems modulo n.
Proposition 2.21. Let n ∈N and k ∈Z. Assume k and n are coprime. If {a1, . . . , an} is a complete residue system modulo n then so is {ka1, . . . , kan}.
Proof. If kai ≡kaj mod n then by the cancellation law for congruences (The-orem 2.11) we have ai ≡aj mod n since gcd(k, n) = 1. Therefore no two (distinct) elements in the set {ka1, . . . , kan} are congruent modulo n. Since there are n elements in this set, it forms a complete residue system.
13 Example 2.22. The set {0, 1, 2, 3, 4} is a complete residue system mod 5. Now gcd(2, 5) = 1 so {2 × 0, 2 × 1, 2 × 2, 2 × 3, 2 × 4} = {0, 2, 4, 6, 8} is also a complete residue system mod 5.
2.3 Linear Congruences The most basic congruences are linear congruences, i.e., those of the form ax ≡b mod n to be solved for x. When the modulus n is small we can just use brute force, i.e., just try every possible value of x mod n. However, this quickly becomes impractical as n increases.
Theorem 2.23 (Linear congruences with exactly one solution). Let a, b ∈Z and let n ∈N. Suppose that a and n are coprime. Then the linear congruence ax ≡b mod n (1) has exactly one solution.
Proof. We need only test the numbers 1, 2, . . . , n since they constitute a complete residue system. Therefore we consider the products a, 2a, . . . , na.
Since a and n are coprime, Proposition 2.21 shows that these numbers also constitute a complete system of residues modulo n. Hence exactly one of the elements of this set is congruent to b modulo n. In other words, there is exactly one x satisfying (1).
Example 2.24. Let a = 160, b = 3 and n = 841. That is, we wish to solve 160x ≡3 mod 841.
(2) Applying the Extended Euclidean Algorithm in this case shows that gcd(160, 841) = 1 = 160 × 205 + 841 × (−39).
(3) (We did this calculation in Example 1.16.) Thus by Theorem 2.23 there is exactly one solution. Moreover, reducing equation (3) mod 841 gives 160 × 205 ≡1 mod 841 and so multiplying through by 3 gives 160 × (205 × 3) ≡160 × 615 ≡3 mod 841.
Therefore x := 205 × 3 = 615 is the unique solution to (2) modulo 841.
14 Theorem 2.25 (Solubility of a linear congruence). Let a, b ∈Z and let n ∈N. Then the linear congruence ax ≡b mod n (4) has one or more solutions if and only if gcd(a, n) | b.
Proof. By definition, the congruence (4) is soluble if and only if n | (b −ax) for some x ∈Z, and this is true if and only if b −ax = ny for some x, y ∈Z.
Hence (4) is soluble if and only if ax + ny = b for some x, y ∈Z.
Therefore the result now follows from Theorem 1.12 (solubility of linear equations in the integers).
Theorem 2.26. Let a, b ∈Z and let n ∈N. Let d = gcd(a, n). Suppose that d | b and write a = da′ , b = db′ and n = dn′. Then the linear congruence ax ≡b mod n (5) has exactly d solutions modulo n. These are given by t, t + n′, t + 2n′, . . . , t + (d −1)n′, (6) where t is the unique solution, modulo n′, of the linear congruence a′x ≡b′ mod n′.
(7) Proof. Every solution of (5) is a solution of (7) and vice versa by Proposition 2.10. Since a′ and n′ are coprime, (7) has exactly one solution t modulo n′ by Theorem 2.23. Thus the d numbers in (6) are solutions of (7) and hence of (5). No two of these are congruent modulo n since the relations t + rn′ ≡t + sn′ mod n with 0 ≤r < d, 0 ≤s < d imply rn′ ≡sn′ mod n, and hence r ≡s mod d, where the last implication follows from Proposition 2.10 (note n/n′ = d).
But 0 ≤|r −s| < d so r = s by Proposition 2.14.
It remains to show that (5) has no solutions other than those listed in (6).
If y is a solution of (5) then ay ≡b mod n. But we also have at ≡b mod n and so ay ≡at mod n.
Thus y ≡t mod n′ by the cancellation law for congruences (Theorem 2.11).
Hence y = t + kn′ for some k ∈Z.
But 15 r ≡k mod d for some r ∈Z such that 0 ≤r < d. Therefore by Proposition 2.10 we have kn′ ≡rn′ mod n, and so y ≡t + rn′ mod n.
Therefore y is congruent modulo n to one of the numbers in (6).
Algorithm 2.27 (How to solve general linear congruences). Let a, b ∈Z and let n ∈N. Suppose we wish to solve the linear congruence ax ≡b mod n.
(8) First apply the Extended Euclidean Algorithm to compute d := gcd(a, n) and find x′, y′ ∈Z such that ax′ + ny′ = d.
(9) If d ∤b then there are no solutions by Theorem 2.25. Otherwise, there are exactly d solutions modulo n by Theorem 2.26, which we can find as follows.
Write a = da′ , b = db′ and n = dn′. Dividing (9) through by d gives a′x′ + n′y′ = 1.
Thus reducing mod n′ gives a′x′ ≡1 mod n′ and multiplying through by b′ gives a′(b′x′) ≡b′ mod n′. Therefore t := b′x′ is the unique solution to a′x ≡ b′ mod n′ (the solution is unique because gcd(a′, n′) = 1). Now by Theorem 2.26 the solutions to (8) are t, t + n′, t + 2n′, . . . , t + (d −1)n′.
Example 2.28. Let a = 33, b = 21 and n = 54. That is, we wish to solve 33x ≡21 mod 54.
(10) We apply the Extended Euclidean Algorithm as follows.
i ri−2 ri−1 qi−1 ri xi yi 0 54 0 1 1 33 1 0 2 54 = 33 × 1 + 21 −1 1 3 33 = 21 × 1 + 12 2 −1 4 21 = 12 × 1 + 9 −3 2 5 12 = 9 × 1 + 3 5 −3 6 9 = 3 × 3 + 0 Therefore gcd(54, 33) = 3 = ax5 + ny5 = 33 × 5 + 54 × (−3).
(11) 16 Thus there are exactly 3 solutions. Moreover, we have a′ = 11, b′ = 7 and n′ = 18 and we may take x′ = 5 and y′ = −3. Hence 1 = a′x′ + n′y′ = 11 × 5 + 18 × (−3).
Reducing mod n′ = 18 gives 11 × 5 ≡1 mod 18 and multiplying through by b′ = 7 gives 11 × (7 × 5) ≡7 mod 18.
Hence t ≡7 × 5 ≡35 ≡17 mod 18 is the unique solution to 11x ≡7 mod 18.
Therefore the set of solutions to (10) modulo 54 is {17, 17 + (1 × 18), 17 + (2 × 18)} = {17, 35, 53}.
2.4 The ring Z/nZ and its units Definition 2.29. Let n ∈N. We write Z/nZ = {[a]n : 0 ≤a ≤n −1} (so that # (Z/nZ) = n). We set [a]n +[b]n := [a+b]n and [a]n[b]n := [ab]n. (Note that Proposition 2.5 shows that these operations are well-defined).
Lemma 2.30. The set Z/nZ, with the above operations, is a commutative ring with 0 = n and 1 = n.
Proof. Omitted and non-examinable.
Definition 2.31. Let n ∈N. We write (Z/nZ)× = {[a]n ∈Z/nZ : ∃[b]n ∈Z/nZ such that [a]n[b]n = n}.
This is the set of units of Z/nZ, and is an abelian group under multiplication (check this!) Proposition 2.32 (Units of Z/nZ). Let n ∈N and a ∈Z. Then [a]n ∈ (Z/nZ)× if and only if gcd(a, n) = 1.
Proof. If [a]n ∈(Z/nZ)× then the linear congruence ax ≡1 mod n has a solution, and so by Theorem 2.25 (solubility of a linear congruence) we have gcd(a, n) | 1. But n > 0 so in fact gcd(a, n) = 1.
Suppose conversely that gcd(a, n) = 1. Then by Theorem 2.23 (linear congruences with exactly one solution), the congruence ax ≡1 mod n has exactly one solution, and so [a]n ∈(Z/nZ)×.
17 Definition 2.33. Let n ∈N and let a ∈Z such that gcd(a, n) = 1. Then the unique solution to ax ≡1 mod n is called the multiplicative inverse of a modulo n and is denoted [a]−1 n = [a−1]n or a−1 mod n.
Example 2.34.
Z 12Z × = {12, 12, 12, 12}.
2.5 Chinese Remainder Theorem Theorem 2.35 (Chinese Remainder Theorem - a special case ). Let n, m ∈N be coprime and let a, b ∈Z be given. Then the pair of linear congruences x ≡ a mod m x ≡ b mod n has a solution x ∈Z. Moreover, if x′ is any other solution then we have x′ ≡x mod mn.
Proof. Since n and m are coprime there exist a′, b′ ∈Z such that a′n ≡ 1 mod m and b′m ≡1 mod n (use Theorem 2.23 or Theorem 2.25). Define x := aa′n + bb′m. Then x ≡aa′n ≡a mod m and x ≡bb′m ≡b mod n.
Hence x is a solution to the given pair of linear congruences.
Suppose x′ ≡a mod m and x′ ≡b mod n. Then m | (x −x′) and n | (x −x′). Since m and n are coprime, it now follows from Proposition 2.12 that mn | (x′ −x). Hence x ≡x′ mod nm.
Remark 2.36. We have used that m and n are coprime twice in the above proof. This hypothesis is necessary because, for example, the pair of congru-ences x ≡2 mod 12, x ≡4 mod 20 has no solution.
Example 2.37. Solve the following system of congruences: x ≡ 2 mod 3, x ≡ 3 mod 7.
Note that 3 and 7 are indeed coprime because they are distinct primes.
Following the proof, we set a = 2, m = 3, b = 3, n = 7. We have mn = 3 × 7 = 21 and 7a′ ≡ 1 mod 3 = ⇒take a′ = 1, 3b′ ≡ 1 mod 7 = ⇒take b′ = 5.
(Note that in more complicated situations we can use the Extended Euclidean Algorithm to compute multiplicative inverses modulo m and n.) Therefore x = aa′n + bb′m = (2 × 1 × 7) + (3 × 5 × 3) = 14 + 45 = 59, and the smallest positive integer solution is 17 ≡59 mod 21 .
18 Theorem 2.38 (Chinese Remainder Theorem). Let n1, n2, . . . , nt ∈N with gcd(ni, nj) = 1 whenever i ̸= j, (i.e. the ni are “coprime in pairs”) and let a1, a2, . . . , at ∈Z be given. Then the system of congruences x ≡ a1 mod n1 .
.
.
x ≡ at mod nt has a solution x ∈Z.
Moreover, if x′ is any other solution, then x′ ≡ x mod N, where N := n1 · · · nt.
Proof. Define Ni := N/ni.
Then gcd(Ni, ni) = 1, since ni is coprime to all the factors of Ni. Hence by Theorem 2.23 (or by Theorem 2.25) there exists xi ∈Z such that Nixi ≡1 mod ni. Define x := Pt i=1 aiNixi. Thus x ≡akNkxk mod nk since nk | Ni for all i ̸= k. Therefore x ≡ak(Nkxk) ≡ ak mod nk for all k.
Suppose x′ ≡ak mod nk for all k. Then x′ ≡x mod nk for all k. Thus nk | (x′ −x) for all k.
Since the ni are pairwise coprime it now follows from Proposition 2.12 that N := n1n2 · · · nt | (x′ −x). This yields x′ ≡ x mod N.
Example 2.39. Solve the following system of congruences: x ≡ 2 mod 3, x ≡ 3 mod 5, x ≡ 2 mod 7.
Note that the 3, 5 & 7 are indeed coprime in pairs because they are distinct primes. Following the proof, we put N := 3×5×7 = 105, N1 = 35, N2 = 21, N3 = 15 and we have 35x1 ≡ 1 mod 3 = ⇒take x1 = 2, 21x2 ≡ 1 mod 5 = ⇒take x2 = 1, 15x3 ≡ 1 mod 7 = ⇒take x3 = 1.
(Note that in more complicated situations we can use the Extended Euclidean Algorithm to compute multiplicative inverses modulo n.) Therefore x = 2N1x1 +3N2x2 +2N3x3 = (2×35×2)+(3×21×1)+(2×15×1) = 233, and the smallest positive integer solution is 23 ≡233 mod 105 .
19 2.6 Euler ϕ-function Definition 2.40. For n ∈N, we define Euler’s totient function, or the ϕ-function, by ϕ(n) := #{a ∈N : 1 ≤a ≤n, gcd(a, n) = 1}.
Remark 2.41. ϕ(1) = 1 and for p prime, ϕ(p) = #{1, 2, 3, . . . , p−1} = p−1.
Remark 2.42. By Proposition 2.32 and the fact that {1, 2, . . . , n} is a com-plete residue system modulo n, we have that ϕ(n) = # (Z/nZ)×. Note that since gcd(0, n) = gcd(n, n) = n for n ∈N, we also have ϕ(n) = #{a ∈Z : 0 ≤a < n, gcd(a, n) = 1}.
Theorem 2.43. Let m, n ∈N be coprime. Then ϕ(mn) = ϕ(m)ϕ(n).
Proof. Let a ∈Z with 0 ≤a < mn and define b, c ∈Z by a ≡b mod m and a ≡c mod n, where 0 ≤b < m and 0 ≤c < n. The Chinese Remainder Theorem tells us that there is a bijective correspondence between choices of a and choices of pairs (b, c). We now show that gcd(a, mn) = 1 ⇔gcd(b, m) = gcd(c, n) = 1.
We shall use Proposition 2.32 (units of Z/nZ) several times.
Suppose gcd(a, mn) = 1. Then the congruence ax ≡1 mod mn has a solution r ∈Z, i.e, ar ≡1 mod mn. By Proposition 2.10 (i) we have ar ≡ 1 mod m since m | mn. Hence br ≡ar ≡1 mod m and so the congruence bx ≡1 mod m is soluble; thus gcd(b, m) = 1. Similarly, gcd(c, n) = 1.
Suppose conversely that gcd(b, m) = gcd(c, n) = 1.
The congruences bx ≡1 mod m and cy ≡1 mod n are soluble so there exist s, t ∈Z such that bs ≡1 mod m and ct ≡1 mod n. Since m and n are coprime, by the Chinese Remainder Theorem there exists r ∈Z such that r ≡s mod m and r ≡t mod n. Hence ar ≡bs ≡1 mod m and ar ≡ct ≡1 mod n and so x = ar is the solution to the simultaneous linear equations x ≡1 mod m and x ≡1 mod n.
By the Chinese Remainder Theorem ar ≡1 mod mn. Hence gcd(a, mn) = 1.
Therefore the number of integers a with 0 ≤a < mn which are coprime to mn, i.e. ϕ(mn), is equal to the number of pairs of integers (b, c) with 0 ≤b < m, gcd(b, m) = 1 and 0 ≤c < n, gcd(c, n) = 1, i.e., ϕ(m)ϕ(n).
Theorem 2.44. Let p be prime and let r ∈N. Then ϕ(pr) = pr −pr−1 = pr−1(p −1).
20 Proof. For all m ∈N, either gcd(pr, m) = 1 or p | m (but not both). Thus ϕ(pr) = #{m ∈N : m ≤pr, p ∤m} = #{m ∈N : m ≤pr} −#{m ∈N : m ≤pr, p | m} = pr −pr−1 = pr−1(p −1).
Examples 2.45. We can use these theorems to compute ϕ(n) as follows: • ϕ(10) = ϕ(2 × 5) = ϕ(2)ϕ(5) = (2 −1)(5 −1) = 1 × 4 = 4.
• ϕ(12) = ϕ(22 × 3) = ϕ(22)ϕ(3) = (22 −2)(3 −1) = 2 × 2 = 4.
• ϕ(100) = ϕ(22 × 52) = ϕ(22)ϕ(52) = (22 −2)(52 −5) = 2 × 20 = 40.
• ϕ(1001) = ϕ(11 × 91) = ϕ(11)ϕ(91) = 10ϕ(7 × 13) = 10ϕ(7)ϕ(13) = 10 × 6 × 12 = 720.
Proposition 2.46. Let n ∈N. By the Fundamental Theorem of Arithmetic, we may write n = pe1 1 · · · per r where the pi’s are distinct primes and each ei ∈N. Then ϕ(n) = r Y i=1 (pi −1)pei−1.
Proof. By Theorems 2.43 and 2.44 we have ϕ(n) = ϕ(pe1 1 · · · per r ) = r Y i=1 ϕ(pei i ) = r Y i=1 (pei i −pei−1 i ) = r Y i=1 (pi −1)pei−1 i .
Corollary 2.47. Let n ∈N. Then ϕ(n) = n Y p|n 1 −1 p where the product runs over all distinct prime divisors of n.
Proof. From the above we have ϕ(n) = r Y i=1 (pi −1)pei−1 i = r Y i=1 pei i (1 −p−1 i ) = n r Y i=1 (1 −p−1 i ) = n Y p|n 1 −1 p .
21 Proposition 2.48. For any n ∈N we have P d|n ϕ(d) = n.
Proof. We classify the elements of {1, 2, . . . , n} according their greatest com-mon divisor with n. Thus {a ∈N : a ≤n} = [ d|n {a ∈N : a ≤n, gcd(n, a) = d} (disjoint union).
Hence n = P d|n Rd where Rd := #{a ∈N : 1 ≤1a ≤n, gcd(n, a) = d}. If d | n then we can write n = dn′ with n ∈N and by the distributive law for gcd’s (Proposition 1.9 (iii)) we have gcd(n, a) = d if and only if a = da′ with gcd(n′, a′) = 1. Moreover, a ≤n if and only if a′ ≤n′. It follows that Rd = #{a′ ∈N : 1 ≤a′ ≤n′, gcd(n′, a′) = 1}, and hence Rd = ϕ(n′). Therefore n = P d|n ϕ( n d). However, when d | n we have n = d · n d; thus when d runs over the positive divisors of n, so does e = n d, and therefore we have n = P e|n ϕ(e).
Example 2.49. For n = 12 we have φ(1) + φ(2) + φ(3) + φ(4) + φ(6) + φ(12) = 1 + 1 + 2 + 2 + 2 + 4 = 12.
2.7 Exponentiation Example 2.50. What is 3k mod 19 as k varies?
k 0 1 2 3 4 5 6 7 8 9 10 3k mod 19 1 3 9 8 5 15 7 2 6 18 16 k 11 12 13 14 15 16 17 18 19 20 3k mod 19 10 11 14 4 12 17 13 1 3 9 Notice that the sequence repeats after a certain point. We can use the fact that 318 ≡1 mod 19 to simplify calculations. For example, by the Division Algorithm we have 100 = 5 × 18 + 10 so 3100 ≡(318)5310 ≡15310 ≡310 ≡16 mod 19.
Proposition 2.51. Let n ∈N and a ∈Z. There exists r ∈N such that ar ≡1 mod n if and only if gcd(a, n) = 1.
Proof. Suppose there exists r ∈N such that ar ≡1 mod n. Then ar−1 is a solution to ax ≡1 mod n and so gcd(a, n) = 1 by Proposition 2.32 (units of Z/nZ). Suppose conversely that gcd(a, n) = 1. There are only finitely 22 many possible values of ak mod n so there exist i, j ∈N with i < j such that ai ≡aj mod n. Since gcd(a, n) = 1 we may apply the cancellation law for congruences (Theorem 2.10) i times to obtain aj−i ≡1 mod n. Thus we may take r = j −i.
Definition 2.52. Let n ∈N, a ∈Z and suppose that gcd(a, n) = 1. Then the least d ∈N such that ad ≡1 mod n is called the order of a mod n, and written ordn(a).
Proposition 2.53. Let n ∈N, a ∈Z and suppose that gcd(a, n) = 1.
For r, s ∈Z we have ar ≡as mod n if and only if r ≡s mod ordn(a).
Proof. Let k = ordn(a). Then ak ≡1 mod n. Assume without loss of gener-ality that r > s. Suppose that r ≡s mod k. Then there exists t ∈N such that r = s + tk. Hence ar = asakt = as(ak)t ≡as mod n.
Suppose conversely that ar ≡as mod n. Since gcd(a, n) = 1 we may apply the cancellation law for congruences (Theorem 2.10) s times to obtain ar−s ≡ 1 mod n. By the Division Algorithm, there exist u, t ∈N ∪{0} such that r −s = tk + u where 0 ≤u < k. Then ar−s = au+tk = au(ak)t ≡au1t ≡au mod n and so au ≡1 mod n. But 0 ≤u < k and k is the least positive integer such that ak ≡1 mod n and so we must have u = 0. Therefore k | (r −s), i.e., r ≡s mod k.
Corollary 2.54. Let n ∈N, a ∈Z and suppose that gcd(a, n) = 1.
Let k ∈Z. Then ak ≡1 mod n if and only if ordn(a) | k.
Proof. Just take r = k and s = 0 in Proposition 2.53.
Corollary 2.55. Let n ∈N, a ∈Z and suppose that gcd(a, n) = 1. Then the numbers 1, a, a2, . . . , aordn(a)−1 are incongruent mod n.
Proof. Combine Propositions 2.14 and 2.53.
23 2.8 Euler-Fermat Theorem Definition 2.56. Let n ∈N. A subset R of Z is said to be a reduced residue system modulo n if (i) R contains ϕ(n) elements, (ii) no two elements of R are congruent modulo n, and (iii) for every r ∈R we have gcd(r, n) = 1.
Remark 2.57. If R is a reduced residue system modulo n then (Z/nZ)× = {[a]n : a ∈R}.
Proposition 2.58. Let n ∈N and let k ∈Z. If {a1, a2, . . . , aϕ(n)} is a re-duced residue system modulo n and gcd(k, n) = 1 then {ka1, ka2, . . . , kaϕ(n)} is also a reduced residue system modulo n.
Proof. If kai ≡kaj mod n then by the cancellation law for congruences (The-orem 2.11) we have ai ≡aj mod n since gcd(k, n) = 1. Therefore no two ele-ments in the set {ka1, . . . , kaϕ(n)} are congruent modulo n. Moreover, since gcd(ai, n) = gcd(k, n) = 1 we have gcd(kai, 1) = 1 (check this!) so each kai is coprime to n.
Theorem 2.59 (The Euler-Fermat Theorem). Let n ∈N, a ∈Z and suppose that gcd(a, n) = 1. Then aϕ(n) ≡1 mod n.
Proof. Let {b1, b2, . . . , bϕ(n)} be a reduced residue system modulo n. Then since gcd(a, n) = 1 the set {ab1, ab2, . . . , abϕ(n)} is also a reduced residue system by Proposition 2.58. Hence the product of all the integers in the first set is congruent modulo n to the product of those in the second set.
Therefore b1 · · · bϕ(n) ≡aϕ(m)b1 · · · bϕ(n) mod n.
Since each bi is coprime to n, we may apply the cancellation law for congru-ences (Theorem 2.11) ϕ(n) times to obtain the desired result.
Corollary 2.60. Let n ∈N, a ∈Z and suppose that gcd(a, n) = 1. Then ordn(a) | ϕ(n).
Proof. Combine the Euler-Fermat Theorem and Corollary 2.54.
Example 2.61. We have ϕ(12) = ϕ(22)ϕ(3) = 2 × 2 = 4. So for every a ∈Z with gcd(a, 12) = 1 we must have ord12(a) = 1, 2 or 4. In fact, ord12(1) = 1 and ord12(5) = ord12(7) = ord12(11) = 2, so working mod 12 there is no element element of order ϕ(12) = 4.
24 Corollary 2.62. Let p be prime and let a ∈Z such that p ∤a.
Then ap−1 ≡1 mod p.
Proof. This follows immediately from the Euler-Fermat Theorem (Theorem 2.59) since ϕ(p) = p −1.
Example 2.63. We saw in Example 2.50 that ord19(3) = 18 = ϕ(19). We also have ord19(8) = 6, which is a factor of 18.
Theorem 2.64 (Fermat’s Little Theorem). Let p be prime and let a ∈Z.
Then ap ≡a mod p.
Proof. If p ∤a this follows easily from Corollary 2.62. If p | a then both ap and a are congruent to 0 mod p.
Remark 2.65. Sometimes the result of Corollary 2.62 is also referred to as Fermat’s Little Theorem.
Remark 2.66. Many of the results in this section and the previous section can thought of in terms of group theory once we recall that (Z/nZ)× is an (abelian) group. For example, ordn(a) is the order of [a]n in (Z/nZ)×.
Moreover, Lagrange’s Theorem in group theory tells us that the order of an element divides the order of the group; so ordn(a) divides ϕ(n) = #(Z/nZ)× which gives the Euler-Fermat Theorem.
2.9 Binary powering algorithm We briefly illustrate this algorithm with an example.
Example 2.67. Suppose that we want to compute 3499 mod 997 efficiently.
Note that 997 is prime so Fermat’s Little Theorem tells us that 3996 ≡1 mod 997. Unfortunately, this doesn’t appear to help us. First we find the binary expansion of 499 as follows: 499 = 28 + 243 = 28 + 27 + 115 = 28 + 27 + 26 + 51 = 28 + 27 + 26 + 25 + 19 = 28 + 27 + 26 + 25 + 24 + 3 = 28 + 27 + 26 + 25 + 24 + 21 + 20.
25 So the binary expansion of 499 is 111110011. Now by squaring the previous term each time, we have 321 ≡9 (mod 997) 322 ≡92 ≡81 (mod 997) 323 ≡812 ≡6561 ≡579 ≡−418 (mod 997) 324 ≡(−418)2 ≡4182 ≡174724 ≡249 (mod 997) 325 ≡2492 ≡62001 ≡187 (mod 997) 326 ≡1872 ≡34969 ≡74 (mod 997) 327 ≡742 ≡5476 ≡491 (mod 997) 328 ≡4912 ≡804 ≡−193 (mod 997).
Therefore 3499 ≡320 × 321 × 324 × 325 × 326 × 327 × 328 (mod 997) ≡3 × 9 × 249 × 187 × 74 × 491 × (−193) (mod 997) ≡27 × 46563 × 36334 × (−193) (mod 997) ≡27 × 701 × 442 × (−193) (mod 997) ≡18927 × (−85306) (mod 997) ≡(−16) × 436 (mod 997) ≡−6976 (mod 997) ≡3 (mod 997).
Note that the advantage of this method is that it minimizes the number of multiplications we need to perform and that each integer we consider has at most twice the number of digits as the modulus.
2.10 Polynomial Congruences Theorem 2.68 (Lagrange’s polynomial congruence theorem). Let f(x) = a0 + a1x + · · · + adxd ∈Z[x] and let p be a prime with p ∤ad. Then f(x) ≡0 mod p has at most d solutions mod p.
Remark 2.69. More generally, any polynomial equation of degree d over a field has at most d solutions (note that Z/pZ = Fp is a field).
26 Proof. The proof is by induction on d. When d = 1 the congruence is linear: a1x + a0 ≡0 mod p.
Since a1 ̸≡0 mod p we have gcd(a1, p) = 1 and so there is exactly one solution by Theorem 2.23 (linear congruences with exactly one solution).
Assume that the theorem is true for polynomials of degree d−1. Suppose for a contradiction that the congruence f(x) ≡0 mod p has d+1 incongruent solutions modulo p, say x0, x1, x2, . . . , xd where f(xk) ≡0 mod p for each k = 0, 1, . . . , d. Recall that for any r ∈N we have the algebraic identity xr −yr = (x −y)(xr−1 + xr−2y + xr−3y2 + · · · + xyr−2 + yr−1).
Thus we also have an algebraic identity f(x) −f(x0) = d X r=1 ar(xr −xr 0) = d X i=1 ar(x −x0)gr(x) where each gr ∈Z[x] is of degree r −1 and has leading coefficient 1. Hence we have f(x) −f(x0) = (x −x0)g(x) where g(x) = Pd r=1 argr(x) is of degree d −1 and has leading coefficient ad. Thus f(xk) −f(x0) = (xk −x0)g(xk) ≡0 mod p, since f(xk) ≡f(x0) ≡0 mod p. But xk −x0 ̸≡0 mod p if k ̸= 0 so we must have g(xk) ≡0 mod p for each k ̸= 0 (we may apply the cancellation law for congruences (Theorem 2.11) because gcd(x −x0, p) = 1). But this means that the congruence g(x) ≡0 mod p has d incongruent solutions modulo p, contradicting the induction hypothesis.
Example 2.70. Note that x2 −1 ≡0 mod 8 has 4 roots, namely 1, 3, 5, 7 mod 8. This is not a counterexample to Theorem 2.68, however, because 8 is not prime (and Z/8Z is not a field).
Corollary 2.71. Let a ∈Z and let p be an odd prime. If a2 ≡1 mod p then a ≡±1 mod p.
Proof. Lagrange’s polynomial congruence theorem (Theorem 2.68) says that a2 ≡1 mod p has at most two solutions. But it is clear that a ≡±1 mod p are solutions and these must be distinct because p is odd. Therefore, we have found all the solutions.
27 Example 2.72. Let p and q be distinct odd primes. Consider the congruence x2 ≡1 mod pq.
It is clear that x ≡±1 mod pq are solutions, but are there other solutions?
By the Chinese Remainder Theorem we have x2 ≡1 mod pq ⇐ ⇒both x2 ≡1 mod p and x2 ≡1 mod q ⇐ ⇒both x ≡±1 mod p and x ≡±1 mod q.
Thus there are four solutions modulo pq. Note that x ≡1 mod pq ⇐ ⇒ x ≡1 mod p x ≡1 mod q and x ≡−1 mod pq ⇐ ⇒ x ≡−1 mod p x ≡−1 mod q which are the “easy” solutions we already mentioned. It remains to solve the two pairs of congruences x ≡1 mod p x ≡−1 mod q and x ≡−1 mod p x ≡1 mod q.
Note that we can use a trick here to save work: if x is the solution to one of these pairs of congruences then −x is the solution to the other congruence.
Consider the following explicit example. We wish to find all solutions to x2 ≡1 mod 145.
Thus it is clear that x ≡±1 mod 145 gives two solutions, but we also want to find the other two solutions. Note that 145 = 5 × 29 and that both 5 and 29 are prime. Thus we want to solve x ≡1 mod 5 x ≡−1 mod 29.
By the Extended Euclidean Algorithm, we have gcd(5, 29) = 1 = 6 × 5 −1 × 29 Thus using the construction of the Chinese Remainder Theorem we may take x ≡(−1) × 6 × 5 + 1 × (−1) × 29 ≡−59 mod 145.
Check that this really is a solution: (−59)2 = 3481 = 1 + 24 × 145 ≡1 mod 145.
Therefore the solutions of x2 ≡1 mod 145 are x ≡±1, ±59 mod 145.
28 2.11 Hensel Lifting The Chinese Remainder Theorem shows that the problem of solving a poly-nomial congruence f(x) ≡0 mod n can be reduced to solving a system of congruences f(x) ≡0 mod pei i (i = 1, . . . , r) where n = pe1 1 · · · per r is the prime factorisation of n. We show that this can be further reduced to congruences with prime moduli together with a set of linear congruences.
Theorem 2.73 (Hensel’s Lemma). Let p be prime. Let f(x) ∈Z[x] and let f ′(x) ∈Z[x] be its formal derivative. If a ∈Z satisfies f(a) ≡0 mod p, f ′(a) ̸≡0 mod p then for each n ∈N there exists an ∈Z such that f(an) ≡0 mod pn and an ≡a mod p.
(12) Moreover, an is unique modulo pn.
Example 2.74. Suppose we want to solve x2 ≡−1 mod 54. This is the same as solving the equation f(x) ≡0 mod 54 where f(x) = x2 + 1. Note that f ′(x) = 2x. An exhaustive search shows that x = ±2 are the solutions to f(x) ≡0 mod 5. Choose a = 2. Then f(a) ≡0 mod 5 and f ′(a) = 2 × 2 = 4 ̸≡0 mod 5. Thus we may apply Hensel’s Lemma. Write a2 = 2+5t1. Then we have f(2 + 5t1) ≡0 mod 52 ⇐ ⇒(2 + 5t1)2 + 1 ≡0 mod 52 ⇐ ⇒4 + 20t1 + 25t2 1 + 1 ≡0 mod 52 ⇐ ⇒5 + 20t1 ≡0 mod 52 ⇐ ⇒1 + 4t1 ≡0 mod 5 ⇐ ⇒4t1 ≡−1 mod 5 ⇐ ⇒t1 ≡1 mod 5.
Thus we may take a2 = 2+5×1 = 7. Check: 72 = 49 ≡−1 mod 25. We could now set a3 = 7+52t2 and find t2 by solving the congruence f(a3) ≡0 mod 53.
29 But instead we can take a short-cut as follows. Write a4 = 7 + 52t3 and try to solve mod 54 directly. Then we have f(7 + 52t3) ≡0 mod 54 ⇐ ⇒(7 + 52t3)2 + 1 ≡0 mod 54 ⇐ ⇒49 + (14 × 25)t3 + 54t2 3 + 1 ≡0 mod 54 ⇐ ⇒50 + (14 × 25)t3 ≡0 mod 54 ⇐ ⇒2 + 14t3 ≡0 mod 52 ⇐ ⇒1 + 7t3 ≡0 mod 52 ⇐ ⇒7t3 ≡−1 mod 52 ⇐ ⇒t3 ≡7 mod 52.
So we have a4 = 7+52 ×7 = 182. Check: 1822 = 33, 124 ≡−1 mod 54. Note that if we had started with the solution a = −2 then we would have ended up with the solution a4 = −182.
Remark 2.75. Even if the hypotheses of Hensel’s Lemma are not satisfied, we can still try to use the same technique to solve the given polynomial equation. However, in this case, the solutions are not guaranteed to exist or to be unique.
Lemma 2.76. Let f ∈Z[X] and let f ′(X) be its formal derivative. Then there exists g ∈Z[X, Y ] satisfying the polynomial identity f(X + Y ) = f(X) + f ′(X)Y + g(X, Y )Y 2.
Proof. This formula comes from isolating the first two terms in the binomial theorem. Writing f(X) = Pd i=0 ciXi we have f(X + Y ) = d X i=0 ci(X + Y )i = c0 + d X i=1 ci(Xi + iXi−1Y + gi(X, Y )Y 2) where gi(X, Y ) ∈Z[X, Y ]. Thus f(X + Y ) = d X i=0 ciXi + d X i=1 iciXi−1Y + d X i=1 cigi(X, Y )Y 2 = f(X) + f ′(X)Y + g(X, Y )Y 2 where g(X, Y ) := Pd i=1 cigi(X, Y ). This gives the desired identity.
Remark 2.77. The identity of Lemma 2.76 is similar to Taylor’s formula: f(x + h) = f(x) + f ′(x)h + (f ′′(x)/2!)h2 + · · · .
30 The problem is that the terms Taylor’s formula have factorials in the denom-inator, which can cause problems when reducing modulo powers of p: think about f ′′(x)/2! mod 2, for example.
Proof of Hensel’s Lemma. We will prove by induction that for each n ∈N there exists a an ∈Z satisfying (12) that is unique mod pn. The case n = 1 is trivial using a1 = a. We now suppose the inductive hypothesis holds for n = k and show it holds for n = k + 1. The idea is to consider ak + pktk and see if tk ∈Z can be chosen in such a way that ak + pktk satisfies the required properties of ak+1.
By Lemma 2.76 with X = ak and Y = pktk there exists zk ∈Z such that f(ak + pktk) = f(ak) + f ′(ak)pktk + zkp2kt2 k ≡f(ak) + f ′(ak)pktk mod pk+1 where the congruence follows since k + 1 ≤2k. In f ′(ak)pktk mod pk+1 the factors f ′(ak) and tk only matter mod p since it already contains a factor of pk and the modulus is pk+1. Thus recalling that ak ≡a mod p we have f ′(a)pktk ≡f ′(ak)pktk mod pk+1. Therefore we have f(ak + pktk) ≡0 mod pk+1 ⇐ ⇒f(ak) + f ′(a)pktk ≡0 mod pk+1 (13) ⇐ ⇒f ′(a)tk ≡−f(ak)/pk mod p, where the ratio −f(ak)/pk is in Z since we have f(ak) ≡0 mod pk by the induction hypothesis, and the last equivalence follows from Proposition 2.10.
But f ′(a) ̸≡0 mod p so gcd(f ′(a), p) = 1 and thus by Theorem 2.23 (linear congruences with exactly one solution) the last congruence (mod p) has a solution tk, which is unique mod p.
We set ak+1 = ak + pktk.
Then we have f(ak+1) ≡0 mod pk+1 and ak+1 ≡ak mod pk, so in particular ak+1 ≡a mod p.
It remains to show uniqueness. Suppose there exists bk+1 ∈Z with f(bk+1) ≡0 mod pk+1 and bk+1 ≡a mod p.
Then we also have f(bk+1) ≡0 mod pk and so by the uniqueness of ak we must have bk+1 ≡ak mod pk. Thus bk+1 = ak + pksk for some sk ∈Z. But (13) and the proceeding discussion shows that sk ≡ tk mod p and thus we must have ak+1 ≡bk+1 mod pk+1, as desired.
Remark 2.78. An adaptation of the above proof shows that under the as-sumptions of Hensel’s Lemma, in principle one can always lift from a solution mod pk to a solution mod p2k. Moreover, for m ≥n ≥1 we always have am ≡an mod pn.
31 2.12 Primitive Roots Recall Corollary 2.60: if n ∈N , a ∈Z with gcd(a, n) = 1 then ordn(a) | ϕ(n).
In this section, we shall be interested in the case that ordn(a) = ϕ(n).
Definition 2.79. Let n ∈N. We say that a ∈Z is a primitive root mod n if and only if gcd(a, n) = 1 and ordn(a) = ϕ(n).
Remark 2.80. This is equivalent to requiring [a]n to be a generator for the abelian group (Z/nZ)×, which must therefore be cyclic.
Example 2.81. Let n = 5 and abbreviate [x]n = [x]5 to [x]. Then we have 0 = , 1 = , 2 = , 3 = = , 4 = = .
Therefore ord5(2) = 4 = ϕ(5) and so 2 is a primitive root of 5.
Remark 2.82. For some values of n there are no primitive roots. For example, every non-trivial element of (Z/8Z)× = {8, 8, 8, 8} has order 2, and so (Z/8Z)× is not cyclic.
Example 2.61 shows that the same is true for (Z/12Z)×.
Lemma 2.83. Let n ∈N and a ∈Z with gcd(a, n) = 1. Then for k ∈Z we have ordn(ak) = ordn(a) gcd(ordn(a), k).
In particular, ordn(ak) = ordn(a) if and only if gcd(ordn(a), k) = 1.
Proof. Let f = ordn(a). The integer ordn(ak) is the least d ∈N such that adk ≡1 mod n. By Corollary 2.54 this is also the least d ∈N such that dk ≡0 mod f. But by the cancellation law for congruences (Theorem 2.11) this last congruence is equivalent to the congruence d ≡0 mod f h where h = gcd(f, k). But it is clear that the least positive solution to this congruence is d = f h and so ordn(ak) = f h, as asserted.
Example 2.84. We saw in Example 2.50 that 3 is a primitive root mod 19, i.e., ord19(3) = ϕ(19) = 18. Thus ord19(33) = ord19(8) = 18/ gcd(18, 3) = 18/3 = 6.
Theorem 2.85. Let p be prime and let d ∈N be a divisor of p −1. Then there are exactly ϕ(d) elements a mod p such that ordp(a) = d. In particular, there are ϕ(p −1) primitive roots modulo p.
32 Proof. Fix a prime p and for any d ∈N such that d | (p −1) define A(d) = {a ∈N : 1 ≤a ≤p −1, ordp(a) = d}.
Let ψ(d) = #A(d) ≥0. We aim to show that ψ(d) = ϕ(d).
Since the sets A(d) partition {1, 2, . . . , p −1} we have X d|(p−1) ψ(d) = p −1.
By Proposition 2.48 we also have X d|(p−1) ϕ(d) = p −1.
Therefore if we can show that ψ(d) ≤ϕ(d) for all d | (p−1) then ψ(d) = ϕ(d) for all such d. (Otherwise, if ψ(d0) < ϕ(d) for some d0, then P d|(p−1) ψ(d) < P d|(p−1) φ(d) - contradiction.) If ψ(d) = 0 then ψ(d) ≤ϕ(d) and so we are done. So we are reduced to considering the case ψ(d) ≥1. Then A(d) ̸= ∅and so a ∈A(d) for some a.
Hence ordp(a) = d and so ad ≡1 mod p. Then (ai)d ≡1 mod p for all i ∈Z.
In particular, the d numbers a, a2, . . . , ad (14) are solutions of the polynomial congruence xd −1 ≡0 mod p.
(15) By Corollary 2.55 the numbers listed in (14) are incongruent mod p since ordp(a) = d. Moreover, (15) has at most d solutions by Lagrange’s polyno-mial congruence theorem (Theorem 2.68). Therefore the d numbers in (14) must be all the solutions of (15) mod p. Hence each number in A(d) must be congruent to ak mod p for some k = 1, . . . , d. By Lemma 2.83 ordp(ak) = d if and only if gcd(k, d) = 1. In other words, among the d numbers in (14) there are ϕ(d) which have order d modulo p. Thus we have shown that ψ(d) = ϕ(d) if ψ(d) ̸= 0, as required.
Example 2.86. There are ϕ(19 −1) = ϕ(18) = 6 primitive roots mod 19.
Thus there are ϕ(19) −6 = 12 elements of (Z/19Z)× that are not primitive roots.
Corollary 2.87. Let p be prime. Then there exists a primitive root g mod-ulo p (note that g need not be unique). In other words, (Z/pZ)× is cyclic.
Moreover, for any a ∈Z with p ∤a there exists k ∈Z such that a ≡gk mod p.
33 Proof. The existence of a primitive root follows from Theorem 2.85 since ϕ(p −1) ≥1.
By definition of primitive root, ordp(g) = p −1 and so 1, g, g2, . . . , gp−2 are congruent modulo p, in some order, to 1, 2, . . . , p −1 (use Corollary 2.55), which gives the last claim.
Theorem 2.88 (The primitive root test). Let n ∈N and let a ∈Z with gcd(a, n) = 1. Then a is a primitive root modulo n if and only if aϕ(n)/q ̸≡1 mod n for every prime q dividing ϕ(n).
Proof. If aϕ(n)/q ≡1 mod n for some prime q dividing ϕ(n) then ordn(a) ≤ ϕ(n)/q < ϕ(n) and so a cannot be a primitive root mod n.
Suppose conversely that aϕ(n)/q ̸≡1 mod n for every prime q dividing ϕ(n). Write ϕ(n) = qr1 1 · · · qrs s where the qi’s are distinct primes and each ri ∈ N. Let m = ordn(a). Then m | ϕ(n) by Corollary 2.60 and so m = qt1 1 · · · qts s where for each i we have 0 ≤ti ≤ri. Suppose m < ϕ(n). Then there exists a j such that tj < rj. Hence m divides qr1 1 · · · q rj−1 j · · · qrs s = (ϕ(n)/qj). But am ≡1 mod n and so aϕ(n)/qj ≡1 mod n, contradicting our hypothesis.
Example 2.89. Find a primitive root modulo 31. Since 31 is prime, we have ϕ(31) = 31 −1 = 30 = 2 × 3 × 5. Thus given a ∈Z with 31 ∤a we need to check that a15 ̸≡1 mod 31, a10 ̸≡1 mod 31, and a6 ̸≡1 mod 31 Try a = 2. Then 210 ≡(25)2 ≡322 ≡12 ≡1 mod 31. Thus 2 is not a primitive root mod31.
Try a = 3. First note that 35 = 243 ≡−5 mod 31. Then we have • 36 = 35 × 3 ≡−5 × 3 ≡−15 ≡16 ̸≡1 mod 31.
• 310 = (35)2 ≡(−5)2 ≡25 ̸≡1 mod 31.
• 315 = 35 × 310 ≡−5 × 25 ≡−125 ≡−1 ̸≡1 mod 31.
Therefore 3 is a primitive root modulo 31.
Theorem 2.90. Let p be an odd prime. If g is a primitive root mod p then g is also a primitive root mod pe for all e ≥1 if and only if gp−1 ̸≡1 mod p2.
Proof. Not examinable (but statement is examinable). See Apostal, Intro-duction to Analytic Number Theory, Chapter 10, for example.
Theorem 2.91. Let n ∈N. Then (Z/nZ)× is cyclic ⇔there exists a prim-itive root modulo n ⇔n = 1, 2, 4, pe, 2pe where e ∈N and p is an odd prime.
Proof. Not examinable (but statement is examinable). See Apostal, Intro-duction to Analytic Number Theory, Chapter 10, for example.
34 2.13 Wilson’s Theorem Theorem 2.92 (Wilson’s Theorem). An integer p ≥2 is prime if and only if (p −1)! ≡−1 mod p.
Example 2.93. For p = 5, we have (5 −1)! = 4! = 24 ≡−1 mod 5; but for p = 6, we have (6 −1)! = 5! = 120 ≡0 mod 6.
Proof. Suppose n is composite. Then there exists d dividing n with 1 < d < n. Therefore d | (n −1)! and d | n. So if (n −1)! ≡−1 mod n then n | ((n−1)!+1) and so d | ((n−1)!+1). Hence d | 1 = ((n−1)!+1)−(n−1)!.
Contradiction. Hence (n −1)! ̸≡−1 mod n.
Suppose p is prime. The case p = 2 is easy, so we can and do assume that p is odd. Each a in {1, 2, . . . , p −1} is coprime to p and therefore has a unique inverse a−1 ∈{1, 2, . . . , p −1} modulo p, that is aa−1 ≡1 mod p.
Note that (a−1)−1 ≡a mod p. If a = a−1 then 1 ≡aa−1 = a2 mod p. By Corollary 2.71 we then have a ≡±1 mod p and so a = 1 or a = p −1. In the product (p −1)! = 1 × 2 × 3 × · · · × (p −2) × (p −1) we pair offeach term, save for 1 and p −1, with its inverse modulo p. We thus have (p −1)! ≡1 × (p −1) ≡−1 mod p.
Example 2.94. As an illustration, consider the case p = 11. Then 10!
= 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 × 10 = 1 × (2 × 6) × (3 × 4) × (5 × 9) × (7 × 8) × 10 ≡ 1 × 1 × 1 × 1 × 1 × 10 = 10 ≡−1 mod 11.
Alternative proof of Wilson’s Theorem using primitive roots. If n is compos-ite we proceed as before. Again, we are reduced to considering the case where p is an odd prime. Let g be a primitive root modulo p (this exists by Corol-lary 2.87). Then the numbers 1, g, g2, . . . , gp−2 are congruent modulo p, in some order, to 1, 2, . . . , p −1. Hence (p −1)! ≡1gg2 · · · gp−2 = g1+2+···+(p−2) mod p.
The sum 1 + 2 + · · · + (p −2) is the sum of an arithmetic progression with p −2 terms, and so equals (p −2)(p −2) + 1 2 = (p −2)(p −1) 2 .
Hence (p −1)! ≡g(p−2)(p−1)/2 mod p.
35 Since p is odd we have p = 2k + 1 for some k ∈N. As k < 2k = p −1 then gk ̸≡1 (mod p) but g2k = gp−1 ≡1 (mod p) because ordp(g) = p −1 by definition of g (or use Fermat’s little theorem). Since (gk)2 = g2k ≡1 mod p we have gk ≡±1 mod p by Corollary 2.71. Hence gk ≡−1 mod p. We finally conclude that (p −1)! ≡g(p−2)(p−1)/2 = g(2k−1)k = (gk)2k−1 ≡(−1)2k−1 = −1 mod p.
3 Quadratic Residues 3.1 Quadratic Residues We shall study the theory of quadratic congruences modulo an odd prime p.
By the familiar technique of completing the square one can reduce any such congruence to the form x2 ≡a mod p.
Example 3.1. Consider the case p = 11.
x 0 1 2 3 4 5 6 7 8 9 10 x2 mod 11 0 1 4 9 5 3 3 5 9 4 1 Notice the symmetry in this table. This is because for any odd prime p and any k ∈Z we have (p−k)2 ≡(−k)2 ≡k2 mod p. For example, 32 ≡(−3)2 ≡ (11 −3)2 ≡82 mod 11. Also notice that x2 ≡a mod 11 has one solution if a ≡0 mod 11, two solutions if a ≡1, 3, 4, 5, 9 mod 11, no solutions if a ≡2, 6, 7, 8, 10 mod 11.
Lemma 3.2. Let p be an odd prime and let a ∈Z. Consider x2 ≡a mod p.
(16) If p | a then (16) is equivalent to x ≡0 mod p. Otherwise, if p ∤a and (16) has a solution x ≡b mod p then p ∤b and x ≡−b is another, different solution.
Proof. If x ≡0 mod p then clearly x2 ≡0 mod p. The converse follows from Euclid’s Lemma for Primes (Theorem 1.23): if x2 ≡0 mod p then p | x2 so we must have p | x, i.e., x ≡0 mod p.
Suppose that p ∤a and b2 ≡a mod p. Then clearly (−b)2 ≡b2 ≡a mod p.
If b ≡−b mod p then 2b ≡0 mod p so b ≡0 mod p by the Cancellation Law for Congruences (Theorem 2.11) since p is odd. But then a ≡b2 ≡0 mod p, contradicting the assumption that p ∤a.
36 Definition 3.3. Let p be an odd prime, and suppose we have a ∈Z such that p ∤a. Then a is a Quadratic Residue of p if there exists x ∈Z such that x2 ≡a mod p, and a is Quadratic Non-Residue if not.
Proposition 3.4. Let p be an odd prime. Then every reduced residue system mod p contains exactly (p −1)/2 quadratic residues and (p −1)/2 quadratic non-residues mod p.
The quadratic residues belong to the residue classes containing the numbers 12, 22, 32, . . . , ((p −1)/2)2.
(17) Proof. First we show that the list of numbers in (17) are distinct mod p.
Indeed, if x2 ≡y2 mod p with 1 ≤x ≤(p −1)/2 and 1 ≤y ≤(p −1)/2 then (x −y)(x + y) ≡0 mod p.
But 1 < (x + y) < p so (x + y) is coprime to p. So by the Cancellation Law for Congruences (Theorem 2.11) we must have (x−y) ≡0 mod p, hence x ≡y mod p and so x = y (by Proposition 2.14). The remaining squares are ((p + 1)/2)2, ((p + 3)/2)2, . . . , (p −2)2, (p −1)2.
Since (p −k)2 ≡(−k)2 ≡k2 mod p for every k ∈Z with 1 ≤k ≤(p −1)/2, these are congruent to ((p −1)/2)2, ((p −3)/2)2, . . . , 22, 12.
These are precisely the numbers in (17). Hence there are precisely (p −1)/2 quadratic residues mod p, and so there are (p −1) −(p −1)/2 = (p −1)/2 quadratic non-residues mod p.
3.2 The Legendre Symbol Definition 3.5. Let p be an odd prime.
For any a ∈Z, we define the Legendre Symbol to be a p = +1, p ∤a and a is a quadratic residue of p, −1, p ∤a and a is a quadratic non-residue of p, 0, p | a.
Remark 3.6. By Lemma 3.2 we see that the congruence x2 ≡a mod p has precisely a p + 1 distinct solutions modulo p.
Remark 3.7. Note that we always have 1 p = 1. Moreover, if a, b ∈Z with a ≡b mod p then a p = b p (this is sometimes known as periodicity).
Examples 3.8.
5 11 = 1, 7 11 = −1, 22 11 = 0.
If m ∈Z with p ∤m then m2 p = 1.
37 3.3 Euler’s Criterion Lemma 3.9. Let p be an odd prime and let g be a primitive root mod p. Let a ∈Z with p ∤a. Then a ≡gk mod p for some k ∈Z and a is a quadratic residue mod p if and only if k is even.
Proof. First note that a primitive root g mod p exists by Corollary 2.87, so a ≡gk mod p for some k ∈Z. Suppose k is even. Then k = 2j for some j ∈Z and so a ≡(gj)2 mod p. Thus a is a quadratic residue mod p. Suppose conversely that a is quadratic residue mod p. Then a ≡b2 mod p for some b ∈Z with p ∤b. Then b ≡gr for some r ∈Z and so gk ≡(gr)2 ≡g2r mod p.
Thus k ≡2r mod p −1 by Proposition 2.53 since ordp(g) = ϕ(p) = p −1. So k ≡2r mod 2 since 2 | (p −1). Hence k ≡0 mod 2, i.e., k must be even.
Theorem 3.10 (Euler’s Criterion). If p is an odd prime and a ∈Z then a p ≡a p−1 2 mod p.
Proof. This is obvious if p | a. So suppose that p ∤a. Let g be a primitive root mod p. Then there exists k ∈Z such that a ≡gk mod p. Since ordp(g) = p−1 we have gp−1 ≡1 mod p and g(p−1)/2 ̸≡1 mod p. But Corollary 2.71 says that g(p−1)/2 ≡±1 mod p. Therefore g(p−1)/2 ≡−1 mod p. Then a(p−1)/2 ≡(gk)(p−1)/2 ≡(g(p−1)/2)k ≡(−1)k mod p.
The result now follows from Lemma 3.9.
Alternative proof of Euler’s Criterion. Again, we may suppose that p ∤a.
Suppose that a p = 1.
Then there exists b ∈Z with p ∤b such that a ≡b2 mod p. Thus by Fermat’s Little Theorem (Corollary 2.62) we have a(p−1)/2 ≡(b2)(p−1)/2 ≡bp−1 ≡1 ≡ a p mod p.
Now suppose that a p = −1 and consider the polynomial f(x) = x(p−1)/2 −1.
Since f(x) has degree (p −1)/2, Lagrange’s polynomial congruence theorem (Theorem 2.68) says that the congruence f(x) ≡0 mod p 38 has at most (p −1)/2 solutions.
But we have shown that the quadratic residues mod p are solutions, and Proposition 3.4 says there are (p −1)/2 of them.
Hence none of the quadratic non residues are solutions and so a(p−1)/2 ̸≡1 mod p. But by Fermat’s Little Theorem (Corollary 2.62) we have a(p−1) ≡1 mod p and so by Corollary 2.71 a(p−1)/2 ≡±1 mod p. Therefore a(p−1)/2 ≡−1 ≡ a p mod p.
This completes the proof.
Theorem 3.11 (Multiplicativity of the Legendre Symbol). Let p be an odd prime. Let a, b ∈Z. Then ab p = a p b p .
Proof. If p | a or p | b then p | ab so ab p = 0 and either a p = 0 or b p = 0.
Hence we have the desired result in this case.
Suppose p ∤a and p ∤b. Then by Euclid’s Lemma for Primes we have p ∤ab. Moreover, by Euler’s Criterion we have ab p ≡(ab)(p−1)/2 = a(p−1)/2b(p−1)/2 ≡ a p b p mod p, and both sides are 1 or −1. If they were different, we would have +1 ≡ −1 mod p and so p | 2, which gives a contradiction as p is odd.
Theorem 3.12. If p is an odd prime then −1 p = (−1) p−1 2 = +1, p ≡1 mod 4, −1, p ≡3 mod 4.
In other words, x2 ≡−1 mod p is soluble if and only if p ≡1 mod 4.
Proof. By Euler’s Criterion we have −1 p ≡(−1) p−1 2 mod p, and both sides are +1 or −1. If they were different, we would have +1 ≡ −1 mod p and so p | 2, which gives a contradiction as p is odd.
39 Example 3.13. Can we solve x2 ≡13 mod 17?
13 17 = −4 17 by periodicity (Remark 3.7) = −1 17 2 17 2 17 by multiplicativity (Theorem 3.11) = −1 17 as (±1)2 = 1 = (−1)(17−1)/2 since 17 ≡1 mod 4 (use Theorem 3.12) = (−1)8 = 1 Hence the congruence is soluble! Note that this proof that a solution exists cannot be adapted to provide a concrete solution. It is purely an existence argument.
Theorem 3.14. There are infinitely many primes p with p ≡1 mod 4.
Proof. It suffices to show that for any N ∈N there exists a prime p with p > N and p ≡1 mod 4. Let M = (2(N!))2 + 1. If p is a prime with p ≤N then M ≡1 mod p and so p ∤M. Let p be a prime factor of M. Then p > N.
As M is odd, p is also odd. Then we have (2(N!))2 ≡−1 mod p and so the congruence x2 ≡−1 mod p is soluble. Therefore p ≡1 mod 4 by Theorem 3.12.
3.4 Gauss’ Lemma Definition 3.15. Let a ∈Z and n ∈N. We write λ(a, n) for the unique integer such that a ≡λ(a, n) mod n and 0 ≤λ(a, n) < n. In other words, λ(a, n) is the remainder when the Division Algorithm is applied to a and n.
(This is not a standard notation, and is intended merely for temporary use in our discussion of quadratic residues.) Theorem 3.16 (Gauss’ Lemma). Let p be an odd prime and let a ∈Z with p ∤a. Then a p = (−1)Λ where Λ := #{j ∈N : 1 ≤j ≤p−1 2 , λ(aj, p) > p 2}.
Example 3.17. Let p = 13 and a = 5.
If j = 1 then λ(aj, p) = λ(5, 13) = 5 < 13/2.
If j = 2 then λ(aj, p) = λ(10, 13) = 10 > 13/2.
If j = 3 then λ(aj, p) = λ(15, 13) = 2 < 13/2.
40 If j = 4 then λ(aj, p) = λ(20, 13) = 7 > 13/2.
If j = 5 then λ(aj, p) = λ(25, 13) = 12 > 13/2.
If j = 6 then λ(aj, p) = λ(30, 13) = 4 < 13/2.
Hence Λ = #{2, 4, 5} = 3 and so 5 13 = (−1)3 = −1.
Proof. Let Sa := {aj : 1 ≤j ≤p−1 2 } and define {r1, . . . , rm} = {λ(aj, p) : aj ∈Sa, 0 < λ(aj, p) < p 2}, {s1, . . . , sn} = {λ(aj, p) : aj ∈Sa, p 2 < λ(aj, p) < p}, so that n = Λ. Note that λ(aj, p) ̸= p 2 since p 2 ̸∈Z and that λ(aj, p) ̸= 0, since p ∤a and p ∤j. Also note that if j1 ̸= j2 then λ(aj1, p) ̸= λ(aj2, p) since λ(aj1, p) = λ(aj2, p) = ⇒ aj1 ≡aj2 mod p = ⇒ j1 ≡j2 mod p (by cancellation law; note p ∤a) = ⇒ j1 = j2 (since 0 < j1, j2 < p).
Hence m + n = #Sa = p−1 2 . We claim that {r1, . . . , rm, (p −s1), . . . , (p −sn)} = {1, 2, . . . , p−1 2 }.
Clearly ri, (p −sj) ∈{1, 2, . . . , p−1 2 } and there are p−1 2 elements ri, (p −sj), so it suffices to show that they are all different. We have already shown that ri ̸= rj and si ̸= sj for i ̸= j. To show that ri ̸= p −sj we argue by contradiction. If ri + sj = p, let ri = λ(aj1, p) and sj = λ(aj2, p). Then ri + sj = p = λ(aj1, p) + λ(aj2, p) ≡aj1 + aj2 ≡a(j1 + j2) mod p.
Hence a(j1 + j2) ≡0 mod p. So by Euclid’s lemma for primes, either p | a or p | (j1 + j2). However, p ∤a by assumption and 2 ≤j1 + j2 ≤p −1 so that p ∤(j1 + j2) - contradiction. Therefore ri ̸= p −sj, which proves the claim.
Now, on the one hand, we have r1r2 · · · rm(p −s1) · · · (p −sn) = 1 × 2 × · · · × p−1 2 = p−1 2 !
≡ r1r2 · · · rms1s2 · · · sn(−1)n mod p.
On the other hand, by the definition of ri, sj, r1r2 · · · rms1s2 · · · sn = p−1 2 Y j=1 λ(aj, p) ≡ p−1 2 Y j=1 (aj) = a p−1 2 p−1 2 ! mod p.
Therefore p−1 2 ! ≡(−1)na p−1 2 p−1 2 ! mod p.
41 Now, since p ∤ p−1 2 !, the cancellation law for congruences shows that 1 ≡(−1)na p−1 2 mod p.
Thus a p−1 2 ≡(−1)n mod p and so a p ≡(−1)n mod p by Euler’s Criterion (Theorem 3.10). Both sides are +1 or −1 and if they were different, we would have +1 ≡−1 mod p and so p | 2, which gives a contradiction as p is odd.
Therefore a p = (−1)n = (−1)Λ as required.
Definition 3.18. For any x ∈R we set ⌊x⌋:= max{n ∈Z : n ≤x}. For example, ⌊3⌋= 3, ⌊π⌋= 3 and ⌊−π⌋= −4.
Corollary 3.19. If p is an odd prime then 2 p = (−1)(p2−1)/8 = +1, p ≡±1 mod 8, −1, p ≡±3 mod 8.
Proof. We shall apply Gauss’ Lemma (Theorem 3.16) for a = 2, so that 2 p = (−1)Λ where Λ = #{1 ≤j ≤p−1 2 : λ(2j, p) > p 2}.
Note that for 1 ≤j ≤p−1 2 we have 2 ≤2j ≤p −1 and so λ(2j, p) = 2j.
Moreover, 2j < p 2 if and only if j < p 4, and p 2 < 2j < p if and only if p 4 < j < p 2. It follows that Λ = #{j ∈N : p 4 < j < p 2}. We have # j : p 4 < j < p 2 = # j ≤p−1 2 −# j < p 4 = p−1 2 − p 4 .
Since p is odd, precisely one of the following cases must occur: (i) p = 8k + 1 = ⇒ p−1 2 = 4k, p 4 = 2k = ⇒ Λ = 2k, (ii) p = 8k + 3 = ⇒ p−1 2 = 4k + 1, p 4 = 2k = ⇒ Λ = 2k + 1, (iii) p = 8k + 5 = ⇒ p−1 2 = 4k + 2, p 4 = 2k + 1 = ⇒ Λ = 2k + 1, (iv) p = 8k + 7 = ⇒ p−1 2 = 4k + 3, p 4 = 2k + 1 = ⇒ Λ = 2k + 2.
Hence (−1)Λ = +1 ⇐ ⇒p = 8k + 1 or 8k + 7 ⇐ ⇒p ≡±1 mod 8.
We note that if p = 8k + r then p2 −1 8 = r2 + 16rk + 64k2 −1 8 = r2 −1 8 + 2(kr + 4k2) ≡r2 −1 8 mod 2.
By checking the cases r = ±1, ±3 we deduce that p2 −1 8 ≡ 0 mod 2, p ≡±1 mod 8, 1 mod 2, p ≡±3 mod 8, and the result follows.
42 Example 3.20. Since 1009 ≡1 mod 8 we have 2 1009 = 1. Since 1997 ≡ −3 mod 8 we have 2 1997 = −1. (Note that 1009 and 1997 are both prime.) Theorem 3.21. There are infinitely many primes p with p ≡−1 mod 8.
Proof. It suffices to show that for any N ∈N there exists a prime p with p > N and p ≡−1 mod 8. Let M = 8(N!)2 −1. If p is a prime with p ≤N then M ≡−1 mod p and so p ∤M.
Let p be a prime factor of M. Then p is odd and p > N. Moreover, (4(N!))2 ≡16(N!)2 ≡2M + 2 ≡2 mod p.
Thus 2 p = +1 and so p ≡±1 mod 8 by Corollary 3.19. But if all prime factors of M were congruent to 1 mod 8, then we would have M ≡1 mod 8, which is not the case. Therefore M must have at least one prime factor p with p ≡−1 mod 8 and p > N.
Lemma 3.22. Let p be an odd prime and let a ∈Z with a odd and p ∤a.
Then a p = (−1)t where t = (p−1)/2 X k=1 ⌊ak/p⌋.
Proof. Recall the notation from the proof of Gauss’ Lemma (Theorem 3.16).
For any j ∈Z we have λ(aj, p) ≡aj mod p, with 0 ≤λ(aj, p) < p. Here λ(aj, p) = aj −pk for some k ∈Z such that 0 ≤aj −pk < p. It follows that k ≤aj p < k + 1, and hence that k = j aj p k . We therefore deduce that λ(aj, p) = aj −p j aj p k . Using this expression we now have m X i=1 ri + n X i=1 si = (p−1)/2 X j=1 λ(aj, p) = (p−1)/2 X j=1 aj −p aj p .
Hence, since a and p are odd, we have (p−1)/2 X j=1 j − (p−1)/2 X j=1 aj p ≡ m X i=1 ri + n X i=1 si mod 2, (∗).
Recall from the proof of Gauss’ Lemma (Theorem 3.16) that {r1, . . . , rm, (p −s1), . . . , (p −sn)} = {1, 2, . . . , p−1 2 }.
43 Thus m X i=1 ri + np + n X i=1 si ≡ (p−1)/2 X j=1 j mod 2, and hence m X i=1 ri + n X i=1 si ≡n + (p−1)/2 X j=1 j mod 2.
Comparing this with (∗), we see that n ≡ (p−1)/2 X j=1 aj p mod 2, and the result follows from Gauss’ Lemma (Theorem 3.16).
3.5 The Law of Quadratic Reciprocity Theorem 3.23 (The Law of Quadratic Reciprocity - LQR). If p and q are distinct odd primes, then p q = q p (−1)( p−1 2 )( q−1 2 ) = + q p , if p ≡1 mod 4 or q ≡1 mod 4, − q p , if p ≡q ≡3 mod 4.
Proof. To prove the Law of Quadratic Reciprocity it suffices, by Lemma 3.22, to show that (p−1)/2 X k=1 qk p + (q−1)/2 X k=1 pk q = p −1 2 × q −1 2 .
We shall count the points in R := (x, y) ∈N × N : 0 < x < p 2, 0 < y < q 2 in two different ways. First note that since p and q are odd, we have #R = #{x : 0 < x < p 2} × #{y : 0 < y < q 2} = p−1 2 × q−1 2 .
We now find another expression for #R. If a point (x, y) were on the line from (0, 0) to (p 2, q 2) we would have y = qx p and hence py = qx. However, then we would have p | qx, which is impossible by Euclid’s lemma for primes, since p ∤q and p ∤x (recall that 0 < x < p/2). Thus there are no points (x, y) of R on the line from (0, 0) to (p 2, q 2).
44 How many points (x, y) of R are there below (or on) the diagonal? For each value of x with 1 ≤x ≤p−1 2 , the pairs (x, y) below the diagonal must satisfy 1 ≤y ≤q px. However, there are ⌊qx p ⌋such values of y. It follows that the total number of points below (or on) the line y = qx/p is (p−1)/2 X k=1 qk p .
Similarly, there are (q−1)/2 X k=1 pk q points above (or on) the line. It follows that #R = (p−1)/2 X k=1 qk p + (q−1)/2 X k=1 pk q .
Comparing the two expressions for #R gives the result.
Example 3.24. What is 29 53 ? In other words, can we solve x2 ≡29 mod 53?
Note that 29 and 53 are both prime. Use LQR: 29 53 = 53 29 (by LQR since 29 ≡1 mod 4) = 24 29 (by periodicity since 53 ≡24 mod 29) = 2 × 2 × 2 × 3 29 = 2 29 3 3 29 (by multiplicativity).
We now use LQR and Corollary 3.19 repeatedly: 2 29 = −1 (by Corollary 3.19 since 29 ≡−3 mod 8) 3 29 = 29 3 (by LQR since 29 ≡−3 mod 4) = 2 3 (by periodicity since 29 ≡2 mod 3) = −1 (by Corollary 3.19 since 3 ≡3 mod 8).
Thus 29 53 = (−1)4 = +1, and hence x2 ≡29 mod 53 is soluble.
45 Example 3.25. Recall that in Example 2.67 we used the binary powering algorithm to show that 3499 ≡3 mod 997.
We now perform this computation in a different way. Note that 997 is a prime and that 997 ≡1 mod 4. Moreover, 997 ≡1 mod 3. Hence by LQR we have 3 997 = 997 3 = 1 3 = 1.
However, Euler’s Criterion (Theorem 3.10) gives 3 997 ≡3(997−1)/2 ≡3498 mod 997.
Hence 3498 ≡1 mod 997 and so 3499 ≡3 mod 997.
Note that we were “lucky” with the choice of exponent here in that is was close to (997−1)/2. In general, if p is prime and a ∈Z with p ∤a then we can use LQR and Euler’s Criterion to compute a(p−1)/2 mod p. (In particular, we must have a(p−1)/2 ≡±1 mod p.) Example 3.26. Determine 3 p where p ≥5 is a prime.
By LQR we have 3 p = p 3 (−1)(p−1)(3−1)/4 = (−1)(p−1)/2 p 3 .
To determine p 3 we need to know the value of p mod 3.
To determine (−1)(p−1)/2 we need to know the value of (p −1)/2 mod 2, or equivalently, the value of p mod 4. Thus there are only four cases to consider, p ≡1, 5, 7 or 11 mod 12. Note that the other cases are excluded because ϕ(12) = 4 and p must be coprime to 12 (since p ≥5).
Case (i): p ≡1 mod 12. In this case p ≡1 mod 3 so p 3 = 1.
Also p ≡1 mod 4 so (p −1)/2 is even. Hence 3 p = 1.
Case (ii): p ≡5 mod 12. In this case p ≡2 mod 3 so p 3 = 2 3 = (−1)(32−1)/8 = −1.
Also p ≡1 mod 4 so (p −1)/2 is even. Hence 3 p = −1.
Case (iii): p ≡7 mod 12. In this case p ≡1 mod 3 so p 3 = 1.
Also p ≡3 mod 4 so (p −1)/2 is odd. Hence 3 p = −1.
Case (iv): p ≡11 mod 12. In this case p ≡2 mod 3 so p 3 = 2 3 = −1.
Also p ≡3 mod 4 so (p −1)/2 is odd. Hence 3 p = 1.
46 Summarising our results, we have 3 p = +1, if p ≡±1 mod 12, −1, if p ≡±5 mod 12.
3.6 The Jacobi Symbol Definition 3.27. Let n be an odd positive integer with prime factorisation n = pe1 1 · · · per r . Then for any a ∈Z we define the Jacobi symbol a n by a n = r Y i=1 a pi ei , where the symbols on the right are Legendre symbols. We also define a 1 = 1.
Theorem 3.28. Let n be an odd positive integer and let a ∈Z.
(i) a n = ±1 if a and n are coprime and a n = 0 otherwise, (ii) a n = b n whenever a ≡b mod n, (iii) ab n = a n b n and a mn = a m a n , (iv) a2 n = 1 whenever a and n are coprime.
Proof. These properties are easily deduced from the corresponding properties of the Legendre Symbol.
Remark 3.29. Let n be an odd positive integer with prime factorisation n = pe1 1 · · · per r . If the congruence x2 ≡a mod n has a solution then a pi = 1 for each i and hence a n = 1. However, the converse is not true since because an even number of factors −1 could appear in the defining product of a n . This is illustrated in the following example.
Example 3.30. We have 2 15 = 2 3 2 5 = (−1)(−1) = 1.
Even though 2 15 = 1 the congruence x2 ≡2 mod 15 is insoluble because x2 ≡2 mod 3 has no solutions.
Theorem 3.31. If n is an odd positive integer then −1 n = (−1)(n−1)/2.
47 Proof. Write n = p1p2 · · · pr where the odd prime factors pi are not necessar-ily distinct. Then we have n = r Y i=1 (1 + pi −1) = 1 + r X i=1 (pi −1) + X i̸=j (pi −1)(pj −1) + · · · .
But each factor pi −1 is even so each sum after the first is divisible by 4.
Hence n ≡1 + r X i=1 (pi −1) mod 4, which gives 1 2(n −1) ≡ r X i=1 1 2(pi −1) mod 2.
Therefore −1 n = r Y i=1 −1 pi = r Y i=1 (−1)(pi−1)/2 = (−1) Pr i=1(pi−1)/2 = (−1)(n−1)/2, which gives the desired result.
Theorem 3.32. If n is an odd positive integer then 2 n = (−1)(n2−1)/8 = +1, n ≡±1 mod 8, −1, n ≡±3 mod 8.
Proof. Write n = p1p2 · · · pr where the odd prime factors pi are not necessar-ily distinct. Then we have n2 = r Y i=1 (1 + p2 i −1) = 1 + r X i=1 (p2 i −1) + X i̸=j (p2 i −1)(p2 j −1) + · · · .
Since each pi is odd, we have p2 i −1 ≡0 mod 8 so n2 ≡1 + r X i=1 (p2 i −1) mod 64 hence 1 8(n2 −1) ≡ r X i=1 1 8(p2 i −1) mod 8.
48 This also holds modulo 2, hence 2 n = r Y i=1 2 pi = r Y i=1 (−1)(p2 i −1)/8 = (−1)(n2−1)/8.
As n is odd we must have n ≡±1, ±3 mod 8 and by checking the cases n = ±1, ±3 we deduce that n2 −1 8 ≡ 0 mod 2, n ≡±1 mod 8, 1 mod 2, n ≡±3 mod 8.
This completes the proof.
Theorem 3.33 (Reciprocity Law for Jacobi symbols). Let m and n be coprime odd positive integers. Then m n n m = (−1)(m−1)(n−1)/4 = +1 if m ≡1 mod 4 or n ≡1 mod 4, −1, if m ≡n ≡3 mod 4.
Proof. Write n = p1p2 · · · pr where the odd prime factors pi are not necessar-ily distinct. Similarly, write m = q1q2 · · · qs where the odd prime factors qj are not necessarily distinct. (Note that since m and n are coprime, we have pi ̸= qj for all i, j.) Then m n n m = r Y i=1 s Y j=1 pi qj qj pi = (−1)t for some t ∈Z. Applying the quadratic reciprocity law (Theorem 3.23) to the first factor of each term pi qj qj pi , we see that we can take t = r X i=1 s X j=1 1 2(pi −1)1 2(qj −1) = r X i=1 1 2(pi −1) s X j=1 1 2(qj −1).
However, the same argument as in the proof of Theorem 3.31 shows that 1 2(n −1) ≡ r X i=1 1 2(pi −1) mod 2 and the corresponding result holds for 1 2(m −1). Therefore t ≡n −1 2 m −1 2 mod 2, which completes the proof.
49 Example 3.34. Determine whether 888 is a quadratic residue or nonresidue of the prime 1999. We have 888 1999 = 2 1999 3 111 1999 = 111 1999 since 1999 ≡−1 mod 8. To calculate 111 1999 using Legendre symbols, we would write 111 1999 = 3 1999 37 1999 and apply the quadratic reciprocity law to each factor on the right. However, the calculation is much simpler with the Jacobi symbol since we have 111 1999 = − 1999 111 = − 1 111 = −1 since 111 ≡1999 ≡3 mod 4 and 1999 ≡1 mod 111. Therefore 888 is a quadratic nonresidue of 1999.
Example 3.35. Determine whether −104 is a quadratic residue or nonresidue of the prime 997. Since 104 = 23 × 13 we have −104 997 = −1 997 2 997 3 13 997 = 2 997 3 13 997 since 997 ≡1 mod 4 = − 13 997 since 997 ≡−3 mod 8 = − 997 13 since 997 ≡1 mod 4 = − 9 13 since 997 ≡9 mod 13 = −1 since 9 is a square.
Therefore −104 is a quadratic nonresidue of 997.
4 Sums of Squares 4.1 Pythagorean triples Definition 4.1. A Pythagorean triple (x, y, z) is a triple of positive integers satisfying x2 + y2 = z2.
50 If gcd(x, y, z) = 1 then (x, y, z) is called a primitive Pythagorean triple.
Remark 4.2. If g = gcd(x, y, z) then (x/g, y/g, z/g) is also a Pythagorean triple. It follows that if g > 1, (x, y, z) can be obtained from the “smaller” primitive Pythagorean triple (x/g, y/g, z/g) by multiplying each entry by g.
Thus it is natural to focus on primitive Pythagorean triples.
It will be useful to note a basic fact about primitive Pythagorean triples.
Theorem 4.3. Let (x, y, z) be a primitive Pythagorean triple. Then gcd(x, y) = gcd(x, z) = gcd(y, z) = 1.
Proof. Suppose gcd(x, y) > 1. Then there is a prime p with p | x and p | y.
Then z2 = x2 +y2 ≡0 (mod p). As p | z2 then by Euclid’s Lemma for primes we have p | z and so p | gcd(x, y, z), contradicting (x, y, z) being a primitive Pythagorean triple. Thus gcd(x, y) = 1.
The proofs that gcd(x, z) = 1 and gcd(y, z) = 1 are similar.
Considering things modulo 4 we can determine the parities of the numbers in a primitive Pythagorean triple.
Theorem 4.4. If (x, y, z) is a primitive Pythagorean triple, then one of x and y is even, and the other odd. (Equivalently, x + y is odd.) Also z is odd.
Proof. Note that if x is even then x2 ≡0 (mod 4) and if x is odd then x2 ≡1 (mod 4). If x and y are both odd then x2 ≡y2 ≡1 (mod 4). Hence z2 ≡x2 + y2 ≡2 (mod 4), which is impossible. If x and y are both even, then gcd(x, y) ≥2 contradicting Theorem 4.3. We conclude that one of x and y is even, and the other is odd.
In any case, z ≡z2 = x2 + y2 ≡x + y (mod 2), so z is odd.
As the rˆ oles of x and y in Pythagorean triples are symmetric, it makes little loss in generality in studying only primitive Pythagorean triples with x odd and y even.
We can now prove a theorem characterizing primitive Pythagorean triples Theorem 4.5. Let (x, y, z) be a primitive Pythagorean triple with x odd.
Then there are r, s ∈N with r > s, gcd(r, s) = 1 and r + s odd, such that x = r2 −s2, y = 2rs and z = r2 + s2.
Conversely, if r, s ∈N with r > s, gcd(r, s) = 1 and r + s odd, then (r2 −s2, 2rs, r2 + s2) is a primitive Pythagorean triple.
51 Proof. Let (x, y, z) be a primitive Pythagorean triple with x odd. Then y is even and z is odd. Let a = 1 2(z −x), b = 1 2(z + x) and c = y/2. Then a, b, c ∈N. Also ab = (z −x)(z + x) 4 = z2 −x2 4 = y2 4 = c2.
Let g = gcd(a, b). Then g | (a + b) and g | (b −a); that is g | z and g | x. As gcd(x, z) = 1, by Theorem 4.3, then g = 1, that is gcd(a, b) = 1.
Let p be a prime factor of a. Then p ∤b, so vp(b) = 0. Hence vp(a) = vp(a) + vp(b) = vp(ab) = vp(c2) = 2vp(c) is even.
Thus a is a square.
Similarly b is a square.
Write a = s2 and b = r2 where r, s ∈N. Then gcd(r, s) | a and gcd(r, s) | b; as a and b are coprime, gcd(r, s) = 1. Now x = b −a = r2 −s2; therefore r > s. Also z = a + b = r2 + s2. As c2 = ab = r2s2, c = rs and so y = 2rs. Finally as x is odd, then 1 ≡x = b −a ≡r2 −s2 ≡r2 + s2 ≡r + s mod 2; that is r + s is odd. This proves the first half of the theorem.
To prove the second part, let r, s ∈N with r > s, gcd(r, s) = 1 and r + s odd. Set x = r2 −s2, y = 2rs and z = r2 + s2. Certainly y, z ∈N and also x ∈N as r > s > 0. Also x2 +y2 = (r2 −s2)2 +(2rs)2 = (r4 −2r2s2 +s4)+4r2s2 = r4 +2r2s2 +s4 = z2.
Hence (x, y, z) is a Pythagorean triple. Certainly y is even, and x = r2−s2 ≡ r −s ≡r + s (mod 2): x is odd.
To show that (x, y, z) is a primitive Pythagorean triple we examine g = gcd(x, z). Since x is odd, g is odd. Also g | (x + z) and g | (z −x), that is g | 2r2 and g | 2s2. As r and s are coprime, then gcd(2r2, 2s2) = 2, and so g | 2. As g is odd g = 1. Hence (x, y, z) is a primitive Pythagorean triple.
We now apply this to the proof of Fermat’s last theorem for exponent 4.
Theorem 4.6. There do not exist x, y, z ∈N with x4 + y4 = z4.
(18) Proof. In fact we prove a stronger result. We claim that there are no x, y, u ∈N with x4 + y4 = u2.
(19) A natural number solution (x, y, z) to (18) gives one for (19), namely (x, y, u) = (x, y, z2). Thus it suffices to prove that (19) is insoluble over N.
52 We use Fermat’s method of descent. Given a solution (x, y, u) of (19) we produce another solution (x′, y′, u′) with u′ < u. This is a contradiction if we start with the solution of (19) minimizing u.
Let (x, y, u) be a solution of (19) over N with minimum possible u. We claim first that gcd(x, y) = 1. If not, then p | x and p | y for some prime p.
Then p4 | (x4 + y4), that is, p4 | u2.
Hence p2 | u.
Then (x′, y′, u′) = (x/p, y/p, u/p2) is a solution of (19) in N with u′ < u. This is a contradiction.
Hence gcd(x, y) = 1.
As gcd(x, y) = 1 then gcd(x2, y2) = 1, and so (x2, y2, u) is a primitive Pythagorean triple by (19). By the symmetry of x and y we may assume that x2 is odd and y2 is even, that is, x is odd and y is even. Hence by Theorem 4.5 there are r, s ∈N with gcd(r, s) = 1 x2 = r2 −s2, y2 = 2rs, u = r2 + s2.
Then x2 + s2 = r2, and as gcd(r, s) = 1 then (x, s, r) is a primitive Py-thagorean triple. As x is odd, there exist a, b ∈N with gcd(a, b) = 1 and x = a2 −b2, s = 2ab, r = a2 + b2 by Theorem 4.5. Then y2 = 2rs = 4(a2 + b2)ab, equivalently (y/2)2 = ab(a2 + b2) = abr. (Recall that y is even.) If p is prime and p | gcd(a, r) then b2 = (a2 +b2)−a2 ≡0 (mod p) and so p | b by Euclid’s Lemma for primes. This is impossible, as gcd(a, b) = 1. Thus gcd(a, r) = 1.
Similarly gcd(b, r) = 1. Now abr is a square. If p | a, then p ∤b and p ∤r.
Thus vp(a) = vp(abr) is even, and so a is a square. Similarly b and r are squares. Write a = x′2, b = y′2 and r = u′2 where x′, y′, u′ ∈N. Then u′2 = a2 + b2 = x′4 + y′4 so (x′, y′, u′) is a solution of (19). Also u′ ≤u′2 = a2 + b2 = r ≤r2 < r2 + s2 = u.
This contradicts the minimality of u in the solution (x, y, u) of (19). Hence (19) is insoluble over N. Consequently (18) is insoluble over N.
4.2 Sums of Squares Definition 4.7. For k ∈N we let Sk = {a2 1 + · · · + a2 k : a1, . . . , ak ∈Z} be the set of sums of k squares. Note that we allow zero; e.g. 1 = 12 + 02 ∈S2.
53 Theorem 4.8. The sets S2 and S4 are closed under multiplication. That is: (i) If m, n ∈S2 then mn ∈S2.
(ii) If m, n ∈S4 then mn ∈S4.
Proof. Let m, n ∈S2. Then m = a2 + b2 and n = r2 + s2 where a, b, r, s ∈Z. By the two-square formula, (a2 + b2)(r2 + s2) = (ar −bs)2 + (as + br)2, it is immediate that mn ∈S2.
Let m, n ∈S4. Then m = a2 + b2 + c2 + d2 and n = r2 + s2 + t2 + u2 where a, b, c, d, r, s, t, u ∈Z. By the four-square formula, (a2 + b2 + c2 + d2)(r2 + s2 + t2 + u2) = (ar −bs −ct −du)2 + (as + br + cu −dt)2 + (at −bu + cr + ds)2 + (au + bt −cs + dr)2, it is immediate that mn ∈S4.
Remark 4.9. The two-square formula comes from complex numbers: (a2 + b2)(c2 + d2) = |a + bi|2|c + di|2 = |(a + bi)(c + di)|2 = |(ac −bd) + (ad + bc)i|2 = (ac −bd)2 + (ad + bc)2.
Similarly the four-square formula comes from the theory of quaternions (if you know what they are).
4.3 Sums of Two Squares We can restrict the possible factorizations of a sum of two squares.
Theorem 4.10. Let p be a prime with p ≡3 (mod 4) and let n ∈N. If n ∈S2 then vp(n) is even.
Proof. Let n = a2 + b2 with a, b ∈Z and suppose p | n. We aim to show that p | a and p | b. Suppose p ∤a. Then there is c ∈Z with ac ≡1 (mod p).
Then 0 ≡c2n = (ac)2 + (bc)2 ≡1 + (bc)2 (mod p).
This implies that −1 p = 1, but we know that −1 p = −1 when p ≡3 (mod 4). This contradiction proves that p | a. Similarly p | b. Thus p2 | (a2 + b2) = n and n/p2 = (a/p)2 + (b/p)2 ∈S2.
54 Let n ∈S2 and k = vp(n). We have seen that if k > 0 then k ≥2 and n/p2 ∈S2. Note that vp(n/p2) = k −2. Similarly if k −2 > 0 (that is if k > 2) then k−2 ≥2 (that is k ≥4) and n/p4 ∈S2. Iterating this argument, we find that if k = 2r + 1 is odd, then n/p2r ∈S2 and vp(n/p2r) = 1, which is impossible. We conclude that k is even.
Remark 4.11. If n ∈N, we can write n = rm2 where m2 is the largest square dividing n and r is squarefree, that is either r = 1 or r is a product of distinct primes. If any prime factor p of r is congruent to 3 modulo 4 then vp(n) = 1 + 2vp(m) is odd, and n / ∈S2. Hence, if n ∈S2, the only possible prime factors of r are p = 2 and the p congruent to 1 modulo 4. Obviously 2 = 12 +12 ∈S2. It would be nice if all primes congruent to 1 modulo 4 were also in S2. Fortunately, this is the case.
Theorem 4.12. Let p be a prime with p ≡1 (mod 4). Then p ∈S2.
Proof. As p ≡1 (mod 4) then −1 p = 1 and so there exists u ∈Z such that u2 ≡−1 (mod p). Let A = {(m1, m2) : m1, m2 ∈Z, 0 ≤m1, m2 < √p} = {(m1, m2) : m1, m2 ∈Z, 0 ≤m1, m2 ≤⌊√p⌋}.
Then A has (1 + ⌊√p⌋)2 elements and so |A| > p.
For m = (m1, m2) ∈R2 define φ(m) = um1 + m2. Then φ : R2 − →R is a linear map, and if m ∈Z2 then φ(m) ∈Z.
As |A| > p, the φ(m) for m ∈A can’t all be distinct modulo p. Hence there are distinct m, n ∈A with φ(m) ≡φ(n) (mod p). Let a = m −n.
Then by linearity φ(a) = φ(m) −φ(n) ≡0 (mod p). Write a = (a, b). Then a = m1 −n1 where 0 ≤m1, n1 < √p so that |a| < √p. Similarly |b| < √p.
Then a2 + b2 < 2p. As m ̸= n then a ̸= (0, 0) and so a2 + b2 > 0. But 0 ≡φ(a) = ua + b (mod p). Hence b ≡−ua (mod p) and so a2 + b2 ≡a2 + (−ua)2 ≡a2(1 + u2) ≡0 (mod p) As a2 + b2 is a multiple of p, and 0 < a2 + b2 < 2p, then a2 + b2 = p. We conclude that p ∈S2.
Alternative proof (constructive). As p ≡1 (mod 4) then −1 p = 1 and so there exists u ∈Z such that u2 ≡−1 (mod p). In other words, there exists m ∈N such that u2 + 1 = mp.
Note that we can assume |u| < p 2, so u2 + 1 < p2 4 + 1 < p2 2 . Thus 1 ≤m < p 2.
55 The idea is as follows. Given a representation a2 + b2 = mp, with 1 ≤ m < p, use this to find another representation c2 + d2 = m′p with 1 ≤m′ < m. Then repeat this process until it terminates (as it must) with m′ = 1, giving the desired solution. Note that the starting point is the representation u2 + 12 = mp of the first paragraph.
So suppose that a2 +b2 = mp for some m ∈N with 1 < m < p. (If m = 1 then we are already done.) Then there exist a′, b′ ∈Z with a ≡a′ mod m, |a′| ≤m 2 and b ≡b′ mod m, |b′| ≤m 2 . Let c = aa′+bb′ m and d = ab′−ba′ m . Now aa′ + bb′ ≡a2 + b2 ≡0 mod m and ab′ −ba′ ≡ab −ba ≡0 mod m, and so c, d ∈Z. Moreover, (a′)2 + (b′)2 ≡a2 + b2 ≡0 mod m and so c2 + d2 = (aa′ + bb′)2 + (ab′ −ba′)2 m2 = (a2 + b2)(a′2 + b′2) m2 = p(a′2 + b′2) m is in fact an integer and a multiple of p. In other words, c2 + d2 = m′p for some m′ ∈Z. Now a′2 ≤m2 4 and b′2 ≤m2 4 , so a′2 + b′2 ≤m2 2 . Thus 0 ≤m′ = a′2 + b′2 m ≤m 2 < m < p.
If m′ = 0 then a′ = b′ = 0. Thus m | a and m | b and so m2 | (a2 + b2) = mp.
Thus m | p.
But p is prime and 1 < m < p, so m ∤p - contradiction.
Therefore 1 ≤m′ < m.
Remark 4.13. In order to make this into an algorithm for finding an expres-sion p = ab + b2 when p is a prime with p ≡1 mod 4, we need to solve the equation u2 ≡−1 mod p. (This is the hard part.) Write p = 4k + 1 where k ∈N. Let g be a primitive root mod p. Then ordp(g) = ϕ(p) = p −1 = 4k and g0, g1, g2, . . . , g4k−1 are congruent to 1, 2, . . . , p −1 in some order. Now x = g2k is a solution to x2 ≡1 mod p and x ̸≡1 mod p, so g2k ≡−1 mod p by Corollary 2.71. If t ≡gr mod p where r is odd then tk is a solution of x2 ≡−1 (mod p) since 2kr ≡2k mod 4k (note that 4k = ordp(g) and use Proposition 2.53). Thus if we pick t ∈{1, . . . , p −1} at random, there is a 50% chance that t ≡gr mod p with r odd. Given such an t, we set u = tk.
Example 4.14. Let p = 1997. Note that p is prime and p ≡1 mod 4. Writing p = 4k + 1 we have k = (p −1)/4 = (1997 −1)/4 = 499. Try t = 2. Then 2499 ≡1585 ≡−412 mod 1997 (one can use the binary powering algorithm to do this). Note that we chose −412 instead of 1585 because | −412| = 412 < 1997/2. Check that (−412)2 ≡−1 mod 1997. Set a = 412 and b = 1.
Then a2 + b2 = 169745 = 85 × 1997, so m = 85.
56 Now 412 ≡−13 mod 85. So take a′ = −13 and b′ = 1. Set c = aa′ + bb′ m = 412 × (−13) + 1 × 1 85 = −63 d = ab′ −ba′ m = 412 × 1 −1 × (−13) 85 = 5.
Now we have 632 + 52 = 3994 = 2 × 1997. Now let a = 63, b = 5 and m = 2.
Then 63 ≡1 mod 2 and 5 ≡1 mod 2. So we take a′ = b′ = 1 and c = aa′ + bb′ m = 63 × 1 + 5 × 1 2 = 34 d = ab′ −ba′ m = 63 × 1 −5 × 1 2 = 29.
Now we have 342 + 292 = 1997, so we are done.
Remark 4.15. In the above example, we need to compute 2499 mod 1997 effi-ciently. The way to do this is to use the binary powering algorithm that was introduced in §2.9.
We now give the computation of 2499 mod 1997 (note that in Example 2.67 we worked mod 997 rather than 1997). First we find the binary expansion of 499 as follows: 499 = 28 + 243 = 28 + 27 + 115 = 28 + 27 + 26 + 51 = 28 + 27 + 26 + 25 + 19 = 28 + 27 + 26 + 25 + 24 + 3 = 28 + 27 + 26 + 25 + 24 + 21 + 20.
So the binary expansion of 499 is 111110011. (This part is exactly the same as in Example 2.67). Now by squaring the previous term each time, we have 221 ≡4 (mod 1997) 222 ≡42 ≡16 (mod 1997) 223 ≡162 ≡256 (mod 1997) 224 ≡2562 ≡65536 ≡1632 ≡−365 (mod 1997) 225 ≡(−365)2 ≡2663424 ≡1423 ≡−574 (mod 1997) 226 ≡(−574)2 ≡329476 ≡1968 ≡−29 (mod 1997) 227 ≡(−29)2 ≡841 (mod 1997) 228 ≡8412 ≡707281 ≡343 (mod 1997).
57 Therefore 2499 ≡220 × 221 × 224 × 225 × 226 × 227 × 228 (mod 1997) ≡2 × 4 × (−365) × (−574) × (−29) × 841 × 343 (mod 1997) ≡(−2920) × 16646 × 288463 (mod 1997) ≡1074 × 670 × 895 (mod 1997) ≡719580 × 670 × 895 (mod 1997) ≡660 × 895 (mod 1997) ≡1585 (mod 1997).
We can now characterize the elements of S2.
Theorem 4.16 (Two-square theorem). Let n ∈N. Then n ∈S2 if and only if vp(n) is even whenever p is a prime congruent to 3 modulo 4.
Proof. If n ∈S2, p is prime and p ≡3 (mod 4) then vp(n) is even by Theorem 4.10.
If vp(n) is even whenever p is a prime congruent to 3 modulo 4 then n = rm2 where each prime factor p of r is either 2 or congruent to 1 modulo 4.
By Theorem 4.12 all primes p with p ≡1 mod 4 lie in S2. Moreover, 2 = 12 + 12 ∈S2. Hence by Theorem 4.8 r ∈S2. Hence r = a2 + b2 where a, b ∈Z and so n = rm2 = (am)2 + (bm)2 ∈S2.
The representation of a prime as a sum of two squares is essentially unique.
Theorem 4.17. Let p be a prime. If p = a2 + b2 = c2 + d2 with a, b, c, d ∈N then either a = c and b = d or a = d and b = c.
Proof. Consider (ac + bd)(ad + bc) = a2cd + abc2 + abd2 + b2cd = (a2 + b2)cd + ab(c2 + d2) = pcd + pab = p(ab + cd).
As p | (ac+bd)(ad+bc) then by Euclid’s lemma for primes either p | (ac+bd) or p | (ad + bc). Assume the former — the latter case can be treated by reversing the rˆ oles of c and d. Now ac + bd > 0 so that ac + bd ≥p. Also (ac + bd)2 + (ad −bc)2 = a2c2 + 2abcd + b2d2 + a2d2 −2abcd + b2c2 = a2c2 + b2d2 + a2d2 + b2c2 = (a2 + b2)(c2 + d2) = p2.
58 As ac+bd ≥p, the only way this is possible is if ac+bd = p and ad−bc = 0.
Then ac2 + bcd = cp and ad2 −bcd = 0, so adding gives a(c2 + d2) = cp, that is ap = cp, so that a = c. Then c2 + bd = p = c2 + d2 so that bd = d2, so that b = d.
Example 4.18. Find two “essentially different” ways of writing 629 = 17 × 37 as the sum of two squares. First note that 17 and 37 are both primes congruent to 1 mod 4, and thus each can be written as the sum of two squares in a unique way. In fact, 17 = 42 + 12 and 37 = 62 + 12. Then 629 = |4 + i|2|6 + i|2 = |(4 + i)(6 + i))|2 = |23 + 10i|2 = 232 + 102 629 = |4 + i|2|6 −i|2 = |(4 + i)(6 −i))|2 = |25 + 2i|2 = 252 + 22.
4.4 Sums of Four Squares We wish to prove the theorem of Lagrange to the effect that all natural numbers are sums of four squares. It is crucial to establish this for primes.
Theorem 4.19. Let p be a prime. Then p ∈S4.
Proof. If p ≡1 (mod 4) then there are a, b ∈Z with p = a2 + b2 + 02 + 02 (Theorem 4.12) so that p ∈S4.
Also 2 = 12 + 12 + 02 + 02 ∈S4 and 3 = 12 + 12 + 12 + 02 ∈S4. We may assume that p > 3 and that p ≡3 (mod 4). As a consequence −1 p = −1.
Let w be the smallest positive integer with w p = −1. (Note that this forces w ≥2.) Then w −1 p = 1 and −w p = −1 p w p = 1.
Hence there are u, v ∈Z with w −1 ≡u2 (mod p) and −w ≡v2 (mod p).
Then 1 + u2 + v2 ≡1 + (w −1) −w ≡0 (mod p).
Let B = {(m1, m2, m3, m4) : m1, . . . , m4 ∈Z, 0 ≤m1, . . . , m4 < √p} = {(m1, m2, m3, m4) : m1, . . . , m4 ∈Z, 0 ≤m1, . . . , m4 < ⌊√p⌋}.
Then B has (1+⌊√p⌋)4 elements. Hence |B| > p2. For m = (m1, m2, m3, m4) define ψ(m) = (um1 + vm2 + m3, −vm1 + um2 + m4). Then ψ : R4 − →R2 is a linear map. If m ∈Z4 then ψ(m) ∈Z2. We write (a, b) ≡(a′, b′) (mod p) if a ≡a′ (mod p) and b ≡b′ (mod p). If we have a list (a1, b1), . . . , (aN, bN) of vectors in Z2 with N > p2, then there must be some distinct i and j with 59 (ai, bi) ≡(aj, bj) (mod p). This happens for the vectors ψ(m) with m ∈B as |B| > p2. Thus there are distinct m, n ∈B with ψ(m) ≡ψ(n) (mod p). Let a = m −n. Then ψ(a) = ψ(m) −ψ(n) ≡(0, 0) (mod p). Let a = (a, b, c, d).
Then a = m1 −n1 where 0 ≤m1, n1 < √p so that |a| < √p. Similarly |b|, |c|, |d| < √p. Then a2 + b2 + c2 + d2 < 4p. As m ̸= n then a ̸= (0, 0, 0, 0) and so a2 + b2 + c2 + d2 > 0.
Now (0, 0) ≡ψ(a) = (ua + vb + c, −va + ub + d) (mod p). Hence c ≡ −ua −vb (mod p) and d ≡va −ub (mod p). Then a2 + b2 + c2 + d2 ≡ a2 + b2 + (ua + vb)2 + (va −ub)2 = (1 + u2 + v2)(a2 + b2) ≡0 (mod p) where the last equality holds because we previously showed that 1+u2+v2 ≡ 0 mod p. As a2 +b2 +c2 +d2 is a multiple of p, and 0 < a2 +b2 +c2 +d2 < 4p, then we must have a2 + b2 + c2 + d2 ∈{p, 2p, 3p}.
When a2 + b2 + c2 + d2 = p then certainly p ∈S4. Alas, we need to consider the bothersome cases where a2 + b2 + c2 + d2 = 2p or 3p.
Suppose that a2 + b2 + c2 + d2 = 2p. Then a2 + b2 + c2 + d2 ≡2 (mod 4) so that two of a, b, c, d are odd and the other two even. Without loss of generality a and b are odd and c and d are even. Then 1 2(a + b), 1 2(a −b), 1 2(c + d) and 1 2(c −d) are all integers, and a simple computation gives a + b 2 2 + a −b 2 2 + c + d 2 2 + c −d 2 2 = a2 + b2 + c2 + d2 2 = p so that p ∈S4.
Finally suppose that a2 + b2 + c2 + d2 = 3p. Then a2 + b2 + c2 + d2 is a multiple of 3 but not 9. As a2 ≡0 or 1 (mod 3) then either exactly one or all four of a, b, c and d are multiples of 3. But the latter case is impossible (for then a2 + b2 + c2 + d2 would be a multiple of 9), so without loss of generality 3 | a and b, c, d ≡±1 (mod 3). By replacing b by −b etc., if necessary, we may assume that b ≡c ≡d ≡1 (mod 3). Then 1 3(b + c + d), 1 3(a + b −c), 1 3(a + c −d), 1 3(a + d −b), are all integers, and a simple computation gives b + c + d 3 2 + a + b −c 3 2 + a + c −d 3 2 + a + d −b 3 2 = a2 + b2 + c2 + d2 3 = p so that p ∈S4.
We can now prove Lagrange’s four-square theorem.
60 Theorem 4.20 (Lagrange). If n ∈N then n ∈S4.
Proof. Either n = 1 = 12 + 02 + 02 + 02 ∈S4, or n is a product of a sequence of primes. By Theorem 4.19, each prime factor of n lies in S4. Then since S4 is closed under multiplication (Theorem 4.8), we have n ∈S4.
61 |
6879 | https://physics.stackexchange.com/questions/544269/find-angle-of-conical-pendulum-with-only-period-and-string-length | homework and exercises - Find angle of conical pendulum with only period and string length - Physics Stack Exchange
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Find angle of conical pendulum with only period and string length [closed]
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I have a lab for Physics which asks to find the angle θ θ given the period and string length of a conical pendulum. We were tasked to construct one of these with a piece of string ranging from 0.4 m 0.4 m to 0.8 m 0.8 m with an object tied to the end (mass does not matter so we were not instructed to weigh the item on the end of the string). The conical pendulum I made had a string length of 0.44 m 0.44 m and had a period of around 1.44 s e c o n d s 1.44 s e c o n d s (14.40 seconds for 10 full spins). The thing is, I'm stumped on how to find the angle with just those two pieces of information.
I realize that finding the radius of the pendulum, plugging that radius into a triangle diagram of the pendulum and then using sin law to find θ θ is a way I can do this, however, with the equations I've seen I don't know how to find the radius with only the period and length. I have attempted to rearrange and combine formulas in an attempt to figure out the radius with the information I know,but to no avail. Any suggestion or hint pointing me in the right direction is highly appreciated.
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edited Apr 16, 2020 at 4:50
Qmechanic♦
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Maybe you should start with a free body diagram showing the forces acting on the weight. The period indicates the centripetal acceleration which leads to the radial force.JohnHoltz –JohnHoltz 2020-04-16 03:17:21 +00:00 Commented Apr 16, 2020 at 3:17
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If you've studied simple harmonic motions (SHMs), you can prove that the time period of a simple pendulum is given by:
T=2 π L cos θ g−−−−−−√T=2 π L cosθ g
Since all these values are known to you (T T and L L), you can find θ θ. This result can also be proved by using dynamics and kinematics (v=2 π R T v=2 π R T) along with the equations of components of the tension along the vertical and the radius (T cos θ=m g T cosθ=m g and T sin θ=m v 2 R T sinθ=m v 2 R).
Usually in phyisics, however, we assume that the angle θ θ is small. Then, by the small angle approximation, we know that for small values of θ θ, cos θ≈1 cosθ≈1. Then,
T≈2 π L g−−√T≈2 π L g
This means the period isn't dependent on the angle θ θ for small values of theta. Note that this is also equal to the time period of a simple pendulum (simple harmonic oscillator) for small amplitudes.
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edited Apr 16, 2020 at 5:15
answered Apr 16, 2020 at 4:21
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6880 | https://math.stackexchange.com/questions/209910/find-the-distance-between-the-line-and-plane | Stack Exchange Network
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Find the distance between the line and plane
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$\begingroup$
Find the distance between the line $x-z=3, x+2y+4z=6$ and the plane $3x+2y+2z=5$.
What I have done so far,
Found the vector by crossing $1,0-1$ and $1,2,4$ from the first line
The line is parallel to plane $\langle 2,-5,2\rangle \cdot \langle 3,2,2 \rangle = 0$
I think my next step is to find a point on Line 1 which satisfies both equations and then insert those values into the plane $3(x)+2(y)+2(z)=5$ and use the formula, $$ \frac{(3(x)+2(y)+2(z)-5)}{(3^2+2^2+2^2)}$$ to find the distance.
The values that I calculuate do not match the posted answer of $7/\sqrt{17}$
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edited Oct 9, 2012 at 15:57
Thomas
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asked Oct 9, 2012 at 15:52
mikemike
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$\endgroup$
3
$\begingroup$ Looking at your question again, I am not sure that I understand what the line is. It looks like your "line" is given by the equations of two planes. $\endgroup$
Thomas
– Thomas
2012-10-09 16:00:05 +00:00
Commented Oct 9, 2012 at 16:00
$\begingroup$ @Thomas any we can descibe lines as intersections of non-parallel planes $\endgroup$
Norbert
– Norbert
2012-10-13 15:09:31 +00:00
Commented Oct 13, 2012 at 15:09
$\begingroup$ @Norbert: Sure, I just wasn't sure that that was what the OP meant... $\endgroup$
Thomas
– Thomas
2012-10-13 17:11:45 +00:00
Commented Oct 13, 2012 at 17:11
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$\begingroup$
Hint: The line and the plane (as you have noted) are parallel. The distance from the plane to the line is therefore the distance from the plane to any point on the line. So just pick any point on the line and use "the formula" to find the distance from this point to the plane.
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answered Oct 9, 2012 at 15:54
ThomasThomas
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$\endgroup$
2
$\begingroup$ Thanks, I just made a simple mathematical error which resulted in an incorrect answer. $\endgroup$
mike
– mike
2012-10-09 16:06:36 +00:00
Commented Oct 9, 2012 at 16:06
1
$\begingroup$ Why the downvote? $\endgroup$
Thomas
– Thomas
2017-04-21 11:39:33 +00:00
Commented Apr 21, 2017 at 11:39
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3
$\begingroup$
If $z=t,$ for the given line $x=z+3=t+3$ and
$x+2y+4z=6$ or, $t+3+2y+4(t)=6$ or, $2y=3-5t$
So, any point on the given line is $(t+3,t,\frac{3-5t}2)$
The distance of $(t+3,t,\frac{3-5t}2)$ from $3x+2y+2z=5$ is $$\left|\frac{3(t+3)+(3-5t)+2(t)-5}{\sqrt{3^2+2^2+2^2}}\right|=\frac 7{\sqrt {17}}$$
Had it not been a constant(which is due to the parallelism), we had to find $t$such that the distance is minimum.
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edited Oct 9, 2012 at 16:15
answered Oct 9, 2012 at 16:07
lab bhattacharjeelab bhattacharjee
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$\endgroup$
1
3
$\begingroup$ But if a line is not parallel to a plane, then they intersect and so distance = 0. $\endgroup$
Ram Keswani
– Ram Keswani
2020-03-13 17:40:59 +00:00
Commented Mar 13, 2020 at 17:40
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1
$\begingroup$
Put $(P_1): x - z -3 = 0$, $(P_2): x+2y+4z=6$, $(P_3): 3x + 2y +2z -5 = 0$ and $\Delta$ is intersection of two plane $(P_1)$ and $(P_2)$. We have, a normal vector of the plane $(P_1)$ is $a = (1,0,-1)$, a normal vector of the plane $(P_2)$ is $b = (1,2,4)$. A direction vector $v$ of $\Delta$ is cross product of $a$ and $b$, therefore $v = (3, 2, 2)$. Another way, $M=(3, \dfrac32, 0)$ lies on $\Delta$ and not belongs to $(P_3)$, thus $\Delta$ is parallel to the plane $(P_3)$. We have, the distance between the line $\Delta$ to the plane $(P_3)$ is the distance from the point $M$ to the plane $(P_3)$. From here, we have answer is $$\dfrac{|3\cdot 3 + 2\cdot\dfrac{3}{2} + 2\cdot 0 - 5|}{\sqrt{3^2 + 2^2 + 2^2}} = \dfrac{7\sqrt{17}}{17}.$$
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answered Oct 9, 2012 at 16:27
minthao_2011minthao_2011
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6881 | https://sb.flleg.gov/nxt/gateway.dll?f=id$id=LAW97-042$t=document-frameset.htm$3.0$p= | Chapter 97-42, 1997 Laws of Florida
CHAPTER 97-42
Committee Substitute for Senate Bill No. 82
An act relating to real estate transactions; amending s. 475.01, F.S.; defining the terms “customer,” “first contact,” and “principal”; redefining the term “transaction broker”; deleting the definitions of the terms “buyer,” “disclosed dual agent,” and “seller”; amending s. 475.25, F.S.; modifying grounds for the imposition of discipline by the Florida Real Estate Commission; conforming a statutory cross-reference; creating ss. 475.2701, 475.272, 475.274, 475.276, 475.278, 475.2801, F.S.; establishing the “Brokerage Relationship Disclosure Act”; providing for notice of nonrepresentation; providing for disclosure of authorized brokerage relationships and the corresponding duties of real estate licensees; authorizing rulemaking by the Florida Real Estate Commission; amending s. 475.5015, F.S.; adding disclosure documents to items to be retained as brokerage business records; amending s. 468.383, F.S.; conforming a statutory cross-reference; providing an effective date.
Be It Enacted by the Legislature of the State of Florida:
Section 1.Subsection (1) of section 475.01, Florida Statutes, is amended to read:
475.01 Definitions.—
(1)As used in this part:
~~(a)“Commission” means the Florida Real Estate Commission.~~
~~(b)“Department” means the Department of Business and Professional Regulation.~~
(a)~~(c)~~“Broker” means a person who, for another, and for a compensation or valuable consideration directly or indirectly paid or promised, expressly or impliedly, or with an intent to collect or receive a compensation or valuable consideration therefor, appraises, auctions, sells, exchanges, buys, rents, or offers, attempts or agrees to appraise, auction, or negotiate the sale, exchange, purchase, or rental of business enterprises or business opportunities or any real property or any interest in or concerning the same, including mineral rights or leases, or who advertises or holds out to the public by any oral or printed solicitation or representation that he is engaged in the business of appraising, auctioning, buying, selling, exchanging, leasing, or renting business enterprises or business opportunities or real property of others or interests therein, including mineral rights, or who takes any part in the procuring of sellers, purchasers, lessors, or lessees of business enterprises or business opportunities or the real property of another, or leases, or interest therein, including mineral rights, or who directs or assists in the procuring of prospects or in the negotiation or closing of any transaction which does, or is calculated to, result in a sale, exchange, or leasing thereof, and who receives, expects, or is promised any compensation or valuable consideration, directly or indirectly therefor; and all persons who advertise rental property information or lists. A broker renders a professional service and is a professional within the meaning of s. 95.11(4)(a). Where the term “appraise” or “appraising” appears in the definition of the term “broker,” it specifically excludes those appraisal services which must be performed only by a state-licensed or state-certified appraiser, and those appraisal services which may be performed by a registered appraiser as defined in part II. The term “broker” also includes any person who is a general partner, officer, or director of a partnership or corporation which acts as a broker. The term “broker” also includes any person or entity who undertakes to list or sell one or more timeshare periods per year in one or more timeshare plans on behalf of any number of persons, except as provided in ss. 475.011 and 721.20.
~~(d)“Salesperson” means a person who performs any act specified in the definition of “broker,” but who performs such act under the direction, control, or management of another person. A salesperson renders a professional service and is a professional within the meaning of s. 95.11(4)(a).~~
(b)~~(e)~~“Broker-salesperson” means a person who is qualified to be issued a license as a broker but who operates as a salesperson in the employ of another.
(c)“Commission” means the Florida Real Estate Commission.
(d)“Customer” means a member of the public who is or may be a buyer or seller of real property and may or may not be represented by a real estate licensee in an authorized brokerage relationship.
(e)“Department” means the Department of Business and Professional Regulation.
(f)“Fiduciary” means a broker in a relationship of trust and confidence between that broker as agent and the seller or buyer as principal. The duties of the broker as a fiduciary are loyalty, confidentiality, obedience, full disclosure, and accounting and the duty to use skill, care, and diligence.
(g)“First contact” means at the commencement of the initial meeting of or communication between a licensee and a seller or buyer; however, the term does not include:
1.A bona fide “open house” or model home showing that does not involve eliciting confidential information, the execution of a contractual offer or an agreement for representation, or negotiations concerning price, terms, or conditions of a potential sale;
2.Unanticipated casual encounters between a licensee and a seller or buyer that do not involve eliciting confidential information, the execution of a contractual offer or an agreement for representation, or negotiations concerning price, terms, or conditions of a potential sale;
3.Responding to general factual questions from a prospective buyer or seller concerning properties that have been advertised for sale; or
4.Situations in which a licensee's communications with a customer are limited to providing general factual information, oral or written, about the qualifications, background, and services of the licensee or the licensee's brokerage firm.
In any of the situations described in subparagraphs 1.-4., “first contact” occurs when the communications between the licensee and the prospective seller or buyer proceed in any way beyond the conditions or limitations described in subparagraphs 1.-4.
(h)“Involuntarily inactive status” means the licensure status that results when a license is not renewed at the end of the license period prescribed by the department.
(i)“Principal” means the party with whom a real estate licensee has entered into a single agent relationship.
(j)~~(f)~~“Real property” or “real estate” means any interest or estate in land and any interest in business enterprises or business opportunities, including any assignment, leasehold, subleasehold, or mineral right; however, the term does not include any cemetery lot or right of burial in any cemetery; nor does the term include the renting of a mobile home lot or recreational vehicle lot in a mobile home park or travel park.
(k)“Salesperson” means a person who performs any act specified in the definition of “broker,” but who performs such act under the direction, control, or management of another person. A salesperson renders a professional service and is a professional within the meaning of s. 95.11(4)(a).
(l)“Single agent” means a broker who represents, as a fiduciary, either the buyer or seller but not both in the same transaction.
~~(g)“Involuntarily inactive status” means the licensure status which results when a license is not renewed at the end of the license period prescribed by the department.~~
~~(h)“Voluntarily inactive status” means the licensure status which results when a licensee has applied to the department to be placed on inactive status and has paid the fee prescribed by rule.~~
~~(i)“Fiduciary” means a broker in a relationship of trust and confidence between that broker as agent and the seller or buyer as principal. The duties of the broker as a fiduciary are loyalty, confidentiality, obedience, full disclosure, and accounting and the duty to use skill, care, and diligence.~~
~~(j)“Disclosed dual agent” means a broker who works as an agent for both the buyer and seller. The broker must obtain the informed consent in writing of all parties to the transaction to be a disclosed dual agent. The disclosed dual agent has all the duties of a fiduciary except full disclosure between the buyer and seller.~~
(m)~~(k)~~“Transaction broker” means a broker who provides limited representation to a buyer, a seller, or both, in a real estate transaction, but does not represent either in a fiduciary capacity or as a single agent.~~facilitates a brokerage transaction between a buyer and a seller. The transaction broker does not affirmatively represent either the buyer or seller as an agent, and no fiduciary duties exist except for the duty of accounting and the duty to use skill, care, and diligence. However, the transaction broker shall treat the buyer and seller with honesty and fairness and shall disclose all known facts materially affecting the value of the property in residential transactions to both the buyer and seller. The broker's role as a transaction broker must be fully disclosed in writing to the buyer and seller.~~
(n)“Voluntarily inactive status” means the licensure status that results when a licensee has applied to the department to be placed on inactive status and has paid the fee prescribed by rule.
~~(l)“Single agent” means a broker who represents, as a fiduciary, either the buyer or seller but not both in the same transaction.~~
~~(m)“Buyer” means a transferee or lessee in a real property transaction and includes a person who executes an offer to purchase or lease real property from a seller.~~
~~(n)“Seller” means the transferor or lessor in a real property transaction and includes an owner who lists real property for sale or lease with a broker, whether or not a purchase agreement or lease results, or who receives an offer to purchase or lease real property.~~
Section 2.Paragraphs (h) and (q) of subsection (1) of section 475.25, Florida Statutes, are amended to read:
475.25 Discipline.—
(1)The commission may deny an application for licensure, registration, or permit, or renewal thereof; may place a licensee, registrant, or permittee on probation; may suspend a license, registration, or permit for a period not exceeding 10 years; may revoke a license, registration, or permit; may impose an administrative fine not to exceed $1,000 for each count or separate offense; and may issue a reprimand, and any or all of the foregoing, if it finds that the licensee, registrant, permittee, or applicant:
(h)Has shared a commission with, or paid a fee or other compensation to, a person not properly licensed as a broker, broker-salesperson, or salesperson under the laws of this state, for the referral of real estate business, clients, prospects, or customers, or for any one or more of the services set forth in s. 475.01(1)(a)~~s. 475.01(1)(c)~~. For the purposes of this section, it is immaterial that the person to whom such payment or compensation is given made the referral or performed the service from within this state or elsewhere; however, a licensed broker of this state may pay a referral fee or share a real estate brokerage commission with a broker licensed or registered under the laws of a foreign state so long as the foreign broker does not violate any law of this state.
(q)Has violated any provision of s. 475.276 or s. 475.278, including the duties owed under those sections.
~~1.Has failed in a single agency to give written notice to all parties to a sale, exchange, purchase, or lease of real property or any interest in real property, revealing the party or parties for whom the licensee is an agent. Disclosure to the party for whom the licensee is an agent must be made at or before the time an agreement for representation is entered into. Disclosure to the party for whom the licensee is not an agent must be made at the time of the first substantive contact.~~
~~2.Has failed in a dual agency to obtain the informed written consent of all parties to a sale, exchange, purchase, or lease of real property or any interest in real property that the licensee intends to operate as a disclosed dual agent. Unless all parties to the transaction grant their written informed consent prior to or at the time of formalization of the dual agency by the licensee, the licensee shall be deemed to be an undisclosed dual agent. The licensee must inform all parties that the licensee is acting as agent for all parties and of the effect of dual agency, including, but not limited to, the fact that, by consenting to the dual agency relationship, the parties are giving up their rights to the undivided loyalty of the licensee, as required by the rules of the commission. When single agency exists, the licensee may change to a disclosed dual agent by making full written disclosure to and obtaining the informed written consent of all the parties. A disclosed dual agent may not disclose among other items:~~
~~a.To the buyer that the seller will accept a price less than the asking or listed price, unless otherwise instructed in writing by the seller;~~
~~b.To the seller that the buyer will pay a price greater than the price submitted in a written offer to the seller, unless otherwise instructed in writing by the buyer; ~~
~~c.The motivation of any party for selling, buying, or leasing a property, unless otherwise instructed in writing by the respective party; or~~
~~d.That a seller or buyer will agree to financing terms other than those offered.~~
~~3.Has failed in a transaction brokerage capacity to give written notice to all parties to a sale, exchange, purchase, or lease of real property or an interest in real property prior to or at the time of the licensee becoming a transaction broker or first substantive contact, whichever occurs first, of the licensee's role as a transaction broker. Unless the buyer and seller are given written notice prior to the licensee's acting in a transaction brokerage capacity, the licensee is deemed to be an agent of either the buyer or seller, or both. The licensee shall treat the buyer and seller honestly and fairly and shall disclose all known facts materially affecting the value of the property in residential transactions to both the buyer and seller.~~
~~For the purposes of this paragraph, the payment or promise of payment of compensation to a licensee does not determine whether an agency or transactional brokerage relationship has been created between any licensee and a seller, landlord, buyer, or tenant. The commission shall implement this paragraph by rule. For purposes of this paragraph, the commission shall also define by rule forms for agency disclosure. The forms provided for in this rule shall be written in plain language and shall provide to the buyer or seller or both, as appropriate, an explanation of the agency relationships and shall offer the buyer or seller or both the explicit right to choose or refuse among these agency relationships.~~
Section 3.Sections 475.2701, 475.272, 475.274, 475.276, 475.278, and 475.2801, Florida Statutes, are created to read:
475.2701 Short title.—Sections 475.2701-475.2801 may be cited as the “Brokerage Relationship Disclosure Act.”
475.272 Purpose.—In order to eliminate confusion and provide for a better understanding on the part of customers in real estate transactions, the Legislature finds that the intent of the Brokerage Relationship Disclosure Act is to provide that:
(1)Disclosed dual agency as an authorized form of representation by a real estate licensee in this state is expressly revoked;
(2)Real estate licensees be required to disclose to customers upon first contact in residential real estate transactions that they are not and will not be represented by a licensee in a real estate transaction unless they engage a real estate licensee in an authorized form of representation, either as a single agent or as a transaction broker;
(3)Disclosure requirements for real estate licensees relating to nonrepresentation and authorized forms of brokerage representation are established;
(4)Florida law provides that real estate licensees will operate as single agents or in a limited representative capacity known as transaction brokers;
(5)Single agents may represent either a buyer or a seller, but not both, in a real estate transaction; and
(6)Transaction brokers provide a limited form of nonfiduciary representation to a buyer, a seller, or both in a real estate transaction.
475.274 Scope of coverage.—The authorized brokerage relationships described in s. 475.278 apply in all brokerage activities as defined in s. 475.01(1)(a). The disclosure requirements of ss. 475.276 and 475.278 apply only to residential sales as defined in s. 475.276.
475.276 Notice of nonrepresentation.—
(1)APPLICABILITY.—
(a)Residential sales.—The real estate licensee disclosure requirements of this section and s. 475.278 apply to all residential sales. As used in this section, the term “residential sales” means the sale of improved residential property of four units or fewer, the sale of unimproved residential property intended for use of four units or less, or the sale of agricultural property of 10 acres or less.
(b)Disclosure limitations.—The real estate licensee disclosure requirements of this section and s. 475.278 do not apply to: nonresidential transactions; the rental or leasing of real property, unless an option to purchase all or a portion of the property improved with four or less residential units is given; auctions; appraisals; and dispositions of any interest in business enterprises or business opportunities, except for property with four or less residential units.
(2)NOTICE REQUIREMENT.—Unless otherwise exempted by this part, all real estate licensees are required to provide to any potential seller or buyer at first contact the notice of nonrepresentation as outlined in subsection (3), except in situations where a licensee knows that the potential seller or buyer is represented by a single agent or a transaction broker. If first contact between a licensee and a customer occurs during the course of a telephone conversation or any other communication in which the licensee is unable to provide the required notice of nonrepresentation, the licensee shall provide an oral notice and thereafter provide the required notice of nonrepresentation at the time of the first face-to-face contact, execution of a brokerage relationship agreement, or execution of a contractual agreement for purchase and sale, whichever occurs first.
(3)CONTENTS OF NOTICE.—
(a)Required information.—The notice required under subsection (2) must contain the following information:
NOTICE OF NONREPRESENTATION
FLORIDA LAW REQUIRES THAT REAL ESTATE LICENSEES PROVIDE THIS NOTICE AT FIRST CONTACT TO ALL POTENTIAL SELLERS AND BUYERS OF REAL ESTATE.
You are hereby notified that .................... (insert name of brokerage firm) and I do not represent you in any capacity. You should not assume that any real estate broker or salesperson represents you unless you agree to engage a real estate licensee in an authorized brokerage relationship, either as a single agent or as a transaction broker. You are advised not to disclose any information you want to be held in confidence until you make a decision on representation.
Your signature below acknowledges receipt of this form and does not establish a brokerage relationship.
..........
Date........................................
(Signature Optional)
........................................
(Signature Optional)
(b)Required format.—The notice required under subsection (2) must be printed as a separate and distinct form on paper no smaller than 8 1/2 inches by 11 inches. Nothing may be added to the form except a brokerage firm logo containing only the firm name, address, and relevant phone numbers. The form title and first sentence are to be in bold typeface of no less than 16-point type. The remainder of the form must be of 12-point type or larger.
475.278 Authorized brokerage relationships; required disclosures.—
(1)AUTHORIZED BROKERAGE RELATIONSHIPS.—A real estate licensee in this state may enter into a brokerage relationship as either a single agent or as a transaction broker with potential buyers and sellers. A real estate licensee may not operate as a disclosed or nondisclosed dual agent. As used in this section, the term “dual agent” means a broker who represents as a fiduciary both the prospective buyer and the prospective seller in a real estate transaction. Once a brokerage relationship is established, this part does not prevent a licensee from changing from one brokerage relationship to the other as long as the buyer or the seller, or both, gives consent as required by subparagraph (3)(c)2. before the change and the appropriate disclosure of duties as provided in this part is made to the buyer or seller. This part does not require a customer to enter into a brokerage relationship with any real estate licensee.
(2)TRANSACTION BROKER RELATIONSHIP.—
(a)Transaction broker - duties of limited representation.—A transaction broker provides a limited form of representation to a buyer, a seller, or both in a real estate transaction but does not represent either in a fiduciary capacity or as a single agent. The duties of the real estate licensee in this limited form of representation include the following:
1.Dealing honestly and fairly;
2.Accounting for all funds;
3.Using skill, care, and diligence in the transaction;
4.Disclosing all known facts that materially affect the value of real property and are not readily observable to the buyer;
5.Presenting all offers and counteroffers in a timely manner, unless a party has previously directed the licensee otherwise in writing;
6.Limited confidentiality, unless waived in writing by a party. This limited confidentiality will prevent disclosure that the seller will accept a price less than the asking or listed price, that the buyer will pay a price greater than the price submitted in a written offer, of the motivation of any party for selling or buying property, that a seller or buyer will agree to financing terms other than those offered, or of any other information requested by a party to remain confidential; and
7.Any additional duties that are mutually agreed to with a party.
(b)Disclosure requirements.—Duties of a transaction broker must be fully described and disclosed in writing to a buyer or seller either as a separate and distinct disclosure document or included as part of another document such as a listing agreement or agreement for representation. The disclosure must be made before, or at the time of, entering into a listing agreement or an agreement for representation. When incorporated into other documents the required notice must be of the same size type, or larger, as other provisions of the document and must be conspicuous in its placement so as to advise customers of the duties of limited representation, except that the first sentence of the information identified in paragraph (c) must be printed in uppercase and bold type.
(c)Contents of disclosure.—The required notice given under paragraph (b) must include the following information in the following form:
FLORIDA LAW REQUIRES THAT REAL ESTATE LICENSEES OPERATING AS TRANSACTION BROKERS DISCLOSE TO BUYERS AND SELLERS THEIR ROLE AND DUTIES IN PROVIDING A LIMITED FORM OF REPRESENTATION.
As a transaction broker, .................... (insert name of Real Estate Firm and its Associates), provides to you a limited form of representation that includes the following duties:
1.Dealing honestly and fairly;
2.Accounting for all funds;
3.Using skill, care, and diligence in the transaction;
4.Disclosing all known facts that materially affect the value of real property and are not readily observable to the buyer;
5.Presenting all offers and counteroffers in a timely manner, unless a party has previously directed the licensee otherwise in writing;
6.Limited confidentiality, unless waived in writing by a party. This limited confidentiality will prevent disclosure that the seller will accept a price less than the asking or listed price, that the buyer will pay a price greater than the price submitted in a written offer, of the motivation of any party for selling or buying property, that a seller or buyer will agree to financing terms other than those offered, or of any other information requested by a party to remain confidential; and
7.Any additional duties that are entered into by this or by separate written agreement.
Limited representation means that a buyer or seller is not responsible for the acts of the licensee. Additionally, parties are giving up their rights to the undivided loyalty of the licensee. This aspect of limited representation allows a licensee to facilitate a real estate transaction by assisting both the buyer and the seller, but a licensee will not work to represent one party to the detriment of the other party.
..........
Date........................................
Signature
........................................
Signature
(3)SINGLE AGENT RELATIONSHIP.—
(a)Single agent - duties.—The duties of a real estate licensee owed to a buyer or seller who engages the real estate licensee as a single agent include the following:
1.Dealing honestly and fairly;
2.Loyalty;
3.Confidentiality;
4.Obedience;
5.Full disclosure;
6.Accounting for all funds;
7.Skill, care, and diligence in the transaction; and
8.Presenting all offers and counteroffers in a timely manner, unless a party has previously directed the licensee otherwise in writing.
(b)Disclosure requirements.—
1.Single agent disclosure.—Duties of a single agent must be fully described and disclosed in writing to a buyer or seller either as a separate and distinct disclosure document or included as part of another document such as a listing agreement or other agreement for representation. The disclosure must be made before, or at the time of, entering into a listing agreement or an agreement for representation. When incorporated into other documents the required notice must be of the same size type, or larger, as other provisions of the document and must be conspicuous in its placement so as to advise customers of the duties of a single agent, except that the first sentence of the information identified in paragraph (c) must be printed in uppercase and bold type.
2.Transition to transaction broker disclosure.—A single agent relationship may be changed to a transaction broker relationship at any time during the relationship between an agent and principal, provided the agent gives the disclosure required under paragraph (2)(b) and the principal gives to the agent consent as required under subparagraph (c)2. before a change in relationship. This disclosure must be in writing to the principal either as a separate and distinct document or included as part of other documents such as a listing agreement or other agreements for representation. When incorporated into other documents the required notice must be of the same size type, or larger, as other provisions of the document and must be conspicuous in its placement so as to advise customers of the duties of limited representation, except that the first sentence of the information identified in subparagraph (c)2. must be printed in uppercase and bold type.
(c)Contents of disclosure.—
1.Single agent duties disclosure.—The notice required under subparagraph (b)1. must include the following information in the following form:
FLORIDA LAW REQUIRES THAT REAL ESTATE LICENSEES OPERATING AS SINGLE AGENTS DISCLOSE TO BUYERS AND SELLERS THEIR DUTIES.
As a single agent, .................... (insert name of Real Estate Entity and its Associates) owe to you the following duties:
1.Dealing honestly and fairly;
2.Loyalty;
3.Confidentiality;
4.Obedience;
5.Full disclosure;
6.Accounting for all funds;
7.Skill, care, and diligence in the transaction; and
8.Presenting all offers and counteroffers in a timely manner, unless a party has previously directed the licensee otherwise in writing.
..........
Date........................................
Signature
2.Transition disclosure.—The notice required under subparagraph (b)2. must include the following information in the following form as well as the information required in paragraph (2)(c):
FLORIDA LAW ALLOWS REAL ESTATE LICENSEES WHO REPRESENT A BUYER OR SELLER AS A SINGLE AGENT TO CHANGE FROM A SINGLE AGENT RELATIONSHIP TO A TRANSACTION BROKERAGE RELATIONSHIP IN ORDER FOR THE LICENSEE TO ASSIST BOTH PARTIES IN A REAL ESTATE TRANSACTION BY PROVIDING A LIMITED FORM OF REPRESENTATION TO BOTH THE BUYER AND THE SELLER. THIS CHANGE IN RELATIONSHIP CANNOT OCCUR WITHOUT YOUR PRIOR WRITTEN CONSENT.
..........I agree that my agent may assume the role and duties of a transaction broker. [must be initialed or signed]
(4)NO BROKERAGE RELATIONSHIP - DUTIES.—A real estate licensee owes to a customer with whom the licensee has no brokerage relationship the following duties:
(a)Dealing honestly and fairly;
(b)Disclosing all known facts that materially affect the value of the property which are not readily observable to the buyer; and
(c)Accounting for all funds entrusted to the licensee.
475.2801 Rules.—The commission may adopt rules establishing disciplinary guidelines, notices of noncompliance, and citations for violations of ss. 475.276 and 475.278.
Section 4.Section 475.5015, Florida Statutes, is amended to read:
475.5015 Brokerage business records.—Each broker shall keep and make available to the department such books, accounts, and records as will enable the department to determine whether such broker is in compliance with the provisions of this chapter. Each broker shall preserve at least one legible copy of all books, accounts, and records pertaining to his real estate brokerage business for at least 5 years from the date of receipt of any money, fund, deposit, check, or draft entrusted to the broker or, in the event no funds are entrusted to the broker, for at least 5 years from the date of execution by any party of any listing agreement, offer to purchase, rental property management agreement, rental or lease agreement, or any other written or verbal agreement which engages the services of the broker. If any brokerage record has been the subject of or has served as evidence for litigation, relevant books, accounts, and records must be retained for at least 2 years after the conclusion of the civil action or the conclusion of any appellate proceeding, whichever is later, but in no case less than a total of 5 years as set above. Disclosure documents required under ss. 475.276 and 475.278 shall be retained by the real estate licensee in all transactions that result in a written contract to purchase and sell real property.
Section 5.Subsection (7) of section 468.383, Florida Statutes, is amended to read:
468.383 Exemptions.—This act does not apply to the following:
(7)Auctions conducted as a part of the sale of real property by a real estate broker, as defined in s. 475.01(1)(a)~~s. 475.01(1)(c)~~.
Section 6.This act shall take effect October 1, 1997.
Approved by the Governor May 7, 1997.
Filed in Office Secretary of State May 7, 1997. |
6882 | https://astr.or.kr/DOIx.php?id=10.4174/jkss.2013.84.5.281 | :: astr.or.kr :: Annals of Surgical Treatment and Research
Open Access, Peer-reviewed
pISSN 2288-6575
eISSN 2288-6796
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Annals of Surgical Treatment and Research
About
About the Journal
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Abstract
INTRODUCTION
METHODS
RESULTS
DISCUSSION
Notes
References
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v.84(5);May 2013
10.4174/jkss.2013.84.5.281
J Korean Surg Soc. 2013 May;84(5):281-286. English.
Published online Apr 24, 2013.
Copyright © 2013, the Korean Surgical Society
Original Article
Afferent loop obstruction following laparoscopic distal gastrectomy with Billroth-II gastrojejunostomy
Dong Jin Kim,Jun Hyun Lee,1 and Wook Kim
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Department of Surgery, Yeouido St. Mary's Hospital, The Catholic University of Korea College of Medicine, Seoul, Korea.
1 Department of Surgery, Bucheon St. Mary's Hospital, The Catholic University of Korea College of Medicine, Bucheon, Korea.
Correspondence to: Wook Kim. Division of Gastrointestinal Surgery, Department of Surgery, Yeouido St. Mary's Hospital, The Catholic University of Korea College of Medicine, 10 63-ro, Yeongdeungpo-gu, Seoul 150-713, Korea. Tel: +82-2-3779-2020, Fax: +82-2-786-0802, Email: kimwook@catholic.ac.kr
Received December 24, 2012; Revised February 12, 2013; Accepted March 05, 2013.
Journal of the Korean Surgical Society is an Open Access Journal. All articles are distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
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AbstractINTRODUCTIONMETHODSRESULTSDISCUSSIONNotesReferences
Abstract
Purpose
Afferent loop (A-loop) obstruction is an uncommon postgastrectomy complication following Billroth-II (B-II) or Roux-en-Y reconstruction. Moreover, its development after laparoscopic gastrectomy has not been reported. Here we report 4 cases of A-loop obstructions after laparoscopic distal gastrectomy (LDG) with B-II reconstruction.
Methods
Among the 396 patients who underwent LDG with a B-II anastomosis between April 2004 and December 2011, 4 patients had A-loop obstruction. Their data were obtained from a prospectively maintained institutional database and analyzed for outcomes.
Results
Four patients (1.01%) developed A-loop obstruction. All were male, and their median age was 52 years (range, 30 to 73 years). The interval between the initial gastrectomies and the operation for A-loop obstruction ranged from 4 to 540 days (median, 33 days). All 4 patients had symptoms of vomiting and abdominal pain and were diagnosed by abdominal computed tomographic (CT) scan. The causes of the A-loop obstructions were adhesions (2 cases) and internal herniations (2 cases) that were treated with Braun anastomoses and reduction of the herniated small bowels, respectively. All patients recovered following the emergency operations.
Conclusion
A-loop obstruction is a rare but serious complication following laparoscopic and open gastrectomy. It should be considered when a patient complains of continuous abdominal pain and/or vomiting after LDG with B-II reconstruction. Prompt CT scan may play an important role in diagnosis and treatment.
Keywords
Afferent loop obstruction; Billroth-II operation; Laparoscopy; Ileus
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AbstractINTRODUCTIONMETHODSRESULTSDISCUSSIONNotesReferences
INTRODUCTION
The incidence of afferent loop (A-loop) obstruction after Billroth-II (B-II) distal gastrectomy has been reported to be 0.3-1.0% . It may be caused by adhesions, kinking, stenosis or internal herniation after open or laparoscopic gastrectomy. A-loop obstruction develops acutely, and its onset can be early or late postoperatively . In some patients, the condition can be rapidly complicated by the development of duodenal stump leakage or perforation followed by peritonitis and can result in sepsis [3, 4]. Therefore, early diagnosis and immediate surgical therapy are mandatory to decrease the high mortality rate associated with this complication. Making this diagnosis is important not only for open gastrectomy but also for laparoscopic gastrectomy. While a number of laparoscopic gastrectomies have been markedly increased, there is no effort to evaluate characteristics of A-loop obstruction following laparoscopic distal gastrectomy (LDG) with a B-II anastomosis. The aim of this study was to elucidate the causes and treatment of A-loop obstruction after LDG with B-II reconstruction for gastric cancer.
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AbstractINTRODUCTIONMETHODSRESULTSDISCUSSIONNotesReferences
METHODS
We retrospectively reviewed a surgical database of 396 gastric cancer patients who underwent LDG with a B-II anastomosis between April 2004 and December 2011. We evaluated the patients' clinicopathologic characteristics and operative findings. The anastomoses were performed by either extracorporeal or intracorporeal methods that were described previously . All anastomoses were performed in the B-II gastrojejunostomy, antecolic and antiperistaltic fashion with 10-15 cm of the usual length of the afferent limb.
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AbstractINTRODUCTIONMETHODSRESULTSDISCUSSIONNotesReferences
RESULTS
Of the 396 patients, 4 (1.01%) developed A-loop obstruction (Table 1). The median age of the patients, each of whom was male, was 45 years old (range, 30 to 73 years). Two of the four patients had advanced gastric cancer, and the remaining two were in the early stage of the disease. All patients had undergone curative resection of their gastric cancer. One patient underwent extracorporeal anastomosis, and the other three underwent intracorporeal anastomoses. All patients were symptomatic with abdominal pain with or without nausea, vomiting and fever. A definitive diagnosis was made by computed tomographic (CT) scan in all cases. The serum amylase level was elevated in two patients, but the serum total bilirubin level did not increase in all patients. The cause of the A-loop obstruction was an adhesion around the afferent limb side of the gastrojejunostomy in cases 1 and 2 (Fig. 1A, B) and internal herniation in the other two patients (Fig. 1C). The internal herniation occurred at the mesenteric gap between the mesocolon and gastrojejunostomy anastomosis. The intervals between the initial gastric resections and the diagnosis of A-loop obstruction ranged from 4 to 540 days. All patients underwent emergency operations before the development of complications such as duodenal stump leakage or peritonitis. The two patients with adhesions underwent adhesiolysis with a Braun anastomosis to avoid rupture of the dilated afferent limb. Two patients with internal herniation underwent reduction of the herniated small bowel into normal position, one via a laparoscopic method and the other via open method. None of the patients died due to the A-loop obstruction, and all four patients are still alive with no evidence of recurrence through the follow-up periods (median, 37 months; range, 18 to 51 months).
Fig. 1
(A) Illustration of afferent obstruction in case 1. (B) Illustration of afferent obstruction in case 2. (C) Illustration of internal herniation in cases 3 and 4.
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Table 1
Characteristics of patients with afferent loop obstructions following laparoscopic distal gastrectomy with Billroth-II reconstruction
Click for larger image
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AbstractINTRODUCTIONMETHODSRESULTSDISCUSSIONNotesReferences
DISCUSSION
According to a nation-wide survey by the Korean Gastric Cancer Association, the proportion of early gastric cancer (EGC) in Korea increased from 47.4% in 2004 to 57.7% in 2009 due to the development of diagnostic instruments as well as increased use of mass and individual screening. Additionally, the use of LDG for EGC has been increasing significantly [6, 7]. In terms of anastomoses, the Billroth I anastomosis is increasingly performed, but the B-II remains one of the main reconstructive options following distal gastrectomy for gastric cancer. The proportions of the reconstruction types following distal gastrectomy in 2009 were 63.4%, 33.1% and 3.3% for B-I, B-II and R-Y, respectively [6, 7]. A-loop obstruction is a possible and important complication following LDG despite its rarity. This report has meaningful value in terms of the incidence of A-loop obstruction in LDG with B-II because the number of laparoscopic gastrectomies has increased and will continue to do so.
A-loop obstruction occurred in 0.3-1.0% as a result of partial or complete obstruction of the afferent limb following B-II or R-Y anastomosis and may be due to adhesions, kinking, internal herniation or stenosis caused by inflammatory changes or malignancy [1, 8]. Because of the relatively low incidence of A-loop obstruction, routine Braun anastomosis for prevention of A-loop obstruction is controversial. We performed it selectively, not in a routine manner.
In our series, the incidence of A-loop obstruction following LDG was similar with previously reported open surgery. We cannot find definite differences between open and laparoscopic gastrectomy regarding the cause of A-loop obstruction. However, there were several specific predisposing factors in A-loop obstruction following LDG such as partial omentectomy and antecolic anastomosis, as found through review of our series. In patient 1, who underwent an extracorporeal anastomosis, an adhesion developed around the proximal afferent limb with remnant omentum (Fig. 1A). Because partial omentectomies were widely performed with laparoscopic gastrectomy, this type of case was not reported in the open surgical group. The other three cases underwent intracorporeal anastomoses; the obstruction in case 2 was caused by an adhesion (Fig. 1B), and the other two were caused by internal herniations (Fig. 1C).
Internal herniation through the jejunal mesenteric gap is a specific morbidity associated with Roux-en-Y reconstruction. This hernia causes twisting of the jejunojejunostomy, thereby causing afferent loop obstruction . To prevent internal herniation, these mesenteric gaps and defects should be tightly closed with stitches. Internal herniation at Petersen's space has also been recognized as a cause of afferent loop obstruction after B-II gastrectomy. In our series, two of the four patients with afferent loop obstructions developed obstructions by this type of internal herniation at the mesenteric gap in the region of the gastrojejunostomy. Therefore, placing a routine stitch to close the mesenteric gap between the afferent limb and mesocolon is recommended to prevent internal herniation (Fig. 2).
Fig. 2
(A) Stitch between the afferent limb and transverse mesocolon. (B) Closure of mesenteric gap following Billroth-II gastrojejunostomy. S, remnant stomach; A, afferent limb; E, efferent limb.
Click for larger image
The intervals between the initial and emergency operations for A-loop obstructions in our series ranged from 4 days to 1.5 years. In our series, adhesions tended to develop early (between 4 and 12 days); otherwise, internal herniation caused late onset of A-loop obstruction (between 53 and 540 days). In the literature, the interval ranges from 3 months and 14 years, regardless of cause [9-11]. Hence, this complication can occur at any time during the postoperative period.
The clinical diagnosis of afferent loop obstruction can be difficult because the symptoms are nonspecific. The usual symptoms are abdominal pain, nausea and vomiting, which occur commonly in any case of small bowel obstruction. Vomitus not containing bile, which suggests complete obstruction of the afferent limb, may be helpful to distinguish between A-loop obstruction and ordinary small bowel obstruction. Stagnation of pancreatic and bile juice in the distended duodenum may lead to elevation of the serum amylase level, which is a frequent finding with acute pancreatitis and can result in an erroneous diagnosis . Elevated serum amylase was observed in only 1 case in this series. Jaundice due to A-loop obstruction is rare. Plain abdominal radiography is not helpful for making the diagnosis because the afferent jejunal limb is fluid-filled and gasless. In case 1 of this series, the patient suffered from abdominal pain and vomiting. However, plain abdominal X-ray showed no definite dilatation of the duodenal c-loop. We therefore treated him using routine conservative treatments of nasogastric tube decompression with fluid and electrolyte supplement. Verification by CT scan was delayed until his symptoms were aggravated. Although a gastrograffin meal series may suggest the diagnosis because of poor filling or nonfilling of the afferent jejunal limb, CT scan is the most useful imaging modality for diagnosing A-loop obstruction. Characteristically, the dilated afferent limb is found in the midabdomen (Fig. 3A), and it is typically located in the horizontal part of the duodenum, placed between the aorta and the superior mesenteric vessels and displacing these vessels ventrally. In the case of internal herniation, a whirling appearance of the mesenteric vessels can typically be observed (Fig. 3B).
Fig. 3
(A) Markedly dilated duodenal c-loop. (B) A whirling appearance of the mesenteric vessels, suggestive of internal herniation through Petersen's space.
Click for larger image
The treatment of choice for A-loop obstructions depends on the cause of obstruction and the length of the afferent limb. A Braun anastomosis is the treatment of choice in cases of severe adhesions and for manual reduction of the herniated small bowel with closure of the mesenteric gap in cases of internal herniation. However, percutaneous transhepatic metal stents with a double-pigtail catheter or endoscopic management can be performed in cases of A-loop obstructions due to cancer recurrence or pancreatitis [13-15]. In our results, there were no candidates for these alternative methods. The reported mortality rate associated with A-loop obstructions is high (30-60%) [4, 16]. Most of these high mortality rates were reported before the development of CT scan or ultrasonography. Delayed diagnosis of afferent loop obstruction may lead to severe complications including duodenal stump leakage and perforation or necrosis of the entire afferent limb, resulting in sepsis and death. In our series, there were no deaths because of our prompt surgical treatment following early diagnosis of the disease by CT scan.
In conclusion, as in an open B-II gastrectomy, the incidence of afferent loop obstruction following LDG with B-II reconstruction was quite low, at 1.01% in our series. When a patient develops progressive abdominal pain and vomiting after LDG with B-II reconstruction with nonspecific findings on plain radiographs with or without an elevated serum amylase, abdominal CT scan should be performed immediately to prevent further disease progression.
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AbstractINTRODUCTIONMETHODSRESULTSDISCUSSIONNotesReferences
Notes
No potential conflict of interest relevant to this article was reported.
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AbstractINTRODUCTIONMETHODSRESULTSDISCUSSIONNotesReferences
References
Aoki M, Saka M, Morita S, Fukagawa T, Katai H. Afferent loop obstruction after distal gastrectomy with Roux-en-Y reconstruction. World J Surg 2010;34:2389–2392.
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6883 | https://artofproblemsolving.com/downloads/printable_post_collections/5158?srsltid=AfmBOoohAbySNV5d1k6QqYTepv2EjgvEwYHARw81_AQY82NdDdxgRv_5 | AoPS Community 1998 All-Russian Olympiad
All-Russian Olympiad 1998
www.artofproblemsolving.com/community/c5158 by v Enhance, behdad.math.math, WakeUp, greentreeroad, JohnPeanuts, stef mol, BaBaK Ghalebi
– Grade level 9
Day 1 1 The angle formed by the rays y = x and y = 2 x (x ≥ 0) cuts off two arcs from a given parabola
y = x2 + px + q. Prove that the projection of one arc onto the x-axis is shorter by 1 than that of the second arc.
2 A convex polygon is partitioned into parallelograms. A vertex of the polygon is called good if it belongs to exactly one parallelogram. Prove that there are more than two good vertices.
3 Let S(x) denote the sum of the decimal digits of x. Do there exist natural numbers a, b, c such that
S(a + b) < 5, S(b + c) < 5, S(c + a) < 5, S(a + b + c) > 50?
4 A maze is an 8 × 8 board with some adjacent squares separated by walls, so that any two squares can be connected by a path not meeting any wall. Given a command LEFT, RIGHT, UP, DOWN, a pawn makes a step in the corresponding direction unless it encounters a wall or an edge of the chessboard. God writes a program consisting of a finite sequence of commands and gives it to the Devil, who then constructs a maze and places the pawn on one of the squares. Can God write a program which guarantees the pawn will visit every square despite the Devil’s efforts?
Day 2 5 We are given five watches which can be winded forward. What is the smallest sum of winding intervals which allows us to set them to the same time, no matter how they were set initially?
6 In triangle ABC with AB > BC , BM is a median and BL is an angle bisector. The line through
M and parallel to AB intersects BL at point D, and the line through L and parallel to BC
intersects BM at point E. Prove that ED is perpendicular to BL .
7 A jeweller makes a chain consisting of N > 3 numbered links. A querulous customer then asks him to change the order of the links, in such a way that the number of links the jeweller must open is maximized. What is the maximum number?
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1AoPS Community 1998 All-Russian Olympiad
8 Two distinct positive integers a, b are written on the board. The smaller of them is erased and replaced with the number ab
|a−b|
. This process is repeated as long as the two numbers are not equal. Prove that eventually the two numbers on the board will be equal.
– Grade level 10
Day 1 1 Two lines parallel to the x-axis cut the graph of y = ax 3 + bx 2 + cx + d in points A, C, E and
B, D, F respectively, in that order from left to right. Prove that the length of the projection of the segment CD onto the x-axis equals the sum of the lengths of the projections of AB and
EF .
2 Two polygons are given on the plane. Assume that the distance between any two vertices of the same polygon is at most 1, and that the distance between any two vertices of different polygons is at least 1/√2. Prove that these two polygons have no common interior points. By the way, can two sides of a polygon intersect?
3 In scalene 4ABC , the tangent from the foot of the bisector of ∠A to the incircle of 4ABC ,other than the line BC , meets the incircle at point Ka. Points Kb and Kc are analogously de-fined. Prove that the lines connecting Ka, Kb, Kc with the midpoints of BC , CA , AB , respec-tively, have a common point on the incircle.
4 Let k be a positive integer. Some of the 2k-element subsets of a given set are marked. Sup-pose that for any subset of cardinality less than or equal to (k + 1) 2 all the marked subsets contained in it (if any) have a common element. Show that all the marked subsets have a common element.
Day 2 5 Initially the numbers 19 and 98 are written on a board. Every minute, each of the two numbers is either squared or increased by 1. Is it possible to obtain two equal numbers at some time?
6 A binary operation ∗ on real numbers has the property that (a ∗ b) ∗ c = a + b + c for all a, b, c.Prove that a ∗ b = a + b.
7 Let n be an integer at least 4. In a convex n-gon, there is NO four vertices lie on a same circle. A circle is called circumscribed if it passes through 3 vertices of the n-gon and contains all other vertices. A circumscribed circle is called boundary if it passes through 3 consecutive vertices, a circumscribed circle is called inner if it passes through 3 pairwise non-consecutive points. Prove the number of boundary circles is 2 more than the number of inner circles.
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2AoPS Community 1998 All-Russian Olympiad
8 Each square of a (2 n − 1) × (2 n − 1) board contains either 1 or −1. Such an arrangement is called successful if each number is the product of its neighbors. Find the number of successful arrangements.
– Grade level 11
Day 1 1 Two lines parallel to the x-axis cut the graph of y = ax 3 + bx 2 + cx + d in points A, C, E and
B, D, F respectively, in that order from left to right. Prove that the length of the projection of the segment CD onto the x-axis equals the sum of the lengths of the projections of AB and
EF .
2 Let ABC be a triangle with circumcircle w. Let D be the midpoint of arc BC that contains A.Define E and F similarly. Let the incircle of ABC touches BC, CA, AB at K, L, M respectively. Prove that DK, EL, F M are concurrent.
3 A set S of translates of an equilateral triangle is given in the plane, and any two have nonempty intersection. Prove that there exist three points such that every triangle in S contains one of these points.
4 yeah you’re right,the official problem is the following one:
there are 1998 cities in Russia, each being connected (in both directions) by flights to three other cities. any city can be reached by any other city by a sequence of flights. the KGB plans to close off 200 cities, no two joined by a single flight. show that this can be done so that any open city can be reached from any other open city by a sequence of flights only passing through open cities.
we begin with some terminology. define a trigraph to be a connected undirected graph in which every vertex has degree at most 3. a trivalent vertex of such a graph is a vertex of degree 3.in this wording, the problem becomes: we have a trigraph G with 1998 vertices, all of which are trivalent. we want to remove 200 vertices, no two of which are adjacent, such that the remaining vertices stay connected. we remove the vertices one at a time. suppose we have deleted k of the 1998 vertices, no two of which are adjacent, such the trigraph G′ induced by the remaining vertices is connected. we will show that if K < 200 , we can always delete a trivalent vertex of G′ such that the graph remains connected. this vertex cannot be adjacent in G to any of the other k deleted vertices, because then its degree in G′ would be less than 3. hence repeating this 200 times gives us the desired set of vertices.
Lemma. let G be a trigraph such that the removal of any trivalent vertex disconnects G. then G
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3AoPS Community 1998 All-Russian Olympiad
is planer. moreover G can be drawn in such a way that every vertex lies on the ”outside” face; in other words, for any point P outside some bounded set, each vertex v of G can be joined to
P by a curve which does not intersect any edges of G (except at v).
Proof. we induct on the number of trivalent vertices of G. if G does not contain any trivalent vertices, then G must be a path or a cycle and the claim is obvious. so suppose G contains
n ≥ 1 trivalent vertices and that every trigraph with fewer trivalent vertices can be drawn as described. if G is a tree the claim is obvious, so suppose G contains a cycle; let v1, . . . , v k
(k ≥ 3) be a minimal cycle. let S = {v1, . . . , v k}, and let T = {i | vi is trivalent }. ( T cannot be empty, because then no vi would be connected to a vertex of degree 3.) for each i ∈ T , let wi
denote the third vertex which is adjacent to vi (other than vi−1 and vi+1 ), and let Si be the set of vertices in G which can be reached from wi without passing though S. (for i 6 ∈ T , let Si = ∅.) we claim that the sets S, S 1, . . . , S k partition the vertices of G. first, note that if v is a vertex of
G not in S then there is a shortest path joining v to a vertex vi of S; the penultimate vertex on this path must be wi, so v ∈ Si. now suppose that v ∈ Si ∩ Sj for some i 6 = j; then vi and vj are trivalent and there exist paths wi → v, wj → v which do not pass though S. we will show that there exists a path from every vertex of G − { vi} to v which does not pass though vi. for k 6 = i
there is a path vk → vj → wj → v; if w ∈ Sk for k 6 = i, then there is a path w → wk → vk → v;if w ∈ Si, there is a path w → wi → v. since S ∪ S1 ∪ . . . ∪ Sk = G, we have shown that the graph obtained from G by deleting vi is connected,a contradiction, as vi is trivalent. therefore
Si ∩ Sj = ∅ for i 6 = j. obviously Si ∩ S is empty for all i; hence S, S 1, . . . , S k partition the vertices of G. let G′, G 1, . . . , G k be the induced subgraphs of S, S 1, . . . , S k in G, respectively. by construction, the only edges in G which are not in one of the graphs G′, G 1, . . . , G k are the edges viwi for i ∈ T . now Gi is a trigraph with fewer than n trivalent vertices, since at least one of the n trivalent vertices in G is in S. hence by the inductive hypothesis, we can draw each
Gi in the plane in such a way that every vertex lies on the outside face. since v1, . . . , v k was a minimal cycle, there are no ”extra” edges between these vertices, so the graph G′ is a k-cycle. now place the vertices of S at the vertices of a small regular k-gon far from all the graphs Gi;then we can draw a curve joining each pair vi, w i. it is east to check that this gives us a drawing of G with the desired properties.
now suppose we have removed k vertices from G, no two of which are adjacent, such that the trigraph G′ induced by the remaining vertices is connected, and suppose that removing any trivalent vertex of G′ disconnects the graph; we must show k ≥ 200 . by the lemma, G′ is planner. we will call a dace other than the outside one a ”proper face”. let F be the number of proper faces of G′; since G′ has 1998 − k vertices and 2997 − 3k edges,
F ≥ 1 − (1998 − k) + (2997 − 3k) = 1000 − 2k.we now show that no two proper faces can share a vertex. observe that each vertex belongs to at most as many faces as its degree; thus vertices of degree 1 lie only on the outside face. no two proper faces can intersect in a vertex of degree 2, or that vertex would not lie on the outside face, contracting the lemma. if two proper daces intersected in a trivalent vertex v, each face would give a path between two of v’s neighbors, so removing v would not disconnect the graph,
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4AoPS Community 1998 All-Russian Olympiad
by an argument similar to that of the lemma. since each proper face contains at least 3 vertices and no two share a vertex, we have 3F ≤
1998 − k. combining this with the previous inequality gives
3000 − 6k ≤ 3F ≤ 1998 − k
so 1002 ≤ 5k and k ≥ 200 , as desired.
Day 2 5 A sequence of distinct circles ω1, ω 2, · · · is inscribed in the parabola y = x2 so that ωn and ωn+1
are tangent for all n. If ω1 has diameter 1 and touches the parabola at (0 , 0) , find the diameter of ω1998 .
6 Are there 1998 different positive integers, the product of any two being divisible by the square of their difference?
7 A tetrahedron ABCD has all edges of length less than 100 , and contains two nonintersecting spheres of diameter 1. Prove that it contains a sphere of diameter 1.01 .
8 A figure Φ composed of unit squares has the following property: if the squares of an m ×
n rectangle ( m, n are fixed) are filled with numbers whose sum is positive, the figure Φ can be placed within the rectangle (possibly after being rotated) so that the sum of the covered numbers is also positive. Prove that a number of such figures can be put on the m×n rectangle so that each square is covered by the same number of figures.
©2019 AoPS Incorporated
5 |
6884 | https://www.nutrient.io/blog/process-mapping-examples/ | Workflow
15 process map examples: Complete guide (2025)
Jonathan D. Rhyne
Updated: August 22, 2025
Table of contents
What is process mapping?
15 process map examples
1. Flowchart process maps
2. Swimlane process maps
3. SIPOC process maps
4. Value stream maps
5. Customer journey maps
6. Service blueprint process maps
7. Cross-functional process maps
8. Decision tree process maps
9. Kanban board process maps
10. RACI chart process maps
11. Gantt chart process maps
12. Process flow diagrams
13. Workflow automation maps
14. Business process model diagrams
15. Integration flow maps
Process map examples by industry
Process mapping tools comparison
Process mapping best practices
Common process mapping mistakes to avoid
Create better business process maps with Nutrient Workflow
FAQ
TL;DR
Explore 15 real-world process map examples across healthcare, manufacturing, IT, finance, and more
Read a complete comparison of tools and when to use each process mapping method
Whether it’s a small startup or a worldwide enterprise, every organization needs established business processes to operate effectively. However, if you don’t clearly understand how your methods work, it’ll be hard to figure out where things are going wrong and where you can improve. For that reason, investing in business process mapping and workflow automation software is critical to keeping up with the intricacies of various workflows that support your organization.
What is process mapping?
A business process map outlines the functions required to complete a process. For example, your HR department may have process map examples that cover employee onboarding workflows and ensure proper documentation completion. These business process map examples serve as visual frameworks that demonstrate how core functions operate within your organization.
Similarly, your accounting department may have specific workflow process maps for invoice processing and accounts payable management that cover every step — from invoice receipt to payment completion — in their accounting systems.
Having business process maps in place benefits your company in many ways, including:
Removing any confusion about the state of your current business processes
Allowing the department to follow a consistent blueprint for getting things done
Helping you spot redundancies in different process steps
Facilitating communications between business areas
Helping you remain in compliance with the legislation applicable to your industry
There are several types of process maps, each with a unique view or application. Most business analysts start with a high-level process map and get more granular during interviews with each team member. They also use mapping tools to version and track their process maps over time to provide historical context. This is especially helpful with complex processes that need to show a high level of detail and identify areas across the organization that are involved.
15 process map examples
The process map examples below demonstrate how different visualization techniques can improve workflow efficiency across your organization. Each process map type serves specific purposes and complexity levels. But before sharing, the following sections will provide an overview of important terminology.
Activity/process
An activity or process represents a step or an activity within a process. A rectangle shape typically represents it.
Flow
The flow shape (sometimes called a process mapping symbol) is typically a line with an arrow on the end. A flow highlights the sequence of execution within a process for the viewer.
Event
An event is what starts, changes, or finishes a process. The events you might want to make visible include errors, messages, cancellations, or links. Events are represented by a circle that contains other symbols, depending on the event type.
Decision
The decision shape, represented by a diamond, indicates that the process must decide moving forward. It can be a binary yes/no decision, or something more complex with multiple choices, like a case statement in a computer program. You should try and capture all options that might impact the flow of a process.
These symbols form the foundation of all process map examples and ensure consistent communication across teams.
1. Flowchart process maps
Flowcharts are the most basic type of process flow. They contain symbols that outline the flow of steps within a process in sequential order. One of the biggest benefits of flowcharts is that you can adapt them to fit pretty much any function, like manufacturing, administration, or even project planning.
The best time to use a flowchart is when you want to provide an easy-to-understand example of how a process works. Flowcharts make spotting where you want to change or improve processes easy. In addition, it’s easier for non-technical users to understand flowcharts versus other business mapping examples.
When creating a flowchart, you should:
Focus on making the process transparent versus making it look perfect.
Capture the key personnel involved in a process, like customers and vendors.
Get the input of individuals involved in completing a process within the business.
2. Swimlane process maps
Swimlane process maps are similar to flowcharts, in that they visually map out the steps of a process. The most significant difference between the two is that swimlane maps line up in lanes that let the viewer know who’s responsible for completing each step, like a department or automation process.
Here’s what to keep in mind when putting together a swimlane process map:
Where are you getting your information from?
Is there anything that might interrupt the swimlane flow?
Does a task need to be assigned to a different person, department, or process?
Will you need to wait for a decision from someone else before moving to the next step?
Will the process require data input to keep the process going?
Will you need to transform the input to a different format?
3. SIPOC process maps
The SIPOC process map is used to outline the intent of a process, along with the key players involved. The acronym SIPOC stands for:
Supplier — The person or organization providing input to the process.
Input — The resource that the supplier adds to the process.
Process — The series of steps involved in converting input into output.
Output — The resource produced by the process.
Customer — The person or entity that receives the output.
SIPOC process maps typically consist of six to eight total steps. It’s an ideal tool for project managers who must provide a high-level view of the functions involved in a process. A SIPOC diagram should convey:
The start and end of the process
Those affected by the process
What goes into the process
The steps involved in completing the process
Creating a SIPOC business process map shows stakeholders the information they need about each step and any dependencies.
4. Value stream maps
A value stream map (VSM) is a detailed flowchart that documents the individual steps within a process, along with the information flow and related data inputs. It may be easier to think of it as a more detailed SIPOC process map. VSMs are useful when you need to show how you can add value to a current process by making functional changes.
VSMs are great for locating and eliminating redundant steps within a business process. For example, you may want to create a VSM of a process’s current state, and then create a future state VSM that shows how it would work with your proposed improvements.
5. Customer journey maps
Customer journey maps visualize the complete experience a customer has with your organization. These process map examples track every touchpoint — from awareness to post-purchase support — making them essential for e-commerce and service businesses.
Key components of customer journey process maps include:
Customer actions and decisions at each stage
Emotional states and pain points
Channels and touchpoints used
Opportunities for improvement
6. Service blueprint process maps
Service blueprints extend customer journey maps by adding internal processes. These process map examples show both customer-facing interactions and behind-the-scenes activities that support service delivery.
Service blueprints are ideal for:
Restaurant service delivery optimization
Healthcare patient experience improvement
Banking and financial services
Hotel and hospitality operations
7. Cross-functional process maps
Cross-functional maps show how processes flow between different departments or teams. These workflow process maps are crucial for product launches, project management, and organizational change initiatives.
Benefits include:
Clear accountability across departments
Identification of handoff points
Communication improvement between teams
Bottleneck identification and resolution
8. Decision tree process maps
Decision trees map out complex decision-making processes with multiple paths and outcomes. These process map examples are perfect for customer support escalation, troubleshooting procedures, and policy compliance.
Use decision trees when:
Multiple decision points exist in a process
Different outcomes require different actions
Training teams on complex procedures
Automating decision-making workflows
9. Kanban board process maps
Kanban boards visualize work flowing through different stages of completion. These process map examples originated in manufacturing, but are now widely used in software development, marketing, and project management.
Kanban process maps show:
Work items in progress
Workflow stages and transitions
Bottlenecks and capacity constraints
Team workload distribution
10. RACI chart process maps
RACI — which stands for Responsible, Accountable, Consulted, Informed — is a kind of chart that clarifies roles and responsibilities for each process step. These process map examples prevent confusion and ensure accountability in team-based workflows.
RACI process maps define:
Who is responsible for completing tasks
Who is accountable for outcomes
Who needs to be consulted
Who should be kept informed
11. Gantt chart process maps
Gantt charts show project timelines and dependencies between tasks. These process map examples are essential for project management and help teams understand sequence and timing requirements.
Gantt process maps include:
Task durations and deadlines
Dependencies between activities
Resource allocation
Critical path identification
12. Process flow diagrams
Process flow diagrams show the sequence of operations in manufacturing and production environments. These business process map examples help optimize production efficiency and quality control.
Common applications:
Manufacturing quality control
Chemical processing
Food production
Assembly line operations
13. Workflow automation maps
Workflow automation maps show how manual processes can be digitized and automated. These process map examples identify automation opportunities and design automated workflows.
Automation maps highlight:
Manual vs. automated tasks
System integrations required
Data flow between systems
Human intervention points
14. Business process model diagrams
Business process model diagrams provide detailed documentation for compliance and training purposes. These process map examples use standardized notation (BPMN) for professional process documentation.
Key features:
Standardized symbols and notation
Exception handling and error flows
System and human interactions
Compliance documentation
15. Integration flow maps
Integration flow maps show how data and information move between different systems and applications. These workflow process maps are critical for IT departments and digital transformation projects.
Integration maps document:
System connections and APIs
Data transformation requirements
Security and authentication points
Error handling and monitoring
Step-by-step process mapping guide
Step 1 — Define your process scope
Identify the process start and end points
Determine which departments are involved
List the main stakeholders
Step 2 — Choose the right process map type
Simple processes: Use flowcharts
Cross-departmental processes: Use swimlane maps
High-level overview: Use SIPOC diagrams
Process improvement: Use value stream maps
Step 3 — Gather information
Interview process participants
Observe the process in action
Document current pain points and bottlenecks
Step 4 — Create your process map
Start with the big picture
Add detail incrementally
Use consistent symbols and notation
Include decision points and exception flows
Step 5 — Validate and refine
Review with process participants
Test the map with real scenarios
Update based on feedback
Document assumptions and constraints
Need help getting started? Our process mapping consultation service provides expert guidance for complex workflows.
Process map examples by industry
Different industries benefit from specific types of business process map examples. Here’s how various sectors apply these workflow process maps to improve operations:
Healthcare process map examples
Patient intake workflow
Registration and insurance verification
Medical history collection
Appointment scheduling
Room assignment process
Medical record management
Chart creation and updates
Provider access controls
Billing information integration
Compliance documentation
Manufacturing process map examples
Quality control workflow
Inspection checkpoints
Defect tracking and reporting
Corrective action procedures
Supplier quality management
Equipment maintenance process
Preventive maintenance scheduling
Breakdown response procedures
Parts inventory management
Performance tracking systems
IT process map examples
Incident response procedure
Alert detection and classification
Escalation pathways
Resolution tracking
Post-incident review process
Software deployment workflow
Code review and testing
Environment promotion
Deployment validation
Rollback procedures
Finance process map examples
Expense approval process
Expense submission workflow
Manager approval chains
Finance review procedures
Payment processing systems
Invoice processing system
Vendor invoice receipt
Approval routing workflow
Payment authorization
Accounting system integration
HR process map examples
Employee onboarding journey
Preboarding preparation
First-day orientation process
Training program workflow
Performance evaluation setup
Performance review process
Goal setting and tracking
Feedback collection workflow
Review meeting scheduling
Development planning process
Sales process map examples
Lead qualification funnel
Lead source tracking
Qualification criteria application
Follow-up scheduling
CRM data management
Contract negotiation workflow
Proposal generation process
Legal review procedures
Customer feedback integration
Final approval workflow
Process mapping tools comparison
Interactive demo available: Try our workflow builder to create your first process map in under five minutes with Nutrient Workflow’s drag-and-drop interface.
| Feature | Nutrient Workflow | Traditional tools | Generic solutions |
--- --- |
| Ease of use | Drag-and-drop builder | Complex setup required | Moderate learning curve |
| Collaboration | Real-time collaboration | Email-based sharing | Limited collaboration |
| Automation | Built-in workflow automation | Manual process only | Basic automation |
| Templates | Industry-specific templates | Generic templates | Limited templates |
| Integration | API and system integrations | Manual data entry | Basic integrations |
| Scalability | Enterprise-ready | Limited scalability | Variable scalability |
| Cost | Competitive pricing | High licensing fees | Variable pricing |
Process mapping best practices
Start with high-level process overview before diving into details.
Include all stakeholders in mapping sessions for complete accuracy.
Use consistent symbols and notation throughout all process maps.
Validate process maps with actual process participants.
Update maps regularly as processes evolve and improve.
Document assumptions and constraints clearly.
Test process maps with real scenarios before implementation.
Focus on value-added steps and eliminate waste.
Common process mapping mistakes to avoid
Creating process maps in isolation without stakeholder input
Over-complicating simple processes with unnecessary detail
Missing exception flows and error-handling procedures
Using inconsistent symbols across different process maps
Failing to update maps when processes change
Not validating maps with actual process users
Focusing only on current state without considering improvements
Ignoring system and technology constraints in process design
Create better business process maps with Nutrient Workflow
Transform your organization’s efficiency with Nutrient Workflow’s comprehensive process mapping and workflow automation platform. Our intuitive drag-and-drop tools make it easy to create, share, and optimize process maps across your entire organization.
Key benefits
Visual process designer — Create professional process maps in minutes
Real-time collaboration — Work together on process improvements
Automation integration — Turn process maps into automated workflows
Industry templates — Start with proven process map examples
Analytics dashboard — Track process performance and identify bottlenecks
Ready to optimize your workflows? Contact Nutrient Workflow today for a demo of our complete process mapping and automation platform.
FAQ
What is the purpose of business process mapping?
Business process mapping helps visualize the steps involved in a process, enabling organizations to identify inefficiencies, redundancies, and areas for improvement.
How can I choose the right process mapping method?
The right method depends on the complexity of the process. For simple processes, flowcharts may be sufficient, while more complex processes may benefit from swimlane or value stream maps.
Why is it important to include key personnel in a process map?
Including key personnel ensures all stakeholders are aware of their responsibilities and helps identify potential bottlenecks or delays in the process.
Can process maps be used for all types of organizations?
Yes. Process maps can be adapted for any organization, regardless of size or industry, to improve workflow, communication, and efficiency.
How often should business process maps be updated?
Business process maps should be updated regularly to reflect changes in workflows, organizational structure, or technology.
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6885 | https://fiveable.me/key-terms/hs-honors-geometry/pythagorean-triple | Pythagorean triple - (Honors Geometry) - Vocab, Definition, Explanations | Fiveable | Fiveable
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Pythagorean triple
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key term - Pythagorean triple
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Definition
A pythagorean triple consists of three positive integers a, b, and c that satisfy the equation $$a^2 + b^2 = c^2$$. This relationship is essential in understanding right triangles, where 'c' represents the length of the hypotenuse while 'a' and 'b' are the lengths of the other two sides. Recognizing pythagorean triples helps in solving problems related to the Pythagorean Theorem and its converse, establishing whether a triangle is right-angled based on its side lengths.
5 Must Know Facts For Your Next Test
Common examples of pythagorean triples include (3, 4, 5), (5, 12, 13), and (8, 15, 17), which all satisfy the Pythagorean Theorem.
Pythagorean triples can be generated using formulas, such as $$m^2 - n^2$$, $$2mn$$, and $$m^2 + n^2$$ for integers m and n with m > n > 0.
The smallest pythagorean triple is (3, 4, 5), which forms a right triangle with integer side lengths.
Not all sets of three numbers will form a pythagorean triple; they must specifically satisfy the condition $$a^2 + b^2 = c^2$$.
Pythagorean triples are used in various applications including construction, navigation, and computer graphics for determining angles and distances.
Review Questions
How can you determine if a set of three numbers forms a pythagorean triple?
To determine if a set of three numbers forms a pythagorean triple, you need to check if they satisfy the equation $$a^2 + b^2 = c^2$$. Here, 'c' should be the largest number among the three since it represents the hypotenuse. If substituting these values into the equation results in a true statement, then those numbers are indeed a pythagorean triple.
Discuss how pythagorean triples relate to identifying right triangles using the Pythagorean Theorem.
Pythagorean triples directly relate to identifying right triangles because they provide specific integer solutions that satisfy the Pythagorean Theorem. By checking if given side lengths correspond to any known pythagorean triples or if they fulfill the equation $$a^2 + b^2 = c^2$$, one can ascertain whether a triangle is right-angled. This connection allows for quick assessments in various mathematical problems.
Evaluate how understanding pythagorean triples can enhance problem-solving skills in geometry involving right triangles.
Understanding pythagorean triples enhances problem-solving skills in geometry by providing shortcuts for calculating side lengths without needing extensive calculations. For instance, if a problem involves finding integer side lengths of a right triangle, recognizing common pythagorean triples enables quick identification of potential solutions. This not only saves time but also builds intuition about relationships between angles and side lengths in geometric contexts.
Related terms
Pythagorean Theorem:
A fundamental principle in geometry stating that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.
Right Triangle:
A triangle that has one angle measuring 90 degrees, where the Pythagorean Theorem applies.
Hypotenuse:
The longest side of a right triangle, opposite the right angle, which is 'c' in the context of a pythagorean triple.
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6886 | https://www.spectroscopyonline.com/view/headspace-raman-spectroscopy | Headspace Raman Spectroscopy
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Spectroscopy
September 1, 2014
Spectroscopy
Spectroscopy-09-01-2014 Volume 29 Issue 9
Headspace Raman Spectroscopy
Columns|Column: Molecular Spectroscopy Workbench
We examine vapor-phase Raman spectroscopy through the acquisition of spectra from gas molecules confined to the headspace.
We examine vapor-phase Raman spectroscopy through the acquisition of spectra from gas molecules confined to the headspace of sealed containers. Studying the Raman spectra of the liquid and vapor phases of compounds with different functional groups, degrees of hydrogen bonding, and polarity provides insight into the energetics of molecular interactions.
You were probably first introduced to rotational and vibrational-rotational spectroscopy as an undergraduate. Perhaps you even applied what you were learning in the lecture hall and from text books in a physical chemistry laboratory involving infrared absorption spectroscopy of gases. If so, that may well have been your last encounter with vibrational spectroscopy of compounds in the vapor phase.
Those of us who perform Raman spectroscopy do so almost entirely with materials in the condensed phase; that is, liquids or solids. However, many of the compounds with which we work have some measure of volatility, particularly liquids that evaporate quite rapidly, such as methanol, isopropyl alcohol, and acetone. Consequently, sealed containers of compounds, solutions, or mixtures will have some number of volatilized molecules in the headspace between the condensed phase and the container seal. A transparent container offers Raman spectroscopists the ability to probe both the condensed and vapor phase in the headspace above the sample.
In this installment, we examine vapor-phase Raman spectroscopy through the acquisition of spectra from gas molecules confined to the headspace. Comparisons of the vapor-phase and condensed-phase spectra reveal the significant effects of molecular interactions, particularly hydrogen bonding, on the vibrational spectrum. The degree to which certain Raman bands differ in energy and bandwidth (primarily from collisional broadening) between the vapor and condensed phases provides insight into the variation of molecular interactions of compounds of different molecular structure and functional groups. We also explore the phenomenon of Fermi resonance and discuss the effect of spin statistics on high spectral resolution vibrational-rotational diatomic molecules.
Carbonated Beverages
Carbonated beverages such as soda, beer, or sparkling water have a substantial amount of CO 2 dissolved in them, which provides the fizz when first opening the bottle or can and the tingle when drinking the contents. We've all experienced drinking a soft drink that has lost most of its CO 2; we say the drink has gone flat. Our own experience with beverages that have gone flat teaches us that CO 2 is not particularly soluble in water at atmospheric pressure and, for that reason, carbonated beverages must be securely sealed. Consequently, carbonated beverages have a high partial pressure of CO 2 in the headspace.
One such carbonated beverage is beer. Fluorescence overwhelms the Raman scattering from the liquid portion of the sealed beer sample when using 532-nm excitation (the excitation wavelength for all spectra shown in this article). However, guiding the laser beam above the liquid through the headspace yields the spectrum of CO 2 shown in Figure 1. One of the most striking features of Raman spectra from the vapor phase are how narrow the bands are relative to those obtained in the liquid phase. The full width at half maximum (FWHM) of the CO 2 bands is 0.9 cm-1 ! The spectrum consists of six bands: one attributed to O 2 and the remaining five bands attributed to CO 2. This may seem a little strange given that group theory predicts only one Raman active mode for CO 2, the ν 1 symmetric stretching mode.
Figure 1: Raman spectrum of CO2 and O2 from the headspace of beer.
A phenomenon called Fermi resonance accounts for the appearance of multiple CO 2 Raman bands. The symmetric stretching mode of CO 2 is of Σ+g symmetry and is expected at approximately 1330 cm-1 (1). The ν 2 doubly degenerate bending mode is Raman forbidden, but does appear in the infrared absorption spectrum at 667 cm-1 (2). However, the overtone (2ν 2) has Σ+g symmetry and is expected at 1334 cm-1. So, we have a fundamental vibrational mode and an overtone of approximately equal energies and the same symmetry. Therefore, these two energy states can interact, and this interaction is called Fermi resonance. This type of resonance can produce quite striking effects with respect to both Raman scattering strength and perturbation of the vibrational states — that is, Raman band positions.
The assignments of the beer headspace CO 2 Raman bands are shown in Table I in accordance with the assignments of Hanf and coworkers (3). The very strong bands at 1284.8 and 1387.5 have been assigned to ν 1 and 2ν 2, respectively. Fermi resonance has caused the ν 1 and 2ν 2 bands of approximately equal energy to split such that the lower energy state shifts lower and the higher energy state shifts higher. The greater the interaction of the two states (that is, the stronger the Fermi resonance), the greater the splitting of the energy states and the observed Raman bands. Here, we see a difference of 103 cm-1. In addition to the band splitting, one also observes a so-called borrowing of intensity with Fermi resonance. Normally, one would expect the 2ν 2 Raman band to be comparatively weak. However, the 2ν 2 band at 1387.5 cm-1 is more intense than the ν 1 fundamental band at 1284.8 cm-1. Also, the strength of the Fermi resonance is so great that isotopic (13 C 16 O 2) and excited state or so-called hot (ν 1 1 and 2ν 2 1) bands appear in the Raman spectrum.
Table I: Assignment of Raman bands in beer headspace spectrum of Figure 1
The Raman spectra obtained from sparkling water offer us an opportunity to compare the spectrum of CO 2 in the vapor phase with that dissolved in water. Raman spectra of CO 2 and O 2 from the headspace of sparkling water and the liquid itself are shown in Figure 2. The headspace Raman spectrum of sparkling water is nearly identical to that from beer. Now, without the overwhelming fluorescence generated by the liquid beer, the Raman spectrum of the CO 2 in liquid sparkling water can be obtained. The liquid spectrum consists of bands at 1272.7 and 1379.6 cm-1 (CO 2) and 1633.6 cm-1 (the bending mode of H 2 O). The effect of the solvent (H 2 O) on the solute (CO 2) can be seen in the positions and widths of the CO 2 ν 1 and 2ν 2 bands. The shifts between the vapor phase and the water solubilized CO 2 ν 1 and 2ν 2 bands are -12.3 and -8.1 cm-1, respectively. Given that the shift in the ν 1 band is 50% greater than that observed for the 2ν 2 band, it is not surprising that the intensity of the solubilized CO 2 ν 1 band is significantly more diminished. Also, the FWHM of the 2ν 2 bands in the vapor phase and water solubilized CO 2 are 0.9 and 9.9 cm-1, respectively. The band width of the solvated CO 2 is 10 times greater than that in the vapor phase, thereby indicating the strength of the interactions of CO 2 with the water solvent. The Raman spectra of the headspace and solvated CO 2 demonstrate that Raman spectroscopy is well suited for the study of molecular interactions of solute and solvent.
Figure 2: Raman spectra of CO2 and O2 from the headspace of sparkling water and the liquid.
Effect of Hydrogen Bonding on the Vibrational Spectrum
In the previous section, we compared the spectra of a gas in the vapor phase and solubilized by water. Here, we compare the spectra of compounds in the liquid and vapor phases and see the effect of hydrogen bonding manifest in the vibrational spectrum of the liquid. Our first example consists of household ammonia purchased at the grocery store. In this case, our headspace Raman spectrum consists of both the solute (NH 3) and solvent (H 2 O). The Raman spectra of the liquid and headspace of household ammonia are shown in Figure 3. The spectrum of the liquid consists of a very broad band ranging from approximately 3000 cm-1 to 3700 cm-1 and consisting of two partially resolved broad peaks. This is the Raman scattering due to water and the enormous bandwidth is attributed to hydrogen bonding. The much narrower peak protruding at 3314 cm-1 is the ν 1 symmetric stretch of NH 3. The Raman bandwidths clearly indicate that the strength of hydrogen bonding among water molecules is far greater than that with ammonia molecules.
Figure 3: Raman spectra of liquid- and vapor-phase household ammonia.
In fact, the effects of hydrogen bonding are made stunningly clear when one compares the liquid spectrum to that of the headspace. The NH 3 symmetric stretch in the vapor phase is much narrower and is shifted relative to the solvated species by 26 cm-1. The absence of solvation by the water molecules reduces the molecular interactions and collisional broadening and thereby causes the NH 3 symmetric stretching mode to shift to higher energy and narrows the distribution of vibrational states. Hydrogen bonding is even more significant in the liquid phase of water, so the difference between OH stretching bands from water in the liquid and vapor phases is even more dramatic. Without the contribution from hydrogen bonding, the OH stretching bands narrow significantly and shift to 3659 cm-1 where a single asymmetric band appears.
Hydrogen bonding plays an important role in the molecular interactions of the polar solvent methanol. Raman spectra of methanol in the liquid and vapor phase are shown in Figure 4. Raman scattering from the CH stretching vibrational modes appear at energies less than 3000 cm-1. The OH stretch of methanol appears as a broad band centered at 3340 cm-1 in the liquid phase; however, it splits into two partially resolved bands at 3684 and 3697 cm-1 in the vapor phase. That is a shift of approximately 350 cm-1, which can be attributed to hydrogen bonding. The CH stretching modes are also affected, but to a substantially lesser degree. The 2832 and 2941 cm-1 bands shift to 2846 and 2956 cm-1, a shift of 14 and 15 cm-1, respectively.
Figure 4: Raman spectra of liquid- and vapor-phase methanol.
These differences in Raman band positions between the liquid and vapor phases provide insight into the energetics of the molecular interactions in the liquid phase. Of course, hydrogen bonding offers the strongest type of molecular interaction that we can expect to see, and compounds bearing OH functional groups can be expected to exhibit hydrogen bonding and have strong molecular interactions. However, as the nonpolar portion of the molecule becomes larger or the polar functional group is removed, we can expect the energetics of molecular interaction to diminish. Compounds of medium polarity or entirely nonpolar can be expected to have smaller differences between the Raman spectra of the liquid and vapor phases.
Organic Compounds with Different Strengths of Molecular Interactions
Acetonitrile is considered a medium polarity solvent that is miscible with many organic solvents (except saturated hydrocarbons) and water. For the purposes of our discussion, we can think of it as having replaced the OH group on methanol with the CN nitrile group. Therefore, we might expect that the differences between the liquid- and vapor-phase CH 3 stretches of acetonitrile might be comparable to those of methanol. However, that expectation is not met in the Raman spectra of the liquid and vapor phase of acetonitrile shown in Figure 5.
Figure 5: Raman spectra of liquid- and vapor-phase acetonitrile.
The Raman spectrum of the liquid phase features two strong bands at 2249.6 and 2940.2 cm-1 attributed to the CN stretch and CH 3 stretch, respectively. Note that the CH 3 stretch bandwidths of both liquid- and vapor-phase acetonitrile are significantly narrower than those observed in either liquid- or vapor-phase methanol spectra. In fact, the acetonitrile CH 3 stretch has a FWHM of only 7.6 cm-1 in the liquid phase and is reduced by approximately 50% to 3.5 cm-1 in the vapor phase. As expected, the CN stretch and CH 3 stretch bands shift to higher energy in the vapor phase, 2262.9 and 2952.8 cm-1. It is noteworthy that the phase-related energy shifts are 13.3 and 12.6 cm-1 for the CN and CH stretches, respectively. Those shifts are approximately equal and slightly less than the 14–15 cm-1 difference that we observed for the CH stretching in methanol. Comparing all of these spectral features and their differences in the acetonitrile and methanol spectra, we can infer that the molecular interactions of methanol are greater than those of acetonitrile.
Our comparison of the Raman spectra of liquid and vapor phases now progresses to the nonpolar solvent benzene. We expected the molecular interactions to be weak compared to those of polar and medium polarity solvents. The Raman spectra of benzene in the liquid and vapor phase shown in Figure 6 confirm that expectation. Here, we focus our attention on the ring breathing mode, which appears at 991.4 cm-1 with a FWHM of 2.1 cm-1 in the liquid spectrum. The Raman band assigned to the ring breathing mode of benzene in the liquid phase is actually narrower than any of the bands from either methanol or acetonitrile in the vapor phase. This is a measure of just how weak the molecular interactions are. Nevertheless, the vapor-phase spectrum reveals a shift of the ring breathing mode band of +0.7 cm-1 to 992.1 cm-1 and a narrowing of the FWHM to 1.1 cm-1. So that even when the molecular interactions in the liquid phase are weak, it is clear that in the vapor phase they are still weaker.
Figure 6: Raman spectra of liquid- and vapor-phase benzene.
Studying the Raman spectra of the liquid and vapor phases of compounds with different functional groups, degrees of hydrogen bonding, and polarity provides insight into the energetics of molecular interactions.
Vibrational-Rotational Raman Spectra of Atmospheric Molecules
We now move from compounds that are liquid at room temperature to atmospheric diatomic molecules. The primary components of air are N 2 and O 2. Therefore, we can expect to find a contribution from these molecules in most headspace Raman spectra. Having no dipole moment, these molecules are of course infrared inactive. However, because the polarizabilities of these molecules are anisotropic, one can observe Raman active rotational transitions in both the rotational and vibrational-rotational spectra. A discussion of the fundamental molecular physics and nuclear spin statistics of rotational and vibrational-rotational Raman spectroscopy is beyond the scope of this installment. Nevertheless, you will need to know just a few facts about the quantum mechanics of vibrational-rotational Raman spectroscopy to appreciate what you observe in the spectra of N 2 and O 2.
Figure 7: Raman spectrum of oxygen in air. Note the rotational side bands to the symmetric stretch at 1556.4 cm-1.
The selection rules for the vibrational-rotational Raman spectrum of a diatomic molecule are given by
where J is the rotational quantum number. Note that the Raman selection rules differ from the rotational infrared active absorption selection rule of Δ J = ±1. An additional consideration is that the rotational wavefunction must have parity such that the overall wavefunction is symmetric. A consequence of these rules is that for O 2 with a total spin of 0, the rotational lines originating from a J state with an even number will be missing. In contrast, N 2 has a total spin of 1 and the rotational lines originating from all J states will be Raman active. We simply state here without detailed explanation that nuclear spin statistics dictate that for N 2, which has a total spin of 1, the intensity ratio of alternate lines is 2:1, whereas O 2 with a total spin of 0 has an intensity ratio of 1:0.
Figure 8: Raman spectrum of oxygen in air featuring the rotational side bands. The FWHM of the rotational side bands are 1.01 cm-1.
Now we are prepared to understand the vibrational-rotational Raman spectra of these atmospheric diatomic gases that are likely to be observed in any Raman headspace measurement. The vibrational-rotational Raman spectrum of O 2 in air is shown in Figure 7. The spectrum consists of the main vibrational symmetric stretch or Q branch at 1556.4 cm-1. The O and S rotational branches can be seen on the low and high energy sides of the Q branch, respectively. Changing the intensity scale of Figure 7, we now have an expanded view of the rotational side bands as seen in Figure 8. The narrow peaks on either side of the Q branch correspond to the vibrational-rotational Raman selection transitions described in the equations above. The side bands of O 2 are all separated by two J states, originate from odd valued J states, and the intensities are all evenly distributed proportional to the populations of the original states. Note how narrow the O 2 sidebands are with a FWHM of 1.01 cm-1.
Figure 9: Raman spectrum of nitrogen in air. Note the rotational side bands to the symmetric stretch at 2330.0 cm-1.
The vibrational-rotational Raman spectrum of N 2 in air is shown in Figure 9. The Q branch of N 2 is significantly stronger than that of O 2, primarily because its concentration in the atmosphere is significantly higher than that of O 2. Note also that the O and S rotational sidebands are significantly weaker than those of O 2. Nevertheless, expansion of the intensity scale as seen in Figure 10 reveals the alternating 2:1 intensity of the sidebands as predicted by nuclear spin statistics. The side bands are all separated by one J state, originate from even and odd valued J states, and the intensities are all evenly distributed proportional to the populations of the original states and with respect to nuclear spin statistics. The O and S rotational sidebands of N 2 with a FWHM of 0.77 cm-1 are even narrower than those of O 2.
Figure 10: Raman spectrum of nitrogen in air featuring the rotational side bands. The FWHM of the rotational side bands are 0.77 cm-1.
Conclusions
We presented vapor-phase Raman spectra of dissolved gases that exhibit Fermi resonance. The liquid- and vapor-phase Raman spectra of nonpolar, medium polarity, and polar solvents were analyzed and the differences demonstrated that Raman spectroscopy is well suited for studying the energetics of molecular interactions. Finally, vibrational-rotational Raman spectroscopy was demonstrated with the atmospheric components of N 2 and O 2.
References
(1) C.N. Banwell, Fundamentals of Molecular Spectroscopy, 3rd ed. (McGraw-Hill, London, 1983), p. 96.
(2) L.A. Woodward,Introduction to the Theory of Molecular Vibrations and Vibrational Spectroscopy (Oxford, London, 1972), p. 348.
(3) S. Hanf, R. Keiner, D. Yan, J. Popp, and T. Frosch, Anal. Chem.86, 5278–5285 (2014).
David Tuschel David Tuschel is a Raman applications manager at Horiba Scientific, in Edison, New Jersey, where he works with Fran Adar. David is sharing authorship of this column with Fran. He can be reached at: david.tuschel@horiba.com
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How Do You Calculate the Median When There are an Even Number of Values in a Distribution?
Data Fluency Data Literacy Training
Author
Chip
👋 Hey guys! Today, we’re talking about how to calculate the median when there are an even number of values in a distribution. This is an important statistical tool used to describe a set of data. Let’s dive in!
First, let’s talk about what the median is. The median is the middle value in a dataset when the values are listed in order. It’s a measure of central tendency that gives us an idea of what a “typical” value in the data set is. 📈
Now, what happens when there’s an even number of values in the dataset? 🤔 In that case, you can’t just pick the middle value – there are two values in the middle! So, to find the median, you take the average of the two middle values.
For example, let’s say you have a dataset of 6 numbers: 1, 3, 5, 7, 9, and 11. To find the median, you would list the values in order: 1, 3, 5, 7, 9, 11. Since there are 6 values, you can’t just pick the middle one. Instead, you take the average of the two middle values, which are 5 and 7. The median would be (5+7)/2 = 6.
So there you have it! 🎉 To find the median when there is an even number of values in a distribution, just take the average of the two middle values. This gives us a better idea of what a “typical” value in the data set is. Keep practicing, guys! 👨🎓👩🎓
Related Tags:
#data fluency#data literacy#exploring central tendency |
6888 | https://www.youtube.com/watch?v=OYGYqRj9-ok | Algebra 2 - Set Equality and Subsets
MyWhyU
195000 subscribers
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Description
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Posted: 27 Apr 2012
Sets can be related to each other in different ways. This chapter describes the set relations of equality, subset, superset, proper subset, and proper superset.
54 comments
Transcript:
Introduction Hello. I'm Professor Von Schmohawk and welcome to Why U. In the previous lecture
we saw how to define sets and their members through the use of set-builder notation. In this lecture, we will learn about several
types of relations between sets. One such relation is "equality". If two sets contain exactly the same elements
they are said to be equal. For example, set A, which consists of
the elements one, two and three and set B consisting of
the elements two, three, and one are equal since they both contain
the same elements. Equality Notice that it doesn't matter that their elements
are listed in different orders. This is because the order in which the elements
of a set are listed is irrelevant. The set consisting of one, two, and three is exactly the same as the set
consisting of two, three, and one. The list does not imply that there is any
particular order to the elements. It only says that every element listed
is a member of the set. So if sets A and B contain the same elements
then they are equal. Subsets Another important relation between sets
is the relation of "subset". If all the elements of set A
are also contained in set B then we say that set A
is a subset of set B. This relation is denoted using
the subset symbol. If two sets, A and B, are equal
then they are subsets of each other. This is because all the elements of A
are members of B and all the elements of B
are also members of A. So any two sets that are equal
are subsets of each other. And since every set is equal to itself
every set is also a subset of itself. So the subset symbol can be used whenever all the elements of A
are also members of B whether A has the same number of elements
as B or has fewer elements. But when A has fewer elements, we can be more
specific and call A a proper subset of B. We denote this using the proper subset symbol which is the subset symbol
without the line underneath. In addition to the relation of subset,
a set can also be a "superset". In any case where set A
is a subset of set B we can also say that set B
is a superset of set A. This relation is denoted using the superset
symbol, which is the subset symbol reversed. Likewise, if A is a proper subset of B
then B is a proper superset of A. But what about the empty set,
the set containing no elements? The empty set is considered to be a subset
of every other set including itself. In fact, the empty set is the only set
which is a subset of every set. So far we have represented sets
by listing their members or using set-builder notation or by drawing little ovals
with the elements inside. In the next lecture, we will see how to
visualize sets using Venn Diagrams. |
6889 | https://math.stackexchange.com/questions/285081/negative-ratio-is-it-possible | elementary number theory - Negative Ratio - Is it possible - Mathematics Stack Exchange
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Negative Ratio - Is it possible
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Asked 12 years, 8 months ago
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I came across this problem recently where
A:B=C:A A:B=C:A
and B=−9 B=−9 and C=−4 C=−4
What is A A then?
I got A 2=36 A 2=36⇒⇒A=±6 A=±6
However, as far as my knowledge goes a ratio is always positive
Therefore, we could easily eliminate A=+6 A=+6 and the answer remains A=−6 A=−6
Is my explanation correct, if not then why so? Can -ve Ratios exist?
elementary-number-theory
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edited Oct 22, 2017 at 23:10
Siong Thye Goh
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asked Jan 23, 2013 at 15:10
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Both +6+6 and −6−6 are correct answers. There is no rule that the ratio has to be positive.
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answered Jan 23, 2013 at 15:22
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Your work is correct, but it's possible to have a negative ratio. Therefore, both A=6 A=6 and A=−6 A=−6 will satisfy the ratio.
It is true that a ratio between natural numbers will always be positive. That's because the natural numbers are the numbers {1,2,3,…}{1,2,3,…}. Since every natural number is positive, every ratio of natural numbers will be as well.
Since you're working with numbers like −9−9 and −4−4, however, it's clear that you're working with the set of integers, which are the numbers {…,−3,−2,−1,0,1,2,3,…}{…,−3,−2,−1,0,1,2,3,…}. Since some of the integers are negative and some are positive, we can definitely have a negative ratio between them.
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edited Dec 9, 2017 at 16:11
Parcly Taxel
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answered Jan 23, 2013 at 15:24
Ashton BakerAshton Baker
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6890 | https://bstockus.wordpress.com/2019/09/26/multiplication-number-talks-using-models/ | Published Time: 2019-09-26T13:00:00+00:00
Multiplication Number Talks Using Models | Teaching to the Beat of a Different Drummer
Teaching to the Beat of a Different Drummer
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Multiplication Number Talks Using Models
5 Replies
(This post is re-blogged frommy other math blog.)
In my previous post I discussed the importance ofplanning number talks with the four stages of using models. I used a 1st grade example in that post, and almost immediately my colleagueHeidi Fessendenshared this wondering.
Great minds think alike, because this is exactly what I’ve been thinking about lately!
In Round Rock ISD, we want our students to learn thinking strategies for multiplication, rather than attempting to memorize facts in isolation. Thinking strategies have the following benefits for our students:
There’sless to memorizebecause there are 5 thinking strategies to learn instead of 121 isolated facts.
They createconsistent languageacross grade levels.
They afford astrategic mindsetaround how we think about computation facts.
Theirutilityextends beyond basic facts to computation with larger numbers.
The thinking strategies we want our students to learn are fromORIGO’s Book of Facts series. (Each strategy is linked to a one-minute video if you’d like to learn more.)
One-Minute Overview Videos
Use-Ten Strategy for Multiplication
Doubling Strategy for Multiplication
Build-Up Strategy for Multiplication
Build-Down Strategy for Multiplication
Use a Rule Strategy for Multiplication – There isn’t a video for this one. This thinking strategy involves multiplying with 0 or 1.
In our curriculum, students learn about these thinking strategies in their core instruction. We have two units in 3rd grade that focus on building conceptual understanding of multiplication and division across a total of 51 instructional days. In between those units, students practice these thinking strategies during daily numeracy time so they can build procedural fluency from their conceptual understanding. My hope is that planning number talks with the four stages of using models will facilitate this rigorous work.
I also hope it supports students in maintaining their fluency at the start of both 4th and 5th grade. Our daily numeracy time at the beginning of both of those grade levels focuses on multiplication and division. Even if every 3rd grade student ended the year fluent, it’s naive to think that fluency will continue into perpetuity without any sort of maintenance.
To help teachers envision what a number talk might look like at different stages of using models, I’ve designed a bank of sample number talks for each thinking strategy.
Sample Number Talks – Use-Ten Strategy for Multiplication
Sample Number Talks – Doubling Strategy for Multiplication
Sample Number Talks – Build-Down Strategy for Multiplication
Sample Number Talks – Build-Up Strategy for Multiplication
Each bank includes a variety of examples from the different stages of using models:
Stage 2Referring to a complete model (Number Talks 1-4)
Stage 3Referring to a partial model (Number Talks 5-8)
Stage 4Solving the problem mentally (Number Talks 9-10)
You’ll notice some “Ask Yourself” questions on many slides. You’re welcome to delete them if you don’t want them visible to students. Ever since readingRoutines for ReasoningbyGrace KelemanikandAmy Lucenta, I’ve been utilizing the same pedagogical strategies they baked into their routines to support emergent bilingual students and students with learning disabilities:
Think-Pair-Share
Ask Yourself Questions
Annotation
Sentence Stems and Sentence Starters
The 4Rs: Repeat, Rephrase, Reword, Record
Since not all of the teachers in our district might be aware of “Ask Yourself” questions, I embedded them on the slides to increase the likelihood they’ll be used by any given teacher utilizing these slides.
Caveat
These sample banks are not designed to be followed in order from Number Talk 1 through Number Talk 10. Student thinking should guide the planning of your number talks.As Kathy Richardson shared in a tweetresponding to my previous post, the four stages of using models are about levels of student thinking, not levels of instruction.
What these number talks afford is different ways of thinking about computation. A traditional number talk that presents a symbolic expression allows students to think and share about the quantities and operations the symbols represent. The teacher supports the students by representing their thinking using pictures, objects, language, and/or symbols.
A number talk that presents models, on the other hand, allows students to think and share about the the quantities shown and the operation(s) implied. The teacher supports the students by representing their thinking with language and/or symbols.
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Trying It Out in the Classroom
For example, I led a number talk in a 5th grade class today, and I started with this image:
A student said she saw10 boxes with 3 dots in each box. I wrote that language down verbatim, and then asked her how we could represent what she said with symbols. She responded with10 × 3.
I asked the 5th graders to turn and talk about why we can use multiplication to represent this model. This was challenging for them! They’ve been multiplying since 3rd grade, but they haven’t necessarily revisited the meaning of multiplication in a while.
They were able to use the model to anchor their understanding. They said it’s because the number 3 repeats. This led us into talking about how there are 10 groups of 3 and how multiplication is a way that we can represent counting equal groups of things.
The number talk continued with this second image:
The first student I called on to defend their answer said,“I know 10 times 3 is 30, so I just took away 3.”
I recorded(10 × 3) – 3 = 27, but I didn’t let the students get away with that. I reminded them that multiplication is about equal groups. If we had 10 groups of 3, then we didn’t just take 3 away, we took away something else.
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One of the students responded,“You took away a group.”
We continued talking which led to me recording(10 groups of 3) – (1 group of 3) = 9 groups of 3under the original equation and then(10 × 3) – (1 × 3) = 27under that.
I have to admit I screwed up in that last equation because I should have written9 × 3instead of 27. Thankfully number talks are an ongoing conversation. Students’ number sense is not dependent on any given day’s number talk, which means they’re forgiving of the occasional mistake.
What we did today is hopefully the start of a series of number talks to get students thinking about how taking away groups is one thinking strategy to help them derive facts they don’t know. Students don’t own that strategy right now, but our conversation today using the model was an excellent start.
Final Thoughts
I’m hoping these samples might inspire you to create number talks of your own based on the kinds of conversations you’re having with your students.Here is a document with dot imagesyou can copy and paste from to create your own number talk images.
If you try out these number talks in your classroom, I’d love to hear how it went. Either tag me in a tweet (@EMathRRISD) or share your experience in the comments.
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This entry was posted in Uncategorized and tagged Kathy Richardson, models, Number Talks on September 26, 2019 by bstockus.
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5 thoughts on “Multiplication Number Talks Using Models”
Natalie Moon (@themathgirl)October 22, 2019 at 7:40 am
Thank you thank you thank you!!!! I’ve done all of these number talks with kids but never have had a “simplified” resource and I appreciate you. I even have google slides labeled similarly but NOT at all done, in fact like 2 slides…ha! Thank you for being so concise. You rock!!!!!
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2. Marie JohnsonNovember 13, 2019 at 10:43 am
Amazing resource! Thanks for sharing!
Reply↓
3. Amy ClineJanuary 1, 2020 at 1:35 pm
I am so excited to come across your site. I have been looking for a way to get started with the number talks that are in my brain. Your resources will be very helpful. This is my first year teaching 3rd grade. I taught 2nd grade for 15 years. Again, thank you.
Reply↓
4. JULIA COCHRAN CHEEKOctober 26, 2021 at 2:32 pm
Thank you so much for sharing these great resources! I’ve been using the doubles strategy with my third graders and they love it!
Reply↓
5. Paula BoyerNovember 17, 2024 at 10:04 am
Hello maate nice blog
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6891 | https://www.youtube.com/watch?v=_wSk6hI9Fvs | Sequence and Series | Understand the relationship between Un, Sn and Sn-1
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Understand the relationship between the Sn, Sn-1 and Un withe ease.
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all right so let's look at the relationship between the sum of the first hand terms and the sum of the firstus one then the M term okay all right so we know that sum of the first end the same as sum of this okay U1 U2 okay the sum of nend terms so U1 U2 u3 up to Infinity so plus up to u n okay so n here it means that Su of the first ter that is X1 is the same as U1 as simple as that so sum of the first two terms will be what so U1 plus U2 okay but U1 okay is the same as what X1 here so I can say X2 the same as so in place of U1 we can put X1 [Music] there okay so let's find U2 from here so my U2 is X2 - X1 okay so let's note this one [Music] down Okay so let's look at some of the first three terms that will be U1 + U2 + u3 okay but U1 + U2 is what is here U1 + U is S2 so X3 X2 + u3 so what is u3 my u3 will be X3 make it the subject [Music] X2 all right so let's observe 1 and two okay so U2 is = to X2 - X1 so u3 the same as what X3 so we can write U 7 as what okay X7 minus what X6 so what is u n okay so my u n will be what so x n - x n - one so that is the relationship between u n xn and then x n minus one all right so thanks for watching and don't forget to subscribe for more exciting videos all right |
6892 | https://www.facebook.com/groups/2502328646536341/posts/7560346384067850/ | People Incorrectly Correcting Other People | Found one in the wild | Facebook
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Mathematical Accuracy of 'Four Times More Likely' vs 'Four Times As Likely'
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Jammie Frye
Ok ok ok. It’s semantics.
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Ignorance is bliss?
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Maybe they’re getting this confused with percentage increase, which folks often get wrong. A 500% increase means a 6x increase, not a 5x increase.
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6893 | https://www.shs-conferences.org/articles/shsconf/pdf/2021/40/shsconf_glob2021_03019.pdf | Use of the cost-benefit analysis method in the risk management process of SMEs Katarina Makka1,, and Katarina Kampova1 1 University of Zilina, Faculty of Security Engineering, Univerzitna 1, 010 26 Zilina, Slovakia. Abstract Research background: Property protection is a worldwide very often used term in the conditions of various sectors. It represents a set of measures that have a preventive effect on the risk of damage to the building. The issue of property protection does not only concern organizations, but also every person who is the owner of a property in which his important interests are located. The protection of buildings is a current topic on a global scale, mainly to ensure the proper functioning through the protection of all important tangible and intangible assets of company. Purpose of the article: The main idea of this article is to approach the issue and create a risk management process, focusing on dealing with risks in the conditions of a particular company, in this step we will use a cost-benefit analysis to help decide on the implementation or rejection of a project to protect the selected object. Methods: Before applying the method of cost-benefit analysis to a specific case of protection of the object of the selected company, it was necessary to characterize the selected company and find out which risks are unacceptable through the creation of a risk management process. The risk management process was created based on structured and unstructured interviews with the company's employees. Findings & Value added: The proposed procedure for risk management and application of the method of cost-benefit analysis in the process of risk management are applicable in the conditions of any other organization in order to create an effective project for the protection of the object. If necessary, the procedure for using the cost-benefit analysis method can be adjusted to suit the needs and conditions of the problem of a particular organization. Keywords: CBA analysis; company; property protection; security property, protection system. JEL Classification: D61; G32; O22 Corresponding author: katarina.makka@uniza.sk SHS Web of Conferences 129, 03019 (2021) Globalization and its Socio-Economic Consequences 2021 © The Authors, published by EDP Sciences. This is an open access article distributed under the terms of the Creative Commons Attribution License 4.0 ( 1 Introduction Micro, small and medium-sized enterprises (SMEs) account for 99% of all enterprises in the EU. They provide two-thirds of jobs in the private sector and contribute more than half of the total added value created by the business in the EU. (Information sheets on the European Union, 2021), (Hudakova et al., 2017). The protection of these entities is therefore a constant topic (Gavurova at al., 2020), (Rowland et al., 2020). One of the important areas that is currently coming to the fore is the area of property protection and security of related assets (Hudakova et al., 2017). The importance of property protection and the security of related assets is increasing on a global scale, on the one hand with increasing value of protected assets and interests and on the other hand with ever-increasing opportunities to overcome elements of protection by intruders. The dynamics of change and increasing the impact on sustainable security, therefore, require a comprehensive systematic management mechanism based on standardized procedures and scientific methodsv (Valaskova et al., 2021). Such a mechanism, which allows not only intuitive, but above all exactly to approach the issue of security related to property, is a system of property protection. One of the possibilities of an effective approach to this system is the use of the Cost benefit analysis method. This method within the system of property protection aims to approach effectively in the selection of appropriate options for protection of buildings and allows SMEs to effectively manage not only the risks associated with property protection, but also to effectively manage their costs (Dvorsky et al., 2021). 2 Methods Cost-benefit analysis is the process of comparing planned or estimated costs and benefits in order to decide whether a project is efficient or ineffective. In the process, it gradually answers the basic question: "What does it bring to and what does the implementation of an investment project take?" Costs indicate the amount of resources needed to implement a given project. Benefits indicate the value of the profit that can be expected in the implementation of a given project. Both costs and benefits are expressed in monetary units (Kampova et al., 2020), (Kampova and Makka, 2018). The basic terms used in the cost-benefit analysis include: - effects of the investment indicate all the impacts that the project has, - costs are all negative effects, - benefits are all positive effects, - beneficiary refers to the entity affected by the effects of the investment, - cash flow is a cash flow that takes the form of income or expenses, - net cash flow indicates the difference between income and expenses, - criteria indicators are indicators that fulfill the function of deciding whether or not a project is acceptable. There are two types of cost-benefit analysis. The first is the ex-ante analysis, which is used in the period before the implementation of the project, when it is needed when deciding on a given project or when deciding on other possible options. The second is the ex-post analysis, which is used in the period after the end of the project for evaluation, while the aim is to specify the impact of the project in comparison with the expected results before its implementation (Masar et al., 2018), (Hudakova and Dvorsky, 2018). The cost-benefit analysis consists of three main parts: the technical part, the financial analysis and the economic analysis (Kampova and Makka, 2018). The technical part is focused on defining the purpose of the project and its technical characteristics. Financial analysis deals with financial costs and benefits and its main goal is to evaluate the project in terms of financial efficiency for the investor. The economic analysis is related to the financial SHS Web of Conferences 129, 03019 (2021) Globalization and its Socio-Economic Consequences 2021 2 analysis and its aim is to evaluate the benefits of the project for all entities involved in it (Kampova and Makka, 2018). The cost-benefit analysis can also be used in the field of protection of buildings, specifically in the process of risk management, after the complete development of the risk assessment process. A graphical representation of the use of this method in the field of property protection can be seen in the following figure (Kampova et al., 2020). Figure 1. The process of cost-benefit analysis within the property protection. Source: Kampova et al., 2020 In the first step, a selected risk enters the process, which needs to be minimized (Boros et al., 2019a). In the second step, it is necessary to evaluate the effectiveness of selected measures to reduce the risk (efficiency, calculation of costs, calculation of revenues). The effectiveness of the measures is expressed in terms of the extent to which the risk is eliminated or reduced through the implementation of the proposed measures (Buganova et al., 2021), (Boros et al., 2019a). The cost of the measures depends on a comparison of the costs and the expected benefits, and it is necessary to assess the effectiveness of the funds spent on the implementation of these measures. The costs of implementing the measures should be quantified as accurately as possible. In the last step, it is necessary to create a cost-benefit analysis for the selected risk. The results of the cost-benefit analysis show how effective the solution would be and whether this project will be implemented or not. 2.1 Application of the cost-benefit analysis method to a specific case of property protection The subject of interest is a company that specializes in the production, sale and service of independent sources of electricity based on internal combustion engines. The company is headquartered near the town of Martin and is one of the medium-sized companies with up to 50 employees. Based on the request of the owner to maintain the anonymity of the company, the work will not publish data such as the name of the company and a more detailed identification of the investigated location of the facility. The company is one of the leading companies in the field of alternative sources of electricity in Central Europe. At present, it covers more than 40% of the Slovak market and exports its products to neighbouring countries, but also to other countries in Europe, Asia and Africa. The main goal of the company is the production of high quality products that are able to work even in the most demanding operating and climatic conditions. The company owns Risk Evaluation of measures: -- efficiency - costs - benefits Cost-benefit analysis Decision support in the field of property protection SHS Web of Conferences 129, 03019 (2021) Globalization and its Socio-Economic Consequences 2021 3 various certificates (eg ISO 9001 and ISO 14001), thanks to which it has become the subject of economic mobilization, which cooperates with the crisis management of the Ministry of Economy of the Slovak Republic. 2.1.1 Design of a project for the property protection of a company The subject of the investment within this project is the creation of the protection of a property of a company. The protection of the property will be implemented mainly through the design of measures to ensure the security of this object. The aim of this project is to reduce security risks to an acceptable level, where the security of the object from the negative effects of real or latent threats is ensured (Lovecek et al., 2017). The beneficiary will be the company in question within this project. The project will be created with an emphasis on efficiency, ie. in order to reduce the risk as much as possible, while the financial costs of implementation will not exceed its benefits. The first step is to create a zero variant that represents the current state. The last incident of burglary after preparation last occurred in 2016, resulting in the theft of € 20,000 worth of company property and damage to the building. The cost of repairing the building was € 4,500. The total financial loss due to this incident was estimated at € 24,500, which represents a very high financial consequence for the company. In addition to the financial impacts already described, the incident also resulted in indirect financial consequences in the form of production restrictions due to the theft of important components and tools for the production of products and fines arising from the contract between the company and customers due to non-compliance with the agreed delivery date. The cause of theft by burglary by the perpetrator after the preparation occurred in 2016 was most likely due to insufficient security of the building. As a result, the company set up a camera system. Apart from the camera system, no other measures have been taken against the occurrence of another incident of this kind. Table 1. Financial consequences in the event of a burglary event by the perpetrator after preparation Impact Impact type The amount € Direct impacts Theft of company assets (eg tools, devices, components, products, cash) 20 000 Damage to company property (eg broken windows, damage to fencing, damage to doors and locks) 4 500 Indirect impacts Restriction or complete cessation of production of products 15 000 Fines arising from the contract between the company and customers 10 000 Total amount 53 500 Source: authors (2020). Based on the performed analysis in the company’s premises to protect property of the organization from burglary, it is necessary to secure the object through (Kampová, et al., 2020), (Siser et al., 2017): - mechanical barriers, - technical security measures, - creation of appropriate regime measures A list of specific security features, including the price for their acquisition and installation, is given in Table 2. SHS Web of Conferences 129, 03019 (2021) Globalization and its Socio-Economic Consequences 2021 4 Table 2. Selected security features for property protection Security feature Producer Type Total price including installation € Security grilles for windows (18 pcs) ADLO Fixed window grille 4 230 Security grilles for doors (4 pcs) ADLO Openable grille with locking for padlocks 1 280 Alarm Detector Infrared Beam Sensor (16 pcs) SATEL Activa-6-BR 6 4 318 Microwave barrier (4 pcs) AVS Electronics BM60M, wireless 4 338 Source: authors (2021) The financial costs for the implementation of the project amount to € 16,691. After installing the security elements, it is necessary to perform regular inspections. These inspections are necessary due to possible complications related to the functionality of individual security elements. The system will be inspected by authorized persons during four years, but at least once every six months. The estimated cost of this activity will be € 4,000. Another part of the proposal for the protection of the property is a comparison of financial impacts. The financial impact is estimated before and after the implementation of security measures. Security measures are proposed for a burglary theft scenario with a ready intruder. The following table (tab. 3) provides an estimate of the financial impact of this scenario after implementation of security measures. Table 3. Estimation of financial impact after implementation of security measures Impact Type of impact Sum € Direct impact Theft of company assets (eg tools, devices, components, products, cash) 5 000 Damage to company property (eg broken windows, damage to fencing, damage to doors and locks) 2 500 Indirect effects Restriction or complete interruption of product production 5 000 Fines arising from the contract between the company and customers 0 Total amount € 12 500 Source: author (2021) 3 Results and discussion Based on the quantification of the costs and benefits of the proposed property protection project, it is possible to calculate the net present value (NPV) of the project for each year. 𝑁𝑃𝑉= ∑ 𝐶𝐹 𝑡 (1+𝑟)𝑡 𝑛 𝑡=0 , (1) where: NPV – Net Present Value, CFt – Cash Flow in period of time t, r – discount rate (Stobierski, 2019). For the project to be effective, the resul of net present value (NPV) must be positive or equal to zero. In this case, the positive values will represent the financial benefits gained through the project. If the final value is zero, this will mean that the project will not bring any financial benefits. SHS Web of Conferences 129, 03019 (2021) Globalization and its Socio-Economic Consequences 2021 5 Table 4. Quantification of costs and benefits in the investment and operational phase of the project The amount v € Investment phase Operational phase Costs 16 691 4 000 Benefits 0 41 000 Source: authors (2021) Costs and benefits per year in the investment phase (zero year) and in the operational phase (4 years) are shown by a graph in the following figures (Figure 2) Figure 2. Cost - benefit comparison Source: author (2021) On the basis of quantified benefits and costs in the investment phase and operational phase, it is possible to calculate the net costs and benefits in each year. Values are obtained as a cost-benefit difference. The final result of the calculation of the net present value is compared with the eligibility criteria - net present value NPV ≥ 0. The net present value (NPV) for this project was calculated € 16,885.53, which is higher than 0. The findings can be considered that the created project is acceptable and its implementation should be carried out. 4 Conclusion The aim of the presented article was to identify the possibilities of using the method of cost-benefit analysis in the process of risk management, in relation to the issue of protection of buildings in the conditions of a particular company. Before applying the cost-benefit analysis method, it was necessary to characterize the selected company and find out which risks are unacceptable by creating a risk management process for this company. The risk management process was created on the basis of structured and unstructured interviews with the company's employees. Through the characteristics of the object and the created risk management process, was found that the object does not have sufficient protection and the highest priority is the risk of theft by burglary by the perpetrator after preparation. The value of this risk was the highest mainly due to the fact that an incident of this kind in the company occurred in 2016. 0 5 000 10 000 15 000 20 000 25 000 30 000 0 1 2 3 4 Costs and benefits of the project Costs [€] Benefits [€] SHS Web of Conferences 129, 03019 (2021) Globalization and its Socio-Economic Consequences 2021 6 Due to the risk of theft by burglary after preparation, a project was created through the method of cost-benefit analysis to reduce this risk. The essence of the project was the proposal of measures to create protection of the building against this kind of risk. The measures were selected on the basis of their suitability for society. At each stage of the project, the steps that had to be taken together with the financial costs incurred were identified. The final evaluation of the project was a financial analysis which showed that the net present value (NPV) of the created project is € 16,885.53. The condition for the effectiveness of the project was that the resulting amount of net present value was higher or equal to zero. The condition of efficiency has been met, which means that its implementation should take place. The developed procedure for risk management and the method of cost-benefit analysis in the process of risk management can be applied in the conditions of any other organization in order to create an effective project for the protection of the property. If necessary, the procedure for using the cost-benefit analysis method can be adjusted to suit the needs and conditions of a particular organization and the problem beeing solved. Acknowledgements Publication of this paper was supported by the Scientific Grant Agency of the Ministry of Education, Science, Research and Sport of the Slovak Republic – VEGA No. 1/0243/20 Integrated risk management system in the conditions of contemporary changes in enterprise environment in Slovakia. References 1. Boros, M., Zvakova, Z., & Halaj, M. (2019a). Required Competencies of Security Managers for Decision-Making. INTED Proceedings, 3918-3923. 2. Boros, M., Zvakova, Z., & Halaj, M. (2019b). Required Competencies of Security Managers for Decision-Making. INTED Proceedings, 3918-3923. 3. Buganova, K., Luskova, M., Kubas, J., Brutovsky, M., & Slepecky, J. (2021). Sustainability of Business through Project Risk Identification with Use of Expert Estimates. Sustainability, 13(11), 6311. 4. Dvorsky, J., Belas, J., Gavurova, B., & Brabenec, T. (2021). Business risk management in the context of small and medium-sized enterprises. Economic Research-Ekonomska Istraživanja, 34(1), 1690-1708. 5. Gavurova, B., Belas, J., Bilan, Y., & Horak, J. (2020). Study of legislative and administrative obstacles to SMEs business in the Czech Republic and Slovakia. Oeconomia Copernicana, 11(4), 689–719. 6. Rowland, Z., Krulicky, T., & Oliinyk, O. (2020). Capital cost quantification model in business activity planning: the evidence of the middle Europe countries. Ekonomicko-manazerske spektrum, 14(1), 30-42. 7. Hudakova, M. & Dvorsky, J. (2018). Assessing the risks and their sources in dependence on the rate of implementing the risk management process in the SMEs. Equilibrium-Quarterly Journal of Economics and Economic Policy, 13(3), 543-567. 8. Hudakova, M . Schonfeld, J., Dvorsky, J. & Luskova, M. (2017). The Market Risk Analysis and Methodology of its More Effective Management in SMEs in the Slovak Republic. Montenegrin Journal of Economics, 13(2), 151-161. SHS Web of Conferences 129, 03019 (2021) Globalization and its Socio-Economic Consequences 2021 7 9. Information sheets on the European Union, Small and medium enterprises (2021, October). 10. Kampova, K., & Makka, K. (2018). Economic Aspects of the Risk Impact on the Fuel Distribution Enterprises. Transport Means - Proceedings of the International Conference, 231-235. 11. Kampova, K., Makka, K., & Zvarikova, K. (2020). Cost benefit analysis within organization security management. SHS Web of Science, 74, 01010. 12. Kampova, K., Lovecek, T., & Rehak, D. (2020). Quantitative approach to physical protection systems assessment of critical infrastructure elements: Use case in the Slovak Republic. International Journal of Critical Infrastructure Protection, 30, 100376. 13. Lovecek, T., Siser, A., & Maris, L. (2017). Use Case of Waterwork Physical Protection System Robustness Evaluation as a Part of Slovak Critical Infrastructure. International Carnahan Conference on Security Technology Proceedings. 14. Masar, M., Hudakova, M. & Luskova, M. (2018). Project Risk Management in Global Environment. In T. Kliestik (Eds.), Globalization and its Socio-Economic Consequences, Slovakia (pp.707-714). 15. Siser, A., Lovecek, T., & Maris., L. (2017). Simulation of possible assault vectors in an attack using a real-life waterworks object as a use case. Procedia Engineering, 192, 794-799. 16. Stobierski, T. (2019, September 5). How to do a Cost-benefit Analysis & why it is important. 17. Valaskova, K., Adamko, P., Frajtova Michalikova, K., & Macek, J. (2021). Quo Vadis, earnings management? Analysis of manipulation determinants in Central European environment. Oeconomia Copernicana, 12(3), 631–669. SHS Web of Conferences 129, 03019 (2021) Globalization and its Socio-Economic Consequences 2021 8 |
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Hydrogen - Thermophysical Properties
Chemical, Physical and Thermal Properties of Hydrogen - H2.
Hydrogen, H2, is a colorless, odorless gas.
Hydrogen is easily ignited. Once ignited it burns with a pale blue, almost invisible flame. The vapors are lighter than air. It is flammable over a wide range of vapor/air concentrations. Hydrogen is not toxic but is a simple asphyxiate by the displacement of oxygen in the air. Under prolonged exposure to fire or intense heat the containers may rupture violently and rocket.
Hydrogen is used to make other chemicals, in petroleum refining and in oxyhydrogen welding and cutting.
The phase diagram of hydrogen is shown below the table.
Chemical, physical and thermal properties of hydrogen:
Values at 25 oC (77 oF, 298 K) and atmospheric pressure
Hydrogen - Thermophysical Properties
| Molecular Weight | 2.016 |
| Specific Gravity, air = 1 | 0.070 |
| Specific Volume (ft3/lb, m3/kg) | 194, 12.1 |
| Density of liquid at atmospheric pressure (lb/ft3, kg/m3) | 4.43, 71.0 |
| Absolute Viscosity (lbm/ft s, centipoises) | 6.05×10-6, 0.009 |
| Sound velocity in gas (m/s) | 1315 |
| Specific Heat - cp - (Btu/lb oF or cal/g oC, J/kgK) | 3.42, 14310 |
| Specific Heat Ratio - cp/cv | 1.405 |
| Gas constant - R - (ft lb/lb oR, J/kg oC) | 767, 4126 |
| Thermal Conductivity (Btu/hr ft oF, W/moC) | 0.105, 0.182 |
| Boiling Point - saturation pressure 14.7 psia and 760 mm Hg - (oF,o K) | -423, 20.4 |
| Latent Heat of Evaporation at boiling point (Btu/lb, J/kg) | 192, 447000 |
| Freezing or Melting Point at 1 atm (oF, oC) | -434.6, -259.1 |
| Latent Heat of Fusion (Btu/lb, J/kg) | 25.0, 58000 |
| Critical Temperature (oF, oC) | -399.8, -240.0 |
| Critical Pressure (psia, MN/m2) | 189, 1.30 |
| Critical Volume (ft3/lb, m3/kg) | 0.53, 0.033 |
| Flammable | yes |
| Heat of combustion (Btu/ft3, Btu/lb, kJ/kg) | 320, 62050, 144000 |
Follow the links below to get values for the listed properties of hydrogen at varying pressure and temperature :
Density and specific weight
Specific heat
Thermal conductivity
See also more about atmospheric pressure, and STP - Standard Temperature and Pressure & NTP - Normal Temperature and Pressure,
as well as Thermophysical properties of: Acetone, Acetylene, Air, Ammonia, Argon, Benzene, Butane, Carbon dioxide, Carbon monoxide, Ethane, Ethanol, Ethylene, Helium, Hydrogen sulfide, Methane, Methanol, Nitrogen, Oxygen, Pentane, Propane, Toluene, Water and Heavy water, D2O.
Hydrogen is a gas at standard conditions. However, at very low temperature and/or high pressures the gas becomes a liquid or a solid.
The hydrogen phase diagram shows the phase behavior with changes in temperature and pressure. The curve between the critical point and the triple point shows the hydrogen boiling point with changes in pressure. It also shows the saturation pressure with changes in temperature.
At the critical point there is no change of state when pressure is increased or if heat is added.
The triple point of a substance is the temperature and pressure at which the three phases (gas, liquid, and solid) of that substance coexist in thermodynamic equilibrium.
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Why are sine values in the 3rd quadrant negative?
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According to Sine - Wolfram Mathworld, Sine is defined as
sin(θ)=opposite(a)hypotenuse(b)sin(θ)=opposite(a)hypotenuse(b)
From that definition, I don't understand why the sine value in the unit circle in the 3rd quadrant is negative; after all, a a and b b should both be negative there and a positive quotient follows from this:
People have already explained to me that b is always positive in this case, since it represents the radius and not a vector like a or c (not drawn in here now).
Which leads me to the final question, how is one supposed to actually know that, based on the above definition?
I mean it must be explicitly mentioned somewhere that b b in this case represents a length and a a, c c vectors that can take on both positive and negative values; the only thing I see on many pages is that a coordinate system is drawn into the unit circle and hence my obviously wrong assumption.
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edited Aug 17, 2023 at 22:40
iwabiwab
asked Aug 15, 2023 at 22:43
iwabiwab
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What is the sign of a a when the distance b b is rotated counterclockwise from the x x-axis into the third quadrant? It's negative. Way overthinking this point.Nij –Nij 2023-08-15 22:48:43 +00:00 Commented Aug 15, 2023 at 22:48
@Nij Well, that is the point. a is negative (but b should be negative too, leading to a positive quotient)iwab –iwab 2023-08-15 22:55:32 +00:00 Commented Aug 15, 2023 at 22:55
2 In this context, b b represents the radius of a circle. b b is positive. When b b is negative, it would go backwards through the circle center, but then you wouldn't be measuring 200∘200∘ or whatever you're taking the sine of.PrincessEev –PrincessEev 2023-08-15 22:57:18 +00:00 Commented Aug 15, 2023 at 22:57
1 @Leonard It's overkill to call a a a "vector." It's just a number. Numbers are vectors, but not really. It's the y y coordinate of the point on the circle. It can be positive or negative. b b is a number too, but it's a length, so it's never negative. You can't proceed without knowing that b b is not negative, so it's pointless to try to imagine a situation where you could deal with this without knowing that.Matt Samuel –Matt Samuel 2023-08-15 23:15:29 +00:00 Commented Aug 15, 2023 at 23:15
2 @Leonard You identified that something doesn't make sense. You asked a question on Math.SE about it, which is a reasonable thing to do. It was clarified for you. It doesn't seem worthwhile to dwell on what would've happened if you didn't do that. Just my perspective.Matt Samuel –Matt Samuel 2023-08-16 00:13:52 +00:00 Commented Aug 16, 2023 at 0:13
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From the site you quote (bold mine):
"The common schoolbook definition of the sine of an angle theta in a right triangle (which is equivalent to the definition just given) is as the ratio of the lengths ..."
An angle θ>π θ>π cannot be seen as an angle in a right triangle.
So you need an extended definion of the trig functions that includes all angles. I suggest you use the following:
Let C C the circle centered at O=(0,0)O=(0,0) and of radius 1 1. Let P∈C P∈C a point such that the segment O P O P forms an angle θ θ with the x x-axis. Then
P=(cos θ,sin θ).P=(cosθ,sinθ).
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answered Aug 16, 2023 at 13:07
Andrea MoriAndrea Mori
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Thanks :-) After your definition no wrong interpretation area remains more on. Maybe I just transferred the triangle from the triangle definition to the unit circle definition too fast. For me it did not seem logical to assume that in the same geometrical object (here the right triangle in the unit circle) 2 parts follow the coordinate system, while 1 part follows only the magnitude. Do you have a good resource for trigonometry that eliminates such confusion or did I just reason wrong?iwab –iwab 2023-08-16 15:33:02 +00:00 Commented Aug 16, 2023 at 15:33
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6897 | https://www.uptodate.com/contents/clinical-manifestations-of-rheumatoid-arthritis#! | Articular manifestations of rheumatoid arthritis - UpToDate
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Articular manifestations of rheumatoid arthritis
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Outline
SUMMARY AND RECOMMENDATIONS
INTRODUCTION
INITIAL CLINICAL PRESENTATION
Patterns of joint involvement
- Typical (classic) rheumatoid arthritis
- Palindromic rheumatism
- Monoarthritis
- Proximal joint arthritis
Morning stiffness
Other symptoms
SPECIFIC JOINT INVOLVEMENT
Upper extremity
- Hands
- Wrists, elbows, and shoulders
Lower extremity
Axial skeleton
Cricoarytenoid joint
LABORATORY FINDINGS
IMAGING
Plain film radiography
Magnetic resonance imaging
Ultrasonography
CLINICAL COURSE
Patterns of progression
Disease activity versus structural damage
Remission
INFORMATION FOR PATIENTS
SUMMARY AND RECOMMENDATIONS
ACKNOWLEDGMENTS
REFERENCES
GRAPHICS
Tables
- Clinical manifestations of RA in disease course
- Assessment of activity RA
Pictures
- RA hand 1
- RA hand 2
- RA boutonniere and Z deformities of hand
- Olecranon bursitis
- Rheumatoid nodules
- Rheumatoid arthritis metatarsal head subluxation
- Popliteal cyst in a child
- Popliteal cyst
Diagnostic Images
- RA of the elbow
- US small popliteal cyst
- US ruptured popliteal cyst
- RA PIP soft tissue swelling x-ray
- RA MCP osteopenia
- RA wrist osteopenia x-ray
- RA ulnar styloid erosion x-ray
- Plain radiograph of a normal wrist
- RA MTP erosions x-ray
- Radiograph left foot normal
- RA MCP mild erosion
- Normal left hand x-ray
- Severe erosive change
- RA PIP erosions
- Normal right hand x-ray
RELATED TOPICS
Patient education: Rheumatoid arthritis symptoms and diagnosis (Beyond the Basics)
Patient education: Rheumatoid arthritis treatment (Beyond the Basics)
Acute phase reactants
Assessment of rheumatoid arthritis disease activity and physical function
Biologic markers in the assessment of rheumatoid arthritis
Carpal tunnel syndrome: Clinical manifestations and diagnosis
Cervical subluxation in rheumatoid arthritis
Clinical manifestations and diagnosis of Felty syndrome
Clinical manifestations and diagnosis of polymyalgia rheumatica
Clinical manifestations and diagnosis of rheumatoid vasculitis
Diagnosis and differential diagnosis of rheumatoid arthritis
Disease outcome and functional capacity in rheumatoid arthritis
Epidemiology of, risk factors for, and possible causes of rheumatoid arthritis
Greater trochanteric pain syndrome (formerly trochanteric bursitis)
Hematologic complications of rheumatoid arthritis
Large granular lymphocyte leukemia in rheumatoid arthritis
Monoarthritis in adults: Etiology and evaluation
Musculoskeletal ultrasonography: Clinical applications
Overview of lower extremity peripheral nerve syndromes
Overview of the management of rheumatoid arthritis in adults
Overview of the systemic and nonarticular manifestations of rheumatoid arthritis
Patient education: Hand pain (The Basics)
Patient education: Rheumatoid arthritis (The Basics)
Popliteal (Baker's) cyst
Rheumatoid nodules
Synovial fluid analysis
Articular manifestations of rheumatoid arthritis
Author:Bryant R England, MD, PhDSection Editor:James R O'Dell, MDDeputy Editor:Philip Seo, MD, MHS
Literature review current through:Aug 2025.
This topic last updated:Jun 20, 2025.
INTRODUCTION
Rheumatoid arthritis (RA) is a chronic, systemic, autoimmune, inflammatory disorder of unknown etiology that primarily involves synovial joints. The arthritis is typically symmetric; if uncontrolled, it may lead to joint deformities due to erosion of cartilage and bone. The disease usually progresses from the periphery to more proximal joints and results in significant disability within 10 to 20 years in patients whose disease does not respond to treatment.
The symptoms of RA can affect patients' capacity to perform the activities of daily living (eg, walking, stairs, dressing, use of a toilet, getting up from a chair, opening jars, doors, typing) and those required in their occupation.
The major clinical features of RA, including the articular manifestations, are reviewed here. The systemic and extraarticular features and the diagnosis and differential diagnosis of RA are discussed in detail separately. (See "Overview of the systemic and nonarticular manifestations of rheumatoid arthritis" and "Diagnosis and differential diagnosis of rheumatoid arthritis".)
INITIAL CLINICAL PRESENTATION
Patterns of joint involvement—Rheumatoid arthritis (RA) most typically presents as polyarticular disease and with a gradual onset, but some patients can present with acute onset, intermittent or migratory joint involvement, monoarticular disease, or a proximal joint arthritis mimicking polymyalgia rheumatica (PMR).
Typical (classic) rheumatoid arthritis
●Time course – Classic RA typically has a gradual onset, but some patients can present with acute onset.
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Disclaimer: This generalized information is a limited summary of diagnosis, treatment, and/or medication information. It is not meant to be comprehensive and should be used as a tool to help the user understand and/or assess potential diagnostic and treatment options. It does NOT include all information about conditions, treatments, medications, side effects, or risks that may apply to a specific patient. It is not intended to be medical advice or a substitute for the medical advice, diagnosis, or treatment of a health care provider based on the health care provider's examination and assessment of a patient's specific and unique circumstances. Patients must speak with a health care provider for complete information about their health, medical questions, and treatment options, including any risks or benefits regarding use of medications. This information does not endorse any treatments or medications as safe, effective, or approved for treating a specific patient. UpToDate, Inc. and its affiliates disclaim any warranty or liability relating to this information or the use thereof. The use of this information is governed by the Terms of Use, available at 2025© UpToDate, Inc. and its affiliates and/or licensors. All rights reserved.
Topic Feedback
Tables
Clinical manifestations of rheumatoid arthritis through the disease courseAssessment of disease activity in rheumatoid arthritis
Clinical manifestations of rheumatoid arthritis through the disease courseAssessment of disease activity in rheumatoid arthritis
Pictures
Synovial thickening of the metacarpophalangeal jointSwelling of the metacarpophalangeal joints of the right hand in rheumatoid arthritisRheumatoid arthritis hand deformities: Boutonniere deformity and Z deformity of the thumbOlecranon bursitisRheumatoid nodulesRheumatoid arthritis in the feet: Metatarsal head subluxationAsymptomatic popliteal (Baker's) cyst in a childPopliteal (Baker's) cyst, knee (photograph and arthrogram)
Synovial thickening of the metacarpophalangeal jointSwelling of the metacarpophalangeal joints of the right hand in rheumatoid arthritisRheumatoid arthritis hand deformities: Boutonniere deformity and Z deformity of the thumbOlecranon bursitisRheumatoid nodulesRheumatoid arthritis in the feet: Metatarsal head subluxationAsymptomatic popliteal (Baker's) cyst in a childPopliteal (Baker's) cyst, knee (photograph and arthrogram)
Diagnostic Images
Rheumatoid arthritis of the elbowSmall popliteal cyst on ultrasound imagingRuptured popliteal cyst on ultrasoundSoft tissue swelling and periarticular osteopenia of proximal interphalangeal joints in rheumatoid arthritisPlain radiograph of osteopenia of metacarpophalangeal joints in rheumatoid arthritisPlain radiograph of osteopenia of the wrist in rheumatoid arthritisPlain radiograph of ulnar styloid erosion in rheumatoid arthritisPlain radiograph of a normal wristPlain radiograph of metatarsophalangeal joint space narrowing and erosions in rheumatoid arthritisPlain radiograph of normal left footPlain radiograph of mild metacarpophalangeal joint erosion rheumatoid arthritisPlain radiograph of normal left hand (AP, lateral, and oblique)Rheumatoid arthritis: Severe erosive changes and hand deformities (plain radiography)Plain radiograph of rheumatoid arthritis proximal interphalangeal joint erosionsPlain radiograph of a normal right hand (AP, lateral, and oblique)
Rheumatoid arthritis of the elbowSmall popliteal cyst on ultrasound imagingRuptured popliteal cyst on ultrasoundSoft tissue swelling and periarticular osteopenia of proximal interphalangeal joints in rheumatoid arthritisPlain radiograph of osteopenia of metacarpophalangeal joints in rheumatoid arthritisPlain radiograph of osteopenia of the wrist in rheumatoid arthritisPlain radiograph of ulnar styloid erosion in rheumatoid arthritisPlain radiograph of a normal wristPlain radiograph of metatarsophalangeal joint space narrowing and erosions in rheumatoid arthritisPlain radiograph of normal left footPlain radiograph of mild metacarpophalangeal joint erosion rheumatoid arthritisPlain radiograph of normal left hand (AP, lateral, and oblique)Rheumatoid arthritis: Severe erosive changes and hand deformities (plain radiography)Plain radiograph of rheumatoid arthritis proximal interphalangeal joint erosionsPlain radiograph of a normal right hand (AP, lateral, and oblique)
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6898 | https://math.stackexchange.com/questions/2128832/proving-mnm-mid-mn-and-mnmn-mid-2m2n-algebrically | combinatorics - Proving $m!(n!)^m \mid (mn)!$ and $m!n!(m+n)! \mid (2m)!(2n)!$ algebrically - Mathematics Stack Exchange
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Proving m!(n!)m∣(m n)!m!(n!)m∣(m n)! and m!n!(m+n)!∣(2 m)!(2 n)!m!n!(m+n)!∣(2 m)!(2 n)! algebrically
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Let m,n m,n be positive integers. Prove the following -
m!(n!)m∣(m n)!m!(n!)m∣(m n)!
m!n!(m+n)!∣(2 m)!(2 n)!m!n!(m+n)!∣(2 m)!(2 n)!
First one is here - Inductive proof
I can give a combinatorial proof. (m n)!m!(n!)m(m n)!m!(n!)m is the way to split m n m n things in m m groups of n n things.
I am looking for algebric proof of both statements.
My Work
Let p p be any prime
Then it suffices to prove (m p)+m(n p)≤(m n p)(m p)+m(n p)≤(m n p) and (m p)+(n p)+(m+n p)≤(2 m p)+(2 n p)(m p)+(n p)+(m+n p)≤(2 m p)+(2 n p) But don't know how to prove these.
It is easy to see if p>m i n(n,m)p>m i n(n,m) then it comes obvious. But how to prove when p≤m i n(n,m)p≤m i n(n,m)
I'd also like to know if their exists any combinatorial prove of the second statement.
P.S: Don't give only hints. I need a full solution. I am messing up things.
combinatorics
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edited Jun 12, 2020 at 10:38
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asked Feb 4, 2017 at 13:50
Rezwan ArefinRezwan Arefin
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4 Why is the tag (legendre-symbol) or (linear-algebra) necessary?S.C.B. –S.C.B. 2017-02-04 13:55:14 +00:00 Commented Feb 4, 2017 at 13:55
@S.C.B. I think we need legendre-symbol while proving it algebrically.Rezwan Arefin –Rezwan Arefin 2017-02-04 13:59:44 +00:00 Commented Feb 4, 2017 at 13:59
2 I think that you are going to need the floor function, not the legendre-symbol, as quadratic residues don't seem very helpful here.S.C.B. –S.C.B. 2017-02-04 14:00:39 +00:00 Commented Feb 4, 2017 at 14:00
1 As far as I have seen, there are no known combinatorial proofs of the second result. The only proof I have seen is based on the inequality ⌊2 x⌋+⌊2 y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋⌊2 x⌋+⌊2 y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋ for all x,y≥0 x,y≥0. The first result, on the other hand, has an easy combinatorial proof.Batominovski –Batominovski 2017-02-04 14:01:42 +00:00 Commented Feb 4, 2017 at 14:01
@S.C.B. I didn't know that there is a floor-function tag :|Rezwan Arefin –Rezwan Arefin 2017-02-04 14:06:07 +00:00 Commented Feb 4, 2017 at 14:06
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One can see that the Bivariate Catalan Number is defined by
C(m,n):=(2 m)!(2 n)!m!n!(m+n)!=(2 m m)(2 n n)(m+n n)C(m,n):=(2 m)!(2 n)!m!n!(m+n)!=(2 m m)(2 n n)(m+n n)
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edited Feb 4, 2017 at 14:42
answered Feb 4, 2017 at 14:26
User8976User8976
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Choosing m m objects from 2 m 2 m and n n objects from 2 n 2 n, I think that should be (2 m m)(2 n n)(2 m m)(2 n n). Can you explain why the divide by (m+n n)(m+n n)?Rezwan Arefin –Rezwan Arefin 2017-02-04 14:39:10 +00:00 Commented Feb 4, 2017 at 14:39
the Bivariate Catalan Number is an integer always so ...User8976 –User8976 2017-02-04 14:43:26 +00:00 Commented Feb 4, 2017 at 14:43
I was talking about the previous edit Rezwan Arefin –Rezwan Arefin 2017-02-04 14:44:34 +00:00 Commented Feb 4, 2017 at 14:44
Why you said Choosing m m objects from 2 m 2 m and n n objects from 2 n 2 n ?Rezwan Arefin –Rezwan Arefin 2017-02-04 14:46:34 +00:00 Commented Feb 4, 2017 at 14:46
How is saying something is an integer an explanation for how it's an integer?anon –anon 2021-08-24 07:10:31 +00:00 Commented Aug 24, 2021 at 7:10
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6899 | https://www.topdoctors.co.uk/medical-dictionary/spermiogenesis/ | Spermiogenesis : what it is, symptoms and treatment | Top Doctors
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Spermiogenesis
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Spermiogenesis
Written in association with: Dr Irfana KoitaFertility specialist in W1G Marylebone London
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Created: 13/11/2012
Edited: 05/06/2023
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What is spermiogenesis?
Spermiogenesis is the last phase in the formation ofspermatozoa(the male sex cells that carry a man's genetic material). During this stage of maturation, the spermatid cells see an increase in the size of the tail and a decrease in the spermatozoa head.
This series of complex transformations result in the final constitution of the spermatozoa, the male cells that transfer DNA to the oocyte.
The mature sperm are released through the seminiferous tubules, which are small tubes inside the testicles, responsible for producing sperm and the hormone testosterone. The approximate duration of this process is two and a half months.
What are the phases of spermiogenesis?
There are four phases in which spermiogenesis occurs. These are:
The Golgi phase - the initial set of events where the spermatid is prepared to acquire the major parts of a sperm cell, the tail and the head. The genetic (DNA) is also tightly packaged, highly condensed and inactive.
Cap phase - this leads to the formation of the acrosomal cap, which is a membrane-bound compartment at the tip of the head of the spermatid.
Formation of the tale- this phase is characterised by the elongation of the microtubules on one of the centrioles of the spermatid to become to tail.
Maturation stage - the final phase where the testis-determining factor, which concentrates testosterone,is formed.
Spermiogenesis occurs in the coiled tubules called seminiferous tubules within the testes.
What is the difference between spermiogenesis and spermatogenesis?
Both spermiogenesis and spermatogenesis are two stages during the formation of sperms. The main difference is that spermatogenesis is the formation of sperm cells whereas spermiogenesis is the maturation of the spermatids into sperm cells.
Spermatogenesis is the complete process of the production of the sperm cells from the cells of the germinal epithelium of males.Spermiogenesis is the final differentiation and maturation process of the spermatids into sperm cells.
The main difference between spermatogenesis and spermiogenesis is the mechanism of each process in the production of the sperm.
What is the importance of spermiogenesis?
Spermiogenesis is an important biological process as it leads to the transformation of spermatids into mature spermatozoa.
During the phases, if there are any abnormalities in the head, midpiece, tail, genetic content, and subcellular structures, it could lead to dysfunctional sperm cells, and thus, to male infertility.
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