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Approximation and Errors

3. Approximation and Errors. Case Study. 3.1 Significant Figures. 3.2 Scientific Notation. 3.3 Approximation and Errors. Chapter Summary. Case Study. I guess there are 120 paper clips. What’s your guess, Linda?. I guess there are 110 clips, John.

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Approximation and Errors

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  1. 3 Approximation and Errors Case Study 3.1 Significant Figures 3.2 Scientific Notation 3.3 Approximation and Errors Chapter Summary

  2. Case Study I guess there are 120 paper clips. What’s your guess, Linda? I guess there are 110 clips, John. Suppose that there are actually 116 paper clips on the plate. Did John or Linda make a better estimation? We need to find the difference between each estimate and the actual number. The smaller the difference is, the better is the estimate. John’s estimate  Actual number  120  116  4 Actual number  Linda’s estimate  116  110  6 Since 4 is less than 6, John made a better estimation.

  3. 3.1 Significant Figures A. Basic Concepts The population of Hong Kong at the end of 2007 is  7 000 000. (correct to the nearest million)  6 950 000. (correct to the nearest ten thousand)  6 953 000. (correct to the nearest thousand)  7  695  6953 The important digits of a number are called significant figures. In most cases, the far left non-zero digit, having the largest place value in a number, is the first significant figure. This is also the most important figure. Subsequent important digits are called the second significant figure, the third significant figure, and so on.

  4. 1 0 4 9 1st significant figure 2nd significant figure 3rd significant figure 4th significant figure Wealsostudysignificant figures in decimals. Forexample,thedecimal 5.307 has 4 significant figures. 3.1 Significant Figures A. Basic Concepts For example, the number 1049 has 4 significant figures. For all numbers, any zeros between 2 non-zero digits are significant figures.

  5. 0. 0 3 8 0 1st significant figure 2nd significant figure 3rd significant figure 3.1 Significant Figures A. Basic Concepts For decimals less than 1, all zeros in front of the first non-zero digit are not significant figures. These zeros are called place holders. For example, the number 0.0380 has only 3 significant figures. For all decimals, any zeros after the last non-zero digit are significant figures.

  6. Numbers Number of Significant Figures 3900 (correct to the nearest integer) has 4 significant figures. But 3900 (correct to the nearest ten) has 3 significant figures. 4008 4 6.092 4 0.7 1 0.002 04 3 9.000 4 0.130 3 3.1 Significant Figures A. Basic Concepts The following table shows some examples. All the significant figures are marked with green colour.

  7. Thephrase‘cor.to3sig. fig.’ is the short form of ‘correct to 3 significant figures’. 3.1 Significant Figures B. Round off to the Required Significant Figures Using the technique of rounding off for estimation, the number 47 054 can be expressed as: (a) 47 100 (cor. to the nearest hundred) (b) 47 000 (cor. to the nearest thousand) (c) 50 000 (cor. to the nearest ten thousand) Alternatively, we can round off a number to a certain number of significant figures. We can express 47 054 as: (a) 47 100 (cor. to 3 sig. fig.) (b) 47 000 (cor. to 2 sig. fig.) (c) 50 000 (cor. to 1 sig. fig.)

  8. Step 1: Determine which digit is to be rounded off. 51 672 3rd significant figure Step 2: Round off according to the next significant figure. 51 672  51 700 (cor. to 3 sig. fig.) Since 7 is larger than 5, we add 1 to the 3rd significant figure. 3.1 Significant Figures B. Round off to the Required Significant Figures The following steps show how to round off 51 672, correct to 3 significant figures.

  9. 3.1 Significant Figures B. Round off to the Required Significant Figures Example 3.1T Round off 472 780 correct to (a) 2 significant figures, (b) 3 significant figures, (c) 4 significant figures. Solution: (a) 470 000 (cor. to 2 sig. fig.) (b) 473 000 (cor. to 3 sig. fig.) (c) 472 800 (cor. to 4 sig. fig.)

  10. 3.1 Significant Figures B. Round off to the Required Significant Figures Example 3.2T Round off 0.300 649 correct to (a) 1 significant figure, (b) 2 significant figures, (c) 5 significant figures. Solution: (a) 0.3 (cor. to 1 sig. fig.) (b) 0.30 (cor. to 2 sig. fig.) (c) 0.300 65 (cor. to 5 sig. fig.)

  11. 3.1 Significant Figures B. Round off to the Required Significant Figures Example 3.3T Round off 9995 correct to (a) 1 significant figure, (b) 2 significant figures, (c) 3 significant figures. Solution: (a) 10 000 (cor. to 1 sig. fig.) (b) 10 000 (cor. to 2 sig. fig.) (c) 10 000 (cor. to 3 sig. fig.)

  12.  $2.1 (cor. to 2 sig. fig.) 3.1 Significant Figures B. Round off to the Required Significant Figures Example 3.4T The total price of 120 oranges is $250. Round off the average price of each orange correct to 2 significant figures. Solution: The average price of each orange  $(250  120)  $2.0833

  13. 3.2 Scientific Notation A. Introduction In science, we often deal with numbers which are very large or small:  Our body produces about 173 000 000 000 red blood cells each day.  The mass of 1 water molecule is about 0.000 000 000 000 000 000 000 03 g. It is convenient to rewrite these numbers as follows:  173 000 000 000  1.73  100 000 000 000  1.73  1011  0.000 000 000 000 000 000 000 03 g  3  1023 g This kind of representation of numbers is called scientific notation. A positive number is expressed in scientific notation if it is in the form a 10n, where 1 a 10 and n is an integer.

  14. aman  am  n Move the decimal point 4 places to the left and then times 104 Move the decimal point 7 places to the right and then times 107  2.2  104  7  107  9.5  108  8  105 3.2 Scientific Notation A. Introduction Example 3.5T Express each of the following numbers in scientific notation. (a) 22 000 (b) 0.000 000 7 (c) 95 000  104 (d) 0.008  102 Solution: (a) 22 000  2.2  10 000 (b) 0.000 000 7  7 0.000 000 1 (c) 95 000  104 (9.5  104)  104 (d) 0.008  102 (8  103)  102

  15.  10 020 000  0.000 06 3.2 Scientific Notation A. Introduction Example 3.6T Convert the following numbers into integers or decimals. (a) 1.002  107 (b) 6  105 Solution: (a) 1.002  107 1.002  10 000 000 (b) 6  105 6  0.000 01

  16. Without using a calculator, evaluate each of the following expressions. Express the answers in scientific notation. (a) (b) (4  103)(3.3  105) 3.2 Scientific Notation A. Introduction Example 3.7T Solution:

  17. In calculator, we should key in ‘1.32 9’ to represent 1.32  109. EXP If we key in ‘4 11 ’, then the calculator would display ‘41011’, which means 4  1011. EXP EXE Students should note thatthekey-insequences of different calculators may be different. 3.2 Scientific Notation A. Introduction The calculator would display ‘1.32E9’ before execution.

  18. 3.2 Scientific Notation B. Applications of Scientific Notation When presenting estimates, we can express numbers in scientific notation and round off the value correct to a certain number of significant figures. For example: 1 228 550 000 m3 of water was consumed in Hong Kong in 2005  2006. 1 228 550 000 m3 1 230 000 000 m3(cor. to 3 sig. fig.)  1.23  109 m3

  19.  3.61  104 3.2 Scientific Notation B. Applications of Scientific Notation Example 3.8T 36 084 people were injured in road accidents last year. Round off the number correct to 3 significant figures and express the answer in scientific notation. Solution: 36 084  36 100 (cor. to 3 sig. fig.)

  20. 3.2 Scientific Notation B. Applications of Scientific Notation Example 3.9T The atomic radii of a helium atom and a gold atom are 3.1  1011 m and 1.35  1010 m respectively. How many times is the atomic radius of a gold atom to that of a helium atom? Give the answer correct to 2 significant figures. Solution: The required ratio (cor. to 2 sig. fig.)

  21. 3.3 Approximation and Errors In Book 1A Chapter 2, we learnt that no measurements give exact values. They are only approximations. Hence, every measurement or estimation has errors.

  22. The absolute error is always positive. 3.3 Approximation and Errors A. Absolute Error The difference between a measured value (or an estimated value) and the actual value is called the absolute error. If the actual value is larger than a measured value, then absolute error  actual value  measured value If a measured value is larger than the actual value, then absolute error  measured value  actual value Example: A story book has 117 pages. Sara guessed that the book has 100 pages. ∴ The absolute error of Sara’s estimation is 17 pages.

  23.  16.6 g 3.3 Approximation and Errors A. Absolute Error Example 3.10T The actual weight of a pack of potato chips is 183.4 g. Find the absolute error if Julie corrects the weight to 1 significant figure. Solution: 183.4 g  200 g (cor. to 1 sig. fig.) The absolute error  (200  183.4) g

  24. 3.3 Approximation and Errors B. Maximum Absolute Error Measuring tools with smaller scale intervals can give estimations with higher accuracy. Therefore, errors of measurements are related to the scale intervals of the tools. In the figure, the scale interval of the ruler is 1 cm. The measured length is 15 cm, correct to the nearest cm. ∴ The minimum and the maximum actual lengths are 14.5 cm and 15.5 cm respectively. The absolute error of this measurement does not exceed 0.5 cm. This figure is called the maximum absolute error.

  25. maximum absolute error   scale interval of the measuring tool 3.3 Approximation and Errors B. Maximum Absolute Error In general, In the figure, the scale interval of the ruler is 1 cm.  Lower limit ( 14.5 cm) Minimum possible value of measurement  Upper limit ( 15.5 cm) Maximum possible value of measurement They can be found by the following formulas. Lower limit  Measured value  Maximum absolute error Upper limit  Measured value  Maximum absolute error

  26. The maximum absolute error  0.01 L The 3rd significant figure of the number 5.38 is 8. Hence the capacity is correct to the nearest 0.01 L.  5.375 L 3.3 Approximation and Errors B. Maximum Absolute Error Example 3.11T The capacity of a bottle is 5.38 L, correct to 3 significant figures. What is the lower limit of the capacity? Solution: The scale interval of measurement  0.01 L  0.005 L The lower limit of the capacity  (5.38  0.005) L

  27. The maximum absolute error of measurement  1 cm  0.5 cm 3.3 Approximation and Errors B. Maximum Absolute Error Example 3.12T Gordon measured the base length and height of a triangle to be 8 cm and 11 cm respectively, both correct to the nearest 1 cm. What is the range of the area of the triangle? Solution: Base length: Height: Lower limit  (8  0.5) cm  7.5 cm Lower limit  10.5 cm Upper limit  (8  0.5) cm  8.5 cm Upper limit  11.5 cm Lower limit of the area Upper limit of the area ∴ The actual area lies between 39.375 cm2 and 48.875 cm2.

  28. 3.3 Approximation and Errors C. Relative Error Consider the following set of data: Both of the absolute errors of the 2 measurements are the same, but they do not reflect the degree of accuracy. To compare the accuracy of 2 estimations, we also have to determine the relative error.

  29. Animal Actual weight (kg) Absolute error (kg) Pig 63.42 0.42 0.006 62 Dog 5.19 0.42 0.0809 Relative error  3.3 Approximation and Errors C. Relative Error The absolute error of the pig’s weight is small (insignificant) when compared to its actual weight. On the other hand, the absolute error of the dog’s weight is significant because it is relatively large when compared to its actual weight. In general, we use the relative error to determine the degree of accuracy of a measurement.

  30. Relative error  3.3 Approximation and Errors C. Relative Error The smaller the relative error is, the higher is the accuracy of a measurement. For example, the degree of accuracy of the measurement of the pig’s weight is higher than that of the dog’s weight. Sometimes it is impossible to find the actual values in some real-life situations. So we can use the maximum absolute error and the measured value instead to find the relative error.

  31. (a) The length of the string  75 cm 3.3 Approximation and Errors C. Relative Error Example 3.13T • The figure below shows the length of a piece of string. Find • (a) the length of the string, • (b) the maximum absolute error of the length, • the relative error of the length, correct to 3 significant figures. Solution: (b) Maximum absolute error (c) Relative error (cor. to 3 sig. fig.)

  32. Louis used a balance to measure the weight of a package of sugar. The maximum absolute error of the balance is 2.5 g. If the relative error of the result is , find the lower limit of the actual result. First find out the measured weight of the package of sugar. ∴ Lower limit of the actual weight  (180 – 2.5) g  177.5 g 3.3 Approximation and Errors C. Relative Error Example 3.14T Solution: Let x g be the weight of a package of sugar. x 2.5  72  180 ∴ The weight of a package of sugar is 180 g.

  33. 3.3 Approximation and Errors D. Percentage Error When the relative error is expressed as a percentage, it is called the percentage error. Percentage error  Relative error  100% For example, since the relative errors of the pig’s and the dog’s weight are 0.006 62 and 0.0809 respectively. The percentage errors of the pig’s and the dog’s weight are 0.662% and 8.09% respectively. The smaller the percentage error is, the higher is the accuracy of a measurement.

  34.  3.34% 3.3 Approximation and Errors D. Percentage Error Example 3.15T A university bought 1258 new computers last year. If the number of computers bought is now correct to the nearest hundred, find the percentage error of the estimation. (Give the answer correct to 3 significant figures.) Solution: 1258  1300 (cor. to the nearest hundred) Absolute error  1300 – 1258  42 Percentage error  (cor. to 3 sig. fig.)

  35. 3.3 Approximation and Errors D. Percentage Error Example 3.16T Sally and Christine measured the capacity of 2 boxes separately. Sally’s result was 300 mL correct to 2 significant figures. Christine’s result was 1250 mL correct to the nearest 50 mL. (a) Find the percentage errors of their measurements. (Give the answer correct to 3 significant figures if necessary.) (b) Hence determine who measured more accurately. Solution: For Christine’s measurement: (a) For Sally’s measurement: Maximum absolute error  50 mL  2  25 mL Maximum absolute error  10 mL  2  5 mL Percentage error Percentage error (cor. to 3 sig. fig.) (b) Since 1.67% is less than 2%, Sally measured more accurately.

  36. Chapter Summary 3.1 Significant Figures • For all numbers, any zeros between 2 non-zero digits are • significant figures. • For all decimals, any zeros after the last non-zero digit are • significant figures.

  37. 3.2 Scientific Notation Chapter Summary All positive numbers can be expressed in the form a10n, where 1 a 10 and n is an integer.

  38. 3.3 Approximation and Errors 2. Maximum absolute error (a) Maximum absolute error  Scale interval of the measuring tool Chapter Summary 1. Absolute error (a) If the actual value is larger than a measured value, then the absolute error  actual value  measured value. (b) If a measured value is larger than the actual value, then the absolute error  measured value  actual value. (b) Lower limit  Measured value  Maximum absolute error (c) Upper limit  Measured value  Maximum absolute error

  39. 3.3 Approximation and Errors Chapter Summary 3. Relative error • Percentage error • Percentage error  Relative error  100%

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