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# 1 MA 1128: Lecture 04 – 1/27/11 Formulas and Applications.

## Presentation on theme: "1 MA 1128: Lecture 04 – 1/27/11 Formulas and Applications."— Presentation transcript:

1 MA 1128: Lecture 04 – 1/27/11 Formulas and Applications

2 Formulas Equations can have more than one variable. A formula is an equation thats used over and over with different numbers in for the variables. For example, the formula for the area of a rectangle, A = lw, has three variables, A, l, and w. The two expressions A and lw are not equivalent, and if we put numbers in for two of A, l, and w, we can solve the equation, and figure out what the third variable is. For example, if we substitute A = 6, and l = 2, (6) = (2)w (Its a good idea to put parentheses around things you substitute for a variable.) We can solve this equation and see that w = 3 (or just see this). This formula gives a relationship between these three attributes of a rectangle. Next Slide

3 Continued It is somewhat surprising that virtually all of the usefulness of mathematics has something to do with the relationships described by equations. As weve already seen, if we do the same thing to both sides of an equation, the relationship between the two sides (equality) remains intact. We can manipulate formulas the same way. For example, if we need to find a rectangle with area A = 10 and width w = 15, what would the length be? Our formula says (10) = l (15). This is a linear equation ( l is the variable instead of x, but this does not matter). Next Slide

4 Continued We should divide both sides by 15. Next Slide We see that l = 10/15 = 2/3. We could also have solved for l before substituting the numbers.

5 Continued The second to last equation is a variation of the A = lw formula that tells us explicitly how to get the length from the area and width. Which is better? Should we substitute first or change the form first? It depends. If youre going to do a similar problem often (you often know A and w, and need to find l ), its probably better to manipulate first. If youre only going to do a problem like this once, its probably easier to substitute first. Its pretty much up to you, however. Next Slide

6 Example Suppose we would like to make a wheel that will roll 8 feet every time it makes one revolution. The outside of the wheel needs to be a circle with circumference 8 feet. The formula we need is C = d. Recall that is a special constant (not a variable). It is, in fact, defined to be the number that makes this formula work. It is known that in decimal form, has an infinite number of digits with no repeating pattern. Your calculator knows the first ten digits or so. = 3.14159265… Using = 3.14 gives a reasonable amount of accuracy, but the more digits you use, the more accuracy youll get. (continued) Next Slide

7 Example (cont.) Back to the problem. We need to find d when C = 8. This time well solve for d before substituting numbers. Next Slide If we substitute C = 8, we have We can use this for the answer, if we want.

8 Example (cont.) If we want a decimal number, we can substitute = 3.14. d = 8/3.14 = 2.547770701 Better yet, we can use the -button on our calculator. This gives us d = 8/ = 2.546479089. The second answer is more accurate, since we used more digits. In the first calculation, we rounded so that we kept three good digits, the 3, the 1 and the 4. Having kept three good digits, I would be confident that the final number has two good digits, 2.5…, although this one worked out better than that. Round-off error, the error introduced by rounding an intermediate number, is not always easy to predict, so I would suggest always keeping all the digits available to you. We may talk about this a bit more later. Next Slide

9 Practice Problems Involving Find the -button on your calculator. You should get 3.14… out to about 9 or 10 places. If I say round to two decimal places, I mean round the number correctly (next digit of 5 or more rounds up) to two places past the decimal point. The number rounded to two decimal places would be 3.14. 1.Round to four decimal places. 2.Round /2 to four decimal places. Recall that the formula for the circumference of a circle is C = d. 3.What is the circumference of a circle that has diameter d = 4? Round your answer to four decimal places. 4.Find the circumference, if the diameter is d = 7? Round to four decimal places. 5.What is the diameter of a circle with circumference C = 10? Round to four decimal places. Answers: 1) 3.1416; 2) 1.5708; 3) 12.5664; 4) 21.9911; 5) 3.1831. Next Slide

10 Practice Problems 1.Solve the formula P = 2l + 2w for l. (This is the formula for the perimeter of a rectangle.) 2.Solve the following formula for F. (This is the conversion formula that relates degrees Celsius with and degrees Fahrenheit.) (Work in reverse of the order of operations – multiply by 9/5 first.) Next Slide Click for answers: 1)2)

11 Solving Application Problems We can use algebra to solve application problems, if we know 1. An equation relating the variables in the problems, and 2. Values for some of the variables. The basic situation is the case where we know values for all of the variables except one, and we want to figure out a value for that one unknown variable. The optional text has the following example. Kate Spence can buy a monthly pass, which entitles the owner to unlimited bus travel, for \$40 per month. Without the pass, each bus ride costs \$1.50. How many rides per month would Kate have to make so that it is less expensive to purchase the pass? Next Slide

12 Example (cont.) The first step is to try to get a feeling for the problem. If she were to take 3 rides per month, and pay for the rides separately, that would cost (3)(1.50) = 4.50. The pass (\$40) would be much more expensive. If she took 100 rides each month, that would cost (100)(1.50) = 150.00. Here the pass would be a much better deal. The basic trick is to find when the one-ride-at-a-time cost is the same as the bus- pass cost. The variables are: The number of rides per month – Lets call this r. The cost per ride – Lets call this c. The total cost per month – Lets call this T. (cont.) Next Slide

13 Example (cont.) Weve already computed a couple of cases, and if Kate pays for the rides one at a time, the total cost is the number of rides times the cost of each ride. r c = T We know that c = 1.50, and we want to find when T = 40.00 (the same as the pass). If we plug these in, we get r(1.50) = 40.00 r = 40.00/1.50 r = 26.6666… Of course, 26.67 rides in one month doesnt necessarily make sense (although it kind of does), but we can see that 26 rides costs less than \$40 bought individually, and 27 rides costs more. Therefore, if Kate is going to ride 27 times a month, or more, she is better off with the pass. Next Slide

14 Practice Problem Lets change the problem a bit. Suppose each ride costs \$1.25, and the monthly bus pass costs \$37.00. 1.How many rides would make the one-at-a-time cost equal to the monthly- pass cost? 2.For the number of rides you found in problem 1 (probably a fraction or decimal number), is the pass a better deal for more rides or fewer rides? 3.So for how many rides or more, or for how many rides or fewer, is the pass a better deal? Click for answers. 1) 29.6 rides makes the two ways cost the same; 2) more rides makes the pass a better deal; 3) 30 rides or more. Next Slide

15 Example In the optional textbook, there is a problem with the formula IR + Ir = E (Im pretty sure this is an electronics formula relating voltage, current, etc.) The problem asks for the equation to be solved for R. That is, we are supposed to rewrite the formula with R by itself on one side. We want to unwrap R, so we work in reverse of the order of operations. R is multiplied by I in the equation and also added to by Ir, So we deal with the term Ir first. Next Slide

16 Example Heres another problem from the book. The yearly cost for joining Sams Club is \$35. The Thygesons estimate that they save an average of 3% when compared to other discount stores that do not have a membership fee. How much would the Thygesons have to spend in a year at Sams Club to recover the yearly membership cost. A couple of particular cases: If they spend \$100, then theyll save about 100 3% = (100)(.03) = 3.00. (not worth the membership fee) If they spend \$2,000, then theyll save about (2000)(.03) = 60.00. (they saved more than the fee) (cont.) Next Slide

17 Example (cont.) The variables (you can choose any letters you want): Amount spent – A Savings rate – s (were assuming this is 3%) Total savings – T (we want this to be more than \$35) 3. Equation: A s = T 4. Solve the break-even equation: A(.03) = 35.00 A = 35.00/.03 A = 1166.666666… The Thygesons need to spend \$1166.67 per year, or more, to justify the membership fee. Next Slide

18 Practice Problems These are taken from the book. 1.Each week Sandy Ivey receives a flat weekly salary of \$240 plus a 12% commission on the total dollar volume of all sales she makes. What must her dollar volume be in a week for her to earn \$540? (Here we only need to find the dollar volume that gets Sandy exactly \$540, which is like the solving the break-even equation.) 2.The Midtown Tennis Club offers two payment plans for its members. Plan 1 is a monthly fee of \$75. Plan 2 has no monthly fee, but court time costs \$18.50 per hour. How many hours would Mrs. Lewis have to play per month so that Plan 1 becomes advantageous, if court time is paid in whole numbers of hours? Click for answers. 1) Rounded to the nearest cent, \$2,500.00; 2) 5 hours or more. End

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