Applications of Rational Functions in Distance, Speed, Time, and Work Problems, Lecture notes of Painting

Download Applications of Rational Functions in Distance, Speed, Time, and Work Problems and more Lecture notes Painting in PDF only on Docsity! Section 7.8 Applications of Rational Functions 731 Version: Fall 2007 7.8 Applications of Rational Functions In this section, we will investigate the use of rational functions in several applications. Number Problems We start by recalling the definition of the reciprocal of a number. Definition 1. For any nonzero real number a, the reciprocal of a is the number 1/a. Note that the product of a number and its reciprocal is always equal to the number 1. That is, a · 1 a = 1. For example, the reciprocal of the number 3 is 1/3. Note that we simply “invert” the number 3 to obtain its reciprocal 1/3. Further, note that the product of 3 and its reciprocal 1/3 is 3 · 1 3 = 1. As a second example, to find the reciprocal of −3/5, we could make the calculation 1 −3 5 = 1÷ ( −3 5 ) = 1 · ( −5 3 ) = −5 3 , but it’s probably faster to simply “invert” −3/5 to obtain its reciprocal −5/3. Again, note that the product of −3/5 and its reciprocal −5/3 is( −3 5 ) · ( −5 3 ) = 1. Let’s look at some applications that involve the reciprocals of numbers. I Example 2. The sum of a number and its reciprocal is 29/10. Find the number(s). Let x represent a nonzero number. The reciprocal of x is 1/x. Hence, the sum of x and its reciprocal is represented by the rational expression x + 1/x. Set this equal to 29/10. x+ 1 x = 29 10 To clear fractions from this equation, multiply both sides by the common denominator 10x. Copyrighted material. See: http://msenux.redwoods.edu/IntAlgText/1 732 Chapter 7 Rational Functions Version: Fall 2007 10x ( x+ 1 x ) = ( 29 10 ) 10x 10x2 + 10 = 29x This equation is nonlinear (it has a power of x larger than 1), so make one side equal to zero by subtracting 29x from both sides of the equation. 10x2 − 29x+ 10 = 0 Let’s try to use the ac-test to factor. Note that ac = (10)(10) = 100. The integer pair {−4,−25} has product 100 and sum −29. Break up the middle term of the quadratic trinomial using this pair, then factor by grouping. 10x2 − 4x− 25x+ 10 = 0 2x(5x− 2)− 5(5x− 2) = 0 (2x− 5)(5x− 2) = 0 Using the zero product property, either 2x− 5 = 0 or 5x− 2 = 0. Each of these linear equations is easily solved. x = 5 2 or x = 2 5 Hence, we have two solutions for x. However, they both lead to the same number- reciprocal pair. That is, if x = 5/2, then its reciprocal is 2/5. On the other hand, if x = 2/5, then its reciprocal is 5/2. Let’s check our solution by taking the sum of the solution and its reciprocal. Note that 5 2 + 2 5 = 25 10 + 4 10 = 29 10 , as required by the problem statement. Let’s look at another application of the reciprocal concept. I Example 3. There are two numbers. The second number is 1 larger than twice the first number. The sum of the reciprocals of the two numbers is 7/10. Find the two numbers. Let x represent the first number. If the second number is 1 larger than twice the first number, then the second number can be represented by the expression 2x+ 1. Thus, our two numbers are x and 2x+1. Their reciprocals, respectively, are 1/x and 1/(2x+ 1). Therefore, the sum of their reciprocals can be represented by the rational expression 1/x+ 1/(2x+ 1). Set this equal to 7/10. Section 7.8 Applications of Rational Functions 735 Version: Fall 2007 d (mi) v (mi/h) t (h) Upstream 60 3 − c ? Downstream 60 3 + c ? Table 1. A distance, speed, and time table. Here is a useful piece of advice regarding distance, speed, and time tables. Distance, Speed, and Time Tables. Because distance, speed, and time are related by the equation d = vt, whenever you have two boxes in a row of the table completed, the third box in that row can be calculated by means of the formula d = vt. Note that each row of Table 1 has two entries entered. The third entry in each row is time. Solve the equation d = vt for t to obtain t = d v . The relation t = d/v can be used to compute the time entry in each row of Table 1. For example, in the first row, d = 60 miles and v = 3− c miles per hour. Therefore, the time of travel is t = d v = 60 3− c . Note how we’ve filled in this entry in Table 2. In similar fashion, the time to travel downstream is calculated with t = d v = 60 3 + c . We’ve also added this entry to the time column in Table 2. d (mi) v (mi/h) t (h) Upstream 60 3 − c 60 3− c Downstream 60 3 + c 60 3 + c Table 2. Calculating the time column entries. To set up an equation, we need to use the fact that the time to travel upstream is twice the time to travel downstream. This leads to the result 60 3− c = 2 ( 60 3 + c ) , or equivalently, 736 Chapter 7 Rational Functions Version: Fall 2007 60 3− c = 120 3 + c . Multiply both sides by the common denominator, in this case, (3− c)(3 + c). (3− c)(3 + c) [ 60 3− c ] = [ 120 3 + c ] (3− c)(3 + c) 60(3 + c) = 120(3− c) Expand each side of this equation. 180 + 60c = 360− 120c This equation is linear (no power of c other than 1). Hence, we want to isolate all terms containing c on one side of the equation. We add 120c to both sides of the equation, then subtract 180 from both sides of the equation. 60c+ 120c = 360− 180 From here, it is simple to solve for c. 180c = 180 c = 1. Hence, the speed of the current is 1 mile per hour. It is important to check that the solution satisfies the constraints of the problem statement. • If the speed of the boat in still water is 3 miles per hour and the speed of the current is 1 mile per hour, then the speed of the boat upstream (against the current) will be 2 miles per hour. It will take 30 hours to travel 60 miles at this rate. • The speed of the boat as it goes downstream (with the current) will be 4 miles per hour. It will take 15 hours to travel 60 miles at this rate. Note that the time to travel upstream (30 hours) is twice the time to travel down- stream (15 hours), so our solution is correct. Let’s look at another example. I Example 6. A speedboat can travel 32 miles per hour in still water. It travels 150 miles upstream against the current then returns to the starting location. The total time of the trip is 10 hours. What is the speed of the current? Let c represent the speed of the current. Going upstream, the boat struggles against the current, so its net speed is 32−cmiles per hour. On the return trip, the boat benefits from the current, so its net speed on the return trip is 32 + c miles per hour. The trip each way is 150 miles. We’ve entered this data in Table 3. Section 7.8 Applications of Rational Functions 737 Version: Fall 2007 d (mi) v (mi/h) t (h) Upstream 150 32 − c ? Downstream 150 32 + c ? Table 3. Entering the given data in a dis- tance, speed, and time table. Solving d = vt for the time t, t = d v . In the first row of Table 3, we have d = 150 miles and v = 32 − c miles per hour. Hence, the time it takes the boat to go upstream is given by t = d v = 150 32− c . Similarly, upon examining the data in the second row of Table 3, the time it takes the boat to return downstream to its starting location is t = d v = 150 32 + c . These results are entered in Table 4. d (mi) v (mi/h) t (h) Upstream 150 32 − c 150/(32− c) Downstream 150 32 + c 150/(32 + c) Table 4. Calculating the time to go upstream and return. Because the total time to go upstream and return is 10 hours, we can write 150 32− c + 150 32 + c = 10. Multiply both sides by the common denominator (32− c)(32 + c). (32− c)(32 + c) ( 150 32− c + 150 32 + c ) = 10(32− c)(32 + c) 150(32 + c) + 150(32− c) = 10(1024− c2) We can make the numbers a bit smaller by noting that both sides of the last equation are divisible by 10. 15(32 + c) + 15(32− c) = 1024− c2 Expand, simplify, make one side zero, then factor. 740 Chapter 7 Rational Functions Version: Fall 2007 Thus, Bill is working at a rate of 1/2 report per hour. Note how we’ve entered this result in the first row of Table 6. Similarly, Maria is working at a rate of 1/4 report per hour, which we’ve also entered in Table 6. We’ve let t represent the time it takes them to write 1 report if they are working together (see Table 5), so the following calculation gives us the combined rate. Rate = Work Time = 1 report t h . That is, together they work at a rate of 1/t reports per hour. This result is also recorded in Table 6. w (reports) r (reports/h) t (h) Bill 1 1/2 2 Maria 1 1/4 4 Together 1 1/t t Table 6. Calculating the Rate entries. In our discussion above, we pointed out the fact that rates add. Thus, the equation we seek lies in the Rate column of Table 6. Bill is working at a rate of 1/2 report per hour and Maria is working at a rate of 1/4 report per hour. Therefore, their combined rate is 1/2 + 1/4 reports per hour. However, the last row of Table 6 indicates that the combined rate is also 1/t reports per hour. Thus, 1 2 + 1 4 = 1 t . Multiply both sides of this equation by the common denominator 4t. (4t) [ 1 2 + 1 4 ] = [ 1 t ] (4t) 2t+ t = 4, This equation is linear (no power of t other than 1) and is easily solved. 3t = 4 t = 4/3 Thus, it will take 4/3 of an hour to complete 1 report if Bill and Maria work together. Again, it is very important that we check this result. • We know that Bill does 1/2 reports per hour. In 4/3 of an hour, Bill will complete Work = 1 2 reports h × 4 3 h = 2 3 reports. That is, Bill will complete 2/3 of a report. • We know that Maria does 1/4 reports per hour. In 4/3 of an hour, Maria will complete Section 7.8 Applications of Rational Functions 741 Version: Fall 2007 Work = 1 4 reports h × 4 3 h = 1 3 reports. That is, Maria will complete 1/3 of a report. Clearly, working together, Bill and Maria will complete 2/3 + 1/3 reports, that is, one full report. Let’s look at another example. I Example 8. It takes Liya 7 more hours to paint a kitchen than it takes Hank to complete the same job. Together, they can complete the same job in 12 hours. How long does it take Hank to complete the job if he works alone? Let H represent the time it take Hank to complete the job of painting the kitchen when he works alone. Because it takes Liya 7 more hours than it takes Hank, let H+ 7 represent the time it takes Liya to paint the kitchen when she works alone. This leads to the entries in Table 7. w (kitchens) r (kitchens/h) t (h) Hank 1 ? H Liya 1 ? H + 7 Together 1 ? 12 Table 7. Entering the given data for Hank and Liya. We can calculate the rate at which Hank is working alone by solving the equation Work = Rate × Time for the Rate, then substituting Hank’s data from row one of Table 7. Rate = Work Time = 1 kitchen H hour Thus, Hank is working at a rate of 1/H kitchens per hour. Similarly, Liya is working at a rate of 1/(H + 7) kitchens per hour. Because it takes them 12 hours to complete the task when working together, their combined rate is 1/12 kitchens per hour. Each of these rates is entered in Table 8. w (kitchens) r (kitchens/h) t (h) Hank 1 1/H H Liya 1 1/(H + 7) H + 7 Together 1 1/12 12 Table 8. Calculating the rates. Because the rates add, we can write 1 H + 1 H + 7 = 1 12 . 742 Chapter 7 Rational Functions Version: Fall 2007 Multiply both sides of this equation by the common denominator 12H(H + 7). 12H(H + 7) ( 1 H + 1 H + 7 ) = ( 1 12 ) 12H(H + 7) 12(H + 7) + 12H = H(H + 7) Expand and simplify. 12H + 84 + 12H = H2 + 7H 24H + 84 = H2 + 7H This last equation is nonlinear, so make one side zero by subtracting 24H and 84 from both sides of the equation. 0 = H2 + 7H − 24H − 84 0 = H2 − 17H − 84 Note that ac = (1)(−84) = −84. The integer pair {4,−21} has product −84 and sums to −17. Hence, 0 = (H + 4)(H − 21). Using the zero product property, either H + 4 = 0 or H − 21 = 0, leading to the solutions H = −4 or H = 21. We eliminate the solution H = −4 from consideration (it doesn’t take Hank negative time to paint the kitchen), so we conclude that it takes Hank 21 hours to paint the kitchen. Does our solution make sense? • It takes Hank 21 hours to complete the kitchen, so he is finishing 1/21 of the kitchen per hour. • It takes Liya 7 hours longer than Hank to complete the kitchen, namely 28 hours, so she is finishing 1/28 of the kitchen per hour. Together, they are working at a combined rate of 1 21 + 1 28 = 4 84 + 3 84 = 7 84 = 1 12 , or 1/12 of a kitchen per hour. This agrees with the combined rate in Table 8.