tag:blogger.com,1999:blog-8231784566931768362.post7069888299012906927..comments2023-09-09T08:21:55.454-04:00Comments on MathNotations: SOMETHING NEW! Instructional Strategy Series: Teaching Average RatesDave Marainhttp://www.blogger.com/profile/13321770881353644307noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-8231784566931768362.post-63790792987000826762008-06-29T11:22:00.000-04:002008-06-29T11:22:00.000-04:00Dave,for the last six years I have been teaching t...Dave,<BR/><BR/>for the last six years I have been teaching traditional word problems in algebra 1 for 9th graders. I have felt a bit embarrassed as I justified this...<BR/><BR/>But lately I have changed my tune. Why drop these problems? <BR/><BR/>1. They are contrived. But so are all of their substitutes, except for things too complicated to correctly model. The artificial context on NY State's exams drives me nuts, and I assume it is bad all over.<BR/><BR/>2. They are hard. Right. Even though the kids know that these problems are variations on R*T=D, they remain challenging. For fractional equations t/r1 + t/r2 = 1 (work problems) are even harder. If we already have tough stuff where kids are taught directly what to do, why yank away that last bit of support and ask them to decide which kind of math to use in unfamiliar situations? Doesn't make sense.<BR/><BR/>Average rate problems are great, and your suggestions for building up to them should be used, as nec'y.<BR/><BR/>By the way, since I still talk about properties and that sort of thing, I find operations on real numbers that may or may not be associative, commutative, or both. I like A max B is the greater, A left B = A, A avg B = (A+B)/2...<BR/><BR/>There's that famous study where many teachers had trouble finding operations that were commutative but not associative...<BR/><BR/>JonathanAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-57389755920952329972008-06-19T18:40:00.000-04:002008-06-19T18:40:00.000-04:00I think your approch makes sense. The natureof the...I think your approch makes sense. The nature<BR/>of the material, availabe teaching time and<BR/>the level of the students seem to be the factors<BR/>that dictate which method should be used.<BR/><BR/>I don't think there is anything wrong with textbooks<BR/>that use method (B) to present material. They are<BR/>a great addition to the in class lecture because<BR/>they allow fast reviewing and training of the<BR/>new material - if the concepts have been understood<BR/>in class. <BR/><BR/>Students should be instructed using both methods<BR/>It shows them that problems can be approached from<BR/>a different angle (=part of studyskills).Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-7432150095381359102008-06-19T06:29:00.000-04:002008-06-19T06:29:00.000-04:00Florian--Your question is central to the whole iss...Florian--<BR/>Your question is central to the whole issue of teaching and learning. <BR/><BR/>For each lesson I would make a decision, based on the group I had, whether to <BR/>(A) guide them through the derivation of the formula/algorithm OR <BR/>(B) to state the method and show them how to apply it. <BR/><BR/>I preferred the former but that was not always possible. In all cases, I attempted to provide some motivation for the formula. Needless to say, I spent considerable time planning each lesson to reduce the possibility of confusion. However, students absolutely need some time to 'mess around' with ideas. On a limited controlled basis this can be done IN the classroom and then continued outside when they grapple with other questions for homework. <BR/><BR/>What I have observed in many classrooms and from what we see in most texts, is almost exclusively (B). I could suggest a few reasons for this but I'll leave it to my readers to speculate...<BR/><BR/>Eric--<BR/>For the upper-level students, particularly those with some physics background, your approach would make sense and, in fact, I stressed this in calculus: <BR/><BR/>The average velocity of a particle (defined as "delta s" over "delta t") is the same as the average value of the velocity function (defined as an integral). <BR/><BR/>The special case of the step function is very nice and I think I would have used it for this type of problem if I had thought of it!<BR/><BR/>Introducing average 'speed' to middle schoolers is a bit different however, IMO.<BR/> <BR/>Eric, Florian:<BR/>Did you think my development would make sense to them or would be too confusing? of course, one could only be sure if we see it 'live'!Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-90512899393256053652008-06-18T22:23:00.000-04:002008-06-18T22:23:00.000-04:00I suggest that you graph speed against time, calli...I suggest that you graph speed against time, calling the function <I>s</I>. You get a step function. Then, define the average of <I>s</I> as the integral of <I>s</I> divided by the integral of 1.<BR/><BR/>You can even discuss this sort of integral without any calculus; you just use the area instead. After all, integrals of step functions are what you need in discussing Lesbegue integrals, but high school students couldn't be expected to know that.<BR/><BR/>You can draw the same sort of pictures in probability theory too.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-28403267862289973912008-06-18T18:05:00.000-04:002008-06-18T18:05:00.000-04:00Dave, what are your thoughts on the idea of sepera...Dave, what are your thoughts on the <BR/>idea of seperating the understanding<BR/>of the problem and the act of applying <BR/>a technique to solve it?<BR/><BR/>It appears to me that many students<BR/>will start to form all kinds of ratios<BR/>until they find a the ones that<BR/>solve the problem. Showing a great deal of<BR/>confusion and lack of understanding <BR/>of the actual solution of the problem. Or not?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-16243136140566524942008-06-18T13:12:00.000-04:002008-06-18T13:12:00.000-04:00Florian--Excellent ideas! I gave serious considera...Florian--<BR/>Excellent ideas! I gave serious consideration to approaching all of these using weighted averages. In the case of different rates and equal distances, we can use the fact that if the rates are in the ratio 40:60 or 2:3, then their respective times are in the ratio 3:2, since the times and rates vary inversely in this case. This also leads to the harmonic mean I mentioned. Developing these ideas into a coherent lesson with guided exercises and strategically posed questions is the real challenge however...<BR/><BR/>Hi Janet! Thanks for joining us. You might look through the index in the right sidebar for any posts relating to trig, advanced algebra, precalculus or particular topics. Not too much on STAT however (other than combinatorial math if that helps). I assume we are talking about high school students here. Are you looking for general strategies for the course or ways to develop specific topics? Using multiple representations (verbal, symbolic, numerical, graphical)? If you'd like you can always email me directly at dmarain at geeemail dot com. There are many experienced educators who visit here, so you may want to be more specific about topics you'd like to see. Do you need this for courses you're teaching in Sep?Dave Marainhttps://www.blogger.com/profile/13321770881353644307noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-52415450484681934472008-06-18T11:05:00.000-04:002008-06-18T11:05:00.000-04:00Apologies for breaking the flow, but I'm checking ...Apologies for breaking the flow, but I'm checking in here as a new subscriber. I'm a math teacher in Texas, teaching precal and stat at the moment, so I'm interested in instructional strategies for math. <BR/>JanetMrs. Ricehttps://www.blogger.com/profile/10382317639211536719noreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-60239016752118680342008-06-18T04:11:00.000-04:002008-06-18T04:11:00.000-04:00Correction: 50 is no solution because m1/q1 != m2/...Correction: 50 is no solution because m1/q1 != m2/q2Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8231784566931768362.post-45963726438675865842008-06-18T04:10:00.000-04:002008-06-18T04:10:00.000-04:00When I first did these problems in highschool I di...When I first did these problems in highschool I disliked them because<BR/>there seemed to be no straightforward or efficient way to solve them.<BR/><BR/>Over time I developed a systematic approach for myself that goes like this:<BR/><BR/>In these types of problems there are always two quantities q1, q2<BR/>and a multiplier for each of them m1, m2. Additionaly the two q have<BR/>a unit f=U/u. Also m1 and m2 are either of unit U or unit u.<BR/><BR/>Now, the problem is always given in two points:<BR/><BR/>a) a worded problem that needs to be translated into an equation like this:<BR/><BR/> q1*f*t1 + q2*f*t2<BR/><BR/>Examples:<BR/><BR/> #2: q1=70, m1=4, q2=90, m2=6, f=[averagepts]/[test]<BR/> #3: q1=40, m1=2, q2=60, m2=2, f=[mi]/[hr]<BR/> #4: q1=40, m1=4, q2=60, m2=2, f=[mi]/[hr]<BR/> #5: q1=40, m1=120, q2=60, m2=120, f=[mi]/[hr]<BR/><BR/>b) a value that needs to be determined.<BR/> <BR/> Case 1: Find the average q for m1+m2 where m1,m2 are of unit u.<BR/> Solution: (q1*m1 + q2*m2) / (m1+m2) <BR/><BR/>Examples:<BR/><BR/> #2: (70*4+90*6)/(4+6) = 820/10 = 82<BR/> #3: (40*2+60*2)/(2+2) = 200/4 = 50<BR/> #4: (40*4+60*2)/(4+2) = 280/6 = 46<BR/><BR/> Case 2: Find the average q for m1+m2 where m1,m2 are of unit U. <BR/> Solution: (m1 + m2) / (m1/q1 + m2/q2)<BR/><BR/>Example:<BR/><BR/> #5: (120+120) / (120/40+120/60) = 240/5 = 48.<BR/><BR/> Clearly, the answer is not 5 because m1 != m2, which means that (q1+q1)/2 is no solution.<BR/><BR/>Please note that I ommited the units in the calculations for simplicity. <BR/>What I like about this approach is that it requires only some thinking<BR/>to form the problem into an equation. The rest of the calculation process<BR/>is then pretty intuitive because all values will fall into place. The units<BR/>will be correct and the solution will have a senseful value and unit. And<BR/>it is easier to find your own error sbecause the translation of the problem<BR/>as an equation and the calculation are almost strictly seperated. Just my 2 cents of course ;)Anonymousnoreply@blogger.com