#$&* course Phy 231 1/23 5pm 001. Rates
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Given Solution: The rate at which you are earning money is the number of dollars per hour you are earning. You are earning money at the rate of 50 dollars / (5 hours) = 10 dollars / hour. It is very likely that you immediately came up with the $10 / hour because almosteveryone is familiar with the concept of the pay rate, the number of dollars per hour. Note carefully that the pay rate is found by dividing the quantity earned by the time required to earn it. Time rates in general are found by dividing an accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003.If you make $60,000 per year then how much do you make per month? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Rate of pay in dollars per year can be converted to dollars per month by converting years to months. One way to set this type of conversion up is to multiply the figure with the original unit (in this case dollars per year) by a fraction relating the original unit to the desired unit (in this case months) and equalling 1. ($60,000/1 year)(1 year/12 months). Since the unit ""year"" appears in both the numerator and the denominator, we can cancel it out, leaving us with ($60,000/1)(1/12months) or $60,000/(12 months). Then we can do simple arithmetic and divide 60,000/12, giving us $5,000/month. This is the way to organize unit conversions that makes the most sense to me; it allows me to convert more than one piece of the puzzle if necessary without getting lost. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Most people will very quickly see that we need to divide $60,000 by 12 months, giving us 60,000 dollars / (12 months) = 5000 dollars / month. Note that again we have found a time rate, dividing the accumulated quantity by the time required to accumulate it. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. Suppose that the $60,000 is made in a year by a small business. Would be more appropriate to say that the business makes $5000 per month, or that the business makes an average of $5000 per month? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It would be more appropriate to say the business makes an average of $5000 each month. Even if revenue came in on a fairly regular monthly schedule (say, a fitness center that charges customers a monthly fee), other variances would be likely. Customers come and go, the electric bill changes seasonally, etc. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Small businesses do not usually make the same amount of money every month. The amount made depends on the demand for the services or commodities provided by the business, and there are often seasonal fluctuations in addition to other market fluctuations. It is almost certain that a small business making $60,000 per year will make more than $5000 in some months and less than $5000 in others. Therefore it is much more appropriate to say that the business makes and average of $5000 per month. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Average rate in miles per hour times travel time in hours equals total distance traveled. So in this case, average rate in miles per hour * 6 hours = 300 miles. Rearranging, I get: average rate in miles per hour = (300 miles)/(6 hours) =50 miles/hour This is an average rate because it does not measure rate of speed at a particular instant, but over a long span of time; it is likely that there are variations in rate of speed at any given moment, as well as starts and stops. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The average rate is 50 miles per hour, or 50 miles / hour. This is obtained by dividing the accumulated quantity, the 300 miles, by the time required to accumulate it, obtaining ave rate = 300 miles / ( 6 hours) = 50 miles / hour. Note that the rate at which distance is covered is called speed. The car has an average speed of 50 miles/hour. We say 'average rate' in this case because it is almost certain that slight changes in pressure on the accelerator, traffic conditions and other factors ensure that the speed will sometimes be greater than 50 miles/hour and sometimes less than 50 miles/hour; the 50 miles/hour we obtain from the given information is clearly and overall average of the velocities. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. If you use 60 gallons of gasoline on a 1200 mile trip, then at what average rate are you using gasoline, with respect to miles traveled? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Rate of gasoline usage would be expressed in this case in gallons per mile. We can calculate this by comparing the total gallons used to the total miles traveled by dividing. rate of gasoline usage = (60 gallons)/(1200 miles) =1/20 gallon/mile or .05 gallons/mile. That sounds strange when we are used to discussing miles per gallon. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The rate of change of one quantity with respect to another is the change in the first quantity, divided by the change in the second. As in previous examples, we found the rate at which money was made with respect to time by dividing the amount of money made by the time required to make it. By analogy, the rate at which we use fuel with respect to miles traveled is the change in the amount of fuel divided by the number of miles traveled. In this case we use 60 gallons of fuel in 1200 miles, so the average rate it 60 gal / (1200 miles) = .05 gallons / mile. Note that this question didn't ask for miles per gallon. Miles per gallon is an appropriate and common calculation, but it measures the rate at which miles are covered with respect to the amount of fuel used. Be sure you see the difference. Note that in this problem we again have here an example of a rate, but unlike previous instances this rate is not calculated with respect to time. This rate is calculated with respect to the amount of fuel used. We divide the accumulated quantity, in this case miles, by the amount of fuel required to cover t miles. Note that again we call the result of this problem an average rate because there are always at least subtle differences in driving conditions that result in more or fewer miles covered with a certain amount of fuel. It's very important to understand the phrase 'with respect to'. Whether the calculation makes sense or not, it is defined by the order of the terms. In this case gallons / mile tells you how many gallons you are burning, on the average, per mile. This concept is not as familiar as miles / gallon, but except for familiarity it's technically no more difficult. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. STUDENT COMMENT Very Tricky! I thought I had a rhythm going. I understand where I messed up. I am comfortable with the calculations. INSTRUCTOR RESPONSE There's nothing wrong with your rhythm. As I'm sure you understand, there is no intent here to trick, though I know most people will (and do) tend to give the answer you did. My intent is to make clear the important point that the definition of the terms is unambiguous and must be read carefully, in the right order. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q007. The word 'average' generally connotes something like adding two quantities and dividing by 2, or adding several quantities and dividing by the number of quantities we added. Why is it that we are calculating average rates but we aren't adding anything? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We aren't adding anything because the premise of these problems has already done the adding for us. When we add up a bunch of pieces of something, or versions of something, or timed trials of something and then dividing by the number of pieces or versions or trials, we are --mathematically speaking-- finding a grand total and dividing by the number of components of that grand total. In many cases, the grand total is not terribly meaningful to what we are trying to understand. But in these cases, the grand total carries some significance: total profits in a year, for instance. But essentially, when we are posed a question that includes a year's worth of profits, that is another way of saying January's profits plus February's profits plus March's profits and so on. That's the ""adding"" part of finding an average. It has already happened. In the case of finding average gas usage during a 1200 mile trip, the problem has already added up gas usage for mile 1 plus gas usage for mile 2, etc., giving us a grand total of gas usage for miles 1 through 1200; it just doesn't phrase it that way. So all that's left for us to find the average is the second half, the division. We divide by months, or by miles, or whatever the problem asks us to. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The word 'average' in the context of the dollars / month, miles / gallon types of questions we have been answering was used because we expect that in different months different amounts were earned, or that over different parts of the trip the gas mileage might have varied, but that if we knew all the individual quantities (e.g., the dollars earned each month, the number of gallons used with each mile) and averaged them in the usual manner, we would get the .05 gallons / mile, or the $5000 / month. In a sense we have already added up all the dollars earned in each month, or the miles traveled on each gallon, and we have obtained the total $60,000 or 1200 miles. Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q008. In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals. The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year. At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We assume both groups could lift more than zero pounds to start with, and we know they could lift an equal number of pounds. But we don't know how many pounds they could lift prior to training. Let's assign x to equal the number of pounds both groups could lift initially. We can then assign y to equal the increase in lifting strength per daily pushup. Then we can express the relationship as follows: x + 10y = 147 (initial lifting strength plus a year's worth of 10 daily pushups led to lifting strength of 147 lbs) x + 50y = 162 (initial lifting strength plus a year's worth of 50 daily pushups led to lifting strength of 162 lbs) Using substitution, we can solve for x in the first equation and plug the result into the second equation x = 147 - 10y x + 50y = 162 (147 - 10y) + 50y = 162 147 + 40y = 162 40y = 162-147 40y = 15 y=15/40 y=.375 lbs per pushup So, each pushup done on a daily basis for a year results in a .375 lb increase in lifting strength. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The second group had 15 pounds more lifting strength as a result of doing 40 more daily pushups than the first. The desired rate is therefore 15 pounds / 40 pushups = .375 pounds / pushup. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. STUDENT COMMENT: I have a question with respect as to how the question is interpreted. I used the interpretation given in the solution to question 008 to rephrase the question in 009, but I do not see how this is the correct interpretation of the question as stated. INSTRUCTOR RESPONSE: This exercise is designed to both see what you understand about rates, and to challenge your understanding a bit with concepts that aren't always familiar to students, despite their having completed the necessary prerequisite courses. The meaning of the rate of change of one quantity with respect to another is of central importance in the application of mathematics. This might well be your first encounter with this particular phrasing, so it might well be unfamiliar to you, but it is important, unambiguous and universal. You've taken the first step, which is to correctly apply the wordking of the preceding example to the present question. You'll have ample opportunity in your course to get used to this terminology, and plenty of reinforcement. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I don't know why I didn't see that the EXTRA 15 lbs per EXTRA 40 pushups was enough to find the rate, considering the problem implied that the rate would be constant. So the second group's pound-per-pushup rate exceeding the first group's performance is the rate, period. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q009. In another part of the study, participants all did 30 pushups per day, but one group did pushups with a 10-pound weight on their shoulders while the other used a 30-pound weight. At the end of the study, the first group had an average lifting strength of 171 pounds, while the second had an average lifting strength of 188 pounds. At what average rate did lifting strength increase with respect to the added shoulder weight? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Here, I suppose the question is implying that we should assume every pound on the shoulders will lead to a predictable and consistent increase in strength. But what seems different here is that we don't know what their performance would be if they did pushups with zero pounds on their shoulders. Using the premise of the first question, though, I will calculate as follows: A 17-pound difference in lifting strength (188-171) results from a twenty-pound difference in shoulder weight (30-10) at the end of the training period. If it's really that simple again, the rate would be 17/20, or .85 pounds of lifting strength for every pound of shoulder weight applied to a 30-daily-pushup regimen. If so, we could calculate that 30 daily pushups without shoulder weight at all would help as follows: 171 - (10*.85) = 171 - 8.5 = 162.5 That seems plausible enough, so I don't think I've encountered anything to make me question the simplicity of this. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight. The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound. The strength advantage was .85 lifting pounds per pound of added weight, on the average. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I'm really having a lot of trouble trusting that the stated conditions are enough to know with certainty what is going on here. Probably, if the problem just said the people walked around all day with varying amounts of weight on their shoulders, it would seem obvious to me. But the mention of a certain number of pushups makes me think it's going to be more complicated: like we'll have to multiply number of pushups with amount of shoulder weight, or something like that. ------------------------------------------------ Self-critique Rating: 2 ********************************************* Question: `q010. During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start. At what average rate was the runner covering distance between those two positions? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The distance travelled is the final position minus the initial position. (In this case, initial position for what we are trying to calculate is not the same as the starting line of the race!) So distance is 200m-100m, or 100 meters of distance covered between the two points we're concerned with. The elapsed time during that same period, final time minus initial time for the period in question, is 22s-12s, or 10 seconds. To find the average rate of speed, we divide distance by elapsed time, or 100m/10s, for an average rate of 10m/s. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The runner traveled 100 meters between the two positions, and required 10 seconds to do so. The average rate at which the runner was covering distance was therefore 100 meters / (10 seconds) = 10 meters / second. Again this is an average rate; at different positions in his stride the runner would clearly be traveling at slightly different speeds. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. STUDENT QUESTION Is there a formula for this is it d= r*t or distance equal rate times time?????????????????? INSTRUCTOR RESPONSE That formula would apply in this specific situation. The goal is to learn to use the general concept of rate of change. The situation of this problem, and the formula you quote, are just one instance of a general concept that applies far beyond the context of distance and time. It's fine if the formula helps you understand the general concept of rate. Just be sure you work to understand the broader concept. Note also that we try to avoid using d for the name of a variable. The letter d will come to have a specific meaning in the context of rates, and to use d as the name of a variable invite confusion. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q011. During a race, a runner passes the 100-meter mark moving at 10 meters / second, and the 200-meter mark moving at 9 meters / second. What is your best estimate of how long it takes the runner to cover the intervening 100 meter distance? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average velocity during that distance is (initial velocity plus final velocity)/2, or (10m/s + 9m/s)/2, or (19m/s)/2, or 9.5m/s. We know the runner is covering 100 meters at that average velocity. distance traveled divided by elapsed time equals average velocity. (100 meters)/elapsed time = 9.5 m/s reorganizing, we get: (100 meters)/(9.5 meters/second) = 10.52 seconds confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: At 10 meters/sec, the runner would require 10 seconds to travel 100 meters. However the runner seems to be slowing, and will therefore require more than 10 seconds to travel the 100 meters. We don't know what the runner's average speed is, we only know that it goes from 10 m/s to 9 m/s. The simplest estimate we could make would be that the average speed is the average of 10 m/s and 9 m/s, or (10 m/s + 9 m/s ) / 2 = 9.5 m/s. Taking this approximation as the average rate, the time required to travel 100 meters will be (100 meters) / (9.5 m/s) = 10.5 sec, approx.. Note that simply averaging the 10 m/s and the 9 m/s might not be the best way to approximate the average rate--for example we if we knew enough about the situation we might expect that this runner would maintain the 10 m/s for most of the remaining 100 meters, and simply tire during the last few seconds. However we were not given this information, and we don't add extraneous assumptions without good cause. So the approximation we used here is pretty close to the best we can do with the given information. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Here, I was actually MORE sure of the precision of my numbers than I should have been. I was assuming that the runner was decelerating at a constant rate. Absent other information, as you said, finding the average is the best guess we can make, but I didn't even realize I was guessing this time. This is a reminder to read closely! ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q012. We just averaged two quantities, adding them and dividing by 2, to find an average rate. We didn't do that before. Why we do it now? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The two steps of finding an average are adding several components of something and then dividing by the number of components. Here, we knew two components of the bigger picture we were looking for and had to add them up first (whereas earlier the bigger picture was already presented to us, and we were trying to deduce something about the components). This time we needed the value of average velocity, and the components we had were already in terms of average velocity, and there were only two of them-- just starting and ending velocity. Adding them up and dividing by two gives us the most accurate average we can get from the information given. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In previous examples the quantities weren't rates. We were given the amount of change of some accumulating quantity, and the change in time or in some other quantity on which the first was dependent (e.g., dollars and months, miles and gallons). Here we are given 2 rates, 10 m/s and 9 m/s, in a situation where we need an average rate in order to answer a question. Within this context, averaging the 2 rates was an appropriate tactic. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. STUDENT QUESTION: I thought the change of an accumulating quantity was the rate? INSTRUCTOR RESPONSE: Quick response: The rate is not just the change in the accumulating quantity; if we're talking about a 'time rate' it's the change in the accumulating quantity divided by the time interval (or in calculus the limiting value of this ratio as the time interval approaches zero). More detailed response: If quantity A changes with respect to quantity B, then the average rate of change of A with respect to B (i.e., change in A / change in B) is 'the rate'. If the B quantity is clock time, then 'the rate' tells you 'how fast' the A quantity accumulates. However the rate is not just the change in the quantity A (i.e., the change in the accumulating quantity), but change in A / change in B. For students having had at least a semester of calculus at some level: Of course the above generalizes into the definition of the derivative. y ' (x) is the instantaneous rate at which the y quantity changes with respect to x. y ' (x) is the rate at which y accumulates with respect to x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I'm solid on this, but I didn't explain it the same way you did, making the distinction between rates and accumulated change. Here's how I understand your explanation about rates: In thinking about a rate as a fraction, we found or were given the average of the numerator only and needed to divide by the given denominator. Here, we are given the information already in fraction form, as a rate. So we find the average of the complete fractions, rather than the numerator only. Is that correct? ------------------------------------------------ Self-critique Rating: 2