#$&* course Mth 158 1/14 12:30 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
.............................................
Given Solution: `aThe graph goes through the x axis when y = 0 and through the y axis when x = 0. The x-intercept is therefore when 0 = 3x - 4, so 4 = 3x and x = 4/3. The y-intercept is when y = 3 * 0 - 4 = -4. Thus the x intercept is at (4/3, 0) and the y intercept is at (0, -4). Your graph should confirm this. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q002. Does the steepness of the graph in the preceding exercise (of the function y = 3x - 4) change? If so describe how it changes. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: No it doesn’t change confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aThe graph forms a straight line with no change in steepness. STUDENT COMMENT Ok, I may not understand what exactly it meant by steepness, I was thinking since it was increasing it would also be getting steeper????? INSTRUCTOR RESPONSE A graph can increase while getting steeper and steeper; or it can increase while getting less and less steep. Or it can increase with no change in steepness. Analogies: When you walk up a hill, typically as you approach the top the slope starts to level off--it gets less steep. When you go up a ramp the steepness stays the same until you get to the end of the ramp. When you start climbing a hill, typically it gets steeper for awhile, the stays at about a constant slope, then gets less steep toward the top. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q003. What is the slope of the graph of the preceding two exercises (the function is y = 3x - 4;slope is rise / run between two points of the graph)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2/1 = 3 what it is saying is that between any two points of the graph the rise / run = 3 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aBetween any two points of the graph rise / run = 3. For example, when x = 2 we have y = 3 * 2 - 4 = 2 and when x = 8 we have y = 3 * 8 - 4 = 20. Between these points the rise is 20 - 2 = 18 and the run is 8 - 2 = 6 so the slope is rise / run = 18 / 6 = 3. Note that 3 is the coefficient of x in y = 3x - 4. Note the following for reference in subsequent problems: The graph of this function is a straight line. The graph increases as we move from left to right. We therefore say that the graph is increasing, and that it is increasing at constant rate because the steepness of a straight line doesn't change. EXPANDED EXPLANATION Any student who has completed Algebra I and Algebra II should be familiar with slope calculations. Most students are. However a number of students appear to be very fuzzy on the concept, and I suspect that not all prerequisite courses cover this concept adequately (though I am confident that it's done well at VHCC). Also a number of students haven't taken a math course in awhile, and might simply be a bit rusty with this idea. In any case the following expanded explanation might be helpful to some students: Slope = rise / run. The rise between two graph points is the change in the y coordinate. The run is the change in the x coordinate. Our function is y = 3 x - 4. When x = 2, we substitute 2 for x to get y = 3 * 2 - 4, which is equal to 2. When x = 8, we substitute 8 for x to get y = 3 * 8 - 4, which is equal to 20. The graph therefore contains the points (2, 2) and (8, 20). You should have made a graph showing these points. If not you should do so now. As you go from point to point your y coordinate goes from 2 to 20. So the 'rise' between the points is 20 - 2 = 18. Your x coordinate goes from 2 to 8. So the 'run' between the points is 8 - 2 = 6. The slope is rise / run = 18 / 6 = 3. The numbers 2 and 8, which were used for the x values, were chosen arbitrarily. Any other two x values would have given you different coordinates, likely with different rise and run. However whatever two x values you use, you will get the same slope. The slope of this graph is constant, and is equal to 3. STUDENT QUESTION Am I not allowed to utilize my calculus tools, yet? Couldn't I have just taken the derivative for the function, y = 3x -4 to obtain 3 as the slope? However, I do know how to do both ways. Which is the more preferred method? INSTRUCTOR RESPONSE This exercise develops a language for describing some aspects of graphs, and does not assume calculus tools. Of course it's fine to use the calculus tools if you have them, as long as you understand the problem at the more basic level as well. Unfortunately, not every student who has had a calculus course would know how to apply those tools to this situation (for example, I've had students from other institutions who have made A's in Applied Calculus courses from other (not particularly reputable) institutions, who didn't understand the concept of a slope). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I knew the answer and I knew how to find the answer but you solution gave me a better understanding of the process. ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q004. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (0,0) (1,1) (2,4) (4,9) to tell if its increasing or decreasing all we need to look at is the y values which are 0, 1, 4, 9 which are all increasing as we move to the right The increase between the points are 1, 3, 5, so with this information we can plainly see that the graph is increasing at an increasing rate. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aGraph points include (0,0), (1,1), (2,4) and (3,9). The y values are 0, 1, 4 and 9, which increase as we move from left to right. The increases between these points are 1, 3 and 5, so the graph not only increases, it increases at an increasing rate STUDENT QUESTION: I understand increasing...im just not sure at what rate...how do you determine increasing at an increasing rate or a constant rate? INSTRUCTOR RESPONSE: Does the y value increase by the same amount, by a greater amount or by a lesser amount every time x increases by 1? In this case the increases get greater and greater. So the graph increases, and at an increasing rate. *&*&. STUDENT QUESTION: I am a little hazy on what the steepness is INSTRUCTOR RESPONSE: The hill analogy I used above might be helpful. Formally, steepness could be defined as the magnitude of the slope, i.e., the absolute value of the slope. Two graphs with respective slopes 4 and -4 would be equally steep; both would have slope of magnitude 4. Both of these graphs would be steeper than, say a graph with slope 3 or -3. NOTE FOR STUDENT WITH CALCULUS BACKGROUND (students who haven't had calculus should ignore this; this explanation is optional even for students who have had calculus) In terms of the calculus, the derivative function is easily seen to be y ' = 2 x, which is positive and increasing, and which therefore implies an increasing slope. Since in this case the slope is positive, which implies that the function is increasing, the increasing slope therefore implies that the value of the function is increasing at an increasing rate. Another terminology which is standard in calculus: If the slope is increasing then the shape of the graph is concave upward. So we could describe this graph as increasing and concave upward. This could also be explained in terms of the second derivative, y '' = 2, which is positive everywhere. The positive second derivative implies that the graph is concave up. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q005. Make a table of y vs. x for y = x^2. Graph y = x^2 between x = -3 and x = 0. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (-3, 9) (-2, 4) (-1, 1) (0, 0) to tell if the graph is increasing or decreasing we just need to look at the values of y which are 9, 4, 1, 0 the numbers are all decreasing which tells us that the graph is decreasing The steepness is also decreasing 5, 3, 1 which so it is decreasing at a decreasing rate. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aFrom left to right the graph is decreasing (points (-3,9), (-2,4), (-1,1), (0,0) show y values 9, 4, 1, 0 as we move from left to right ). The magnitudes of the changes in x from 9 to 4 to 1 to 0 decrease, so the steepness is decreasing. Thus the graph is decreasing, but more and more slowly. We therefore say that the graph on this interval is decreasing at a decreasing rate. NOTE FOR STUDENT WITH CALCULUS BACKGROUND (students who haven't had calculus should ignore this; this explanation is optional even for students who have had calculus) In terms of the calculus, the derivative function is easily seen to be y ' = 2 x, which is positive and increasing, and which therefore implies an increasing slope. Since in this case the slope is negative, which implies that the function is decreasing, the increasing slope therefore implies that the rate of decrease is decreasing. The value of the function is therefore decreasing at a decreasing rate. Another terminology which is standard in calculus: If the slope is increasing then the shape of the graph is concave upward. So we could describe this graph as decreasing and concave upward. This could also be explained in terms of the second derivative, y '' = 2, which is positive everywhere. The positive second derivative implies that the graph is concave up. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q006. Make a table of y vs. x for y = `sqrt(x). [note: `sqrt(x) means 'the square root of x']. Graph y = `sqrt(x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (0,0) (1,1) (2,1.414) (3, 1.732) first we need to look at our y values 0, 1, 1.414, 1.732 the graph is increasing at a decreasing rate confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aIf you use x values 0, 1, 2, 3, 4 you will obtain graph points (0,0), (1,1), (2,1.414), (3. 1.732), (4,2). The y value changes by less and less for every succeeding x value. Thus the steepness of the graph is decreasing. The graph would be increasing at a decreasing rate. If the graph represents the profile of a hill, the hill starts out very steep but gets easier and easier to climb. You are still climbing but you go up by less with each step, so the rate of increase is decreasing. If your graph doesn't look like this then you probably are not using a consistent scale for at least one of the axes. If your graph isn't as described take another look at your plot and make a note in your response indicating any difficulties. STUDENT QUESTION: I am still unsure why the steepness is decreasing, I see why going from right to left, but the graph looks linear? INSTRUCTOR RESPONSE: The y value increases, but it changes by less and less for every succeeding x value. So the graph is increasing, but by less and less with each step. It's increasing but at a decreasing rate. The graph does not look linear. If it does, then it's probably because your x and/or y axis is not scaled in equal increments. NOTE FOR CALCULUS-PREPARED STUDENTS (students who haven't had calculus should ignore this; this explanation is optional even for students who have had calculus) In terms of the calculus, the derivative function is easily seen to be y ' = 1 / (2 sqrt(x)), which is positive but decreasing, and which therefore implies a decreasing slope. Since in this case the slope is positive, which implies that the function is increasing, the decreasing slope therefore implies that the rate of increase is decreasing. The value of the function is therefore increasing at a decreasing rate. Another terminology which is standard in calculus: If the slope is increasing then the shape of the graph is concave upward. So we could describe this graph as increasing and concave downward. This could also be explained in terms of the second derivative, y '' = -1 / (4 x^(3/2)), which is negative on this interval. The negative second derivative implies that the graph is concave down. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q007. Make a table of y vs. x for y = 5 * 2^(-x). Graph y = 5 * 2^(-x) between x = 0 and x = 3. Would you say that the graph is increasing or decreasing? Does the steepness of the graph change and if so, how? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First we have to remember that we have to get rid of the negative exponent and we do that like this y = 5 * 1/(2^x) (0, 5) (1,2.5) (2,1.25) (3, .625) now we take a look at our y values 5, 2.5, 1.25, .625 and we can see that it is decreasing at a decreasing rate. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** From basic algebra recall that a^(-b) = 1 / (a^b). So, for example: 2^-2 = 1 / (2^2) = 1/4, so 5 * 2^-2 = 5 * 1/4 = 5/4. 5* 2^-3 = 5 * (1 / 2^3) = 5 * 1/8 = 5/8. Etc. The decimal equivalents of the values for x = 0 to x = 3 will be 5, 2.5, 1.25, .625. These values decrease, but by less and less each time. The graph is therefore decreasing at a decreasing rate. ** STUDENT QUESTION I don’t understand how the graph decreases at a decreasing rate because it decreases by half every time. The ˝ is constant. INSTRUCTOR RESPONSE The values decrease by a factor of 1/2 every time. That means each number would be multiplied by 1/2 to get the next. As a result the numbers we are halving keep decreasing. Half of 5 is 2.5; half of 2.5 is 1.25; half of 1.25 is .625. The decreases from one number to the next are respectively 2.5, 1.25 and .625. If the y values 5, 2.5, 1.25, .625 are placed at equal x intervals, it should be clear that the graph is decreasing at a decreasing rate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q008. Suppose you stand still in front of a driveway. A car starts out next to you and moves away from you, traveling faster and faster. If y represents the distance from you to the car and t represents the time in seconds since the car started out, would a graph of y vs. t be increasing or decreasing? Would you say that the graph is increasing at an increasing rate, increasing at a constant rate, increasing at a decreasing rate, decreasing at an decreasing rate, decreasing at a constant rate, or decreasing at a decreasing rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph would be increasing at an increasing rate because it says that it is moving away for you traveling faster and faster which would say that since the rate in which the car is traveling is increasing and the distance is also getting greater the slope would have to increase on the graph. confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** The speed of the car increases so it goes further each second. On a graph of distance vs. clock time there would be a greater change in distance with each second, which would cause a greater slope with each subsequent second. The graph would therefore be increasing at an increasing rate. ** STUDENT COMMENT I don’t fully understand a distance vs. time graph. INSTRUCTOR RESPONSE If y represents the distance from you to the car and t represents the time in seconds since the car started out, then the graph of y vs. t is a graph of distance vs. clock time. The car is speeding up, so in any series of equal time intervals it moves further with each new interval. The distance it moves on an interval is represented by the difference between the y coordinates, so if it move further during an interval the 'rise' of the graph on that interval will be greater. If the intervals are equally spaced along the t axis, the result is an increasing graph with increasing slope. This is best understood by sketching the graph according to this description. STUDENT QUESTION I understand the clock time but could you give me some examples of numbers to sketch a graph. I am drawing a blank to how to make myself understand.????? INSTRUCTOR RESPONSE If the car's velocity for the first second averages 1 ft / sec, then in subsequent second 3 ft / sec, then 5 ft / sec, then 7 ft / sec, it will move 1 foot during the first second, 3 feet during the next, 5 feet during the next and 7 feet during the next. A graph of velocity vs. clock time would be a straight line, since the velocity increases by the same amount every second. However the positions of the car, as measured from the starting point, would be position 1 foot after 1 second position 4 feet after 2 seconds (the position changes by 3 feet, started this second at 1 ft, so the car ends up with position 4 feet) position 9 feet after 3 seconds (the position changes by 5 feet, started this second at 4 ft, so the car ends up with position 9 feet) position 16 feet after 4 seconds (the position changes by 7 feet, started this second at 9 ft, so the car ends up with position 16 feet) So the graph of position vs. clock time has positions 0, 1, 4, 9 and 16 feet after 0, 1, 2, 3 and 4 seconds, respectively. The position vs. clock time graph is therefore increasing at an increasing rate. Let me know if this doesn't answer your question. STUDENT QUESTION I still don’t totally understand why it would necessarily be increasing at an increasing rate. Couldn’t it be a decreasing or even a standard rate as I mentioned above? INSTRUCTOR RESPONSE If the car speeds up then its distance from its starting position increases at an increasing rate. Its speed might be increasing at an increasing, constant or decreasing rate, but not its position. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #(*!
.............................................
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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q003.If you make $60,000 per year then how much do you make per month? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: $60,000 per year 12 months in a year $60,000 / 12 = $5,000 per month confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): Ok ------------------------------------------------ Self-critique Rating: OK ********************************************* 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 $5000 per month? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It would have to be an average for the total is spread out over months with possible fluctuations. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 $5000 per month. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q005. If you travel 300 miles in 6 hours, at what average rate are you covering distance, and why do we say average rate instead of just plain rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 300 miles total distance 6 hours total time 300 / 6 = 50 miles per hour Average is said because there are instances where you will go slower and faster. It does not say you maintain the same speed. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: 1200 total miles 60 total gallon of gas 1200 / 60 = 20 miles per gallon with respect to gas 60 / 1200 = 0.05 gallons with respect to miles confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: Because the problem are spread out for months, miles, dollars and minutes. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 $60,000 or 1200 miles. Thus when we divide by the number of months or the number of gallons, we are in fact calculating an average rate. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q008. In a study of how lifting strength is influenced by various ways of training, a study group was divided into 2 subgroups of equally matched individuals. The first group did 10 pushups per day for a year and the second group did 50 pushups per day for year. At the end of the year to lifting strength of the first group averaged 147 pounds, while that of the second group averaged 162 pounds. At what average rate did lifting strength increase per daily pushup? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2 groups 10 pushups for first group 50 pushups for second group 147 lbs group 1 162 lbs group 2 50 – 10 = 40 pushups difference 162 – 147 = 15 lbs difference 15 / 40 = 0.375 lbs average per daily pushup confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: 2 groups 30 pushups per day for both groups 10 lb weight for group 1 30 lb. weight group 2 171 lbs lifting strength group 1 188 lbs lifting strength group 2 30 – 10 = 20 lbs weight difference 188 – 171 = 17 lifting lbs difference 20 / 17 = 0.85 lbs lifting strength increase confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The difference in lifting strength was 17 pounds, as a result of a 20 pound difference in added weight. The average rate at which strength increases with respect added weight would therefore be 17 lifting pounds / (20 added pounds) = .85 lifting pounds / added pound. The strength advantage was .85 lifting pounds per pound of added weight, on the average. You need to make note of anything in the given solution that you didn't understand when you solved the problem. If new ideas have been introduced in the solution, you need to note them. If you notice an error in your own thinking then you need to note that. In your own words, explain anything you didn't already understand and save your response as Notes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q010. During a race, a runner passes the 100-meter mark 12 seconds after the start and the 200-meter mark 22 seconds after the start. At what average rate was the runner covering distance between those two positions? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 12 + 22 = 34 total seconds 200 meters total 200 / 34 = 5.88 seconds per meter confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): I counted the entire 200 meters and not just between the two points. I see where I went wrong. ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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: 300 – 100 = 200 meters total distance 10 + 9 = 19 meters per second 200 / 19 = 10.5 second average confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: We did not have enough information about the period confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): Right, we were given quantities not rates. ------------------------------------------------ Self-critique Rating: 3 "