#$&* course Phy 232 6/7 9:15am 005. Calculus *********************************************
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Given Solution: `aSlope = rise / run. Between points (7, 17) and (10, 29) we get rise / run = (29 - 17) / (10 - 7) =12 / 3 = 4. The slope between points (3, 5) and (7, 17) is 3 / 1. (17 - 5) / (7 -3) = 12 / 4 = 3. The segment with slope 4 is the steeper. The graph being a smooth curve, slopes may vary from point to point. The slope obtained over the interval is a specific type of average of the slopes of all points between the endpoints. 2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001 and 2.0001, what values are taken by the expression 1 / (x - 2)? 1. As the process continues, with x getting closer and closer to 2, what happens to the values of 1 / (x-2)? 2. Will the value ever exceed a billion? Will it ever exceed one trillion billions? 3. Will it ever exceed the number of particles in the known universe? 4. Is there any number it will never exceed? 5. What does the graph of y = 1 / (x-2) look like in the vicinity of x = 2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = 2.1, 1/(x-2) = 1/(2.1-2) = 1/0.1 = 10 x = 2.01, 1/(x-2) = 1/(2.01-2) = 1/(0.01) = 100 x = 2.001, 1/(x-2) = 1/(2.001-2) = 1/(0.001) = 1000 x = 2.0001, 1/(x-2) = 1/(2.0001-2) = 1/(0.0001) = 10000 (1) As x gets closer and closer to 2, the values get higher and higher. The closer the value to 2, the higher the value becomes. (4)Technically you can infinitely add zeroes to make the value closer and closer to 2, so therefore the value can be infinitely large. (2, 3) It can exceed one billion, one trillion, and the number of particles in the universe; it is infinite. There is no number it won’t exceed; the value will go to infinity. (5) The graph of y = 1/(x-2) does not exist at the point x = 2. At x = 2, the value would be 1/(2-2) which equals 1/0. Since you cannot divide 1 by 0, the point of x = 2 does not exist on this graph. This graph would therefore contain a vertical asymptote, since x=2 does not exist but the numbers before and after do. Even the smallest change closer to 2 (i.e. 2.0001 to 2.00001) will create a bigger and bigger change in the value. Approaching 2 from the right, the value will go to infinity, and approaching 2 from the left the value will go to negative infinity. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aFor x = 2.1, 2.01, 2.001, 2.0001 we see that x -2 = .1, .01, .001, .0001. Thus 1/(x -2) takes respective values 10, 100, 1000, 10,000. It is important to note that x is changing by smaller and smaller increments as it approaches 2, while the value of the function is changing by greater and greater amounts. As x gets closer in closer to 2, it will reach the values 2.00001, 2.0000001, etc.. Since we can put as many zeros as we want in .000...001 the reciprocal 100...000 can be as large as we desire. Given any number, we can exceed it. Note that the function is simply not defined for x = 2. We cannot divide 1 by 0 (try counting to 1 by 0's..You never get anywhere. It can't be done. You can count to 1 by .1's--.1, .2, .3, ..., .9, 1. You get 10. You can do similar thing for .01, .001, etc., but you just can't do it for 0). As x approaches 2 the graph approaches the vertical line x = 2; the graph itself is never vertical. That is, the graph will have a vertical asymptote at the line x = 2. As x approaches 2, therefore, 1 / (x-2) will exceed all bounds. Note that if x approaches 2 through the values 1.9, 1.99, ..., the function gives us -10, -100, etc.. So we can see that on one side of x = 2 the graph will approach +infinity, on the other it will be negative and approach -infinity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q003. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average height of the first trapezoid (points (3,5) and (7,9)) is about 7 units ((5+9)/2 = 7), while the width is 4 units. The average height of the second trapezoid is about 2 units ((4+2)/2=2), while the width is 40 units. The second trapezoid obviously has a greater area because while the height of the first one is a little more than three times bigger than the height of the second one, the width of the second one is 10 times bigger than the width of the first. The difference in width is a lot larger than the difference in height, so the width determines which has a greater area, and in this case the second trapezoid would be the greater. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aYour sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area. To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area. This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q004. If f(x) = x^2 (meaning 'x raised to the power 2') then which is steeper, the line segment connecting the x = 2 and x = 5 points on the graph of f(x), or the line segment connecting the x = -1 and x = 7 points on the same graph? Explain the basis of your reasoning. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since f(x) = x^2, then: x = 2, y = 2^2 = 4 Point: (2,4) x = 5, y = 5^2 = 25 Point: (5, 25) x = -1, y = (-1)^2 = 1 Point: (-1, 1) x = 7, y = 7^2 = 49 Point: (7, 49) Slope of line connecting (2,4) and (5,25): (25-4)/(5-2) = 21/3 = 7 Slope of line connecting (-1, 1) and (7, 49): (49-1)/(7-(-1)) = 48/8 = 6 The line segment connecting x = 2 to x = 5 is steeper because after finding the y values using the equation, I found the slope using rise over run and this segment had a higher slope. A higher slope means the steepness is great. This segment had a slope of 7, while the other segment had a slope of 6, making the first segment steeper. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe line segment connecting x = 2 and the x = 5 points is steeper: Since f(x) = x^2, x = 2 gives y = 4 and x = 5 gives y = 25. The slope between the points is rise / run = (25 - 4) / (5 - 2) = 21 / 3 = 7. The line segment connecting the x = -1 point (-1,1) and the x = 7 point (7,49) has a slope of (49 - 1) / (7 - -1) = 48 / 8 = 6. The slope of the first segment is greater. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q005. Suppose that every week of the current millennium you go to the jeweler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time ( this is so), that the the gold remains undisturbed (maybe, maybe not so), that no other source adds gold to your backyard (probably so), and that there was no gold in your yard before.. 1. If you construct a graph of y = the number of grams of gold in your backyard vs. t = the number of weeks since Jan. 1, 2000, with the y axis pointing up and the t axis pointing to the right, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly? 2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week. 3. Answer the same question assuming that every week you bury half the amount you did the previous week. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1. Since the number of gold is increasing at the same rate each week, this graph would be a rising straight line. It would be rising because more gold is added each week, and straight because the same amount is added each week. 2. Since the number of gold added each week is increasing every week, it would be a line that rises faster and faster. This is because each week more gold than the last week is added, so the change in the y axis is exponential and will get bigger as the x axis (time) gets bigger. 3. Although less gold is being bought each week, the amount of gold in the yard is increasing because gold is still being added each week. The line will be rising, but more and more slowly. Each week the number of gold being added gets less and less, so the line will increase less and less per week. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a1. If it's the same amount each week it would be a straight line. 2. Buying gold every week, the amount of gold will always increase. Since you buy more each week the rate of increase will keep increasing. So the graph will increase, and at an increasing rate. 3. Buying gold every week, the amount of gold won't ever decrease. Since you buy less each week the rate of increase will just keep falling. So the graph will increase, but at a decreasing rate. This graph will in fact approach a horizontal asymptote, since we have a geometric progression which implies an exponential function. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q006. Suppose that every week you go to the jeweler and obtain a certain number of grams of pure gold, which you then place in an old sock and bury in your backyard. Assume that buried gold lasts a long, long time, that the the gold remains undisturbed, and that no other source adds gold to your backyard. 1. If you graph the rate at which gold is accumulating from week to week vs. the number of weeks since Jan 1, 2000, will the points on your graph lie on a level straight line, a rising straight line, a falling straight line, a line which rises faster and faster, a line which rises but more and more slowly, a line which falls faster and faster, or a line which falls but more and more slowly? 2. Answer the same question assuming that every week you bury 1 more gram than you did the previous week. 3. Answer the same question assuming that every week you bury half the amount you did the previous week. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: These graphs will take into account rates, which in this case is the slope of the previous graphs. 1. This graph would be a level straight line since the amount of gold added each week is the same, so the rate of gold being added is constant. Over time, the rate will always be the same because the amount added will always be the same. 2. This graph will be a rising straight line, because the rate of gold being added gets higher and higher each week. Each week more and more gold is being added, so in turn, each week the rate of gold being added is increased. This would create a straight line, but one that is rising upwards to signify the increase in gold. 3. The graph would be a straight line that is falling because the rate of gold being added gets lower and lower each week. Since the amount of gold being bought each week is being decreased, the rate would therefore be decreasing. This would create a straight line that is falling to signify a decreasing rate. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThis set of questions is different from the preceding set. This question now asks about a graph of rate vs. time, whereas the last was about the graph of quantity vs. time. Question 1: This question concerns the graph of the rate at which gold accumulates, which in this case, since you buy the same amount eact week, is constant. The graph would be a horizontal straight line. Question 2: Each week you buy one more gram than the week before, so the rate goes up each week by 1 gram per week. You thus get a rising straight line because the increase in the rate is the same from one week to the next. Question 3. Since half the previous amount will be half of a declining amount, the rate will decrease while remaining positive, so the graph remains positive as it decreases more and more slowly. The rate approaches but never reaches zero. STUDENT COMMENT: I feel like I am having trouble visualizing these graphs because every time for the first one I picture an increasing straight line INSTRUCTOR RESPONSE: The first graph depicts the amount of gold you have in your back yard. The second depicts the rate at which the gold is accumulating, which is related to, but certainly not the same as, the amount of gold. For example, as long as gold is being added to the back yard, the amount will be increasing (though not necessarily on a straight line). However if less and less gold is being added every year, the rate will be decreasing (perhaps along a straight line, perhaps not). FREQUENT STUDENT RESPONSE This is the same as the problem before it. No self-critique is required. INSTRUCTOR RESPONSE This question is very different that the preceding, and in a very significant and important way. You should have self-critiqued; you should go back and insert a self-critique on this very important question and indicate your insertion by preceding it with ####. The extra effort will be more than worth your trouble. These two problems go to the heart of the Fundamental Theorem of Calculus, which is the heart of this course, and the extra effort will be well worth it in the long run. The same is true of the last question in this document. STUDENT COMMENT Aha! Well you had me tricked. I apparently misread the question. Please don’t do this on a test! INSTRUCTOR RESPONSE I don't usually try to trick people, and wasn't really trying to do so here, but I was aware when writing these two problems that most students would be tricked. My real goal: The distinction between these two problems is key to understanding what calculus is all about. I want to at least draw your attention to it early in the course. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ``q007. If the depth of water in a container is given, in centimeters, by 100 - 2 t + .01 t^2, where t is clock time in seconds, then what are the depths at clock times t = 30, t = 40 and t = 60? On the average is depth changing more rapidly during the first time interval or the second? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: t = 30, d(30) = 100 - 2 t + .01 t^2 = 100 - 2 (30) + .01 (30)^2 = 49 t = 40, d(40) = 100 - 2 t + .01 t^2 = 100 - 2 (40) + .01 (40)^2 = 36 t = 60, d(60) = 100 - 2 t + .01 t^2 = 100 - 2 (60) + .01 (60)^2 = 16 You can calculate the change in average depth by calculating the slope, which in this case would be the difference in depth divided by the difference in time. Slope for first interval: (36-49)/(40-30) = -13/10 = 13 cm/ 10 seconds = 1.3 cm/second Slope for second interval: (16-36)/(60-40) = -20/20 = 20 cm/ 20 seconds = 1 cm/second The depth is changing more rapidly during the first time interval because the slope is higher at 1.3 cm/second when compared to the second slope of 1 cm/second. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt t = 30 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 30 + .01 * 30^2 = 49. At t = 40 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 40 + .01 * 40^2 = 36. At t = 60 we get depth = 100 - 2 t + .01 t^2 = 100 - 2 * 60 + .01 * 60^2 = 16. 49 cm - 36 cm = 13 cm change in 10 sec or 1.3 cm/s on the average. 36 cm - 16 cm = 20 cm change in 20 sec or 1.0 cm/s on the average. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q008. If the rate at which water descends in a container is given, in cm/s, by 10 - .1 t, where t is clock time in seconds, then at what rate is water descending when t = 10, and at what rate is it descending when t = 20? How much would you therefore expect the water level to change during this 10-second interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Rate when t = 10: 10 - .1 (10) = 9 cm/second Rate when t = 20: 10 - .1 (20) = 8 cm/second In this 10 second interval, the rate goes from 9 cm/second to 8 cm/second. In order to find an overall rate, you can find the average rate of change for the system. This would be: (9 + 8)/2 = 8.5 cm/second Since the average rate of change is 8.5 cm/second, we can multiply the rate by the amount of seconds of the time interval to cross out the seconds and be left with just the cm changed. The time interval would be the first time subtracted from the second time. 20 seconds - 10 seconds = 10 seconds (8.5 cm/second) x (10 seconds) = 85 cm 85 cm is the change in water confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aAt t = 10 sec the rate function gives us 10 - .1 * 10 = 10 - 1 = 9, meaning a rate of 9 cm / sec. At t = 20 sec the rate function gives us 10 - .1 * 20 = 10 - 2 = 8, meaning a rate of 8 cm / sec. The rate never goes below 8 cm/s, so in 10 sec the change wouldn't be less than 80 cm. The rate never goes above 9 cm/s, so in 10 sec the change wouldn't be greater than 90 cm. Any answer that isn't between 80 cm and 90 cm doesn't fit the given conditions.. The rate change is a linear function of t. Therefore the average rate is the average of the two rates, or 8.5 cm/s. The average of the rates is 8.5 cm/sec. In 10 sec that would imply a change of 85 cm. STUDENT RESPONSES The following, or some variation on them, are very common in student comments. They are both very good questions. Because of the importance of the required to answer this question correctly, the instructor will typically request for a revision in response to either student response: • I don't understand how the answer isn't 1 cm/s. That's the difference between 8 cm/s and 9 cm/s. • I don't understand how the answer isn't 8.5 cm/s. That's the average of the 8 cm/s and the 9 cm/s. INSTRUCTOR RESPONSE A self-critique should include a full statement of what you do and do not understand about the given solution. A phrase-by-phrase analysis of the solution is not unreasonable (and would be a good idea on this very important question), though it wouldn't be necessary in most situations. An important part of any self-critique is a good question, and you have asked one. However a self-critique should if possible go further. I'm asking that you go back and insert a self-critique on this very important question and indicate your insertion by preceding it with ####, before submitting it. The extra effort will be more than worth your trouble. This problem, along with questions 5 and 6 of this document, go to the heart of the Fundamental Theorem of Calculus, which is the heart of this course, and the extra effort will be well worth it in the long run. You should review the instructions for self-critique, provided at the link given at the beginning of this document. STUDENT COMMENT The question is worded very confusingly. I took a stab and answered correctly. When answering, """"How much would you therefore expect the water level to change during this 10-second interval?"""" It is hard to tell whether you are asking for what is the expected change in rate during this interval and what is the changing """"water level."""" But now, after looking at it, with your comments, it is clearer that I should be looking for the later. Thanks! INSTRUCTOR RESPONSE 'Water level' is clearly not a rate. I don't think there's any ambiguity in what's being asked in the stated question. The intent is to draw the very important distinction between the rate at which a quantity changes, and the change in the quantity. It seems clear that as a result of this question you understand this and will be more likely to make such distinctions in your subsequent work. This distinction is at the heart of the calculus and its applications. It is in fact the distinction between a derivative and an integral. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q009. Sketch the line segment connecting the points (2, -4) and (6, 4), and the line segment connecting the points (2, 4) and (6, 1). The first of these lines if the graph of the function f(x), the second is the graph of the function g(x). Both functions are defined on the interval 2 <= x <= 6. Let h(x) be the function whose value at x is the product of the values of these two functions. For example, when x = 2 the value of the first function is -3 and the value of the second is 4, so when x = 2 the value of h(x) is -3 * 4 = -12. What is the value of h(x) when x = 6? Is the value of h(x) ever greater than its value at x = 6? What is your best description of the graph of h(x)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In order to find the value of h(x), we must know the values for f(x) and g(x). To find the values of the functions, we must know the equation of the line. The equation of the line can be found by first finding the slope using rise/run, and then using point slope form [ (y-y1) = m(x - x1) ] = with one of the points and changing it into the standard form (y=mx+b). So, Line segment connecting points (2, -4) and (6, 4): Slope = (4 - (- 4))/(6-2) = 8/4 = 2 (y - (-4)) = 2(x-2) y+4 = 2x -4 y = 2x - 8 Line segment connecting points (2, 4) and (6, 1): Slope = (1-4)/(6-2) = -3/4 (y - 4) = (-3/4) (x-2) y - 4 = (-3/4)x + 3/2 y = (-3/4)x + 11/2 So when x = 6, f(x) = 2(6) -8 = 4 g(x) = (-3/4) (6) + 11/2 = -9/2 + 11/2 = 1 h(x) = f(x) * g(x) = 4 * 1 = 4 The value of h(x) is never as great as its value at x = 6 because first of all it is only confirmed to be defined from x = 2 to x = 6, and if you look on a graph the lines cross and make a lopsided x. The highest value of h(x) will have the highest values of f(x) and g(x), and the values are indicated by the height. At x = 6, this is the value at which the product of f(x) and g(x) would create the highest number between the defined region. My best description of the graph of h(x) is a graph that is increasing at a steady rate since it is just multiplication of two straight lines. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q010. A straight line segment connects the points (3,5) and (7,9), while the points (3, 9) and (7, 5) are connected by a curve which decreases at an increasing rate. From each of the four points a line segment is drawn directly down to the x axis, so that the first line segment is the top of a trapezoid and the second a similar to a trapezoid but with a curved 'top'. Which trapezoid has the greater area? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Both of these shapes have the same width of 4, and they would have the same height if they the two points were connected by the same type of line. If each was connected by a straight line segment then they would have exactly the same area. So in this case, the points connected by the straight line segment (points (3,5) and (7,9)) have a greater area just because of the shape of the line. This line is straight, but the other line is decreasing at an increasing rate. Some space is taken out of the shape by making the top dip inward to show that it is decreasing at an increasing rate. Therefore, the first trapezoid with the straight line segment has a larger area. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q011. Describe the graph of the position of a car vs. clock time, given each of the following conditions: • The car coasts down a straight incline, gaining the same amount of speed every second • The car coasts down a hill which gets steeper and steeper, gaining more speed every second • The car coasts down a straight incline, but due to increasing air resistance gaining less speed with every passing second Describe the graph of the rate of change of the position of a car vs. clock time, given each of the above conditions. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Position vs. time description: While the car coasts down a straight incline and gains speed every second, if coasting down the incline is gaining x-position then the graph would be a parabolic shape. It would be increasing at a faster and faster rate. While the car coasts down a hill which gets steeper and steeper and gains more speed every second, the graph would be an increasing parabolic shape that increasing a lot during each second. This graph would be steeper than the first. While the car coasts down a straight incline and gains less speed with every second, the graph would still increase but at a decreasing rate. It is slowing down so the distance achieved after each second will be slowly decreasing. Rate of change description: While the car coasts down a straight incline and gains speed every second, the graph would be a straight line rising upwards. This is because the same amount of speed is being gained each second, so the rate is steadily increasing and will be a straight line. While the car coasts down a hill which gets steeper and steeper and gains more speed every second, the graph would be parabolic in shape and be increasing. Since more speed is being gained each second, the rate will increase at an increasing rate and therefore be parabolic. While the car coasts down a straight incline and gains less speed with every second, the graph would be decreasing at an increasing rate because it’s getting slower and slower. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q012. If at t = 100 seconds water is flowing out of a container at the rate of 1.4 liters / second, and at t = 150 second the rate is 1.0 liters / second, then what is your best estimate of how much water flowed out during the 50-second interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average rate between these two time intervals would be: (1.4 liters/second + 1.0 liters/second)/2 = 1.2 liters/second In order to find the amount of water that flowed out in the 50 second interval, take the average rate of change (1.2 liters/second) and multiply it by the specified time interval in seconds (50 seconds). (1.2 liters/second) x (50 seconds) = 60 liters So the amount of water that flowed out during the 50 second interval is about 60 liters. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!